[
    {
        "Q": "At 7:15,Sal says that PV=k.KE(system).In the last video,he only said that PV=k .So,how come it is k.KE now.I don't get it.Please help.Thanks in advance.",
        "A": "Well,I understand your point.I also think of it this way-the units of PV =the units of KE (k is a constant) PV(FV/A)=Nm^3/A=Nm^3/m^2=Nm=kgm^2/s^2 KE(mv^2/2)=kgm^2/s^2 Thus,the units are just the same.The only difference is the constant. Is it because of this that PV=kKE?Am I right?",
        "video_name": "x34OTtDE5q8",
        "timestamps": [
            435
        ],
        "3min_transcript": "I think you also have the sense that-- what would have more energy? A 100 degree cup of tea, or a 100 degree barrel of tea. I want to make them equivalent in terms of what they're holding. I think you have a sense. Even though they're the same temperature, they're both pretty warm-- let's say this is 100 degrees Celsius, so they're both boiling-- that the barrel, because there's more of it, is going to have more energy. It's equally hot, and there's just more molecules there. That's what temperature is. Temperature, in general, is a measure roughly equal to some energy-- per molecule. So the average kinetic energy of the system divided by the total number of molecules we have. Another way we could talk about is, temperature is essentially energy per molecule. So something that has a lot of molecules, where N is the number of molecules. Another way we could view this is that the kinetic energy of the system is going to be equal to the number of molecules times the temperature. This is just a constant-- times 1 over K, but we don't even know what this is, so we could say that's still a constant-- so the kinetic energy of the system is going to be equal to some constant times the number of particles We don't know what this is, and we're going to figure this out later. This is another interesting concept. We said that pressure times volume is proportional to the kinetic energy of the system-- the aggregate, if you take all of the molecules and combine their kinetic energies. These aren't the same K's-- I could put another constant here and call that K1. And we also know that the kinetic energy of the system is equal to some other constant times the number of molecules I have times the temperature. If you think about it, you could also say that this is proportional to this, and this is proportional to this. You could say that pressure times volume is proportional"
    },
    {
        "Q": "at 7:00 why is Newtons the number of particles?? isnt newton kg*m/s^2\nat 6:15 Sal say its number of molecules. how is that possible if newton is kg*m/s^2?",
        "A": "There are no Newtons in this video. N stands for Number, as in Number of molecules. It s a variable, not a unit.",
        "video_name": "x34OTtDE5q8",
        "timestamps": [
            420,
            375
        ],
        "3min_transcript": "I think you also have the sense that-- what would have more energy? A 100 degree cup of tea, or a 100 degree barrel of tea. I want to make them equivalent in terms of what they're holding. I think you have a sense. Even though they're the same temperature, they're both pretty warm-- let's say this is 100 degrees Celsius, so they're both boiling-- that the barrel, because there's more of it, is going to have more energy. It's equally hot, and there's just more molecules there. That's what temperature is. Temperature, in general, is a measure roughly equal to some energy-- per molecule. So the average kinetic energy of the system divided by the total number of molecules we have. Another way we could talk about is, temperature is essentially energy per molecule. So something that has a lot of molecules, where N is the number of molecules. Another way we could view this is that the kinetic energy of the system is going to be equal to the number of molecules times the temperature. This is just a constant-- times 1 over K, but we don't even know what this is, so we could say that's still a constant-- so the kinetic energy of the system is going to be equal to some constant times the number of particles We don't know what this is, and we're going to figure this out later. This is another interesting concept. We said that pressure times volume is proportional to the kinetic energy of the system-- the aggregate, if you take all of the molecules and combine their kinetic energies. These aren't the same K's-- I could put another constant here and call that K1. And we also know that the kinetic energy of the system is equal to some other constant times the number of molecules I have times the temperature. If you think about it, you could also say that this is proportional to this, and this is proportional to this. You could say that pressure times volume is proportional"
    },
    {
        "Q": "Great video, very interesting. When you talk about the Bacteriophages, ( 17:50 ) that inject their DNA though the harder cell walls, how does the DNA then go on to alter the cell if it is just loose genetic material within the cell?",
        "A": "Various viruses can do it differently. Sometimes the viral genome comes with proteins that will splice it into the host genome. Sometimes the viral genome is made of RNA that can do the splicing on its own. There are numerous other mechanisms.",
        "video_name": "0h5Jd7sgQWY",
        "timestamps": [
            1070
        ],
        "3min_transcript": "So maybe most of it was the viral DNA, but it might have, when it transcribed and translated itself, it might have taken a little bit-- or at least when it translated or replicated itself-- it might take a little bit of the organism's previous DNA. So it's actually cutting parts of DNA from one organism and bringing it to another organism. Taking it from one member of a species to another member of But it can definitely go cross-species. So you have this idea all of a sudden that DNA can jump between species. It really kind of-- I don't know, for me it makes me appreciate how interconnected-- as a species, we kind of imagine that we're by ourselves and can only reproduce with each other and have genetic variation within But viruses introduce this notion of horizontal transfer via transduction. Horizontal transduction is just the idea of, look when I replicate this virus, I might take a little bit of the organism that I'm freeloading off of, I might take a little 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."
    },
    {
        "Q": "At 17:45, Sal mentions that the bacteria could be far worse for the virus. Could someone explain how the bacteria could potentially harm the virus?",
        "A": "I personally think that the bacteria could be more harm to the virus because the virus simply attaches itself to the bacteria, injects everything inside of it, and then it might just kind of sit there and become inactive. I don t believe that things inside of the bacteria can take some of the bacteria s membrane a form a new bacteria. I could be wrong though.",
        "video_name": "0h5Jd7sgQWY",
        "timestamps": [
            1065
        ],
        "3min_transcript": "So maybe most of it was the viral DNA, but it might have, when it transcribed and translated itself, it might have taken a little bit-- or at least when it translated or replicated itself-- it might take a little bit of the organism's previous DNA. So it's actually cutting parts of DNA from one organism and bringing it to another organism. Taking it from one member of a species to another member of But it can definitely go cross-species. So you have this idea all of a sudden that DNA can jump between species. It really kind of-- I don't know, for me it makes me appreciate how interconnected-- as a species, we kind of imagine that we're by ourselves and can only reproduce with each other and have genetic variation within But viruses introduce this notion of horizontal transfer via transduction. Horizontal transduction is just the idea of, look when I replicate this virus, I might take a little bit of the organism that I'm freeloading off of, I might take a little 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."
    },
    {
        "Q": "at13:17, Why isn't there any cure/vaccine for AIDS?",
        "A": "The virus mutate itself in no time",
        "video_name": "0h5Jd7sgQWY",
        "timestamps": [
            797
        ],
        "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": "At 19:05, I saw this big red thing on the white blood cell. What was it?",
        "A": "This question has been asked several times before, and I ll say what I said before - I m not positive, but I believe that it is a platelet.",
        "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": "so from 00:01 to 23:17 he talking about the common cold and flu (influenza)? :|",
        "A": "He s talking about Viruses in general and not about a specific one. At the beginning he says that because he has a cold that he s going to talk about Viruses. In between those times he covers how a most viruses interact with living cells and also how a retrovirus might.",
        "video_name": "0h5Jd7sgQWY",
        "timestamps": [
            1,
            1397
        ],
        "3min_transcript": "Considering that I have a cold right now, I can't imagine a more appropriate topic to make a video on than a virus. And I didn't want to make it that thick. A virus, or viruses. And in my opinion, viruses are, on some level, the most fascinating thing in all of biology. Because they really blur the boundary between what is an inanimate object and what is life? I mean if we look at ourselves, or life as one of those things that you know it when you see it. If you see something that, it's born, it grows, it's constantly changing. Maybe it moves around. Maybe it doesn't. But it's metabolizing things around itself. It reproduces and then it dies. You say, hey, that's probably life. And in this, we throw most things that we see-- or we throw in, us. We throw in bacteria. We throw in plants. I mean, I could-- I'm kind of butchering the taxonomy system here, but we tend to know life when we see it. information inside of a protein. Inside of a protein capsule. So let me draw. And the genetic information can come in any form. So it can be an RNA, it could be DNA, it could be single-stranded RNA, double-stranded RNA. Sometimes for single stranded they'll write these two little S's in front of it. Let's say they are talking about double stranded DNA, they'll put a ds in front of it. But the general idea-- and viruses can come in all of these forms-- is that they have some genetic information, some chain of nucleic acids. Either as single or double stranded RNA or single or double stranded DNA. And it's just contained inside some type of protein structure, which is called the capsid. And kind of the classic drawing is kind of an icosahedron type looking thing. Let me see if I can do justice to it. It looks something like this. And not all viruses have to look exactly like this. And we're really just scratching the surface and understanding even what viruses are out there and all of the different ways that they can essentially replicate themselves. We'll talk more about that in the future. And I would suspect that pretty much any possible way of replication probably does somehow exist in the virus world. But they really are just these proteins, these protein capsids, are just made up of a bunch of little proteins put together. And inside they have some genetic material, which might be DNA or it might be RNA. So let me draw their genetic material. The protein is not necessarily transparent, but if it was, you would see some genetic material inside of there. So the question is, is this thing life? It seems pretty inanimate. It doesn't grow. It doesn't change. It doesn't metabolize things. This thing, left to its own devices, is just It's just going to sit there the way a book on a table just sits there. It won't change anything."
    },
    {
        "Q": "At 6:32, is it possible that the ice may slow down due to hydrogen bonding between the molecules?",
        "A": "Assume there is no friction, thus now intermolecular bonds between Hs which creates friction",
        "video_name": "CQYELiTtUs8",
        "timestamps": [
            392
        ],
        "3min_transcript": "We've never seen, at least in our human experience, it looks like everything will eventually stop. So this is a very unintuitive thing to say, that something in motion will just keep going in motion indefinitely. Everything in human intuition says if you want something to keep going in motion, you have to keep putting more force, keep putting more energy into it for it to keep going. Your car won't go forever, unless you keep, unless the engine keeps burning fuel to drive and consuming energy. So what are they talking about? Well, in all of these examples-- and I think this is actually a pretty brilliant insight from all of these fellows is that-- all of these things would have gone on forever. The ball would keep going forever. This ice block would be going on forever, except for the fact that there are unbalanced forces acting on them to stop them. So in the case of ice, even though ice on ice doesn't have a lot of friction, there is some friction between these two. And so you have, in this situation, the force of friction is going to be acting against the direction of the movement of the ice. so if you have the actual water molecules in a lattice structure in the ice cube, and then here are the water molecules in a lattice structure on the ice, on the actual kind of sea of ice that it's traveling on-- they do kind of bump and grind into each other. Although they're both smooth, there are imperfections here. They bump and grind. They generate a little bit of heat. And they'll, essentially, be working against the movement. So there's a force of friction that's being applied to here. And that's why it's stopping. Not only a force of friction, you also have some air resistance. The ice block is going to be bumping into all sorts of air particles. It might not be noticeable at first, but it's definitely going to keep it from going on forever. Same thing with the ball being tossed to the air. Obviously, at some point, it hits the ground because of gravity. So that's one force acting on it. But even once it hits the ground, it doesn't keep rolling forever, once again, because of the friction, especially if there's grass 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."
    },
    {
        "Q": "At 3:10, the ice has 2 equal forces acting upon it at opposite sides, but wouldn't the ice just go upwards instead of remaining stationary because of friction?",
        "A": "No. Friction, in this case, points sideways, so it can t make it go up. The reaction force does point upwards, but it s as big as gravity, so it also can t do that.",
        "video_name": "CQYELiTtUs8",
        "timestamps": [
            190
        ],
        "3min_transcript": "I can keep observing that rock. And it is unlikely to move, assuming that nothing happens to it. If there's no force applied to that rock, that rock will just stay there. So the first part is pretty obvious. So, \"Every body persists in a state of being at rest\"-- I'm not going to do the second part-- \"except insofar as there's some force being applied to it.\" So clearly a rock will be at rest, unless there's some force applied to it, unless someone here tries to push it or roll it or do something to it. What's less intuitive about the first law is the second part. \"Every body persists in,\" either, \"being in a state of rest or moving uniformly straight forward, except insofar as it is compelled to change its state by force impressed.\" So this Newton's first law-- and I think I should do a little aside here, because, this right here is Newton. And if this is Newton's first law, Well, the reason is is because Newton's first law is really just a restatement of this guy's law of inertia. And this guy, another titan of civilization really, this is Galileo Galilei. And he is the first person to formulate the law of inertia. And Newton just rephrased it a little bit and packaged it with his other laws. But he did many, many, many other things. So you really have to give Galileo credit for Newton's first law. So that's why I made him bigger than here. But I was in the midst of a thought. So we understand if something is at rest, it's going to stay at rest, unless there's some force that acts on it. And in some definitions, you'll see unless there's some unbalanced force. And the reason why they say unbalanced is, because you could have two forces that act on something and they might balance out. For example, I could push on this side of the rock with a certain amount of force. And if you push on this side of the rock with the exact same amount of force, the rock won't move. And the only way that it would move if there's a lot more so if you have an unbalanced force. So if you have a ton of-- and maybe the rock is a bad analogy. Let's take ice, because ice is easier to move, or ice on ice. So there's ice right here. And then, I have another block of ice sitting on top of that ice. So once again, we're familiar with the idea, if there's no force acting on it that ice won't move. But what happens if I'm pushing on the ice with a certain amount of force on that side, and you're pushing on the ice on that side with the same amount of force? The ice will still not move. So this right here, this would be a balanced force. So the only way for the ice to change its condition, to change its restful condition is if the force is unbalanced. So if we add a little bit of force on this side, so it more than compensates the force pushing it this way, then you're going to see the ice block start to move, start to really accelerate in that direction."
    },
    {
        "Q": "What I don't understand, is that wouldn't a body not be in a state of rest because of gravity? Sal uses the rock (1:10) as an example, and he says there is no force applied to that rock. Wouldn't gravity be a force on the rock? I mean, wouldn't the rock fly out of the atmosphere from the smallest amount of gravity somewhere else? What I'm thinking is that Earth's gravity would keeping the rock there, but the rock wouldn't be at \"rest.\" So, in this definition there would rarely/never be rest?",
        "A": "Force of gravity is balanced by other forces (by restoring force I think but I m not sure), so rock is in state of rest. Many forces are applied to that rock but they balance each other and that s why rock doesn t move.",
        "video_name": "CQYELiTtUs8",
        "timestamps": [
            70
        ],
        "3min_transcript": "In this video, I want to talk a little bit about Newton's First Law of Motion. And this is a translation from Newton's Principia from Latin into English. So the First Law, \"Every body persists in a state of being at rest, or moving uniformly straightforward, except insofar as it is compelled to change its state by force impressed.\" So another way to rephrase what they're saying is, that if there's something-- every body persists-- so everything will stay at rest, or moving with a constant velocity, unless it is compelled to change its state by force. Unless it's acted on by a force, especially an unbalanced force. and I'll explain that in a second. So if I have something that's at rest, so completely at rest. So I have-- and this is something that we've seen before. Let's say that I have a rock. Let's say that I have a rock someplace I can keep observing that rock. And it is unlikely to move, assuming that nothing happens to it. If there's no force applied to that rock, that rock will just stay there. So the first part is pretty obvious. So, \"Every body persists in a state of being at rest\"-- I'm not going to do the second part-- \"except insofar as there's some force being applied to it.\" So clearly a rock will be at rest, unless there's some force applied to it, unless someone here tries to push it or roll it or do something to it. What's less intuitive about the first law is the second part. \"Every body persists in,\" either, \"being in a state of rest or moving uniformly straight forward, except insofar as it is compelled to change its state by force impressed.\" So this Newton's first law-- and I think I should do a little aside here, because, this right here is Newton. And if this is Newton's first law, Well, the reason is is because Newton's first law is really just a restatement of this guy's law of inertia. And this guy, another titan of civilization really, this is Galileo Galilei. And he is the first person to formulate the law of inertia. And Newton just rephrased it a little bit and packaged it with his other laws. But he did many, many, many other things. So you really have to give Galileo credit for Newton's first law. So that's why I made him bigger than here. But I was in the midst of a thought. So we understand if something is at rest, it's going to stay at rest, unless there's some force that acts on it. And in some definitions, you'll see unless there's some unbalanced force. And the reason why they say unbalanced is, because you could have two forces that act on something and they might balance out. For example, I could push on this side of the rock with a certain amount of force. And if you push on this side of the rock with the exact same amount of force, the rock won't move. And the only way that it would move if there's a lot more"
    },
    {
        "Q": "3:23 sal says that while ice sits on ice theres no external force acting upon it but as i understand it the very fact that ice sits there is probabaly because gravity acts on the blocks ,could anyone please explain it to me",
        "A": "To get the ice moving, you would need an outside force. As said above, gravity pulls downward, but that is an acceleration. If you want to talk about forces, the weight of the ice is what keeps the ice grounded. The normal force pushes upwards to balance the weight force. Static friction is the reason that is stays still side to side.",
        "video_name": "CQYELiTtUs8",
        "timestamps": [
            203
        ],
        "3min_transcript": "I can keep observing that rock. And it is unlikely to move, assuming that nothing happens to it. If there's no force applied to that rock, that rock will just stay there. So the first part is pretty obvious. So, \"Every body persists in a state of being at rest\"-- I'm not going to do the second part-- \"except insofar as there's some force being applied to it.\" So clearly a rock will be at rest, unless there's some force applied to it, unless someone here tries to push it or roll it or do something to it. What's less intuitive about the first law is the second part. \"Every body persists in,\" either, \"being in a state of rest or moving uniformly straight forward, except insofar as it is compelled to change its state by force impressed.\" So this Newton's first law-- and I think I should do a little aside here, because, this right here is Newton. And if this is Newton's first law, Well, the reason is is because Newton's first law is really just a restatement of this guy's law of inertia. And this guy, another titan of civilization really, this is Galileo Galilei. And he is the first person to formulate the law of inertia. And Newton just rephrased it a little bit and packaged it with his other laws. But he did many, many, many other things. So you really have to give Galileo credit for Newton's first law. So that's why I made him bigger than here. But I was in the midst of a thought. So we understand if something is at rest, it's going to stay at rest, unless there's some force that acts on it. And in some definitions, you'll see unless there's some unbalanced force. And the reason why they say unbalanced is, because you could have two forces that act on something and they might balance out. For example, I could push on this side of the rock with a certain amount of force. And if you push on this side of the rock with the exact same amount of force, the rock won't move. And the only way that it would move if there's a lot more so if you have an unbalanced force. So if you have a ton of-- and maybe the rock is a bad analogy. Let's take ice, because ice is easier to move, or ice on ice. So there's ice right here. And then, I have another block of ice sitting on top of that ice. So once again, we're familiar with the idea, if there's no force acting on it that ice won't move. But what happens if I'm pushing on the ice with a certain amount of force on that side, and you're pushing on the ice on that side with the same amount of force? The ice will still not move. So this right here, this would be a balanced force. So the only way for the ice to change its condition, to change its restful condition is if the force is unbalanced. So if we add a little bit of force on this side, so it more than compensates the force pushing it this way, then you're going to see the ice block start to move, start to really accelerate in that direction."
    },
    {
        "Q": "At 8:12 Sal says \"if gravity disappeared, and you had no air...\". What's the relation between gravity and air?",
        "A": "He uses gravity and air as two examples of forces that occur naturally all around us.",
        "video_name": "CQYELiTtUs8",
        "timestamps": [
            492
        ],
        "3min_transcript": "so if you have the actual water molecules in a lattice structure in the ice cube, and then here are the water molecules in a lattice structure on the ice, on the actual kind of sea of ice that it's traveling on-- they do kind of bump and grind into each other. Although they're both smooth, there are imperfections here. They bump and grind. They generate a little bit of heat. And they'll, essentially, be working against the movement. So there's a force of friction that's being applied to here. And that's why it's stopping. Not only a force of friction, you also have some air resistance. The ice block is going to be bumping into all sorts of air particles. It might not be noticeable at first, but it's definitely going to keep it from going on forever. Same thing with the ball being tossed to the air. Obviously, at some point, it hits the ground because of gravity. So that's one force acting on it. But even once it hits the ground, it doesn't keep rolling forever, once again, because of the friction, especially if there's grass 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."
    },
    {
        "Q": "At around 5:05 Sal says everything will eventually stop, if so, how is the earth constantly orbiting the sun? What is causing the earth to move?",
        "A": "The Earth continues to spin upon its axis because there are no outside forces acting to stop its rotation.",
        "video_name": "CQYELiTtUs8",
        "timestamps": [
            305
        ],
        "3min_transcript": "so if you have an unbalanced force. So if you have a ton of-- and maybe the rock is a bad analogy. Let's take ice, because ice is easier to move, or ice on ice. So there's ice right here. And then, I have another block of ice sitting on top of that ice. So once again, we're familiar with the idea, if there's no force acting on it that ice won't move. But what happens if I'm pushing on the ice with a certain amount of force on that side, and you're pushing on the ice on that side with the same amount of force? The ice will still not move. So this right here, this would be a balanced force. So the only way for the ice to change its condition, to change its restful condition is if the force is unbalanced. So if we add a little bit of force on this side, so it more than compensates the force pushing it this way, then you're going to see the ice block start to move, start to really accelerate in that direction. This, you know, something that's at rest will stay at rest, unless it's being acted on by an unbalanced force. What's less obvious is the idea that something moving uniformly straightforward, which is another way of saying something having a constant velocity. What he's saying is, is that something that has a constant velocity will continue to have that constant velocity indefinitely, unless it is acted on by an unbalanced force. And that's less intuitive. Because everything in our human experience-- even if I were to push this block of ice, eventually it'll stop. It won't just keep going forever, even assuming that this ice field is infinitely long, that ice will eventually stop. Or if I throw a tennis ball. That tennis ball will eventually stop. It'll eventually grind to a halt. We've never seen, at least in our human experience, it looks like everything will eventually stop. So this is a very unintuitive thing to say, that something in motion will just keep going in motion indefinitely. Everything in human intuition says if you want something to keep going in motion, you have to keep putting more force, keep putting more energy into it for it to keep going. Your car won't go forever, unless you keep, unless the engine keeps burning fuel to drive and consuming energy. So what are they talking about? Well, in all of these examples-- and I think this is actually a pretty brilliant insight from all of these fellows is that-- all of these things would have gone on forever. The ball would keep going forever. This ice block would be going on forever, except for the fact that there are unbalanced forces acting on them to stop them. So in the case of ice, even though ice on ice doesn't have a lot of friction, there is some friction between these two. And so you have, in this situation, the force of friction is going to be acting against the direction of the movement of the ice."
    },
    {
        "Q": "At 6:20, Sal said that water has a \"lattice structure\". What does that mean?",
        "A": "Hi Amol Chavan, Sal refers to a lattice structure or crystal structure is an arrangement of atoms or molecules in a crystalline solid or a liquid. This term is often used to illustrate the bonding of the hydrogen and oxygen atoms in water. Hope that helps! - JK",
        "video_name": "CQYELiTtUs8",
        "timestamps": [
            380
        ],
        "3min_transcript": "This, you know, something that's at rest will stay at rest, unless it's being acted on by an unbalanced force. What's less obvious is the idea that something moving uniformly straightforward, which is another way of saying something having a constant velocity. What he's saying is, is that something that has a constant velocity will continue to have that constant velocity indefinitely, unless it is acted on by an unbalanced force. And that's less intuitive. Because everything in our human experience-- even if I were to push this block of ice, eventually it'll stop. It won't just keep going forever, even assuming that this ice field is infinitely long, that ice will eventually stop. Or if I throw a tennis ball. That tennis ball will eventually stop. It'll eventually grind to a halt. We've never seen, at least in our human experience, it looks like everything will eventually stop. So this is a very unintuitive thing to say, that something in motion will just keep going in motion indefinitely. Everything in human intuition says if you want something to keep going in motion, you have to keep putting more force, keep putting more energy into it for it to keep going. Your car won't go forever, unless you keep, unless the engine keeps burning fuel to drive and consuming energy. So what are they talking about? Well, in all of these examples-- and I think this is actually a pretty brilliant insight from all of these fellows is that-- all of these things would have gone on forever. The ball would keep going forever. This ice block would be going on forever, except for the fact that there are unbalanced forces acting on them to stop them. So in the case of ice, even though ice on ice doesn't have a lot of friction, there is some friction between these two. And so you have, in this situation, the force of friction is going to be acting against the direction of the movement of the ice. so if you have the actual water molecules in a lattice structure in the ice cube, and then here are the water molecules in a lattice structure on the ice, on the actual kind of sea of ice that it's traveling on-- they do kind of bump and grind into each other. Although they're both smooth, there are imperfections here. They bump and grind. They generate a little bit of heat. And they'll, essentially, be working against the movement. So there's a force of friction that's being applied to here. And that's why it's stopping. Not only a force of friction, you also have some air resistance. The ice block is going to be bumping into all sorts of air particles. It might not be noticeable at first, but it's definitely going to keep it from going on forever. Same thing with the ball being tossed to the air. Obviously, at some point, it hits the ground because of gravity. So that's one force acting on it. But even once it hits the ground, it doesn't keep rolling forever, once again, because of the friction, especially if there's grass"
    },
    {
        "Q": "at 6:15 he used H2O as a base can we use the HSO4- from the process where we generate electrophile?",
        "A": "HSO\u00e2\u0082\u0084\u00e2\u0081\u00bb can act as a base, but H\u00e2\u0082\u0082O is stronger and it is present in much larger amounts.",
        "video_name": "rC165FcI4Yg",
        "timestamps": [
            375
        ],
        "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": "At 10:30 he says that the Egg gets all the organelles. Does that mean that we get all of the organelles from our mother? (I know about the maternal inheritance of mitochondria.) If so, what happens with the organelles in the sperm cells?",
        "A": "No such thing. Organelles are made as instructed by DNA. The part of mitochondrial genome that is still within it is maternally inherited. All others are synthesised using simple molecules, as instructed by maternal and paternal DNA in the nucleus.",
        "video_name": "TX7-Kdn6lJQ",
        "timestamps": [
            630
        ],
        "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 9:30 he talked about the organ of corti that is comprised of both a basilar and tectorial membrane. What is the functionality of the two membranes and where are they located within the organ of corti? Thank you.",
        "A": "the basilar membrane is within the cochlea of the ear and is quite stiff . There are two fluid filled tubes that are in the coil and the function of the membrane is to keep these two fluids away from each other as they are very different. the tectorial membrane is a gel with 97% being water. it is parallel to the basilar membrane. its exact function has not yet been found but it is understood that it is essential for normal functioning of the ear.",
        "video_name": "6GB_kcdVMQo",
        "timestamps": [
            570
        ],
        "3min_transcript": "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 The reason that the fluid doesn't move back to the oval window when it goes to the very tip of the cochlea is because in between, in the very middle of the cochlea, is a membrane. Let me use a different color. There's actually a membrane. I'm gonna use this black line to demarcate the membrane that runs along the length of the cochlea. This membrane is something known as the organ of corti. Let me just write that down here - organ of corti. This organ of corti is actually composed of two different things. It's composed of something known as the basilar membrane, and another membrane known as the tectorial membrane. tectorial membrane. One final thing that I just want to touch upon is a general classification of the different parts of the ear."
    },
    {
        "Q": "At 3:21 he mentioned frequency as how close the peaks are, instead that distance between two peaks is called wavelength. Frequency is just the number of waves in a specific time period. Feel free to comment, I am not 100% sure!",
        "A": "Yes ,you are correct.",
        "video_name": "6GB_kcdVMQo",
        "timestamps": [
            201
        ],
        "3min_transcript": "and it makes a very distinct sound. Let's imagine that these two lines right here are your hands. When you clap your hands, the lines move towards each other, so your hands are moving towards each other. They're moving towards each other fairly quickly. In between your hands are a whole bunch of little air molecules, which I'm drawing. These air molecules, which I'm drawing right now, let's imagine that they're these little purple dots. So, in between your hands are a whole bunch of these air molecules. They're just floating around doing their thing. Then all of a sudden, the hands are moving towards each other, and all of a sudden, this space that these air molecules occupy gets a lot smaller. A little bit later in time, as the hands are moving towards each other - so here we are just drawing the hands almost about to touch. What happened was all these air molecules that are just floating around, they had all this space.. Now all of a sudden, they're really compressed, so they're really, really close together, They're very compacted now. You can imagine that as your hands are even closer together, that the air molecules get even more compacted. Basically, what is effectively going on is the air molecules here are getting pressurized. As you bring your hands together, you're actually adding all the molecules up, and it creates this pressure. This area of pressure actually tries to escape. It tries to escape and it kinda goes this way. It tries to escape out wherever it can. As it's escaping, it creates these areas of high and low pressure. That's what I'm representing here by these lines. These areas of high and low pressure are known as sound waves. We can have different types of sound waves. We can have sound waves that are really, really close together, or really far away from each other. If we draw this graphically, Basically, what I'm drawing here is, up here would be an area of high pressure. Over here would be an area of low pressure. Basically, there are just areas of high and low pressure. How close these peaks are together is the frequency. If I clap my hands even faster together, or if there's something else that's a higher frequency, a higher-pitched sound, the sound waves would be closer to one another, and it would look something like this. Depending on the frequency of the sound wave, it's perceived to be a different noise. Let's imagine that this sound wave right here is F1, and that this one over here is F2. Sound waves of lower frequency actually travel further. This actually happens in the ear, so these lower frequency sound waves actually travel further, and they actually penetrate deeper into the cochlea,"
    },
    {
        "Q": "Does the equation at 4:56 imply that there is no magnetic force when a charge isn't moving? If so, how does a paperclip feel a magnetic force towards a magnet when both objects are held stationary?",
        "A": "Go to youtube and search for veritasium how do magnets work and watch the pair of videos.",
        "video_name": "NnlAI4ZiUrQ",
        "timestamps": [
            296
        ],
        "3min_transcript": "So that's fine, you say, Sal, that's nice. You drew these field lines. And you've probably seen it before if you've ever dropped metal filings on top of a magnet. They kind of arrange themselves But you might say, well, that's kind of useful. But how do we determine the magnitude of a magnetic field at any point? And this is where it gets interesting. The magnitude of a magnetic field is really determined, or it's really defined, in terms of the effect that it has on a moving charge. So this is interesting. I've kind of been telling you that we have this different force called magnetism that is different than the electrostatic force. But we're defining magnetism in terms of the effect that it has on a moving charge. And that's a bit of a clue. And we'll learn later, or hopefully you'll learn later as you advance in physics, that magnetic force or a magnetic field is nothing but an electrostatic field moving at a very high speed. Or you could almost view it as they are the same thing, just from different frames of reference. I don't want to confuse you right now. But anyway, back to what I'll call the basic physics. So if I had to find a magnetic field as B-- so B is a vector and it's a magnetic field-- we know that the force on a moving charge could be an electron, a proton, or some other type of moving charged particle. And actually, this is the basis of how they-- you know, when you have supercolliders-- how they get the particles to go in circles, and how they studied them by based on how they get deflected by the magnetic field. But anyway, the force on a charge is equal to the magnitude of the charge-- of course, this could be positive or negative-- times, and this is where it gets interesting, the velocity of the charge cross the magnetic field. So you take the velocity of the charge, you could either multiply it by the scalar first, or you could take the cross product then multiply it by the scalar. this isn't a vector. But you essentially take the cross product of the velocity and the magnetic field, multiply that times the charge, and then you get the force vector on that particle. Now there's something that should immediately-- if you hopefully got a little bit of intuition about what the cross product was-- there's something interesting going on here. The cross product cares about the vectors that are perpendicular to each other. So for example, if the velocity is exactly perpendicular to the magnetic field, then we'll actually get a number. If they're parallel, then the magnetic field has no impact on the charge. That's one interesting thing. And then the other interesting thing is when you take the cross product of two vectors, the result is perpendicular to both of these vectors. So that's interesting. A magnetic field, in order to have an effect on a charge, has to be perpendicular to its you velocity. And then the force on it is going to be perpendicular to"
    },
    {
        "Q": "what does Q stand for at 5:03",
        "A": "The magnitude of the charge.",
        "video_name": "NnlAI4ZiUrQ",
        "timestamps": [
            303
        ],
        "3min_transcript": "So that's fine, you say, Sal, that's nice. You drew these field lines. And you've probably seen it before if you've ever dropped metal filings on top of a magnet. They kind of arrange themselves But you might say, well, that's kind of useful. But how do we determine the magnitude of a magnetic field at any point? And this is where it gets interesting. The magnitude of a magnetic field is really determined, or it's really defined, in terms of the effect that it has on a moving charge. So this is interesting. I've kind of been telling you that we have this different force called magnetism that is different than the electrostatic force. But we're defining magnetism in terms of the effect that it has on a moving charge. And that's a bit of a clue. And we'll learn later, or hopefully you'll learn later as you advance in physics, that magnetic force or a magnetic field is nothing but an electrostatic field moving at a very high speed. Or you could almost view it as they are the same thing, just from different frames of reference. I don't want to confuse you right now. But anyway, back to what I'll call the basic physics. So if I had to find a magnetic field as B-- so B is a vector and it's a magnetic field-- we know that the force on a moving charge could be an electron, a proton, or some other type of moving charged particle. And actually, this is the basis of how they-- you know, when you have supercolliders-- how they get the particles to go in circles, and how they studied them by based on how they get deflected by the magnetic field. But anyway, the force on a charge is equal to the magnitude of the charge-- of course, this could be positive or negative-- times, and this is where it gets interesting, the velocity of the charge cross the magnetic field. So you take the velocity of the charge, you could either multiply it by the scalar first, or you could take the cross product then multiply it by the scalar. this isn't a vector. But you essentially take the cross product of the velocity and the magnetic field, multiply that times the charge, and then you get the force vector on that particle. Now there's something that should immediately-- if you hopefully got a little bit of intuition about what the cross product was-- there's something interesting going on here. The cross product cares about the vectors that are perpendicular to each other. So for example, if the velocity is exactly perpendicular to the magnetic field, then we'll actually get a number. If they're parallel, then the magnetic field has no impact on the charge. That's one interesting thing. And then the other interesting thing is when you take the cross product of two vectors, the result is perpendicular to both of these vectors. So that's interesting. A magnetic field, in order to have an effect on a charge, has to be perpendicular to its you velocity. And then the force on it is going to be perpendicular to"
    },
    {
        "Q": "i dont get what a magnetic mono-pole is ( 0:42 - 0:46 ) ?",
        "A": "Thanx Matt :)",
        "video_name": "NnlAI4ZiUrQ",
        "timestamps": [
            42,
            46
        ],
        "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": "at2:01 why does sal substitute the value of acceleration due to gravity?",
        "A": "Because this is equal to the centripetal acceleration at this point as the only force acting on the object at the top of the circle is Fg (draw the free body diagram)",
        "video_name": "4SQDybFjhRE",
        "timestamps": [
            121
        ],
        "3min_transcript": "What I want to do now is figure out, what's the minimum speed that the car has to be at the top of this loop de loop in order to stay on the track? In order to stay in a circular motion. In order to not fall down like this. And I think we can all appreciate that is the most difficult part of the loop de loop, at least in the bottom half right over here. The track itself is actually what's providing the centripetal force to keep it going in a circle. But when you get to the top, you now have gravity that is pulling down on the car, almost completely. And the car will have to maintain some minimum speed in order to stay in this circular path. So let's figure out what that minimum speed is. And to help figure that out, we have to figure out what the radius of this loop de loop actually is. And it actually does not look like a perfect circle, based on this little screen shot that I got here. It looks a little bit elliptical. But it looks like the radius of curvature right over here is actually smaller than the radius That if you made this into a circle, it would actually be maybe even a slightly smaller circle. But let's just assume, for the sake of our arguments right over here, that this thing is a perfect circle. And it was a perfect circle, let's think about what that minimum velocity would have to be up here at the top of the loop de loop. So we know that the magnitude of your centripetal acceleration is going to be equal to your speed squared divided by the radius of the circle that you are going around. Now at this point right over here, at the top, which is going to be the hardest point, the magnitude of our acceleration, this is going to be 9.81 meters per second squared. And the radius, we can estimate-- I copied and pasted the car, and it looks like I can get it to stack on itself four times to get And I looked it up on the web, and a car about this size is going to be about 1.5 meters high from the bottom of the tires to the top of the car. And so it looks like-- just eyeballing it based on these copying and pasting of the cars, that the radius of this loop de loop right over here is 6 meters. So this right over here is 6 meters. So you multiply both sides by 6 meters. Or actually, we could keep it just in the variables. So let me just rewrite it-- just to manipulate it so we can solve for v. We have v squared over r is equal to a. And then you multiply both sides by r. You get v squared is equal to a times r. And then you take the principal square root of both sides. You get v is equal to the principal square root of a times r. And then if we plug in these numbers, this velocity that we have to have in order to stay in the circle is going to be the square root of 9.81"
    },
    {
        "Q": "at 9:00, you said the cyclobutadiene is antiaromatic, but you didn't really mention this molecule is planar.\nonly at the start when you talk about this molecule, you mentioned the p orbitals may overlap each other.\nbut may doesn't mean definitely",
        "A": "If a compound doesn t follow Huckel s rule it can t be aromatic. In fact cyclobutadiene has 4n pi electrons which would make it antiaromatic. And yes it is planar. In fact what happens to avoid this issue: cyclobutadiene will distort into a rectangle with 2 long sides and 2 short sides, this will make more sense if you draw the frost circle for the rectangle. Hopefully that makes sense.",
        "video_name": "yg0XJWHPqOA",
        "timestamps": [
            540
        ],
        "3min_transcript": "are going to give me four molecular orbitals. And I can draw in my frost circle right here. So once again, we're going to draw a line through the center to separate my bonding from my antibonding molecular orbitals. You always start at the bottom of your frost circle. Four-membered ring. So we're going try to draw a four-sided figure in our frost circle. So I'm going to attempt to draw this square in here. And once again, where our polygon intersects with our circle represents the energy level of our molecular orbitals. So I have a total of four molecular orbitals. And if I go over here, I have a molecular orbital below the line. So that's my bonding molecular orbital. I have a molecular orbital above the center line there. So that's my antibonding molecular orbital. And this time I have two molecular orbitals that are right on the line, which represents non-bonding molecular orbitals. When I go ahead and put in my 4 pi electrons, So that takes care of 2 pi electrons. And now I have two more. And so here I have the non-bonding molecular orbitals are on the same energy level. And so if you remember Hund's rule from electron configurations, you can't pair up these last two pi electrons because the orbitals are of equal energy. And so this is the picture that we get. And you can see that I have two unpaired electrons. And two unpaired electrons implies that this molecule is extremely reactive. And experimentally, it is. So cyclobutadiene will actually react with itself. So it's experimentally extremely reactive which tells you that it's not extra stable. It's not aromatic. So you don't have 4n plus 2 pi electrons. And so the second criteria is not fulfilled, but this compound does satisfy the first criteria. And so the term for this compound is anti-aromatic, which means it fulfills the first criteria, but does not satisfy the second criteria. It does not have 4n plus 2 pi electrons. It has 4n pi electrons. And so we say that is anti-aromatic. And there are actually very few examples of anti-aromatic compounds. But cyclobutadiene is considered to be anti-aromatic. In the next video, we're going to look at a few more examples of aromatic stability."
    },
    {
        "Q": "At 0:43 where was the particle before expansion? What were its surroundings?",
        "A": "there re no surroundings. The universe is only expanding on the inside because there is no outside. If there were an outside of the universe, it s the same size it s always been.",
        "video_name": "eUF59jCFcyQ",
        "timestamps": [
            43
        ],
        "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": "At 2:40, if the Universe is finite or has no edge then would you be able to get outside or would it be impossible?",
        "A": "it would be impossible",
        "video_name": "eUF59jCFcyQ",
        "timestamps": [
            160
        ],
        "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": "At 10:30 , if three points are in the universe , and after some significant amount of time , the universe expands and the three points get separated farther apart. So , is the EARTH getting any farther apart from the SUN and the MOON getting farther apart from EARTH??",
        "A": "No, gravitational forces overpower the force of expansion at closer ranges.",
        "video_name": "eUF59jCFcyQ",
        "timestamps": [
            630
        ],
        "3min_transcript": "means that if you go up and you just keep going up, you'll eventually come back from the bottom. So if you keep going all the way up, you'll eventually come back to the point that you were. It might be an unbelievably large distance, but you'll eventually get back where you were. If you go to the right, you'll eventually come back all the way around to the point where you were. And if you were to go into the page-- so if you were to go into the page-- let me draw it that way-- if you go into the page, you would eventually come back from above the page and come back to the point that you are. So that's what this implication would be. That you would eventually get back to where you are. So let's go back to the question of an expanding universe, a expanding universe that's not expanding into any other space. That is all of the space, but it's still expanding. Well, this is the model. So you could imagine shortly after the Big Bang, our four-dimensional sphere looked like this. Maybe right at the Big Bang, it was like this little unbelievably small sphere. Then a little bit later, it's this larger sphere. Let me just shade it in to show you that it's kind of popping out of the page, that's it's a sphere. And then at a later time, the sphere might look like this. The sphere might look like this. Now, your temptation might be to say, wait, Sal, isn't this stuff outside of this sphere, isn't that some type of a space that it's expanding into? Isn't that somehow part of the universe? And I would say if you're talking in three dimensions, no, it's not. The entire universe is this surface. It is this surface of this four-dimensional sphere. If you start talking about more dimensions, then, yes, you could talk about maybe things outside of our three-dimensional universe. So as this expands in space/time-- so one way to view the fourth dimension getting further and further apart. And I'll talk about more evidence in future videos for why the Big Bang is the best theory we have out there right now. But as you could imagine, if you have two points on this sphere that are that far apart, as this sphere expands, this four-dimensional sphere, as this bubble blows up or this balloon blows up, those two points are just-- let me draw three points. Let's say those are three points. Those three points are just going to get further and further apart. And that's actually one of the main points that-- or one of the first reasons why it made sense to believe the Big Bang-- is that everything is expanding, not from some central point. But everything is expanding from everything. That if you go in any direction from any point in the universe, everything else is expanding away. And the further away you go, it looks like the faster it's expanding away from you. So I'll leave you there, something for you to kind of think about a little bit. And then we'll build on some of this to think about what it means to kind of observe the observable universe."
    },
    {
        "Q": "in 5:52, If time was the 4th dimension, wouldn't the universe experience a cycle through time?",
        "A": "The sphere as a fourth dimension would be infinitely expanding, so even time cannot make it back to the point of origin.",
        "video_name": "eUF59jCFcyQ",
        "timestamps": [
            352
        ],
        "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 10:14 Sal talks about the sphere getting bigger and the points are getting further away from each other. So, if the universe is a sphere and it is continually expanding, Will we (earth) ever possibly get further away from the sun? Will the sun engulf us all before that happens?",
        "A": "Earth is held near the sun by gravity, which keeps them together even as the space they are in stretches.",
        "video_name": "eUF59jCFcyQ",
        "timestamps": [
            614
        ],
        "3min_transcript": "means that if you go up and you just keep going up, you'll eventually come back from the bottom. So if you keep going all the way up, you'll eventually come back to the point that you were. It might be an unbelievably large distance, but you'll eventually get back where you were. If you go to the right, you'll eventually come back all the way around to the point where you were. And if you were to go into the page-- so if you were to go into the page-- let me draw it that way-- if you go into the page, you would eventually come back from above the page and come back to the point that you are. So that's what this implication would be. That you would eventually get back to where you are. So let's go back to the question of an expanding universe, a expanding universe that's not expanding into any other space. That is all of the space, but it's still expanding. Well, this is the model. So you could imagine shortly after the Big Bang, our four-dimensional sphere looked like this. Maybe right at the Big Bang, it was like this little unbelievably small sphere. Then a little bit later, it's this larger sphere. Let me just shade it in to show you that it's kind of popping out of the page, that's it's a sphere. And then at a later time, the sphere might look like this. The sphere might look like this. Now, your temptation might be to say, wait, Sal, isn't this stuff outside of this sphere, isn't that some type of a space that it's expanding into? Isn't that somehow part of the universe? And I would say if you're talking in three dimensions, no, it's not. The entire universe is this surface. It is this surface of this four-dimensional sphere. If you start talking about more dimensions, then, yes, you could talk about maybe things outside of our three-dimensional universe. So as this expands in space/time-- so one way to view the fourth dimension getting further and further apart. And I'll talk about more evidence in future videos for why the Big Bang is the best theory we have out there right now. But as you could imagine, if you have two points on this sphere that are that far apart, as this sphere expands, this four-dimensional sphere, as this bubble blows up or this balloon blows up, those two points are just-- let me draw three points. Let's say those are three points. Those three points are just going to get further and further apart. And that's actually one of the main points that-- or one of the first reasons why it made sense to believe the Big Bang-- is that everything is expanding, not from some central point. But everything is expanding from everything. That if you go in any direction from any point in the universe, everything else is expanding away. And the further away you go, it looks like the faster it's expanding away from you. So I'll leave you there, something for you to kind of think about a little bit. And then we'll build on some of this to think about what it means to kind of observe the observable universe."
    },
    {
        "Q": "Whats finite at 07:12",
        "A": "finite means it has limits unlike infinite",
        "video_name": "eUF59jCFcyQ",
        "timestamps": [
            432
        ],
        "3min_transcript": "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. Let's imagine this sphere. Once again, if you're on this surface of this four-dimensional sphere-- I obviously can not draw a four-dimensional sphere. But if you're on the surface of this four-dimensional sphere, if you go in any direction, you'll come back out and come back to where you started. If you go that way, you'll come back around here. Now, the universe is super huge. So even light, maybe light itself will take an unbelievable amount of time to traverse it. And if this sphere itself is expanding, it might be expanding so fast that light might not ever be able to come back around it. But in theory, if something were fast enough, if something were to keep going around, it could eventually go back to this point. Now, when we talk about a three-dimensional surface-- it's a three-dimensional surface of a four-dimensional sphere-- that means that any of the three dimensions-- over here, on the surface, I can only draw two. But that means if this is true, if the universe is"
    },
    {
        "Q": "if the percentage of mercury to chlorine is higher, why is there a 1:2 ratio of mercury?",
        "A": "An atom of mercury has a lot more mass than an atom of chlorine. Compare the relative atomic masses from the periodic table, mercury is 200.59 while chlorine is only 35.45...so 1 atom of mercury has the mass of about 5.6 chlorine atoms. This is why we need to use moles!!",
        "video_name": "NM0WycKCCDU",
        "timestamps": [
            62
        ],
        "3min_transcript": "Sal: What I want to do in this video is start with mass composition and see if we can figure out the empirical formula of the molecule that we're dealing with based on the mass composition. Let's say that we have a bag and we're able to measure that this bag is 73 percent, it's 73 percent mercury and it is, the remainder of the bag, 27 percent chlorine. Based just on this can we figure out the likely empirical formula for the molecule that we have in that bag? I encourage you to pause the video and try to see if you can figure it out on your own. Well one way to think about it is let's just assume a number. This is all the information we have, let's just assume we have 100 grams of it. We could assume a thousand grams or 10,000 grams or 57 grams, but I'll pick 100 grams because it will make the numbers easy to work with in our head. Let's just assume, let me make it clear that I'm assuming this. I'm going to assume that I have 100 grams of this molecule If I assume that that means that the 73 percent that is mercury is going to be 73 grams and the 27 percent that is chlorine is going to be 27 grams of chlorine. Let me make it clear this is mercury and this is chlorine. Now I just need to think about, well how many moles of mercury is 73 grams and how many moles of chlorine is 27 grams. To do that I'll look up the periodic table right here. I have the atomic weight which is the weighted average of the atomic masses that's found in nature. The atomic weight here for mercury is 200.59. That means, let me write this right over here. One mole of, one mole of mercury is, Similarly we could look up the atomic weight for chlorine. Chlorine right over here, 35.453. We could say one mole of chlorine, and once again this is a weighted average of all of the isotopes of chlorine as found in nature. I guess we'll just go with that number. One mole of chlorine is going to be 35.453, 35.453, 35.453 grams. Given this information right over here how many moles of mercury is this, roughly, and how many moles of chlorine is this, roughly? I say roughly because getting an empirical formula from measurements of mass composition is going to be necessarily a messy affair, it's not going to come out completely,"
    },
    {
        "Q": "at 6:16 for calculating the ratio, is the bigger number of moles always divided by the smallest number of moles, regardless of the differences in mass percentages?",
        "A": "Yes. Always divide all the moles you have calculated by the smallest moles, that gives you the ratio between the atoms which is essentially the empirical formula.",
        "video_name": "NM0WycKCCDU",
        "timestamps": [
            376
        ],
        "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": "at 3:57, I thought that each element had a set number of electrons. So is that a hypothetical question orrr...",
        "A": "In a NEUTRAL atom the number of protons is equal to the number of electrons. But this does not always have to be the case. Atoms can and do gain or lose electrons. This is the whole point of the video. An ion is an atom that does not have the same number of electrons and protons, so it has a charge.",
        "video_name": "zTUnjPALX_U",
        "timestamps": [
            237
        ],
        "3min_transcript": "what element you're dealing with, so now if you look at what element has five protons we're dealing with boron. So this is going to be boron. Neutral boron would have five protons and five electrons. But this one has one extra electron, so it has one extra negative charge. So you can write it like this, one minus. Or you could just say it has a negative charge. So this is a boron ion right over here. As soon as you have an imbalance between protons and electrons you no longer would call it an atom, you would call it an actual ion. Now let's do an example question dealing with this. So our question tells us... Our question ... our question tells us ... So let's just look up platinum on our periodic table. Platinum is sitting right over here if you can see it. So an atom of platinum has a mass number of 195. And 195 looks pretty close to that atomic mass we have there. And it contains 74 electrons. 74 electrons. How many protons and neutrons does it contain and what is its charge? Alright, so let's think about this a little bit. So we're dealing with platinum. So by definition platinum has 78 protons, so we know that. It has 78 protons. They're telling us it has 74 electrons. 74 electrons. protons than electrons. So you're going to have a positive four charge. Four more of the positive thing than you have of the negative things. So you could write this as platinum with a plus four charge. This is a platinum ion, a positive platinum ion. The general term when we're talking about a positive ion, we're talking about a cation. That is a positive ion. Up there when we talked about boron being negative, a negative ion, that is an anion. This is just to get ourselves used to some of the terminology. But we're not done answering the question. They say an atom of platinum has a mass number of 195"
    },
    {
        "Q": "Hey Sal! at 5:03, you mentioned that there's a 'propyl' functional group on Carbon-3...just to clarify, it's an 'ethyl' group, not propyl (C2H5) :)",
        "A": "people with exceptional talent are prone to commit minor mistakes more often - my maths tution teacher",
        "video_name": "GFiizJ-jGVw",
        "timestamps": [
            303
        ],
        "3min_transcript": "So this molecule is zusammen. Which on some levels, you can think of as the same thing as cis, but cis and trans stops applying when you start having more than two functional groups. In this case, we have three. So we would call this Z-4-methylhept-3-ene. And that's because the higher priority functional groups are on the same side of the double bond. Now let's do this one over here. And someone pointed out, rightly, that I had misspelled zusammen in the last video. It's actually spelled like this, zusammen. I had spelled it with two s's and one m. I guess you can forgive me. I don't speak German. But anyway, I thought I would point that out. Now let's try to label this thing right over here. So the first thing, this once again is an alkene. Let's identify the longest carbon chain here. So it looks like one, two, three, four, five, six, seven, eight carbons. Double bonds are closer to the left hand side. One, two, three, four, five, six, seven, eight. So just the main chain is oct-- let me make sure I have some space here-- it is oct-3-ene. And then we have, well we have one functional group sitting off of the main chain. We have this bromine sitting right over there on the third carbon. So we would call this 3-bromooct-3-ene. And now we have to figure out is it entgegen or zusammen. So if we look on the carbon on the right hand side, it's pretty obvious that this is the only functional group. We just have a hydrogen there. So let me circle it in the magenta. And then on the left hand side we have two functional groups. We have this [UNINTELLIGIBLE] bromo or we have a bromine sitting right there. And then we have this propyl group. tempted to say it takes higher priority. But remember, in the Cahn-Ingold-Prelog system, you give higher priority to the atom that has a higher atomic number. Bromine has an atomic number of 35. Carbon has an atomic number of only 6. So Bromine is actually higher priority. So this is the higher priority functional group right over here. So now for deciding whether it's entgegen or zusammen, we see that our higher priority groups are apart. They're on opposite sides of the double bond. This one is on top. This one is below. We are apart. So this is entgegen. Or we would write this is E-3-bromooct-3-ene. And E is for-- just as a bit of a refresher-- it's for entgegen, a word that I enjoy saying, entgegen."
    },
    {
        "Q": "What would the naming be for the last molecule if there were 2 bromines attached to the left hand side carbon in the last molecule (\"3:52\")? Would the E-Z naming apply at all in this case?",
        "A": "If there s two then you won t have to convey which bromine is where using the E-Z convention because you have no choice but to draw both in the correct position.",
        "video_name": "GFiizJ-jGVw",
        "timestamps": [
            232
        ],
        "3min_transcript": "And what we need to do to identify the highest priority group is to use the Cahn-Ingold-Prelog namings or priority scheme that we learned several videos ago. And there you literally go from this carbon, you look at what it's bonded to, and you compare the atomic numbers. But in both cases it's a carbon to a carbon. This is a carbon to a carbon. So their atomic numbers are the same. So then you go one bond further away and you see which one is bonded to a higher atomic number atom. This carbon bonds to a carbon, which is a higher atomic number. This carbon only bonds to three hydrogens. This one does two hydrogens and one carbon. Because it's bonded to another carbon, it takes priority. This propyl group is a higher priority functional group. So now when we're trying to decide whether it is entgegen or zusammen, we look at these 2 groups. And we see that they are sitting on the same side of the double bond. They are both above the carbons. So this molecule is zusammen. Which on some levels, you can think of as the same thing as cis, but cis and trans stops applying when you start having more than two functional groups. In this case, we have three. So we would call this Z-4-methylhept-3-ene. And that's because the higher priority functional groups are on the same side of the double bond. Now let's do this one over here. And someone pointed out, rightly, that I had misspelled zusammen in the last video. It's actually spelled like this, zusammen. I had spelled it with two s's and one m. I guess you can forgive me. I don't speak German. But anyway, I thought I would point that out. Now let's try to label this thing right over here. So the first thing, this once again is an alkene. Let's identify the longest carbon chain here. So it looks like one, two, three, four, five, six, seven, eight carbons. Double bonds are closer to the left hand side. One, two, three, four, five, six, seven, eight. So just the main chain is oct-- let me make sure I have some space here-- it is oct-3-ene. And then we have, well we have one functional group sitting off of the main chain. We have this bromine sitting right over there on the third carbon. So we would call this 3-bromooct-3-ene. And now we have to figure out is it entgegen or zusammen. So if we look on the carbon on the right hand side, it's pretty obvious that this is the only functional group. We just have a hydrogen there. So let me circle it in the magenta. And then on the left hand side we have two functional groups. We have this [UNINTELLIGIBLE] bromo or we have a bromine sitting right there. And then we have this propyl group."
    },
    {
        "Q": "At 4:40 what is Thermogenesis?",
        "A": "the production of heat, especially in a human or animal body.",
        "video_name": "f_Z1zsR9lFM",
        "timestamps": [
            280
        ],
        "3min_transcript": "So what tends to make a paracrine hormone work regionally is that the high concentration of the receptors are very close to the site of synthesis. And the same with autocrine, is often they're made, and there's a very high concentration of the receiving end right at that cell, right next to that cell. SALMAN KHAN: And this might be a silly question, but it's called endocrinology. Are there paracrinologists? NEIL GESUNDHEIT: Well, it's a good point. I don't think so. I think we just, perhaps because the paracrine function of hormones was discovered later, we still carry this all under the umbrella of endocrinology. SALMAN KHAN: Right. So all of hormones is endocrinology, even though endocrine hormones are the ones that act at far distances. NEIL GESUNDHEIT: That's right. I think that's a good way to summarize it. Now I like the diagram that you created here because it illustrates some of the major endocrine organs, the ones we'll be focusing on in later lectures. So the first one that you showed very nicely in the head, at the base of the brain, is that orange structure. And that would be the pituitary gland. That's right. because from the pituitary, we make hormones that work on yet other organs. So I'll give you an example. One of the hormones that's made by the pituitary is called thyroid stimulating hormone, or TSH. And after it leaves the pituitary, it goes into the circulation and it acts on the thyroid gland, where there are high receptors for TSH on the surface of the thyroid cells. And it stimulates the thyroid gland to make thyroid hormone, typically thyroxine T4 or triodothyronine, T3. Those would be the two main circulating thyroid hormones. SALMAN KHAN: And what do those do? NEIL GESUNDHEIT: Those regulate metabolism, they regulate appetite, they regulate thermogenesis, they regulate muscle function. They have widespread activities on other parts of the body. SALMAN KHAN: But it kind of upregulates or downregulates the entire body and the metabolism. NEIL GESUNDHEIT: That's right. So someone with hyperthyroidism would have very high metabolism. You may know the classic picture someone with a high heart rate, rapid metabolism, weight loss. That would be someone with excess amounts And then you see pretty much the inverse picture when someone has a deficiency of thyroid hormone and someone with hypothyroidism. So it's critical to maintain just the right amount of almost all of these hormones, and the thyroid hormones are good examples of this. But the ultimate regulation is from that pituitary gland. SALMAN KHAN: This is kind of the master one. It sends a signal there, and then that kind of does the-- NEIL GESUNDHEIT: That's right. And we'll talk later about feedback loops, because how does the pituitary know when to stop making TSH? And basically, like a thermostat, it can sense the levels of thyroid hormone. And when those levels are just at the right level, and not too high, it'll decrease the amount of TSH it makes. If the levels are too low, it'll increase TSH to try to stimulate the thyroid gland to make yet more thyroid hormone. SALMAN KHAN: Very cool. And what else do we have here? So the other hormones, some of the major ones. The pituitary, in addition to making the thyroid stimulating hormones, it makes a hormone called ACTH, adrenal corticotrophic hormone, which acts on the adrenal cortex. And the adrenal is that gland exactly"
    },
    {
        "Q": "What does that ATTACK word signifies at 5:16 ?\nPlease help. Thank you.",
        "A": "Form a bond. The electrons of the O bond to the C.",
        "video_name": "Z4F88tTx9-8",
        "timestamps": [
            316
        ],
        "3min_transcript": "So let me show the pi bond here, and pi bonds are regions of high electron density so this pi bond can act like a nucleophile in an organic chemistry mechanism. Now let's look at electrophiles. So an electrophile is electron-loving and since electrons are negatively charged we're gonna think about an electrophile as having a region of low electron density so it could have a full positive charge on it because a positive charge would be attracted to electrons, or you could be talking about a partial positive charge. So first let's look at this compound. We know that chlorine is more electronegative than carbon. So chlorine is going to withdraw some electron density and if chlorine is withdrawing electron density away from this carbon, this carbon is partially positive. Next let's look at acetone. So oxygen is more electronegative than carbon so oxygen is going to withdraw some electron density away from this carbon here and this carbon would be partially positive, so this carbon is the electrophilic portion of this compound. Next let's look at a carbocation where there's a full positive charge on this carbon so this carbon has only three bonds to it which gives it a full positive charge. Obviously a full positive charge is going to love electrons. Opposite charges attract, so this carbon is the electrophilic portion of this ion. And finally let's look at this compound, right. We know that oxygen is more electronegative than carbon so oxygen withdraws some electron density away from structure here, so let me take these pi electrons and move them out onto the oxygen, so let's draw a resonance structure so I put in my double bond. Now if I'm showing those pi electrons moving off onto the oxygen I would need three lone pairs of electrons on that top oxygen giving it a negative one formal charge. I took a bond away from this carbon in magenta which is this carbon which gives it a plus one formal charge, so that's one of the possible resonance structures that you can draw and of course we know the carbon in magenta is an electrophilic center, but I could draw another resonance structure so let me go ahead and do that, put in my brackets over here. I could take these pi electrons, I'll show it on this one actually, these pi electrons and move them over to here, so let's draw the resulting resonance structure. So I'd have a double bond here now"
    },
    {
        "Q": "At 6:17, wouldn't the lack of Hydrogen in the middle Carbon (Carbocation) make it negative instead of positive?",
        "A": "No. The middle carbon could be positive, negative, or a radical. It all depends on whether the carbon has lost an H\u00e2\u0081\u00ba, H\u00e2\u0081\u00bb, or H.",
        "video_name": "Z4F88tTx9-8",
        "timestamps": [
            377
        ],
        "3min_transcript": "Next let's look at acetone. So oxygen is more electronegative than carbon so oxygen is going to withdraw some electron density away from this carbon here and this carbon would be partially positive, so this carbon is the electrophilic portion of this compound. Next let's look at a carbocation where there's a full positive charge on this carbon so this carbon has only three bonds to it which gives it a full positive charge. Obviously a full positive charge is going to love electrons. Opposite charges attract, so this carbon is the electrophilic portion of this ion. And finally let's look at this compound, right. We know that oxygen is more electronegative than carbon so oxygen withdraws some electron density away from structure here, so let me take these pi electrons and move them out onto the oxygen, so let's draw a resonance structure so I put in my double bond. Now if I'm showing those pi electrons moving off onto the oxygen I would need three lone pairs of electrons on that top oxygen giving it a negative one formal charge. I took a bond away from this carbon in magenta which is this carbon which gives it a plus one formal charge, so that's one of the possible resonance structures that you can draw and of course we know the carbon in magenta is an electrophilic center, but I could draw another resonance structure so let me go ahead and do that, put in my brackets over here. I could take these pi electrons, I'll show it on this one actually, these pi electrons and move them over to here, so let's draw the resulting resonance structure. So I'd have a double bond here now so let me put that in here, draw in the hydrogen, put in my brackets, and I removed a bond, we took a bond away from, let me use blue for this, from this carbon, so this carbon now has a plus one formal charge, so the carbon in blue is this carbon over here, so let me draw in a plus one formal charge, so that is also electrophilic, right, a full positive charge is going to be attracted to a negative charge, so this compound actually has, this compound actually has two electrophilic centers, so this carbon here and also this carbon."
    },
    {
        "Q": "At 11:06, What is Molar Concentration?",
        "A": "Molar concentration is defined as the number of moles of solute per liter of solvent. For instance, adding 1 mol of NaCl to make a 1 L solution makes a 1 molar solution of NaCl.",
        "video_name": "qbCZbP6_j48",
        "timestamps": [
            666
        ],
        "3min_transcript": "through beer. The Beer-Lambert law tells us that the absorbance is proportional-- I should write it like this-- the absorbance is proportional to the path length-- so this would be how far does the light have to go through the solution. So it's proportional to the path length times the concentration. And usually, we use molarity for the concentration. Or another way to say it is that the absorbance is equal to some constant-- it's usually a lowercase epsilon like that-- and this is dependent on the solution, or the solute in question, what we actually have in here, and the temperature, and the pressure, and all of that. Well it's equal to some constant, times the length it has to travel, times the concentration. This thing right here is concentration. And the reason why this is super useful is, you can imagine, if you have something of a known concentration-- let me draw right here. So let's say we have an axis right here, that's axis. And over here I'm measuring concentration. This is our concentration axis. And we're measuring it as molarity. And let's say the molarity starts at 0. It goes, I don't know, 0.1, 0.2, 0.3, so on and so forth. And over here you're measuring absorbance, in the vertical axis you measure absorbance. You measure absorbance just like that. Now let's say you have some solution and you know the So let me write down M for molar. And you measure its absorbance, and you just get some number here. So you measure its absorbance and you get its absorbance. So this is a low concentration, it didn't absorb that much. You get, I don't know, some number here, so let's say it's 0.25. And then, let's say that you then take another known concentration, let's say 0.2 molar. And you say that, oh look, it has an absorbance of 0.5. So let me do that in a different color. It has an absorbance, right here, at 0.5. And I should put a 0 in front of these, 0.5 and 0.25. What this tells you, this is a linear relationship. That for any concentration, the absorbance is going to be on a line. And if you want a little review of algebra, this epsilon is actually going to be the slope of that line. Well actually, the epsilon times the length will be the slope. I don't want to confuse you too much."
    },
    {
        "Q": "Where does the term \"spectrophotometry\"come from? 0:04",
        "A": "Spectro- meaning to look photo- meaning light -metry- meaning measurement",
        "video_name": "qbCZbP6_j48",
        "timestamps": [
            4
        ],
        "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": "Near 6:40, why does he write twiceT=I2/I0?",
        "A": "Because he is mentioning about the second case, but it should be noted in order to not to be confused that it is not twice , it is just a subscript 2 identifying the second case.",
        "video_name": "qbCZbP6_j48",
        "timestamps": [
            400
        ],
        "3min_transcript": "You're getting the most light into your eye. This would be a slightly darker color, and this would be the darkest color. That makes complete sense. If you dissolve something, if you dissolve a little bit of something in water, it will still be pretty transparent. If you dissolve a lot of something in water, it'll be more opaque. And if the cup that you're dissolving in, or the beaker that you're in gets even longer, it'll get even more opaque. So hopefully that gives you the intuition behind spectrophotometry. And so the next question is, well what is it even good for? Why would I even care? Well you could actually use this information. You could see how much light is transmitted versus how much you put in to actually figure out the concentration of a solution. That's why we're even talking about it in a chemistry context. So before we do that-- and I'll show you an example of that in the next video-- let me just define some terms of ways of measuring how concentrated this is. Or ways of measuring how much light is transmitted versus how much was put in. And so when the people who defined it said, well you know, what we care about is how much is transmitted versus how much went in. So let's just define transmittance as that ratio, the amount that gets through. So in this example, the transmittance of number 1 would be the amount that got through over the amount that you put in. Over here, the transmittance would be the amount that you got out over the amount that you put in. And as we see, this one right here will be a lower number. I2 is lower than I1. So this will have a lower transmittance than number 1. So let's call this transmittance 2. This is transmittance 1. And transmittance 3 is the light that comes out, that And this is the smallest number, followed by that, followed by that. So this will have the least transmittance-- it's the most opaque-- followed by that, followed by that. Now another definition-- which was really kind of a derivative of the-- not in the calculus sense, this is just derived from transmittance and we'll see it has pretty neat properties-- is the notion of absorbance. And so here, we're trying to measure how good is it at absorbing? This is measuring how good are you at transmitting? A higher number says your transmitting a lot. But absorbance is how good you're absorbing. So it's kind of the opposite. If you're good at transmitting, that means you're bad at absorbing, you don't have a lot to absorb. If you're good at absorbing, that means you're not transmitting much. So absorbance right here. And that is defined as the negative log of transmittance. And this logarithm is base 10."
    },
    {
        "Q": "At 3:12, couldn't you just round the numbers like 1.0079 to just 1.01?",
        "A": "No, that is not being done correctly. Unlike what Sal does in many of his videos, you cannot round to whatever amount you find convenient. You MUST respect the number of significant digits you have. Almost any chemistry will mark a problem wrong if you do not round in such a way that your number of significant digits requires. You cannot round less than that, you cannot round more than that.",
        "video_name": "UPoXG1Z3sI8",
        "timestamps": [
            192
        ],
        "3min_transcript": "this is going to be 12.011 atomic mass units, and the contribution from hydrogen, the contribution from hydrogen, let me do that in yellow... The contribution from hydrogen, from the six hydrogens, is going to be six times the atomic weight of hydrogen, which is 1.0079, by this periodic table that I have right here, once again, the weighted average of all the isotopes, et cetera, et cetera, 1.0079. 1.0079, and so, the molecular weight of this whole thing, a typical benzene molecule, if you were to take the weighted average of all of its molecular masses, based on the prevalence of the different isotopes on Earth, well, you would just say it's just going to be the sum of these two things. It is going to be six times 12.011 is that going to give us? Well, let's see, let's get a calculator out, so let me clear that, so 12.011 times six gives us 72.066, so this right over here. is 72.066, and actually, I probably could have done that in my head, well anyway, let's look at what we have from the hydrogen, so the hydrogen are going to be six times 1.0079 gets us to 6.0474, plus 6.0474, but since I only go to the thousandth place in terms of precision here, if I care about significant figures, significant digits, then I'm only going to go three spaces to the right of the decimal, but let's add these two together, so I'm going to get 6.0474, plus 72.066 equals, and I'm just going to go three to the right, so 78.113, so this is equal to, 78.113 atomic mass units, that's the molecular mass of a molecule of benzene, now what percentage is from the carbon? Well, it's going to be equal to the 72.066 over the 78.113 which is equal to, all right, so let me just clear this."
    },
    {
        "Q": "02:23 - If you caught the flu between Jan 7th and 10th, why did the symptoms show up only on 11th and not as soon as you caught it ?\n\nI do understand that it takes some time for the virus to start affecting the body, but isnt 3-4 days too long?",
        "A": "It takes some time for the tiny amount of virus that you actually pick up to infiltrate cells, reproduce, and infiltrate more cells. Some of the symptoms are from the effects of the virus invading the cells, and some are side effects from your immune system attacking the virus.",
        "video_name": "6vy5CX6vK0I",
        "timestamps": [
            143
        ],
        "3min_transcript": "So today's January 11, 2013. I'm just going to circle the date on my little calendar here. And let's say I came home from work and I just felt really lousy, just awful. Fevers, sore throat, cough, body aches, you name it. This is the first day I've been feeling like this. And up until now, since the new year started, I was feeling really great. I had no symptoms. I was going to work. Feeling myself. So all these days I was feeling good. And then all of a sudden the 11th hit and I just got, all of a sudden, these symptoms. So I'm already suspecting the flu based on what we know. It started abruptly. I've got the symptoms for it. And a few questions start popping up in my head. The first question I want to know is, when can I expect to start feeling better? That's usually the first thing people want to know. So let's think about what we know about the flu in terms of how long the symptoms usually last. Because that's going to help us predict when I can expect to start feeling myself again. So we know that usually symptoms last for three to seven days. going to expect to feel kind of the same maybe. I might start feeling a little bit better by the 16th or 17th. But that would be seven days. So these are the days I can expect to feel kind of lousy. And on average I should start feeling myself again maybe by the 18th and 19th. I should start feeling the way I normally do. So according to this calendar I would start feeling better by January 18. That would be my target date. And this isn't exact. This is just a rough idea. So what's the next thing that people usually try to figure out about the flu? They always want to know, who did they get it from? They always want to figure out who the culprit was. Who gave the flu to them. So I'm no different. I want to know who did I get it from. And so I think back and I say, well, I felt good on the 10th and I felt pretty lousy on the 11th. And your instinct might be to say, well, On Thursday. But, in fact, you have to go a little bit further back. Sometimes you can get it back as far as four days. So I'm going to circle the days that I could've potentially gotten the flu from somebody. And it turns out it could've been any time this week. So I'm going to write that down. So January 7th to January 10th. That's the window in which somebody gave me the flu. Now how do I know that I got it from somebody? Maybe I got it from the doorknob. Or maybe I got it from the remote control that someone was touching. And those kinds of environmental objects, sometimes you can get diseases from there. But with the flu you generally get it from another person. And the reason is, is because you've got this RNA that's protected by an envelope. Remember this green layer here, this double layer, is a lipid or a fat bilayer."
    },
    {
        "Q": "Based on the pneumonia videos, pneumonia is the infection of the flu virus in the lungs in the situation described at 10:04, and pneumonitis (inflammation of alveolar walls) is secondary to this infection. At 10:04 when he describes inflammation of the alveolar walls, is that an instance where pneumonitis and pneumonia are interchangeable? Or, should inflammation of alveolar walls always be called pneumonia if it is caused by pneumonia?",
        "A": "The disease entity is called pneumonia clinically and usually there is further description such as bacterial pneumonia, viral pneumonia, fungal pneumonia, etc. At the microscopic, pathologic level, the changes seen are called pneumonitis. When there is no infection, sometimes the disease is described based upon the pathologic changes hence things like aspiration pneumonitis.",
        "video_name": "6vy5CX6vK0I",
        "timestamps": [
            604,
            604
        ],
        "3min_transcript": "These are high risk individuals. So why do we care so much about these high risk individuals? Well, it's because they develop complications of flu. And this is what it really boils down to. You remember we initially talked about all the hundreds of thousands of people in the US and around the world that get hospitalized for the flu. And then the numbers of people that die from the flu. Well, overwhelmingly it's people in this group. This high risk group. And the things that they get, the kind of complications they get, are many. Actually, flu leads to many different types of complications. And I'm going to draw out just a few of them for you. I don't want to give you an exhaustive list. But I want you to at least get an appreciation for the kinds of things we're talking about. So, for example, let's say these are your lungs. I'm drawing two branches of your lungs. And this is going to your left lung So this is your trachea splitting up. And you know that the flu, the influenza, is going to affect the cells in your respiratory tree. So it's going to affect these cells and it's going to cause inflammation. You're going to get a big immune response. And if that response is really big, let's say you have a big response, and if it's around these airways here, these bronchioles-- let me actually extend this out a little bit, so you can at least appreciate where the arrow's going. If the response is really strong in the bronchioles, we call that bronchitis. So someone might actually develop bronchitis as a result of getting the flu. Now someone else might actually have a big inflammatory reaction in these little air sacs. Your lungs end in thousands and thousands of little air sacs. And if that happens, then you might call that pneumonia. You might say, well, this person has pneumonia. Actually, lots and lots of it. Smooth muscle that wraps around the bronchioles. And sometimes with the flu you actually can trigger twitichiness of that smooth muscle. It starts spasming. And when that happens we know that sometimes as an asthma attack. So you can actually get an asthma attack related to the flu. So all sorts of things like this can happen. And it's awful. These are things that can actually land you in the hospital. Or can cause death, as well. So these are the kinds of complications. And there are other ones. Things like ear infections and sinus infections and many, many other things as well. But here I just wanted to show you a few of the complications that people get. And show you and remind you that it's usually the high risk people that you have to worry about."
    },
    {
        "Q": "What would happen if in 1:45 you push on both pistons with the same amount of force? Would the water just displace? What if there is no possibility for the water to displace? Then could you no longer push on the pistons?",
        "A": "I think that if you push on both the pistons with the same force, the piston with larger area will displace the liquid as more pressure is applied on it. if there is no possibility for water to displace, then you can no longer push on the pistons.",
        "video_name": "lWDtFHDVqqk",
        "timestamps": [
            105
        ],
        "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": "At around 9:30...how do the -1, 0, 1, relate to the px, py, and pz?",
        "A": "This is just a convention that is used. However, the three p orbitals are equivalent to one another and only acquire the x, y and z suffixes based on how the three axes are drawn. Therefore, it is arbitrary which p orbital is linked to -1, 0 and 1 and it would be equally valid to, say, relate -1, 0 and 1 to pz, px and py, or to any other permutation. All you need to know is that if l =1 then this means there are three p orbitals, and that these three p orbitals are along different axes.",
        "video_name": "KrXE_SzRoqw",
        "timestamps": [
            570
        ],
        "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": "At 4:10, if the electron is not found in the sphere of the s orbital, than where else would it be found?",
        "A": "That sphere is just the most likely place for an electron to be found, it could be on the other side of the universe but the probability of that is immensely low.",
        "video_name": "KrXE_SzRoqw",
        "timestamps": [
            250
        ],
        "3min_transcript": "For n is equal to one, let's say the average distance from the nucleus is right about here. Let's compare that with n is equal to two. n is equal to two means a higher energy level, so on average, the electron is further away from the nucleus, and has a higher energy associated with it. That's the idea of the principal quantum number. You're thinking about energy levels or shells, and you're also thinking about average distance from the nucleus. All right, our second quantum number is called the angular momentum quantum number. The angular momentum quantum number is symbolized by l. l indicates the shape of the orbital. This will tell us the shape of the orbital. Values for l are dependent on n, so the values for l go from zero so it could be zero, one, two, or however values there are up to n minus one. For example, let's talk about the first main energy level, or the first shell. n is equal to one. There's only one possible value you could get for the angular momentum quantum number, l. n minus one is equal to zero, so that's the only possible value, the only allowed value of l. When l is equal to zero, we call this an s orbital. This is referring to an s orbital. The shape of an s orbital is a sphere. We've already talked about that with the hydrogen atom. Just imagine this as being a sphere, so a three-dimensional volume here. The angular momentum quantum number, l, since l is equal to zero, that corresponds to an s orbital, so we know that we're talking about an s orbital here 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."
    },
    {
        "Q": "At 5:15, why would you not want to take one of these devices apart? What could be in items like this that you would need to be afraid of?",
        "A": "Underneath the bottom cover you find the electric heater and water hoses. This is a dangerous combination (water and electricity). The manufacturer wants to keep you away from the chance of messing up the connections and putting a defective coffeemaker back in service. In this demo video, the coffeemaker takes a one-way trip to full disassembly. It will never make another cup.",
        "video_name": "XQTIKNXDAao",
        "timestamps": [
            315
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "At 2:50, when Sal started talking about velocity, was he actually meaning instantaneous velocity?",
        "A": "Yes, he was meaning instantaneous velocity. There are two kinds of velocity: average velocity and instantaneous velocity. By just saying velocity , we have to think about which one it means through the context.",
        "video_name": "ITA1rW5UraU",
        "timestamps": [
            170
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 12:05, what are some other things that can cause Cardiac Arrest?",
        "A": "Many. Among others, Brugada syndrome.",
        "video_name": "vYnreB1duro",
        "timestamps": [
            725
        ],
        "3min_transcript": "It's less likely but sometimes a plaque could also go downstream, kind of form a thromboembolism. It would be this thrombogenic material, the clots around it. It would actually go and block the artery further downstream and be embolism. That can also block the artery and cause tissue to die. But the main cause is this intense clotting that can occur pretty quickly and completely obstruct the artery. There is one last word i want to touch sometimes mixed in with all the other words, that is cardiac arrest. That's because sometimes we use them in the same context. one thing can lead to another. Heart attack is not cardiac arrest. Cardiac arrest is the actual dying of the heart. Some part infarct, that's what they called myocardial infarction. Myocardial means the tissue of heart or the muscle of the heart that's dying. Sometimes it is called myocardial infarction. That is not cardiac arrest. Cause you can have some of your heart tissue die and you can survive. Your heart would be impaired. But you will continue to live. Cardiac arrest is literally your heart stopping. This would obviously cause someone to die. If you have a bad enough heart attack, if you have enough of the tissue get starved of oxygen so that it dies, infarction occurs. Then it could lead to cardiac arrest. It always won't lead to cardiac arrest. Frankly, heart attack is not the only thing that can cause cardiac arrest. Cardiac arrest is heart stopping. Heart failure is essentially just saying that heart can not provide all of the needs for the body."
    },
    {
        "Q": "At about 3:00 Sal talks about plaque and how it can damage your heart. So, if you do get plaque in your blood vessel, how do you get rid of it?",
        "A": "Presumably exercise, a low fat, healthy diet and stopping smoking if one is doing so already.",
        "video_name": "vYnreB1duro",
        "timestamps": [
            180
        ],
        "3min_transcript": "so this was my ....this is what I thought people were talking about when they were saying clogging of the arteries and maybe when they got clogged enough, the stopped blood flow to the rest of the body somehow and that would actually kill the person. I want to make it very clear right now. Those are not the arteries that people are talking about getting clogged, when people talk about heart disease or heart attacks. The arteries that they are talking about are the arteries that actually provide blood to the heart. Remember the heart itself is a muscle. It itself needs oxygen. So you have these arteries right over here, the red tubes. Those are arteries. and then the blue ones are veins. They're taking the de-oxygenated blood away from the tissue of the heart. And these are called coronary arteries. And this one over here at least from the point of view of me or you looks like it's on the right. This right over here is called the left coronary artery or LCA. And this right over here in red is called the right coronary arteries or the RCA. And so when people talk about arteries getting blocked or getting clogged, they're talking about the coronary arteries. They're talking about the things that supply blood to the heart. So let's zoom in on one of them....Maybe we can zoom in right over here, that part of the artery. That's the tube....clear where I am zooming in. I am zooming in right over here. So over time, I am not going into the details how this happened. It is subject for another video. You can have these plaques build up along the walls of the artery. So over time if a person doesn't have the right diet, or maybe they just have a predisposition to it, 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,"
    },
    {
        "Q": "at 9:35 why is the carbon cation sp2 hybridized? shouldn't it be sp3?",
        "A": "No. A carbocation only has 3 electron domains so it is planar and sp2 hybridized.",
        "video_name": "KPh60w6McPI",
        "timestamps": [
            575
        ],
        "3min_transcript": "so we form a carbo cation. So this is a stable carbo cation. This is a tertiary carbo cation. So that is why this tertiary alcohol reacts via a SN-one type mechanism. The stability of the carbo cation. And so in our final step, we have the nucleophile is going to attack our electrophile. So the nucleophile attacks our electrophile. The chloride anion attacks our carbo cation, attacks that carbon there and so we form our final product, which is tertbutyl chloride. So let me go ahead and draw in these electrons and lets highlight some again. The electrons in blue. These electrons form the bond, So they bonded right here. And so let me go ahead and highlight this carbon in red. So this carbon in red is this one right here. So we form tertbutyl chloride and we lost water in the process. So this is a very easy reaction to do. It occurs at room temperature and just take and just shake them together and you can form your final product this way. So thinking about stereochemistry and SN-one type mechanism, the carbon in red right here is not a chiral center and so we don't have to worry about what kind of stereochemical outcome that we would predict for the product. So this is our final product. There is no stereochemistry. Just to refresh your memory, for a SN-one type mechanism because this formation of this carbo cation, this carbon and the carbo cation is SP-two hybridized, and so it is planar. And so when we draw out that carbon in red here. Lets say that's that carbon in red. It's SP-two hybridized, which means that the carbons that are directly bonded to it lie on the same plane. So these carbons lie on the same plane. SP-two hybridized with a p orbital. So there is a p orbital. Lets sketch that in. So there is a plus one formal charge in this carbon. So when your nucleophile attacks, your nucleophile could attack from either side of that plane. or it could attack from this side. So if your final product has a chiral center, you need to think about stereochemistry. But not in this case. In this case we don't have one. So we lucked out. This one was a little bit straightforward. So that's a couple of examples of SN-one and SN-two reactions of alcohols."
    },
    {
        "Q": "at 08:22 , Sal says that the [ln V] is evaluated over final \"velocity\" to starting \"velocity\". I strongly think it should be replaced with \"volume\".\nPls. correct me if I'm wrong.\n\nthank you.",
        "A": "He misspoke. It is volume. After all, the video is on volume...",
        "video_name": "ixRtSV3CXPA",
        "timestamps": [
            502
        ],
        "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": "at 11:16 why is it (Tf/Ts)^2/3(Vf/Vs)=1????\nwhy is it 1???",
        "A": "To start with, ln [(Tf/Ts)^3/2(Vf/Vs)]=0. Now, ln (natural log) is logarithm base e (a constant), so log e [(Tf/Ts)^3/2(Vf/Vs)] = 0. From this, and by the property of logarithm, e to the power of 0 (e^0) = [(Tf/Ts)^3/2(Vf/Vs)]. Property of logarithm says whatever number to the 0th power equals 1. So since (e^0) = [(Tf/Ts)^3/2(Vf/Vs)], this whole [(Tf/Ts)^3/2(Vf/Vs)] expression equals 1.",
        "video_name": "ixRtSV3CXPA",
        "timestamps": [
            676
        ],
        "3min_transcript": "Plus, for the same reason, the natural log of VF over the natural log of VS. When you evaluate this, it's the natural log of VF minus the natural log of VS, which can be simplified this to this, just from our logarithmic properties. So this equals 0. And now we can-- this coefficient out front, we can use our logarithmic properties. Instead of putting a 3/2 natural log of this, we can rewrite this as the natural log of TF over TS to the 3/2. Now we can keep doing our logarithm properties. You take the log of something plus the log of something. That's equal to the log of their product. So this is equal to-- I'll switch colors-- The natural log of TF over TS to the 3/2 power, times the natural log of VF over VS. And this is a fatiguing proof. Now what can we say? Well, we're saying that e to the 0 power-- the natural log is log base e-- e to the 0 power is equal to this thing. So this thing must be 1. E to the 0 power is 1. So we can say-- we're almost there-- that TF, our final temperature over our starting temperature to the 3/2 power, times our final volume over our starting volume is equal to 1. Now let's take this result that we worked reasonably hard to produce. Remember all of this, we just said, we're dealing with an adiabatic process, and we started from the principle of just what the definition of internal energy is. And then we substitute it with our PV equals nRT formulas. point is equal to 3/2 times PV. And then we integrated over all the changes, and we said, look, this is adiabatic. So the total change-- the sum of all of our change in internal energy and work done by the system has to be 0, then we use the property of log to get to this result. Now let's do these for both of these adiabatic processes over here. So the first one we could do is this one where we go from volume B at T1 to volume C at T2. Watch the Carnot cycle video, if you forgot that. This was the VB All of these things up here were at temperature 1. All of the things down here were at temperature 2 So we're at temperature 1 up here, and temperature 2 down here, volume C. So let's look at that. So on that right part, that right process, our final temperature was temperature 2. So let me write it down. Temperature 2."
    },
    {
        "Q": "Around 2:00, for the equation for the very first question, why is molarity used instead of the number of moles present?",
        "A": "HCl will dissociate completely and form 0.500 moles of H3O+. Molarity is the number of moles present, i.e. the concentration.",
        "video_name": "JoGQYSTlOKo",
        "timestamps": [
            120
        ],
        "3min_transcript": "- [Voiceover] Let's say we're doing a titration and we start with 20 mL of .500 molar HCl. So we're starting with a strong acid, and to the strong acid, we're going to add a solution of a strong base. We're going to add a .500 molar solution of NaOH, and as we add the base, the pH is going to increase, and we can show this on our titration curve. So we put the pH on the y-axis, and on the x-axis we put the volume of base that we are adding. So in part A, our goal is to find the pH before we've added any of our sodium hydroxide, so the addition of 0.0 mL of our base. So if we haven't added any base, the only thing we have present is acid. We know that HCl is a strong acid, so it is going to ionize 100%. So if HCl donates a proton to H2O, then we get H3O plus, and we would get the conjugate base to HCl, which is Cl minus. let's go ahead and write that, .500 molar HCl, and since we know this is 100% ionization because we have a strong acid, that's the same concentration of hydronium ions that we'll have in solution, so we have .500 molar for the concentration of hydronium ions. Now it's easy to find the pH because we know that the pH is equal to the negative log of the concentration of hydronium ions. So let's go ahead and do the calculation. We take the negative log of the concentration of hydronium which is .500, so that gives us .301. So we plug this into here, and we get a pH equal to 0.301. So we can find that point on our titration curve. We've added 0.0 mL of base, and our pH looks like it's just above zero here on our titration curve, and we calculated it to be .301. Let's find another point on our titration curve, this time after we add 10 mL of our base. So we find 10 mL of our base right here, and we're trying to find this point. So what is the pH after we add 10 mL? Well it looks like it's pretty close to one. Let's see if we can calculate what the pH is. So if we're adding base, we know that the base that we're adding, the hydroxide ions that we're adding, are going to neutralize the hydronium ions that are already present. So first, let's calculate how many moles of hydronium ions that we had present here. So the concentration of hydronium ions is .500."
    },
    {
        "Q": "At 1:20, shouldn't it be hydroxonium not hydronium?",
        "A": "Hydroxonium and hydronium mean the same thing and both terms are in use.",
        "video_name": "JoGQYSTlOKo",
        "timestamps": [
            80
        ],
        "3min_transcript": "- [Voiceover] Let's say we're doing a titration and we start with 20 mL of .500 molar HCl. So we're starting with a strong acid, and to the strong acid, we're going to add a solution of a strong base. We're going to add a .500 molar solution of NaOH, and as we add the base, the pH is going to increase, and we can show this on our titration curve. So we put the pH on the y-axis, and on the x-axis we put the volume of base that we are adding. So in part A, our goal is to find the pH before we've added any of our sodium hydroxide, so the addition of 0.0 mL of our base. So if we haven't added any base, the only thing we have present is acid. We know that HCl is a strong acid, so it is going to ionize 100%. So if HCl donates a proton to H2O, then we get H3O plus, and we would get the conjugate base to HCl, which is Cl minus. let's go ahead and write that, .500 molar HCl, and since we know this is 100% ionization because we have a strong acid, that's the same concentration of hydronium ions that we'll have in solution, so we have .500 molar for the concentration of hydronium ions. Now it's easy to find the pH because we know that the pH is equal to the negative log of the concentration of hydronium ions. So let's go ahead and do the calculation. We take the negative log of the concentration of hydronium which is .500, so that gives us .301. So we plug this into here, and we get a pH equal to 0.301. So we can find that point on our titration curve. We've added 0.0 mL of base, and our pH looks like it's just above zero here on our titration curve, and we calculated it to be .301. Let's find another point on our titration curve, this time after we add 10 mL of our base. So we find 10 mL of our base right here, and we're trying to find this point. So what is the pH after we add 10 mL? Well it looks like it's pretty close to one. Let's see if we can calculate what the pH is. So if we're adding base, we know that the base that we're adding, the hydroxide ions that we're adding, are going to neutralize the hydronium ions that are already present. So first, let's calculate how many moles of hydronium ions that we had present here. So the concentration of hydronium ions is .500."
    },
    {
        "Q": "3:55 What sets off that movement/What causes it?\n\nThe other two steps in the molozonide breaking apart make sense, but this first one doesn't.",
        "A": "The instability of oxygen-oxygen bonds and electrostatic attraction. The bonds between oxygen are very weak and are prone to breaking. Once that bond breaks, the electrons will be attracted to a positive charge. Since there is a slight partial positive charge on the carbon, the electrons move into carbon s valence shell. However, this violates the octet rule and thus breaks an adjacent C-C bond in order to have only 8 electrons associated with carbon, setting off a chain of bonds breaking/electron movement.",
        "video_name": "bFj3HpdC4Uk",
        "timestamps": [
            235
        ],
        "3min_transcript": "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 between the oxygen and the carbon. And I'm going to say that these pi electrons here in red, those are the ones that formed this bond between this carbon and this oxygen. And then finally, these electrons-- I'm saying these are my pi electrons in here-- are going to move off onto this top oxygen. So you could say that those magenta electrons would be right there. And so now we have this structure. Now oxygen oxygen bonds are relatively weak. So they're unstable and so we have two oxygen oxygen bonds in this molecule. And so one of those oxygen oxygen bonds is going to break in the next step of the mechanism. And so I could pick either one since they're symmetrical. I'm just going to say that--- I'm just going to say that these electrons over here-- so I'm going to say that this oxygen oxygen bond is going to break in the next part of our mechanism. And so, if we think about these electrons and this oxygen moving in here, that would break this bond between the two to the carbon on the right. So this bond is going to break, push those electrons into here, and then these electrons in green are going to come off onto the top oxygen [INAUDIBLE]. So our unstable oxygen oxygen bond breaks. And so let's go ahead and draw what we would get. Well, on the left side, we have a carbon. And now that carbon is going to have two bonds to the oxygen. And the oxygen is going to have two lone pairs of electrons. It would make a carbonyl compound. On the right, the carbon on the right is bonded to two other things and now has a double bond to this oxygen here. And now this oxygen has only one lone pair of electrons on it. And then this oxygen is bonded to the other oxygen, and that oxygen now has three lone pairs of electrons, which would give it a negative 1 formal charge. So we have a negative 1 formal charge here, this oxygen right here gets a positive 1 formal charge, and we form a carbonyl oxide on the right. So this charge compound on the right's"
    },
    {
        "Q": "Why do seismic waves travel faster through denser material (4:45) ?",
        "A": "You can think that in a dense material the molecules are very closer together and when a wave hits one molecule it takes less time for it to reach the next one and the wave then travles faster. In air the molecules are far apart and the wave than travles slower.",
        "video_name": "yAQSucmHrAk",
        "timestamps": [
            285
        ],
        "3min_transcript": "All of a sudden, the waves were reaching there faster. The slope of this line changed. It took less time for each incremental distance. So for some reason, the waves that we're going at these farther stations, the stations that were more than 200 kilometers away, somehow they were accelerated. Somehow they were able to move faster. And he's the one that realized that this was because the waves that were getting to these further stations must have traveled through a more dense layer of the earth. So let's just think about it. So if we have a more dense layer, it will fit this information right over here. So if we have a layer like this, which we now know to be the crust, and then you have a denser layer, which we now know to be the mantle, then what you would have is-- so you have your earthquake right over here, closer by, while you're still within the crust, it would be proportional. And then let's say that this is exactly, this right here is 200 kilometers away. But then if you go any further, the waves would have to travel. They would travel, so they would go like this. And then they would get refracted even harder. So they would get refracted. So they would be a little bit curved ahead of time. But then they're going to a much denser material. Or it's not gradually dense, it's actually kind of a all of a sudden a considerably more dense material, so it will get refracted even more. And then it'll go over here. And since it was able to travel all of this distance in a denser material, it would have traveled faster along this path. And so it would get to this distance on the surface that's more than 200 kilometers away, it would get there faster. And so he said that there must be a denser layer that those waves are traveling through, which we now know to be the mantle. And the boundary between what we now know to be the crust and this denser layer, It's called the Mohorovicic discontinuity. And sometimes this is called the Moho for short. So that boundary between the crust and the mantle is now named for him. But this was a huge discovery, because not only was he able to tell us, based on the data-- based on, kind of, indirect data, just based on earthquakes happening, and measuring when the earthquakes reach different points of the earth-- that there probably is a denser layer. And if you do the math, under continental crust that denser layer is about 35 kilometers down. He was able to tell us that there is that layer. But even more importantly, he was able to give the clue that just using information from earthquakes, we could essentially figure out the actual composition of the earth. Because no one has ever dug that deep. No one has ever dug into the mantle, much less the outer core or the inner core. In the next few videos, we're going to kind of take this insight, that we can use information from earthquakes, to actually think about how we know that there is an outer liquid core"
    },
    {
        "Q": "At 2:33 how is the molecule cyclic? Does that mean it can form a ring structure? if so, i'm having trouble picturing how it would do so.",
        "A": "The molecule is cyclic because when you follow the bonds from one atom to another, you come back in a loop.to your starting point. The loop is not round like a circle but, because it is a closed loop, it is described as cyclic and as a ring ..",
        "video_name": "FaOOx6IZxV8",
        "timestamps": [
            153
        ],
        "3min_transcript": "with a partial positive charge. The bottom halogen is now an electrophile, so it wants electrons. It's going to get electrons from those pi electrons here, which are going to move out and nucleophilic attack that partially positively charged halogen atom. And then this lone pair of electrons is going to form a bond with this carbon at the same time these blue electrons move out onto the halogen. So when we draw the result of all those electrons moving around, we're going to form a bond between the carbon on the right and the halogen, and we use the magenta electrons to show that. So there's now a bond there. And so we used red electrons before to show these electrons in here forming a bond with the carbon on the left. That halogen had two lone pairs of electrons still on it, like that, which gives that halogen a plus 1 formal charge. in an earlier video. And if I think about that cyclic halonium ion, I think about the halogen being very electronegative. It's going to attract, I'll say the electrons in magenta again just to be consistent, closer towards it. So it's going to take away a little bit of electron density from this carbon right down here. So I'm going to say this carbon is partially positive. It's going to have some partial carbocationic character. So in the next step of the mechanism, water's going to come along. And water's going to function as a nucleophile. So one of the lone pairs of electrons on water is going to nucleophilic attack our electrophile, which is this carbon right here. And so when that lone pair of electrons on oxygen attacks this carbon, that's going to kick the electrons in magenta off onto your halogen. And so let's go ahead and draw the product. We're going to have, on the left carbon, this halogen now used to have two lone pairs of electrons. so now it looks like that. On the right, we still have the carbon on the right bonded to other things, except now it's bonded to what used to be our water molecule. So the oxygen is now bonded to the carbon. And there's still one lone pair of electrons on that oxygen, giving it a plus 1 formal charge. So let's go ahead and highlight these electrons here in blue. Those electrons in blue are the ones that formed this bond between the carbon and the oxygen. So we're almost done. The last step of the mechanism would just be an acid-base reaction. So another water molecule comes along, and one of the lone pairs of electrons on the water molecule is going to function as a base and take this proton, leaving these two electrons behind on the oxygen. And we are finally done. We have formed our halohydrin, right? So I have my halogen on one side. And then I now have my OH on the opposite side, like that."
    },
    {
        "Q": "At 6:50, how does the oxygen attach to the tertiary carbocation instead of bromine? Isn't the bromine atom more electronegative than oxygen to hold onto the bond?",
        "A": "Water is much more electrophilic than Br\u00e2\u0081\u00bb, and there is much more water available to attack the cyclic bromonium ion.",
        "video_name": "FaOOx6IZxV8",
        "timestamps": [
            410
        ],
        "3min_transcript": "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? And the bromine is going to swing over to the carbon on the left. It's now going to have three lone pairs of electrons around it, like that. And the methyl group that was down relative to the plane is going to be pushed up when that water nucleophilic attacks. And now the methyl group is up, and this oxygen is now going to be bonded to this carbon. And so we still have our two hydrogens attached to it, like that. And there's a lone pair of electrons on this oxygen, giving it a plus 1 formal charge. So once again, let's go ahead and highlight those electrons. I'll draw them in blue here. These electrons right here, those are the ones that formed this bond. So let's go back and let's think about the formation of that cyclic bromonium ion in a different way here. So on the left I showed the bromine adding from the top. If I think about the alkene portion of my starting material, well, there's"
    },
    {
        "Q": "At 4:27, Jay numbers one of the carbons as Beta Carbon 3. I don't understand why Carbon 3 is a Beta-carbon. It's not connected to the alpha carbon.",
        "A": "The \u00ce\u00b1-carbon is the carbon bearing the leaving group (C-2). So the \u00ce\u00b2-carbons are the ones next to it (C-1 and C-3).",
        "video_name": "uCW6154hPkc",
        "timestamps": [
            267
        ],
        "3min_transcript": "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, we have the yellow chlorine up and axial. When we go to the beta one carbon, the hydrogen in green is the only one that's anti-periplanar to the halogen. If I turn to the side here, it's easier to see that we have all four of those atoms in the same plane, so the green proton is anti-periplanar to the chlorine. The other hydrogen, the one in white, is not anti-periplanar, so it will not participate in our E two mechanism. We go to the beta two carbon, and this hydrogen in white is not anti-periplanar, and when we look at the down axial position, it's occupied by a methyl group, so that is where a hydrogen would need to be if it were to participate in an E two mechanism. For E two elimination in cyclohexanes, the halogen must be axial, so here is our halogen that's axial, and when the halogen is axial in this chair conformation, is this one in green as we saw in the video, so if a strong base comes along, and takes the proton in green, the electrons in here would move in to form our double bond, and these electrons come off onto the chlorine, so a double bond forms between the alpha and the beta one carbons, which would give us this as our only product, so we don't get a double bond forming between our alpha and our beta two carbon because we would need to have a hydrogen where our methyl group is, so this, if we did have a hydrogen here, this hydrogen would be anti-periplanar to our halogen, but in this case, we get only one product, so this is the only product observed, and we figured that out because we drew our chair conformation. Let's do another E two mechanism for a substituted cyclohexane, and I'll start by numbering my cyclohexane, so that's carbon one, and this is carbon two, this is carbon three, and this is carbon four."
    },
    {
        "Q": "at 9:52, why does he say that because the iso-propyl group is axial can't participate in the mechanism?",
        "A": "In order to get elimination of HCl, the Cl onC2 and the \u00ce\u00b2 H must be in a trans diaxial conformation. If the Cl is axial, the isopropyl group on C1 is also axial and the \u00ce\u00b2 H on C1 is equatorial. There is no axial H on C1, because the isopropyl group has replaced it. The axial \u00ce\u00b2 H must therefore come from C3.",
        "video_name": "uCW6154hPkc",
        "timestamps": [
            592
        ],
        "3min_transcript": "and next to that would be a beta carbon, and this beta hydrogen is anti-periplanar to this halogen. We have another beta carbon over here with another hydrogen that's anti-periplanar to this halogen. Finally, let's draw our two products, so let's take a proton from the beta one position first, so our base, let me draw it in here, so our strong base is going to take this proton, and so these electrons would move into here to form our double bond, and these electrons come onto the chlorine to form the chloride anion as our leaving group. So this would form a double bond between what I called carbons two and three 'cause this is carbon one, this is carbon two, this is carbon three, and this is carbon four, so we have a methyl group that's up at carbon one, so let me draw in our methyl group up at carbon one, so we put that on a wedge, so if that's carbon one, then this is carbon two, and this is carbon three, and that's where our double bond forms, so the double bond forms between carbons two and three, so let me go ahead and put that in, going away from us, we put that on a dash, so that's one product. If our base took this proton, then the electrons would move into here, and these electrons would come off onto the chlorine, so if we took a proton from the beta two carbon, we would form a double bond between carbons three and four, so here's carbons three and four, so I put a double bond in there. I still have a methyl group that's going up at what I called carbon one, and my isopropyl group is at carbon four, but since now, this carbon is sp two hybridized, I need to draw in this isopropyl group on a straight line, so sometimes, students would put this isopropyl group in on a wedge or a dash, but you're trying to show the planar geometry around this carbon, so a straight line is what you need, and so those are the two products. Let's do one more, and you can see this substrate in the previous example. The only difference is this time, the chlorine is on a wedge instead of a dash, so if I number my ring one, two, three, four, I've already put in both chair conformations to save time, so that's carbon one, this is carbon two, this is carbon three, and this is carbon four. On the other chair conformation, this is carbon one, two, three, and four, and notice for the chair conformation on the left, we have the chlorine in the axial position, so this would be the alpha carbon, and the carbons next to the alpha carbon would be the beta carbons, so this one on the right is a beta carbon, and the one on the left is a beta carbon. We need a proton that's anti-periplanar, and the only one that fits would be this hydrogen right here, so if a base takes that proton, so let me draw in our strong base, taking this proton, these electrons would move into here to form our double bond, the electrons come off onto our chlorine, and a double bond forms between carbons two and three,"
    },
    {
        "Q": "At 1:52 someone mentioned protozoans. What are protozoans",
        "A": "In some systems of biological classification, the Protozoa are a diverse group of unicellular eukaryotic organisms. Historically, protozoa were defined as single-celled organisms with animal-like behaviours, such as motility and predation.",
        "video_name": "1aJBToJrlvA",
        "timestamps": [
            112
        ],
        "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 6:17, why is the pz, px, py etc. used and what do the subscripts stand for?",
        "A": "In the p subshell there are three p orbitals: the px, py, and pz orbitals. These three orbitals are identical, except that they point in different directions (they are orthogonal to each other). The subscripts distinguish the p orbitals based on their orientation; if you draw an imaginary x-y-z axis with the origin at your atom of interest, then the px orbital points along the x-axis, the py orbital points along the y axis, and the pz orbital points along the z axis.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            377
        ],
        "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 1:57, the electron went way far out, how far out can that get?",
        "A": "There is no limit, but the probability of being very far away is infinitesimal.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            117
        ],
        "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": "In 11:50, he explains about the structure of Nitrogen. It is 1s^2, 2s^2,2p^3. How can there be 3 atoms in p, where it is not possible when its configuration is s?",
        "A": "An orbital can have at most 2 electrons. In the s subshell, there is only one orbital, thus an s subshell is full with 2 electrons. In the p subshell, there are three orbitals, each can have at most 2 electrons, for a total maximum of 6 electrons. Similarly, the d subshell has 5 orbitals for a maximum of 10 electrons. The f subshell has 7 orbitals for a maximum of 14 electrons. Thus, in nitrogen each of the three p orbitals in the p subshell has one electron, for a total of 3 electrons in the 2p subshell.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            710
        ],
        "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": "When you were in the second period at 8:01 to 8:07 you only count B for boron before making it 2p. Why didnt you count Li lithium and Be Berrilium?",
        "A": "The valence electrons of Li and Be occupy the 2s orbital, not the 2p. Thus, they don t have anything to do with the 2p.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            481,
            487
        ],
        "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": "Around 12:55 it explains silicone and i dont get how you calculate the electron configuration. I find it terrible confusing. how you have added up the numbers. which numbers do i add up to make the electron configuration?",
        "A": "Firstly silicon is not the same as silicone (be careful on exams!). For a neutral atom the number of electrons = the number of protons. For silicon that is 14. The fill order is: 1s 2s, 2p 3s, 3p 4s (google orbital fill diagram) An s orbital can take 2 electrons, a p orbital can take 6 electrons. We just keep filling until we get to 14. 1s2 (12 left), 2s2 (10 left), 2p6 (4 left), 3s2 (2 left), 3p2 (none left). Sorry can t do the superscripts here.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            775
        ],
        "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 05:38 it shows wierd looking things, what are those?",
        "A": "Those weird looking things are orbitals, the s, p, d and f orbitals. These orbitals are the mathematical functions of the regions of space about the nucleus in which the electrons can be found, maximum of 2 electrons per orbital.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            338
        ],
        "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": "What are the diagrams at 04:49 , I don't know them even though Sal mentioned them.",
        "A": "These diagrams show the spacial geometry of atomic orbitals. The s orbitals (first column) are actually spheres, but they are shown here as cross-sections to show their nodes (places where the probability of finding an electron is 0). The red shows where the formula describing the orbital (called a wavefunction, from quantum mechanics) has a positive value, and the blue is where it has a negative value.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            289
        ],
        "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": "At 11:17, why is Be 2s one when Li is 1s two? They are right beside each other.",
        "A": "He was only doing lithium. The complete electron configuration of Li is 1s2 2s1. The electron configuration of Be is 1s2 2s2.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            677
        ],
        "3min_transcript": "It will look like this. You take a bunch of observations, and you say, wow, it's a lot more likely to find that electron in kind of the dumbell, in that dumbbell shape. But you could find it out there. You could find it there. You could find it there. This is just a much higher probability of finding it in here than out here. And that's the best way I can think of to visualize it. Now what we were doing here, this is called an electron configuration. And the way to do it-- and there's multiple ways that are taught in chemistry class, but the way I like to do it -- is you take the periodic table and you say, these groups, and when I say groups I mean the columns, these are going to fill the s subshell or the s orbitals. You can just write s up here, just right there. These over here are going to fill the p orbitals. Actually, let me take helium out of the picture. 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"
    },
    {
        "Q": "In 4:50, does the red and blue represent + 1/2 and -1/2 spin?",
        "A": "No, the different colours do not represent the spin as if you look at the 4 lobes of the d orbital (for example the dxz sub-shell), there are two red and two blue sections, but at most it will only contain 2 electrons, which can be found in any of the 4 lobes and not one in each section or even pair of sections.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            290
        ],
        "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": "how can we place helium[inert gas]in the second group \"10:20\"?",
        "A": "becouse it has two electrons in its outer shell just like the other elements of the second group. the rest of the noblegasses have 8 electrons in their outer shell",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            620
        ],
        "3min_transcript": "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, It will look like this. You take a bunch of observations, and you say, wow, it's a lot more likely to find that electron in kind of the dumbell, in that dumbbell shape. But you could find it out there. You could find it there. You could find it there. This is just a much higher probability of finding it in here than out here. And that's the best way I can think of to visualize it. Now what we were doing here, this is called an electron configuration. And the way to do it-- and there's multiple ways that are taught in chemistry class, but the way I like to do it -- is you take the periodic table and you say, these groups, and when I say groups I mean the columns, these are going to fill the s subshell or the s orbitals. You can just write s up here, just right there. These over here are going to fill the p orbitals. Actually, let me take helium out of the picture. 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."
    },
    {
        "Q": "I dont get how to use the chart at 4:46?",
        "A": "The chart show the shapes of each of the orbitals in a given energy level. For example, for n = 4, the chart shows in turn the shapes of the orbitals: 4s, 4px, 4py, 4pz, the five 4d orbitals, and some of the 4f orbitals.Except for levels 1 to 3, you do not have to memorize the shapes. It is more of a reference tool when you want to know what the shapes look like.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            286
        ],
        "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": "great vid Sal. But wouldn't @11:11 Lithium be 2s^1 then 2s^2?",
        "A": "You would have to fill up the lowest orbital, which means the electron configuration always starts with 1s^1, 1s^2 and so forth. When writing it s like taking the electron configuration of the rightmost element immediately above the period of that element and then writing out the next configuration of the period it is in.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            671
        ],
        "3min_transcript": "It will look like this. You take a bunch of observations, and you say, wow, it's a lot more likely to find that electron in kind of the dumbell, in that dumbbell shape. But you could find it out there. You could find it there. You could find it there. This is just a much higher probability of finding it in here than out here. And that's the best way I can think of to visualize it. Now what we were doing here, this is called an electron configuration. And the way to do it-- and there's multiple ways that are taught in chemistry class, but the way I like to do it -- is you take the periodic table and you say, these groups, and when I say groups I mean the columns, these are going to fill the s subshell or the s orbitals. You can just write s up here, just right there. These over here are going to fill the p orbitals. Actually, let me take helium out of the picture. 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"
    },
    {
        "Q": "configuration of berilium is 2s1 or 2s2 at 11:30",
        "A": "Beryllium has four electrons, so its configuration is 1s\u00c2\u00b22s\u00c2\u00b2.",
        "video_name": "FmQoSenbtnU",
        "timestamps": [
            690
        ],
        "3min_transcript": "It will look like this. You take a bunch of observations, and you say, wow, it's a lot more likely to find that electron in kind of the dumbell, in that dumbbell shape. But you could find it out there. You could find it there. You could find it there. This is just a much higher probability of finding it in here than out here. And that's the best way I can think of to visualize it. Now what we were doing here, this is called an electron configuration. And the way to do it-- and there's multiple ways that are taught in chemistry class, but the way I like to do it -- is you take the periodic table and you say, these groups, and when I say groups I mean the columns, these are going to fill the s subshell or the s orbitals. You can just write s up here, just right there. These over here are going to fill the p orbitals. Actually, let me take helium out of the picture. 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"
    },
    {
        "Q": "At 00:42 Sal says that the reaction is \"spontaneous\", but I don't exactly understand what that means. How did he figure that out?",
        "A": "Spontaneous means that somethings starting energy is higher than its ending, it comes from something called Gibbs free energy. This means it releases energy and doesn t need energy to proceed.",
        "video_name": "-KE7jTXwNYs",
        "timestamps": [
            42
        ],
        "3min_transcript": "- We've talked a lot about ATP being the energy currency of cells, but I want to dig a little bit deeper into that in this video. And as we'll see when we go from ATP to ADP, ADP + a Phosphate group, we have a release of free energy. If we look at just the system, ATP's free energy is over here, but once hydrolysis has taken place and now it's ADP + a Phosphate group, the free energy has dropped by roughly 30.5 KJ/mol our Delta G is -30.5 Kilojoules per mole and if you watch the videos on Gibbs Free Energy this tells us that this is a spontaneous reaction. Delta, Delta G is less than zero, which tells us that this is going to be spontaneous. Now when I first learned this I was like, well if it's going to be spontaneous why wouldn't all of the ATP just spontaneously and all of the water turn into ADP and just release it's energy as heat or whatever else? And the key is, it has to get over this hump. it has to go uphill a little bit if there's no enzyme to catalyze it. And the reason why we have this uphill hump is the way that ATP gets broken up is that you have to, if we're talking about a water molecule doing the hydrolysis which is typically what people think about hydrolysis, although it can sometimes be done by a different molecule, if you think about the water molecule what needs to happen is, you need this lone pair of electrons on this oxygen to be able to do what we call a nucleophilic attack on this Phosphorous in this phosphate group and if that happens, it forms this bond and then these electrons can be taken back by this oxygen which gives its negative charge right over there. Now you might say, \"Okay, this makes a lot of sense\", but you have to rememeber this, electrons are negative and they're surrounded by these negative charges. So they have to overcome getting close to these things so as they're approaching these negative charges, they want to repel each other, so you have to overcome that. And the way that it's overcome is a class of enzymes And what they do is, remember these enzymes are these big protein complexes and the ATP molecule combined in the right place. And they essentially try and surround the ATP molecule with some positive ions. So let's say there's a positive ion here. So it can keep these electrons busy while the water, or whatever is doing the nucleophilic attack, can not have to worry about these electrons over here. So it might have a positive ion here. And remember, if we think in three dimensions, this thing is all wrapped around in different ways around the ATP molecule, so this is the ATPase. And so by having the enzyme over here, you're lowering the activation energy. And so it might end up looking something more like this. And so the reaction can actually occur. So the reason why you don't just see this happening all the time without an enzyme is why you have to overcome this hump."
    },
    {
        "Q": "At 2:06 NH4Cl is called an acid, but isn't it a salt?",
        "A": "It is a salt, but NH4+ is ammonium, which is the conjugate acid of ammonia (NH3).",
        "video_name": "lsHq5aqz4uQ",
        "timestamps": [
            126
        ],
        "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": "how can i identify that solution is buffer solution ? And at 4:35 how does he know that whole of the NH4Cl is going to dissociate into 0.20M of NH4+",
        "A": "You need to identify the conjugate acids and bases, and I presume that comes with practice. The same way you know that HCl dissolves to form H+ and Cl-, or H2SO4 form 2H+ and (SO4)2-. In this example with NH4Cl, the conjugate acids and bases are NH4+ and Cl-. Notice how also the way the formula is written will help you identify the conjugate acids and bases (acids come first on the left, bases on the right)",
        "video_name": "lsHq5aqz4uQ",
        "timestamps": [
            275
        ],
        "3min_transcript": "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. the same amount of ammonium over here. So we're gonna be left with, this would give us 0.19 molar for our final concentration of ammonium. Hydroxide we would have zero after it all reacts, And then the ammonium, since the ammonium turns into the ammonia, if we lose this much, we're going to gain the same concentration of ammonia. So over here we put plus 0.01. So the final concentration of ammonia would be 0.25 molar. And now we can use our Henderson-Hasselbalch equation. So let's go ahead and plug everything in. So ph is equal to the pKa. We already calculated the pKa to be 9.25. And then plus, plus the log of the concentration of base, all right, that would be NH three. So the concentration of .25."
    },
    {
        "Q": "At 8:48, why is it that you first have to react HCL with H2O to get H3O, and in turn, react the with NH3 instead of directly reacting HCL with HN3 as you did (with NH4 and NaOH) in the first example? I would like to know, for future examples, what instances I would have to utilize this method instead of the first method.",
        "A": "The HCl exists in water almost exclusively as H\u00e2\u0082\u0083O\u00e2\u0081\u00ba and Cl\u00e2\u0081\u00bb. So it is more correct chemically to write the reaction using H\u00e2\u0082\u0083O\u00e2\u0081\u00ba. In practice, you can do it either way. What matters is the concentration of NH\u00e2\u0082\u0084\u00e2\u0081\u00ba, not the equations you use to produce it.",
        "video_name": "lsHq5aqz4uQ",
        "timestamps": [
            528
        ],
        "3min_transcript": "So this shows you mathematically how a buffer solution resists drastic changes in the pH. Next we're gonna look at what happens when you add some acid. So we're still dealing with our same buffer solution with ammonia and ammonium, NH four plus. But this time, instead of adding base, we're gonna add acid. So we add .03 moles of HCl and let's just pretend like the total volume is .50 liters. And our goal is to calculate the pH of the final solution here. So the first thing we could do is calculate the concentration of HCl. So that would be moles over liters. So that's 0.03 moles divided by our total volume of .50 liters. And .03 divided by .5 gives us 0.06 molar. And HCl is a strong acid, so you could think about it as being H plus and Cl minus. And since this is all in water, H plus and H two O would give you H three O plus, or hydronium. So .06 molar is really the concentration of hydronium ions in solution. And so the acid that we add is going to react with the base that's present in our buffer solution. So this time our base is going to react and our base is, of course, ammonia. So let's write out the reaction between ammonia, NH3, and then we have hydronium ions in solution, H 3 O plus. So this reaction goes to completion. And if ammonia picks up a proton, it turns into ammonium, NH4 plus. And if H 3 O plus donates a proton, we're left with H 2 O. So we write H 2 O over here. For our concentrations, we're gonna have .06 molar And the concentration of ammonia is .24 to start out with. So we have .24. And for ammonium, it's .20. So we write 0.20 here. So all of the hydronium ion is going to react. So we're gonna lose all of it. So we're left with nothing after it all reacts. So it's the same thing for ammonia. So that we're gonna lose the exact same concentration of ammonia here. So we're gonna lose 0.06 molar of ammonia, 'cause this is reacting with H 3 O plus. And so after neutralization, we're left with 0.18 molar for the concentration of ammonia. And whatever we lose for ammonia, we gain for ammonium since ammonia turns into ammonium. So we're going to gain 0.06 molar"
    },
    {
        "Q": "At 3:52 what life could survive without nitrogen?",
        "A": "Nothing. Nitrogen is one of the building blocks of protein. As all living things need protein, all life would die - plants and animals.",
        "video_name": "6rwoktPmqpY",
        "timestamps": [
            232
        ],
        "3min_transcript": "you'll notice here that nitrogen gas is made up of two nitrogen atoms stuck together with a triple bound. It's one thing to break apart a single covalent bond, but three? As you can imagine those two nitrogen atoms are a total pain to pry apart, but that molecule has to be split in order for a plant to get at the pieces. In fact, plants can assimilate a bunch of different forms of nitrogen, nitrates, nitrites to a lesser extent and even ammonium, which is what you get when you mix ammonia with water, but all that darn nitrogen gas in the atmosphere is beyond their powers of assimilation. So, plants need help taking advantage of this ocean of nitrogen that we're all swimming in, which is why they need to have that nitrogen fixed so that they can use it. Even though plants aren't wile enough to wrangle those two nitrogen atoms apart, certain nitrogen fixing bacteria are. These bacteria hang out in soil or water or even form symbiotic relationships with the root nodules of some plants, most of which are legumes. That's a pretty big family of plants, soybeans, clover, peanuts, kudzu, all legumes. which then becomes ammonium when it's mixed with water which can be used by plants. They do this with a special enzyme called nitrogenase which is the only biological enzyme that can break that crazy triple bond. Ammonia can also be made by decomposers, fungi, protists; other kinds of bacteria that munch on your proteins and DNA after you die, but they're not picky they like poop and urine too. Then once this has happened, other bacteria known as nitrifying bacteria can take this ammonia and convert it into nitrates; three oxygen atoms attached to a single nitrate atom and nitrite; two oxygens attached to a nitrogen. Those are even easier than ammonium for plants to assimilate. So, the take home here is, if it wasn't for these bacteria there would be a whole lot less of biologically available nitrogen hanging around and as a result there would be a lot fewer living things on the planet. As usual, thanks bacteria we owe you one, but I should mention that it's not just bacteria who can wrangle those two nitrogen atoms apart, lightning of all things has enough energy to break bonds between nitrogens, And in the 20th century, smarty pants humans also figured out various ways to synthetically fix a ton of nitrogen all at once, which is why we have synthetic fertilizers now, there's so much food growing all over the place. Once the atmospheric nitrogen is converted into a form that plants can use to make DNA and RNA and amino acids, organic nitrogen takes off up the food chain, animals eat the plants and use all that sweet, sweet, bio-available nitrogen to make our own amino acid. Then we pee or poop it out or die and the decomposers go to town on it, breaking it down into ammonia and it just keeps going, until one day, that organic nitrogen finds itself in denitrifying bacteria who's job it is to metabolize the nitrogen oxides and turn them back into nitrogen gas using a special enzyme called nitrate reductase. These guys do their business and then release the N2 back in the atmosphere. That my friends is the nitrogen cycle, if you remember nothing else remember that a) you owe bacteria a solid because they were smart enough to make an enzyme"
    },
    {
        "Q": "I have a problem with how we include friction in this problem. If the entire premise of the easy style of doing these problems is to treat the system as a single object, shouldn't we use 20 kg for the friction component, and not 12?\n\nIt makes sense to me that the box is being pressed down on the table by the additional force, and the normal force pressing back upwards would also be 20 kg * 9.8, rather than 12 kg *9.8.\n\nThis occurs at 4:40",
        "A": "How is the box pressing down with 20 * 9.8 N of force? The force from the 3 and 5 kg blocks on the 12 kg block comes from the ropes that are going over the pulleys so it is horizontal and not not down so it doesn t add to the normal force.",
        "video_name": "ibdidr-bEvI",
        "timestamps": [
            280
        ],
        "3min_transcript": "to reduce the acceleration? Yeah, there's this force of gravity over here. This force of gravity on the three kilogram mass is trying to prevent the acceleration because it's pointing opposite the direction of motion. The motion of this system is upright and down across this direction but this force is pointing opposite that direction. This force of gravity right here. So, I'm gonna have to subtract three kilograms times 9.8 meters per second squared. Am I gonna have any other forces that try to prevent the system from moving? You might think the force of gravity on this 12 kilogram box, but look, that doesn't really, in and of itself, prevent the system from moving or not moving. That's perpendicular to this direction. I've called the direction of motion, this positive direction. If it were a force this way, if it were a force this way or a force that way it'd try to cause acceleration of the system. by the normal force, so I don't even have to worry about that force. So, are there any forces associated with the 12 kilogram box that try to prevent motion? It turns out there is. There is going to be a force of friction between the table because there's this coefficient of kinetic friction. So, I've got a force this way, this kinetic frictional force, that's gonna be, have a size of Mu K times f n. That's how you find the normal force and so this is gonna be minus, the Mu K is 0.1 and the normal force will be the normal force for this 12 kilogram mass. So, I'll use 12 kilograms times 9.8 meters per second squared. You might object, you might say, \"Hey, hold on, 12 times 9.8, that's the force of gravity. \"Why are you using this force? \"I thought you said we didn't use it?\" Well, we don't use this force by itself, but it turns out this force of friction depends on this force. So, we're really using a horizontal force, which is why we've got this negative sign here, but it's a horizontal force. It just so happens that this horizontal force depends on a vertical force, which is the normal force. And so that's why we're multiplying by this .1 that turns this vertical force, which is not propelling the system, or trying to stop it, into a horizontal force which is trying to reduce the acceleration of the system. That's why I subtracted and then I divide by the total mass and my total mass is gonna be three plus 12 plus five is gonna be 20 kilograms. Now, I can just solve. If I solve this, I'll get that the acceleration of this system is gonna be 0.392 meters per second squared. So, this is a very fast way. Look it, this is basically a one-liner. If you could put this together right, it's a one-liner. There's much less chance for error than when you're trying to solve three equations with three unknowns. This is beautiful."
    },
    {
        "Q": "At 0:13 Sal uses a sign to represent the angle of the launch. I wanted to know what that sign means? Thanks.",
        "A": "It s the Greek letter theta, which stands for angle (it s often used in place of an unknown angle). Hope this helps!",
        "video_name": "RhUdv0jjfcE",
        "timestamps": [
            13
        ],
        "3min_transcript": "Let's say we're going to shoot some object into the air at an angle. Let's say its speed is s and the angle at which we shoot it, the angle above the horizontal is theta. What I want to do in this video is figure out how far this object is going to travel as a function of the angle and as a function of the speed, but we're going to assume that we're given the speed. That that's a bit of a constant. So if this is the ground right here, we want to figure out how far this thing is going to travel. So you can imagine, it's going to travel in this parabolic path and land at some point out there. And so if this is at distance 0, we could call this distance out here distance d. Now whenever you do any problem like this where you're shooting something off at an angle, the best first step is to break down that vector. Remember, a vector is something that has magnitude and direction. The magnitude is s. Maybe feet per second or miles per hour. So if you have s and theta, you're giving me a vector. And so what you want to do is you want to break this vector down into its vertical and horizontal components first and then deal with them separately. One, to help you figure out how long you're in the air. And then, the other to figure out how far you actually travel. So let me make a big version of the vector right there. Once again, the magnitude of the vector is s. So you could imagine that the length of this arrow is s. And this angle right here is theta. And to break it down into its horizontal and vertical components, we just set up a right triangle and just use our basic trig ratios. So let me do that. So this is the ground right there. I can drop a vertical from the tip of that arrow to set up a right triangle. And the length of the-- or the magnitude of the vertical component of our velocity is going to be this That is going to be-- you could imagine, the length of that is going to be our vertical speed. So this is our vertical speed. Maybe I'll just call that the speed sub vertical. And then, this right here, the length of this part of the triangle-- let me do that in a different color. The length of this part of the triangle is going to be our horizontal speed, or the component of this velocity in the horizontal direction. And I use this word velocity when I specify a speed and a direction. Speed is just the magnitude of the velocity. So the magnitude of this side is going to be speed horizontal. And to figure it out, you literally use our basic trig ratios. So we have a right triangle. This is the hypotenuse. And we could write down soh cah toa up here."
    },
    {
        "Q": "at 3:23 shouldn't the hypothesis be in an if, then statement?",
        "A": "You can set up a hypothesis as an if/then statement, but it isn t always required. Hope that helps!",
        "video_name": "N6IAzlugWw0",
        "timestamps": [
            203
        ],
        "3min_transcript": "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. freezing point. Lower freezing point, so it takes colder temperatures to freeze it than fresh water. Than fresh water. So, this, right over here, this would be a good hypothesis. It doesn't matter whether the hypothesis is actually true or not. We haven't actually run the experiment, but it's a good one, because we can construct an experiment that tests this very well. Now, what would be an example of a bad hypothesis or of something that you couldn't even necessarily consider as part of the scientific method? Well, you could say that there is a fairy that blesses that blesses, let's say that performs magic, performs magic on the pond to freeze it faster."
    },
    {
        "Q": "oxygen and carbon are having a triple bond at 6:17 the valency of oxygen is 2 how is it possible?",
        "A": "I think this is because the new third bond is a dative covalent bond , when both electrons of electron pair shared to form a bond is from ONE atom (in this case, oxygen) only. It can be wrong so plz look it up :)",
        "video_name": "vFfriC55fFw",
        "timestamps": [
            377
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 2:59, what is a photon?",
        "A": "Its the basic unit of light. Its the smallest amount of light that you can play around(create, reflect, refract) with.",
        "video_name": "y55tzg_jW9I",
        "timestamps": [
            179
        ],
        "3min_transcript": "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. I actually like to think of it as kind of a, as a bit of a vehicle, and to imagine that, let's imagine that I had a car. So let me draw a car. So we're looking at the top of a car. So this is the passenger compartment, and it has four wheels on the car. We're looking at it from above. And let's say it's traveling on a road. It's traveling on a road. On a road, the tires can get good traction. The car can move pretty efficiently, and it's about to reach an interface it's about to reach an interface where the road ends and it will have to travel on mud. It will have to travel on mud. Now on mud, obviously, the tires' traction will not be as good. The car will not be able to travel as fast. So what's going to happen? Assuming that the car, the steering wheel isn't telling it to turn or anything, the car would just go straight in this direction. But what happens right when--which wheels are going to reach the mud first? Well, this wheel. This wheel is going to reach the mud first."
    },
    {
        "Q": "At 6:30, don't you need to take the inverse of that expression to get the equivalent resistance of the 2 resistors in parallel?",
        "A": "There are two formulas for computing 2 parallel resistors. The one I used is Rp = (R1 R2)/(R1+R2). The other formula is the one with all the reciprocals. That s the one you are thinking of: 1/Rp = 1/R1 + 1/R2. Both give the same answer.",
        "video_name": "j-iR7puLj6M",
        "timestamps": [
            390
        ],
        "3min_transcript": "If you look here, I have two batteries that are hooked up, their inputs, their positive side is hooked up together and their negative side is hooked up together, so they're actually just acting like one, big battery, so let me draw that. I'm going to draw the circuit again so it looks like this. Here's my combined big battery. And it goes to... Relabel these again so we don't get mixed up. Okay. This is R1. This is R2. So this circuit looks a little simpler, and I'm gonna look at it again, see if I can do any more simplification, so what I recognize right here, right in this area right here, R1 and R2 are in parallel. They have the same voltage on their terminals. That means they're in parallel. I know how to simplify parallel resistors. We'll just use the parallel resistor equation that I have in my head, and that looks like this, let's go to this color here. Okay, so parallel resistors, R1 in parallel with R2. I made up this symbol, two vertical lines, that means they're in parallel, and the formula for two parallel resistors is R1 times R2, over R1 plus R2. Now I'll plug in the values. over here at our schematic, they're the same value, and that has a special thing when in parallel resistors so it's actually R R over 2R. Because those resistors are the same. And you can see I can cancel that and I can cancel that and two parallel resistors, if the resistors are equal, is equal to half the resistance. And let's plug in the real values, 1.4 ohms over 2 equals 0.7 ohms. That's the equivalent resistance of these two resistors in parallel. So this is a good time to redraw this circuit again. Let's do it again. Here's our battery. This time I'm going to draw the equivalent resistance. Then we have R3,"
    },
    {
        "Q": "at 7:30,what are microtubules?",
        "A": "mi\u00c2\u00b7cro\u00c2\u00b7tu\u00c2\u00b7bule A microscopic tubular structure present in numbers in the cytoplasm of cells, sometimes aggregating to form more complex structures",
        "video_name": "X1bmedVziGw",
        "timestamps": [
            450
        ],
        "3min_transcript": "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 near the cell membranes. After being pulled in this direction and that, the chromosomes line up right down the middle of the cell and that brings us to the latest installment of biolo-graphy. (music) So how do chromosomes line up like that? We know that there are motor proteins involved, but like, how? What are they doing? Well, remember when I said earlier that there were a lot of things that we don't totally understand about mitosis? It's sort of weird that we don't because we can literally watch mitosis happening under microscopes, but chromosome alignment is a good example of a small detail that is only very recently been figured out. And it was a revelation like 130 years in the making. Mitosis was first observed by a German biologist by the name of Walther Flemming, who in 1878 was studying the tissue of salamander gills and fins when he saw cell's nuclei split in two and migrate away from each other to form two new cells. He called this process mitosis because of the messy jumble of chromatin, a term he also coined that he saw on the nuclei. But Flemming didn't pick up on the implications of this discovery for genetics, which was still a young discipline and over the next century generations of scientists started piecing together the mitosis puzzle by determining the role of microtubules, say, or identifying motor proteins. And the most recent contribution to this research was made by a postdoctoral student named Tomomi Kiyomitsu at MIT. He watched the same process that Flemming watched and figured out how, at least one, of the motor proteins helps snap the chromosomes into line. He was studying the motor protein called dynein which sits on the inside of the membrane. Think of the microtubules as tug-of-war ropes with the chromosomes as the flag in the middle. What Kiyomitsu discovered was that the dynein plays tug-of-war with itself. Dynein grabs on to one end of the microtubules and pulls the tubules and chromosomes towards one end of the cell. When the ends of the microtubules come too close to the cell membrane they release a chemical signal that punts the dynein"
    },
    {
        "Q": "At 9:07 does mol mean molecule or is that what it is called.",
        "A": "mol means mole (which is Avagadro s number of an atom or molecule).",
        "video_name": "-QpkmwIoMaY",
        "timestamps": [
            547
        ],
        "3min_transcript": "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 And this makes sense. This liter will cancel out with that liter. That atmospheres cancels out with that atmospheres. I'm about to multiply it by temperature right here in kelvin. We'll cancel out there. And then we'll have a 1 over moles in the denominator. A 1 over moles in the denominator will just be a moles because you're going to invert it again. So that gives us our answer in moles. And so finally our temperature-- and you've got to remember you've got to do it in kelvin. So 25 degrees Celsius-- let me right it over here-- 25 degrees Celsius is equal to, you just add 273 to it, so this is equal to 298 kelvin. So times 298 kelvin. And now we just have to calculate this. So let's do that."
    },
    {
        "Q": "At 13:26 I don't understand why Sal multiplied the H2O g with the density of water that's stated in the question above. I would love an explanation.",
        "A": "Density is mass per volume of a substance. So Density=Mass/Volume (which is why the unit is grams/mL). We have the mass for water = 980grams and density of water (at the same temp) as 0.997g/ml. so substituting the value we can get the volume of water as 980g/0.997g/mL which gives us the volume in mL.",
        "video_name": "-QpkmwIoMaY",
        "timestamps": [
            806
        ],
        "3min_transcript": "So this is equal to 54.4 moles. And we could see this liters cancels out with that liters. Kelvin cancels out with kelvin. Atmospheres with atmospheres. You have a 1 over mole in the denominator. So then 1 over 1 over moles is just going to be moles. Now, this is going to be 54.4 moles of water vapor in the room to have our vapor pressure. If more evaporates, then more will condense-- we will be beyond our equilibrium. So we won't ever have more than this amount evaporate in that room. So let's figure out how much liquid water that actually is. Let me do it over here. So 54.4 moles-- let me write it down-- moles of H2O. That's going to be in its vapor form and its going to evaporate. So what is the molar mass of water? Well, it's roughly 18. I actually figured it out exactly. It's actually 18.01 if you actually use the exact numbers on the periodic table, at least one that I used. So we could say that there's 18.01 grams of H2O for every 1 mole of H2O. And obviously, you can just look up the atomic weight of hydrogen, which is a little bit over 1, and the atomic weight of oxygen, which is a little bit below 16. So you have two of these. So 2 plus 16 gives you pretty close to 18. So this right here will tell you the grams of water that can evaporate to get us to that equilibrium pressure. So let's get the calculator out. So we have the 54.4 times 18.01 is equal to 970-- well, this 0.7, it becomes 980. So this is 980 grams of H2O needs to evaporate for us to get to our equilibrium pressure, to our vapor pressure. So let's figure out how many milliliters of water this is. So they tell us the density of water right here. 0.997-- let me do this in a darker color-- 0.997 grams per millileter. Or another way you could view this is for everyone 1 milliliter you have 0.997 grams of water at 25 degrees Celsius. So for every milliliter-- this is grams per milliliter-- we want milliliters per gram because we want this and this to cancel out. So we're essentially just going to divide 980 by 0.997."
    },
    {
        "Q": "At 8:30, why is the volume of the room used instead of the volume of water?",
        "A": "As the liquid water sits in the container, it releases water vapour which spreads throughout the room until it is at an equilibrium with the liquid water. There will be equal amounts of water vapour in every spot of the room and Sal wants to know how many total molecules of vapour there are. So, he takes the entire volume of the room and not just that of the liquid water.",
        "video_name": "-QpkmwIoMaY",
        "timestamps": [
            510
        ],
        "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": "10:50 \"You could treat a Covalent Bond like an Ionic Bond\", this confuses me. Does an electron actually move into the other atoms shell in an Ionic Bond ? or is the Ionic Bond just a Covelant Bond, where the probability of finding the shared electron is much higher around the more Electroniegative atom ?",
        "A": "It s somewhere in between. In a completely ionic bond, the electron would move into the other atom s shell and not be shared at all. In most, if not all, actual ionic bonds, the probability of not finding the shared electron in an orbital around the more electronegative atom is very low, so we just say that the electron moved to the more electronegative atom.",
        "video_name": "126N4hox9YA",
        "timestamps": [
            650
        ],
        "3min_transcript": "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. You could draw the dot structure above, and this would be considered be correct. You could draw it like this. Or you could treat it like an ionic bond down here. This is relatively close to the cutoff. So this is an overview of electronegativity. And even though we've been dealing with numbers in this video, in future videos, we don't care so much about the numbers. We care about the relative differences in electronegativity. So it's important to understand something as simple as oxygen is more electronegative than carbon. And that's going to help you when you're doing organic chemistry mechanisms."
    },
    {
        "Q": "At 3:58 he says that carbon and hydrogen have an electromagnetism difference of 0.4 and that still is considered a non polar covalent bond. At what difference would it be considered a polar covalent bond? 0.5, 0.8, 1?",
        "A": "You must have meant electronegativity difference The auto-correct changes that to electromagnetism :) generally 0.5 works. Some textbooks say other values, so I don t believe there is a standard value.",
        "video_name": "126N4hox9YA",
        "timestamps": [
            238
        ],
        "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": "i dont understand 1.7 means what at 9:53",
        "A": "If the difference in electronegativities of the two atoms is greater than 1.7, the bond between them is considered to be ionic.",
        "video_name": "126N4hox9YA",
        "timestamps": [
            593
        ],
        "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": "at 10:44, carbon forms an ionic bond with lithium. But as we have seen earlier carbon ALWAYS FORMS COVALENT BONDS. So how come it is forming an ionic bond with lithium?",
        "A": "Bonds vary all the way from 100 % ionic to 100 % covalent. The C-Li bond is about 43 % ionic and 57 % covalent. The bond is highly polar covalent. It behaves in many reactions as if it were ionic.",
        "video_name": "126N4hox9YA",
        "timestamps": [
            644
        ],
        "3min_transcript": "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. You could draw the dot structure above, and this would be considered be correct. You could draw it like this. Or you could treat it like an ionic bond down here. This is relatively close to the cutoff. So this is an overview of electronegativity. And even though we've been dealing with numbers in this video, in future videos, we don't care so much about the numbers. We care about the relative differences in electronegativity. So it's important to understand something as simple as oxygen is more electronegative than carbon. And that's going to help you when you're doing organic chemistry mechanisms."
    },
    {
        "Q": "At 4:19, Sal started listing momentums and positions for the atoms. But doesn't that violate the Heisenberg Uncertainty Principle?",
        "A": "Yes, it does. He can t be certain of both at the same time, however, his point was not to show the momentums and positions but rather to demonstrate the difference between micro and macrostates.",
        "video_name": "5EU-y1VF7g4",
        "timestamps": [
            259
        ],
        "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 9:03 Sal says that it all takes place in space . So if it takes place in space how would the rock move downwards . I mean there is no gravitational force there acting vertically downwards .",
        "A": "By space, he meant vacuum. He didn t mean the space where there is no gravity",
        "video_name": "5EU-y1VF7g4",
        "timestamps": [
            543
        ],
        "3min_transcript": "macro level. That's often a very important thing to think about. And we'll go into concepts like entropy and internal energy, and things like that. And you can rack your brain, how does it relate to atoms? And we will relate them to atoms and molecules. But it's useful to think that the people who first came up with these concepts came up with them not really being sure of what was going on at the micro level. They were just measuring everything at the macro level. Now I want to go back to this idea here, of equilibrium. Because in order for these macrostates to be defined, the system has to be in equilibrium. And let me explain what that means. If I were to take a cylinder. And we will be using this cylinder a lot, so it's good to get used to this cylinder. And it's got a piston in it. move up and down. This is the roof of the cylinder. The cylinder's bigger, but let's say this is a, kind of a roof of the cylinder. And we can move this up and down. And essentially we'll just be changing the volume of the cylinder, right? I could have drawn it this way. I could have drawn it like a cylinder. I could have drawn it like this, and then I could have drawn the piston like this. So there's some depth here that I'm not showing. We're just looking at the cylinder front on. And so, at any point in time, let's say the gas is between the cylinder and the floor of our container. You know, we have a bunch of molecules of gas here, a huge number of molecules. And let's say that we have a rock on the cylinder. We're doing this in space so everything above the piston is a vacuum. Actually just let me erase everything above. Let me just erase this stuff, just so you see. Just let me write that down. So all of this stuff up here is a vacuum, which essentially says there's nothing there. There's no pressure from here, there's no particles here, just empty space. And in order to keep this-- we know already, we've studied it multiple times, that this gas is generating, you know things are bumping into the wall, the floor of this piston all the time. They're bumping into everything, right? We know that's continuously happening. So we would apply some pressure to offset the pressure being generated by the gas. Otherwise the piston would just expand. It would just move up and the whole gas would expand. So let's just say we stick a big rock or a big weight on top of-- let me do it in a different color-- We put a big weight on top of this piston, where the force-- completely offsets the force being applied by the gas."
    },
    {
        "Q": "At 10:01, if you're doing this in space, the rock on top would not be able to offset the pressure inside of the cylinder, right? Since there's no gravity, it would render the rock weightless. I guess I'm being nit-picky, haha.",
        "A": "That bothered me a little too. I believe it s in space so that there s no extra air pressure acting on the system from outside, or anything like that, but he may have forgotten about that when he talked about the rock. In the right setup, you could still experience enough gravity in space to make this work. I settled for, It s on the moon, as my explanation. EDIT: And it seems other possibilities are covered in other people s questions and answers!",
        "video_name": "5EU-y1VF7g4",
        "timestamps": [
            601
        ],
        "3min_transcript": "move up and down. This is the roof of the cylinder. The cylinder's bigger, but let's say this is a, kind of a roof of the cylinder. And we can move this up and down. And essentially we'll just be changing the volume of the cylinder, right? I could have drawn it this way. I could have drawn it like a cylinder. I could have drawn it like this, and then I could have drawn the piston like this. So there's some depth here that I'm not showing. We're just looking at the cylinder front on. And so, at any point in time, let's say the gas is between the cylinder and the floor of our container. You know, we have a bunch of molecules of gas here, a huge number of molecules. And let's say that we have a rock on the cylinder. We're doing this in space so everything above the piston is a vacuum. Actually just let me erase everything above. Let me just erase this stuff, just so you see. Just let me write that down. So all of this stuff up here is a vacuum, which essentially says there's nothing there. There's no pressure from here, there's no particles here, just empty space. And in order to keep this-- we know already, we've studied it multiple times, that this gas is generating, you know things are bumping into the wall, the floor of this piston all the time. They're bumping into everything, right? We know that's continuously happening. So we would apply some pressure to offset the pressure being generated by the gas. Otherwise the piston would just expand. It would just move up and the whole gas would expand. So let's just say we stick a big rock or a big weight on top of-- let me do it in a different color-- We put a big weight on top of this piston, where the force-- completely offsets the force being applied by the gas. the area of the piston-- over some areas so that we could figure out its pressure. And that pressure will completely offset the pressure But the pressure of the gas, just as a reminder, is going in every direction. The pressure on this plate is the same as the pressure on that side, or on that side, or on the bottom of the container that we're dealing with. Now let's say that we were to just evaporate this-- well let's not say that we evaporate the rock. Let's say that we just evaporate half of the rock immediately. So all of a sudden our weight that's being pushed down, or the force that's being pushed down just goes to half immediately. Let me draw that. So I have-- maybe I would be better off just cut and pasting this right here. So if I copy and paste it. So now I'm going to evaporate half of that rock magically. So let me take my eraser tool. And I just evaporate half of it."
    },
    {
        "Q": "As at 1:38 minutes it is said that this emission spectrum is unique to hydrogen atom , which means we have different emission spectrum's for different atoms , so does that in turn mean that we have different energies for same energy levels in different atoms ?",
        "A": "Yes. Some even swap places.",
        "video_name": "Kv-hRvEOjuA",
        "timestamps": [
            98
        ],
        "3min_transcript": "- [Voiceover] I'm sure that most of you know the famous story of Isaac Newton where he took a narrow beam of light and he put that narrow beam of light through a prism and the prism separated the white light into all the different colors of the rainbow. And so if you did this experiment, you might see something like this rectangle up here so all of these different colors of the rainbow and I'm gonna call this a continuous spectrum. It's continuous because you see all these colors right next to each other. So they kind of blend together. So that's a continuous spectrum If you did this similar thing with hydrogen, you don't see a continuous spectrum. So, if you passed a current through a tube containing hydrogen gas, the electrons in the hydrogen atoms are going to absorb energy and jump up to a higher energy level. When those electrons fall down to a lower energy level they emit light and so we talked about this in the last video. This is the concept of emission. If you use something like a prism or a defraction grading to separate out the light, for hydrogen, you don't get a continuous spectrum. So, since you see lines, we call this a line spectrum. So this is the line spectrum for hydrogen. So you see one red line and it turns out that that red line has a wave length. That red light has a wave length of 656 nanometers. You'll also see a blue green line and so this has a wave length of 486 nanometers. A blue line, 434 nanometers, and a violet line at 410 nanometers. And so this emission spectrum is unique to hydrogen and so this is one way to identify elements. And so this is a pretty important thing. And since line spectrum are unique, this is pretty important to explain where those wavelengths come from. And we can do that by using the equation we derived in the previous video. So I call this equation the Balmer Rydberg equation. And you can see that one over lamda, is equal to R, which is the Rydberg constant, times one over I squared, where I is talking about the lower energy level, minus one over J squared, where J is referring to the higher energy level. For example, let's say we were considering an excited electron that's falling from a higher energy level n is equal to three. So let me write this here. So we have an electron that's falling from n is equal to three down to a lower energy level, n is equal to two. All right, so it's going to emit light when it undergoes that transition. So let's look at a visual representation of this. Now let's see if we can calculate the wavelength of light that's emitted. All right, so if an electron is falling from n is equal to three to n is equal to two, I'm gonna go ahead and draw an electron here. So an electron is falling from n is equal to three energy level down to n is equal to two, and the difference in those two energy levels"
    },
    {
        "Q": "He explains the process of the oxidation of water (at about 15:26) but where does the energy for this oxidation come from?",
        "A": "From the sun, every process in photosynthesis is fueled indirectly through light. During the light reaction ATP (energy) and NADPH2 (reduction power) is produced which can be later used for reducing CO2 to sugar. The NADPH2 contains the electrons which were taken from the water oxidation.",
        "video_name": "GR2GA7chA_c",
        "timestamps": [
            926
        ],
        "3min_transcript": "the stroma to the lumen. Then the hydrogen protons want to go back. They want to-- I guess you could call it-- chemiosmosis. They want to go back into the stroma and then that drives ATP synthase. Right here, this is ATP synthase. ATP synthase to essentially jam together ADPs and phosphate groups to produce ATP. Now, when I originally talked about the light reactions and dark reactions I said, well the light reactions have two byproducts. It has ATP and it also has-- actually it has three. It has ATP, and it also has NADPH. NADP is reduced. It gains these electrons and these hydrogens. So where does that show up? Well, if we're talking about non-cyclic oxidative photophosphorylation, or non-cyclic light reactions, the final electron acceptor. energy states, the final electron acceptor is NADP plus. So once it accepts the electrons and a hydrogen proton with it, it becomes NADPH. Now, I also said that part of this process, water-- and this is actually a very interesting thing-- water gets oxidized to molecular oxygen. So where does that happen? So when I said, up here in photosystem I, that we have a chlorophyll molecule that has an electron excited, and it goes into a higher energy state. And then that electron essentially gets passed from one guy to the next, that begs the question, what can we use to replace that electron? And it turns out that we use, we literally use, the electrons in water. So over here you literally have H2O. So you can kind of imagine it donates two hydrogen protons and two electrons to replace the electron that got excited by the photons. Because that electron got passed all the way over to photosystem I and eventually ends up in NADPH. So, you're literally stripping electrons off of water. And when you strip off the electrons and the hydrogens, you're just left with molecular oxygen. Now, the reason why I want to really focus on this is that there's something profound happening here. Or at least on a chemistry level, something profound is happening. You're oxidizing water. And in the entire biological kingdom, the only place where we know something that is strong enough of an oxidizing agent to oxidize water, to literally take away electrons from water. Which means you're really taking electrons away from oxygen. So you're oxidizing oxygen. The only place that we know that an oxidation agent is"
    },
    {
        "Q": "at 10:04 sal says 'hydrogen protons,what is the difference between an hydrogen proton and a simple proton?",
        "A": "protons are protons. He s just explaining how they came to be there.",
        "video_name": "GR2GA7chA_c",
        "timestamps": [
            604
        ],
        "3min_transcript": "Photosystem II. You have maybe another complex. And these are hugely complicated. I'll do a sneak peek of what photosystem II actually looks like. This is actually what photosystem II looks like. So, as you can see, it truly is a complex. These cylindrical things, these are proteins. These green things are chlorophyll molecules. I mean, there's all sorts of things going here. And they're all jumbled together. I think a complex probably is the best word. It's a bunch of proteins, a bunch of molecules just jumbled together to perform a very particular function. We're going to describe that in a few seconds. So that's what photosystem II looks like. Then you also have photosystem I. And then you have other molecules, other complexes. You have the cytochrome B6F complex and I'll draw this in a different color right here. I don't want to get too much into the weeds. Because the most important thing is just to understand. So you have other protein complexes, protein molecular 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"
    },
    {
        "Q": "At 15:30, Sal explains that water is hydrolyzed to resupply photosystem II with electrons it transferred to photosystem I. I always learned this reaction was carried out by a photoactive enzyme on the phospholipid bilayer but Sal seems to indicate its actually carried out by the photosystem. Any ideas?",
        "A": "The reaction is carried out by a poorly understood Oxygen Evolving Complex (OEC) which is very integrally attached to the photosystem II and its working is strongly coupled with the working of PSII and therefore it is actually considered to be single complex.",
        "video_name": "GR2GA7chA_c",
        "timestamps": [
            930
        ],
        "3min_transcript": "the stroma to the lumen. Then the hydrogen protons want to go back. They want to-- I guess you could call it-- chemiosmosis. They want to go back into the stroma and then that drives ATP synthase. Right here, this is ATP synthase. ATP synthase to essentially jam together ADPs and phosphate groups to produce ATP. Now, when I originally talked about the light reactions and dark reactions I said, well the light reactions have two byproducts. It has ATP and it also has-- actually it has three. It has ATP, and it also has NADPH. NADP is reduced. It gains these electrons and these hydrogens. So where does that show up? Well, if we're talking about non-cyclic oxidative photophosphorylation, or non-cyclic light reactions, the final electron acceptor. energy states, the final electron acceptor is NADP plus. So once it accepts the electrons and a hydrogen proton with it, it becomes NADPH. Now, I also said that part of this process, water-- and this is actually a very interesting thing-- water gets oxidized to molecular oxygen. So where does that happen? So when I said, up here in photosystem I, that we have a chlorophyll molecule that has an electron excited, and it goes into a higher energy state. And then that electron essentially gets passed from one guy to the next, that begs the question, what can we use to replace that electron? And it turns out that we use, we literally use, the electrons in water. So over here you literally have H2O. So you can kind of imagine it donates two hydrogen protons and two electrons to replace the electron that got excited by the photons. Because that electron got passed all the way over to photosystem I and eventually ends up in NADPH. So, you're literally stripping electrons off of water. And when you strip off the electrons and the hydrogens, you're just left with molecular oxygen. Now, the reason why I want to really focus on this is that there's something profound happening here. Or at least on a chemistry level, something profound is happening. You're oxidizing water. And in the entire biological kingdom, the only place where we know something that is strong enough of an oxidizing agent to oxidize water, to literally take away electrons from water. Which means you're really taking electrons away from oxygen. So you're oxidizing oxygen. The only place that we know that an oxidation agent is"
    },
    {
        "Q": "Why is photosystem II before photosystem I? 9:10",
        "A": "Photosystem I was discovered before photosystem II, even though photosystem I is the one that comes first. So they re named a bit counterintuitively, but that s only because of the order they were discovered in.",
        "video_name": "GR2GA7chA_c",
        "timestamps": [
            550
        ],
        "3min_transcript": "The important thing from the photosynthesis point of view is that it's this membrane. And on the outside of the membrane, right here on the outside, you have the fluid that fills up the entire chloroplast. So here you have the stroma. And then this space right here, this is the inside of your thylakoid. So this is the lumen. So if I were to color it pink, right there. This is your lumen. Your thylakoid space. And in this membrane, and this might look a little bit familiar if you think about mitochondria and the electron transport chain. What I'm going to describe in this video actually is an electron transport chain. Many people might not consider it the electron transport chain, but it's the same idea. Same general idea. So on this membrane you have these proteins and these complexes of proteins and molecules that span this membrane. So let me draw a couple of them. So maybe I'll call this one, photosystem II. Photosystem II. You have maybe another complex. And these are hugely complicated. I'll do a sneak peek of what photosystem II actually looks like. This is actually what photosystem II looks like. So, as you can see, it truly is a complex. These cylindrical things, these are proteins. These green things are chlorophyll molecules. I mean, there's all sorts of things going here. And they're all jumbled together. I think a complex probably is the best word. It's a bunch of proteins, a bunch of molecules just jumbled together to perform a very particular function. We're going to describe that in a few seconds. So that's what photosystem II looks like. Then you also have photosystem I. And then you have other molecules, other complexes. You have the cytochrome B6F complex and I'll draw this in a different color right here. I don't want to get too much into the weeds. Because the most important thing is just to understand. So you have other protein complexes, protein molecular 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"
    },
    {
        "Q": "At 12:00, is the ATP synthase drawn backwards? Shouldn't the rotor be on bottom if it is pumping the electrons from the lumen to the stroma and producing ATP?",
        "A": "Sal drew the ATP synthase the correct way round - upside down relative to in a mitochondrion. In the thylakoid, protons go from inside (the lumen) to outside (the stroma), and ATP is generated in the stroma. In a mitochondrion, protons go from outside to the inside, and generate ATP within the mitochondrion. In some bacteria, the ATP synthase is used in reverse, so ATP is hydrolysed to pump protons out of the cell, but this is not what s happening here.",
        "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 1:26 it says that the red light reflected of the rose enters our eye and falls on one of the red colored cones, we can see that the rose is red. What would happen if this ray of light falls on a blue or green cone, is this situation even possible?",
        "A": "Different receptors are built to respond to different stimuli. The other cones would not respond to the light of a different frequency. Think about sound, people lose their high range because these specific receptors are damaged and die. Think about the skin, some receptors respond to touch, different ones respond to heat. It is amazing.",
        "video_name": "0ugcw7wOZBg",
        "timestamps": [
            86
        ],
        "3min_transcript": "When you're looking at an object, it's necessary to break it down into its component features in order to make sense of what you're looking at. This is known as feature detection. There are many components that make up feature detection. So let's go into these. When you're looking at a rose, not only do you have to look and decide, OK, what color am I looking at, you also have to figure out, OK, what shape is the rose. So you have to get a little bit of information about the form of the rose. You also need to get information about motion. So is the rose moving? Am I throwing it across a room? What's going on? So whenever we look at any object, we need to get information about color, we need to get information about the form of the object, and information about motion. So let's go into each one of these different features. So our ability to sense color actually arises from the presence of cones within our retina. Cones are extremely important, because they're So we have three major types of cones. There are red comes, which make up 60% of the cones in your eye. There are green cones, which make up 30% of the cones in your eye. And there are blue cones, which make up about 10% of the cones in your eye. Now, we divide them into red, green, and blue cones because the red cones are extremely sensitive to red light. So if we're looking at this rose, the petals are actually reflecting red light. And so this is red light enters your eye. And if it happens to hit a red cone, then the cone will activate, and it will fire an action potential. And this action potential will reach your brain, and your brain will recognize that you're looking at something that is red. So this basically happens regardless of what we're looking at. And this has come to be known as the trichromatic theory of color vision. So what other features do we need to take into consideration when we're looking at this rose? 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."
    },
    {
        "Q": "At 9:28 why exactly does he write 3d6 when he should be writing 4d6?",
        "A": "He writes 3d\u00e2\u0081\u00b6 because the 3d subshell id filled after the 4s subshell.",
        "video_name": "NYtPw0WiUCo",
        "timestamps": [
            568
        ],
        "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": "At 5:34, why does carbon have 4 valence electrons instead of 2 when the 2s2 shell is filled already?",
        "A": "To expand on Just Keith s answer and clarify a bit more, only the level 1 shell is filled in carbon. The 2s sublevel is filled, but the electrons in it are still much closer in energy to the 2p sublevel than they are to those in the filled level 1.",
        "video_name": "NYtPw0WiUCo",
        "timestamps": [
            334
        ],
        "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": "At 5:00, why am I not supposed to write the electron confu-thingy as Be 2p2?\nIts confusing.",
        "A": "Be is not a noble gas, the square bracket notation is only used with noble gasses",
        "video_name": "NYtPw0WiUCo",
        "timestamps": [
            300
        ],
        "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 9:38 when Sal says, \"the 4th energy subshell,\" is subshell interchangeable with energy level? I thought the subshell related to the shape of the orbital.",
        "A": "He should have said the 4s subshell. Energy levels can split into subshells which are composed of orbitals that correspond with the type of subshell. The way we name these are a bit confusing, because there is a difference between say the 2p subshell and a 2p orbital. It is very easy to use the wrong word when describing this stuff.",
        "video_name": "NYtPw0WiUCo",
        "timestamps": [
            578
        ],
        "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": "9:28 Why is it 3d^2 and not 4d^2? Why is it written [Ne] when he's talking about iron (Fe)?",
        "A": "Because it is? The 4s and 3d orbitals are similar in energy so they are filled around the same time, but the 4d orbitals are quite a bit higher than those two. Why is what written [Ne]? That element in square brackets notation represents the electron configuration of the noble gas from the row above, it saves time when you get to much heavier elements by removing redundant information. Argon is the noble gas in the row above iron so you use [Ar] to represent the following: 1s^2 2s^2 2p^6 3s^2 3p^6",
        "video_name": "NYtPw0WiUCo",
        "timestamps": [
            568
        ],
        "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": "At 5:20, is the esophagus a part of the lower or upper respiratory tract, or is it not included because it is meant for food and water?",
        "A": "It is not included because it is meant for food and water, it is part of the digestive tract.",
        "video_name": "Z-yv3Yq4Aw4",
        "timestamps": [
            320
        ],
        "3min_transcript": "is kind of the more medical word, I guess. And sitting over the larynx is the epiglottis. And the epiglottis is basically like a lid kind of protecting the larynx from making sure that food and water don't go into it. Now, there's another tube I just alluded to, and it's sitting right here, and this purple tube, is our esophagus So the esophagus is basically, it's fantastic for things like food and water. You want food and water to go down the esophagus because it's going to lead to the stomach. So you want food and water to go that way, but you don't want food and water to go into the larynx. And so you want to make sure that the epiglottis, that lid, is working really well. And if you're swallowing food and water, this epiglottis will literally just kind of close up and protect your larynx. But in this case, that's not happening. We're not actually food and water, we're a little molecule of oxygen, Let's see what happens to it. I'm going to drag up the canvas a little bit. Let's make a little bit of space, and I want to just stop it right there because I want to show you that the air molecule, the oxygen molecule has already kind of made an interesting crossroads. It's actually kind of broken an important boundary, and that's this boundary right here. And on the top of this boundary, I've included the larynx and of course, all the other stuff we just talked about-- the mouth and the nose-- and this is considered our upper respiratory tract. So anything above this dashed line is our upper respiratory tract, and then, of course, you can then guess that anything below the line must then be our lower respiratory tract. So this is an important boundary because people will talk about the upper and lower tract, and I want to make sure you know what is on which side. So on the top of it, is the larynx and everything Let me label that here. The trachea is right here, the wind pipe or the trachea, and everything below that, which, of course, mainly includes things like the lungs, but as we'll see a few other structures that we're going to name. So I'm going to keep moving down, but now you know that important boundary exists. So now let me just make a little bit more space you can see the entire lungs. You can see the molecule is going to go through the trachea, and actually, I have my left lung incompletely drawn. Let me just finish it off right there. So we have our right and left lung, right? These are the two lungs, and our air is going to just kind of slowly pass down-- our molecule of oxygen is going to pass down, and it's going to go either into the right lung or the left lung. Now here, I want to make sure I just take a quick pause and show you the naming structure. And the important word of the day"
    },
    {
        "Q": "At 11:05, Rishi said respiratory bronchials, did he mean respiratory bronchides?",
        "A": "I think so since there s alveoli on bottom.",
        "video_name": "Z-yv3Yq4Aw4",
        "timestamps": [
            665
        ],
        "3min_transcript": "the left and right lung. And the other clue we said was the lobes, so of course, the right one has the upper lobe, the middle lobe, and the lower lobe, and the left lung only has the upper and lower. So that's an important clue. I just want to make sure we don't forget our little tricks that we've learned for telling apart the lungs. So I'm going to take a little pause there, and now, I'm going to show you in a sped up version all of the different branch points. So for example, here we have just a couple of branch points, one, two, and getting into this segmental bronchi would be the third branch point. But I'm going to speed things along and show you how many more branch points there actually are before we get to the final part of our lung where the gas exchange actually happens, so enjoy. [MUSIC - NIKOLAI RIMSKY-KORSAKOV, \"FLIGHT OF THE BUMBLEBEE\"] we start with kind of a bronchi, and we said that there is a naming structure for how to name the bronchi, but that's really just the first three branches. And then after the first three branches, all of the orange stuff, all those branches going from branch point 4 all the way down to about branch point 20, those are the conducting bronchials. So that's the name we give them. They're no longer bronchi, they're bronchials. And so if you see that word just keep that in mind, that we're a little bit further along in the lungs. And then after the conducting bronchials, and we call them the respiratory bronchials. And actually the final I should mention this the final conducting bronchial, sometimes you'll see this called the terminal bronchial. It's that kind of a bad name because terminal sounds like we're done, but actually we're just done with the conducting bronchials, and we're still kind of going into the respiratory bronchials. I guess if I'm only point to one, I should just probably make it singular. And then finally, we get into the alveolar ducts and the alveolar sac, which is kind of a few alveoli put together. And if it's plural as alveoli, then singular just talking about one little part of that sac would be alveolus. So that's where our little molecule of oxygen ends up, and this is kind of where it ends up before it's going to participate in gas exchange. Now, this entire area, going from respiratory bronchials on downwards, this is all called the respiratory zone,"
    },
    {
        "Q": "I'm pretty sure at 2:17 the C has only 2 methyl groups on it; the third one on the right is supposed to be the bond that was connected to X, right?",
        "A": "Don t they have to be hydrogens? SN2 can t occur with a tertiary carbon, or occurs so little to be negligible.",
        "video_name": "X9ypryY7hrQ",
        "timestamps": [
            137
        ],
        "3min_transcript": "One way to make ethers is to use the Williamson ether synthesis, which is where you start with an alcohol, and you add a strong base to deprotonate the alcohol. Once you deprotonate the alcohol, you add an alkyl halide, and primary alkyl halides work the best. We'll talk about why in a minute. And what happens is you end up putting the R prime group from your alkyl halide on to what used to be your alcohol to form your ether like that. So let's look at the mechanism for the Williamson ether synthesis, where you start with your alcohol. We know that alcohols can function as weak acids. So if you react an alcohol with a strong base, something like sodium hydride, we know that the hydride portion of the molecule is going to function as a strong base. This lone pair of electrons is going to take that proton, which is going to kick these electrons off onto the oxygen. So if we're drawing the product of that acid-based reaction, we now have an oxygen with three lone pairs of electrons around it, giving it a negative 1 formal charge. And we call that an alkoxide anion, which charged sodium ion floating around. So there's some electrostatic or ionic interaction between those opposite charges. And here's where you introduce your alkyl halide. So if we draw our alkyl halide, it would look like this. And we know that there's an electronegativity difference between our halogen and our carbon, where our halogen is going to be partially negative, and our carbon is going to be partially positive. Partially positive carbon means that that carbon wants electrons. It's going to function as an electrophile in the next step of the mechanism. And a lone pair of electrons in the oxygen is going to function as a nucleophile. So opposite charges attract. A lone pair of electrons on our nucleophile are going to attack our electrophile, our carbon. At the same time, the electrons in the bond between the carbon the halogen are going to kick off onto the halogen like that. So this is an SN2-type mechanism, because that has the decreased steric hindrance compared to other alkyl halides. So what will happen is, after nucleophilic attack, we're going to attach our oxygen to our carbon like that, and we form our ether. So if we wanted to, we could just rewrite our ether like this to show it as we added on an R prime group like that. Let's look at an example of the Williamson ether synthesis. So if I start with a molecule over here on the left, and it's kind of an interesting-looking molecule. It's called beta-naphthol. And so beta-naphthol has two rings together like this, and then there's an OH coming off of one of the rings, So that's beta-naphthol. And in the first part, we're going to add potassium hydroxide as our base."
    },
    {
        "Q": "At about 6:28, Sal says that he should only have 3 spots to the right of the decimal, but shouldn't he actually end up expressing his answer as 18.0149 because that still has the same amount of significant figures as the 15.999 because the 0 to the right of the decimal point before the other digits in 18.0149 should not be considered significant.",
        "A": "That 0 is significant",
        "video_name": "_WXndBGQnyI",
        "timestamps": [
            388
        ],
        "3min_transcript": "have an atomic mass of one. If you have two times one, it's just gonna be two atomic mass units. Then 16 from the oxygen. Two plus 16, which is going to get us... Let me do this in another color, since I've been using... It's going to give us 18 atomic mass units. Now, if you wanted to be more precise or if you wanted to say, \"Well, I have this big bag \"of water, I'm not looking exactly at one water molecule. \"What is going to be, on average, \"the mass of those water molecules?\" It might be helpful to get a little it more precise. Then it is helpful, especially if you're talking about a large number of molecules and you really just wanna take the weighted average of all of those molecules. Then it makes sense to say, \"Well, let's \"use the atomic weight.\" So for hydrogen, the atomic weight was 1.0079. We call it atomic weight but it's really just the weighted average, it's not weight in kind of the physics sense of measuring a force. So, 1.0079 atomic mass units and the oxygen is is 15.999 atomic mass units and if you wanted to get a more precise number here, let's get a calculator out. I got my calculator. I'm gonna have two times 1.0079 is equal to, and then to that I'm gonna add, plus 15.999. Gets us to... Let's see, I should go no more than three decimal places to the right. Since I added this, if I don't wanna If you wanna a review of that, you should look at the video on significant figures. Since I added something with just three decimals to to the right I shouldn't have more than three decimals in my answer so, 18.01, I'll round in the thousandths place, 015. So the real one is 18.015 atomic mass units. You could actually consider this the molecular weight. Because once again, we're using atomic weights. We're using weighted average and you could think of, maybe the weighted average of water molecules would be a little bit closer to this. These two numbers are very close. So we might call this molecular weight. Either way, these ideas are very closely related. When we're thinking on the atomic scale, when people use the word weight, they're not using it like a force like you use in physics that you would measure in newtons or pounds."
    },
    {
        "Q": "9:13, when did the first oceans come from , how did they form",
        "A": "We re not sure, but water is fairly common in the solar system. The water on earth might have come mostly from comets striking the earth.",
        "video_name": "nYFuxTXDj90",
        "timestamps": [
            553
        ],
        "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": "what does it mean for the orbit to go outward at 1:40 ?",
        "A": "They got further away from the Sun.",
        "video_name": "nYFuxTXDj90",
        "timestamps": [
            100
        ],
        "3min_transcript": "We finished off the last video in the Hadean Eon. It was named for Hades or the ancient Greek underworld. Hades is also the name of the god that ran the Greek underworld, Zeus's oldest brother. And it was an appropriate name, although the idea of, the ancient Greek notion of the underworld isn't exactly the more modern notion of Hell. But it was a hellish environment. You had all this lava flowing around. You had things impacting the Earth from space. And as far as we can tell right now, it was completely inhospitable to life. And to make matters worse, even though the Earth started to cool down a little bit, maybe the crust became a little bit more solid. Maybe the collisions started to happen less and less, as we started to go a few hundred million years fast After the Theia rammed into the early earth and formed the moon, there was something called the Late Heavy Bombardment. And right now, the consensus is that life, whatever we are descended from, would have 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"
    },
    {
        "Q": "At 2:17, if Earth had life only for part of its history (starting in ~3.8 - 4 Ga), then does that mean that other planets might have had life at one point in their histories?",
        "A": "Sure, that s possible. There are some tantalizing hints that Mars may have had bacteria-like life at one point, but we don t know.",
        "video_name": "nYFuxTXDj90",
        "timestamps": [
            137
        ],
        "3min_transcript": "We finished off the last video in the Hadean Eon. It was named for Hades or the ancient Greek underworld. Hades is also the name of the god that ran the Greek underworld, Zeus's oldest brother. And it was an appropriate name, although the idea of, the ancient Greek notion of the underworld isn't exactly the more modern notion of Hell. But it was a hellish environment. You had all this lava flowing around. You had things impacting the Earth from space. And as far as we can tell right now, it was completely inhospitable to life. And to make matters worse, even though the Earth started to cool down a little bit, maybe the crust became a little bit more solid. Maybe the collisions started to happen less and less, as we started to go a few hundred million years fast After the Theia rammed into the early earth and formed the moon, there was something called the Late Heavy Bombardment. And right now, the consensus is that life, whatever we are descended from, would have 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"
    },
    {
        "Q": "What do the arrows below the reaction arrow represent? (At 3:18)",
        "A": "They are just lines connecting the Br\u00c3\u00b8nsted acids to their conjugate bases, and the Br\u00c3\u00b8nsted bases to their conjugate a",
        "video_name": "jIL333CKE9A",
        "timestamps": [
            198
        ],
        "3min_transcript": "And then what do you get if you add an H plus to OH minus? You would get H2O or water. So let me go ahead and draw water in here. And I'll put in my lone pairs of electrons. Let's follow our electrons along so the two electrons, right, this lone pair right here on the hydroxide anion picked up this proton. So let's say those two electrons in magenta are these two electrons, and this was the proton that they picked up. And then we also need to follow the electrons, the electrons in here, I'll make them blue. So these electrons in blue come off onto the oxygen. So let's say those electrons in blue are right here, which gives the oxygen a negative one formal charge. So this is an acid-based reaction. And we can even identify conjugate acid-based pairs here. So on the left, right, on the left this was acetic acid. This was our Bronsted-Lowry acid. What is the conjugate base to acetic acid? Well, that would be over here, right. and this would be the conjugate base. Let me identify this as being the conjugate base. This is the acetate anion, right. So this is our conjugate base. For hydroxide, hydroxide on the left side functioned as a base, right. So the conjugate acid must be on the right side. So if you add a proton to OH minus, you get H2O. So water is the conjugate, oops. I'm writing conjugate base here, but it's really the conjugate acid, right. So we've identified our conjugate acid-base pairs. All right, the biggest mistake that I see when students are drawing acid-base mechanisms is they mess up their curved arrows. So the biggest mistake I see, and I'll do this in red so it'll remind you not to do it, is they show this proton right here moving to the hydroxide anion. And that is incorrect, all right. That's a very common mistake, because curved arrows show the movement of electrons, right. These two electrons up here in magenta are the ones that are taking that acetic proton. So this is incorrect. Don't draw your acid-base mechanisms like this. Let's do one more acid-base mechanism for some extra practice here. So on the left we have acetone and on the right we have the hydronium ion, H3O plus. So the hydronium ion is gonna function as our Bronsted-Lowry acid. It's going to donate a proton to acetone, which is going to be our Bronsted-Lowry base. Remember when you're drawing an acid-base mechanism, your curved arrows show the movement of electrons. So if acetone functions as our base, a lone pair of electrons on this oxygen could take this proton right here and leave these electrons behind on this oxygen. So let's show the results of our acid-base mechanism. So on the left, right, the lone pair on the left of the oxygen didn't do anything."
    },
    {
        "Q": "At 6:32 why did the empirical formula was H2SO4 and why not H2O4S",
        "A": "The sulfate anion is just more formally written as SO\u00e2\u0082\u0084 rather than as O\u00e2\u0082\u0084S. Many other elements form similar ions, and they re also written with the general formula of YO\u00e2\u0082\u0093.",
        "video_name": "sXOIIEZh6qg",
        "timestamps": [
            392
        ],
        "3min_transcript": "There's a city in Louisiana which we used to drive to all the time -- I think we had some family friends there --called Port Sulfur. And they did a lot of sulfur processing there. And lucky for the residents, at least at the time -- I apologize if they fixed the issue -- it smelled like sulfur, which smells like rotten eggs. But anyway, one mole of sulfur. One mole of sulfur. So sulfur's atomic mass is 32 atomic mass units per sulfur atom. So a whole mole of it is going to have a mass of 32 grams. So a mole of sulfur-- not a mule. Maybe I should invent a new unit called the mule. So a mole of sulfur is 32 grams. So how many moles do we have? We have a little bit more than one, but let's be precise here, because everything else is a little bit of a decimal. So if we have 32.65 grams of sulfur -- divided by 32 grams per mole we have 1.02 moles of sulfur. This was hydrogen up here. So here, you should hopefully see a pretty good ratio, here. For every one sulfur atom-- I mean, the ratio worked exactly out and that's because I did this problem before. I actually made up this problem before we worked, so I made it so the numbers worked out. But one mole of sulfur, for every mole of sulfur, so for every 6.02 times 10 to the 23 sulphur atoms, you have two moles of hydrogen, right? This ratio is 1:2. Two times 1.02 is 2.04. And then, for every one mole of sulfur, you have four moles of oxygen. Right? Literally, if you multiply this times four you get 4.08. So the ratio of hydrogen to sulfur to oxygen we have two hydrogens and we have four oxygens. So the empirical formula of this is H2. And then we have one sulfur. And then we have four oxygens. And this is sulfuric acid, one of the things you would least like poured on you most of the time. Anyway, hope you found that useful."
    },
    {
        "Q": "0:34: Isn't 2.04% + 65.3% +32.65% = 99.99% and not 100% so wouldn't that statement be incorrect?",
        "A": "he rounded a tiny bit, that .01% is not enough to make the statement incorrect",
        "video_name": "sXOIIEZh6qg",
        "timestamps": [
            34
        ],
        "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": "at 6:31, can I write H2O4S instead of H2SO4 like Sal?",
        "A": "yes most people go alphabetically when arranging molecular formulas.",
        "video_name": "sXOIIEZh6qg",
        "timestamps": [
            391
        ],
        "3min_transcript": "There's a city in Louisiana which we used to drive to all the time -- I think we had some family friends there --called Port Sulfur. And they did a lot of sulfur processing there. And lucky for the residents, at least at the time -- I apologize if they fixed the issue -- it smelled like sulfur, which smells like rotten eggs. But anyway, one mole of sulfur. One mole of sulfur. So sulfur's atomic mass is 32 atomic mass units per sulfur atom. So a whole mole of it is going to have a mass of 32 grams. So a mole of sulfur-- not a mule. Maybe I should invent a new unit called the mule. So a mole of sulfur is 32 grams. So how many moles do we have? We have a little bit more than one, but let's be precise here, because everything else is a little bit of a decimal. So if we have 32.65 grams of sulfur -- divided by 32 grams per mole we have 1.02 moles of sulfur. This was hydrogen up here. So here, you should hopefully see a pretty good ratio, here. For every one sulfur atom-- I mean, the ratio worked exactly out and that's because I did this problem before. I actually made up this problem before we worked, so I made it so the numbers worked out. But one mole of sulfur, for every mole of sulfur, so for every 6.02 times 10 to the 23 sulphur atoms, you have two moles of hydrogen, right? This ratio is 1:2. Two times 1.02 is 2.04. And then, for every one mole of sulfur, you have four moles of oxygen. Right? Literally, if you multiply this times four you get 4.08. So the ratio of hydrogen to sulfur to oxygen we have two hydrogens and we have four oxygens. So the empirical formula of this is H2. And then we have one sulfur. And then we have four oxygens. And this is sulfuric acid, one of the things you would least like poured on you most of the time. Anyway, hope you found that useful."
    },
    {
        "Q": "At 6:35, Sal writes the empirical formula as H2 S O4. Is there usually a specific order we need to write empirical or molecular formulas?",
        "A": "In H2SO4 H2 is the positively charged ion while the SO4 is negatively charged. While writing the molecular formula, the most electropositive element is written first. Hope this helps :)",
        "video_name": "sXOIIEZh6qg",
        "timestamps": [
            395
        ],
        "3min_transcript": "There's a city in Louisiana which we used to drive to all the time -- I think we had some family friends there --called Port Sulfur. And they did a lot of sulfur processing there. And lucky for the residents, at least at the time -- I apologize if they fixed the issue -- it smelled like sulfur, which smells like rotten eggs. But anyway, one mole of sulfur. One mole of sulfur. So sulfur's atomic mass is 32 atomic mass units per sulfur atom. So a whole mole of it is going to have a mass of 32 grams. So a mole of sulfur-- not a mule. Maybe I should invent a new unit called the mule. So a mole of sulfur is 32 grams. So how many moles do we have? We have a little bit more than one, but let's be precise here, because everything else is a little bit of a decimal. So if we have 32.65 grams of sulfur -- divided by 32 grams per mole we have 1.02 moles of sulfur. This was hydrogen up here. So here, you should hopefully see a pretty good ratio, here. For every one sulfur atom-- I mean, the ratio worked exactly out and that's because I did this problem before. I actually made up this problem before we worked, so I made it so the numbers worked out. But one mole of sulfur, for every mole of sulfur, so for every 6.02 times 10 to the 23 sulphur atoms, you have two moles of hydrogen, right? This ratio is 1:2. Two times 1.02 is 2.04. And then, for every one mole of sulfur, you have four moles of oxygen. Right? Literally, if you multiply this times four you get 4.08. So the ratio of hydrogen to sulfur to oxygen we have two hydrogens and we have four oxygens. So the empirical formula of this is H2. And then we have one sulfur. And then we have four oxygens. And this is sulfuric acid, one of the things you would least like poured on you most of the time. Anyway, hope you found that useful."
    },
    {
        "Q": "Shouldn't a mole of different gases have different volumes not all the same of 22.4 at 11:50? Is it because in ideal gases we disregard the gas's volume?",
        "A": "Most gases approximate ideal gas behavior as long as the pressure is not to high and the temperature is not too low.",
        "video_name": "GwoX_BemwHs",
        "timestamps": [
            710
        ],
        "3min_transcript": "Pressure is 1 atmosphere, but remember we're dealing with atmospheres. 1 atmosphere times volume-- that's what we're solving for. I'll do that and purple-- is equal to 1 mole times R times temperature, times 273. Now this is in Kelvin; this is in moles. We want our volume in liters. So which version of R should we use? Well, we're dealing with atmospheres. We want our volume in liters, and of course, we have moles and Kelvin, so we'll use this version, 0.082. So this is 1, so we can ignore the 1 there, the 1 there. So the volume is equal to 0.082 times 273 degrees Kelvin, So if I have any ideal gas, and all gases don't behave ideally ideal, but if I have an ideal gas and it's at standard temperature, which is at 0 degrees Celsius, or the freezing point of water, which is also 273 degrees Kelvin, and I have a mole of it, and it's at standard pressure, 1 atmosphere, that gas should take up exactly 22.4 liters. And if you wanted to know how many meters cubed it's going to take up, well, you could just say 22.4 liters times-- now, how many meters cubed are there -- so for every 1 meter cubed, you have 1,000 liters. I know that seems like a lot, but it's true. Just think about how big a meter cubed is. If you have something at 1 atmosphere, a mole of it, and at 0 degrees Celsius. Anyway, this is actually a useful number to know sometimes. They'll often say, you have 2 moles at standard temperature and pressure. How many liters is it going to take up? Well, 1 mole will take up this many, and so 2 moles at standard temperature and pressure will take up twice as much, because you're just taking PV equals nRT and just doubling. Everything else is being held constant. The pressure, everything else is being held constant, so if you double the number of moles, you're going to double the volume it takes up. Or if you half the number of moles, you're going to half the volume it takes up. So it's a useful thing to know that in liters at standard temperature and pressure, where standard temperature and pressure is defined as 1 atmosphere and 273 degrees Kelvin, an ideal gas will take up 22.4 liters of volume."
    },
    {
        "Q": "At 4:59, there's a model of a molecular structure. What kind of molecule is it modeling?",
        "A": "That is nitroglycerin. The black balls represent carbons, white is hydrogen, blue is nitrogen and red is oxygen.",
        "video_name": "Rd4a1X3B61w",
        "timestamps": [
            299
        ],
        "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 5:26 why did it become 16? and where do 4 come from? why is it plus not minus not like the other example?",
        "A": "CO2 C has 4 valence electrons, O has 6 valence electrons 4 + 6 + 6 = 16 valence electrons",
        "video_name": "97POZGcfoY8",
        "timestamps": [
            326
        ],
        "3min_transcript": "And we'll go ahead and move on to the next step. So let me go ahead and put in my lone pairs of electrons around my chlorine here. So we have our dot structure. Next, we're going to count the number of electron clouds that surround the central atom. And I like to use the term electron cloud. You'll see many different terms for this in different textbooks. You'll see charge clouds, electron groups, electron domains, and they have slightly different definitions depending on which textbook you look in. And really the term of electron cloud helps describe the idea of valence electrons in bonds and in lone pairs of electrons occupying these electron clouds. And you could think about them as regions of electron density. And, since electrons repel each other, those regions of electron density, those clouds, want to be as far apart from each other as they possibly can. And so let's go ahead and analyze our molecule here. So surrounding the central atom. surrounding our central atom. So we could think about those as being an electron cloud. And then over here we have another electron cloud. So we have two electron clouds for this molecule, and those electron clouds are furthest apart when they point in opposite directions. And so the geometry or the shape of the electron clouds around the central atom, if they're pointing in opposite directions, it's going to give you a linear shape here. So this molecule is actually linear because we don't have any lone pairs to worry about here. So we're going to go ahead and predict the geometry of the molecule as being linear. And if that's linear, then we can say the bond angle-- so the angle between the chlorine, the beryllium, and the other chlorine-- is 180 degrees. So just a straight line. All right. So that's how to use VSEPR to predict the shape. Let's do another example. So CO2-- so carbon dioxide. Carbon has four valence electrons. Oxygen has six. And we have two of them. So 6 times 2 gives us 12. 12 plus 4 gives us 16 valence electrons to deal with for our dot structure. The less electronegative atom goes in the center, so carbon is bonded to oxygen, so two oxygens like that. We just represented four valence electrons. Right? So two here and two here, so that's four. So 16 minus 4 gives us 12 valence electrons left. Those electrons are going to go on our terminal atoms, which are oxygens if we are going to follow the octet rule. So each oxygen is surrounded by two electrons. So, therefore, each oxygen needs six more valence electrons. I'll go ahead and put in six more valence electrons on our oxygen. Now you might think we're done, but, of course, we're not because carbon is going to follow the octet rule. Carbon does not have a formal charge of 0 in this dot structure, so even though we've represented all of our valence electrons now,"
    },
    {
        "Q": "At 1:56 how come we don't put 2 more lone pairs of electrons on Be?",
        "A": "Total number of electrons in BeCL2 = 16 Which are all used up. ( 6 each to Cl + 4 electrons as bonds)",
        "video_name": "97POZGcfoY8",
        "timestamps": [
            116
        ],
        "3min_transcript": "This next set of videos, we're going to predict the shapes of molecules and ions by using VSEPR, which is an acronym for valence shell electron pair repulsion. And really all this means is that electrons, being negatively charged, will repel each other. Like charges repel, and so when those electrons around a central atom repel each other, they're going to force the molecule or ion into a particular shape. And so the first step for predicting the shape of a molecule or ion is to draw the dot structure to show your valence electrons. And so let's go ahead and draw a dot structure for BeCl2. So you find beryllium on the periodic table. It's in group 2, so two valence electrons. Chlorine is in group 7, and we have two of them. So 2 times 7 is 14. And 14 plus 2 gives us a total of 16 valence electrons that we need to account for in our dot structure. So you put the less electronegative atom So beryllium goes in the center. We know it is surrounded by two chlorines, And we just represented four valence electrons. So here's two valence electrons. And here's another two for a total of four. So, instead of 16, we just showed four. So now we're down to 12 valence electrons that we need to account for. So 16 minus 4 is 12. We're going to put those left over electrons on our terminal atoms, which are our chlorines. And chlorine is going to follow the octet rule. Each chlorine is already surrounded by two valence electrons, so each chlorine needs six more. So go ahead and put six more valence electrons on each chlorine. And, since I just represented 12 more electrons there, now we're down to 0 valence electron. So this dot structure has all of our electrons in it. And some of you might think, well, why don't you keep going? Why don't you show some of those lone pairs of electrons in chlorine moving in to share them with the beryllium to give it an octet of electrons? And the reason you don't is because of formal charge. So let's go ahead and assign a formal charge So remember each of our covalent bonds consists of two electrons. So I go ahead and put that in. And if I want to find formal charge, I first think about the number of the valence electrons in the free atom. And that would be two, four-- four berylliums. So we have two electrons in the free atom. And then we think about the bonded atom here, so when I look at the covalent bond, I give one of those electrons to chlorine and one of those electrons to beryllium. And I did the same thing for this bond over here, and so you can see that it is surrounded by two valence electrons. 2 minus 2 gives us a formal charge of 0. And so that's one way to think about why you would stop here for the dot structure. So it has only two valence electrons, so even though it's in period 2, it doesn't necessarily have to follow the octet rule. It just has to have less than eight electrons. And so, again, formal charge helps you understand why you can stop here for your dot structure. Let me go ahead and redraw our molecule"
    },
    {
        "Q": "At 6:45 Sal says that the base must have the same number of moles as the weak acid. Doesn't that statement assume the acid is monoprotic (donates 1 H+)? How would the calculation change for a diprotic acid (donates 2 H+)? The moles for the acid would just be doubled, right?",
        "A": "You re right, so the normality of the base must equal the normality of the acid. Therefore, for diprotic molecules you would need twice as much base.",
        "video_name": "BBIGR0RAMtY",
        "timestamps": [
            405
        ],
        "3min_transcript": "And when we're adding more hydrogens, we're getting really acidic really fast. But we have a lot of the conjugate acid there in the solution already. So we're going to have an acidic equivalence point. Now, let me give you an actual problem, just to hit all the points home. Because everything I've done now has been very hand-wavey, and no numbers. So let me draw one. Let me draw a weak acid. And you'll recognize it because you're good at this now. But I'll deal with some real numbers here. So let's say that's a pH of 7. We're going to titrate it. It starts off at a low pH because it's a weak acid. And as we titrate it, it's pH goes up. And then it hits the equivalence point and it goes like that. The equivalence point is right over here. And let's say our reagent that we were adding is sodium hydroxide. And let's say it's a 0.2 molar solution. I'll use 700 milliliters of sodium hydroxide is our equivalence point. Right there. So the first question is how much of our weak acid did we have? So what was our original concentration of our weak acid? This is just a general placeholder for the acid. So original concentration of our weak acid. Well, we must have added enough moles of OH at the equivalent point to cancel out all of the moles of the weak acid in whatever hydrogen was out there. But the main concentration was from the weak acid. This 700 milliliters of our reagent must have the same number of moles as the number of moles of weak acid we started off with. And let's say our solution at the beginning was 3 liters. 3 liters to begin with, before we started titrating. the solution. But let's just say that in the beginning, we started with 3 liters. So how many moles have we sopped up? Well, how many moles of OH are there in 700 milliliters of our solution? Well, we know that we have 0.2 moles per liter of OH. And then we know that we don't have-- times 0.7 liters, right? 700 milliliters is 0.7 liters. So how many moles have we added to the situation? 2 times 7 is 14. And we have 2 numbers behind the decimal. So it's 0.14. So 700 milliliters of 0.2 molar sodium hydroxide, and we have 700 milliliters of it, or 0.7 liters. We're going to have 0.14 moles of, essentially, OH that we"
    },
    {
        "Q": "A magnetic field creates a force on a moving charge and a moving charge creates a magnetic field which would then create a force on a moving charge..., right? How does stationary charge create its own static electric charge like Sal is talking about at 9:01? Did I misunderstand him? I am trying to grasp how electricity and magnetism are the same. Thanks for your help! :-)",
        "A": "Electric fields originate from any charge (moving or not) and changing magnetic fields. Magnetic fields originate from moving charge (ie. current) and changing electric fields. A charge will not interact with the field it generates itself.",
        "video_name": "Ri557hvwhcM",
        "timestamps": [
            541
        ],
        "3min_transcript": "It's 3 meters. So 2 pi times 3. So it equals the permeability of free space. The 2 and the 2 cancel out over 3 pi. So how do we calculate that? Well, we get out our trusty TI-85 calculator. And I think you'll be maybe pleasantly surprised or shocked to realize that-- I deleted everything just so you can see how I get there-- that it actually has the permeability of free space stored in it. So what you do is you go to second and you press constant, which is the 4 button. It's in the built-in constants. Let's see, it's not one of those. You press more. It's not one of those, press more. Oh look at that. Mu not. The permeability of free space. That's what I need. And I have to divide it by 3 pi. Divide it by 3-- and then where is pi? There it is. Divided by 3 pi. It equals 1.3 times 10 to the negative seventh. It's going to be teslas. The magnetic field is going to be equal to 1.3 times 10 to the minus seventh teslas. So it's a fairly weak magnetic field. And that's why you don't have metal objects being thrown around by the wires behind your television set. But anyway, hopefully that gives you a little bit-- and just so you know how it all fits together. We're saying that these moving charges, not only can they be affected by a magnetic field, not only can a current be affected by a magnetic field or just a moving charge, it actually creates them. And that kind of creates a little bit of symmetry in your head, hopefully. Because that was also true of electric field. A charge, a stationary charge, is obviously pulled or pushed by a static electric field. And it also creates its own static electric field. Because if you keep studying physics, you're going to actually prove to yourself that electric and magnetic fields are two sides of the same coin. And it just looks like a magnetic field when you're in a different frame of reference, When something is whizzing past you. While if you were whizzing along with it, then that thing would look static. And then it might look a little bit more like an electric field. But anyway, I'll leave you there now. And in the next video I will show you what happens when we have two wires carrying current parallel to each other. And you might guess that they might actually attract or repel each other. Anyway, I'll see you in the next video."
    },
    {
        "Q": "Why does the acid and base stay liquid and turn pink but not form H20 (water) and NaCl (Salt) as the formula stated at about 1:40 in the video?",
        "A": "It turns pink because you add an indicator, which has a pink colour in basic solutions. And they reaction has produced water en NaCl, but the salt stays dissolved in the solution. (NaCl is easily soluble).",
        "video_name": "d1XTOsnNlgg",
        "timestamps": [
            100
        ],
        "3min_transcript": "Titration is a procedure for determining the concentration of a solution. And so let's say we're starting with an acidic solution. So in here let's say we have some hydrochloric acid. So we have come HCl. And we know the volume of HCL, let's say we're starting with 20.0 milliliters of HCl. But we don't know the concentration right? So question mark here for the concentration of HCl. We can find out that concentration by doing a titration. Next we need to add a few drops of an acid base indicator. So to this flask we're also going to add a few drops of an acid base indicator. We're gonna use phenolphthalein. And phenolphthalein is colorless in acid but turns pink in the presence of base. And since we have our phenolphthalein in acid right now we have a clear solution. There's no color to it. Up here we're gonna have our standard solution right? We're gonna have a known concentration of sodium hydroxide. So let's say we have a solution of sodium hydroxide And we're ready to start our titration. So we allow the sodium hydroxide to drip into our flask containing our HCl and our indicator. And the acid in the base will react, right? So we get an acid base neutralization reaction. HCl plus NaOH right? If we think about the products, this would be OH minus, this would be H plus, H plus and OH minus give us H2O. And our other product we would have Na plus and Cl minus, which give us NaCl, or sodium chloride. So let's say we add a certain volume of base right? So now this would be higher, and we see our solution turn light pink. Alright so let's say we see our solution turn light pink and it stays light pink. That means that all of the acid And we have a tiny amount of excess base present, and that's causing the acid base indicator to remain pink. So a tiny excess of base means we've neutralized all of the acid present. And where the indicator changes color, this is called the end point of a titration, alright? So when our solution changes color, that's the end point of our titration. And here we stop and we check and see the volume of base that we used in our titration. So if we started right here, if we started with that much base, let's say we ended down here, alright? So we still have a little bit of base left. And this would be the volume of base that we used in the titration. Alright so we have a change in volume here, and let's say that it's 48.6 milliliters. So it took 48.6 milliliters of our base to completely neutralize the acid that we had present."
    },
    {
        "Q": "At 4:53, how do you know that there's a 1-1 mole ratio? And what does he mean by that?",
        "A": "From the balanced equation; the unwritten (1) before the element/compound is the balance. (1)NaOH+(1)HCl==>(1)NaCl+(1)H2O",
        "video_name": "d1XTOsnNlgg",
        "timestamps": [
            293
        ],
        "3min_transcript": "Alright so let's go ahead and do that, and let's start with the concentration of sodium hydroxide. Alright we know that we started with point one zero zero molar solution of sodium hydroxide. So point one zero zero molar. And molarity is equal to mols over liters. Alright so this is equal to mols over liters. And our goal is to figure out how many mols of base that we used to neutralize the acid that was present. Alright so we can take our volume here, 48.6 mililiters and we can convert that into liters. Alright so just move your decimal place three places to the left. So one, two, three. So that's point zero four eight six liters. So this is equal to mols over And so let's get some more space. Alright let me just rewrite this really quickly. Zero point one zero zero is equal to X over zero point zero four eight six. So we're just solving for X, and X represents the mols of sodium hydroxide that were necessary to neutralize the acid that we had present. Alright so when you solve for X, you get zero point zero zero four eight six mols of sodium hydroxide used in our titration. Next you look at the balanced equation for what happened . So if I look at my balanced equation alright there's a one here and there's a one here. So we have a one to one mol ratio. And the equivalence point is where just enough of your standard solution has been added to completely react with the solution that's being titrated. all of the acid has been neutralized. Right? So it's completely reacted. And since we have a one to one mol ratio, if I used this many mols of sodium hydroxide, that must be how many mols of HCl that we had present in our original solution. So therefore, I can go ahead and write that I must have had zero point zero zero four eight six mols of HCl present in the flask before we started our titration. Right and I knew that because of the one to one mol ratio. Remember our goal was to find the concentration of HCl. The original concentration. And concentration, molarity is equal to mols over liters. So now I know how many mols of HCl I had, and my original volume of HCl was 20 milliliters right?"
    },
    {
        "Q": "At around 3:45, David said that the momentums were positive 5 and negative 10. Could you choose that moving to the left was instead positive and moving to the right was negative? Making the equation: (0.2kg)(-5m/s) - (0.2kg)(10m/s)?",
        "A": "Try it and see what happens. Good way to learn.",
        "video_name": "uMYAc04D0ak",
        "timestamps": [
            225
        ],
        "3min_transcript": "the impulse from all forces on an object, like this ball, that should just equal the change in momentum of that object, like the change in momentum of this ball. So if we can figure out the change in momentum of this ball, we can figure out the net impulse on this ball. And since it's the net impulse, and this formula appears also true, this is equivalent, which is saying that it's the net force, multiplied by the time duration, during which that net force is acting. This is hard for people to remember, sometimes my students like to remember it as Jape Fat. So, if you look at this, it looks like J-A-P, this kinda looks like an E, F-A-T. So if you need a way, a pneumonic device, to remember this, Jape Fat is a way to remember how impulse, change in momentum, force, and time, are all related. So let's do it. We can't use force because we don't know it yet, but I can figure out the change in momentum 'cause I know the velocities. So, we know that the change in momentum is gonna be P final, the final momentum, What's my final momentum? My final momentum is M times V, so it's gonna be mass times V final, minus mass times V initial, and my mass is .2, so I've got a mass of 0.2 kilograms. My final velocity is five, because the ball recoiled to the right with positive five. Positive five 'cause it's moving to the right. I'm gonna assume rightward is positive. Then minus, the mass is .2 again, so 0.2 kilograms. My initial velocity is not 10. This is 10 meters per second to the left, and momentum is a vector, it has direction, so you have to be careful with negative signs here. This is the most common mistake. People just plug in positive 10, then get the wrong answer. But this ball changed directions, so the two velocities here have to have two different sides, so this has to be a negative 10 meters per second, if I'm assuming rightward is positive. This leftward velocity, and this leftward initial velocity, And, if you didn't plug that in, you'd get a different answer, so you gotta be careful. So, what do I get here if I multiply this all out? I'm gonna get zero, no, sorry, I'm gonna get one kilogram meters per second, minus a negative two kilogram meters per second, and that's gonna give me positive three kilogram meters per second is the impulse, and that should make sense. The impulse was positive. The direction of the impulse, which is a vector, is the same direction as the direction of the force. So, which way did our face exert a force on the ball? Our face exerted a force on the ball to the right. That's why the impulse on the ball is to the right. The impulse on this person's face is to the left, but the impulse on the ball is to the right, because the ball was initially going left and it had a force on it to the right that made it recoil and bounce back to the right. That's why this impulse has a positive direction to it."
    },
    {
        "Q": "how does bacteria get its energy 1:40",
        "A": "Bacteria obtain energy by either ingesting other organisms and organic compounds or by producing their own food. The bacteria that produce their own food are called autotrophs. Bacteria that must consume other organic molecules for energy are called heterotrophs.",
        "video_name": "dQCsA2cCdvA",
        "timestamps": [
            100
        ],
        "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": "2:07\nhow much DNA do we have?",
        "A": "Humans have 46 chromosomes that contain all of the genetic information, and there are over 25,000 genes in the human genome. Genes are composed of DNA, and it is predicted that there are over 3 billion base pairs in the human genome. Humans have approximately 10 trillion cells, so if you were to line all of the DNA found in every cell of a human body it would stretch from the earth to the sun 100 times!",
        "video_name": "dQCsA2cCdvA",
        "timestamps": [
            127
        ],
        "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": "At 11:20:\n\nWhat did 12 ever do?\nHow was it activated?",
        "A": "Factor 12 is the first factor that is activated in the intrinsic pathway. It is activated by a called Kallikrein. Factor 12 then simply becomes a catalyst to convert 11 from its inactive form to its active form.",
        "video_name": "FNVvQ788wzk",
        "timestamps": [
            680
        ],
        "3min_transcript": "which is actually one of these little yellow guys. And that tissue factor activates VII, which activates X, so you get a shot - a spark that shoots down this way and activates a little bit of X. And then X will activate a little bit of thrombin, and then thrombin will get the intrisic workhorse going. And how will thrombin do that? Well thrombin actually activates a whole bunch of these guys, and to remember the ones that it activates, you just need to take the five odd numbers starting at five. So what is that? That's V, VII, IX, XI, and XIII. Actually, this is just almost right, but it actually turns out that it's not IX, it's VIII because it couldn't be quite that easy. So those are the five that it activates. So let's draw that in here in our drawing. So let's draw that in the form of blue arrows because thrombin is blue. We said it's going to activate VII. We said it's going to activate not IX, but VIII, so this will be an awkward arrow to draw. We said it's going to activate XI, and we said it's going to activate XIII. Where's our XIII? Well, we haven't actually drawn it in yet, so let's quickly chat about that. The end goal of this whole cascade is to get these fibrin molecules, and these fibrin molecules together will form some strands. It actually turns out that there's one more step, which is to connect these strands together. So we're going to want to connect these strands together with some cross links. These cross links will just hold them together so that they actually form a tight mesh. It turns out that it's this step right here, which is enabled by factor XIII. So let's draw the final thrombin activity, which is to activate XIII. it's going to activate all the necessary things in this intrinsic pathway to get it going. You might actually be wondering about XII up there because thrombin is not hitting him, and actually it turns out that if you remove a person's factor XII, they can still clot pretty well. So it's clear that XII is not a totally necessary part of this intrinsic pathway. And to be clear again, with our use of arrows, this green arrow here is different from these white arrows in the sense that here, we are saying that fibrin is going to become fibrin strands, which is going to become interlaced fibrin strands. So if this was all there was to the story, then every time you had a little bit of damage to your endothelium, you would cause the extrinsic pathway to fire. So you'd create a little activated VII. You would activate some X, which would activate some II, which is thrombin, which would start to create fibrin from fibrinogen. And moreover, the thrombin would have"
    },
    {
        "Q": "At 4:08 why would it be an SN2 if that is a tertiary carbon and SN2 rxn only happens in primary and secondary carbons?",
        "A": "its not a tertiary carbon, its given at the start that its a secondary or primary carbon",
        "video_name": "LccmkSz-Y-w",
        "timestamps": [
            248
        ],
        "3min_transcript": "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. and it's going to attack our partially positive carbon. An SN2 type mechanism. So as the chloride attacks, this stuff on the right is going to leave. So the electrons in magenta are actually going to move in here, and then these electrons are going to kick off onto that chlorine. So when we draw the product, we can go ahead and show the chlorine has now added on to our carbon on the left. And on the right, if you follow the movement of those electrons, they're going to form sulfur dioxides. So SO2. And also the chloride anions, so the Cl minus, like that. And so we've done it. We've substituted our chlorine atom for the OH and formed an alkyl halide. So this is just a better way of forming an alkyl chloride from an alcohol. So if we look at an example, we'll just take something like ethanol here."
    },
    {
        "Q": "Hey, at 5:40, is there any reason we use PBr3 instead of H-Br? Thanks.",
        "A": "The phosphate ester that is made when you use PBr3 provides a better leaving group than OH on its own. Jay makes this point in the video.",
        "video_name": "LccmkSz-Y-w",
        "timestamps": [
            340
        ],
        "3min_transcript": "and it's going to attack our partially positive carbon. An SN2 type mechanism. So as the chloride attacks, this stuff on the right is going to leave. So the electrons in magenta are actually going to move in here, and then these electrons are going to kick off onto that chlorine. So when we draw the product, we can go ahead and show the chlorine has now added on to our carbon on the left. And on the right, if you follow the movement of those electrons, they're going to form sulfur dioxides. So SO2. And also the chloride anions, so the Cl minus, like that. And so we've done it. We've substituted our chlorine atom for the OH and formed an alkyl halide. So this is just a better way of forming an alkyl chloride from an alcohol. So if we look at an example, we'll just take something like ethanol here. had some pyridine as our base. We're going to replace the OH with our chlorine like that. And so once again, if we look at our alcohol, this is a primary alcohol. And so primary alcohols work the best because there's decreased steric hindrance. And we don't have to worry about stereochemistry, since we don't have any chirality centers in our product. Let's look at a way to form alkyl bromide. So we just formed an alkyl chloride. Let's look at the general reaction for forming an alkyl bromide here. So I go ahead and have my alcohol. And I react that with phosphorus tribromide, PBr3. The OH group is going to leave and I'm going to put a bromine in its place. And once again, this mechanism is an SN2 type mechanism. So primary or secondary alcohols only. And possible inversion of configuration present or not. So another SN2 mechanism. And again, we need to use phosphorus tribromide because the OH group is not the best leaving group. When we look at this mechanism here, let's go ahead and show that lone pair of electrons better, like that. We have phosphorus tribromide. So I'm going to go to draw the dot structure. So we would have these bromines here with lone pairs of electrons. And there's three of them. So I'll go ahead and put in those bromines. And we still have two more valence electrons to account for, and those would go on our phosphorus like that. So the first step, it's analogous to our previous mechanism. Lone pair of electrons on oxygen are going to for a bonds with phosphorus. And that would kick these electrons off onto one of the bromines, so I just chose that one. It doesn't matter which one you choose. And so when we show the result of that, we would now have our oxygen bonded to a phosphorus."
    },
    {
        "Q": "at 1:35 you said that a bullet goes as fast as a jet. how fast does a jet go in miles?",
        "A": "Legally around 300MPH for the big commercial jets, Regular Jets can go around 600MPH legally, but when we talking Illegally were talking 1000MPH+",
        "video_name": "GZx3U0dbASg",
        "timestamps": [
            95
        ],
        "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": "At 6:15, when you begin to talk about AUs, the measurement is about equal to the distance between the earth and the sun. Was this done on purpose?",
        "A": "yes, that s the definition of the AU",
        "video_name": "GZx3U0dbASg",
        "timestamps": [
            375
        ],
        "3min_transcript": "So it's 4,360 hours to circumnavigate the sun, going at the speed of a bullet or a jetliner. And so that is-- 24 hours in the day-- that is 181 days. It would take you roughly half a year to go around the sun at the speed of a jetliner. Let me write this down. Half a year. The sun is huge. Now, that by itself may or may not be surprising--and actually let me give you a sense of scale here, because I have this other diagram of a sun. And we'll talk more about the rest of the solar system in the next video. But over here, at this scale, the sun, at least on my screen-- if I were to complete it, it would probably be about 20 inches in diameter. 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."
    },
    {
        "Q": "At 3:32, how do you know the speed of a bullet.",
        "A": "The muzzle velocity of standard rounds for guns are well known. The manufacturing is well controlled so that they will have predictable trajectories when fired.",
        "video_name": "GZx3U0dbASg",
        "timestamps": [
            212
        ],
        "3min_transcript": "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. So it's 4,360 hours to circumnavigate the sun, going at the speed of a bullet or a jetliner. And so that is-- 24 hours in the day-- that is 181 days. It would take you roughly half a year to go around the sun at the speed of a jetliner. Let me write this down. Half a year. The sun is huge. Now, that by itself may or may not be surprising--and actually let me give you a sense of scale here, because I have this other diagram of a sun. And we'll talk more about the rest of the solar system in the next video. But over here, at this scale, the sun, at least on my screen-- if I were to complete it, it would probably be about 20 inches in diameter."
    },
    {
        "Q": "At around 1:21, Sal mentions that the Earth is approximately 40,000 km. About how many miles is that?",
        "A": "To convert to miles just divide by 1.6; so 25,000",
        "video_name": "GZx3U0dbASg",
        "timestamps": [
            81
        ],
        "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": "in 6:10 why does sal multiply both sides of the equation by 2?",
        "A": "to remove the 2 times acceleration which is in the denominator in the RHS of the equation.",
        "video_name": "2ZgBJxT9pbU",
        "timestamps": [
            370
        ],
        "3min_transcript": "that goes by-- this is our change in velocity. So elapsed time is the same thing. I write it over here-- is our change in velocity divided by our acceleration. And just to make sure you understand this, it just comes straight from the idea that acceleration-- or let me write it this way-- that change in velocity is just acceleration times time. Or I should say, acceleration times change in time. So if you divide both sides of this equation by acceleration, you get this right over here. So that is what our displacement-- Remember, I want an expression for displacement in terms of the things we know and the one thing that we want to find out. Well, for this example right over here, we know a couple of things. Well actually, let me take it step by step. We know that our initial velocity is 0. the average velocity is going to be our final velocity divided by 2, since our initial velocity is 0. Our change in velocity is the same thing as final velocity minus initial velocity. And once again, we know that the initial velocity is 0 here. So our change in velocity is the same thing as our final velocity. So once again, this will be times. Instead of writing change in velocity here, we could just write our final velocity because we're starting at 0. Initial velocity is 0. So times our final velocity divided by our acceleration. Final velocity is the same thing as change in velocity because initial velocity was 0. And all of this is going to be our displacement. And now it looks like we have everything written in things we know. or both sides of this equation by 2 times our acceleration on that side. And we multiply the left-hand side by-- I'll do the same colors-- 2 times our acceleration. On the left hand side, we get 2 times our acceleration times our displacement is going to be equal to, on the right hand side, the 2 cancels out with the 2, the acceleration cancels out with the acceleration-- it will be equal to the velocity, our final velocity squared. Final velocity times final velocity. And so we can just solve for final velocity here. So we know our acceleration is negative 9.8 meters"
    },
    {
        "Q": "At 9:48, wouldn't it be more accurate to express Vf as Vf = - |sqrt(19.6h)| m/s?",
        "A": "Yes, but by convention when we use the square root sign we mean the positive root of the value.",
        "video_name": "2ZgBJxT9pbU",
        "timestamps": [
            588
        ],
        "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": "At 4:00, how does he get (final velocity + initial velocity) / 2 ?",
        "A": "That s how you calculate an average of two numbers. You add them up and divide by 2.",
        "video_name": "2ZgBJxT9pbU",
        "timestamps": [
            240
        ],
        "3min_transcript": "that positive velocity means upwards, or a positive vector means up, a negative vector means down. So we're going to have an initial velocity over here of 0. And then at the bottom we're going to have some final velocity here that is going to be a negative number. So it's going to have some negative value over here. So this is going to be negative. This is going to be a negative number right over there. And we know that the acceleration of gravity for an object on free fall, an object in free fall near the surface of the earth. We know it, and we're going to assume that it's constant. So our constant acceleration is going to be negative 9.8 meters per second squared. and given that their initial velocity is 0 and that our acceleration is negative 9.8 meters per squared, we want to figure out what our final velocity is going to be right before we hit the ground. We're going to assume that this h is given in meters, right over here. And we'll get an answer in meters per second for that final velocity. So let's see how we can figure it out. So we know some basic things. And the whole point of these is to really show you that you can always derive these more interesting questions from very basic things that we know. So we know that displacement is equal to average velocity times change in time. And we know that average velocity-- if we assume acceleration is constant, which we are doing-- average velocity is the final velocity plus the initial velocity over 2. that goes by-- this is our change in velocity. So elapsed time is the same thing. I write it over here-- is our change in velocity divided by our acceleration. And just to make sure you understand this, it just comes straight from the idea that acceleration-- or let me write it this way-- that change in velocity is just acceleration times time. Or I should say, acceleration times change in time. So if you divide both sides of this equation by acceleration, you get this right over here. So that is what our displacement-- Remember, I want an expression for displacement in terms of the things we know and the one thing that we want to find out. Well, for this example right over here, we know a couple of things. Well actually, let me take it step by step. We know that our initial velocity is 0."
    },
    {
        "Q": "At 9:13 when both sides are square rooted, how come the 19.6 doesnt become 4.42?",
        "A": "It is true that you could write 4.42 instead of \u00e2\u0088\u009a19.6, but you would lose the accuracy of the number. 19.6 is not equal to 4.42, rather 4.4271887242357310647984509622058. And yet, though it is more accurate, it is still not. If you get an irrational number, just keep it in the \u00e2\u0088\u009a form.",
        "video_name": "2ZgBJxT9pbU",
        "timestamps": [
            553
        ],
        "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 3:34 how come it is called the swimmers view?",
        "A": "Because one arm is raised up by the patient s head as if they were swimming the freestyle stroke.",
        "video_name": "cbkTTluHaTw",
        "timestamps": [
            214
        ],
        "3min_transcript": "is blocking it. DR. MAHADEVAN: Exactly. You can see that that big white thing there is the shoulder that's gotten in the way. And it's making it hard to see. SAL KHAN: They shouldn't have worn those lead shoulder pads. DR. MAHADEVAN: [LAUGHING] And it's making it hard to see whether there's something going on down there right now. So it's really a mystery, as you've shown. SAL KHAN: So how do you solve this problem? DR. MAHADEVAN: If you look over at the other film, it's what we call a swimmer's view. And what we've asked the patient to do is raise one arm up and lower the other. And in doing so, you kind of clear that lower cervical spine and allow better visualization of the entire spine. SAL KHAN: I see. And you're taking it from the direction of the raised arm, on the side of raised arm. DR. MAHADEVAN: You take it from the side. And you can see. SAL KHAN: This is the raised arm right over here. DR. MAHADEVAN: Exactly, that's the raised arm. SAL KHAN: I see. And the other arm on the further side of the patient is down. And that's what allows us to get to the shoulder in a position, so it doesn't block like it does in this left view. DR. MAHADEVAN: Exactly, exactly. SAL KHAN: I see. And over here, it is much clearer And OK. So let me see. So we can count. This is number one right up here. DR. MAHADEVAN: That's one. SAL KHAN: One, two, three, four, five, six-- yeah, we already got to six. We didn't see six over here. And then we got seven. DR. MAHADEVAN: Exactly. SAL KHAN: OK. And so you would call this is an adequate view for what we're trying-- of the neck, because now we can look at all the way DR. MAHADEVAN: Absolutely. We can get all the way down to seven. And ideally, you want to see the top of one, which comes-- actually, in this counting system, we go one through seven. And then we start back at one again, because we're starting with the thoracic vertebrae. SAL KHAN: Oh, look at that. It's like with those streets, where they restart numbering. And you can't find it. So it becomes one again. DR. MAHADEVAN: Exactly. SAL KHAN: Did I number that right? And again, we're looking more to the front. You've got your numbers perfectly on every spinous process, the little bump that you can feel, if you press on the back of your neck. But what we're really interested is the alignment of the front of the vertebral bodies. SAL KHAN: So this is one. This is two, three, four, five, six, seven. Where's the top of one? DR. MAHADEVAN: If you just continue down right there. And it sometimes is difficult to see. But exactly, you want to see that there's alignment right in front of-- I'll assume that there's something here that I can't really see. But you're an expert. So maybe you see things that I don't. OK, so now what do we do with this? DR. MAHADEVAN: Now, we've shown you that you can get a swimmer's view. And it can show you all the way down to C7, T1. But on the original view, as you've shown, we can't see that. So what we did for this patient was get a swimmer's view. SAL KHAN: I see. So this is adequate. And we have this other slide right over here. We have this other one right over here. And why is this one interesting? DR. MAHADEVAN: This is the same patient. And now we've taken that same view that we talked about before, the swimmer's view, where you've got-- SAL KHAN: This is the same patient as this patient right over here, not this patient over here SAL KHAN: Because that one looked overall pretty healthy. DR. MAHADEVAN: That was a normal swimmer's view. But here is an abnormal swimmer's view."
    },
    {
        "Q": "3:40 why did carbon have 'too many bonds' isnt it perfectly stable with 4 bonds and 0 formal charge?",
        "A": "You re forgetting the implied hydrogen on each carbon, benzene has the formula C6H6, each carbon is bonded to 2 other carbons and 1 hydrogen",
        "video_name": "oxf0LMJTklg",
        "timestamps": [
            220
        ],
        "3min_transcript": "on this oxygen. I'll make it magenta. That lone pair is next to the pi bond. The one in red and so, we can go ahead and draw a resonance structure and we take these electrons in magenta and move then into here. That would mean too many bonds to this carbon. We take the electrons in red and we push them off onto this carbon. Let's go ahead and draw our resonance structure. We have our ring here and we have now a double bond between the oxygen and the carbon. Only two lone pairs of electrons on this oxygen now. The electrons in magenta move in here to form a pi bond and the electrons in red move off onto this carbon right here. That's gonna give that carbon a -1 formal charge. Let's go ahead and draw a -1 formal charge here. The electrons in blue have not moved and the electrons in green, We put those in there like that. All right, next, we have the exact same pattern that we did before. We have a lone pair of electrons next to a pi bond. The lone pair of electrons are the electrons in red right here. Next to a pi bond, the electrons in blue. Let's go ahead and draw another resonance structure. We could take these electrons in red, push them into here. That would mean too many bonds to this carbon. If you take these electrons in blue and push them off onto this carbon. Let's draw that resonance structure. Once again, we have our carbon double bonded to an oxygen up here. We said that these electrons were the ones in magenta and the electrons in red moving here to form a pi bond. The electrons in blue move off onto this carbon and that gives this carbon a -1 formal charge. This carbon has a -1 formal charge. that has the blue electrons on it. We still have our electrons in green over here and we have the exact same pattern. We have a lone pair next to a pi bond. The lone pair are the ones in blue and this time, the pi bond are the electrons in green here. We can draw yet another resonance structure. We could take the electrons in blue, move them into here. That would mean too many bonds to this carbon. To take the electrons in green and push them off onto that carbon and let's draw that resonance structure. Once again, we have our ring. We draw our ring in here. We have this double bond up here. Put in lone pairs of electrons on the oxygen and these electrons were the ones in magenta and we go around the ring. We had our electrons in red right here. The electrons in blue move into here and finally, the electrons in green move off onto this carbon. This carbon right here in green."
    },
    {
        "Q": "at 8:38 aren't the both of the products same molecule flipped around ?",
        "A": "I don t know which way you want to flip, but the answer is No . They are different molecules. You can flip them any way you want, and you will not be able to make all the bonds coincide with each other.",
        "video_name": "fSk1Crn3R2E",
        "timestamps": [
            518
        ],
        "3min_transcript": "All I have to do is take away that double bond and I'm done. Well sometimes that's true. But in this case, we actually formed two new chirality centers, right? So this top carbon here is a chirality center, and this bottom carbon here is also a chirality center. So sometimes it's not quite that simple. We need to think about the syn addition of those hydrogens when you think about the possible products that would result. So we're going to get two products here. Let's look at the one on the left. Well, one possibility is I can add those two hydrogens on the same side as a wedge, right? So I have one hydrogen as a wedge, the other hydrogen as a wedge. That's our syn addition. And that means that this top carbon here, this ethyl group, must be going away from me. And down here at the bottom carbon, the methyl group must be going away from me. So that's one possible product. The other possibility, instead of having my two hydrogens So there's a hydrogen and then here's a dash, and there's a hydrogen. So at this top carbon here, now my ethyl group is coming out at me. And at this bottom carbon now my methyl group is coming out at me, like this. So I have two possibilities. And if I look at these two products, I can see that they are enantiomers, they are mirror images of each other. So these two would be my enantiomers, and these would be the products of my reaction. So be very careful when thinking about syn additions here. Let's do one more example of a hydrogenation reaction. Let's do a bridged bicyclic compound. So let's look at a famous bridged bicyclic compound. Let's see if we can draw it here. 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."
    },
    {
        "Q": "At 9:32, are we supposed to just know that charcoal (C) is needed with our metal (Pd), or is there some rule being followed?",
        "A": "This is just something to know. When the Pd is spread over carbon, it greatly increases the available surface area, and thus the efficiency, of the catalyst. Thus, practically, a palladium hydrogenation is usually palladium on carbon (written as Pd/C).",
        "video_name": "fSk1Crn3R2E",
        "timestamps": [
            572
        ],
        "3min_transcript": "So there's a hydrogen and then here's a dash, and there's a hydrogen. So at this top carbon here, now my ethyl group is coming out at me. And at this bottom carbon now my methyl group is coming out at me, like this. So I have two possibilities. And if I look at these two products, I can see that they are enantiomers, they are mirror images of each other. So these two would be my enantiomers, and these would be the products of my reaction. So be very careful when thinking about syn additions here. Let's do one more example of a hydrogenation reaction. Let's do a bridged bicyclic compound. So let's look at a famous bridged bicyclic compound. Let's see if we can draw it here. 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."
    },
    {
        "Q": "ok, wait. for statement two, or 3:30, an unbalanced force doesn't necessarily mean tethering the moving object. what if the object hit a wall? then it would slow down, right? or if it rubbed against a surface, it might change direction, but it would also change speed. how is statement 2 wrong?",
        "A": "hitting a wall and rubbing against a surface are unbalanced forces and they change the motion of the object",
        "video_name": "D1NubiWCpQg",
        "timestamps": [
            210
        ],
        "3min_transcript": "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, they were traveling in when they let go. They'll keep going on in that direction. And if we assume very, very, very small frictions from the ice skating rink, they'll actually have the same speed. So the force, the inward force, the tension from the rope pulling on the skater in this situation, would have only changed the skater's direction. So and unbalanced force doesn't necessarily have to impact the object's speed. It often does. But in that situation, it would have only impacted the skater's direction. Another situation like this-- and once again, this involves centripetal acceleration, inward forces, inward acceleration-- is a satellite in orbit, or any type of thing in orbit. So if that is some type of planet, and this is one of the planet's moons right over here, the reason why it stays in orbit is because the pull of gravity keeps making the object change its direction, but not its speed."
    },
    {
        "Q": "At 3:58,Sal said that the satellite changes its direction but not its speed.So if that same planet is moving around the sun, shouldn't the speed of the satellite vary if it has to go around the planet?",
        "A": "I think when the satellite is in the geo stationary orbit its velocity is constant other wise the satellite velocity changes like a planet around the sun",
        "video_name": "D1NubiWCpQg",
        "timestamps": [
            238
        ],
        "3min_transcript": "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, they were traveling in when they let go. They'll keep going on in that direction. And if we assume very, very, very small frictions from the ice skating rink, they'll actually have the same speed. So the force, the inward force, the tension from the rope pulling on the skater in this situation, would have only changed the skater's direction. So and unbalanced force doesn't necessarily have to impact the object's speed. It often does. But in that situation, it would have only impacted the skater's direction. Another situation like this-- and once again, this involves centripetal acceleration, inward forces, inward acceleration-- is a satellite in orbit, or any type of thing in orbit. So if that is some type of planet, and this is one of the planet's moons right over here, the reason why it stays in orbit is because the pull of gravity keeps making the object change its direction, but not its speed. So this was its speed right here. If the planet wasn't there, it would just keep going on in that direction forever and forever. But the planet right over here, there's an inward force of gravity. And we'll talk more about the force of gravity in the future. But this inward force of gravity is going to accelerate this object inwards while it travels. And so after some period of time, this object's velocity vector-- if you add the previous velocity with how much it's changed its new velocity vector. Now this is after its traveled a little bit-- its new velocity vector might look something like this. And it's traveling at the exact right speed so that the force of gravity is always at a right angle to its actual trajectory. It's the exact right speed so it doesn't go off into deep space and so it doesn't plummet into the earth. And we'll cover that in much more detail. But the simple answer is, unbalanced force on a body"
    },
    {
        "Q": "at 2:56, it says \"and they could hold on to the rope, and as long as they hold on to the rope, they'll keep going in circles. wound't they eventually stop from the friction on the ice?",
        "A": "yes but he is imagining ice that is so slippery that it has no friction, so that we can get the idea of what it means for an object to have no net force on it",
        "video_name": "D1NubiWCpQg",
        "timestamps": [
            176
        ],
        "3min_transcript": "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, they were traveling in when they let go. They'll keep going on in that direction. And if we assume very, very, very small frictions from the ice skating rink, they'll actually have the same speed. So the force, the inward force, the tension from the rope pulling on the skater in this situation, would have only changed the skater's direction. So and unbalanced force doesn't necessarily have to impact the object's speed. It often does. But in that situation, it would have only impacted the skater's direction. Another situation like this-- and once again, this involves centripetal acceleration, inward forces, inward acceleration-- is a satellite in orbit, or any type of thing in orbit. So if that is some type of planet, and this is one of the planet's moons right over here, the reason why it stays in orbit is because the pull of gravity keeps making the object change its direction, but not its speed."
    },
    {
        "Q": "At 1:10, why the first statement is true?",
        "A": "Why wouldn t it be true? All the statement is doing is rephrasing Newton s First Law.",
        "video_name": "D1NubiWCpQg",
        "timestamps": [
            70
        ],
        "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": "Once Sal has simplified the equation (12:51) to be 3PV=mv^2 N he divides the equation by two. This takes on the form 3/2PV=(mv^2/2 N). Why doesn't the number of molecules in the system also get divided by two when he divides the two halves of the equation by two? It seems to be exempt from his division of two on both sides of the equation. Shouldn't the equation look like 3/2PV=(mv^2 N)/2?",
        "A": "If you remember order of operations, division and multiplication can be done in either order. He wrote it as (mv^2)/2 *N so that you could see that 1/2 * mv^2 is kinetic energy",
        "video_name": "qSFY7GKhSRs",
        "timestamps": [
            771
        ],
        "3min_transcript": "This is the pressure from one particle. If I wanted the pressure from all of the particles on that wall-- so the total pressure on that wall is going to be from N over 3 of the particles. The other particles aren't bouncing off that wall. So we don't have to worry about them. So if we want the total pressure on that wall-- I'll just write, pressure sub on the wall. Total pressure on the wall is going to be the pressure from one particle, mv squared, over our volume, times the total number of particles hitting the wall. The total number of particles is N divided by 3, because only 3 will be going in that direction. So, the total pressure on that wall is equal to mv squared, over our volume of our container, times the total particles divided by 3. Let's see if we can manipulate this thing a little bit. So if we multiply both sides by-- let's see what we can do. to mv squared, times N, where N is the number of particles. Let's divide both sides by N. So we get 3pv over-- actually, no, let me leave the N there. Let's divide both sides of this equation by 2. So we get, what do we get? We get 3/2 pv is equal to-- now this is interesting. It's equal to N, the number of particles we have, times mv squared over 2. Remember, I just divided this equation right here by 2 to get this. And I did this for a very particular reason. mv squared over 2 is the kinetic energy of that little particle we started off with. That's the formula for kinetic energy. Kinetic energy is equal to mv squared over 2. So this is the kinetic energy of one particle. Now, we're multiplying that times the total number of particles we have, times N. So N times the kinetic energy of one particle is going to be the kinetic energy of all the particles. And, of course, we also made another assumption. I should state that I assumed that all the particles are moving with the same velocity and have the same mass. In a real situation, the particles might have very different velocities. But this was one of our simplifying assumptions. So, we just assumed they all have that. So, if I multiply N times that-- this statement right here-- is the kinetic energy of the system."
    },
    {
        "Q": "At 2:35, it is said that experimental studies have shown that the structure of the amide is planar, which means the molecule's sp2 hybridized resonance structure contributes more towards the resonance hybrid. But the stability of the first structure (with sp3 hybrid N) is greater due to zero formal charge. Why is that the stable structure does not contribute more to the resonance hybrid?",
        "A": "Although the first structure does not have any formal charges, the electronegativity of the oxygen atom does give it a partial negative charge. The resonance on the second structure is a stabilizing factor, and its sp2 hybridization is more stable than the sp3 hybridization of the first structure. (Covered later in the section on acid-base chemistry.)",
        "video_name": "kQCS1AhAnMI",
        "timestamps": [
            155
        ],
        "3min_transcript": "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 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"
    },
    {
        "Q": "What is the proper usage of the spelling (5:35), blastocoele vs. blastocoel? My textbook says blastocoele, and I am just curious. Thanks!",
        "A": "both are right",
        "video_name": "-yCIMk1x0Pk",
        "timestamps": [
            335
        ],
        "3min_transcript": "to start filling in some of this gap between the embryoblast and the trophoblast, so you're going to start having some fluid that comes in there, and so the morula will eventually look like this, where the trophoblast, or the outer membrane, is kind of this huge sphere of cells. And this is all happening as they keep replicating. Mitosis is the mechanism, so now my trophoblast is going to look like that, and then my embryoblast is going to look like this. Sometimes the embryoblast-- so this is the embryoblast. Sometimes it's also called the inner cell mass, so let me write that. And this is what's going to turn into the organism. And so, just so you know a couple of the labels that are organism, and we are mammals, we call this thing that the morula turned into is a zygote, then a morula, then the cells of the morula started to differentiate into the trophoblast, or kind of the outside cells, and then the embryoblast. And then you have this space that forms here, and this is just fluid, and it's called the blastocoel. A very non-intuitive spelling of the coel part of blastocoel. But once this is formed, this is called a blastocyst. That's the entire thing right here. Let me scroll down a little bit. This whole thing is called the blastocyst, and this is the case in humans. Now, it can be a very confusing topic, because a lot of times in a lot of books on biology, you'll say, hey, you go from the morula to the blastula or the Let me write those words down. So sometimes you'll say morula, and you go to blastula. Sometimes it's called the blastosphere. And I want to make it very clear that these are essentially the same stages in development. These are just for-- you know, in a lot of books, they'll start talking about frogs or tadpoles or things like that, and this applies to them. While we're talking about mammals, especially the ones that are closely related to us, the stage is the blastocyst stage, and the real differentiator is when people talk about just blastula and blastospheres. There isn't necessarily this differentiation between these outermost cells and these embryonic, or this embryoblast, or this inner cell mass here. But since the focus of this video is humans, and really that's where I wanted to start from, because that's what we are and that's what's interesting, we're going to"
    },
    {
        "Q": "At 6:00 what is the difference between MORULA and BLASTOCYST??",
        "A": "A morula is a special kind of blastocyst, it has 16 cells assembled in a solid block of cells and looks like a mulberry/ Latin: morula. A proper blastocyst is a few days older, has more cell and forms a hollow sphere.",
        "video_name": "-yCIMk1x0Pk",
        "timestamps": [
            360
        ],
        "3min_transcript": "to start filling in some of this gap between the embryoblast and the trophoblast, so you're going to start having some fluid that comes in there, and so the morula will eventually look like this, where the trophoblast, or the outer membrane, is kind of this huge sphere of cells. And this is all happening as they keep replicating. Mitosis is the mechanism, so now my trophoblast is going to look like that, and then my embryoblast is going to look like this. Sometimes the embryoblast-- so this is the embryoblast. Sometimes it's also called the inner cell mass, so let me write that. And this is what's going to turn into the organism. And so, just so you know a couple of the labels that are organism, and we are mammals, we call this thing that the morula turned into is a zygote, then a morula, then the cells of the morula started to differentiate into the trophoblast, or kind of the outside cells, and then the embryoblast. And then you have this space that forms here, and this is just fluid, and it's called the blastocoel. A very non-intuitive spelling of the coel part of blastocoel. But once this is formed, this is called a blastocyst. That's the entire thing right here. Let me scroll down a little bit. This whole thing is called the blastocyst, and this is the case in humans. Now, it can be a very confusing topic, because a lot of times in a lot of books on biology, you'll say, hey, you go from the morula to the blastula or the Let me write those words down. So sometimes you'll say morula, and you go to blastula. Sometimes it's called the blastosphere. And I want to make it very clear that these are essentially the same stages in development. These are just for-- you know, in a lot of books, they'll start talking about frogs or tadpoles or things like that, and this applies to them. While we're talking about mammals, especially the ones that are closely related to us, the stage is the blastocyst stage, and the real differentiator is when people talk about just blastula and blastospheres. There isn't necessarily this differentiation between these outermost cells and these embryonic, or this embryoblast, or this inner cell mass here. But since the focus of this video is humans, and really that's where I wanted to start from, because that's what we are and that's what's interesting, we're going to"
    },
    {
        "Q": "at around 3:56 , does this mean that oxygen can have an oxidation state of 1+ when bonded with any other element in it's group ?",
        "A": "No, oxygen is more electronegative than the other elements in its group so it will be -2 with all of them.",
        "video_name": "R2EtXOoIU-E",
        "timestamps": [
            236
        ],
        "3min_transcript": "And if they were hypothetically ionic bonds, what would happen? Well, if you had to give these electrons to somebody, you would give them to the oxygen, the electrons in this period, give them to the oxygen, giving it an oxidation state of negative 1. With the hydrogen having these electrons taken away, it's going to have an oxidation state of positive 1. And the same thing's going to be true for that oxygen and that hydrogen right over there. So this is fascinating, because this is an example where oxygen has an oxidation state not of negative 2, but an oxidation state of negative 1. So this is already kind of interesting. Now it gets even more interesting when we go to oxygen difluoride. Why is this more interesting? Because fluorine is the one thing on this entire table that is more electronegative than oxygen. This is a covalent bond, but in our hypothetical ionic bond, if we had to give these electrons to one So the fluorine, each of them would have an oxidation state of negative 1. And the oxygen here-- now, you could imagine, this is nuts for oxygen. The oxidation state for oxygen, it's giving up these electrons. It would be a positive 2. And we talk about oxidation states when we write this little superscript here. We write the sign after the number. And that's just the convention. But it has an oxidation state of positive 2. Oxygen, the thing that likes to oxidize other things, it itself has been oxidized by fluorine. So this is a pretty dramatic example of how something might stray from what's typical oxidation state or it's typical oxidation number. And in general, oxygen will have an oxidation state or oxidation number in most molecules of negative 2. But unless it's bonded with another oxygen or it's bonded to fluorine, which is a much more but it's the only atom that is more electronegative than-- or the only element is more electronegative than oxygen."
    },
    {
        "Q": "At 6:33, Sal says vector v is the sum of is the sum the other two component vector and he has even proved it in his previous video. But, doesn't that break down the traditional Pythagorean Theorem?\nEven the result at the end of the video doesn't sum up to 10(value of vector v) but when we use the Pythagorean Theorem, the equation does satisfy. Anyone please explain .",
        "A": "When you use the Pythagoras theorem, you find the magnitude of the vector. But, in this video, Sal was talking about the vectors, not just their magnitude. Hope this helps :)",
        "video_name": "2QjdcVTgTTA",
        "timestamps": [
            393
        ],
        "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": "At 2:15 Sal mentions an \" important caveat \" , what exactly is that ?",
        "A": "A caveat often means an exeption or a problem, ex. It is fun to ride a bike but there s a caveat, you might fall over and hurt yourself",
        "video_name": "zA0fvwtvgvA",
        "timestamps": [
            135
        ],
        "3min_transcript": ""
    },
    {
        "Q": "6:35 Isn't the inaccuracy in finding the work in each measured area coming from the fact that you are only measured the area of the rectangle instead of including the triangle above it as well? Assuming that the pressure changes form a straight line, in order to get an accurate value, couldn't you get the area of the rectangle, then use an equation to find the area of a right triangle to find the area above, then add both to find the correct area?",
        "A": "I suggest that you watch the calculus videos, it will help to make sense of finding the areas under curves. You are right, when dV is not infinitely small, then it is not exactly a perfect rectangle. But using calculus, we can make it so that we are adding up all the rectangles where dV is so infinitely small that the little triangle at the top doesn t exist. I m sure Sal can explain this a lot better than me, in his Calculus videos :)",
        "video_name": "M5uOIy-JTmo",
        "timestamps": [
            395
        ],
        "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": "At 13:54 the amount of work needed for the system to reach from state 1 to state 2 is larger than the amount of the whole cycle. Is it because when the system was returning from state 2 to state 1 it was doing work?",
        "A": "You have it in reverse. From state 1 to state 2, the system is doing work on the surroundings. (Volume is increasing (work of expansion)). From state 2 to state 1, the surroundings are doing work on the system. (Volume is decreasing (work of compression)). Because the area under 1 to 2, is greater than the area under 2 to 1, the net effect is more work being done on the surroundings than is being done on the system.",
        "video_name": "M5uOIy-JTmo",
        "timestamps": [
            834
        ],
        "3min_transcript": "State 1. And I do something, you know, I'm in a quasi-static process and it, you know, it's doing something weird, and I get to state 2 here. And it's going in this direction. So my volume is increasing. So in this situation, what is the work done by the system? Easy enough, it's the area under this curve. Now let's say that I keep doing some type of quasi-static process, but it takes a different path. I'm doing something else, other than adding the marbles directly back. So my new path looks something like this to get back to state 1. So these arrows are going back. So now what is the work done to the system? well my volume is decreasing, so it's the area under the The area under the second curve is the work done to the system. So if I want to know what the net work the system did, going from state 1 to state 2, and then going back to state 1-- remember, this is a pressure and volume diagram-- what is it? Well the work that the system did was this whole area under this brown curve. And then it had some work done to it, which is the area under this magenta curve. So the net work it did is essentially the white, the whole area, minus this red area. So the net work it did would be essentially just the area inside this loop. And hopefully you don't have to know calculus to do this, although calculus you would actually use to compute these areas. But I just want to give you that intuition, that the area inside this closed loop is actually the amount of work that our system has done. And what's important is the direction that it's going. of this clockwise motion. This is the work that our system has done, which, I don't know, to me is a pretty interesting thing. And later we can use this notion to come up with some other ideas behind our state variables I'll make one little aside here. Remember, our state variable pressure volume, we did stuff to it then we went back to that state. That stayed the same. And I want to say another thing. For our purposes, when we're dealing with ideal gases, where the internal energy is essentially the kinetic energy of the system, if we go and do all sorts of crazy stuff and come back, our internal energy hasn't changed. So the internal energy is always going to be the same at this point. So if I said, I did all of this stuff and came back here, what is my change in internal energy? It's 0. The change is 0. Now if I said I went from here to here, I would have a different internal energy and my change would be something real. But since this is a state function, it doesn't care how I got there."
    },
    {
        "Q": "At 8:22 how does thermal radiation interact with our skin and transfer energy?Is it the magnetic and electric field that interacts or the photon of light that transfers heat?",
        "A": "I was confused about this as well, seeing as the apparent cold air still exists between you and the fire as you are warming up. I m not sure that is true once the fire is active. You are correct in assuming that the photon transfers the heat. Photons can carry heat energy, even though they have no mass. This phenomenon also explains how you warm up when you step into the sunlight on a chilly day :)",
        "video_name": "8GQvMt-ow4w",
        "timestamps": [
            502
        ],
        "3min_transcript": "charged particles, are made up of protons and electrons. And so as you accelerate these and the more that you accelerate these the more radiation you are going to release. And you might say, Okay, you said I'm observing that in fire, where am I observing radiation? Well just the very fact that you can see the fire, the light emitted from the fire, that is electromagnetic radiation, that is electromagnetic radiation. It's just the electromagnetic radiation in the wavelengths that your eye considers to be visible light, or that your eye considers to be light. Even the particles up here, so even the particles up here that are still quite hot, they are also emitting electromagnetic radiation because they're all bumping into each other and their electrons and their protons are all getting accelerated in different ways, they're also releasing electromagnetic radiation, but it is at a slower wavelength than your eye is capable of perceiving as light. If you had an infrared camera you would see the flames being much larger. And if you were to look at the flame closely you would see down here right where the combustion reaction is happening, the flame looks blue. And that's because the blue light is higher energy light, and that's because the particles are being accelerated more down here, and then it goes from blue to kind of a white to a yellow, to a red or to an orange, to a red, and then it, to your regular eyes, it disappears. But everything, everything that has some temperature is releasing electromagnetic radiation. And you're like, Okay, well that's all fine, I can see it but how is that a form of energy transfer? Well, if you're ever sat next to a flame you will feel the heat. In fact, even if the air between you and the flame is cold you would still feel like you're getting warm. So if this is a flame right over here, so that is fire, and let's say you have cold air, cold air, let's say you're at a campfire right over here, if you are standing right over here you would still feel heat, you would still feel like you're getting warmed up. And that's because that electromagnetic radiation is being emitted from the air particles that we perceive as fire and then that can actually excite particles on your skin and it will transfer energy to your skin and so you feel like you are actually getting warmed up. I remember once, this is kind of a strange story, but I was on the highway and there was a car on fire and I was literally, we drove to the far lane because it was on fire, we're three lanes away from it and it kind of exploded. I don't think anyone was in it, hopefully no one was in it, but I remember right when it exploded it was an intense, immediate heat that we felt through the window of the car, and that was electromagnetic radiation. That was thermal radiation being released by these accelerated particles in the air around that explosion,"
    },
    {
        "Q": "At 3:20 why 7 times?",
        "A": "Just an arbitrary amount of times, he could have done it more or less. The amount of windings would effect the current produced.",
        "video_name": "jabo8iTesqQ",
        "timestamps": [
            200
        ],
        "3min_transcript": "Of course, the cap is what we're going to make our winding around. So the washers that are going to support the winding need to be pretty close to the cap. And the permanent magnets also need to be pretty close. So we're just making our marks here so that we can position everything. And it's pretty important that things are lined up fairly accurately, otherwise the winding may not turn smoothly. So we determined that we're going to measure off of the center point and about an inch away from the center of the square. So now that we have our intersection points marked, we're going to go ahead and drill our holes. We've taped the drill about 1/2 an inch up, so we want the hole to go in about 1/2 an inch. And we're using a 3/16 inch bit on the sides. And then we'll use a 1/8 inch bit for our-- the other sides. And the other sides are going to hold our bearings. So these are the bearing screws. so that's what these screws are from-- or for. These actually came from an Ikea bed. And then these are just some scrap screws that we had laying around. You can use any screws or nails. What's important is that the screws are lined up right and that they're the same level. So that will improve the efficiency of our motor. So we're just checking the level and making sure everything's lined up correctly. And now we're going to take some hot glue and get our permanent magnets. Make sure that they are oriented so that they are attracted to one another. So we want their opposite poles facing one another. And so we just put some hot glue on the screw and then set the magnet in there. And the main reason we did it this way is that it's really easy to change the position of the magnets if we need to. And we may need to do that. So we've got a washer here. in a vertical orientation. And the hot glue allows for easy adjustments. And so now we're going to take our bottle cap that we had talked about earlier. We're going to wrap our field coil around it. And we're going to go around it seven times. And then once we've got the coil a fairly good length, we have about 3 inches of wire on either side. We'll loop the wire through and tie it off so that it stays in a consistent loop. There we go. And then we'll do the other side as well. I'll just loop the wire through. And that just kind of holds the loop together and gives us something to connect with our washers. All right, now we're just going to tape the edges of the wire to keep it together."
    },
    {
        "Q": "0:30 How do you know that Co2 is going to be your cation? Sorry, I'm a bit behind. (And confused)",
        "A": "In the formula of an ionic compound (at least in one that only has 2 different elements) the first element is the cation and the second is the anion.",
        "video_name": "vVTwzjvWySs",
        "timestamps": [
            30
        ],
        "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": "Around 9:30, Sal is trying to find the legs of the right triangle. Why does he use \"soh\" and \"cah\"? How do either of those relate to the length of the side?\nAt 9:55, he says \"5 sine, of 36.899, is equal to...\" What is \"sine\" and \"cosine\"? And how are they used here?\nThanks in advance.",
        "A": "soh, cah and toa are mnemonics (look it up) for Sine = Opposite side / Hypotenuse ==> SOH Cosine = Adjacent side / Hypotenuse ==> CAH Tangent = Opposite side / Adjacent side ==> TOA",
        "video_name": "xp6ibuI8UuQ",
        "timestamps": [
            570,
            595
        ],
        "3min_transcript": "It's 36.8699 degrees. So I'm picking that particular number for a particular reason. Now what I wanna do is I wanna figure out this vector's horizontal and vertical component. So I wanna break it down into something that's going straight up or down and something that's going straight right or left. So how do I do this? Well, one, I could just draw them, visually, see what they look like. So its vertical component would look like this. It would start... Its vertical component would look like this. And its horizontal component would look like this. Its horizontal component would look like this. The horizontal component, the way I drew it, it would start where vector A starts and go as far in the X direction as vector A's tip, but only in the X direction, and then you need to, to get back to the head of vector A, you need to have its vertical component. we could call the vertical component over here A sub Y, just so that it's moving in the Y direction. And we can call this horizontal component A sub X. Now what I wanna do is I wanna figure out the magnitude of A sub Y and A sub X. So how do we do that? Well, the way we drew this, I've essentially set up a right triangle for us. This is a right triangle. We know the length of this triangle, or the length of this side, or the length of the hypotenuse. That's going to be the magnitude of vector A. And so the magnitude of vector A is equal to five. We already knew that up here. So how do we figure out the sides? Well, we could use a little bit of basic trigonometry. If we know the angle, and we know the hypotenuse, how do we figure out the opposite side to the angle? So this right here, this right here is the opposite side to the angle. And if we forgot some of our basic trigonometry we can relearn it right now. Soh-cah-toa. Cosine is adjacent over hypotenuse. Tangent is opposite over adjacent. So we have the angle, we want the opposite, and we have the hypotenuse. So we could say that the sine of our angle, the sine of 36.899 degrees, is going to be equal to the opposite over the hypotenuse. The opposite side of the angle is the magnitude of our Y component. ...is going to be equal to the magnitude of our Y component, the magnitude of our Y component, over the magnitude of the hypotenuse, over this length over here, which we know is going to be equal to five. Or if you multiply both sides by five, you get five sine of 36.899 degrees, is equal to the magnitude of the vertical component"
    },
    {
        "Q": "In 6:25, Sal writes ||a||. What do the double lines mean?",
        "A": "Those lines indicate that we are taking the magnitude of the vector between them. They are sort of like a vector version of the absolute value sign in math.",
        "video_name": "xp6ibuI8UuQ",
        "timestamps": [
            385
        ],
        "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 12:30 What does he mean by saying making this a 2 component velocity",
        "A": "Sal said: break that down into two component velocities . These velocities are horizontal and vertical components of the given velocity.",
        "video_name": "xp6ibuI8UuQ",
        "timestamps": [
            750
        ],
        "3min_transcript": "Now before I take out the calculator and figure out what this is, let me do the same thing for the horizontal component. Over here we know this side is adjacent to the angle. And we know the hypotenuse. And so cosine deals with adjacent and hypotenuse. So we know that the cosine of 36.899 degrees is equal to... Cosine is adjacent over hypotenuse. So it's equal to the magnitude of our X component over the hypotenuse. The hypotenuse here has... Or the magnitude of the hypotenuse, I should say, which has a length of five. Once again, we multiply both sides by five, and we get five times the cosine of 36.899 degrees is equal to the magnitude of our X component. So let's figure out what these are. Let me get the calculator out. I wanna make sure it's in degree mode. So let me check. Yep, we're in degree mode right over there. Don't wanna... Make sure we're not in radian mode. Now let's exit that. And we have the vertical component is equal to five times the sine of 36.899 degrees, which is, if we round it, right at about three. So this is equal to... So the magnitude of our vertical component is equal to three. And then let's do the same thing for our horizontal component. So now we have five times the cosine of 36.899 degrees, is, if once again we round it to, I guess, our hundredths place, we get it to being four. So we see here is a situation where we have... This is a classic three-four-five Pythagorean triangle. The magnitude of our horizontal component is four. The magnitude of our vertical component, right over here, is equal to three. And once again, you might say, Sal, why are we going through all of this trouble? that if we say something has a velocity, in this direction, of five meters per second, we could break that down into two component velocities. We could say that that's going in the upwards direction at three meters per second, and it's also going to the right in the horizontal direction at four meters per second. And it allows us to break up the problem into two simpler problems, into two one-dimensional problems, instead of a bigger two-dimensional one."
    },
    {
        "Q": "At 3:41, what is a nuetron",
        "A": "A nuetron is a sub-atomic particle that is inisde an atom. In a nuclues (which is in an atom) there are two sub-atomic particles - protons and neutrons. Neutrons have a mass of 1 (relative to the mass of a proton) and a charge of 0.",
        "video_name": "i-NNWI8Ccas",
        "timestamps": [
            221
        ],
        "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 5:00,it is said that hydrogen will turn to deuterium and then to helium , why cant deuterium change to TRITIUM and then to helium ?",
        "A": "You start with a bunch of Hydrogen-1, which is basically free protons. Some of the protons fuse, and immediately decay to form deuterium. Where do the free neutrons, which only have a 15 minute half life, come from to then fuse to form tritium?",
        "video_name": "i-NNWI8Ccas",
        "timestamps": [
            300
        ],
        "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": "I was reading that a neutron is heavier than a proton. Then how could a proton and a neutron have a smaller mass than 2 protons (as Sal mentioned at around 4:00)? To convert a proton into a neutron, one would have to add the mass difference in the form of energy. Then how would the reaction be exothermic?",
        "A": "The Deuteron D is a stable particle composed of a proton p and a neutron n. To separate D into its components you have to provide energy in order to overcome the strong binding forces. This energy you have to add corresponds to a mass according to Einsteins formula E=mc^2 Therefore the products (p+n) are heavier than the particle D you startet from.",
        "video_name": "i-NNWI8Ccas",
        "timestamps": [
            240
        ],
        "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": "As far as i know a neutron is composed of one proton and one electron but Sal at 3:53 said that when two protons come very very close to each other, a proton degrades into a neutron.\nSo, how is it possible for the creation of a neutron without a proton?",
        "A": "A neutron is not composed of a proton and electron. A proton decays into a neutron when one of its constituent up quarks decays into a down quark. A neutron can also decay into a proton via the reverse process.",
        "video_name": "i-NNWI8Ccas",
        "timestamps": [
            233
        ],
        "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": "During the video on the birth of stars, Sal said at about 0:17 that gravity would bring the hydrogen atoms together. But excluding the clouds of hydrogen, there is really nothing in space. Why won't the hydrogen atoms just diffuse and get evenly distributed in the universe?",
        "A": "Even individual atoms have a tiny tiny bit of gravitational force that attracts them into forming clouds. If a certain cluster of hydrogen atoms form a dense enough cloud, they attract even more hydrogen atoms until eventually you get the kind of mass associated with stars.",
        "video_name": "i-NNWI8Ccas",
        "timestamps": [
            17
        ],
        "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": "In the beginning Sal starts with hydrogen atoms...\nQ1.Were did that hydrogen come from in the first place?\nQ2.How did it become a cloud 0:03? (Why were the hydrogen atoms slightly close in the beginning)",
        "A": "The hydrogen came from the big bang (supposedly), just like all of the other matter. It became a cloud because the atoms were and are attracted to each other, and they like to form together.",
        "video_name": "i-NNWI8Ccas",
        "timestamps": [
            3
        ],
        "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": "0:30 wouldn't the hydrogen atoms accelerate since the gravatational force begins acting over a shorter and shorter distance therefore becoming stronger",
        "A": "I would imagine that since as they move in more, they heat up more, and so the gas would also try to expand, creating a counter force that would balance it out.",
        "video_name": "i-NNWI8Ccas",
        "timestamps": [
            30
        ],
        "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": "What's a nuclei? {2:48}",
        "A": "More than one nucleus.",
        "video_name": "i-NNWI8Ccas",
        "timestamps": [
            168
        ],
        "3min_transcript": "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 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."
    },
    {
        "Q": "7:10- What does he say?",
        "A": "At 7:10, he says, ...there s a huge step temperature...",
        "video_name": "i-NNWI8Ccas",
        "timestamps": [
            430
        ],
        "3min_transcript": "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 So they are generating some of their heat. Or there are even smaller objects that just get to the point there's a huge temperature and pressure, but fusion is not actually occurring inside of the core. And something like Jupiter would be an example. And you could go several masses above Jupiter where you get something like that. So you have to reach a certain threshold where the mass, where the pressure and the temperature due to the heavy mass, get so large that you start this fusion. And-- but the smaller you are above that threshold, the slower the fusion will occur. But if you're super massive, the fusion will occur really, really fast. So that's the general idea of just how stars get formed and why they don't collapse on themselves and why they are these kind of balls of fusion reactions existing in the universe. In the next few videos, we'll talk about what happens once that hydrogen fuel in the core starts to run out."
    },
    {
        "Q": "So at 7:16 he mentioned that Jupiter was a example of not fusing into a star. If there was enough mass, could Juipter become a star?",
        "A": "Jupiter would have to engulf all of the other planets in the solar system 20 times over in order to be massive enough to star fusing its hydrogen.",
        "video_name": "i-NNWI8Ccas",
        "timestamps": [
            436
        ],
        "3min_transcript": "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 So they are generating some of their heat. Or there are even smaller objects that just get to the point there's a huge temperature and pressure, but fusion is not actually occurring inside of the core. And something like Jupiter would be an example. And you could go several masses above Jupiter where you get something like that. So you have to reach a certain threshold where the mass, where the pressure and the temperature due to the heavy mass, get so large that you start this fusion. And-- but the smaller you are above that threshold, the slower the fusion will occur. But if you're super massive, the fusion will occur really, really fast. So that's the general idea of just how stars get formed and why they don't collapse on themselves and why they are these kind of balls of fusion reactions existing in the universe. In the next few videos, we'll talk about what happens once that hydrogen fuel in the core starts to run out."
    },
    {
        "Q": "at 0:30, Sal says that the gravity interacts with \"atoms\" in a huuge distance, Doesn't gravity has a limitation? Sorry for my english.",
        "A": "No, as far as we understand it, gravity has unlimited range, although the strength of the field declines with the square of the distance from the mass that generates the field.",
        "video_name": "i-NNWI8Ccas",
        "timestamps": [
            30
        ],
        "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 0:33 , why sal took theta between 0 and 90 degrees ?\nPlease explain .\nThank you .",
        "A": "At 0 or 90 degrees, the given equation for distance (d=(2s^2/g)*cosX*sinX) is 0 (sin 90=0, cos 0=0). This can be seen in the graph at 1:26, where 0 and 90 are x-axis intercepts. Since we re looking for optimal distance, these values can be immediately discarded. This makes sense in a real world way, because the horizontal distance traveled if the projectile is shot straight up or straight into the surface will be 0.",
        "video_name": "snw0BrCBQYQ",
        "timestamps": [
            33
        ],
        "3min_transcript": "Now that we have distance explicitly as a function of the angle that we're shooting the object at, we can use a little bit of calculus to figure out the optimal angle, the angle that's going optimize our distance. And since we only care about angles from 0 degrees to really 90 degrees, let's constrain ourselves. So we're going to optimize things for angles between 0 degrees. So theta is going to be greater than or equal to 0 and less than or equal to 90. So let's see how we can do it. And just to get an idea of what we're even conceptually doing with the calculus, remember when you take a derivative, you are finding the slope of a line, an instantaneous slope of a line. And if you were to graph this-- and I encourage you to graph it on your own, maybe with a graphing calculator-- it will look something like this over the interval. It will look like this where that is the distance as a function of theta axis and then this would And we care about angles between 0 and 90 degrees. So if you were to graph this thing, so this is 0 degrees, this is maybe 90 degrees right here. The graph of this function will look like this. It'll look something like this. It will look something like that. And what we want to do is find the angle, there's some angle here that gives us the optimal distance. So this is, right here, this is the optimal distance. And what we want to do is find that out. And when you look at the graph, and you could do it on a graphing calculator if you like, what happens to the instantaneous slope at that optimal distance? Well it's flat. The slope there is 0. So what we need to do is take the derivative of this function and then just figure out at what angle is the function equal to 0? And then we're done We will know this mystery angle, this optimal angle, to shoot the object at. So let's take the derivative. So the derivative, we'll just use our derivative rules here. The derivative of-- I will call it d prime I guess, or we could say the derivative of the distance with respect to theta is equal to-- we're assuming that s and g are constants, so we don't have to worry about them right now. We could just put them out front since we're assuming they're constants. And then we can do the product rule to take the derivative of this part with respect to theta. In the product rule, we take the derivative of the first function times the second function. So the derivative of cosine of theta is negative sine of theta. And we're going to multiply that times the second function. So that's times the sine of theta."
    },
    {
        "Q": "At 5:30, in reference to the formula, how can one tell what the graph will look like?",
        "A": "Calculate a few points on it, and make a sketch",
        "video_name": "2GQTfpDE9DQ",
        "timestamps": [
            330
        ],
        "3min_transcript": "Here's why I'm taking the absolute value of the product, well, if they're different charges, this will be a negative number, but we just want the overall magnitude of the force. So we could take, it's proportional to the absolute value of the product of the charges and it's inversely proportional to not just the distance between them, not just to r, but to the square of the distance. The square of the distance between them. And what's pretty neat about this is how close it mirrors Newton's law of gravitation. Newton's law of gravitation, we know that the force, due to gravity between two masses, remember mass is just another property of matter, that we sometimes feel is a little bit more tangible because it feels like we can kind of see weight and volume, but that's not quite the same, or we feel like we can feel or internalize things like weight and volume which are related to mass, but in some ways it is just another property, another property, especially as you get into more of a kind of fancy physics. Our everyday notion of even mass starts to But Newton's law of gravitation says, look the magnitude of the force of gravity between two masses is going to be proportional to, by Newton's, by the gravitational concept, proportional to the product of the two masses. Actually, let me do it in those same colors so you can see the relationship. It's going to be proportional to the product of the two masses, m one m two. And it's going to be inversely proportional to the square of the distance. The square of the distance between two masses. Now these proportional personality constants are very different. Gravitational force, we kind of perceive this is as acting, being strong, it's a weaker force in close range. But we kind of imagine it as kind of what dictates what happens in the, amongst the stars and the planets and moons. While the electrostatic force at close range is a much stronger force. It can overcome the gravitational force very easily. But it's what we consider happening at either an atomic level or kind of at a scale But needless to say, it is very interesting to see how this parallel between these two things, it's kind of these patterns in the universe. But with that said, let's actually apply let's actually apply Coulomb's law, just to make sure we feel comfortable with the mathematics. So let's say that I have a charge here. Let's say that I have a charge here, and it has a positive charge of, I don't know, let's say it is positive five times 10 to the negative three Coulombs. So that's this one right over here. That's its charge. And let's say I have this other charge right over here and this has a negative charge. And it is going to be, it is going to be, let's say it's negative one... Negative one times 10 to the negative one Coulombs."
    },
    {
        "Q": "At 3:03 how does the OH molecule gets 1+ve formal charge? I'm not able to understand......",
        "A": "The O atom in the oxonium ion has two electrons in a lone pair, plus one electron from each of the \u00cf\u0083 bonds (5 valence electrons). That\u00e2\u0080\u0099s one less valence electron than in an isolated O atom. \u00e2\u0088\u00b4 Formal charge = 6 - 5 = +1.",
        "video_name": "wCspf85eQQo",
        "timestamps": [
            183
        ],
        "3min_transcript": "which is an excellent leaving group. Let's draw what we would form. Let's sketch in our carbon chain here and we know that a bond forms between the oxygen and the carbon in red, so the carbon in red is this carbon and let's make these electrons magenta. So those electrons form a bond between the oxygen and the carbon in red. The oxygen is still attached to this ethyl group here, so let's draw in those two carbons. And the oxygen is still bonded to this hydrogen. So let's put in this hydrogen. We still have a lone pair of electrons left on this oxygen, so I'll put in that lone pair right here. And that gives us a plus one formal charge on the oxygen. Next, we need to make a neutral molecule for our product, so we need to have another molecule of ethanol come along, so let's draw that in here. So, ethanol is our solvent. is going to function as a base. We need to take this proton here and these electrons are left behind on the oxygen. Let's draw our final product. We sketch in our carbon chain. We have our oxygen. We have these two carbons. And now we have two lone pairs of electrons on this oxygen. Let's make these electrons blue. So, our second step is an acid-base reaction where we take a proton and these are, these electrons in blue here, to form our final product which is an ether. Notice we don't have to worry about any stereochemistry for our final product. We don't have any chiral centers to worry about. Let's look at another reaction. For this reaction, we're starting with a secondary alkyl halides. If I look at my summary over here with a secondary substrate, we could have either an S N 2 mechanism or an S N 1 mechanism, First, let's look at the nucleophile. This is N a plus and SH minus, so let me draw in the SH minus here which that is going to be our nucleophile and that's a strong nucleophile. A negative charge on a sulfur would make a strong nucleophile. And for our solvent, we saw in an earlier video, DMSO is a polar A product solvent which favors an S N 2 reaction. So with a strong nucleophile and a polar A product solvent, we need to think about in S N 2 mechanism. So we know our nucleophile attacks at the same time that we get loss of our leaving group, so our nucleophile is going to attack this carbon. So again, I'll make this carbon red. At the same time that we get loss of leaving groups, so these electrons are gonna come off onto the bromine to form the bromide anion. For this reaction, we need to think about the stereochemistry of our S N 2 reaction. Our nucleophile has to attack from the opposite side"
    },
    {
        "Q": "At 4:32 It says that only one cell will be a successful sex cell while the other three are polar bodies- what do these polar bodies do?",
        "A": "In humans, they really don t do anything. However, in plants, they also become sex cells.,",
        "video_name": "IQJ4DBkCnco",
        "timestamps": [
            272
        ],
        "3min_transcript": "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. going and going, going. This cell is just like this cell, while these sex cells are differeent than this one right over here. Now, where does this happen in the body? We've talked about this in previous videos. These are your somatic cells right over here. These are the ones that make up the bulk of your body, somatic cells. And where is this happening? Well, this is happening in germ cells, As we mentioned, if you're male it's in your tesis and if you're female it's in your ovaries. And germ cells actually can undergo mitosis to produce other germ cells that have a diploid number of chromosomes, or they can undergo meiosis in order to produce sperm or egg cells in order to produce gametes."
    },
    {
        "Q": "at 0:19, what is a diploid number?",
        "A": "Diploid number means the number of chromosomes in the body cells of a diploid organism. In humans, 46 is the diploid number (body cells) and 23 is the haploid number (egg and sperm). Note: A diploid organism has paired sets of chromosomes in a cell. In humans, there are 23 homologous pairs of chromosomes where one set is inherited from each parent. Hope it helps.",
        "video_name": "IQJ4DBkCnco",
        "timestamps": [
            19
        ],
        "3min_transcript": "- [Voiceover] Before we go in-depth on meiosis, I want to do a very high level overview comparing mitosis to meiosis. So, in mitosis, this is all a review, if you've watched the mitosis video, in mitosis, we start with a cell, that has a diploid number of chromosomes. I'll just write 2n to show it has a diploid number. For human beings, this would be 46 chromosomes. 46 for humans, you get 23 chromosomes from your mother, 23 chromosomes from your father or you can say you have 23 homologous pairs, which leads to 46 chromosomes. Now after the process of mitosis happens and you have your cytokinesis and all the rest, you end up with two cells that each have the same genetic information as the original. So you now have two cells that each have the diploid number of chromosomes. So, 2n and 2n. 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,"
    },
    {
        "Q": "4:37 If Meiosis is occurring in the respective reproductive organs than why does he speak about mixing our parents dna, I know that DNA get's passed on, but how does the whole \"mixing process\" occur in our own bodies? And to what extent?",
        "A": "During sexual reproduction, the male and female gametes (The sperm and the egg) fuse, leading to the formation of a zygote. The zygote s DNA is a mixture of both the parents DNA, which comes from the respective gametes.",
        "video_name": "IQJ4DBkCnco",
        "timestamps": [
            277
        ],
        "3min_transcript": "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. going and going, going. This cell is just like this cell, while these sex cells are differeent than this one right over here. Now, where does this happen in the body? We've talked about this in previous videos. These are your somatic cells right over here. These are the ones that make up the bulk of your body, somatic cells. And where is this happening? Well, this is happening in germ cells, As we mentioned, if you're male it's in your tesis and if you're female it's in your ovaries. And germ cells actually can undergo mitosis to produce other germ cells that have a diploid number of chromosomes, or they can undergo meiosis in order to produce sperm or egg cells in order to produce gametes."
    },
    {
        "Q": "Sorry if this seems like an awfully basic question, but why does O get a negative charge at 4:01?",
        "A": "Well oxygen is a very electronegative atom hence it can remain stable holding an additional electron/negative charge.",
        "video_name": "nv2kfBFkv4s",
        "timestamps": [
            241
        ],
        "3min_transcript": "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. what we call in organic chemistry a nucleophilic attack on this carbon right over here. And when it does that, if we were doing a more formal reaction mechanism, we could say, \"Hey, well, maybe one \"of the double bonds goes back, \"the electrons in it go back to this oxygen, \"and then that oxygen would have a negative charge.\" But then that lone pair from that double bond could then reform, and as that happens, this oxygen that's in the hydroxyl group will take back both of these electrons. Would take back both of those electrons, and now it's going to have an extra lone pair. Let me do that by erasing this bond and then giving it an extra lone pair. It already had two lone pairs, and then when it took that bond, it's gonna have a third lone pair. And then it's going to have a negative charge. to grab a hydrogen proton someplace. And now it could just grab any hydrogen proton, but probably the most convenient one would be this one, because if this nitrogen is going to use this lone pair to form a bond with carbon, it's going to have a positive charge, and it might wanna take these electrons back. So you could imagine where one of these lone pairs is used to grab this hydrogen proton, and then the nitrogen can take these electrons, can take these electrons back. So hopefully you didn't find this too convoluted, but I always like to think, what could actually happen here? And so you see, this lone pair of electrons from the nitrogen forms this orange bond with the carbon. Let me do that in orange color if I'm going to call it an orange bond. It forms this orange bond. What we call this orange bond, we could call this a peptide bond, or a peptide linkage. Peptide bond, sometimes called a peptide..."
    },
    {
        "Q": "at 0:44, Sal describes the first law of thermodynamics. however, isn't it the law of conservation of energy?",
        "A": "Indeed the first law of thermodynamics is a version of the law of conservation of energy.",
        "video_name": "Xb05CaG7TsQ",
        "timestamps": [
            44
        ],
        "3min_transcript": "I've now done a bunch of videos on thermodynamics, both in the chemistry and the physics playlist, and I realized that I have yet to give you, or at least if my memory serves me correctly, I have yet to give you the first law of thermodynamics. And I think now is as good a time as any. The first law of thermodynamics. And it's a good one. It tells us that energy-- I'll do it in this magenta color-- energy cannot be created or destroyed, it can only be transformed from one form or another. So energy cannot be created or destroyed, only transformed. So let's think about a couple of examples of this. And we've touched on this when we learned mechanics and bunch of this in the chemistry playlist as well. So let's say I have some rock that I just throw as fast as I can straight up. Maybe it's a ball of some kind. So I throw a ball straight up. That arrow represents its velocity vector, right? it's going to go up in the air. Let me do it here. I throw a ball and it's going to go up in the air. It's going to decelerate due to gravity. And at some point, up here, the ball is not going to have any velocity. So at this point it's going to slow down a little bit, at this point it's going to slow down a little bit more. And at this point it's going to be completely stationary and then it's going to start accelerating downwards. In fact, it was always accelerating downwards. It was decelerating upwards, and then it'll start accelerating downwards. So here its velocity will look like that. And here its velocity will look like that. Then right when it gets back to the ground, if we assume negligible air resistance, its velocity will be the same magnitude as the upward but in the downward direction. tons in the projectile motion videos in the physics playlist, over here we said, look, we have some kinetic And that makes sense. I think, to all of us, energy intuitively means that you're doing something. So kinetic energy. Energy of movement, of kinetics. It's moving, so it has energy. But then as we decelerate up here, we clearly have no kinetic energy, zero kinetic energy. So where did our energy go? I just told you the first law of thermodynamics, that energy cannot be created or destroyed. But I clearly had a lot of kinetic energy over here, and we've seen the formula for that multiple times, and here I have no kinetic energy. So I clearly destroyed kinetic energy, but the first law of thermodynamics tells me that I can't do that. So I must have transformed that kinetic energy. I must have transformed that kinetic energy into something else."
    },
    {
        "Q": "heat isn't a form of energy is it? @ 7:28 I thought heat was the process or transfer of energy from system to surroundings?",
        "A": "Heat is a form of thermal energy.",
        "video_name": "Xb05CaG7TsQ",
        "timestamps": [
            448
        ],
        "3min_transcript": "And the first law says, oh, Sal, it all turned into potential energy up here. And you saw it turned into potential energy because when the ball accelerated back down, it turned back into But then I say, no, Mr. First Law of Thermodynamics, look, at this point I have no potential energy, and I had all kinetic energy and I had a lot of kinetic energy. Now at this point, I have no potential energy once again, but I have less kinetic energy. My ball has fallen at a slower rate than I threw it to begin with. And the thermodynamics says, oh, well that's because you have air. And I'd say, well I do have air, but where did the energy go? And then the first law of thermodynamics says, oh, when your ball was falling-- let me see, that's the ball. Let me make the ball yellow. So when your ball was falling, it was rubbing up against air particles. It was rubbing up against molecules of air. And right where the molecules bumped into the wall, there's Friction is just essentially, your ball made these molecules that it was bumping into vibrate a little bit faster. And essentially, if you think about it, if you go back to the macrostate/ microstate problem or descriptions that we talked about, this ball is essentially transferring its kinetic energy to the molecules of air that it rubs up against as it falls back down. And actually it was doing it on the way up as well. And so that kinetic energy that you think you lost or you destroyed at the bottom, of here, because your ball's going a lot slower, was actually transferred to a lot of air particles. It was a lot of-- to a bunch of air particles. Now, it's next to impossible to measure exactly the kinetic energy that was done on each individual air particle, because we don't even know what their microstates were to begin with. But what we can say is, in general I transferred some heat to these particles. I raised the temperature of the air particles that the kinetic energy. Remember, temperature is just a measure of kinetic-- and temperature is a macrostate or kind of a gross way or a macro way, of looking at the kinetic energy of the individual molecules. It's very hard to measure each of theirs, but if you say on average their kinetic energy is x, you're essentially giving an indication of temperature. So that's where it went. It went to heat. And heat is another form of energy. So that the first law of thermodynamics says, I still hold. You had a lot of kinetic energy, turned into potential, that turned into less kinetic energy. And where did the remainder go? It turned into heat. Because it transferred that kinetic energy to these air particles in the surrounding medium. Fair enough. So now that we have that out of the way, how do we measure the amount of energy that something contains? And here we have something called the internal energy. The internal energy of a system."
    },
    {
        "Q": "At 11:06, Sal used the full decimal for the atomic weight of Chlorine, but rounded for the atomic weight of Phosphorus. Won't this throw off some of his data?",
        "A": "He will have to round his final answer to the same number of sig figs as his chlorine number.",
        "video_name": "jFv6k2OV7IU",
        "timestamps": [
            666
        ],
        "3min_transcript": ""
    },
    {
        "Q": "8:00 , how do we know what ratio goes on top or bottom",
        "A": "The compound you have goes on the bottom, the one you want goes on top. In that example he needs to convert moles P4 to moles Cl2 so in the ratio moles P4 goes in the denominator so it will cancel, and moles Cl2 goes in the numerator.",
        "video_name": "jFv6k2OV7IU",
        "timestamps": [
            480
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 4:05 is there a difference between \"mole\" and \"mol\" ?",
        "A": "Using mol for mole is like using m for meter",
        "video_name": "jFv6k2OV7IU",
        "timestamps": [
            245
        ],
        "3min_transcript": ""
    },
    {
        "Q": "i thought at 3:10 ~ 3:20 sal said atomic weight but ment atomic mass unit\nwhat is a atomic weight? or is this a typo?",
        "A": "Atomic weight is an outdated term that we no longer use, as we once did, to refer to atomic mass.",
        "video_name": "jFv6k2OV7IU",
        "timestamps": [
            190,
            200
        ],
        "3min_transcript": ""
    },
    {
        "Q": "why did sal used 1.45g p4 at time 4:47?",
        "A": "If you go back to 00:00 you ll see it s given in the question",
        "video_name": "jFv6k2OV7IU",
        "timestamps": [
            287
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 7:04, why 6 moles multiplied in the conversion equation instead of 12 moles?",
        "A": "See the balanced equation,in that 6 is the stoichiometric coefficient of cl2 . this meant 6 moles of cl2 . this was why 6 was multiplied.",
        "video_name": "jFv6k2OV7IU",
        "timestamps": [
            424
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 9:20, why did he use 35.453 instead of rounding up to 36 or down to 35 like he would have in the other videos?",
        "A": "Because Chlorine exists as three different isotopes with atomic weights of 35, 36 and 37, precision chemistry uses an average atomic weight based on the proportion of the isotopes in nature. Early chemistry just uses the Cl 35 for simplicity as it is the most abundant.",
        "video_name": "jFv6k2OV7IU",
        "timestamps": [
            560
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 4:24, does the unit like grams or moles effect the problem?",
        "A": "Is it different to have 2 kg of socks instead of 2 socks? Yes, right? A mole is a number. Gram is a measure of mass. Not interchangeable.",
        "video_name": "jFv6k2OV7IU",
        "timestamps": [
            264
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 3:59 you wrote mol. do you mean mole?",
        "A": "mol is the standard abbreviation for mole.",
        "video_name": "jFv6k2OV7IU",
        "timestamps": [
            239
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At around 9:20, why did Sal not round the atomic weight of chlorine to 35? You can't have a chlorine atom that has 18.453 neutrons right?",
        "A": "No but you can have trillions and trillions of atoms that average out to that weight.",
        "video_name": "jFv6k2OV7IU",
        "timestamps": [
            560
        ],
        "3min_transcript": ""
    },
    {
        "Q": "About 2:40, when he's describing the configuration for nickel, if the 3d8 is at the end then it's organized by energy state, right? And if the 4s2 is last with the 3d8 behind it then it's organized by distance from the nucleus?",
        "A": "yes",
        "video_name": "YURReI6OJsg",
        "timestamps": [
            160
        ],
        "3min_transcript": "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 from the nucleus than these right here. Now, another way to figure out the electron configuration for nickel-- and this is covered in some chemistry classes, although I like the way we just did it because you look at the periodic table and you gain a familiarity with it, which is important, because then you'll start having an intuition for how different elements react with each other -- is to just say, oK, nickel has 28 electrons, if it's neutral. It has 28 electrons, because that's the same number of protons, which is the atomic number. Remember, 28 just tells you how many protons there are. This is the number of protons. We're assuming it's neutral. So it has the same number of electrons. That's not always going to be the case. But when you do these electron configurations, that tends to be the case. So if we say nickel has 28, has an atomic number of 28, so it's electron configuration we can do it this way, too. We can write the energy shells. So one, two, three, four."
    },
    {
        "Q": "is D orbitals always like the chart that khan made in 5:53",
        "A": "Yes the d orbitals have more energy, but backfill previous shells.",
        "video_name": "YURReI6OJsg",
        "timestamps": [
            353
        ],
        "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": "at 7:23, where are noble gases on the periodic table and what are they?",
        "A": "They are all the way over on the right side, and they are gases that generally will not engage in chemical reactions. They are inert.",
        "video_name": "YURReI6OJsg",
        "timestamps": [
            443
        ],
        "3min_transcript": "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 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."
    },
    {
        "Q": "2:30 , why is isn't 4D 8 , why do we subtract 1 from the D block",
        "A": "there is a trend with the respective D and S orbitals once you are referring to Copper and Chromium. The trend is that when 4s orbitals have a lower energy than the 3d orbitals so they give off an electron and have only one electron on its orbital instead of the required electron pair. When referring to 4D and 5s orbitals its the same trend.",
        "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": "At 02:43, shouldn't the 3d8 come before the 4s2?",
        "A": "4s comes before 3d because, before the orbitals are filled, the 4s is lower in energy than the 3d so is occupied first.",
        "video_name": "YURReI6OJsg",
        "timestamps": [
            163
        ],
        "3min_transcript": "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 from the nucleus than these right here. Now, another way to figure out the electron configuration for nickel-- and this is covered in some chemistry classes, although I like the way we just did it because you look at the periodic table and you gain a familiarity with it, which is important, because then you'll start having an intuition for how different elements react with each other -- is to just say, oK, nickel has 28 electrons, if it's neutral. It has 28 electrons, because that's the same number of protons, which is the atomic number. Remember, 28 just tells you how many protons there are. This is the number of protons. We're assuming it's neutral. So it has the same number of electrons. That's not always going to be the case. But when you do these electron configurations, that tends to be the case. So if we say nickel has 28, has an atomic number of 28, so it's electron configuration we can do it this way, too. We can write the energy shells. So one, two, three, four."
    },
    {
        "Q": "In the first video in this series on 'Orbitals' Sal instructed us to visualize that those electrons furthest away from the nucleus posses the greatest potential energy; they are in the highest energy state. On this logic, how is it that at around 3:20 in this video Sal lets us know that within Nickel [Ni] the 3D8 electrons are higher energy than the 4S2; how can this be when the 4S2 electrons are further away from the nucleus than 3D8?",
        "A": "The 3d electrons in Ni have to fit in between the negative charges of all of the 3s and 3p electrons. The 4s shell doesn t have as much repulsion, so it takes less energy for an e- to live in a 4s orbital than in a 3d orbital.",
        "video_name": "YURReI6OJsg",
        "timestamps": [
            200
        ],
        "3min_transcript": "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 from the nucleus than these right here. Now, another way to figure out the electron configuration for nickel-- and this is covered in some chemistry classes, although I like the way we just did it because you look at the periodic table and you gain a familiarity with it, which is important, because then you'll start having an intuition for how different elements react with each other -- is to just say, oK, nickel has 28 electrons, if it's neutral. It has 28 electrons, because that's the same number of protons, which is the atomic number. Remember, 28 just tells you how many protons there are. This is the number of protons. We're assuming it's neutral. So it has the same number of electrons. That's not always going to be the case. But when you do these electron configurations, that tends to be the case. So if we say nickel has 28, has an atomic number of 28, so it's electron configuration we can do it this way, too. We can write the energy shells. So one, two, three, four."
    },
    {
        "Q": "At 00:21 Sir said that helium belongs to 's' block but all noble gases belong to 'p' block. This can disturb the configuration of electrons in Ni atom. Why is 'He' considered to be in 's' block?",
        "A": "He has no p orbitals at all. It just has a completely full 1s\u00c2\u00b2 orbital and that is all.",
        "video_name": "YURReI6OJsg",
        "timestamps": [
            21
        ],
        "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": "At 1:35, are there more than one type of current if there is electron current? I thought that the only thing moving was the electrons moving from atom to atom, or are the atoms actually moving, because protons can't move on their own.",
        "A": "Current is a general concept of moving charge. There are two types of charge, but we don t divide that into two types of current. One + charge moving to the left is the same current as one - charge moving to the right.",
        "video_name": "17EhKw2tsu4",
        "timestamps": [
            95
        ],
        "3min_transcript": "- [Voiceover] When we start to study electricity, we need to get an idea of what is Current and what is Voltage and in two earlier videos, I talked about the idea of current and voltage, current and voltage and, and what they meant. And when we talked about current, it's easiest to describe current when we talk about wires. Let's say we have a copper wire. We talked about a copper wire and inside it was, there was electrons in it, and they have a negative charge, we know they have a negative charge, and if we put a voltage on them, those electrons would move in some direction like that. So if I put a plus voltage over here and a minus voltage over here, the electrons are repelled by the minus voltage and they're attracted to the positive voltage. That is called an Electron Current. in terms of what's actually happening inside a wire makes some sense, it's easier to understand current and that these electrons are moving around. And whenever we talk about this, we'll talk about it specifically that there's an electron current going on here. Now at the same time, what I said in that video, and I'll say again, is the convention for describing current is this. This is called the Conventional Current Direction. The convention we've had for hundreds of years is that current is the direction that a positive charge would move if there was a positive charge there. So, whenever we talk about current from now on, it'll always be conventional current, and in fact, we don't even need to mention conventional any more, it's just current. Current is the direction that positive charges would move. the word, electron current. Now, as a reminder, when we talked about voltage, uh, this was built up by analogy and the analogy was to electrons rolling down a mountaintop, so here's our mountain, remember this? And I built a battery or another voltage source like this, and we said, that what a battery does is it pumps out energetic electrons, and they go down a hill. Roll downhill and go back into the, back into the positive terminal of the battery. And when we design circuits, what we do, is we, we put stuff in the way of this electron on its path, and this is where we build our circuits. So the, the electron current is going in this direction here down the hill. The conventional current direction or the current direction is this way."
    },
    {
        "Q": "9:50 When you said 1c=6.24x10^18, can you show me step by step to convert it to electrons? how did you get 1.6x10^-19?",
        "A": "1 C = 6.24E18 electrons. Divide both sides bye 6.24E18 and you will have coulombs per electron.",
        "video_name": "IDQYakHRAG8",
        "timestamps": [
            590
        ],
        "3min_transcript": "much, much, much, much smaller mass than a proton, most of the mass of an atom is from the protons and the neutrons. So an electron has a much, much smaller mass than the protons and the neutrons, but it has the same but opposite charge as a proton. So sometimes the convention is to write negative e, or maybe even negative one e, sometimes depending on whether you view this as a kind of the actual charge or whether you view this as a unit, but here I'll view this as the actual charge. You could view negative e as the charge, as the charge of an electron. And something that has no charge, like a neutron, we say they're neutral, and actually that is why they are called neutrons, because they are neutral, they don't have charge. So that right over there, that over there is, is a neutron. Now when we start to get on kind of a larger scale, not on a sub-atomic scale anymore, in general the unit of charge that we typically use is the coulomb, is the coulomb. Coulomb, it's named for Charles Augustin de Coulomb, so if we're talking about the guy, and he was an 18th Century French physicist, we would use capital C, but if we're talking about the units, we would use lowercase c, the coulomb, the coulomb. And the coulomb is defined, so one coulomb, let me write it right over here, one coulomb and it uses the abbreviation uppercase C, is equal, or I'll say approximately equal to, we're going to round here, it's approximately equal to 6.24, 6.24 times 10 to the eighteenth e, or you could say, in magnitude wise, it's equal to the charge of 6.24 times 10 to the eighteenth protons, it would be the opposite if you're talking about electrons, it would be 6.24 times 10 to the eighteenth electrons. Now if you want to go the other way around, what is the charge of, the magnitude of the charge of say a proton in terms of coulombs, well you would just take the inverse of this. So you could say that e is approximately equal to the inverse of this which is 1.60, I guess you could say the reciprocal of this, 1.60 times 10 to the negative 19, times 10 to the negative 19 coulombs. So hopefully this gives you an appreciation for, I guess at a base level, what charge is. And in some ways it's like it's this everyday thing, you're used to it, we're used to dealing with electricity and we'll talk much more about that in depth. But at some levels it is this thing, one of the mysteries of the universe, how did these two particles know to attract each other, you know it looks like they're at a distance, how do they immediately exert a force on each other. how do these particles know immediately to repel each other,"
    },
    {
        "Q": "At 6:55, Sal says that your hair is attracted to the balloon because the negative charge of the balloon has repelled the electrons in your hair, leaving a disproportionate number of protons. But wouldn't it also be the case (and simpler to explain) that you hair is positively charged simply because it's missing the electrons that were \"stolen\" by the balloon?",
        "A": "But the balloon attracts hair that it did not rub. How does it do that?",
        "video_name": "IDQYakHRAG8",
        "timestamps": [
            415
        ],
        "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 6:16, why does the balloon grab electrons only? Why not protons?",
        "A": "Protons are very heavy and are tightly bound in the nucleus of the atoms. Electrons are light and are loosely bound to their atoms",
        "video_name": "IDQYakHRAG8",
        "timestamps": [
            376
        ],
        "3min_transcript": "But if we run with this model, we can imagine at the atomic scale, the nuclei of atoms are composed of protons and neutrons. So if you have some protons, and then you have some neutrons, I'll do two of each, you have some neutrons, and based on this framework, protons have a positive charge. Protons have a positive charge. Now once again, this convention of calling them positive and putting a plus on it, it's not like protons have a little plus sign tattooed onto them somehow. We could have called those, we could have said they have a red charge, or we could have even said, we wouldn't of had to even use the word charge, this is just a convention that we have decided to use. And so we say protons have positive charge and then, kind of buzzing around the nucleus of an atom, you often, or usually, or often have electrons. Electrons have a lot less mass. 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."
    },
    {
        "Q": "6:19\nThis might be an arbitrary detail, but why is the balloon grabbing electrons from your hair and not protons? Is there a reason for this?\nThanks.",
        "A": "Atoms have electrons on the outside, and nuclei on the inside. The electrons are very small compared to the nucleus.",
        "video_name": "IDQYakHRAG8",
        "timestamps": [
            379
        ],
        "3min_transcript": "But if we run with this model, we can imagine at the atomic scale, the nuclei of atoms are composed of protons and neutrons. So if you have some protons, and then you have some neutrons, I'll do two of each, you have some neutrons, and based on this framework, protons have a positive charge. Protons have a positive charge. Now once again, this convention of calling them positive and putting a plus on it, it's not like protons have a little plus sign tattooed onto them somehow. We could have called those, we could have said they have a red charge, or we could have even said, we wouldn't of had to even use the word charge, this is just a convention that we have decided to use. And so we say protons have positive charge and then, kind of buzzing around the nucleus of an atom, you often, or usually, or often have electrons. Electrons have a lot less mass. 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."
    },
    {
        "Q": "10:03 i thought coulombs were only for electrons/positrons",
        "A": "Coulomb is the unit of charge. Whatever has charge, the charge can be measured in coulombs.",
        "video_name": "IDQYakHRAG8",
        "timestamps": [
            603
        ],
        "3min_transcript": "much, much, much, much smaller mass than a proton, most of the mass of an atom is from the protons and the neutrons. So an electron has a much, much smaller mass than the protons and the neutrons, but it has the same but opposite charge as a proton. So sometimes the convention is to write negative e, or maybe even negative one e, sometimes depending on whether you view this as a kind of the actual charge or whether you view this as a unit, but here I'll view this as the actual charge. You could view negative e as the charge, as the charge of an electron. And something that has no charge, like a neutron, we say they're neutral, and actually that is why they are called neutrons, because they are neutral, they don't have charge. So that right over there, that over there is, is a neutron. Now when we start to get on kind of a larger scale, not on a sub-atomic scale anymore, in general the unit of charge that we typically use is the coulomb, is the coulomb. Coulomb, it's named for Charles Augustin de Coulomb, so if we're talking about the guy, and he was an 18th Century French physicist, we would use capital C, but if we're talking about the units, we would use lowercase c, the coulomb, the coulomb. And the coulomb is defined, so one coulomb, let me write it right over here, one coulomb and it uses the abbreviation uppercase C, is equal, or I'll say approximately equal to, we're going to round here, it's approximately equal to 6.24, 6.24 times 10 to the eighteenth e, or you could say, in magnitude wise, it's equal to the charge of 6.24 times 10 to the eighteenth protons, it would be the opposite if you're talking about electrons, it would be 6.24 times 10 to the eighteenth electrons. Now if you want to go the other way around, what is the charge of, the magnitude of the charge of say a proton in terms of coulombs, well you would just take the inverse of this. So you could say that e is approximately equal to the inverse of this which is 1.60, I guess you could say the reciprocal of this, 1.60 times 10 to the negative 19, times 10 to the negative 19 coulombs. So hopefully this gives you an appreciation for, I guess at a base level, what charge is. And in some ways it's like it's this everyday thing, you're used to it, we're used to dealing with electricity and we'll talk much more about that in depth. But at some levels it is this thing, one of the mysteries of the universe, how did these two particles know to attract each other, you know it looks like they're at a distance, how do they immediately exert a force on each other. how do these particles know immediately to repel each other,"
    },
    {
        "Q": "At 10:40 Why does he say -19 not -18?",
        "A": "As when we divide 1 by 6.24x10 to the power 18,then we get a value 0.1609x10 to the power -18,and shifting one decimal backwards,th value becomes 1.609x 10 to the power -19.",
        "video_name": "IDQYakHRAG8",
        "timestamps": [
            640
        ],
        "3min_transcript": "in general the unit of charge that we typically use is the coulomb, is the coulomb. Coulomb, it's named for Charles Augustin de Coulomb, so if we're talking about the guy, and he was an 18th Century French physicist, we would use capital C, but if we're talking about the units, we would use lowercase c, the coulomb, the coulomb. And the coulomb is defined, so one coulomb, let me write it right over here, one coulomb and it uses the abbreviation uppercase C, is equal, or I'll say approximately equal to, we're going to round here, it's approximately equal to 6.24, 6.24 times 10 to the eighteenth e, or you could say, in magnitude wise, it's equal to the charge of 6.24 times 10 to the eighteenth protons, it would be the opposite if you're talking about electrons, it would be 6.24 times 10 to the eighteenth electrons. Now if you want to go the other way around, what is the charge of, the magnitude of the charge of say a proton in terms of coulombs, well you would just take the inverse of this. So you could say that e is approximately equal to the inverse of this which is 1.60, I guess you could say the reciprocal of this, 1.60 times 10 to the negative 19, times 10 to the negative 19 coulombs. So hopefully this gives you an appreciation for, I guess at a base level, what charge is. And in some ways it's like it's this everyday thing, you're used to it, we're used to dealing with electricity and we'll talk much more about that in depth. But at some levels it is this thing, one of the mysteries of the universe, how did these two particles know to attract each other, you know it looks like they're at a distance, how do they immediately exert a force on each other. how do these particles know immediately to repel each other, that they're communicating somehow, or I guess once you get to quantum mechanical, an argument can be made that they are communicating somehow. But in our everyday, kind of logical sense, it's like well at a distance, how do these things actually know to repel or attract, and what is this charge anyway? You know we've put all these names around it but to kind of help us think about it and have a framework and predict what will happen. But do we really know what this charge thing is. So on one level it's kind of plain and mundane, and it deals with balloons and hair, but on another level it's this deep thing about this universe, it's a deep property of matter that we can manipulate and we can predict, but it is still this very fundamental and somewhat mysterious thing."
    },
    {
        "Q": "Starting with 7:45, Sal multiplied the number of the mole( ~6 * 10 ^ 23 molecules / mole) with the 1 * 10 ^ -7 moles / liter.\nI know what the term \"mole\" is.\nHowever, I just don't get the part why would you multiply the mole on the already made(shown) [H3O]?",
        "A": "I didn t watch the video, but he must have wanted to know how many molecules there are per liter (molecules / mole) * (moles / liter ) = molecules / liter",
        "video_name": "NUyYlRxMtcs",
        "timestamps": [
            465
        ],
        "3min_transcript": "giving that a positive charge. And so you might say, \"Well how frequently would \"I find hydronium ions in water?\" Well the concentration, let me actually draw a little tub of water here, let's say this is a liter of water. This is a liter, this is a liter of water. The concentration of hydronium in typical water, the concentration of H three O, the concentration of H three O in typical water, and you put brackets around something to denote \"concentration,\" is one times ten to the negative seven molar. And molar, this just means \"moles per liter.\" This is the same thing as one times ten to the negative seven moles, moles per liter. And now you might be saying, \"Well, what's a mole?\" Well I encourage you to watch the video on what a mole is, but a mole is a quantity. It's like saying, \"a dozen.\" But it's a much larger, a dozen is equal to 12 of something. A mole is roughly equal to, let me write it. A mole is approximately equal to 6.02 times ten to the 23rd. And you're typically talking about molecules. A mole of a substance means approximately 6.022, it actually keeps going, times ten to the 23rd molecules of that thing. So you might say, \"Hey, one times ten to the negative seven \"times 6.02 times ten to the 23rd, \"that would still get us,\" well let's see, this, One times ten to the negative seven moles per liter, times, times, I'll do it this way. Times six, I'll just go with six, since we're gonna go approximately. So approximately, six times ten to the 23rd. Six times 23rd molecules, molecules per mole. Molecules per mole, well these two would cancel out, and you would multiply these two numbers, you would get six times, let's see. that's still gonna be ten to the 16th power, molecules per liter. Molecules per liter, so your first reaction is, \"Oh my God!\" \"I'm gonna have six times,\" or roughly, I'll say roughly. \"Approximately six times ten to the 16th \"molecules of hydronium in this?\" \"That's a lot, we should see it all the time.\" But we have to remind ourselves. There's just a lot of molecules of water in there as well. In fact, a liter of water is roughly, so one liter of H two O, contains, contains approximately 56, 56 moles, moles of H two O. So one way to think about it is, \"I have one, I have one times,\" and if I'm thinking about a liter of water, I have, I'll do it over here, \"I have one times ten to the negative seven \"moles of, moles of H three O for every,"
    },
    {
        "Q": "4:05, what are ions?",
        "A": "Ions are atoms that have either lost or gained an electron, giving them an overall charge since they have an unequal number of protons (positive charge) and electrons (negative charge).",
        "video_name": "NUyYlRxMtcs",
        "timestamps": [
            245
        ],
        "3min_transcript": "and the partial positive charge over here. So these would be attracted to this partial positive charge. Remember there, you have a partial negative charge over here, this is actually what's forming the hydrogen bond. And it actually could bond to the hydrogen proton, while both of these electrons, including one of these electrons that used to be part of this hydrogen, or you could consider used to be part of that hydrogen, are nabbed, are nabbed by this oxygen. And in this circumstance, and I'm not saying that this happens all the time, but under just the right conditions, this actually can happen, and what would result, so let me, what result is, this thing over here, instead of just being a neutral water molecule, would look like this. So you have your oxygen, you have, not only your two hydrogens now, you now have a third hydrogen. You now have a third hydrogen. So you have theses two covalent bonds, these two covalent bonds, this lone pair. And now this lone pair, which I have circled in blue, This electron right over here of the hydrogen got nabbed by this oxygen. So now you've formed another covalent bond. And now this character over here, he's lost the hydrogen proton, but he's kept all of the electrons. So this character over here's gonna look like this. You're gonna have your oxygen, and now it's only going to only be bonded to one hydrogen, only bonded to one hydrogen. Has these two original lone pairs. These two original lone pairs right over here. And then took both of the electrons from this covalent bond. And took both of the electrons from this covalent bond. And so it has another lone pair. So this molecule gained just a proton without getting any electrons. So if you do that, you're now going to have a net positive charge for this one over here. And this molecule over here, actually let me, let me, ugh, let me just write it. I wanna write it a little bit neater. And this molecule over here, so we have this molecule plus this one, this one So it now has a negative charge. So just like that, you went from two neutral water molecules, to two ions. And these ions, this one over here, the one on the left, the one that is now H three O, H three O, H three O, and it now has a positive charge, positive charge, actually I put that O in a different color. H three, H three O. It's a positive charge, this is called the \"hydronium ion.\" Hydronium, hydronium. And this one over here, that is OH minus, so it's OH, O, let me get the colors right. OH minus. This is called the \"hydroxide ion,\" or since it's negative you can just call it an \"anion.\""
    },
    {
        "Q": "How can that one Hydrogen at 3:37 take 2 electrons if it only has one proton (+1) charge?",
        "A": "No hydrogens took any electrons at any point here Both the electrons that once were in an oxygen-hydrogen bond on the right water molecule have become a third lone pair of electrons on that right product (hydroxide)",
        "video_name": "NUyYlRxMtcs",
        "timestamps": [
            217
        ],
        "3min_transcript": "and the partial positive charge over here. So these would be attracted to this partial positive charge. Remember there, you have a partial negative charge over here, this is actually what's forming the hydrogen bond. And it actually could bond to the hydrogen proton, while both of these electrons, including one of these electrons that used to be part of this hydrogen, or you could consider used to be part of that hydrogen, are nabbed, are nabbed by this oxygen. And in this circumstance, and I'm not saying that this happens all the time, but under just the right conditions, this actually can happen, and what would result, so let me, what result is, this thing over here, instead of just being a neutral water molecule, would look like this. So you have your oxygen, you have, not only your two hydrogens now, you now have a third hydrogen. You now have a third hydrogen. So you have theses two covalent bonds, these two covalent bonds, this lone pair. And now this lone pair, which I have circled in blue, This electron right over here of the hydrogen got nabbed by this oxygen. So now you've formed another covalent bond. And now this character over here, he's lost the hydrogen proton, but he's kept all of the electrons. So this character over here's gonna look like this. You're gonna have your oxygen, and now it's only going to only be bonded to one hydrogen, only bonded to one hydrogen. Has these two original lone pairs. These two original lone pairs right over here. And then took both of the electrons from this covalent bond. And took both of the electrons from this covalent bond. And so it has another lone pair. So this molecule gained just a proton without getting any electrons. So if you do that, you're now going to have a net positive charge for this one over here. And this molecule over here, actually let me, let me, ugh, let me just write it. I wanna write it a little bit neater. And this molecule over here, so we have this molecule plus this one, this one So it now has a negative charge. So just like that, you went from two neutral water molecules, to two ions. And these ions, this one over here, the one on the left, the one that is now H three O, H three O, H three O, and it now has a positive charge, positive charge, actually I put that O in a different color. H three, H three O. It's a positive charge, this is called the \"hydronium ion.\" Hydronium, hydronium. And this one over here, that is OH minus, so it's OH, O, let me get the colors right. OH minus. This is called the \"hydroxide ion,\" or since it's negative you can just call it an \"anion.\""
    },
    {
        "Q": "at 11:13, how are you getting 116 J? what is the conversion?",
        "A": "1/2 * 0.145 kg * (40 m/s)\u00c2\u00b2 = 116 kg*m\u00c2\u00b2/s\u00c2\u00b2 = 116 J",
        "video_name": "o7_zmuBweHI",
        "timestamps": [
            673
        ],
        "3min_transcript": "If an object is translating and rotating and you want to find the total kinetic energy of the entire thing, you can just add these two terms up. If I just take the translational one half M V squared, and this would then be the velocity of the center of mass. So you have to be careful. Let me make some room here, so let me get rid of all this stuff here. If you take one half M, times the speed of the center of mass squared, you'll get the total translational kinetic energy of the baseball. And if we add to that the one half I omega squared, so the omega about the center of mass you'll get the total kinetic energy, both translational and rotational, so this is great, we can determine the total kinetic energy altogether, rotational motion, translational motion, from just taking these two terms added up. So what would an example of this be, let's just get rid of all this. Let's say this baseball, someone pitched this thing, and the radar gun shows that this baseball was hurled So it's heading toward home plate at 40 meters per second. The center of mass of this baseball is going 40 meters per second toward home plate. Let's say it's also, someone really threw the fastball. This thing's rotating with an angular velocity of 50 radians per second. We know the mass of a baseball, I've looked it up. The mass of a baseball is about 0.145 kilograms and the radius of the baseball, so a radius of a baseball is around seven centimeters, so in terms of meters that would be 0.07 meters, so we can figure out what's the total kinetic energy, well there's gonna be a rotational kinetic energy and there's gonna be a translational kinetic energy. The translational kinetic energy, gonna be one half the mass of the baseball times the center of mass speed of the baseball squared which is gonna give us one half. The mass of the baseball was 0.145 and the center of mass speed of the baseball is 40, that's how fast the center of mass of this baseball is traveling. translational kinetic energy. How much rotational kinetic energy is there, so we're gonna have rotational kinetic energy due to the fact that the baseball is also rotating. How much, well we're gonna use one half I omega squared. I'm gonna have one half, what's the I, well the baseball is a sphere, if you look up the moment of inertia of a sphere cause I don't wanna have to do summation of all the M R squareds, if you do that using calculus, you get this formula. That means in an algebra based physics class you just have to look this up, it's either in your book in a chart or a table or you could always look it up online. For a sphere the moment of inertia is two fifths M R squared in other words two fifths the mass of a baseball times the raise of the baseball squared. That's just I, that's the moment of inertia of a sphere. So we're assuming this baseball is a perfect sphere. It's got uniform density, that's not completely true. But it's a pretty good approximation. Then we multiply by this omega squared,"
    },
    {
        "Q": "5:57 When we are numbering the \"shortest path\" as stated in the video, i.e. the carbons between the bridge carbons, do we automatically number away from carbon number 1? So if there were 2 carbons between bridge carbons 1 and 4, how would we number them?",
        "A": "If it s symmetrical then it doesn t really matter If it was subsituted then you would number the carbon with the functional group first",
        "video_name": "cM-SFbffb7k",
        "timestamps": [
            357
        ],
        "3min_transcript": "bicyclo[4.2.0.]octane. Let's do another one following our general formula here for naming a bicyclo compound. So let's take a cyclohexane ring. And let's make this a bridgehead carbon. And we're actually going to put a carbon between our bridgehead carbons like that. Now, this is hard to see when it's drawn this way. So usually, you'll see it drawn a little bit differently. Usually you'll see it drawn with a little bit more three-dimensionality to it. So hopefully we can do that here. So it looks something like this. So let's find our bridgehead carbons here. So we know that this is a bridgehead carbon, this is a bridgehead carbon. Those correspond to these guys over here. And how many cuts would it take to turn this into an open chain alkane? Well, let's go ahead and look at the drawing on the left. we can see it's now an open chain compound. So it took two cuts, so it's bicyclo. Let's go ahead and put those bonds back in there like that. So it's a bicyclo compound. So we can go ahead and start naming it as a bicyclo compound. We need to go ahead and number it, so let's start with our bridgehead carbon. So we'll start with this is our bridgehead carbon. Find the longest path. Well, it doesn't matter if I go left or right here, since it's the same length each way around. So I'll just go to the right. So 1, 2, 3, 4. Second longest path, so I'm going to keep on going this way, so 5 and 6. And then finally, the shortest path, which is the one carbon in between my two bridgehead carbons, which would then get a 7, so seven total carbons. All right, so when I'm naming this, I need to put my brackets in here. And I put the number of carbons in the longest path, So how many carbons were my longest path, excluding Here's 1 and here is 2, so I'm going to go ahead and put a 2 in here. And then I go to my second longest path. Well, that was over on this side. And there are also two carbons on my second longest path. Obvious, it's the same length as the other one. So it's 2, 2. And then my shortest path, how many carbons are there in my shortest path, not counting my bridgehead carbon? There's one. So I go ahead and put a 1 here. And so I have bicyclo[2.2.1]. And then my total number of carbons is 7. So this is a bicycloheptane molecules. So bicyclobicyclo[2.2.1]heptane would be the official IUPAC name for this molecule. This molecule turns up a lot in nature. So much so, that it actually has its own common name. This is also called norborane. So let's go ahead and write that here. So this is norborane, again, a structure"
    },
    {
        "Q": "9:43. Looking at the bicyclo on the left...why couldn't you start from the bridge carbon and then go left and make the 6th carbon into the second? I'm guessing Counting on the longest carbon chain takes precedence over having the methyl on a lower carbon?",
        "A": "Correct. Counting on the longest carbon chain takes precedence over having the methyl on a lower numbered carbon. But if the 6-methyl had been on C-7, the name would still be 6,8-dimethylbicyclo[3.2.1]octane. We would still start numbering by going around the largest bridge, but we would use the numbering in the crossed-out structure in order to get the lowest numbers.",
        "video_name": "cM-SFbffb7k",
        "timestamps": [
            583
        ],
        "3min_transcript": "I could start numbering with either of these be number one. So let's just make this number 1 here. So if that's number 1, I next go to my longest path. So that would be going to the right here. So this would give me a 2 for this carbon, a 3 for this carbon, a 4 for this carbon, a 5 for this carbon. Next, the second longest path. Well, I can just keep going, make that a 6, make this a 7. And then finally, my shortest path, which is up here at the top. So that would be 8 total carbons for my parent name. On the dot structure on the left, again, the exact same dot structure. This time, I'm going to start with the opposite bridgehead carbons. So I'm going to start with this bridgehead carbon. And then follow the same rule. So the longest path, which would be to the right here. So 2, 3, 4, 5, 6, 7, and then 8. So which one of these numbering systems is the correct one? Remember, our goal is to get the lowest And if I look at the example on the right, I have a methyl group in the 7 position and a methyl group in the 8 position. The example on the left has a methyl group in the 6 position and a methyl group in the 8 position. Since I want to give my lowest number possible to my substituent, I'm going to not number it the way on the right. So I'm going to go with the way on the left here. So let's go ahead and start naming it. We already know this is a bicyclo compound. And we have two methyl groups, one at 6 and one at 8. So we could start naming it by saying 6,8-dimethyl. And then we know this is a bicyclo compound, so bicyclo. And then in the brackets, we're going to do the longest path first, excluding the bridgehead carbon. So the longest path is on the right. How many carbons are there in the longest path? There are three excluding the bridgehead carbon, Next, the second longest path, excluding the bridgehead carbon. Well, that was these two right here. So next, there'll be a 2. And then finally, the shortest path excluding our bridgehead carbon. There's only one carbon in our shortest path. So we put that in there. And then, there were 8 total carbons, so it is octane. So the final IUPAC name for this molecule is 6,8-dimethylbicyclo[3.2.1.]octane."
    },
    {
        "Q": "At 1:48, how is the OH coming out at us in space",
        "A": "That picture of the model with black carbons and a red oxygen is showing how the model looks in 3D. If you put the carbon backbone in the same plane, then the groups coming off the carbon are either coming towards you or going away. It may help if you get your hands on a molecular modelling kit.",
        "video_name": "ZAgQH2azx3w",
        "timestamps": [
            108
        ],
        "3min_transcript": "- [Voiceover] In the video on bond line structures, we started with this Lewis dot structure on the left, and I showed you how to turn this Lewis dot structure into a bond line structure. So here's the bond line structure that we drew in that video. Bond line structures contain the same information as a Lewis dot structure, but it's obviously much easier, much faster, to draw the bond line structure on the right than the full Lewis dot structure on the left. What about three dimensional bond line structure? So how could you represent this molecule in three dimensions, using a flat sheet of paper? Well, on the left here, is a picture where I made a model of this molecule, and this is going to help us draw this molecule in three dimensions. So we have a flat sheet of paper, how could we represent this picture on our flat sheet of paper? Let's start with the carbon in the center, so that's our carbon in magenta, so that's this one on our Lewis dot structure, this one on our bond line structure. Well, the carbon in magenta is SP3 hybridized, 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."
    },
    {
        "Q": "7:13 - 7:20 | Is there not also some Carbon-13 and Carbon-11?",
        "A": "Most carbon is carbon-12, followed by a small about of carbon-13. All other isotopes of carbon are radioactive and only exist in insignificant trace amounts. However, carbon-11 is important for medical purposes.",
        "video_name": "NG-rrorZcM8",
        "timestamps": [
            433,
            440
        ],
        "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": "At 6:50 Sal says \"one proton or one neutron is very close to 1 amu\". He had said before that the electron has really small mass. Does that mean that 1 amu is essentially the combined mass of 1 proton or neutron plus 1 electron?",
        "A": "No. An individual proton is 1.007276466812 u. An individual neutron is 1.00866491600 u. And an individual electron is 0.00054857990946 u. The amu is defined as 1/12th the mass of Carbon-12, which includes 6 protons, neutrons, electrons. It also has binding energy, which converts some of the mass into energy to hold the nucleus together.",
        "video_name": "NG-rrorZcM8",
        "timestamps": [
            410
        ],
        "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": "At 3:10, can't we write HOH as H2O?",
        "A": "You can, but HOH is more commonly used in university or higher level organic chemistry when you deal with reaction mechanism.",
        "video_name": "XEPdMvZqCHQ",
        "timestamps": [
            190
        ],
        "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 5:07 (CH)3COCH3, Is the Centered Carbon primary,secondary or tertiary?",
        "A": "Given that the definitions for primary through quaternary carbons are based on how many other carbon atoms are attached - what do you think the answer is? Look at the Lewis structure 6:09, there are three bonds to the carbons of CH3 groups and one to oxygen. That means the central carbon is a tertiary carbon. Note that the condensed structure is actually (CH3)3COCH3.",
        "video_name": "XEPdMvZqCHQ",
        "timestamps": [
            307
        ],
        "3min_transcript": "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. is this carbon. The carbon in red is also bonded to an oxygen all right, so we need to draw in an oxygen next. So now we have our oxygen. Notice the carbon in red now has an octet of electrons around it. The oxygen is bonded to another CH three group. So the oxygen is bonded to another CH three and let's draw that in so we have our CH three and since we're doing a partially condensed I won't draw in those bonds. We have CH three like that. I could put in my lone pairs of electrons on the oxygen to give the oxygen an octet of electrons. And now we have our partially condensed structure. If we want to expand it even more, and draw the full Lewis structure, again we start with the carbon in red. So here's the carbon in red, and that carbon is bonded to another carbon and this carbon is bonded to three hydrogens, So this CH three group that I just drew is this one. All right, next, we have a CH three group on the left side, so I need to draw in a CH three on the left, hopefully I have enough room to do that. I'll squeeze it in here, so we have our hydrogens, and that's our second CH three group. So let me circle it in green here. So here's a CH three. And then finally we have let me make this blue down here, we have another CH three group, so I'll draw that one in. So we have another CH three, we'll make room for all these hydrogens here. And that's the one in blue. So we're drawing out all of the bonds now in our full Lewis structure. Next we have an oxygen, so we have an oxygen right in here, with two lone pairs of electrons on the oxygen. The oxygen is bonded to another CH three. So let me let me pick a color here for that one. So we have another CH three on the right,"
    },
    {
        "Q": "At 1:25, instead of having two double bonds in the sulphuric acid shouldn't have two dative bonds?",
        "A": "As far as I know, either way, it should mean the same thing.",
        "video_name": "dbdVMThH1n8",
        "timestamps": [
            85
        ],
        "3min_transcript": "Let's think about what might happen if we had a solution of this carboxylic acid here. We might as well name it just to get some practice. We have one, two, three, four, five, six, seven carbons. So this is heptan-, and then we don't write heptane, because this is a carboxylic acid. It is heptanoic acid. So let's see what happens if we have heptanoic acid reacting with-- this is one, two carbons, and then it has an OH group, so this is ethanol. That's what the OH group does. It makes this an alcohol. And it's in the presence of a sulfuric acid catalyst. This right here is sulfuric acid, one of the stronger acids. I'll actually draw its structure, because I always find it frustrating when people just write just the formula here without the actual structure, because the structure actually shows you why it's so acidic. So it has a double bond to an oxygen, another double bond to an oxygen, and then it has a single bond to an OH group, and then it has another single bond to an OH group. And notice, it has one, two, three, four, five, six valence electrons. Now, the reason why this is such a strong acid, is that if either of these oxygens take the electron from this proton, so actually give away the proton to the solution, there's a ton of resonance structures here. And maybe I'll make a whole video on sulfuric acid, just to show you all of the resonance structures. But, in general, whenever you see a reaction when they say it's catalyzed by an acid, all you have to do is realize it's just going to make the surrounding solution a lot more acidic, just a ton more acidic. Maybe we're in a solution of ethanol, and if we are in a solution of ethanol, it'll just add protons to the ethanol itself. So you can imagine this guy right over here. Let me draw these oxygen-hydrogen bonds, so you have this oxygen and it is bonded to a hydrogen there. This is floating around the solution. You have your ethanol that looks like this, so two carbons and then bonded to an oxygen, and then that oxygen is bonded to a hydrogen. The oxygen has two lone pairs, just like that. And so this guy really is good at getting rid of the protons. So you have a situation where this electron can be taken back by this oxygen, and then it can actually be given here, and there's all these resonance structures. But it's just very good at taking that electron. And that can happen at the exact same time that one of the ethanols capture the hydrogen proton, at the exact same time that this oxygen captures that hydrogen proton. And if you just look at this part right here, this will just result in-- this part alone will just result in a"
    },
    {
        "Q": "I am actually looking at an updated organic textbook right now and I believe 13:53 should be be two different steps. Deprotanation occurs after the leaving group leaves hope that helps all you guys understand the mechanism better but thanks Kahn for doing most of it right!",
        "A": "No. Actually you are mistaken. You are a step premature....there must be a proton transfer between OH and OR before the leaving group leaves. He has the correct mechanism...I have no idea what you re specifically referring to.",
        "video_name": "dbdVMThH1n8",
        "timestamps": [
            833
        ],
        "3min_transcript": "And so if that happened, then the next step in our reaction-- and remember, these can all go in either direction. The next step in our reaction will look like this. You would have your two, three, four, five, six, seven carbons, single bond to an oxygen, and then you have your bond to this oxygen, which is bound to two carbons, one, two carbons, just like that. And this guy on top is bonded to an oxygen. He's got two lone pairs. And this guy over here grabbed a proton. He gave an electron to a hydrogen proton. He had two, but now he gave one of them to this proton. And so he gave it to that hydrogen. That hydrogen is now neutral. It gained an electron. This oxygen is now positive, because it It still has this other lone pair over here. It is now positive, and frankly, it is now a good leaving group. So, in the next step, you can have someone else. Remember, other people need protons earlier in this reaction. So this proton might get lost, maybe by one of the other ethanol molecules, so let me draw that. Let me do a color I haven't used yet. I'll use orange. So maybe one of the other ethanol molecules, or one of the other intermediaries in this whole reaction. So ethanol, I'll just do ethanol because it's easier to draw. It might give an electron to just the nucleus. And then this guy can take the electron back. This guy could take that hydrogen's electron back and give it to this carbon that was the carbonyl carbon several steps ago. And then since he's got that electron, he can then give back this electron to this, what was an OH group, but now back to him. And then the resulting products will look like this. And this is all in equilibrium. We now have two, three, four, five, six, seven carbons. Now, this guy has a double bond again. So you have an oxygen right there. It is now a double bond. I'll draw this newly formed double bond in magenta. This guy has left as water, so I can just draw that, so now you have this other OH group that is bonded to this hydrogen over here. He has now left as water. And now you have this other thing over here. You have what was that ethanol group has now attached itself. That ethanol has lost its hydrogen. It has attached itself to what was a carboxylic acid, so now it looks like this. It now is bonded to an oxygen and is bonded to one, two carbons, just like that. And this whole reaction that I showed you is called"
    },
    {
        "Q": "At 3:36, can you give me an example of a good mutation?",
        "A": "Bacterial flagella, antibiotic resistance, and the ability to metabolize many sugars for food. hope that helps :)",
        "video_name": "tzqZsPjHFVQ",
        "timestamps": [
            216
        ],
        "3min_transcript": "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 Genetic drift does not cause individuals to be more fit, just different. Finally, when it comes to allele game changers you got to respect the gene flow which is when individuals with different genes find their way into a population and spread their alleles all over the place. Immigration and emigration are good examples of this. As with genetic drift, its effects are most easily seen in small populations. Again, our factors: Natural selection; alleles for fitter organisms become more frequent. Sexual selection; alleles for more sexually attractive organisms become more frequent. Mutation; new alleles popping up due to mistakes in DNA. Genetic drift; changes an allele frequency due to random chance. Gene flow; changes in allele frequency due to mixing with new genetically different populations. Now that you know all that, in order to explain specifically how these processes influence populations we're going to have to completely forget about them. This is what's called the Hardy-Weinberg principle. Godfrey Hardy and Wilhelm Weinberg were two scientists in 1908"
    },
    {
        "Q": "around 3:02 John talks about sexual selection choosing a mate. I thought it was endorphins in the brain that caused this. Correct me if i am wrong",
        "A": "Half to half. Yes, sexual selection does help choose a mate, but endorphins help.",
        "video_name": "tzqZsPjHFVQ",
        "timestamps": [
            182
        ],
        "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": "On 6:30, what does he mean by \"to nab on to a hydrogen proton\"?",
        "A": "To nab on to a hydrogen proton basically means that the oxygen can bond to a hydrogen proton.",
        "video_name": "L677-Fl0joY",
        "timestamps": [
            390
        ],
        "3min_transcript": "and then these hydrogen will just be grabbed by another water molecule or something so the proton will be let go. That's why we call it an acid. If it wasn't in a solution it would have the hydrogens but it would be very acidic as soon as you put it into a neutral solution it's going to lose those hydrogens. The phosphate groups are what make it, are what make it an acid but it's confusing sometimes because usually when you see it depicted, you see it with these negative charges and that's because it has already lost its hydrogen proton. You're actually depicting the conjugate base here but that's where it gets its acidic name from because it starts protonated or it gets in this acid form, it's protonated but it readily loses it. And so that's why it has its, that's where it gets the name acid form from. Each of these nucleotides they have a phosphate group. Now the next thing you might notice, the next thing you might notice is. The next thing you might notice is this group right over here. It is a cycle, it is a ring and that's because it is a sugar. This sugar is based on, it's a five-carbon sugar. What I have depicted here, this sugar, this is ribose. This sugar right over here is ribose. This is when it's just as a straight chain and like many sugars, it can take a cyclical form. Actually it can take many different cyclical forms but the one that's most typically described is when you have that. Let me number the carbons because carbon numbering is important when we talk about DNA. But if we start carbonyl group right over here we call that the one carbon or the one prime carbon. One prime, two prime, three prime, four prime and five prime. That's the five prime carbon. You form the cyclical form of ribose as if you have the oxygen. You have the oxygen right over here on the four prime carbon. It uses one of its lone pairs. It uses one of its lone pairs to form a bond. With the one prime carbon and I drew it that way because it kind of does bend. The whole molecule's going to have to bend that way to form this structure. And then when it forms that bond the carbon can let go of one of these double bonds and then that can, then the oxygen, the oxygen can use that. The oxygen can use those electrons to go grab a hydrogen proton from some place. To nab on to a hydrogen proton. When it does that you're in this form and this form, just to be clear of what we're talking about, this is the one prime carbon. One prime, two prime, three prime, four prime and five prime carbon. Where we see this bond, this is the one prime carbon. it was part of a carbonyl. Now it lets go of one of those double bonds so that this oxygen can form a bond with a hydrogen proton. It let go of a double bond there so that this could form a bond with a hydrogen proton."
    },
    {
        "Q": "At about 7:40 what is hydronium?",
        "A": "Hydronium is H30. A water molecule is H20. When some molecular structure release a hydrogen ion (hydroxide), the water molecule, being electronegative (hoging electrons), take the hydroxide, thus forming Hydronium, with three Hydrogen molecules and one Oxygen molecule.",
        "video_name": "L677-Fl0joY",
        "timestamps": [
            460
        ],
        "3min_transcript": "With the one prime carbon and I drew it that way because it kind of does bend. The whole molecule's going to have to bend that way to form this structure. And then when it forms that bond the carbon can let go of one of these double bonds and then that can, then the oxygen, the oxygen can use that. The oxygen can use those electrons to go grab a hydrogen proton from some place. To nab on to a hydrogen proton. When it does that you're in this form and this form, just to be clear of what we're talking about, this is the one prime carbon. One prime, two prime, three prime, four prime and five prime carbon. Where we see this bond, this is the one prime carbon. it was part of a carbonyl. Now it lets go of one of those double bonds so that this oxygen can form a bond with a hydrogen proton. It let go of a double bond there so that this could form a bond with a hydrogen proton. 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."
    },
    {
        "Q": "At 7:37 Sal says the Hydrogen might have lost an electron. How does that happen?",
        "A": "Atoms lose electrons in chemical reactions or when they come in contact with other atoms whether of the same element or another. So when the hydrogen atom comes in contact with another atom, it may lose or also gain electrons (from its outermost shell). Thus an Hydrogen ion is formed.",
        "video_name": "TStjgUmL1RQ",
        "timestamps": [
            457
        ],
        "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": "Around 6:50, what is meant by 'imbalance of electrons or protons'? How does this imbalance occur?",
        "A": "Imbalance of protons and electrons means that it will either have more electrons (and less protons) or more protons (and less electrons). This imbalance causes it to have a positive or negative charge.",
        "video_name": "TStjgUmL1RQ",
        "timestamps": [
            410
        ],
        "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": "At 6:55 Sal talks about a positive \"+\" Hydrogen ion.\nWouldn't that just be a proton, as Hydrogen has only 1 electron and 1 proton as an atom?",
        "A": "Yes a hydrogen ion is a proton.",
        "video_name": "TStjgUmL1RQ",
        "timestamps": [
            415
        ],
        "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": "At 3:51 Sal says the atoms are bouncing around and have a lot of energy,but where does that energy come from",
        "A": "Energy comes from the universe. Just like matter, energy can not be created or destroyed, only transferred. Sometimes energy is transferred through heat, or it could be seen as movement. Energy comes from the atoms surroundings, and it will in turn release energy to the surroundings, depending on whether or not the reaction is exothermic or endothermic.",
        "video_name": "TStjgUmL1RQ",
        "timestamps": [
            231
        ],
        "3min_transcript": "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 So whenever you see these reactions in biology or chemistry class, keep that in mind. It looks all neat and organized but in a real system, these are all of these things just bouncing around in all different crazy ways. And that's why energy's an important thing here. Because the more energy you apply to the system, the more that they're going to bounce around, the more that they're going to interact with each other. The more reactants you put in, the more chance they're going to bounce around and be able to react with each other. Now I'm gonna introduce another word that you're gonna see in chemistry a lot. This water, these two... We see we have two water molecules here. We could call them molecules, but since they are actually made up of two or more different elements, we could also call this a compound. So water, water is, you could call it a molecule, or you could call it a compound. So this is a molecule or compound, while this molecular hydrogen, you would not call this a compound. And this molecular oxygen, of course it's a molecule,"
    },
    {
        "Q": "Around 3:20, Sal mentions that the formation of water produces a lot of energy, but how can a chemical reaction lead to the production of energy. Doesn't it violates the law of Conservation of Energy?",
        "A": "no. Potential energy gets converted to another form (thermal, or KE)",
        "video_name": "TStjgUmL1RQ",
        "timestamps": [
            200
        ],
        "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": "are the two products that were formed(b1 and b2) at 7:00 exactly the same or are they enantiomers?",
        "A": "They are identical, there are no chiral centers on the molecule, and the double bond is in the same spot",
        "video_name": "MDh_5n0OO2M",
        "timestamps": [
            420
        ],
        "3min_transcript": "And so, let me draw in a proton there. And we think about a weak base coming along and taking that proton, so I'll draw in my weak base here. So it takes this proton and these electrons move into here, so let's draw this product. So we would have our alkene that looks like this. So let's follow those electrons. I'll make them dark blue. Electrons in this bond move into here to form our double bond. And so, now we've gone through the complete mechanism, and we have two products. So let me circle our two products, so this is really just one product, and then this would be our second product. For this reaction, we actually get 90% of the alkene on the right and 10% of the alkene on the left. And so, let's look at the degree of substitution of our two products, 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": "at 7:33 could these three (two) molecules be considered isoprenes? Does the location of the double bond matter in an isoprene?",
        "A": "Isoprene is 2-methylbuta-1,3-diene. These molecules are not isoprenes, although they have an isoprene skeleton (C atoms joined as in 2-methylbutane). When we say that molecules contain isoprene units, we mean that they consist of 5-carbon isoprene units joined together. The location of any double bonds doesn t matter.",
        "video_name": "MDh_5n0OO2M",
        "timestamps": [
            453
        ],
        "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": "At 3:26 and 3:42, how does he know that both the products are either trisusbtitual or disubstitual?",
        "A": "You count the number of C atoms that are directly attached to the alkene carbons. In 1-methylcyclohexene, the alkene carbons are C1 and C2. The carbon atoms directly attached to C1 are C6 and the methyl carbon. The carbon atom directly attached to C2 is C3. This makes a total of three carbons, so the alkene is trisubstituted. In 3-methylcyclohexene, the carbon atom directly attached to C1 is C6. The carbon atom directly attached to C2 is C3. This makes a total of two carbons, so the alkene is disubstituted.",
        "video_name": "MDh_5n0OO2M",
        "timestamps": [
            206,
            222
        ],
        "3min_transcript": "and a plus one formal charge on the oxygen. So the loan pair, let's say this lone pair here in magenta, picks up a proton from sulfuric acid to form this bond, and that gives us water as a leaving group. And we know water is a good leaving group. The electrons in this bond can come off onto the oxygen to form H2O. And when that happens, we take a bond away from this carbon in red. So we're gonna form a carbocation. So we take away a bond from the carbon in red. Let me go ahead and draw in our carbocation here. So the carbon in red would be this carbon. That carbon would have a plus one formal charge. This is a tertiary carbocation, because the carbon in red is directly bonded to three other carbons. So this one, this one, and this one. So this is a stable carbocation. Next, let's think about the next step of an E1 mechanism. and take a proton from one of the beta carbons. And let's start with beta two. So let's thinking about a proton on the carbon, so let me draw on in here. And our weak base comes along and takes this proton, which would leave this electrons to move into here to form a double bond. So let's draw that product. So we would have our double bond forming right here. Let me draw in the rest of the molecule. So our electrons in, let me make these blue here. So the electrons in light blue are going to move in to form our double bond, and so we would get this out alkene. So that's beta two. Let's think about what would happen if we took a proton away from beta one. So let's draw that one in next. So I'm gonna draw the carbocation again. Let me draw that in here. And beta one would be up here, so let me put in a proton on beta one. and taking this proton, so our base is probably water. And these electrons would move into here this time. So if that happens, let's draw that product. Our double bond would form up here, and let me draw in the rest of this molecule. So let me use red for those electrons. So electrons in red are going to move into here to form this alkene. Notice that the two alkenes that we just drew is really the same molecule. This is the same compound. So we haven't formed two different products. If you take a proton away from the beta one or the beta two carbon, you're gonna make the same alkene. But what about the beta three carbon? So that's our last example. And let me go ahead. I forgot to in a plus one formal charge on our carbocation. Let me draw one more carbocation, the same one at tertiary carbocation. The difference is this time"
    },
    {
        "Q": "At 5:49 you named the molecule Cyclohexane Carbaldehyde but you did not explain why that particular compound is called a Carbaldehyde.",
        "A": "A carbaldehyde is an aldehyde that is attached to another entity which is often a ring system.",
        "video_name": "JMsqu236bZo",
        "timestamps": [
            349
        ],
        "3min_transcript": "So if you see benzaldehyde, you can use that in your iupac name. And so if we look at it over here on the right you can see that benzaldehyde can form the base for the name for this molecule. So let's go ahead and write that over here. So we have \"benzaldehyde.\". And then we're going to number to give this carbon on our ring, number one. Right so we have two choices we can go around the ring clockwise or counterclockwise. And we know we want to give the lowest number possible to our substituents and so that would be by going around clockwise. So let's do that, so two, three and then four. So we have two substituents, right. We have a methoxy substituent here at carbon three. We have a hydroxy substituent here at carbon four. And so we need to put those into alphabetical order so H comes before M and so we're going to start with the hydroxies, so let's, hopefully we'll have enough room here. So \"four-hydroxy\" and then we have a \"three-methoxy So vanillin of course is where we get our vanilla flavor from. So a fantastic vanilla smell so these are white crystals which have an amazing vanilla scent to them. So also a lot of fun to do labs with. Alright so a lot of aldehydes have nice smells to them. Let's look at two more aldehydes. Alright so let's look at the one on the left here right. So it's not benzaldehyde, we don't have benzene ring here anymore, we have a cyclohexene, so this one is going to be called \"cyclohexane carbaldehyde.\" So, \"cyclohexane carbaldehyde\" is the iupac name for this. So, \"carbaldehyde.\" Alright, let's look at this molecule over here on the right so this is two aldehydes Alright so let's look at this, so it'll be one, two, three, four and five total carbons. And so it's two aldehydes, so it's going to be \"dial\" and it's five carbons total so it's going to be \"pentane\", so we could write \"pentane.\" We could write \"pentanedial\" here. And if you wanted to put where the numbers are, so it'll be one, five, so you can put that in there. So \"one-five-pentanedial\" for two aldehydes. Alright, let's look at ketone nomenclature next and so just to remind you of the general structure of ketones. So once again you have a carbonyl, a carbon double bonded to an oxygen. This time you have two alkyler groups right, so you have two alkyler groups here and they can be the same or they can be different right. So I can write \"R prime\" here and either one would still be a ketone. So once again we looked at butane earlier, so our four-carbon alkane, \"butane.\""
    },
    {
        "Q": "From 6:56-7:08, is there an example that would make what was said about the relationship about mass and acceleration easier to understand? At 7:08, what do you mean by \"the harder it is to change it's constant velocity?\"",
        "A": "the harder it is to change it s constant velocity A velocity is changed by accelerating or decelarating. Sal is saying that if you apply the same force to an object with a larger mass, it will accelerate less than an object with lower mass. This happens when you strike a tennis ball and a bowling ball with the same amount of force. The tennis ball will accelerate more than the bowling ball, because the mass of a tennis ball is lower than a bowling ball.",
        "video_name": "ou9YMWlJgkE",
        "timestamps": [
            416,
            428,
            428
        ],
        "3min_transcript": "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 and what do we get? [cancels out] 20 divided by 2 is 10 kilograms cancel kilograms and so we have the acceleration, in this situation is equal to 10 metres per second squared--is equal to the acceleration. So when we doubled the force, we went from 10 Newtons to 20 Newtons, the acceleration doubled. We went from 5 metres per second squared to 10 metres per second squared. So we see that they are directly proportional and the mass is how proportional they are. And so you can imagine what happens if we double the mass. If we double in, let's say in this situation, with 20 Newtons then we won't be dividing by 2 kilograms anymore we'll be dividing by 4 kilograms. And so then we'll have 20 divided by 4 which will be 5 and it'll be metres per second squared. So if you make the mass larger, if you double it then your acceleration would be half as much. you need to accelerate it or for a given force, the less that it will accelerate it. The harder it is to change it's constant velocity."
    },
    {
        "Q": "at 1:10 Sal says that Force equals mass times acceleration. Isn't \"net force\" equals mass times acceleration?",
        "A": "Yes it is, but f=ma is just an easy-to-understand, basic version of the equation. The real equation is: The sum of the forces on an object equals the mass of the object times its acceleration \u00ce\u00a3F = m * a",
        "video_name": "ou9YMWlJgkE",
        "timestamps": [
            70
        ],
        "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": "In 5:07 why do we divide both sides by 2kg?",
        "A": "We divide both sides by 2kg so that the mass gets neutralized on one side so that we can find the acceleration.",
        "video_name": "ou9YMWlJgkE",
        "timestamps": [
            307
        ],
        "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 1:16, Sal said something about a vector quantity. Um, what is a vector quantity?",
        "A": "A quantity which has magnitude and direction both",
        "video_name": "ou9YMWlJgkE",
        "timestamps": [
            76
        ],
        "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": "At 0:39, Sal says methanol is a protic solvent and releases H+ ions into the solution, but at 7:05, he says that methanol is a weak base. If methanol releases H+, it should be an acid but here it also acts like a base...so which one is it?",
        "A": "Methanol is like water. It can act as either an acid or a base.",
        "video_name": "MtwvLru62Qw",
        "timestamps": [
            39,
            425
        ],
        "3min_transcript": "Let's think about what type of reaction might occur if we have this molecule right over here. I won't go through the trouble of naming it. It would take up too much time in this video. But it's dissolved in methanol. When we talk about what type of reactions, we're going to pick between Sn2, Sn1, E2 and E1 reactions. Now, maybe the first place to start or the place I like to start is to just look at the solvent itself. And when we're trying to decide what type of reaction will occur, the important thing to think about is, is the solvent protic or aprotic? And if you look at this solvent right here, this is methanol. It is protic. And in case you don't remember what protic means, it means that there are protons flying around in the solvent, that they can kind of go loose and jump around from one molecule to another. And the reason why I know that methanol is protic is because you have hydrogen bonded to a very electronegative atom in oxygen. oxygen can steal hydrogen's electron. And then the hydrogen itself, without the electron, the hydrogen proton will be flying around because it doesn't have a neutron. So this is a protic solvent. Now, you might say, well does anything with a hydrogen, would that be protic? And the answer is no. If you have a bunch of hydrogens bonded to just carbons, that is not protic. Carbon is not so electronegative that it could steal a hydrogen's electron and have the hydrogens So a big giveaway is hydrogen bonded to a very electronegative atom like oxygen. So this is protic. And when we think about protic out of all of the reactions we studied, that favors-- well, even a better way to think about it is it disfavors. So it tells us that it's unlikely to have an Sn2 or an E2 reaction. And the logic there is an Sn2 reaction needs a strong An E2 reaction needs a strong base. Now, if you have protons flying around, the nucleophile or the base is likely to react with the proton. It would not be likely to react with the substrate itself. So a protic solution, you're likely to have an Sn2 or E2. What you are likely to have is an Sn1 or an E1 reaction. Both of these need the leaving group to leave on its own, and actually, having protons around might help to stabilize the leaving group to some degree. So it makes Sn2, E2 unlikely, Sn1, E1 a little more likely. So far, these are our good candidates. Now, the next thing to think about is to just look at the leaving group itself, or see if there is even a leaving group. And over here, everything we see on this molecule is either a carbon or a hydrogen, except for this iodine right here."
    },
    {
        "Q": "At 7:53 he says 'a meagenta electron has been donated to the carbocation', what is a 'magenta electron'?",
        "A": "He is simply referring to the colour of the dot that he uses on the screen to make it clear which is the moving electron. All the other dots in the structure of methanol on the screen are blue. There is no such thing as a magenta or a blue electron. Electrons are too small to have any colour.",
        "video_name": "MtwvLru62Qw",
        "timestamps": [
            473
        ],
        "3min_transcript": "So this step right here is common to both Sn1 and E1 reaction. The leaving group has to leave. Now, after this, they start to diverge. In an Sn1, the leaving group essentially gets substituted with a weak nucleophile. In an E1, a weak base strips off one of the beta hydrogens and forms an alkene. So let's do them separately. So over here, I'm going to do the Sn1. And on the right-hand side, I will do the E1 reaction. So let me start over here. So the Sn1 is starting over here at this step. I'll just redo this step over here. So this has a positive charge. That has a positive charge here. The iodide has left. I don't have to draw all its valence electrons anymore. 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."
    },
    {
        "Q": "At 4:00, Sal talks about friction between the 'piston' and the 'wall' of the cylinder. Now my question is more of an observation. Sal doesn't seem to mention anything about the friction generated from the molecules within the cylinder, assuming they have KE during this process. Therefore, wouldn't the heat generated from this also influence the result in this video? Like I said, just an observation and wondered whether there might be a reason for sal not saying anything about it.",
        "A": "Ideal gases have no friction between molecules. Most of the time when we are doing introductory level thermo we are assuming an ideal gas, which is not a bad assumption because many gases behave close to ideal as long as the pressure is not too high and the temperature is not too low",
        "video_name": "PFcGiMLwjeY",
        "timestamps": [
            240
        ],
        "3min_transcript": "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 here, and then always-- instead of having one big weight on top that I took off or took on, because it would throw the system out of equilibrium-- I did it in really small increments. I just moved grains of sand, so that the system was always really close to equilibrium. And that's called quasistatic. and I've defined that before. And that means that you're always in kind of a quasi-equilibrium. So your state variables are always defined. But that, by itself, does not give you reversability. You have to be quasistatic and frictionless in order to be reversible. Now, what do we mean by frictionless? Well, I think you know what frictionless means. Is that like you see in this system right here, if I make this piston a little bit bigger, that when this piston rubs against the side of this wall, in kind of our real Those molecules start bumping against each other, and then they start making them vibrate, so they transfer some kinetic energy. From just by rubbing into each other, they start generating some kinetic energy, or some heat. So you normally have some heat generated from friction. Now, if you have some heat generated from friction, when I remove a pebble-- first all, when I remove that first pebble, it might not even do anything. Because it might not even overcome-- you can kind of view it as the force of friction. But let's say I remove some pebbles, and this thing moves up a little bit. But because some of the, I guess you could say, the force differential, the pressure differential between the pebbles and the gas inside, and the pressure of the gas, was used to generate heat as opposed to work, when I add the pebbles back, if I have friction, I'm not going to get back to the same point that I was before. Because friction is always resisting the movement."
    },
    {
        "Q": "At 0:03, Sal said \"nucleuses\". Is it nuclei?",
        "A": "Yes, the plural of nucleus is nuclei.",
        "video_name": "lJX8DxoPRfk",
        "timestamps": [
            3
        ],
        "3min_transcript": ""
    },
    {
        "Q": "I have a question at 10:44, when Sal starts to draw the pi bonds - those p-Orbitals overlap two times, so in total the C Atoms would have 3 bonds, two pi bonds and including the sigma bond, right? Shouldn't it be only one overlap of the p-Orbitals, as each C Atom has only one more electron to share in the p-Orbital? Or is it irrelevant, how many times this p-Orbital overlaps, because it only has one more electron inside?",
        "A": "While the drawing shows two overlaps, of the pi bond it is actually a theoretical drawing of the possibilities of where the electrons can be. The electron can not be in two places at one time, so though it appears to overlap two times, you are correct to think that it is irrelevant how many times it overlaps because there is only one electron inside. Short answer, though the drawing overlaps two times, it is only representing one bond.",
        "video_name": "lJX8DxoPRfk",
        "timestamps": [
            644
        ],
        "3min_transcript": ""
    },
    {
        "Q": "This might be a stupid question, but at 11:00 Sal divided the products over the reactants, and this is really the equiiibrium constant formula. But what difference would it make if we took the reactants and divided by the products? (Ok, we would get the inverse value, but what difference would it make?)",
        "A": "As long as the inverse of the equilibrium constant is used, there is no difference. It is normally just easier to use the products divided by the reactants because then the inverse is not needed.",
        "video_name": "ONBJo7dXJm8",
        "timestamps": [
            660
        ],
        "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": "At 2:30, it's being said that energy of photon =E3 - E1\nBut in reality we always do final energy shell - initial energy shell so over here also it should be E1 - E3....?na..??",
        "A": "Which way you write it depends on whether you are referring to the excitation of the electron from E1 to E3, or the dropping back down of the electron from E3 to E1. The magnitude of the two will be the same.",
        "video_name": "AznXSVx2xX0",
        "timestamps": [
            150
        ],
        "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": "At about 4:46, he says that the back panel was injection molded. What exactly does it mean if something is injection molded?",
        "A": "It means that a material with a high melting point (usually metal) is used to make a mold in which molten plastic or polypropylene or other materials can be injected. After the material cools off, the mold is opened and the object can be taken out. Injection molding is usually used because it s quite cheap",
        "video_name": "qLMsZKx_a8s",
        "timestamps": [
            286
        ],
        "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 4:06, why doesn't the ethylene glycol react with the carboxylic acid as well as the ketone?",
        "A": "It could, but I believe the reaction with the ketone is faster because there is less steric hindrance. Therefore, this will be your major product.",
        "video_name": "fY_ejjMRYg0",
        "timestamps": [
            246
        ],
        "3min_transcript": "we have the structure of our alcohol product here. So it would be a four-carbon alcohol, so one, two, three, and four, and the exact same thing down here. So our other product would be butanol. And we have two equivalents of it. So that'd be one, two, three, four carbons. So, by looking at our acetal and thinking about hydrolyzing it and thinking about where those portions came from, we can easily come up with the products of this hydrolysis of this acetal. So let's see how we could use this reaction in, if we were trying to synthesize this molecule over here on the right. And so, if we're trying to synthesize this molecule from this molecule, first you might be able to think you could figure it out by looking at the functional groups. Alright, so we have a ketone over here on the left and we want a ketone over here for our product. We also have a carboxylic acid over here to the left, and then we have an alcohol over here on the right. this kind of transformation, you might think to reduce the carboxylic acid to form your alcohol. And so, you could do that with something like lithium aluminum hydride. So you might think, and you could just add some lithium aluminum hydride here, in the first step. And the second step, add a source of protons to protonate your alkoxide anion to form your alcohol as your product. The only problem with trying to do this in one step is lithium aluminum hydride is also going to reduce your ketone here, to form a secondary alcohol. And so, this will not work. The first thing you have to do is protect the ketone, and then you can use your lithium aluminum hydride. So, we're going to use, we're going to protect our ketone using an acetal, of course. And we're going to react it with ethylene glycol. So if we react our starting compound here with ethylene glycol, and we use an acid Alright, so the ethylene glycol is going to react with the ketone portion of the molecule to form an acetal, specifically a cyclic acetal. Alright, so over here on the left we still have our carboxylic acid. And over here on the right, now we're going to have an oxygen bonded to this carbon, another oxygen bonded to this carbon, and they are, of course, connected. And so, that's our cyclic acetal, which came from this oxygen, this carbon, this carbon, and this oxygen. Right, so that's this oxygen, this carbon, this carbon, and this oxygen. So acetals are stable in basic conditions. And so, now we can add our lithium aluminum hydride, so we add our lithium aluminum hydride here in the first step. And that is going to reduce our carboxylic acid. And so in the second step we can add a source of protons, and we can add some excess water."
    },
    {
        "Q": "At 2:37 isn't the angle supposed to be 60degrees? Because if there's a right angled triangle, then the right angle would be 90degrees and theta would be 30 so 180-90-30=60, no?",
        "A": "Since the is an interior angle alternate to the angle theta, it is 30 degrees. Sal does a great job of explaining this in the previous video, Inclined Plane Force Components, starting at about 4:24.",
        "video_name": "Mz2nDXElcoM",
        "timestamps": [
            157
        ],
        "3min_transcript": "times the gravitational field times 9.8 meters per second squared. So it's going to be 98 newtons downward. So this is 98 newtons downward. I just took 10 kilograms. Let me write it out. So the force due to gravity is going to be equal to 10 kilograms times 9.8 meters per second squared downward. This 9.8 meters per second squared downward, that is the field vector for the gravitational field of the surface of the earth, I guess is one way to think about it. Sometimes you'll see the negative 9.8 meters per second squared. And then that negative is giving you the direction implicitly because the convention is normally that positive is upward and negative is downward. We'll just go with this right over here. So the magnitude of this vector is 10 times 9.8, which is 98 kilogram meters per second squared, which So the magnitude here is 98 newtons and it is pointing downwards. Now what we want to do is break this vector up into the components that are perpendicular and parallel to the surface of this ramp. So let's do that. So first, let's think about perpendicular to the surface of the ramp. So perpendicular to the surface of the ramp. So this right over here is a right angle. And we saw in the last video, that whatever angle this over here is, that is also going to be this angle over here. So this angle over here is also going to be a 30-degree angle. And we can use that information to figure out the magnitude of this orange vector right over here. And remember, this orange vector is the component of the force of gravity that is perpendicular to the plane. And then there's going to be some component that is parallel to the plane. I'll draw that in yellow. Some component of the force of gravity that is parallel to the plane. And clearly this is a right angle, And this is parallel to the plane. If it's perpendicular to the plane, it's also perpendicular to this vector right over here. So we can use some basic trigonometry, like we did in the last video, to figure out the magnitude of this orange and this yellow vector right over here. This orange vector's magnitude over the hypotenuse is going to be equal to the cosine of 30. Or you could say that the magnitude of this is 98 times the cosine of 30 degrees newtons. 98 times the cosine of 30 degrees newtons. And if you want the whole vector, it's in this direction. And the direction going into the surface of the plane. And, based on the simple trigonometry-- and we go into this in a little bit more detail in the last video-- we know that the component of this vector that is parallel to the surface of this plane is going to be 98 sine of 30 degrees."
    },
    {
        "Q": "At 1:15, Sal says that the force is 98 Newtons. Because it is downward, wouldn't it be -98?",
        "A": "Sal is probably referring to the magnitude of the force, but yes, if you define up to be positive direction, then down would be negative.",
        "video_name": "Mz2nDXElcoM",
        "timestamps": [
            75
        ],
        "3min_transcript": "Let's say that I have a ramp made of ice. Looks like maybe a wedge or some type of an inclined plane made of ice. And we'll make everything of ice in this video so that we have negligible friction. So this right here is my ramp. It's made of ice. And this angle right over here, let's just go with 30 degrees. And let's say on this ramp made of ice, I have another block of ice. So this is a block of ice. It is a block of ice, it's shiny like ice is shiny. And it has a mass of 10 kilograms. And what I want to do is think about what's going to happen to this block of ice. So first of all, what are the forces that we know Well if we're assuming we're on Earth, and we will, and we're near the surface, then there is the force of gravity. There's the force of gravity acting on this block of ice. And the force of gravity is going to be equal to-- it's going to be in the downward direction, times the gravitational field times 9.8 meters per second squared. So it's going to be 98 newtons downward. So this is 98 newtons downward. I just took 10 kilograms. Let me write it out. So the force due to gravity is going to be equal to 10 kilograms times 9.8 meters per second squared downward. This 9.8 meters per second squared downward, that is the field vector for the gravitational field of the surface of the earth, I guess is one way to think about it. Sometimes you'll see the negative 9.8 meters per second squared. And then that negative is giving you the direction implicitly because the convention is normally that positive is upward and negative is downward. We'll just go with this right over here. So the magnitude of this vector is 10 times 9.8, which is 98 kilogram meters per second squared, which So the magnitude here is 98 newtons and it is pointing downwards. Now what we want to do is break this vector up into the components that are perpendicular and parallel to the surface of this ramp. So let's do that. So first, let's think about perpendicular to the surface of the ramp. So perpendicular to the surface of the ramp. So this right over here is a right angle. And we saw in the last video, that whatever angle this over here is, that is also going to be this angle over here. So this angle over here is also going to be a 30-degree angle. And we can use that information to figure out the magnitude of this orange vector right over here. And remember, this orange vector is the component of the force of gravity that is perpendicular to the plane. And then there's going to be some component that is parallel to the plane. I'll draw that in yellow. Some component of the force of gravity that is parallel to the plane. And clearly this is a right angle,"
    },
    {
        "Q": "You found out force acting parallel to the plane is 49N while force acting perpendicular to the plane if 49*(1.732) [49root3]. Therefore force along vertical is more than horizontal. Therefore the body should not move. But the body will move with acceleration 4.9m/s^2 as mentioned in 10:05. Why ?",
        "A": "Why do you think the body should not move? There is a force of 49 N pushing it down the plane. Why would that force not cause acceleration? I think maybe you should watch the video again. You might need to also watch some of the earlier videos about Newton s laws",
        "video_name": "Mz2nDXElcoM",
        "timestamps": [
            605
        ],
        "3min_transcript": "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. That's two bars. Or maybe I'll write parallel. Parallel downwards to the surface. Now I'm going to leave you there, and I'll let you think about another thing that I'll address in the next video is, what if you had this just standing still? If it wasn't accelerating downwards, if it wasn't accelerating and sliding down, what would be the force that's keeping it in a kind of a static state? We'll think about that in the next video."
    },
    {
        "Q": "At 1:05, you talked of the asteroid that wiped off the dinosaurs. What is the name of this asteroid? And what are the other theories of the beginning of the universe except the big bang?",
        "A": "The asteroid impact crater linked to the end of the age of dinosaurs is found at Chicxulub, Mexico. The big bang theory is the only currently successful scientific theory for a aging universe. It replaced the solid state theory after the discovery, by Erwin Hubble, of the universe s expansion.",
        "video_name": "DRtLXagrMHw",
        "timestamps": [
            65
        ],
        "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": "9:51 wouldn't the line be more of a curve?",
        "A": "Hi, The line would be straight because in this we assume the velocity is constant.",
        "video_name": "T0zpF_j7Mvo",
        "timestamps": [
            591
        ],
        "3min_transcript": "as it went up, but it's actually slowing it down it's pulling it, it's accelerating it in downward direction so that's why you have a negative right over there, that was out convention at the beginning of last video, up is positive, down is negative, so let's focus So this part right over here, negative 4.9 m/s*s times delat t square times delta t, times delta t square, this will make it a little bit easier Although it still, let me get the calculator out So when one second has past, I let my trusty TI-85 out now when 1s past it's 19.6 times 1, well that's just 19.6 minus 4.9 times 1 square So that's just minus 4.9, mius 4.9, gives us 14.7 meters So 14.7 meters So after 1s, the ball has travel 14.7 meters in the air I'll do the same agenda, so after 2s, our velocity is 19.6 minus 9.8 times 2, times 2 this is 2s has gone by well 9.8 times 2, 9.8 times 2 square seconds gives us 19.6 m/s so these just cancel out, so we get out velocity now is zero So after two seconds our velocity now is zero let me make it so this thing should more look like a line I don't get a sense, so this is let me just draw the line like this so our velocity now is zero after 2 seconds what is our displacement? So literally we at the point with the ball has no velocity and right for that exact moment of time it's stationary and then what do we have going on in our displacement? We have 19.6 let me get the calculator out for this We can do it by hand but for the sake of quickness 19.6 times 2 minus 4.9 times 2s squared, this is 2s squared so that's times four, so that gives us 19.6 meters So we have 19, we are at 19.6 meters after 2 seconds we are 19.6 meters in the air now let's go to 3s, so after 3s, our velocity now"
    },
    {
        "Q": "At 08:10, you state that for the molecule, you couldn't get a tertiary carbocation, and I was wondering why not. Couldn't you move one of the hydrogens from the second carbon?",
        "A": "You could, but then the second carbon would still be secondary, not tertiary.",
        "video_name": "iEKA0jUstPs",
        "timestamps": [
            490
        ],
        "3min_transcript": "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 Let's do another one. So we have our cyclohexane ring like that. And then, we have our ethyl group attached to this carbon. And then, our chlorine attached to that carbon. So that's going to be our major product. All right. Let's look at the stereochemistry of this reaction really fast. So let's look at what happens if we react this alkene with hydrochloric acid. So what will we get it? All right. First step-- pi electrons take the proton. Kick the electrons off onto the chlorine. So once again, which side do we add our proton? So two possibilities. I could add to the left side of the double bond or add to the right side of the double bond. It makes sense to add it to the right side of the double bond because that gives us a more stable carbocation. So if I add it to the right side of the double bond, this carbon ends up being positively charged. What kind of a carbocation is that? That carbon is bonded to two other carbons. So it's a secondary carbocation here. There's no kind of rearrangement that we could get here to get a tertiary carbocation. So a secondary carbocation is as stable as we're going to get. Now, carbocations are carbons with three bonds to them, meeting that carbon is sp2 hybridized. So let's redraw this carbocation here. So I'm going to say that this carbon right here represents my carbocation. What's bonded to it? Well, on the left side, there is an ethyl group-- CH2, CH3 right here. And I know that there's a methyl group bonded to it. I'm going to put the methyl group going back in space here. And then there's also a hydrogen attached to that carbon. We just didn't draw it in on our carbocation. So like this. I know this carbon is sp2 hybridized, meaning there's an untouched, unhybridized p orbital on this carbon. All right. So let me draw my p orbital in there like that. And let me go ahead and make sure that everyone realizes this is my carbocation. So sp2 hybridized carbon means the atoms"
    },
    {
        "Q": "At 6:26, I don't understand why that tertiary carbon has a positive charge.. Didn't it lose a proton? Therefore shouldn't it be negatively charged?",
        "A": "No it lost a hydride, the H took both electrons that were in the C-H bond. That carbon now has 3 bonds and 0 lone pairs Formal charge = valence electrons - lone pair electrons - bonds 4 - 0 - 3 = +1",
        "video_name": "iEKA0jUstPs",
        "timestamps": [
            386
        ],
        "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 5:52, Isn't it 2-methylbutyl? Shouldn't the methyl get the lowest possible number(Count Carbons from the right)? Oh, maybe when counting the carbons in a substituent, I have to count from the carbon ATTACHED to the main chain? Am I right?",
        "A": "Yes when you re numbering something like this you number from where the group attaches to the main chain.",
        "video_name": "joQd0qVnX4M",
        "timestamps": [
            352
        ],
        "3min_transcript": "Well, I have a methyl group coming off of carbon 2 this time, so it would be 2-methylpropyl for this complex substituent. The common name for this is isobutyl. So butyl again because there are four total carbons in this complex substituent. Iso because, once again, you have these two methyl groups, so they're like the same, so you get that Y formation. So that's isobutyl. The next one, longest carbon chain, there are two carbons in my longest carbon chain so that would be ethyl. And when I number my longest carbon chain, I can see that I have two methyl groups, and each of those methyl groups is coming off of carbon 1. So I would say this would be 1,1-dimethylethyl. So 1,1-dimethylethyl would be the IUPAC name for this complex You will also see tert-butyl. So tert-butyl is probably used even more frequently. Again, butyl because there are a total of four carbons here. So those are the three possibilities for a complex substituent with a total of four carbons. Let's look at just a few of the possibilities for complex substituents that have five carbons. There are actually much more than this, but these are the ones that are most commonly used. So let's just focus in on these two. So once again, we'll draw our zigzag line to represent the fact that this is actually connected to some straight-chain alkane. And once again, we find our longest carbon chain-- 1, 2, 3, 4. So that would be butyl. And when I number that carbon chain-- 1, 2, 3, 4-- I can see that I have a methyl group coming off of carbon 3. So it would be 3-methylbutyl for the IUPAC name. So you could say isopentyl since there are five carbons now, and iso, because again, you have this methyl group and this methyl group looking like a Y. They're like the same thing. Or I've seen this called isoamyl before. So isoamyl or isopentyl are acceptable IUPAC names as well. What about this one on the right? Longest carbon chain-- 1, 2, 3. So that would be propyl. And numbering it 1, 2, 3, immediately it is obvious that you have two methyl groups coming off of carbon 2. So it would be 2,2-dimethylpropyl, otherwise known as neopentyl since once again you have five carbons for these. So again, there are many more, and we'll stop with those. And so that gives you an idea about how to approach naming complex substituents."
    },
    {
        "Q": "At 6:50, NeoPentyl can also be IsoPentyl, because the C is touching only one other C, am I right?",
        "A": "No. Neopentyl has two methyl groups on C2 of a three-carbon chain. Isopentyl has one methyl group on C3 of a four-carbon chain. They are two completely different things.",
        "video_name": "joQd0qVnX4M",
        "timestamps": [
            410
        ],
        "3min_transcript": "You will also see tert-butyl. So tert-butyl is probably used even more frequently. Again, butyl because there are a total of four carbons here. So those are the three possibilities for a complex substituent with a total of four carbons. Let's look at just a few of the possibilities for complex substituents that have five carbons. There are actually much more than this, but these are the ones that are most commonly used. So let's just focus in on these two. So once again, we'll draw our zigzag line to represent the fact that this is actually connected to some straight-chain alkane. And once again, we find our longest carbon chain-- 1, 2, 3, 4. So that would be butyl. And when I number that carbon chain-- 1, 2, 3, 4-- I can see that I have a methyl group coming off of carbon 3. So it would be 3-methylbutyl for the IUPAC name. So you could say isopentyl since there are five carbons now, and iso, because again, you have this methyl group and this methyl group looking like a Y. They're like the same thing. Or I've seen this called isoamyl before. So isoamyl or isopentyl are acceptable IUPAC names as well. What about this one on the right? Longest carbon chain-- 1, 2, 3. So that would be propyl. And numbering it 1, 2, 3, immediately it is obvious that you have two methyl groups coming off of carbon 2. So it would be 2,2-dimethylpropyl, otherwise known as neopentyl since once again you have five carbons for these. So again, there are many more, and we'll stop with those. And so that gives you an idea about how to approach naming complex substituents. you have to use them when you're naming straight-chain or cycloalkane molecules. So let's look at a cycloalkane molecule, and let's see how to name this guy. Well, I have four carbons in my ring, and I have four carbons in this group. So tie goes to the cycloalkane. So remember from the last video, if you have an equal number of carbons in your ring as with your chain, you're going to name it as an alkyl cycloalkane. The cycloalkane wins the tie. So there are four carbons, so this'll be cyclobutane. So let's go ahead and write cyclobutane here. And once you've determined that you're going to name it as a cycloalkane, then you have to look at this complex substituent and say, OK, well, that's 1, 2, 3, so that would be propyl. And then when you number that complex substituent 1,"
    },
    {
        "Q": "At 1:00 when 1-methylethyl is being named, why is the longest carbon chain ethyl which has two carbons rather than propyl which would be a three carbon chain?",
        "A": "This is because you need to be clear which carbon the group of connected by. You CAN call it a butyl group, but you still must specify which carbon the group is attached to, in this case, it is a sec-butyl.",
        "video_name": "joQd0qVnX4M",
        "timestamps": [
            60
        ],
        "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 8:16, why is the carbon compound called 1-methylpropylcyclobutane but not butylcyclobutane?",
        "A": "A butyl group consists of a consists of a chain of four carbon atoms. The longest chain (starting from the ring attachment!) is only three carbon atoms long, so it cannot be named butyl, even though it has four carbon atoms in total.",
        "video_name": "joQd0qVnX4M",
        "timestamps": [
            496
        ],
        "3min_transcript": "So you could say isopentyl since there are five carbons now, and iso, because again, you have this methyl group and this methyl group looking like a Y. They're like the same thing. Or I've seen this called isoamyl before. So isoamyl or isopentyl are acceptable IUPAC names as well. What about this one on the right? Longest carbon chain-- 1, 2, 3. So that would be propyl. And numbering it 1, 2, 3, immediately it is obvious that you have two methyl groups coming off of carbon 2. So it would be 2,2-dimethylpropyl, otherwise known as neopentyl since once again you have five carbons for these. So again, there are many more, and we'll stop with those. And so that gives you an idea about how to approach naming complex substituents. you have to use them when you're naming straight-chain or cycloalkane molecules. So let's look at a cycloalkane molecule, and let's see how to name this guy. Well, I have four carbons in my ring, and I have four carbons in this group. So tie goes to the cycloalkane. So remember from the last video, if you have an equal number of carbons in your ring as with your chain, you're going to name it as an alkyl cycloalkane. The cycloalkane wins the tie. So there are four carbons, so this'll be cyclobutane. So let's go ahead and write cyclobutane here. And once you've determined that you're going to name it as a cycloalkane, then you have to look at this complex substituent and say, OK, well, that's 1, 2, 3, so that would be propyl. And then when you number that complex substituent 1, of carbon 1. So you would write 1-methylpropyl. And if you wanted to, you could identify that 1-methylpropyl as coming off of carbon 1 of your cyclobutane. So you could put this in parentheses and write 1-(1-methylpropyl)cyclobutane. Or you could just leave the one off, and say (1-methylpropyl)cyclobutane, because, again, it is implied. What is the common name for this complex substituent? So 1-methylpropyl, and we go back up here, and we find 1-methylpropyl was also called sec-butyl. So we could also have named this molecule sec-butylcyclobutane. So let's go ahead and write that. So sec-butylcyclobutane is a perfectly acceptable IUPAC name as well."
    },
    {
        "Q": "At 6:25, where do you get the Oxaloacetic Acid from?",
        "A": "At the end of krebs cycle oxalo acetate is formed from malate in the presence of the enzyme malate dehydrogenase. the same oxalo acetic acid is used to combine with Acetyl Co enzyme-A The whole thing s a cycle",
        "video_name": "juM2ROSLWfw",
        "timestamps": [
            385
        ],
        "3min_transcript": "So we have this kind of preparation step for the Krebs Cycle. We call that pyruvate oxidation. And essentially what it does is it cleaves one of these carbons off of the pyruvate. And so you end up with a 2-carbon compound. You don't have just two carbons, but its backbone of carbons is just two carbons. Called acetyl-CoA. And if these names are confusing, because what is acetyl coenzyme A? These are very bizarre. You could do a web search on them But I'm just going to use the words right now, because it will keep things simple and we'llget the big picture. So it generates acetyl-CoA, which is this 2-carbon compound. And it also reduces some NAD plus to NADH. And this process right here is often given credit-- or the Krebs cycle or the citric acid cycle gets But it's really a preparation step for the Krebs cycle. Now once you have this 2-carbon chain, acetyl-Co-A right here. you are ready to jump into the Krebs cycle. This long talked-about Krebs cycle. And you'll see in a second why it's called a cycle. Acetyl-CoA, and all of this is catalyzed by enzymes. And enzymes are just proteins that bring together the constituent things that need to react in the right way so that they do react. So catalyzed by enzymes. This acetyl-CoA merges with some oxaloacetic acid. A very fancy word. But this is a 4-carbon molecule. These two guys are kind of reacted together, or merged together, depending on how you want to view it. I'll draw it like that. It's all catalyzed by enzymes. And this is important. Some texts will say, is this an enzyme catalyzed reaction? Everything in the Krebs cycle is an And they form citrate, or citric acid. Which is the same stuff in your lemonade or your orange juice. And this is a 6-carbon molecule. You have a 2-carbon and a 4-carbon. You get a 6-carbon molecule. And then the citric acid is then oxidized over a bunch of steps. And this is a huge simplification here. But it's just oxidized over a bunch of steps. Again, the carbons are cleaved off. Both 2-carbons are cleaved off of it to get back to oxaloacetic acid. And you might be saying, when these carbons are cleaved off, like when this carbon is cleaved off, what happens to it? It becomes CO2. It gets put onto some oxygen and leaves the system. So this is where the oxygen or the carbons, or the carbon dioxide actually gets formed. And similarly, when these carbons get cleaved off, it forms CO2."
    },
    {
        "Q": "@12:27 its shows that 10 NADH molecules are formed from one glucose molecule, however at the end of the video we count 8NADH molecules. Which one is correct?",
        "A": "You should have 10NADH at the very end of cellular respiration. 2 from glycolysis, 2 from bridging reaction and 6 from TCA/Kreb/Citric acid cycle.",
        "video_name": "juM2ROSLWfw",
        "timestamps": [
            747
        ],
        "3min_transcript": "We already accounted for the glycolysis right there. Two net ATPs, two NADHs. Now, in the citric acid cycle, or in the Krebs cycle, well first we have our pyruvate oxidation. That produced one NADH. But remember, if we want to say, what are we producing for every glucose? This is what we produced for each of the pyruvates. This NADH was from just this pyruvate. But glycolysis produced two pyruvates. So everything after this, we're going to multiply by two for every molecule of glucose. So I'll say, for the pyruvate oxidation times two means that we got two NADHs. And then when we look at this side, the formal Krebs cycle, what do we get? We have, how many NADHs? One, two, three NADHs. this cycle for each of the pyruvates produced from glycolysis. So that gives us six NADHs. We have one ATP per turn of the cycle. That's going to happen twice. Once for each pyruvic acid. So we get two ATPs. And then we have one FADH2. But it's good, we're going to do this cycle twice. This is per cycle. So times two. We have two FADHs. Now, sometimes in a lot of books these two NADHs, or per turn of the Krebs cycle, or per pyruvate this one NADH, they'll give credit to the Krebs cycle for that. So sometimes instead of having this intermediate step, they'll just write four NADHs right here. And you'll do it twice. Once for each puruvate. So they'll say eight NADHs get produced from the Krebs cycle. But the reality is, six from the Krebs cycle two from the Now the interesting thing is we can account whether we get to the 38 ATPs promised by cellular respiration. We've directly already produced, for every molecule of glucose, two ATPs and then two more ATPs. So we have four ATPs. Four ATPs. How many NADHs do we have? 2, 4, and then 4 plus 6 10. We have 10 NADHs. And then we have 2 FADH2s. I think in the first video on cellular respiration I said FADH. It should be FADH2, just to be particular about things. And these, so you might say, hey, where are our 38 ATPs? We only have four ATPs right now. But these are actually the inputs in the electron transport chain. These molecules right here get oxidized in the electron transport chain. Every NADH in the electron transport chain produces three ATPs."
    },
    {
        "Q": "At 2:35 a correction appears saying he meant to say glycolysis occurs in the cytosol not the cytoplasm. I learned in my class it does occur in the cytoplasm, what's the difference between these two terms?",
        "A": "Cytosol is a part of cytoplasm. It is the part containing all the water and organic molecules. Cytoplasm consist of the cytosol and the organelles suspended in it.",
        "video_name": "juM2ROSLWfw",
        "timestamps": [
            155
        ],
        "3min_transcript": "And I always say the net there, because remember, it used two ATPs in that investment stage, and then it generated four. So on a net basis, it generated four, used two, it gave us two ATPs. And it also produced two NADHs. That's what we got out of glycolysis. And just so you can visualize this a little bit better, let me draw a cell right here. Maybe I'll draw it down here. Let's say I have a cell. That's its outer membrane. Maybe its nucleus, we're dealing with a eukaryotic cell. That doesn't have to be the case. It has its DNA and its chromatin form all spread around like that. And then you have mitochondria. And there's a reason why people call it the power We'll look at that in a second. So there's a mitochondria. It has an outer membrane and an inner I'll do more detail on the structure of a mitochondria, maybe later in this video or maybe I'll do a whole video on them. That's another mitochondria right there. And then all of this fluid, this space out here that's between the organelles-- and the organelles, you kind of view them as parts of the cell that do specific things. Kind of like organs do specific things within our own bodies. So this-- so between all of the organelles you have this fluidic space. This is just fluid of the cell. And that's called the cytoplasm. And that's where glycolysis occurs. So glycolysis occurs in the cytoplasm. Now we all know-- in the overview video-- we know what the next step is. The Krebs cycle, or the citric acid cycle. And that actually takes place in the inner membrane, or I should say the inner space of these mitochondria. Let me draw a mitochondria here. So this is a mitochondria. It has an outer membrane. It has an inner membrane. If I have just one inner membrane we call it a crista. If we have many, we call them cristae. This little convoluted inner membrane, let me give it a label. So they are cristae, plural. And then it has two compartments. Because it's divided by these two membranes. This compartment right here is called the outer compartment. This whole thing right there, that's the outer compartment. And then this inner compartment in here, is called the matrix. Now you have these pyruvates, they're not quite just ready for the Krebs cycle, but I guess-- well that's a good intro into how do you make them ready for the Krebs cycle? They actually get oxidized. And I'll just focus on one of these pyruvates. We just have to remember that the pyruvate, that this"
    },
    {
        "Q": "at 5:23, sal was trying to find out the work required to move the charger a distance of 5 meter closer. shoudnt he have multiplied the force by the distance before integrating?",
        "A": "well he already did that when he multiplied it by dr. And he also mentioned it at about 5:50 that it is the work done in moving the particle by a distance dr and he is going to sum up the work done by using integration. She already has multiplied the force by the infinitiesmally small distance and then adding them up using integral.",
        "video_name": "CqsYCIjSm9A",
        "timestamps": [
            323
        ],
        "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": "At 7:00 , could anyone do a brief explanation of why integration is important of it works applying it in this case?",
        "A": "integration here is important as the electric field is not uniform. it keeps changing with every point, so by applying the formula we do not get a clear idea of the work done throughout, so we do integration of small distance",
        "video_name": "CqsYCIjSm9A",
        "timestamps": [
            420
        ],
        "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": "in 3:37, Sal says that the arteries are blue. Is that true, because I thought the veins were blue?",
        "A": "Sal might have said that arteries are blue but may be it meant pulmonary arteries are blue.",
        "video_name": "QhiVnFvshZg",
        "timestamps": [
            217
        ],
        "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": "At 9:28, Sal said that Veins carry deoxygenated blood but before it he labeled that veins carry oxygenated blood.",
        "A": "The one Sal labelled before, so he was talking about the pulmonary vein. And where he labelled on 9:28 that was not dealing with lungs, thus it was labelled a vein that caries deoxygenated blood. Remember, when one is talking about pulmonary , it refers to that one which which is dealing with the lungs. The lungs have an opposite effect as compared to the other body; the veins carry oxygenated while the arteries carry deoxygenated blood.",
        "video_name": "QhiVnFvshZg",
        "timestamps": [
            568
        ],
        "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": "At 10:18 sal says that the blood from the pulmonary artery goes to the heart but shouldn't it be going to the lungs?",
        "A": "Yes, it should go to the lungs. \u00f0\u009f\u0098\u008a",
        "video_name": "QhiVnFvshZg",
        "timestamps": [
            618
        ],
        "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 8:52, he mentions something about a hydronium ion being produced. Does this ion affect the solution or organism such that it causes harm to them?",
        "A": "The hydronium ion is just a water molecule that has been protonated. Water is H2O and the hydronium ion is H3O+ (ie, water with an extra H, giving rise to a positive charge). It occurs naturally in water, even at neutral pH, and is present at increasingly higher concentrations as the pH of water is reduced by adding acid. It is only harmful if the pH of the solution is very low although bear in mind that the pH of stomach acid is very low (pH 1.5 to 3.5).",
        "video_name": "-Aj5BTnz-v0",
        "timestamps": [
            532
        ],
        "3min_transcript": "and in the double bond right [over] here it could let go of one of the bonds the electrons in one of the bonds and then that can be taken back by the oxygen or even better that can be used by that oxygen to capture a Hydrogen proton in the solution and actually probably part of a Hydronium Molecule But let me just draw it this way this would just be used to capture a Hydrogen proton that would just be a hydrogen a hydrogen atom without its electron. It's just a hydrogen ion It would just be a hydrogen proton and that would form this bond That would form this bond right over here and let me let me just be very clear this carbon this carbon right over here is This carbon right over there this oxygen this oxygen is This oxygen is that oxygen right over there, and so hopefully you see how it forms a cyclone. You're probably saying Oh, wait wait isn't the way I've drawn it looks like there's an extra hydrogen over here, and then that would leave this guy with a Positive charge we leave with a positive charge, but you can imagine we're in a solution of water then hey I have some I have another water molecule right over here And you know these things are all bouncing around and interacting in different ways But it could use let me do that in the right color it could use So that's oxygen it could use one of its lone pairs instead of this you know this will become positive temporarily But then it can use it can do it can use one of its lone pairs to grab just the hydrogen proton which would allow Which will allow this character to take its to take its Electrons to take these electrons back and turn into this character and just be neutral and then this this guy Would have gained so we have a proton going into the solution you have hi But we took a proton from [the] solution We took a proton we gave a proton to the solution And so you could end up with this so the whole reason I did This is something that's really valuable to get very very familiar [with] because you're going to see Glucose and other sugars in many many many different molecules throughout your academic career"
    },
    {
        "Q": "When Sal talks about the atoms 'ionizing' (1:30), where do the electrons that were knocked off go? Could there be He-, like there is He+?",
        "A": "The electrons just jump off into the space around the atom, but they are no longer orbiting it. Sort of like the nucleus was swinging them around on a string; then the string broke. Another atom might pick up the electron or it could just float around between atoms. And He- can exist but it s not very happy so that extra electron tends to not stick around for very long.",
        "video_name": "X_3QAB3o4Vw",
        "timestamps": [
            90
        ],
        "3min_transcript": "- [Voiceover] In the last video we learned that there are a class of stars called Cepheid variables. And these are the super giant stars, as much as 30000 times as bright as the sun. A mass, as much as 20 times the mass of the sun. And what's neat about them is, one, because they're so large and so bright, you can see them really really far away. And what's even neater about them is that they're variable, that they pulsate. And because their pulsations are related to their actual luminosity, you know if you see a cepheid variable star in some distant galaxy, you know what it's luminosity actually is if you were kind of at the star, because you know you can see how it's period of pulsation. And so if you know it's actual luminosity, and you know it's obviously apparent luminosity, you know how much it's gotten dimmed. And the more dim it's gotten from its actual state, you know the farther away it is. So that's the real value of them. What I want to do in this video is to try to explain why they're variables. Why they pulsate. And to do that, to do that, 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,"
    },
    {
        "Q": "At 2:03, why the voltage drop is +10 V and voltage rise is -10V?\nshouldn't be in a opposite way?",
        "A": "Hello Moon, It s a play on words. To drop an object is to give it a negative rise. Same thing in each case. Regards, APD",
        "video_name": "Bt6V7D5av9A",
        "timestamps": [
            123
        ],
        "3min_transcript": "- [Voiceover] Now we're ready to start hooking up our components into circuits, and one of the two things that are going to be very useful to us are Kirchhoff's laws. In this video we're gonna talk about Kirchhoff's voltage law. If we look at this circuit here, this is a voltage source, let's just say this is 10 volts. We'll put a resistor connected to it and let's say the resistor is 200 ohms. Just for something to talk about. One of the things I can do here is I can label this with voltages on the different nodes. Here's one node down here. I'm going to arbitrarily call this zero volts. Then if I go through this voltage source, this node up here is going to be at 10 volts. 10 volts. So here's a little bit of jargon. We call this voltage here. The voltage goes up as we go through the voltage source, and that's called a voltage rise. at this point in the circuit right here and we went from this node down to this node, like that, the voltage would go from 10 volts down to zero volts in this circuit, and that's called a voltage drop. That's just a little bit of slang, or jargon that we use to talk about changes in voltage. Now I can make an observation about this. If I look at this voltage rise here, it's 10 volts, and if I look at that voltage drop, the drop is 10 volts. I can say the drop is 10 volts, or I could say the rise on this side is minus 10 volts. A rise of minus 10. These two expressions mean exactly the same thing. It meant that the voltage went from 10 volts to zero volts, sort of going through this 200 ohm resistor. which is, v-rise minus v-drop equals what? Equals zero. I went up 10 volts, back down 10 volts, I end up back at zero volts, and that's this right here. This is a form of Kirchhoff's voltage law. It says the voltage rises minus the voltage drops is equal to zero. So if we just plug our actual numbers in here what we get is 10 minus 10 equals zero. I'm gonna draw this circuit again. Let's draw another version of this circuit. This time we'll have two resistors instead of one. We'll make it..."
    },
    {
        "Q": "0:47 Why is it weak?\nIs it weak because both oxygens have the same electronegitivity?",
        "A": "Not because of the electronegativity but because of how alcohols are more stable than cyclic peroxides and can easily form from a reaction with water.",
        "video_name": "KfTosrMs5W0",
        "timestamps": [
            47
        ],
        "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": "around 7:25 - why would the H2O nucleophilic attack the partially positive carbons when the oxygen in the epoxide has a formal positive charge?",
        "A": "Formal charge is not necessarily the same thing as the actual charge. In this case much of the actual positive charge is on the carbons. Also, the result of that interaction would be a peroxide with a highly unstable (high energy) bond between two oxygens.",
        "video_name": "KfTosrMs5W0",
        "timestamps": [
            445
        ],
        "3min_transcript": "So I'm going to put my oxygen right here, and then that's bonded to our two carbons like this. And then we see if we can draw the rest of the ring. And so in the back here, here is the rest of my cyclohexane ring like that. And we'll go ahead and put in our lone pairs of electrons. So this is the same exact drawing above here. Now I have my H3O plus in here like this, so my hydronium ion is present with a lone pair of electrons, giving us a plus one formal charge like that. So the oxygen on our epoxide is going to act as a base. It's going to take a proton. So this lone pair of electrons is going to take this proton right here, which would kick these electrons in here off onto my oxygen. So let's draw the result of that acid-base reaction. So I'm going to make a protonated epoxide. and it's connected to those carbons down here. So I'll go ahead and draw the rest of my ring in the back here like that. And then one lone pair of electrons didn't do anything, so it's still there. One lone pair of electrons is the one that formed the bond on that proton, so this is my structure now. And this would give this oxygen a plus one formal charge, so it's positively charged now. So this is the same structure that we saw in the earlier videos, like with our cyclic halonium ion. And just like the cyclic halonium ion in those earlier videos-- check out the halohydrin video-- you're going to get a partial carbocation character with these carbons down here. So the resonance hybrid is going to give these carbons some partial positive character. So when water comes along as a nucleophile, are going to be attracted to those carbons. So opposite charges attract. These two blue carbons are partially positive. The negative electrons are attracted to the partially positive carbon, and you're going to get nucleophilic attacks. So let's say this lone pair of electrons attacks right here. Well, that would kick the electrons in this bond off onto your oxygen. So let's go ahead and draw the result of an attack on the carbon on the left. So let's get some more room here. So what would happen in that instance? Well, let's go ahead and draw our cyclohexane ring back here. So here is our cyclohexane ring. The oxygen attacked the carbon on the left. So there is the oxygen that did the nucleophilic attack, so it has two hydrogens on it. It has one lone pair of electrons now, and it formed a plus one formal charge. Our epoxide opened."
    },
    {
        "Q": "6:58 Where does the water come from?",
        "A": "The H2O is deprotentated (minus 1 hydrogen) H3O.",
        "video_name": "KfTosrMs5W0",
        "timestamps": [
            418
        ],
        "3min_transcript": "So I'm going to put my oxygen right here, and then that's bonded to our two carbons like this. And then we see if we can draw the rest of the ring. And so in the back here, here is the rest of my cyclohexane ring like that. And we'll go ahead and put in our lone pairs of electrons. So this is the same exact drawing above here. Now I have my H3O plus in here like this, so my hydronium ion is present with a lone pair of electrons, giving us a plus one formal charge like that. So the oxygen on our epoxide is going to act as a base. It's going to take a proton. So this lone pair of electrons is going to take this proton right here, which would kick these electrons in here off onto my oxygen. So let's draw the result of that acid-base reaction. So I'm going to make a protonated epoxide. and it's connected to those carbons down here. So I'll go ahead and draw the rest of my ring in the back here like that. And then one lone pair of electrons didn't do anything, so it's still there. One lone pair of electrons is the one that formed the bond on that proton, so this is my structure now. And this would give this oxygen a plus one formal charge, so it's positively charged now. So this is the same structure that we saw in the earlier videos, like with our cyclic halonium ion. And just like the cyclic halonium ion in those earlier videos-- check out the halohydrin video-- you're going to get a partial carbocation character with these carbons down here. So the resonance hybrid is going to give these carbons some partial positive character. So when water comes along as a nucleophile, are going to be attracted to those carbons. So opposite charges attract. These two blue carbons are partially positive. The negative electrons are attracted to the partially positive carbon, and you're going to get nucleophilic attacks. So let's say this lone pair of electrons attacks right here. Well, that would kick the electrons in this bond off onto your oxygen. So let's go ahead and draw the result of an attack on the carbon on the left. So let's get some more room here. So what would happen in that instance? Well, let's go ahead and draw our cyclohexane ring back here. So here is our cyclohexane ring. The oxygen attacked the carbon on the left. So there is the oxygen that did the nucleophilic attack, so it has two hydrogens on it. It has one lone pair of electrons now, and it formed a plus one formal charge. Our epoxide opened."
    },
    {
        "Q": "so at 9:44 the picture has two purple dun bell shaped p orbtials are they individual or is it just one long p orbital for C2H4",
        "A": "At 2:30 he shows the shape of a p orbital. Every p orbital is made out of two halfs. :) In ethin every C atom has ONE p orbital.",
        "video_name": "ROzkyTgscGg",
        "timestamps": [
            584
        ],
        "3min_transcript": "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. Steric number is equal to the number of sigma bonds plus number of lone pairs of electrons. If my goal was to find the steric number for this carbon. I count up my number of sigma bonds. That's one, two and then I know when I double bond one of those is sigma and one of those is pi. One of those is a sigma bond. A total of three sigma bonds. I have zero lone pairs of electrons around that carbon. Three plus zero, gives me a steric number of three. I need three hybrid orbitals and we've just seen in this video that three SP2 hybrid orbitals form if we're dealing with SP2 hybridization. If we get a steric number of three, you're gonna think about SP2 hybridization. One S orbital and two P orbitals hybridizing. That carbon is SP2 hybridized and of course, this one is too. Both of them are SP2 hybridized. Let's do another example. Let's do boron trifluoride. 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."
    },
    {
        "Q": "@10:55 there are 9 lone pairs right? then y it is taken 0",
        "A": "those pairs are of Flourine (F) not of (B) so it doens t count",
        "video_name": "ROzkyTgscGg",
        "timestamps": [
            655
        ],
        "3min_transcript": "Steric number is equal to the number of sigma bonds plus number of lone pairs of electrons. If my goal was to find the steric number for this carbon. I count up my number of sigma bonds. That's one, two and then I know when I double bond one of those is sigma and one of those is pi. One of those is a sigma bond. A total of three sigma bonds. I have zero lone pairs of electrons around that carbon. Three plus zero, gives me a steric number of three. I need three hybrid orbitals and we've just seen in this video that three SP2 hybrid orbitals form if we're dealing with SP2 hybridization. If we get a steric number of three, you're gonna think about SP2 hybridization. One S orbital and two P orbitals hybridizing. That carbon is SP2 hybridized and of course, this one is too. Both of them are SP2 hybridized. Let's do another example. Let's do boron trifluoride. 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"
    },
    {
        "Q": "How can he tell its going to be a sigma bond at 10:40?",
        "A": "Because a single covalent bond is a sigma bond.",
        "video_name": "ROzkyTgscGg",
        "timestamps": [
            640
        ],
        "3min_transcript": "Steric number is equal to the number of sigma bonds plus number of lone pairs of electrons. If my goal was to find the steric number for this carbon. I count up my number of sigma bonds. That's one, two and then I know when I double bond one of those is sigma and one of those is pi. One of those is a sigma bond. A total of three sigma bonds. I have zero lone pairs of electrons around that carbon. Three plus zero, gives me a steric number of three. I need three hybrid orbitals and we've just seen in this video that three SP2 hybrid orbitals form if we're dealing with SP2 hybridization. If we get a steric number of three, you're gonna think about SP2 hybridization. One S orbital and two P orbitals hybridizing. That carbon is SP2 hybridized and of course, this one is too. Both of them are SP2 hybridized. Let's do another example. Let's do boron trifluoride. 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"
    },
    {
        "Q": "Like it said at 11:40, in overall, is it you always fill valence electrons into sp2 orbitals first then free p orbital right?",
        "A": "Yes, however this is a very unusual situation because boron does not usually follow the octet rule. Ordinarily, you would have the sp\u00c2\u00b2 and the p all filled (counting the electrons shared with the other atoms).",
        "video_name": "ROzkyTgscGg",
        "timestamps": [
            700
        ],
        "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 10:45 why the number of lone pairs are taken as zero?",
        "A": "Because we are only counting the electrons on boron and it has no lone pairs.",
        "video_name": "ROzkyTgscGg",
        "timestamps": [
            645
        ],
        "3min_transcript": "Steric number is equal to the number of sigma bonds plus number of lone pairs of electrons. If my goal was to find the steric number for this carbon. I count up my number of sigma bonds. That's one, two and then I know when I double bond one of those is sigma and one of those is pi. One of those is a sigma bond. A total of three sigma bonds. I have zero lone pairs of electrons around that carbon. Three plus zero, gives me a steric number of three. I need three hybrid orbitals and we've just seen in this video that three SP2 hybrid orbitals form if we're dealing with SP2 hybridization. If we get a steric number of three, you're gonna think about SP2 hybridization. One S orbital and two P orbitals hybridizing. That carbon is SP2 hybridized and of course, this one is too. Both of them are SP2 hybridized. Let's do another example. Let's do boron trifluoride. 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"
    },
    {
        "Q": "When you squared that negative number in 2:37, does it become non negative?",
        "A": "Yes. The square of a negative number is positive. The cube, however, will be negative.",
        "video_name": "gluN2wxqES0",
        "timestamps": [
            157
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 5:20, it is assumed that tension is not equal to the force of gravity on the 3kg box because the 3kg box is accelerating. How can you assume that the 3kg box is accelerating at that point? How can you know that the box isn't standing still?",
        "A": "The surface that the 5kg box is sitting on is assumed to be frictionless---meaning any force that is applied to it will move it.",
        "video_name": "QKXeZFwFPS0",
        "timestamps": [
            320
        ],
        "3min_transcript": "You always draw a force diagram. So what forces do I have on the 5 kg mass? I'm gonna have a force of gravity, I'll draw that straight down. FG, and there's gonna be an equal force, normal force upward. So this normal force up should be equal to the force of gravity and magnitude because this box is probably not gonna be accelerating vertically. There's no real reason why it should be if this table is rigid. And there's one more force on this box. There's a force to the right. That's gonna be the force of tension. And if there's no friction on this table, then I have no leftward forces here. I'm ignoring air resistance since we usually ignore air resistance. So that's is, the only horizontal force I've got is T, tension. And I divide by the mass of the 5 kg box, which is 5 kg. But we got a problem. Look it, we don't know the acceleration of the 5 kg mass, and we don't know the tension. I can't solve this. Normally what you do in this case, is you go to the vertical direction, the other direction in other words. That's just gonna tell me that the normal force is gonna be equal to the force of gravity. And we kind of already knew that. So that doesn't help. So what do we do? Well, you might note, this is only the equation for the 5 kg mass. And so now I have to do this for the 3 kg mass. So let's come over here, let's say that the acceleration of the 3 kg mass is gonna be equal to the net force on the 3 kg mass, divided by the mass of the 3 kg mass. And again, which direction should we pick? Well this acceleration over here is gonna be vertical. So let's solve this for the vertical direction. I'm gonna add one more sub-script, Y, to remind myself. And you should do this too so you remember which direction you're picking. So what forces do I plug in here? You figure that out with a force diagram. I'm gonna have a force of gravity on this 3 kg mass, and then I'm gonna have the same size of friction, or sorry, the same as tension, that I had over here. So the tension on this side of the rope, it's gonna be the same as the tension on this side. or friction. So assuming that its mass is negligible, there's basically no friction, then I'm gonna have a tension. That tension is gonna be the same size. So I'll draw that coming upward. But it's not gonna be as big as the force of gravity is on this 3 kg mass. I've got the force of gravity here. This tension is gonna be smaller, and the reason is, this 3 kg mass is accelerating downwards. So these forces can't be balanced. The upward force of tension has gotta be smaller than the force of gravity on this 3 kg mass. But this tension here should be the same as this tension here. So I'll plug those in. So let's plug this in. A of the 3 kg mass, in the Y direction is gonna be equal to, I've got two vertical forces. I've got tension up, so I'll make that positive, 'cause we usually treat up as positive. I've got gravity down, and so I'm gonna have negative, 'cause it's downward force of 3 kg times the acceleration"
    },
    {
        "Q": "At 2:53, isn't it 5,5 diethyldecane if you count from the closest sides to each ethyl group?",
        "A": "i think you need to review your counting mechanics, because I do not see two adjacent ethyl groups on any 5,5 carbons",
        "video_name": "q_Q9C1Ooofc",
        "timestamps": [
            173
        ],
        "3min_transcript": "I think we're ready now to tackle some more, or even more complicated examples. So let's draw something crazy here. Let me draw a chain. Let me draw it like that. And so like we've done in all of the examples, you'll want to find the longest chain. We can count from here one, two, three, four, five, six, seven, Or maybe it's one, two, three, four, five, six, seven. Or maybe it's 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. That is our longest chain. Let me just make that in green. So our longest chain here is in green. So this backbone has 10 carbons in it. The prefix for 10 is dec. It is a alkane since it has all single bonds. So we can write decane for the backbone. And then it has a group right here. And this group consists of one, two carbons attached The prefix for two carbons is eth, so this is an ethyl group. The y-l is because it's a group attached to the main alkane chain. So we call this ethyl decane. But we have to specify where the ethyl group is attached, and we want to give it as low of a number as possible so we start counting on the side closest to it. So it's one, two, three, four, five. So this is 5-ethyldecane. Now let's complicate this a little bit more. So let me just copy and paste this. So I have pasted it there, and let me complicate this molecule Let me add another ethyl group to it. So let's say we have another ethyl group over there. Now what is this going to be? to be that thing in green. So it's still going to be a decane. But now we have two ethyl groups. One on the five carbon, one, two, three, four, five, and then one on the six carbon. You might be tempted to write 5-ethyl 6-ethyldecane, which really wouldn't be wrong, but it would just be maybe more letters than you want to write. Instead you write 5 comma 6-diethyldecane. The 5, 6, tells us the two carbons on the main backbone that the ethyl groups are attached to, and the di says that we have two ethanes, or two ethyl groups I should say, not ethane groups, two ethyl groups, one over here, and one right over here."
    },
    {
        "Q": "9:05 Positives in left rod attract positives in right rod?",
        "A": "Like charges repel, they don t attract.",
        "video_name": "ZgDIX2GOaxQ",
        "timestamps": [
            545
        ],
        "3min_transcript": "Let's say you have two conducting rods. Say these are made out of metal. One of them has a net amount of negative charge on it which is going to reside on the outside edge because that's what net charge does on a conductor, but this other rod, this other metal conducting rod, does not have any net charge on it. What would happen if I took this first rod touched it to the second rod? You probably guessed, charges want to get as far away from each other as possible so these negatives realize \"Hey, if we spread out, \"some of us go on to this rod and some of us stay here, \"we can spread out even father away from each other.\" That's what they would do. If these rods were the same size, you'd have equal amounts on each. If the second rod was bigger, more of them would go on to this second one because that would allow them to spread out even more. Some would stay on the smaller one. That's charged by just touching something. That's easy. You can charge something also, you can get clever. You can do something called charge, you can charge something by induction it's called. 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?"
    },
    {
        "Q": "At 12:33, how do we know that whether it's the ceiling or rather the balloon that reorients or polarise its atoms in order to create an attractive force?",
        "A": "We put the charge on the balloon., not on the ceiling, right? Rubber is an insulator, so it is difficult for charge to move around.",
        "video_name": "ZgDIX2GOaxQ",
        "timestamps": [
            753
        ],
        "3min_transcript": "but some of the electrons will leave which means that this rod, which used to be uncharged now has a net amount of positive charge in it. I've charged this rod without even touching it because I let the negative electrons leave. If I'm clever, what I can do is I can just cut this wire before I take away the thing that induced the charge. If I remove this now and move it far away, what these negatives would have done is they would have said \"Shoot, okay, \"I am glad that that's over. \"Now I can rejoin. \"I'm attracted to this positive again. \"I'm going to rejoin my positives.\" and this thing will become uncharged again but now they can't get back. They're stuck. There's no way for these to get back because you've cut the cord here and you've permanently charged this piece of metal without even touching it. It's called charge by induction. It's a quick way we charge something up. Everyone's tried this. You take a balloon. What happens? How do you charge it up? You rub it against your hair. It steals electrons from your hair and the balloon becomes negatively charged. What do you do with it? You know what you do with it. You take this thing and you put it near a wall or a ceiling and if you're lucky, it sticks there, which is cool! How does it work? Well, remember, this is an insulating material rubber. The ceiling is an insulating material. Electrons aren't getting transferred but even in an insulating material, the atom can reorient or polarize by shifting. The negatives in that atom can shift to one side and the other side becomes a little more positive and what that does, it causes a net force between the ceiling and the balloon because these positives are a little closer. These positives are attracting negatives and the negatives are attracting the positives with a little bit greater a force than these negatives are repelling the other negatives in the ceiling. is also attracting the balloon and the balloon is attracting the ceiling with greater force than the negatives are repelling the balloon, the balloon can stick because of the insulating material's ability to polarize and cause and electric attraction. This is what I said earlier. Even if it's an insulator, sometimes it can interact with something electric because the atom can shift and polarize."
    },
    {
        "Q": "At 11:24, if the left rod was positively charged, would the positive charges on the right rod leave to the ground instead?",
        "A": "The positive charges were the nuclei, and nuclei can t really move in a solid.",
        "video_name": "ZgDIX2GOaxQ",
        "timestamps": [
            684
        ],
        "3min_transcript": "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. but some of the electrons will leave which means that this rod, which used to be uncharged now has a net amount of positive charge in it. I've charged this rod without even touching it because I let the negative electrons leave. If I'm clever, what I can do is I can just cut this wire before I take away the thing that induced the charge. If I remove this now and move it far away, what these negatives would have done is they would have said \"Shoot, okay, \"I am glad that that's over. \"Now I can rejoin. \"I'm attracted to this positive again. \"I'm going to rejoin my positives.\" and this thing will become uncharged again but now they can't get back. They're stuck. There's no way for these to get back because you've cut the cord here and you've permanently charged this piece of metal without even touching it. It's called charge by induction. It's a quick way we charge something up."
    },
    {
        "Q": "At 9:10 why did he assumes protons are attracted by the electron? isn't more easy for the electrons to move to protons?",
        "A": "The point is, they are attracted. It does not matter who moves.",
        "video_name": "ZgDIX2GOaxQ",
        "timestamps": [
            550
        ],
        "3min_transcript": "Let's say you have two conducting rods. Say these are made out of metal. One of them has a net amount of negative charge on it which is going to reside on the outside edge because that's what net charge does on a conductor, but this other rod, this other metal conducting rod, does not have any net charge on it. What would happen if I took this first rod touched it to the second rod? You probably guessed, charges want to get as far away from each other as possible so these negatives realize \"Hey, if we spread out, \"some of us go on to this rod and some of us stay here, \"we can spread out even father away from each other.\" That's what they would do. If these rods were the same size, you'd have equal amounts on each. If the second rod was bigger, more of them would go on to this second one because that would allow them to spread out even more. Some would stay on the smaller one. That's charged by just touching something. That's easy. You can charge something also, you can get clever. You can do something called charge, you can charge something by induction it's called. 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?"
    },
    {
        "Q": "At 9:08, he says \"these positives in this charge rod are attracting these positives\" when explaining charge by induction. But since like charges repel, shouldn't they repel instead of attracting?",
        "A": "If he said that, he misspoke.",
        "video_name": "ZgDIX2GOaxQ",
        "timestamps": [
            548
        ],
        "3min_transcript": "Let's say you have two conducting rods. Say these are made out of metal. One of them has a net amount of negative charge on it which is going to reside on the outside edge because that's what net charge does on a conductor, but this other rod, this other metal conducting rod, does not have any net charge on it. What would happen if I took this first rod touched it to the second rod? You probably guessed, charges want to get as far away from each other as possible so these negatives realize \"Hey, if we spread out, \"some of us go on to this rod and some of us stay here, \"we can spread out even father away from each other.\" That's what they would do. If these rods were the same size, you'd have equal amounts on each. If the second rod was bigger, more of them would go on to this second one because that would allow them to spread out even more. Some would stay on the smaller one. That's charged by just touching something. That's easy. You can charge something also, you can get clever. You can do something called charge, you can charge something by induction it's called. 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?"
    },
    {
        "Q": "Is displacement and distance traveled the same thing? because I thought distance between two points is displacement. In here, I noticed Mr.Khan using 'distance traveled' and 'displacement' interchangeably. Like on 14:07 he has the equation he used to find 's' set up equaled to 'D'",
        "A": "They are not quite the same. Displacement is a vector whereas distance is a scalar. Displacement and distance are often equivalent, but there are many cases where they aren t. For example, after completing one lap around a circular track, the distance would be the circumference of the track whereas the displacement would be 0 (because you ended up back where you started).",
        "video_name": "MAS6mBRZZXA",
        "timestamps": [
            847
        ],
        "3min_transcript": "so this is interesting, the distance we travel is equal to one half of the initial velocity plus the final velocity so this is really if you just took this quantity right over here it's just the arithmetic, I have trouble saying that word it's the arithmetic mean of these two numbers, so I'm gonna define, this is something new, I'm gonna call this the average velocity we have to be very careful with this this right here is the average velocity but the only reason why I can just take the starting velocity and ending velocity and adding them together and divide them by two since you took an average of two thing it's some place over here and I take that as average velocity it's because my acceleration is constant but it's not always the assumption but if you do have a constant acceleration like this you can assume that the average velocity is gonna be the average of the initial velocity and the final velocity if this is a curve and the acceleration is changing you could not do that but what is useful about this is if you wanna figure out the distance that was travelled, you just need to know the initial velocity and the final velocity, average their two and multiply the times it goes by so in this situation our final velocity is 13m/s our initial velocity is 5m/s so you have 13 plus 5 is equal to 18 you divided that by 2, you average velocity is 9m/s if you take the average of 13 and 5 and 9m/s times 4s gives you 36m so hopefully it doesn't confuse you I just wannt show you but you shouldn't memorize they can all be deduced"
    },
    {
        "Q": "at 7:30 , why did sal subtracted initial velocity from the final velocity to find height? instead, he could multiply constant acceleration with delta time. ex} 4s * 2m/s^2 = 8 m/s",
        "A": "in this case only the Vi-Vf x t = a x t^2 in other cases it may not be the same.",
        "video_name": "MAS6mBRZZXA",
        "timestamps": [
            450
        ],
        "3min_transcript": "after 3s we will be 11m/s, and after 4s we will be at 13 m/s so you multiply how much time pass times acceleration this is how much faster we are gonna be going, we are already going 5m/s 5 plus how much faster? 13 m/s so this right up here is 13m/s So I will take a little pause here hopefully intuitive and the whole play of that is to show you this formula you will see in many physics book is not something that randomly pop out of there it just make complete common sense Now the next thing I wanna talk about is what is the total distance that we would have travel? and we know from the last video that distance is just the area under this curve right over here, so it's just the area under this curve 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"
    },
    {
        "Q": "At 0:30,u assigned right to positive and left to negative.Is the right direction always the positive one and the left always negative.Or is it something we can choose?",
        "A": "You can choose it however you want.",
        "video_name": "MAS6mBRZZXA",
        "timestamps": [
            30
        ],
        "3min_transcript": "The goal of this video is to explore some of the concept of formula you might see in introductional physics class but more importantly to see they are really just common sense ideas So let's just start with a simple example Let's say that and for the sake of this video keep things that magnitudes and velocities that's the direction of velocity etc. let's just assume that if I have a positive number that it means for example postive velocity that it means I'm going to the right let's say I have a negative number we won't see in this video let's assume we are going to the left In that way I can just write a number down only operating in one dimension you know that by specifying the magnitude and the direction if I say velocity is 5m/s that means 5m/s to the right if I say negative 5m/s that means 5m/s to the left let's say just for simplicitiy, say that we start with initial velocity we start with an initial velocity of 5m/s once again I specify the magnitude and the direction because of this let's say we have a constant acceleration we have a constant acceleration 2m/s^2 or 2m per second square and once again since this is positive it is to the right and let's say that we do this for a duration so my change in time, let's say we do this for a duration of 4 I will just use s, second and s different places so s for this video is seconds So I want to do is to think about how far do we travel? and there is two things how fast are we going? after 4 seconds and how far have we travel over the course of those 4 seconds? so let's draw ourselves a little diagram here So this is my velocity axis, and this over here is time axis 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"
    },
    {
        "Q": "Could you simplify the equation at 12:55 e4ven further and say 1/2t(vi+vf)?",
        "A": "Not unless the amount of time between Vi and Vf is 1 unit of time. The average of 2 velocities is still a velocity not a distance.",
        "video_name": "MAS6mBRZZXA",
        "timestamps": [
            775
        ],
        "3min_transcript": "minus one half of the original something so what is vi minus one half vi? so anything minus its half is just a positive half left so these two terms, this term and this term will simplify to one half vi one half initial velocity plus one half times the final velocity plus one half times the final velocity and all of that is being multiplied with the change in time the time that has gone by and this tells us the distance, the distance that we travel another way to think about it, let's factor out this one half you get distance that is equal to change in time times factoring out the one half so this is interesting, the distance we travel is equal to one half of the initial velocity plus the final velocity so this is really if you just took this quantity right over here it's just the arithmetic, I have trouble saying that word it's the arithmetic mean of these two numbers, so I'm gonna define, this is something new, I'm gonna call this the average velocity we have to be very careful with this this right here is the average velocity but the only reason why I can just take the starting velocity and ending velocity and adding them together and divide them by two since you took an average of two thing it's some place over here and I take that as average velocity it's because my acceleration is constant but it's not always the assumption but if you do have a constant acceleration like this you can assume that the average velocity is gonna be the average of the initial velocity and the final velocity if this is a curve and the acceleration is changing you could not do that but what is useful about this is if you wanna figure out the distance that was travelled, you just need to know the initial velocity and the final velocity, average their two and multiply the times it goes by so in this situation our final velocity is 13m/s our initial velocity is 5m/s so you have 13 plus 5 is equal to 18 you divided that by 2, you average velocity is 9m/s if you take the average of 13 and 5 and 9m/s times 4s gives you 36m so hopefully it doesn't confuse you I just wannt show you"
    },
    {
        "Q": "At 1:23, he says that ONE photon hits ONE electron. How do you know if it is just one photon with an energy of say, x or multiple photons with a cumulative energy of x which dislodges the electron ? Is it just an assumption Einstein took or is there a scientific reason behind it ?\nAlso we are talking about Classical Mechanics isn't the photons motion supposed to be studied by Quantum Mechanics ?",
        "A": "they could have possibly altered the number of photons in experimental apparatus and observed the same result.",
        "video_name": "vuGpUFjLaYE",
        "timestamps": [
            83
        ],
        "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": "At 12:23 Why is it said that the structure on the left is the right one? ALSO why is it that we have only two resonance structures for the molecule at 12:15?",
        "A": "These are isomers, not resonance structures. The five orbitals have a trigonal bipyramidal geometry. There are only two places where the lone pair can go \u00e2\u0080\u0094 the axial ot the equatorial location. The see-saw geometry is more stable because the lone pair in the axial location has less total repulsion from the other electrons.",
        "video_name": "0na0xtIHkXA",
        "timestamps": [
            743,
            735
        ],
        "3min_transcript": ""
    },
    {
        "Q": "How do you find the conjugate acid? They were mentioned around 7:55 but it was not explained how he knew those were the conjugate bases.",
        "A": "A conjugate acid/base pair are chemicals that are different by a proton or electron pair. For instance, the strong acid HCl has a conjugate base of Cl-. Remember that acids donate protons (H+) and that bases accept protons. So each conjugate pair essentially are different from each other by one proton. There s a lot of info in the acid base section too!",
        "video_name": "7BgiKyvviyU",
        "timestamps": [
            475
        ],
        "3min_transcript": "this negative-one formal charge is not localized to this oxygen; it's de-localized. And so, because we can spread out some of that negative charge, that increases the stability of the anion here, so this is relatively stable, so increased stability, due to de-localization. So, the fact that we can draw an extra resonance structure, means that the anion has been stabilized. And so, this is called, \"pushing electrons,\" so we're moving electrons around, and it's extremely important to feel comfortable with moving electrons around, and being able to follow them. So, the only way to get good at this is to do a lot of practice problems, so please do that; do lots of practice problems in your textbook. If we compare that to the ethoxide anion, so over here, if we try to do the same thing, if we try to take a lone pair of electrons on this oxygen, and move it into here, already has four bonds; so it's already bonded to two hydrogens, and then we have this bond, and this bond. And so, moving those electrons in, trying to de-localize those electrons, would give us five bonds to carbon, and so we can't do that; we can't draw a resonance structure for the ethoxide anion. So those electrons are localized to this oxygen, and so this oxygen has a full, negative-one formal charge, and since we can't spread out that negative charge, or it's going to destabilize this anion. So this is not as stable, so decreased stability, compared to the anion on the left, because we can't draw a resonance structure. If we think about the conjugate acids to these bases, so the conjugate acid to the acetate anion would be, of course, acetic acid. So we go ahead, and draw in acetic acid, like that. The conjugate acid to the ethoxide anion So we go ahead, and draw in ethanol. And we think about which one of those is more acidic. We know that acetic acid is more acidic; it's more likely to donate a proton, because the conjugate base is more stable, because, you could think about resonance, or de-localization of electrons. If you're looking at ethanol, ethanol's not as likely to donate its proton, because the conjugate base, the ethoxide anion is not as stable, because you can't draw any resonance structures for it. The negative charge is not able to be de-localized; it's localized to that oxygen. So this is just one application of thinking about resonance structures, and, again, do lots of practice. In the next video, we'll talk about different patterns that you can look for, and we talked about one in this video: We took a lone pair of electrons, so right here in green, and we noticed this lone pair of electrons was next to a pi bond, and so we were able to draw another resonance structure for it. We don't have that situation with ethoxide:"
    },
    {
        "Q": "At 8:36 Jay states that the ethanol molecule is not as likely to donate its proton because its conjugate base is not as stable. He also states the negative charge is localized to the oxygen. What exactly is it that prevents ethanol from donating this proton? Is it the partial negative dipole on the oxygen that causes a strong attraction to the hydrogen?",
        "A": "He s comparing ethanol to acetic acid here, that s what we need to keep in mind. In acetic acid the negative charge can be spread over two oxygen atoms whereas in ethanol it s stuck on one. The ethanol isn t being prevented from donating its proton, it still happens somewhat, but when you have a resonance stabilised conjugate base like there is in acetic acid we find they are better able to act as an acid.",
        "video_name": "7BgiKyvviyU",
        "timestamps": [
            516
        ],
        "3min_transcript": "already has four bonds; so it's already bonded to two hydrogens, and then we have this bond, and this bond. And so, moving those electrons in, trying to de-localize those electrons, would give us five bonds to carbon, and so we can't do that; we can't draw a resonance structure for the ethoxide anion. So those electrons are localized to this oxygen, and so this oxygen has a full, negative-one formal charge, and since we can't spread out that negative charge, or it's going to destabilize this anion. So this is not as stable, so decreased stability, compared to the anion on the left, because we can't draw a resonance structure. If we think about the conjugate acids to these bases, so the conjugate acid to the acetate anion would be, of course, acetic acid. So we go ahead, and draw in acetic acid, like that. The conjugate acid to the ethoxide anion So we go ahead, and draw in ethanol. And we think about which one of those is more acidic. We know that acetic acid is more acidic; it's more likely to donate a proton, because the conjugate base is more stable, because, you could think about resonance, or de-localization of electrons. If you're looking at ethanol, ethanol's not as likely to donate its proton, because the conjugate base, the ethoxide anion is not as stable, because you can't draw any resonance structures for it. The negative charge is not able to be de-localized; it's localized to that oxygen. So this is just one application of thinking about resonance structures, and, again, do lots of practice. In the next video, we'll talk about different patterns that you can look for, and we talked about one in this video: We took a lone pair of electrons, so right here in green, and we noticed this lone pair of electrons was next to a pi bond, and so we were able to draw another resonance structure for it. We don't have that situation with ethoxide: but we don't have a pi bond next to it, And so, more in the next video on that."
    },
    {
        "Q": "I didnt understan the part when he says that the rate of the reaction is equal to the rate of O2 (time 06:37). How do we know that?\nThanks.",
        "A": "The rate of reaction is equal to the, R = rate of formation of any component of the reaction / change in time. Here in this reaction O2 is being formed, so rate of reaction would be the rate by which O2 is formed.",
        "video_name": "8wIodo1HD4Y",
        "timestamps": [
            397
        ],
        "3min_transcript": "And let's say that oxygen forms at a rate of 9 x 10 to the -6 M/s. So what is the rate of formation of nitrogen dioxide? Well, if you look at the balanced equation, for every one mole of oxygen that forms four moles of nitrogen dioxide form. So we just need to multiply the rate of formation of oxygen by four, and so that gives us, 3.6 x 10 to the -5 Molar per second. So, NO2 forms at four times the rate of O2. What about dinitrogen pentoxide? So, N2O5. Look at your mole ratios. For every one mole of oxygen that forms we're losing two moles of dinitrogen pentoxide. So if we're starting with the rate because our mole ratio is one to two here, we need to multiply this by 2, and since we're losing dinitrogen pentoxide, we put a negative sign here. So this gives us - 1.8 x 10 to the -5 molar per second. So, dinitrogen pentoxide disappears at twice the rate that oxygen appears. All right, let's think about the rate of our reaction. So the rate of our reaction is equal to, well, we could just say it's equal to the appearance of oxygen, right. We could say it's equal to 9.0 x 10 to the -6 molar per second, so we could write that down here. The rate is equal to the change in the concentration of oxygen over the change in time. All right, what about if we wanted to express this in terms of the formation of nitrogen dioxide. was 3.6 x 10 to the -5. All right, so that's 3.6 x 10 to the -5. So you need to think to yourself, what do I need to multiply this number by in order to get this number? Since this number is four times the number on the left, I need to multiply by one fourth. Right, so down here, down here if we're talking about the change in the concentration of nitrogen dioxide over the change in time, to get the rate to be the same, we'd have to multiply this by one fourth. All right, finally, let's think about, let's think about dinitrogen pentoxide. So, we said that that was disappearing at -1.8 x 10 to the -5. So once again, what do I need to multiply this number by in order to get 9.0 x 10 to the -6?"
    },
    {
        "Q": "At 5:55, We are losing N2O5 because it's a reactant and not a product, right?",
        "A": "Sort of. N2O5 [reactant] is being consumed/used up to generate NO2 and O2 [products]. I usually think of it as the number of moles or the concentration of the reactants become less as products are being formed in the process.",
        "video_name": "8wIodo1HD4Y",
        "timestamps": [
            355
        ],
        "3min_transcript": "and the formation of B, and we could make this a little bit more general. We could say that our rate is equal to, this would be the change in the concentration of A over the change in time, but we need to make sure to put in our negative sign. We put in our negative sign to give us a positive value for the rate. So the rate is equal to the negative change in the concentration of A over the change of time, and that's equal to, right, the change in the concentration of B over the change in time, and we don't need a negative sign because we already saw in the calculation, right, we get a positive value for the rate. So, here's two different ways to express the rate of our reaction. So here, I just wrote it in a little bit more general terms. Let's look at a more complicated reaction. Here, we have the balanced equation for the decomposition of dinitrogen pentoxide And let's say that oxygen forms at a rate of 9 x 10 to the -6 M/s. So what is the rate of formation of nitrogen dioxide? Well, if you look at the balanced equation, for every one mole of oxygen that forms four moles of nitrogen dioxide form. So we just need to multiply the rate of formation of oxygen by four, and so that gives us, 3.6 x 10 to the -5 Molar per second. So, NO2 forms at four times the rate of O2. What about dinitrogen pentoxide? So, N2O5. Look at your mole ratios. For every one mole of oxygen that forms we're losing two moles of dinitrogen pentoxide. So if we're starting with the rate because our mole ratio is one to two here, we need to multiply this by 2, and since we're losing dinitrogen pentoxide, we put a negative sign here. So this gives us - 1.8 x 10 to the -5 molar per second. So, dinitrogen pentoxide disappears at twice the rate that oxygen appears. All right, let's think about the rate of our reaction. So the rate of our reaction is equal to, well, we could just say it's equal to the appearance of oxygen, right. We could say it's equal to 9.0 x 10 to the -6 molar per second, so we could write that down here. The rate is equal to the change in the concentration of oxygen over the change in time. All right, what about if we wanted to express this in terms of the formation of nitrogen dioxide."
    },
    {
        "Q": "At 07:18 he says that increase in temperature results in increase in solubility. how is that so?",
        "A": "the water molecules move around when heated, then gives the salts the opportunity wedge in between them.",
        "video_name": "zjIVJh4JLNo",
        "timestamps": [
            438
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 4:27 Sal says that the NaCl molecule breaks, isn't it that a molecule needs a lot of energy to break? Where does that energy come from?",
        "A": "When the NaCl ions dissociate, the attraction between the ions is replaced by the attraction of the ions to the water molecules. Nevertheless, the process is slightly endothermic but the dissolution is favoured by an increase in entropy as the system becomes more disordered. Solutions always have higher entropy than crystalline solids.",
        "video_name": "zjIVJh4JLNo",
        "timestamps": [
            267
        ],
        "3min_transcript": ""
    },
    {
        "Q": "Hi :)\n\nAt 7:24, it is mentioned that the molecules have the same kinetic energy. How so? If the volume of the box was decreased then the molecules kinetic energy would increase, wouldn't it?\n\nOr does it mean that all the molecules inside this smaller box all have the same average kinetic energy? But this kinetic energy is still higher than the kinetic energy of the molecules on the bigger box, isn't it?\n\nThank you!",
        "A": "Me thinks the average K.E of molecules remains the same but the extra energy they gained by virtue of reduced volume is transferred as increase in heat of collision which is very minute.",
        "video_name": "tQcB9BLUoVI",
        "timestamps": [
            444
        ],
        "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": "In the 3rd example, while drawing the final products, why arent there any stereochemistry with the left carbon? (5:10 - 5:20)",
        "A": "There is no stereochemistry to worry about in the left carbon because two identical groups i.e. methyl groups are attached to the carbon. Hence it is NOT a chiral carbon.",
        "video_name": "b_qDLacdkFg",
        "timestamps": [
            310,
            320
        ],
        "3min_transcript": "That carbon, that carbon I just circled is a chiral center. So, we do need to worry about stereochemistry. Alright this carbon over here on the left is not a chiral center. So we don't have to worry about the carbon in red. We do have to worry about the carbon in magenta. So let's think about the stereochemistry with a syn addition Adding the H and the OH to the same sides. Alright so we need to think about this alkene here and these carbons that I've marked in yellow, alright those are sp2 hybridized carbons. So the geometry around those carbons is plainer, and we add our H and OH, right. We're turning those into sp3 hybridized carbons. Alright so let's go ahead and sketch in one of our products here. So one of the possible products would be to add the H and the OH on the same side. I'm gonna draw them as wedges. So here's our hydrogen right as a wedge, and then here's the OH as a wedge. because we already figured that out, right. We know the OH adds to this carbon, the one on the right. We know the H added to the one over here in red. Alright, so I'm showing a syn addition of my hydrogen, of my H and of my OH here and so those are wedges, and so let's think about what's attached to those carbons. So let me go back up to here. This is the carbon in magenta. That's this carbon down here. What else is attached to that carbon? Well there's a hydrogen. Alright, and therefore if I think about the stereochemistry that hydrogen must be going away from me now for this product. We're going from an sp2 hybridized carbon in our alkene to a an sp3 hybridized carbon. So that hydrogen is going away from us. Let's think about, let's think about this other carbon. Let me make this other carbon over here red. So this carbon right here. Alright, this carbon. What else is attached to this carbon? Well there's a methyl group. Alright, and if this hydrogen is coming out at me So we have a methyl group going away from us in space, and so that's one of our possible products. Alright, I could draw this out the way we're used to seeing it where we would show the OH coming out at us in space. Alright, the OH is coming out at us. So that's one of our possible products. Alright our other products, I could show the H and the OH adding on the same side but I can show them adding as dashes. Alright so I could draw in our carbons here, and I can show this time, I can show the H adding as a dash, and the OH adding as a dash. Now that means that this hydrogen right here would have to be a wedge, right. So that hydrogen would have to be a wedge. So I draw in the hydrogen here as a wedge, and then for the other carbon, the one in red, so this one. This methyl group would have to be coming out at us now."
    },
    {
        "Q": "AT 4:10, you said \"now we divide by the mass that's 3kg.\" But in your previous video while doing the chalkboard question you didn't divide it by the mass, even though it was given as 3Kg. Why is that so? Why you didn't divided it in the vertical acceleration while here you did?",
        "A": "He does divide by the mass in the chalkboard question, go back and take another look. The only difference is that he just writes the mass as m and doesn t plug in a value for it. In both problems, the next step is to multiply both sides by this mass. Since the left hand side is zero, and zero times anything is zero, the masses (in both problems) vanish anyway.",
        "video_name": "EEnzhdQJUYA",
        "timestamps": [
            250
        ],
        "3min_transcript": "It's hard to say, we've got forces vertical, we've got forces horizontal. There's only two directions to pick, X or Y in this problem. We're gonna pick the vertical direction, even though it doesn't really matter too much. But because we know one of the forces in the vertical direction, we know the force of gravity. Force of gravity is 30 Newtons. Usually that's a guod strategy, pick the direction that you know something about at least. So we're gonna do that here. We're gonna say that the acceleration vertically equals to the net force vertically over the mass. And so now we plug in. If this can is just sitting here, if there's no acceleration, if this is in not an elevator transporting these peppers up or down, and it's not in a rocket, if it's just sitting here with no acceleration, our acceleration will be zero. That's gonna equal the net force and the vertical direction. So what are we gonna have? So what are the forces in the vertical direction here? One force is this 30 Newton force of gravity. This points down, we're gonna assume upward is positive, that means down in a negative. I could have written -MG, but we already knew it was 30 Newtons, so I'll write -30 Newtons. Then we've got T1 and T2. Both of those point up. But they don't completely point up, they partially point up. So part of them points to the right, part of them points upward. Only this vertical component, we'll call it T1Y, is gonna get included into this calculation, 'cause this calculation only uses Y directed forces. And the reason is only Y directed forces, vertical forces, affect the vertical acceleration. So this T1Y points upward, I'll do plus T1 in the Y direction. And similarly, this T2. It doesn't all point vertically, only part of it points vertically. So I'll write this as T2 in the Y direction. And that's also upward, so since that's up, I'll count it as plus T2 in the Y direction. And that's it, that's all our forces. Notice we can't plug in the total amount T2 in this formula, 'cause only part of it points up. of the T1 force because only part of it points vertically. And then we divide by the mass, the mass is 3 kilograms. But we're gonna multiply both sides by 3 kilograms, and we're gonna get zero equals all of this right here, so I'll just copy this right here. We use this over again, that comes down right there. But now there's nothing on the bottom here. So what do we do at this point? Now you might think we're stuck. I mean, we've got two unknowns in here. I can't solve for either one, I don't know either one of these. I know they have to add up to 30, so I'd do fine, if I added 30 to both sides, I'd realize that these two vertical components of these tension forces added up have to add up to 30, and that makes sense. They have to balance the force downward. But I don't know either of them, so how do I solve here? Well, let's do this. If you ever get stuck on one of the force equations for a single direction, just go to the next equation. Let's try A in the X direction."
    },
    {
        "Q": "at 2:40 you said there was a primary carbon. Would'nt that be a methyl carbon because there is only one carbon?",
        "A": "primary carbon is carbon attached to only one carbon",
        "video_name": "_-I3HdmyYfE",
        "timestamps": [
            160
        ],
        "3min_transcript": "That gives this carbon a negative one formal charge, So we form a carbanion here, also called an alkynide anion for this portion. So this is an alkynide anion, a carbanion in here. It's a relatively stable conjugate base, because the electrons, these two electrons here, are housed in an SP hybridized orbital, which has a lot of S character to it, so it's relatively small. So those negatively charged electrons are held a little bit more closely to the positively charged nucleus of this carbon here. So that somewhat stabilizes the conjugate base, which is the reason why a terminal alkyne can function as an acid. So once we formed our alkynide anion, we can use that alkynide anion to do an alkylation reaction. So let's go ahead and redraw that alkynide anion here. So I'm going to draw this portion, this R And then we have our carbon triple bonded to another carbon, a negative charge And this can now function as a nucleophile. So a negatively charged anion can function as a nucleophile. And if we react this alkyline anion with an alkyl halide-- let's go ahead and draw an alkyl halide here. So I'm going to put hydrogen there, and I'll put my halogen over here on the right, so putting my lone pairs of electrons. Let's draw in one R group right here and then a hydrogen over here. So here is my alkyl halides. And if you react a strong nucleophile with an alkyl halide that is not very sterically hindered-- this is a primary alkyl halide right here-- you're going to get an SN2 reaction. So think about this being an SN2 reaction. We have an alkyl halide, which has a polarized bond between the carbon and the halogen. So if the halogen is more electronegative, it's going to pull the electrons and the bond between it and carbon closer to itself. So this halogen ends up being partially negative. This carbon, therefore, will be partially positive, like that. This carbon right here is partial positive charge. It wants electrons. Of course, our nucleophile has those electrons. So the lone pair of electrons on our carbon can attack our electrophile, so nucleophile attacks electrophile. And an SN2 mechanism, remember, is a concerted mechanism, meaning the nucleophile attacks the electrophile at the same time you're leaving group is leaving here. So these electrons are going to kick off onto your halogen. So let's go ahead and draw the product. So now we would have an R group, carbon triple bonded to another carbon. And now this is bonded to yet another carbon. So we formed a carbon carbon bond here in this reaction. And then this hydrogen is up here. This R group is still here, coming out at us, and then the hydrogen going away from us like that in space. And then we have our halogen over here with now four lone pairs of electrons, a negative 1 formal charge."
    },
    {
        "Q": "At around 5:25, David makes a destructive interference wave by moving one wave source a half a wavelength forward. Later he also notes that this placement of the two wave sources will produce no sound because the amplitudes cancel each other out. But what about that one half wavelength of the sound wave at the start, that wave does not get cancelled out right? So technically isn't there a sound being produced for that half of the wavelength?",
        "A": "The destructive interference occurs at a particular point in space, not everywhere at the same time.",
        "video_name": "oTjTXS40pqs",
        "timestamps": [
            325
        ],
        "3min_transcript": "Are there any applications of this? Well yeah. So imagine you're sitting on an airplane and you're listening to the annoying roar of the airplane engine in your ear. It's very loud and it might be annoying. You put on your noise canceling headphones, and what those noise canceling headphones do? They sit on your ear, they listen to the wave coming in. This is what they listen to. This sound wave coming in, and they cancel off that sound by sending in their own sound, but those headphones Pi shift the sound that's going into your ear. So they match that roar of the engine's frequency, but they send in a sound that's Pi shifted so that they cancel and your ear doesn't hear anything. Now it's often now completely silent. They're not perfect, but they work surprisingly well. They're essentially fighting fire with fire. They're fighting sound with more sound, and they rely on this idea of destructive interference. They're not perfectly, totally destructive, but the waves I've drawn here are totally destructive. If they were to perfectly cancel, we'd call that total destructive interference, And it happens because this wave we sent in was Pi shifted compared to what the first wave was. So let me show you something interesting if I get rid of all this. Let me clean up this mess. If I've got wave source one, let me get wave source two back. So this was the wave that was identical to wave source one. We overlap 'em, we get constructive interference because the peaks are lining up perfectly with the peaks, and these valleys or troughs are matching up perfectly with the other valleys or troughs. But as I move this wave source too forward, look at what happens. They start getting out of phase. When they're perfectly lined up we say they're in phase. They're starting to get out of phase, and look at when I move it forward enough what was a constructive situation, becomes destructive. Now all the peaks are lining up with the valleys, they would cancel each other out. And if I move it forward a little more, it lines up perfectly again and you get constructive, move it more I'm gonna get destructive. Keep doing this, I go from constructive to destructive over and over. So in other words, one way to get constructive interference and just put them right next to each other. And a way to get destructive is to take two wave sources that are Pi shifted out of phase, and put them right next to each other, and that'll give you destructive 'cause all the peaks match the valleys. But another way to get constructive or destructive is to start with two waves that are in phase, and make sure one wave gets moved forward compared to the other, but how far forward should we move these in order to get constructive and destructive? Well let's just test it out. When they're right next to each other we get constructive. If I move this second wave source that was initially in phase all the way to here, I get constructive again. How far did I move it? I moved it this far. The front of that speaker moved this far. So how far was that? Let me get rid of this. That was one wavelength. So look at this picture. From peak to peak is exactly one wavelength. We're assuming these waves have the same wavelength. So notice that essentially what we did, we made it so that the wave from wave source two"
    },
    {
        "Q": "What is that ocean creature called at 9:20?",
        "A": "It is an Opabinia. It lived during the Cambrian Era and had five eyes. It s not part of any living phylum today.",
        "video_name": "MS7x2hDEhrw",
        "timestamps": [
            560
        ],
        "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": "at 1:40, what does velocity mean?",
        "A": "Velocity is a measurement which includes a speed and a direction: (ie: 40 km/h south)",
        "video_name": "oRKxmXwLvUU",
        "timestamps": [
            100
        ],
        "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": "At 8:33, Sal mentions dimensional analysis. What is dimensional analysis?",
        "A": "Dimensional analysis is basically just a fancy name for unit conversion by canceling like dimensions...",
        "video_name": "oRKxmXwLvUU",
        "timestamps": [
            513
        ],
        "3min_transcript": "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 And then in the units, in the numerator, you have meters, and in the denominator, you have hours. Meters per hour. And so this is equal to 5,000 meters per hour. And you might say, hey, Sal, I know that 5 kilometers is the same thing as 5,000 meters. I could do that in my head. And you probably could. But this canceling out dimensions, or what's often called dimensional analysis, can get useful once you start doing really, really complicated things with less intuitive units than something But you should always do an intuitive gut check right here. You know that if you do 5 kilometers in an hour, that's a ton of meters. So you should get a larger number if you're talking about meters per hour. And now when we want to go to seconds, let's do an intuitive gut check. If something is traveling a certain amount in an hour, it should travel a much smaller amount in a second, 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."
    },
    {
        "Q": "Does anyone have any idea as to why the letter s is used to represent displacement instead of d? (at time 1:51)\nMy best guess is that s stands for Science... Not or that it is used to be the SUM of the displacement vectors...",
        "A": "d is delta in calculus",
        "video_name": "oRKxmXwLvUU",
        "timestamps": [
            111
        ],
        "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": "At 0:38 he said that 5km was a magnitude, but then at 0:48 he said that 5km was the distance. Which it is?",
        "A": "Both, the distance is the magnitude.",
        "video_name": "oRKxmXwLvUU",
        "timestamps": [
            38,
            48
        ],
        "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": "At 1:20 does the arrow over the V indicate *which* direction, or is it just to show that it has a direction at all?",
        "A": "It is to indicate that the value represented by the letter is a vector, not a scalar. I.e. to show that it has a direction at all.",
        "video_name": "oRKxmXwLvUU",
        "timestamps": [
            80
        ],
        "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": "At 1:50, Sal says that the symbol for displacement is S, but I have been learning in my science class that displacement is X. Is one right and one wrong, or are there just other variables?",
        "A": "Both x and s are used for representing displacement",
        "video_name": "oRKxmXwLvUU",
        "timestamps": [
            110
        ],
        "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": "What is the sign Sal uses at 4:46 to show a partial positive or negative charge?",
        "A": "The symbol is a a lower case delta . Delta (a triangle, related to the letter D) is used for change, so a lower case delta means a small (partial) change.",
        "video_name": "Rr7LhdSKMxY",
        "timestamps": [
            286
        ],
        "3min_transcript": "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, around the atoms that make up the atoms. And so, in this type of a covalent bond, the electrons, the two electrons that this bond represents, are going to spend more time around the oxygen then they are going to spend around the hydrogen. And these, these two electrons are gonna spend more time around the oxygen, then are going to spend around the hydrogen. And we know that because oxygen is more electronegative, and we'll talk about the trends in a second. This is a really important idea in chemistry, and especially later on as you study organic chemistry. Because, because we know that oxygen is more electronegative, and the electrons spend more time around oxygen then around hydrogen, it creates a partial negative charge on this side, and partial positive charges on this side right over here, which is why water has many of the properties that it does, and we go into much more in depth in that in other videos. And also when you study organic chemistry, a lot of the likely reactions that are or a lot of the likely molecules that form can be predicted based on elecronegativity. And especially when you start going into oxidation numbers and things like that, electronegativity will tell you a lot. So now that we know what electronegativity is, let's think a little bit about what is, as we go through, as we start, as we go through, as we go through a period, as say as we start in group one, and we go to group, and as we go all the way all the way to, let's say the halogens, all the way up to the yellow column right over here, what do you think is going to be the trend for electronegativity? And once again, one way to think about it is to think about the extremes. Think about sodium, and think about chlorine, and I encourage you to pause the video and think about that. Assuming you've had a go at it, and it's in some ways the same idea,"
    },
    {
        "Q": "At 6:45, does this mean that the elements in groups 1 & 2 are more likely to be in ionic bonds than covalent bonds because it's easier to just give an electron and be positively charged and bonded or does it not matter?",
        "A": "That is exactly what it means",
        "video_name": "Rr7LhdSKMxY",
        "timestamps": [
            405
        ],
        "3min_transcript": "or a lot of the likely molecules that form can be predicted based on elecronegativity. And especially when you start going into oxidation numbers and things like that, electronegativity will tell you a lot. So now that we know what electronegativity is, let's think a little bit about what is, as we go through, as we start, as we go through, as we go through a period, as say as we start in group one, and we go to group, and as we go all the way all the way to, let's say the halogens, all the way up to the yellow column right over here, what do you think is going to be the trend for electronegativity? And once again, one way to think about it is to think about the extremes. Think about sodium, and think about chlorine, and I encourage you to pause the video and think about that. Assuming you've had a go at it, and it's in some ways the same idea, Something like sodium has only one electron in it's outer most shell. It'd be hard for it to complete that shell, and so to get to a stable state it's much easier for it to give away that one electron that it has, so it can get to a stable configuration like neon. So this one really wants to give away an electron. And we saw in the video on ionization energy, that's why this has a low ionization energy, it doesn't take much energy, in a gaseous state, to remove an electron from sodium. But chlorine is the opposite. It's only one away from completing it's shell. The last thing it wants to do is give away electron, it wants an electron really, really, really, really badly so it can get to a configuration of argon, so it can complete it's third shell. So the logic here is that sodium wouldn't mind giving away an electron, while chlorine really would love an electron. So chlorine is more likely to hog electrons, while sodium is very unlikely to hog electrons. So this trend right here, when you go from the left to the right, your getting more electronegative. More electro, electronegative, as you, as you go to the right. Now what do you think the trend is going to be as you go down, as you go down in a group? What do you think the trend is going to be as you go down? Well I'll give you a hint. Think about, think about atomic radii, and given that, what do you think the trend is? Are we gonna get more or less electronegative as we move down? So once again I'm assuming you've given a go at it, so as we know, from the video on atomic radii, our atom is getting larger, and larger, and larger, as we add more and more and more shells. And so cesium has one electron in it's outer most shell, in the sixth shell, while, say, lithium has one electron. Everything here, all the group one elements, have one electron in it's outer most shell, but that fifty fifth electron,"
    },
    {
        "Q": "what i the sign made by Sal at 4:53 to denote the partial positivity ?",
        "A": "It s a lowercase delta (Greek letter): \u00ce\u00b4",
        "video_name": "Rr7LhdSKMxY",
        "timestamps": [
            293
        ],
        "3min_transcript": "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, around the atoms that make up the atoms. And so, in this type of a covalent bond, the electrons, the two electrons that this bond represents, are going to spend more time around the oxygen then they are going to spend around the hydrogen. And these, these two electrons are gonna spend more time around the oxygen, then are going to spend around the hydrogen. And we know that because oxygen is more electronegative, and we'll talk about the trends in a second. This is a really important idea in chemistry, and especially later on as you study organic chemistry. Because, because we know that oxygen is more electronegative, and the electrons spend more time around oxygen then around hydrogen, it creates a partial negative charge on this side, and partial positive charges on this side right over here, which is why water has many of the properties that it does, and we go into much more in depth in that in other videos. And also when you study organic chemistry, a lot of the likely reactions that are or a lot of the likely molecules that form can be predicted based on elecronegativity. And especially when you start going into oxidation numbers and things like that, electronegativity will tell you a lot. So now that we know what electronegativity is, let's think a little bit about what is, as we go through, as we start, as we go through, as we go through a period, as say as we start in group one, and we go to group, and as we go all the way all the way to, let's say the halogens, all the way up to the yellow column right over here, what do you think is going to be the trend for electronegativity? And once again, one way to think about it is to think about the extremes. Think about sodium, and think about chlorine, and I encourage you to pause the video and think about that. Assuming you've had a go at it, and it's in some ways the same idea,"
    },
    {
        "Q": "At 13:13, he says that \"x\" represents the concentration of Hydronium ion [H3O]. But the -log of x was pH. I don't understand why. Can someone please explain this for me?\nThanks :3",
        "A": "To you figure out the pH of any substance, the formula says: pH = -log([H3O+]) , or pH = -log([H+]) If x = [H3O+], so pH = -log(x); It s all the same thing. Sorry if my english ain t so good, I am brazillian. I hope it helped you.",
        "video_name": "XZWoMXVANww",
        "timestamps": [
            793
        ],
        "3min_transcript": "So, NH4+ and NH3 are a conjugate acid-base pair. We're trying to find the Ka for NH4+ And again, that's not usually found in most text books, but the Kb value for NH3, is. It's: 1.8 times 10 to the negative five. So for a conjugate acid-base pair, Ka times Kb is equal to Kw. We're trying to find Ka. We know Kb is 1.8 x 10-5 This is equal to: 1.0 times 10 to the negative 14. So we can once again find Ka on our calculator. So, 1.0 x 10-14... We divide that by 1.8 x 10-5 So if we get some room down here, we say: Ka = 5.6 x 10-10 This is equal to: so it'd be X squared over here... And once again, we're going to assume that X is much, much smaller than .050 So we don't have to worry about X right here, but it's an extremely small number, .050 - X is pretty much the same as .050 So we plug this in and we have: .050, here. So we need to solve for X. We get out the calculator, and we're going to take 5.6 x 10-10, and we're going to multiply by .05 and then we're gonna take the square root of that to get us what X is. So: X = 5.3 x 10-6 X represents the concentration of hydronium ions, so this is a concentration, right? This is the concentration of hydronium ions, so to find the pH, all we have to do is take the negative log of that. So, the pH is equal to the negative log of the concentration of hydronium ions. So we can just plug that into here: 5.3 x 10-6, and we can solve; and let's take the - log(5.3 x 10-6) And so we get: 5.28, if we round up, here. So let me get a little more room... So we're rounding up to 5.28 for our final pH."
    },
    {
        "Q": "hey but in case of CH3OH ( at 5:00 ) the total no of valence electrons dont match with the no of bonds as shown at 6:29 ..??",
        "A": "I think you should also count the dots. CH3OH has a total number of 14 valence electrons. At 6:29, the structure shown has a total of 5 bonds, and 4 dots on the Oxygen. It means that the structure has 14 valence electrons and is correct.",
        "video_name": "BIZNBfBuu1w",
        "timestamps": [
            300,
            389
        ],
        "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": "I noticed at 2:40 he writes the formula for Methylamine as CH_3NH_2 as opposed to CH_5N, does the expression of the formula vary based on context?",
        "A": "There can be variations in a general formula of an organic compound. This happens due to a phenomenon which we have named as isomerism. Simply put, there can be various structures of a general formula if not specified.",
        "video_name": "BIZNBfBuu1w",
        "timestamps": [
            160
        ],
        "3min_transcript": "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, 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."
    },
    {
        "Q": "When drawing the diagrams, do you choose the element that has the least number of atoms in that bond for the central atom? Secondly, how do you know how to place the atoms in larger bonds (ex. CH3 NH2 at 2:45) (how to configure the diagram)?",
        "A": "H atoms must always be external atoms. That leaves C and N as candidates for the central atom. The central atom is the *least electronegative atom:* C. So, C is the central atom. Attach N to the C. Then add H atoms and electrons to give each atom its octet.",
        "video_name": "BIZNBfBuu1w",
        "timestamps": [
            165
        ],
        "3min_transcript": "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, 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."
    },
    {
        "Q": "At 9:10, it states that there would be a high potential energy assuming that q and Q are positive. Is the same true for when both charges are negative? What if q is positive, and Q is negative (vise versa as well)?",
        "A": "The formula is the same and always works, as long as you put all the signs in correctly. So if Q is positive, it creates a positive electric potential V. If Q is neg, it creates a neg electric potential V. But then if you put a positive q in a positive V, it will have positive potential ENERGY (PE = qV). Same for a neg q near a neg Q (thus a neg V). But a positive q in a neg V or a neg q in a positive V will both have NEG potential energy.",
        "video_name": "ks1B1_umFk8",
        "timestamps": [
            550
        ],
        "3min_transcript": "when you're doing physics with constants. Look at this is in terms of meters, so I've got to use meters here, so nine centimeters is .09 meters. And if I multiply all this out what you'll get is, 10 to the negative ninth cancels this 10 to the ninth, the powers are 10, these just go away, and then I have nine divided by .09, that gets equal 100. So I chose this so that we got the same answer down there. Okay, 100 Joules per Coulomb, you might be like, where the Joules comes from? And how is this Joules per Coulomb? Well let's look at it, if we took, look at, one of these meters cancels one of these meters, and one of these Coulombs cancels one of those Coulombs, what are we left with? We're left with Newton times meter over Coulomb, but Newton times meter, that's force times distance, that's Joules, that's where we get Joules per Coulomb. So this really does give us the number of Joules 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,"
    },
    {
        "Q": "At 2:37, how can we even predict that the Eagle Nebula doesn't exist right now if the photons we are seeing currently depict it as it was 7000 years ago, and nothing travel faster than light, i.e., these photons? Shouldn't the 7000 year old photons be our most recent image of this nebula? How could we know of something that happened after this point in time if light can't reach us that fast?",
        "A": "Actually, right now we can see a hint of a destructive energy burst (probably from a nearby supernova) going towards the Eagle Nebula. And what we are seeing IS the Eagle Nebula as it was 7,000 years ago. So actually, the Eagle Nebula as we know it does not exist today. It is either gone or completely disfigured.",
        "video_name": "w3IKEa_GOYs",
        "timestamps": [
            157
        ],
        "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": "At 7:31 in the video, Sal stated that the chain goes from 5' to 3'. Is it possible for it to go from 3' to 5'? If not, why?",
        "A": "DNA polymerase only works in the 5 to 3 direction. This is down to the specificity of this enzyme. If you think about it, DNA replication could result in a real mess if DNA polymerase could work in both directions.",
        "video_name": "0CQ5ls3Uc2Q",
        "timestamps": [
            451
        ],
        "3min_transcript": "One prime. Two prime. Three prime. Four prime. Five prime. And then when you look at it as a ring, this was the one prime. This is the two prime. This is the three prime. This is the four prime. This is the five prime. Or if you were to number them on this diagram right over here, actually in the DNA molecule, this is the one prime. This is the two prime carbon. This is the three prime carbon. This is the four prime carbon. And this is the five prime carbon. And so one way to think about it is we'll go phosphate group and it's connected with what we call phosphodiester linkages. Phosphodiester linkages, that's what's essentially allowing these backbones to link up. But we're going from phosphate to five prime carbon and then through the sugar we go to the three prime carbon. And then we go to another phosphate. Then we go to the five prime carbon. Let me label that, this is the five prime carbon. And that just comes straight out of just numbering these starting with the carbon that was a number one carbon. When it's straight chain form, it's part of the carbonyl group. But you see we're going from five. We go phosphate, five prime, three prime, phosphate, five prime, three prime, phosphate. So one way to describe the orientation is saying, \"Hey, we're going in the direction \"from five prime to three prime.\" So we could say that we're going from five prime to three prime, that way on the left-hand chain. And what are we doing on the right hand chain? Well, let's number them again. So this is the one prime carbon. Now this thing relative to this is upside down, it's inverted. So one prime. Two prime. Three prime. Four prime. Five prime. I could do it up here. One prime carbon. Two prime carbon. Three prime carbon. Four prime carbon. Five prime carbon. Here you're going from phosphate, three prime, five prime, phosphate, So the way that the sugars are oriented if you're going from top to bottom you're going from three prime to five prime. So on the right hand side, it's three prime, five prime. And so if you wanted to draw an arrow from five prime to three prime, you could look at it like that. And so you could say these are parallel but since they are essentially pointing in different directions even though they are actually parallel, we would call this structure of DNA antiparallel. So this would be an anti... antiparallel structure of DNA. So these two strands, they're complementary. They're defined by each other. The thymine bonds with the adenine, the cytosine bonds with guanine. They are attracted to each other through these hydrogen bonds. But the two backbones, they're pointed in different directions. And now another interesting thing to think about, since we're talking about the molecular structure of DNA, is how do these things form? How did these things know to orient in this way?"
    },
    {
        "Q": "At 1:53, Why is the fact that nitrogenous bases are forming hydrogen bonds offset their basic property? Somehow making them \"happy\" or something?",
        "A": "as he said in 1:35, it is done in order to take more hydrogen protons",
        "video_name": "0CQ5ls3Uc2Q",
        "timestamps": [
            113
        ],
        "3min_transcript": "- [Voiceover] In the video on the molecular structure of DNA we saw that DNA is typically made up of two strands where the backbone of each of the strands is made up of phosphate alternating between a-- Do some different colors. A phosphate group and then you have a sugar. You have a phosphate group. And then you have a sugar. And then you have a phosphate group. And then you have a sugar. And so I could draw the strand something like this. So phosphate and then we have a sugar. Oops, let me just draw all the phosphates ahead of time. So you have the phosphates on that end and then you have the sugars. And you see the same thing on the other strand as well. Where we have phosphate with a sugar then another phosphate then a sugar then another phosphate. Let me circle the sugars as well. and then you have the sugar there as well. So on the other strand it's also going to look like this. So let me draw the phosphates. I'm just abstracting them now. So the phosphate and then you have the sugars in between the phosphates. And what links them, you can think of them as the rungs on the ladder. These are the complementary nitrogenous bases. And the reason why we call them nitrogenous bases, I actually forgot to talk about it in the last videos, is that these nitrogens are really electronegative and they can take up more hydrogen protons. They have an extra lone pair. The nitrogens have an extra lone pair that can be used up under the right conditions to potentially sop up more hydrogen protons. Now, a lot of people ask, \"Well, if you have these nitrogenous bases here, \"why is DNA called an acid?\" Why is it called an acid?\" Well the first thing is that the basic properties of the nitrogenous base are offset to a good degree based on the fact that they're able to hydrogen bond with each other. And that's what actually forms when these complimentary nitrogenous bases form these hydrogen bonds with each other. But even more, the reason why we call it an acid is the phosphate groups, when they're protonated, are acids. Now the reason why we tend to draw them deprotonated is they're so acidic that if you put them in a neutral solution, they're going to be deprotonated. So this is the form that you're more likely to find it in the nucleus of an actual cell. Once it's actually already deprotonated. But in general, phosphate groups are considered acidic. And if I were to draw kind of a more pure phosphate group, and I talked about this already in the last video, I would have it protonated and so I wouldn't draw that negative charge like that. So that's just a review of last time. Since I already started abstracting it, let's abstract further. So let's draw the nitrogenous bases a little bit. So I have thymine here. And I will do thymine in this green color."
    },
    {
        "Q": "At 1:32, Sal says that \"as we get more into physics, we'll see that maybe we shouldn't necessarily think of time as driving; maybe position, in some ways, is driving time.\" What is Sal referring to? Is there a relevant Khan Academy video or Wikipedia page?",
        "A": "Spacetime diagrams when you study relativity have time on the vertical axis. In relativity you cannot untangle space from time. if you change position very quickly (travel near the speed of light, for example), you move through time more slowly.",
        "video_name": "PRx_R9iIWk4",
        "timestamps": [
            92
        ],
        "3min_transcript": "- [Voiceover] Let's say this is me right over here, and I'm drifting through space at a constant velocity relative to any other inertial frame of reference, and so I am, I am in an inertial frame of reference myself, and in fact I'm going to define my frame of reference by me, I'm gonna say I'm at the origin of my frame of reference. So at all times, I consider myself to be stationary, and I am at the point, 'x' equals zero, and we're gonna focus on just the 'x' dimension, to simplify our discussion, and I have, you know, my oxygen and everything, and food, so no need to worry about me. Now, what I've drawn here are some axes, so that I can plot the path of things as time progresses in my frame of reference. And one thing that many of ya'll might have noticed is that I have plotted time in seconds on the vertical axis, and our position, our 'x' position, in meters on the horizontal axis. And that might be a little bit counter-intuitive especially with math backgrounds, in fact, it's a little bit uncomfortable for me, we often prefer to put time on the horizontal axis, and position on the vertical axis, so this is a little bit counter-intuitive. But this is, you know, our choice is a little bit arbitrary, in math class, we liked to think of our independent variable on our horizontal axis, and we think of time as somehow driving position, so that's why we tend to put time there than we'd position there, but you can flip them around. And as we'll get more into physics, we'll see, well maybe we shouldn't necessarily think of time always driving position, or maybe position in some ways are driving time. But we'll get into that. But let's just first get a little bit comfortable with this. So what would be my position over time on this diagram right over here? And you might notice these numbers, one second, two second, three second, and then in meters I have three times to the eighth, six times to the eighth, nine times to the eighth, so these are massive, massive numbers, and you could guess where they are coming from. that the speed of light is approximately 3x10 to the eighth meters per second. But we'll get to that in a second. But let's just think about my position over time, in my frame of reference here. So, time equals zero, my 'x' position in this frame of reference is zero, I consider myself to be stationary. After one second, well my position is still zero, after two seconds my position is still 'x' equals zero, after three seconds my position is still 'x' equals zero, so my, I guess you could almost consider my path on this diagram, where I'm plotting time and space, at least in the 'x' direction, is going to look like, is going to look like, whoops, I can do a better job than that. It's going to look, it's going to look like, it's going to look like that. That would be my path on my little time and space axes,"
    },
    {
        "Q": "at 9:00 it is written vt+1/2atsquare .and delta t is a very small quantityand delta t multilied by delta t gives a much more small quantity and should not be taken into consideration.then why it is writtem as t square",
        "A": "The t in that equation is not the delta t of calculus, it s the time during which the acceleration took place.",
        "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": "At 6:05 wouldn't the final velocity be zero after it hits the ground?",
        "A": "Final velocity refers to the speed it has when it first hits the ground, not after it lands. It would not be very useful or interesting to say that everything has a final velocity of 0.",
        "video_name": "wlB0x9W-qBU",
        "timestamps": [
            365
        ],
        "3min_transcript": "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? your initial velocity plus your acceleration times change in time. If you are starting at 10m/s and you are accelerated at 1m/s^2 then after 1 second you will be going 1 second faster than that. (11m/s) So this right here is your final velocity. Let me make sure that these are all vector quantities...(draws vector arrows) All of these are vector quantities. Hopefully it is ingrained in you that these are all vector quantities, direction matters. And let's see how we can simplify this Well these two terms (remember we are just dealing with the average velocity here) These two terms if you combine them become 2 times initial velocity (2vi). two times my initial velocity and then divided by this 2 plus all of this business divided by this 2."
    },
    {
        "Q": "7:11 I'm confused. You are shifting the numbers around, assumably to find the specific variable, but you boxed in blue the Vi+Vf( which has been moved to another part of the equation, extended and broken down in yellow) into a translation that read 2Vi, which is just Vi+Vi, right? Am I missing something, or how did Vf become a second Vi, and where did it come from?",
        "A": "Sal re-wrote Vf, which he brackets at 6:25, as initial velocity plus acceleration times change in time. He then collects the two Vi s together (2Vi) and divides them by two, and also divides the acceleration times change in time by two. (Remember, we re just re-writing the V avg. part of the S = V avg. x delta T equation so it s in terms of initial velocity.) Hope that s clear :)",
        "video_name": "wlB0x9W-qBU",
        "timestamps": [
            431
        ],
        "3min_transcript": "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? your initial velocity plus your acceleration times change in time. If you are starting at 10m/s and you are accelerated at 1m/s^2 then after 1 second you will be going 1 second faster than that. (11m/s) So this right here is your final velocity. Let me make sure that these are all vector quantities...(draws vector arrows) All of these are vector quantities. Hopefully it is ingrained in you that these are all vector quantities, direction matters. And let's see how we can simplify this Well these two terms (remember we are just dealing with the average velocity here) These two terms if you combine them become 2 times initial velocity (2vi). two times my initial velocity and then divided by this 2 plus all of this business divided by this 2. 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"
    },
    {
        "Q": "At 3:52 Sal said that if something is moving up, it's given a positive sign, but if it's moving down, it's given a negative sign. What if it's being affected by the moon's gravity? Then, it's moving up relative to earth, but down relative to the moon.",
        "A": "positive being up and negative being down is just a conventional way of doing things. We can choose the y-axis to point in any direction, so long as it is perpendicular to the x-axis. It just seems natural to choose y to point upwards. So when the Earth is pulling something down, we give that a negative value. And using that same convention, then the moon above us would pull things up with a positive value.",
        "video_name": "wlB0x9W-qBU",
        "timestamps": [
            232
        ],
        "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": "At 4:03, Sal stated that a vector is positive if an object goes up and negative if it goes down. What about left and right?",
        "A": "It can be whatever way you want, just make sure you stick to it for the whole problem. Personally if I m working on a problem and I have a lot of vectors going in a certain direction, I consider that direction positive and the opposite direction negative. Sometime this is right, sometimes it s left.",
        "video_name": "wlB0x9W-qBU",
        "timestamps": [
            243
        ],
        "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": "In my physics class we use the equation x = xi + (vi)(\u00ce\u0094t) + 1/2 a(\u00ce\u0094t)^2\nBut at the end of the video (by 9:27) I see Sal using the equation x = (vi)(\u00ce\u0094t) + 1/2 a(\u00ce\u0094t)^2\nWhy is the equation used in my class different?",
        "A": "xi is just a measure of where you are starting out. Sal s is assigning the starting point a coordinate of zero. But if you decided to say you were starting from, say 10 feet, then you would have to add that 10 feet if you wanted to calculate your distance from zero.",
        "video_name": "wlB0x9W-qBU",
        "timestamps": [
            567
        ],
        "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": "stupid question, at 8:35 sal multiplies change in time by change in time but the first change in time is placed over 2...so why is it not half change in time multiplied by change in time? does that make sense?",
        "A": "That does make sense, but that s because they will both give them same answer. Half of change in time multiplied by time written as (1/2 t)*t will give you the same answer as half of change in time squared ( (1/2 t^2) ). It s just easier to remember 1/2 at^2. You could test that by putting some numbers in for time. example if t = 5: 1/2 * 5 = 5/2 5/2 * 5 = 25/2 or 12.5 5^2 = 25 1/2 * 25 = 25/2 or 12.5",
        "video_name": "wlB0x9W-qBU",
        "timestamps": [
            515
        ],
        "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": "Sal shows that a=g (where a is a vector quantity displayed at (1:30)). My question is, would gravity ALSO be a vector quantity? If so, why isn't there an arrow placed above it?\n\nThanks :)",
        "A": "Every force is a vector. If you are talking about a force of gravity, then it is a vector.",
        "video_name": "wlB0x9W-qBU",
        "timestamps": [
            90
        ],
        "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": "@ 9:20... Shouldn't the drawing only have one methyl group branching off the 1st carbon on each of the ehtyl groups branching off of first and third carbons of the ring? Doesn't the \"di\" in 1,1-dimethylethyl imply that there are only two methyl groups (instead of four like is shown in your drawing)? Such as how you only have two methyl groups in 2,2-dimethyl-hexane. Or is it different because it is an alkyl group branching off another alkyl group instead of the parent chain?",
        "A": "It s sort of like an order of operations problem from arithmetic. You have to pay attention to the parenthesis and consider everything within them for each case. Solve 1,1-dimethylethyl first, then put that in place at 1 and 3 on the cyclopentane.",
        "video_name": "6BR0Q5e74bs",
        "timestamps": [
            560
        ],
        "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": "Shouldnt the 3,6,9,9-tetramethyldodecane be named as a 4,4,7,10-tetramathyldodecane\n\n06:37",
        "A": "it could be 1,4,7,7-tetramethyldodecane",
        "video_name": "6BR0Q5e74bs",
        "timestamps": [
            397
        ],
        "3min_transcript": "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, nine, so at the three, the six and twice at the nine spot, we have methyl groups, and we have four methyl groups. That's all the tetramethyl is saying so it's We know we have four of them here: 3,6,9,9. We have methyl groups at each of those places. We have one methyl group at three, and then that is bonded with the third carbon on the dodecane chain. We have one at six bonded to the six carbon on the dodecane chain. We have two at nine, so that's one at nine and then we have another one at nine bonded to the nine carbon on the dodecane chain. That's it. That's 3,6,9,9- tetramethyldodecane. Let's do another one. 1,3-bis(1,1-dime thylethyl)cyclopentane. So once again, just kind of breathe slowly. 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"
    },
    {
        "Q": "hey i think there is a mistake at 5:26 in the naming of this molecule the sum of the methyl prefix is 3+6+9+9 =27 but if we named the molecule 4,4,7,10-Tetremethylcyclododecane the actual sum is equal to 25 or am i wrong ?",
        "A": "You don t sum them, that isn t the rule despite what your teacher may have said. You use the set of numbers that has a lower number at the first point of difference. 3 is less than 4 so the video is correct.",
        "video_name": "6BR0Q5e74bs",
        "timestamps": [
            326
        ],
        "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 6:38, Wouldn't that be called 3,3,6,9 instead of 3,6,9,9? I mean shouldn't we start where there is more branching at a smaller no.?",
        "A": "I know this question was asked 2 years ago but for anyone reding though in the future: No. Count again from the right hand side and you will see that would put the methyls on carbon #4 not carbon #3. 3 is a lower number than 4, the name given in the video is correct.",
        "video_name": "6BR0Q5e74bs",
        "timestamps": [
            398
        ],
        "3min_transcript": "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, nine, so at the three, the six and twice at the nine spot, we have methyl groups, and we have four methyl groups. That's all the tetramethyl is saying so it's We know we have four of them here: 3,6,9,9. We have methyl groups at each of those places. We have one methyl group at three, and then that is bonded with the third carbon on the dodecane chain. We have one at six bonded to the six carbon on the dodecane chain. We have two at nine, so that's one at nine and then we have another one at nine bonded to the nine carbon on the dodecane chain. That's it. That's 3,6,9,9- tetramethyldodecane. Let's do another one. 1,3-bis(1,1-dime thylethyl)cyclopentane. So once again, just kind of breathe slowly. 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"
    },
    {
        "Q": "At 5:36, if the penny was thrown up at a velocity of 30 m/s, then it would go up, reach a velocity of 0m/s and by the time it comes down to the same point from which it was thrown, wouldn't the velocity at that point then be different than 30 m/s?",
        "A": "It would be -30 m/s, since velocity is a vector. The speed at the bottom would be the same on the way down as on the way up. It has to be, because the acceleration is constant the whole time, and the distance up is the same as the distance down. (ignoring air resistance)",
        "video_name": "emdHj6WodLw",
        "timestamps": [
            336
        ],
        "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": "What is beta- plus decay ?\nIt was spoken by sal at 1:38 .",
        "A": "Beta decay is a kind of radioactive decay, which occurs so that the nucleus becomes more stable. In beta- decay, a neutron emits an electron to become a proton, so that the nucleus becomes more stable. In beta+ decay, also known as positron emission, a proton emits a positron to become a neutron, to make the nucleus more stable.",
        "video_name": "FEF6PxWOvsk",
        "timestamps": [
            98
        ],
        "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 2:47 sal said anti electron neutrino, well i don't understand what neutrino is?",
        "A": "Neutrinos are subatomic particles produced by the decay of radioactive elements and are elementary particles that lack an electric charge.",
        "video_name": "FEF6PxWOvsk",
        "timestamps": [
            167
        ],
        "3min_transcript": "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 associate with the other forces. Now, the next strongest force-- and just to give a sense of how weak gravity is even relative to the weak interaction, the weak interaction is 10 to the 25th times the strength of gravity. And you might be saying, if this is so strong, how come this does it operate on planets or us relative to the Earth? Why doesn't this apply to intergalactic distances the way gravity does? And the reason is the weak interaction really applies to very small distances, very, very small distances. So it can be much stronger than gravity, but only over very, very-- and it really only applies on the subatomic scale. You go anything beyond that, it kind of disappears as an actual force, as an actual interaction."
    },
    {
        "Q": "At 6:03, how do you know which block it is? (ex. D block)",
        "A": "Groups 1-2 are the s block Groups 13-18 are the p block Groups 3-12 are the d block The 2 rows under the main periodic table are the f block The block generally (not always) tells you which orbitals electrons are being filled in to",
        "video_name": "UXOcWAfBdZg",
        "timestamps": [
            363
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "At 7:57, how are there 18 noble gases.\nAren't they just 6(+1)?!",
        "A": "Sal says So the noble gasses, that s the other name for the group 18 elements, noble gasses , not that there are 18 noble gasses. Groups in the periodic table are the vertical columns that are numbered from 1-18.",
        "video_name": "UXOcWAfBdZg",
        "timestamps": [
            477
        ],
        "3min_transcript": "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. of not being reactive. Why don't they react? Because they have eight valence electrons. They have filled their outermost shell. They don't find the need. They're noble. They're kind of above the fray. They don't find the need to have to react with anyone else."
    },
    {
        "Q": "At around 5:30, why would it be 4s2?",
        "A": "b coz 4s has lower energy than 3d as it is inner orbital more close to the nucleus hence it must get fille first",
        "video_name": "UXOcWAfBdZg",
        "timestamps": [
            330
        ],
        "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": "At 5:35, Sc would have 2, not 3 valence electrons? I'm curious as to why you would not add the 2 and 1 together, since you do that with Carbon (and you add the 2s together to get 4 valence electrons).",
        "A": "You do not add the 2 and 1 together because the 1 electron is part of the d block and therefore cannot be a valence electron. The d blocks cannot be used as a valence electron because they are not one of the highest energy furthest out electrons, but p blocks can be used as valence electrons. With carbon you get 4 valence electrons because you add the s and p blocks together.",
        "video_name": "UXOcWAfBdZg",
        "timestamps": [
            335
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "at 1:01 Sal marks the 1st column but he doesn't mark HYDROGEN . Can some tell me the reason for this ?",
        "A": "This is because H is not an alkali metal and it doesn t share all the properties of the alkali metals. It doesn t perfectly fit anywhere in the periodic table but it s usually put in Group 1.",
        "video_name": "UXOcWAfBdZg",
        "timestamps": [
            61
        ],
        "3min_transcript": "- Let's talk a little bit about groups of the periodic table. Now, a very simple way to think about groups is that they just are the columns of the periodic table, and the standard convention is to number them. This is the first column, so that's group one, second column, third group, fourth, fifth, sixth, seventh, eighth, group nine, group 10, 11, 12, 13, 14 15, 16, 17, and 18. As some of ya'll might be thinking, what about these F block elements over here? If we were to properly do the periodic table we would shift all of these that everything from the D block and P block all are right words and make room for these F block elements, but the convention is is that we don't number them. But what's interesting? Why do we go to the trouble about calling one of these columns, about calling these columns a group? This is what's interesting about the periodic table is that all of the elements in a column, 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."
    },
    {
        "Q": "At 6:59, Sal says that Carbon has four valence electrons, but the subscript for the p shell is only 2. I'm confused as to why Sal added on the electrons in the 2s shell to the 2p.",
        "A": "Because the 2s electrons are also carbon s valence electrons. For main group elements it will be the electrons in the highest energy shell, both 2s and 2p are in carbon s highest energy shell.",
        "video_name": "UXOcWAfBdZg",
        "timestamps": [
            419
        ],
        "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": "At 6:31, for the cycle pentane, shouldn't there be 12 hydrogens, not 10 (because of the alkane formula Cn H2n+2)?",
        "A": "For a normal pentane it would be 12 hydrogens, like: CH3CH2CH2CH2CH3 if you would now want to make a ring out of this ordinary pentane, you would have to bond the two CH3 s together. In order to do so, both would hove to lose a bond first, which are both bonds to hydrogen atoms. So that leaves 10 hydrogens for cyclic pentane.",
        "video_name": "NRFPvLp3r3g",
        "timestamps": [
            391
        ],
        "3min_transcript": "be pentane. But if I have five carbons and they form a ring, so let me draw it. So it's one, two, three, four, five carbons and it forms a ring. Let me make the drawing a little bit better. So it's really, I'm just drawing a pentagon. But notice, this has five carbons on it. I can draw the carbons here. Carbon, carbon, carbon, carbon, carbon. And just as a review, what you don't see is the hydrogens Each of these guys have two bonds, so they must have two bonds with something else and those are going to be with And I'lll draw it here just as a bit of a review, but you notice very quickly, the drawing gets extremely messy when you draw the two hydrogens on each of these carbons. So it's a little bit over-- maybe I shouldn't be doing that. So it becomes very messy when you draw the hydrogens, so it's better to just assume that they're there. If we don't draw all four bonds of the carbon, the other two bonds are going to be with hydrogen. So here, you might say, OK, this is an alkane, because I All of these are single bonds with the carbon. I have five carbons, so you might say this is pentane, but you have to think about one more thing. It's in a ring, so we add the prefix cyclo- to it. So this is, because it's a ring, we write cyclopentane. So let me just break that apart. This tells us that we're dealing with a ring. You see that this is a ring right there. This tells us that we're dealing with five carbons, and then this tells us right here, the -ane part, that tells us that they are all single bonds. All carbon-carbon single bonds. No double or triple bonds. All single bonds. Let's start with the word and let's see if we can figure out what the actual structure would look like. So what is this telling me? This tells me I'm dealing with a ring. That is a ring. It's going to have a ring structure. It's going to have nine carbons, nine C's, and then it's an alkane, so they're all going to be single bonds. So if I want to draw it, I want to draw nine carbons in a ring, it's not a trivial thing to draw. I'll try my best, so let's see, that's one, two, three, four, five, six, seven, eight. Let's see, let me draw it. I'll try a little a better shot at it. So, let's see, you have one, two, three, four, five, six,"
    },
    {
        "Q": "at around 10:00,\nwhy doesn't the fluoride anion react with H+ and form HF??",
        "A": "The Fluoride is a very small anion, and indeed, the smallest anion. So, It s electrons will be held very near to it s nucleus. We have to supply high amount of energy to make it bond.It ll be less reactive than Iodide, chloride and bromide which are considerably bigger than Fluoride and they can share their electrons to bond using lesser Energy supply.",
        "video_name": "Z9Jh-Q59xso",
        "timestamps": [
            600
        ],
        "3min_transcript": "Sn2-type reaction. In a protic solvent, what happens is that the things that are really electronegative and really small, like a fluoride anion-- let me draw a fluoride anion. In a protic solvent, what's going to happen is it's going to be blocked by hydrogen bonds. It's very negative, right? It has a negative charge. And it's also tightly packed. As you can see right here, its electrons are very close, tied in. It's a much smaller atom or ion, in this case. If we looked at iodide, iodide has 53 electrons, many orbitals. Actually, iodide would have 54. It would have the same as iodine plus one. Fluoride will have 10 electrons, nine from fluorine plus it gains another one, so it's a much smaller atom. So when you have water hanging around it, let's say you have That has a negative charge. Water is polar. Actually, both of these are polar, so I should write down polar for both of these. This is a polar protic solvent. This is a polar aprotic solvent. In this case, water is still more electronegative than the carbon, so it still has a partial negative charge. These parts still have a partial negative. Water still has a partial negative charge. The hydrogen has a partial positive charge so it is going to be attracted to the fluorine. This is going to happen all around the fluorine. And if these waters are attracted to the fluorine in kind of forming a tight shell around it, it makes it hard for fluorine to react. So it's a worse nucleophile than, say, iodide or hydroxide in a polar protic solvent. Hydroxide has the same issue. compare them, iodide is much bigger. Maybe I'll draw it like this. I'll draw its valence shell like this. It's a much bigger ion. It has all these electrons in here. And so, it still will form hydrogen bonds with the water. It still will form hydrogen bonds with the hydrogen end of the water because they're partially positive, but it's going to be less tightly packed. And on top of that, iodide is more polarizable, which means that its electron cloud is so big and the valence electrons are so far away from the nucleus that they can be influenced by things and then be more likely to react. So let's say this iodide is getting close to a carbon that has a partial positive charge."
    },
    {
        "Q": "When you order the halides in order of decreasing nucleophilicity, where would OH- go? 12:33",
        "A": "Good question, Sal actually answers this in the next video. In short it depends on whether the solvent is protic or not. In short, it is a stronger nucleophile than all the halides in an aprotic solvent, and not quite as strong as iodide in a protic solvent.",
        "video_name": "Z9Jh-Q59xso",
        "timestamps": [
            753
        ],
        "3min_transcript": "it's attached to three hydrogens. We've seen that this will have a partial negative charge. It's more electronegative than the carbon, which will have a partial positive charge. When this guy, this big guy with the electrons really far away, gets close to this, more of the electron cloud is going to be attracted to the partial positive charge. It'll get distorted a little bit and so it is more likely to react in a polar protic solvent. Fluoride, on the other hand, is very tightly packed, blocked by the hydrogen bonds. It's less likely to react. If you were to look at the Periodic Table, if you look at just the halogens in a polar protic solvent, the halides-- this would be the ion version of the halogens-- the halide's iodide will be the best nucleophile. let me write this down-- we have a situation where the iodide is the best nucleophile, followed by bromide, followed by chloride, and then last of all is the fluoride. The exact opposite is true in an aprotic solvent. In an aprotic solvent, the fluoride, which is-- fluorine is far more electronegative. Fluoride is more basic. It will be more stable if it is able to form a bond with something than iodide. Iodide is pretty stable. If you look at a hydrogen iodide, it's actually a highly So iodide itself, the conjugate base of hydrogen iodide, is going to be a very bad base. When you're dealing with an aprotic solvent, you go in the direction of basicity. We're going to learn in the next video that actually basicity and nucleophilicity are related, but they aren't the same concept. We're going to talk about that in a little bit. If you're in an aprotic solvent, you're not reacting with the solvent as much. And then in this situation, fluoride is actually the best nucleophile, followed by chloride, followed by bromide, followed by iodide. So here, you're going in the direction of basicity. This is the best. This is the worse in an aprotic solvent. If it was in a protic solvent, this is flipped around. This becomes the best and this becomes the worst."
    },
    {
        "Q": "At around 12:35 the Professor talks about transposons and conjugation. During which he said trasposons can jump from cell to cell, what I would like to know is if this process has somethig in relation with synapsis.",
        "A": "No, this is a direct linking of bacteria through pili to transfer genetic material, while the signals sent across synapsis are chemical. Neuro transmitters are released from the axon terminals (end of the neurons axon) of one neuron, which then travel across the synapse to the dendrites of the recieving neuron.",
        "video_name": "TDoGrbpJJ14",
        "timestamps": [
            755
        ],
        "3min_transcript": "by the time you have a million bacteria, you'll have a thousand mutations. So they have mutations,but they also have this form. I don't want to call it sexual reproduction, because it's not sexual reproduction. They don't form gametes and the gametes don't fertilize each other and then produces a zygote. But two bacteria can get near each other and then one of their piluses--I'll do that right here. So the piluses are these little structures on the side of the bacteria They're these little tubes,really. One of the piluses can connect from one bacteria to another, and then essentially you have a mixing of what's inside one bacteria with another. So let me draw their nucleoids. And then they have these other pieces of just DNA that hangs out called plasmids. Maybe this guy has got this extra neat plasmid. He got it from someplace, and it's making him able to do things that this guy couldn't do. Maybe this is the R plasmid,which is known for making a bacteria resistant to a lot of antibiotics. And what happens is,that bacteria--and actually, there's mechanisms where the bacteria know that,hey, this guy doesn't have the R plasmid. And we're just beginning to understand how it actually works, but this will actually replicate itself and give this guy a version of the R plasmid. You could also have these things,transposons, and I should make a whole video on this because we have transposons,too. But there's parts of DNA that can jump from one part of a fragment of DNA to another, and these can also end up in the other one. So what you have is kind of--it's not formal sexual reproduction, but what you essentially have is a connection, and these bacteria are just constantly swapping DNA with each other and DNA is jumping back and forth,so you can imagine all sorts of combinations of DNA happen even within what you used to call one bacterial species and become resistant to different things. If this makes it resistant to an antibiotic, then it can kind of spread the information to produce those resistant proteins or whatever to the other bacteria. So this is kind of a form of introducing variation. And so when you transfer stuff via this pilus,or the plural is pili, this is called conjugation,bacterial conjugation. Now,the last thing I want to talk about, because it's something that you've heard a lot about,are antibiotics. A lot of people,they get sick. The first thing they want to get is an antibiotic. And an antibiotic is just a whole class of chemicals and compounds, some of them naturally derived,some of them not,that kill bacteria. So now if someone is undergoing a surgery and they get a cut, instead of them having to worry about getting an infection,"
    },
    {
        "Q": "At 8:25 the Professor talks about Archaea cells, what type of environment are these unique cells found in ?",
        "A": "Thank You that makes everything clear",
        "video_name": "TDoGrbpJJ14",
        "timestamps": [
            505
        ],
        "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": "Isn't the direction of angular velocity supposed to be perpendicular to both linear velocity and the radius? At 4:10 David says that the direction of angular velocity is anti-clockwise?",
        "A": "The angular velocity vector is perpendicular to the linear velocity and the radius But that still leaves two possible directions for it to point. You need to know the direction of the spin in order to know which of those two directions is the one. Specifying the direction of the spin is the same as specifying which direction the velocity vector points.",
        "video_name": "garegCgMxxg",
        "timestamps": [
            250
        ],
        "3min_transcript": "in a straight line, the regular displacement was a defined b, the final position minus the initial positions, which we called delta x. And that's just usually called the displacement, which is measured in meters. Okay, so now we know how to quantify the amount of angle that this ball has rotated through, but another quantity that might be useful is the rate at which it's traveling through that angle. Just like up here, knowing about the displacement is good, but you might want to know about the rate that it's being displaced. In terms of regular linear quantities that was called the velocity of the ball, and it was defined to be the displacement per time. So down here we'll define a similar quantity, but it's going to be the angular velocity, which is defined analogously to the regular velocity. If regular velocity is displacement per time, the angular velocity is going to be the angular displacement per time. And the symbol we used to represent angular velocity is the Greek letter omega, which looks like a w, but it's really the Greek letter omega. are going to be radians per seconds. Since delta theta, the angular displacement is in radians, and the time is in seconds. Just like how regular velocity had units of meters per second, angular velocity has units of radians per second. What is angular velocity mean? What is this omega? It represents the rate at which an object is changing its angle in time. So let's say the tennis ball starts here, and it's going through a circle at this leisurely rate, that means the rate at which it's changing its angle is very small and it has a very small omega. Whereas if you had this tennis ball going through a circle very fast, the rate at which it's going in a circle would be large and that means the angular velocity and omega would also be large. So the velocity and the angular velocity are related, they're not equal because the velocity gives you how many meters per second something is going through, and the angular velocity gives you how many radians per second it's going through, but if it's got a larger angular velocity, it's going to have a larger velocity as well. angular velocity is also a vector, so I'll put an arrow over this omega. Which way does it point? Technically speaking, you'd use the same right hand rule you use to determine the direction of the angular displacement. But again if it's rotating counter clockwise, we can just consider that to be positive, and if it's rotating clockwise, we can consider that to be a negative omega, or a negative angular velocity. So let me get rid of these, and let's define our last angular motion variable. You can probably guess what it is. There's regular displacement and there's angular displacement. There's regular velocity and there's angular velocity. And then the next logical step in this motion variable sequence would be the acceleration, which was defined for regular variables to be the change in velocity over the change in time. So we'll define an analogous angular quantity that would be the angular acceleration. And it's going to be defined to be, instead of change in velocity over change in time, it's going to be the change in the angular velocity over the change in time. And the letter we use to denote angular acceleration"
    },
    {
        "Q": "At 6:46 you mentioned that Si can make 5 bonds because it has d orbitals. Being in the 3rd period and above the d block, how does Si have d orbitals?",
        "A": "It is because in the 3rd shell there is 3s2 3p6 3d10 in a full shell and even before the transition metals there are d orbitals since while the 3s and 3p orbitals fill the 3d orbitals form from the electrons having more places to go. This makes all elements past neon hypervalent and thus Iron is able to have a +7 oxidation state instead of just +3 like you would expect from an oxide.",
        "video_name": "KsdZsWOsB84",
        "timestamps": [
            406
        ],
        "3min_transcript": "So I'm going to go ahead and draw in a fluoride anion here, which is normally an extremely poor nucleophile. So it's actually selective for silicon. So if the fluoride functions as a nucleophile, it's going to attack the silicon here. And it could do this for a couple of reasons. So let's talk about those reasons here. So first of all, the silicon is bonded to some carbons. And silicon is bigger than carbon, if you look at where it is in the periodic table. And so the silicon carbon bonds are longer than we're used to seeing. And that means that there's decreased steric hindrance. So the silicon is a little bit more exposed, and that allows the fluoride anion to attack it a little more. So another factor that allows this is silicon is in the third period on the periodic table. So it has vacant d orbitals. between the fluorine and the silicon. So let me go ahead and draw what we would get after the fluoride attacks the silicon. So we would have this portion of the molecule. And we would have our oxygen bonded to our silicon in this intermediate. And now we could show the fluorine bonded to the silicon, And the silicon is still bonded to two methyl groups and also a tert-butyl group, like that. This will give the silicon a negative 1 formal charge. And it looks a little bit weird, because we see silicon has five bonds to it. But that's, again, it's OK because of where silicon is on the periodic table. It has those has those d orbitals, and so forming five bonds for an intermediate is OK. It's OK for it to have an expanded octet. Another reason why fluoride can attack the silicon very well is because the bond that forms between fluorine and silicon happens to be very strong. And we can finish up by kicking these electrons back onto the oxygen and protonating and forming our target compound. So we would go ahead and form our target compound here. So we would get back our alcohol, like that. And we also successfully added on this portion of the molecule on the right. And then we would also form-- we would now have the fluorine bonded to the silicon, like that. So we selectively removed our protecting group and we formed our target compound. And so that's the idea of a protecting group. It allows you to protect one area of the molecule and react with another area of the molecule. And it's also nice to have it easily removed to get back your target molecule."
    },
    {
        "Q": "at 6:01 , why did sal draw the graph as \"discontinuous function\" ?!\ngenerally , when i draw a graph like this case how can i decide if i'll draw it as continuous or discontinuous function ?",
        "A": "The given function was defined piece-wise with a jump discontinuity at t=2. Since it was defined that way (with the jump discontinuity), you can only draw it that way.",
        "video_name": "6FTiHeius1c",
        "timestamps": [
            361
        ],
        "3min_transcript": "I'll use on this yellow. Let's say on a rate function, that is... Let me make it a little bit interesting. Let's say it's one meter per second, for our time is zero is less than or equal to time which is less than or equal to two seconds. Honestly, this is all in seconds where we're of time. There's two meters per second, for t is greater than two seconds. What's that going to look like? Actually, try to graph it yourself, and just say, \"Well, what is the total change in distance \"over the first, let's say five seconds?\" We want to do delta t over, not the first four seconds but the first five seconds. Well, let's graph it. Let's graph it. So this is one meter per second. One meter per second. That is two meters per second. That's in meters per second. That's my rate axis. One, they're obviously not of the same scale, three, four, five. What is this rate function look like? Well, my rate is one meter per second between time is zero and two, including two seconds, and then the rate jumps. Nothing can accelerate instantly like this. You'd need an infinite force or an infinitely small mass I guess to, or maybe there's something that's thinking about... Anyway, I only get two complex here but this is unrealistic mode. It's not typical for something to just have spontaneous velocity increase like that but let's just go with it. Then after the two seconds, we are at a constant rate of two meters per second. Now, what is our total change in distance over the first five seconds here? or we can break up the problem. We could say, \"Well, over the first two seconds.\" Change in time is two seconds, times our constant rate over those two seconds. It's going to be two seconds times one meter per second. Well, that's going to give us two meters. So this here is going to be, shall we do that in orange color, that's going to give us two meters there, and then we look at the next section. Our change in time here is three seconds, and then we multiply that, times our constant two meters per second. That's going to give us an area of six. If we look at the units, in both cases, we're multiplying seconds times meters per second which is going to give us meters. This is going to be two plus six meters or eight meters. So hopefully this is giving you the intuition that the area under the rate curve or the rate function is"
    },
    {
        "Q": "At 7:55, Khan says that if the radius goes down, then the tangential velocity goes up.\n\nConsidering a ball on a string wrapping itself around and around, its velocity is increasing even though there is no torque acting upon it. But according to the Law of Conservation of Momentum, the momentum is conserved. But if the tangential velocity increased, then mv increases and so does the linear momentum.\n\nCan someone explain to me why this is? How can the linear momentum change without torque?",
        "A": "Torque is rotational and linear momentum is linear, so torque can t change linear momentum, it can only change angular momentum. To apply conservation of momentum you have to consider the system of the ball and the object to which it is tied. The total momentum of those two is conserved. If you ignore the object to which it is tied, then you can consider the string to be exerting an outside force on the ball. The string is accelerating the ball and changing its momentum.",
        "video_name": "nFSMu3bxXVA",
        "timestamps": [
            475
        ],
        "3min_transcript": "we'll do torque in pink. If torque is equal to zero, if there's no net torque going on here, if the magnitude of torque is equal to zero, then we will have no change. No change in angular momentum. And we will look at that mathematically in a few seconds. But just from this there's a very interesting thing that arises. And something that you might have observed at even the Olympics or in other things. And this is the idea that you can, by changing your radius, you could actually change your tangential velocity. And as we've seen in previous videos, tangential velocity is closely related to angular velocity. So let's explore that a little bit. So when we write it in the world where, well actually you see it straight out of this, if L is constant, if r went down, So let me rewrite it over here. So L, whoops. L is equal to mass times tangential velocity, or actually well yeah, tangential velocity, or the velocity that's perpendicular to the radius, times the radius. Now what happens, if we assume that this is constant, if we assume that there's no torque, so we're in this world. So this over here is going to be constant. So what happens if we were to reduce r? Somehow this wire started to reel in a little bit or started to wrap around here, and that's actually a reasonable thing, you could imagine as it rotates it starts to wrap around this thing so the wire gets shorter. So if r goes down, and this is constant, the mass isn't going to change, Well if L is constant, mass isn't changing, or the velocity that's perpendicular to the radius is going to go up. And if we wanna think about it, we can think about it in terms of angular velocity, we know that angular velocity, which we would measure in radians per second, we would use the symbol omega, omega is defined, and we go into much more depth in this in other videos, as tangential velocity, the magnitude of the velocity that is perpendicular to the radius, divided by the radius. Or if you solve for tangential velocity, you get v is equal to is equal to omega r. And so if you substitute back into this, really this definition for angular momentum,"
    },
    {
        "Q": "does p always mean -log? 12:17",
        "A": "Well, when it s in front of something and it s a LOWERCASE p, then yes, it s always -log. Hope I answered your question well! ;D",
        "video_name": "LJmFbcaxDPE",
        "timestamps": [
            737
        ],
        "3min_transcript": "With hydrogen and its conjugate base. We know that there's an equilibrium constant for this. We've done many videos on that. The equilibrium constant here is equal to the concentration of our hydrogen proton times the concentration of our conjugate base. When I say concentration, I'm talking molarity. Moles per liter divided by the concentration of our weak acid. Now. Let's solve for hydrogen concentration. Because what I want to do is I want to figure out a formula, and we'll call it the Hendersen-Hasselbalch Formula, which a lot of books want you to memorize, which I don't think you should. I think you should always just be able to go from this kind of basic assumption and get to it. But let's solve for the hydrogen so we can figure out a relationship between pH and all the other stuff that's in this formula. 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."
    },
    {
        "Q": "9:38, in the correction, isn't the conjugate of a weak acid always a weak base? Like HCN (weak acid) has conjugate CN^- (weak base)",
        "A": "HCl is a strong acid. The conjugate base, the chloride ion, is a very weak base, so no. However, the ionisation of HCl in water is not a reversible reaction (it goes to completion), so its not appropriate for a discussion of buffers, as they involve equilibrium reactions..",
        "video_name": "LJmFbcaxDPE",
        "timestamps": [
            578
        ],
        "3min_transcript": "turning into a weak acid and producing more OH. So the pH won't go down as much as you would expect if you just threw this in water. This is going to lower the pH, but then you have more OH that could be produced as this guy grabs more and more hydrogens from the water. So the way to think about it is it's kind of like a cushion or a spring in terms of what a strong acid or base could do to the solution. And that's why it's called a buffer. Because it provides a cushion on acidity. If you add a strong base to water, you immediately increase its pH. Or you decrease its acidity dramatically. But if you add a strong base to a buffer, because of Le Chatelier's Principal, essentially, you're not going to affect the pH as much. Same thing. If you add and acid to that same buffer, it's not going to affect the pH if you had thrown that acid in water because the equilibrium reaction can always kind of refill the amount of OH that you lost if you're adding acid, or it can refill the amount of hydrogen you lost if you're adding a base. And that's why it's called buffer. It provides a cushion. So it give some stability to the solution's pH. The definition of a buffer is just a solution of a weak acid in equilibrium with its conjugate weak base. That's what a buffer is, and it's called a buffer because it provides you this kind of cushion of pH. It's kind of a stress absorber, or a shock absorber for the acidity of a solution. Now, with that said, let's explore a little bit the math of a buffer, which is really just the math of a weak acid. So if we rewrite the equation again, so HA is in equilibrium. With hydrogen and its conjugate base. We know that there's an equilibrium constant for this. We've done many videos on that. The equilibrium constant here is equal to the concentration of our hydrogen proton times the concentration of our conjugate base. When I say concentration, I'm talking molarity. Moles per liter divided by the concentration of our weak acid. Now. Let's solve for hydrogen concentration. Because what I want to do is I want to figure out a formula, and we'll call it the Hendersen-Hasselbalch Formula, which a lot of books want you to memorize, which I don't think you should. I think you should always just be able to go from this kind of basic assumption and get to it. But let's solve for the hydrogen so we can figure out a relationship between pH and all the other stuff that's in this formula."
    },
    {
        "Q": "At 4:30, why do you decrease the poh when you increase the oh?",
        "A": "Because pH + pOH = 14.00 at 25 Celcius",
        "video_name": "LJmFbcaxDPE",
        "timestamps": [
            270
        ],
        "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": "3:15 Sal says a force is needed to be applied in order to move the particle towards the positive side. How would this force be applied?",
        "A": "Well, there might be another electrical force being applied from the other side. There might be something pushing it. There is simply some form of a force. It s abstract. If you were supposed to know it, Sal would tell you.If you need to know it in some other scenario, you ll be given some way to find it out. Unless it s in life. That s one game where no variables are necessarily provided.",
        "video_name": "zqGvUbvVQXg",
        "timestamps": [
            195
        ],
        "3min_transcript": "Only that particle can have energy. Electrical potential, or electric potential, this is associated with a position. So, for example, if I have a charge and I know that it's at some point with a given electric potential, I can figure out the electric potential energy at that point by just multiplying actually this value by the charge. Let me give you some examples. Let's say that I have an infinite uniformly charged plate. So that we don't have to do calculus, we can have a uniform electric field. Let's say that this is the plate. I'll make it vertical just so we get a little bit of change of pace, and let's say it's positively charged plate. And let's say that the electric field is constant, right? It's constant. No matter what point we pick, these field vectors should all 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"
    },
    {
        "Q": "at 5:15 you say the charge closer to the plate has a higher PE. Why is this if the E field is uniform? if F= qE then wouldnt the force be the same at both pts?",
        "A": "Hi, The force on the charge of course would not vary since F=qE and Sal did not say that the Force is higher. What he said that the P.E is higher. WE know that P.E is the work done in taking the charge to that point. Also Work done is Force times the Distance. Therefore is force is constant the distance will determine the P.E. And from the video it s clear that the particle will be at a higher potential near the plate since it has to travel a greater distance to reach there. Hope it helps Cheers",
        "video_name": "zqGvUbvVQXg",
        "timestamps": [
            315
        ],
        "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": "at 4:50, what does meters have to do with electricity?",
        "A": "The equation for work is Force times Distance equals work. The 2 meters represents the distance of the charge.",
        "video_name": "zqGvUbvVQXg",
        "timestamps": [
            290
        ],
        "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": "At 10:54, why did Linneas develop the division in the first place? What was his purpose?",
        "A": "He simply wanted to organize the cluttered jumble of species.",
        "video_name": "oHvLlS_Sc54",
        "timestamps": [
            654
        ],
        "3min_transcript": "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 It could be just as rich or richer than everything I've drawn over here. This is just a small fraction of the entire tree of life, but let's go even broader than that. So if you go even broader than that, you say, well, what's a kind of life form that isn't Eukarya, that wouldn't have these more complex cell structures, the mitochondria in the cells, the cell nucleuses? Then you just have to think about something like bacteria. And if you want to go even broader, there's things like viruses that you could even debate whether they really even are life, because they are dependent on other life forms for their actual reproduction. But they do have genetic material, like everything else. And that, to me, is kind of a mind-blowing idea. As different as a plant is-- look at a house plant that is in your house right now or the tree when you walk home or bacteria or this jellyfish. There is a commonality in that we all have DNA. And that DNA, for the most part, replicates in a very, very, very, similar way. So it's actually crazy that we actually even are related or that we even do have a common ancestor with some"
    },
    {
        "Q": "what is sal saying at 6:21 ? how we can find kinetic energy from position time graph ?",
        "A": "He s saying that the weight slows down until it reaches the maximum displacement, where it changes direction, and at that point (instant) the velocity is 0. And since KE depends on velocity, at that point KE is also 0.",
        "video_name": "Nk2q-_jkJVs",
        "timestamps": [
            381
        ],
        "3min_transcript": "It will have compressed all the way over here. So at T over 2, it'll have been here. And then at the points in between, it will be at x equals 0, right? It'll be there and there. Hopefully that makes sense. So now we know these points. But let's think about what the actual function looks like. Will it just be a straight line down, then a straight line up, and then the straight line down, and then a straight line up. That would imply-- think about it-- if you have a straight line down that whole time, that means that you would have a constant rate of change of your x value. Or another way of thinking about that is that you would have a constant velocity, right? Well do we have a constant velocity this entire time? We know that at this point right here you have a very high velocity, right? You have a very high velocity. We know at this point you have a very low velocity. So you're accelerating this entire time. actually accelerating at a decreasing rate. But you're accelerating the entire time. And then you're accelerating and then you're decelerating this entire time. So your actual rate of change of x is not constant, so you wouldn't have a zigzag pattern, right? And it'll keep going here and then you'll have a point here. So what's happening? When you start off, you're going very slow. Your change of x is very slow. And then you start accelerating. And then, once you get to this point, right here, you start decelerating. Until at this point, your velocity is exactly 0. So your rate of change, or your slope, is going to be 0. And then you're going to start accelerating back. Your velocity is going to get faster, faster, faster. It's going to be really fast at this point. And then you'll start decelerating at that point. So at this point, what does this point correspond to? You're back at A. So at this point your velocity is now 0 again. And now you're going to start accelerating. Your slope increases, increases, increases. This is the point of highest kinetic energy right here. Then your velocity starts slowing down. And notice here, your slope at these points is 0. So that means you have no kinetic energy at those points. And it just keeps on going. On and on and on and on and on. So what does this look like? Well, I haven't proven it to you, but out of all the functions that I have in my repertoire, this looks an awful lot like a trigonometric function. And if I had to pick one, I would pick cosine. Well why? Because when cosine is 0-- I'll write it down here-- cosine of 0 is equal to 1, right? So when t equals 0, this function is equal to A. So this function probably looks something like A cosine of-- and I'll just use the variable omega t-- it probably"
    },
    {
        "Q": "At 6:25, I now that the spring does not infinitely go on from experience as a kid. Help?",
        "A": "That s because of friction. If there were no friction, it would go on forever.",
        "video_name": "Nk2q-_jkJVs",
        "timestamps": [
            385
        ],
        "3min_transcript": "It will have compressed all the way over here. So at T over 2, it'll have been here. And then at the points in between, it will be at x equals 0, right? It'll be there and there. Hopefully that makes sense. So now we know these points. But let's think about what the actual function looks like. Will it just be a straight line down, then a straight line up, and then the straight line down, and then a straight line up. That would imply-- think about it-- if you have a straight line down that whole time, that means that you would have a constant rate of change of your x value. Or another way of thinking about that is that you would have a constant velocity, right? Well do we have a constant velocity this entire time? We know that at this point right here you have a very high velocity, right? You have a very high velocity. We know at this point you have a very low velocity. So you're accelerating this entire time. actually accelerating at a decreasing rate. But you're accelerating the entire time. And then you're accelerating and then you're decelerating this entire time. So your actual rate of change of x is not constant, so you wouldn't have a zigzag pattern, right? And it'll keep going here and then you'll have a point here. So what's happening? When you start off, you're going very slow. Your change of x is very slow. And then you start accelerating. And then, once you get to this point, right here, you start decelerating. Until at this point, your velocity is exactly 0. So your rate of change, or your slope, is going to be 0. And then you're going to start accelerating back. Your velocity is going to get faster, faster, faster. It's going to be really fast at this point. And then you'll start decelerating at that point. So at this point, what does this point correspond to? You're back at A. So at this point your velocity is now 0 again. And now you're going to start accelerating. Your slope increases, increases, increases. This is the point of highest kinetic energy right here. Then your velocity starts slowing down. And notice here, your slope at these points is 0. So that means you have no kinetic energy at those points. And it just keeps on going. On and on and on and on and on. So what does this look like? Well, I haven't proven it to you, but out of all the functions that I have in my repertoire, this looks an awful lot like a trigonometric function. And if I had to pick one, I would pick cosine. Well why? Because when cosine is 0-- I'll write it down here-- cosine of 0 is equal to 1, right? So when t equals 0, this function is equal to A. So this function probably looks something like A cosine of-- and I'll just use the variable omega t-- it probably"
    },
    {
        "Q": "At 8:06 we made the carbocation gets attacked by water . why we couldnot we do it in the starting instead of making it atacked by hydronium ion?",
        "A": "Water is too weak to attack when there is no positive charge.",
        "video_name": "O_yeKo6-qIg",
        "timestamps": [
            486
        ],
        "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": "At 0:36, He says work is energy transferred by force, Using this definition shouldn't the formula be Work = Energy times Force ?",
        "A": "No. Work IS energy. when you exert a force over a distance, you transfer energy to the object you are exerting the force on. The amount of energy you transfer is force*distance.",
        "video_name": "2WS1sG9fhOk",
        "timestamps": [
            36
        ],
        "3min_transcript": "Welcome back. I'm now going to introduce you to the concepts of work and energy. And these are two words that are-- I'm sure you use in your everyday life already and you have some notion of what they mean. But maybe just not in the physics context, although they're not completely unrelated. So work, you know what work is. Work is when you do something. You go to work, you make a living. In physics, work is-- and I'm going to use a lot of words and they actually end up being kind of circular in their definitions. But I think when we start doing the math, you'll start to get at least a slightly more intuitive notion of what they all are. So work is energy transferred by a force. So I'll write that down, energy transferred-- and I got this from Wikipedia because I wanted a good, I guess, relatively intuitive definition. Energy transferred by a force. And that makes reasonable sense to me. But then you're wondering, well, I know what a force is, you know, force is mass times acceleration. But what is energy? And then I looked up energy on Wikipedia and I found this, But it also I think tells you something that these are just concepts that we use to, I guess, work with what we perceive as motion and force and work and all of these types of things. But they really aren't independent notions. They're related. So Wikipedia defines energy as the ability to do work. So they kind of use each other to define each other. Ability to do work. Which is frankly, as good of a definition as I could find. And so, with just the words, these kind of don't give you much information. So what I'm going to do is move onto the equations, and this'll give you a more quantitative feel of what these words mean. So the definition of work in mechanics, work is equal to force times distance. different color just because this yellow might be getting tedious. And I apply a force of-- let's say I apply a force of 10 Newtons. And I move that block by applying a force of 10 Newtons. I move that block, let's say I move it-- I don't know-- 7 meters. So the work that I applied to that block, or the energy that I've transferred to that block, the work is equal to the force, which is 10 Newtons, times the distance, times 7 meters. And that would equal 70-- 10 times 7-- Newton meters. So Newton meters is one, I guess, way of describing work. And this is also defined as one joule."
    },
    {
        "Q": "When writing the overall formula at 5:55, what if the force that was applied had a anticlockwise torque? (i.e if the force was applied on top of the board, n the same end as worked out in the video) Doesn't the formula change to - Torque = F.r.cos (theta)?",
        "A": "Typically, clockwise torques are negative and counterclockwise/anticlockwise torques are positive. This is because clockwise torques would cause an object to rotate in the negative direction, giving theta a negative value. Counterclockwise/anticlockwise torques would be positive for a similar reason. Thus, the formula is still \u00cf\u0084=Fr sin(\u00ce\u00b8)",
        "video_name": "ZQnGh-t25tI",
        "timestamps": [
            355
        ],
        "3min_transcript": "in order to get the door to rotate. In other words, only perpendicular components of the force, that is perpendicular to the R, are going to exert a torque. So it's only this component of the 10 newtons that's going to contribute toward the torque, and we can find that. If this was 30 degrees, this is an alternate interior angle. That means that this is also 30 degrees. So these angles are identical based on geometry. That means this component that's perpendicular, I can write as well let's see, it's the opposite side. That side is opposite of this 30 degrees, so I can say that it's gonna be 10 newtons, the hypotenuse would be 10 newtons times sine of 30. And 10 newtons times sine of 30 degrees is five newtons. So finally I can say that the torque exerted on this door by this force of 10 newtons at 30 degrees would be the perpendicular component, which is five newtons, times how far that force was applied from the axis, is gonna be 10 newton meters. So at this point I wouldn't blame you if you weren't like, see, this is why I hate torque. I've gotta remember that this two meters is from the axis to the point where the force is applied. I've gotta remember that I'm only supposed to take the perpendicular component, and I'm supposed to remember that perpendicular means perpendicular to this R vector. You might wonder, is there an easier way to do this? Is there a formula that makes it so I don't have to carry so much cognitive load when I'm trying to solve these problems? And there is. Since this force component is always the component that's perpendicular to the R, we can take that into account when writing down the formula. In other words, the way you find that perpendicular component is by taking the magnitude of the total force, the 10 newtons, and you multiply by sine of the angle between the R vector and the F vector. That's what we did to get this five newtons, so why not just write down this formula explicitly in terms of the total force times sine theta? That's gonna be the perpendicular component, So what this represents is this here, this F sine theta is F perpendicular, and then you multiply by R just like we always do. Now in most textbooks you'll see it written like this. They like putting the sine theta at the end. Looks a little cleaner. So if we do F times R times sine theta, now we can just plug in the entire 10 newtons in for the force, the entire two meters in for the R, and this theta would be the angle between the force and the R vector, but that's crucial. You gotta remember if you're gonna use this formula instead of this formula, you've gotta remember that this angle here is always the angle between the force vector and the R vector, which is the vector from the axis to the point where the force is applied. Which in this case, was 30 degrees. So sometimes it's not obvious. How do you figure out the angle between F and R? Well first you identify the direction of F and the direction of R. The safest way to figure it out would be to imagine taking this F vector and just moving it so its tail is at the tail of the R vector."
    },
    {
        "Q": "At 5:36, why did he move the hydrogen to the other side of the periodic table?",
        "A": "Every element wants to satisfy its Electron or Valence Shell. The first shell occupies 2 electrons, so Hydrogen only needs 1 more to satisfy its shell. The same goes for the elements in Group 7, they only need one more electron to satisfy their shells, so he says in theory it could be moved there.",
        "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": "If hydroxide is a polyatomic atom that is represented as OH- then how come at the minute 5:20 you write it as HO- ?",
        "A": "It doesn t really matter which way around it is written, although OH- is much more common.",
        "video_name": "CCsNJFsYSGs",
        "timestamps": [
            320
        ],
        "3min_transcript": "These are typically oxidized. Now, we could keep going. If we were to go right over here to the Group 5 elements, typical oxidation state is negative 3. And so you see a general trend here. And that general trend-- and once again, it's not even a hard and fast rule of thumb, even for the extremes, but as you get closer and closer to the middle of the periodic table, you have more variation in what these typical oxidation states could be. Now, I mentioned that I put hydrogen aside. Because if you really think about it, hydrogen, yes, hydrogen only has one electron. And so you could say, well, maybe it wants to give away that electron to get to zero electrons. That could be a reasonable configuration for hydrogen. But you can also view hydrogen kind of like a halogen. So you could kind of view it kind of like an alkali metal. But in theory, it could have been put here on the periodic table as well. You could have put hydrogen here, because hydrogen, 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."
    },
    {
        "Q": "At 2:00, what is the typical oxidation state of a halogen and why?",
        "A": "The typical oxidation state, or oxidation number, of a halogen is -1, mainly because halogens are so willing to gain another electron in order to fill out its octet.",
        "video_name": "CCsNJFsYSGs",
        "timestamps": [
            120
        ],
        "3min_transcript": "Let's see if we can come up with some general rules of thumb or some general trends for oxidation states by looking at the periodic table. So first, let's just focus on the alkali metals. And I'll box them off. We'll think about hydrogen in a second. Well, I'm going to box-- I'm going to separate hydrogen because it's kind of a special case. But if we look at the alkali metals, the Group 1 elements right over here, we've already talked about the fact they're not too electronegative. They have that one valence electron. They wouldn't mind giving away that electron. And so for them, that oxidation state might not even be a hypothetical charge. These are very good candidates for actually forming ionic bonds. And so it's very typical that when these are in a molecule, when these form bonds, that these are the things that are being oxidized. They give away an electron. So they get to-- a typical oxidation state for them would be positive 1. If we go one group over right over here to the alkaline earth So they're likely to fully give or partially give away two electrons. So if you're forced to assign an ionic-- if you were to say, well, none of this partial business, just give it all away or take it, you would say, well, these would typically have an oxidation state of positive 2. In a hypothetical ionic bonding situation, they would be more likely to give the two electrons because they are not too electronegative, and it would take them a lot to complete their valence shell to get all the way to 8. Now, let's go to the other side of the periodic table to Group 7, the halogens. The halogens right over here, they're quite electronegative, sitting on the right-hand side of the periodic table. They're one electron away from being satisfied from a valence electron point of view. So these are typically reduced. And I keep saying typically, because these are not going to always be the case. There are other things that could happen. But this is a typical rule of thumb that they're likely to want to gain an electron. If we move over one group to the left, Group 6-- and that's where the famous oxygen sits-- we already said that oxidizing something is doing to something what oxygen would have done, that oxidation is taking electrons away from it. So these groups are typically oxidized. And oxygen is a very good oxidizing agent. Or another way of thinking about it is oxygen normally takes away electrons. These like to take away electrons, typically two electrons. And so their oxidation state is typically negative 2-- once again, just a rule of thumb-- or that their charge is reduced by two electrons."
    },
    {
        "Q": "at 5:28 you talk about having a blood vessel break in a certain part of the brain. since we have different sections in the brain for different subjects and topics such as music,personality,vision, etc. would the stroke effect your relationship with that certain subject?",
        "A": "If the braintissue dies due to lack of oxygen or due to the compression of the bloodflow in that part of the brain, then those functions are lost or damaged. So it s possible you go blind, you lose the possibility to speak or understand what s bein said et cetera et cetera...",
        "video_name": "xbyfeEW56Nc",
        "timestamps": [
            328
        ],
        "3min_transcript": "moving through the blood that eventually blocks a blood vessel. So you can actually have a thromboembolism, you can actually have a blood clot that gets broken off -- so let me ignore this for now-- let me paint over it a little bit so that this isn't the main cause of blockage-- but you could actually have a blood clot that breaks off, becomes an embolus, and since it's an embolus due to a blood clot, you call it a thrombembolus -- I always have trouble saying all of these words-- and eventually it blocks an artery over here. So this right here is an embolism, but either way, you're blocking the blood flow further down the brain, [which] could cause infarction, that brain tissue will die, and whatever that brain tissue did for mental function, or whatever, for this person who is experiencing this stroke to do those things. that's called a silent stroke, but damage is occurring. The person experiencing the stroke-- and I'm not a doctor, so take all of this with a grain of salt-- the person experiencing the stroke could be anywhere from - well, one, they may not even notice that damage is occurring, they might have a headache, or it might be more severe, they might actually not be able to properly move a side of their body, or a side of their face, or properly be able to speak, so it really depends on what part of the brain is being damaged. But in either of these situations, an ischemic stroke is caused by some type of restriction or blockage that causes things downstream to not get proper oxygen, and then, so you can imagine, cells over here aren't going to get their oxygen, and then they might actually die. A hemorrhagic stroke - to hemorrhage means to bleed, it's literally just a fancy word for bleeding- and so in a hemorrhagic stroke you have a situation where a blood vessel I'm actually trying to draw the same blood vessel- where it actually breaks. We'll talk more in the future of why a blood vessel might break - strongly related to high blood pressure and other risk factors, but I don't want to get into that right now - but you could imagine if a blood vessel breaks, you have all this blood spewing into the brain in, kind of, an uncontrolled way. So let's say this little diagram [that] I drew right here, that part of the brain, if you have a hemorrhagic stroke, you have all of this blood that's flowing into the brain, and all of that uncontrolled blood will mess up that part of the brain, that causes those neurons and brain tissue to malfunction and maybe causes some of them to die, and it would also cause the blood flow further dowstream to be impaired, so the stuff downstream aren't going to get the blood they need because all the blood is being released everywhere else. And since 87% of strokes are ischemic strokes, the remainder are hemorrhagic, so this is the remaining 13% of strokes."
    },
    {
        "Q": "At 6:31, is that also called latent heat?",
        "A": "Yes, you noticed right! That s precisely what latent heat is.:-)",
        "video_name": "lsXcKgjg8Hs",
        "timestamps": [
            391
        ],
        "3min_transcript": "times the amount of ice. That's what we're solving for. So times x. Times the change in temperature. So this is a 10 degree change in Celsius degrees, which is also a 10 degree change in Kelvin degrees. We can just do 10 degrees. I could write Kelvin here, just because at least when I wrote the specific heat units, I have a Kelvin in the denominator. It could have been a Celsius, but just to make them cancel out. This is, of course, x grams. So the grams cancel out. So that heat absorbed to go from minus 10 degree ice to zero degree ice is 2.05 times 10 is 20.5. So it's 20.5 times x joules. Now, once we're at zero degrees, the ice can even absorb more energy before increasing in temperature as it melts. Remember, when I drew that phase change diagram. The ice gains some energy and then it levels out as it melts. As the the bonds, the hydrogen bonds start sliding past each other, and the crystalline structure breaks down. So this is the amount of energy the ice can also absorb. Let me do it in a different color. Zero degree ice to zero -- I did it again --to zero degree water. Well the heat absorbed now is going to be the heat of fusion of ice. Or the melting heat, either one. That's 333 joules per gram. It's equal to 333.55 joules per gram 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."
    },
    {
        "Q": "9:29 What is a sigma bond? What is a pi bond? Is pi bond in all double bonds or multiple bonds? Is a sigma bond in every single bond?",
        "A": "All single bonds are sigma bonds. However, a double bond consists of 1 sigma bond and 1 pi bond. You can rotate around a sigma bond (single bond), but not a pi bond (involved in double and triple bonds), which I m sure will be important in upcoming videos. A tripple bond consists of 1 sigma bond and 2 pi bonds.",
        "video_name": "u1eGSL6J6Fo",
        "timestamps": [
            569
        ],
        "3min_transcript": "electron would start kind of jumping around or moving around, depending how you want to think about it, in that orbital over there, 2px. Then you'd have the next electron jumping around or moving around in the 2py orbital, so it would be moving around like this. If you went just off of this, you would say, you know what? These guys, this guy over here and that guy over there is lonely. He's looking for a opposite spin partner. This would be the only places that bonds would form. You would expect some type of bonding to form with the x-orbitals or the y-orbitals. Now, that's what you would expect if you just straight-up kind of stayed with this model of how things fill and how orbitals look. The reality of carbon, and I guess the simplest reality of carbon, is if you look at a methane molecule, is very different than what you would expect here. First of all, what you would expect here is that carbon But we know carbon forms four bonds and it wants to pretend like it has eight electrons. Frankly, almost every atom wants to pretend like it has eight electrons. So in order for that to happen, you have to think about a different reality. This isn't really what's happening when carbon bonds, so not what happens when carbon bonds. What's really happening when carbon bonds, and this will kind of go into the discussion of sp3 hybridization, but what you're going to see is it's not that complicated of a topic. It sounds very daunting, but it's actually pretty straightforward. What really happens when carbon bonds, because it wants to form four bonds with things, is its configuration, you could imagine, looks more like this. So you have 1s. We have two electrons there. Then you have your 2s, 2px, 2py and 2pz. It has four electrons that are willing to pair up with electrons from other molecules. In the case of methane, that other molecule is a hydrogen. So what you could imagine is that the electrons actually-- maybe the hydrogen brings this electron right here into a higher energy state and puts it into 2z. That's one way to visualize it. So this other guy here maybe ends up over there, and then these two guys are over there and over there. Now, all of a sudden, it looks like you have four lonely guys and they are ready to bond, and that's actually more accurate of how carbon bonds. It likes to bond with four other people. Now, it's a little bit arbitrary which electron ends up in each of these things, and even if you had this type of bonding, you would expect things to bond along the x, y, and z axis. The reality is, the reality of carbon, is that these four"
    },
    {
        "Q": "At 5:12, why did Sal say straight VANILLA s and p orbitals??",
        "A": "It s American slang for normal or original . It refers to the original flavour of ice cream (vanilla). Since then, many new, more exciting flavours have been invented. The phrase plain vanilla would then refer to the original or unhybridized atomic orbitals.",
        "video_name": "u1eGSL6J6Fo",
        "timestamps": [
            312
        ],
        "3min_transcript": "Remember, these are really just probability clouds, but it's helpful to kind of visualize them as maybe a little bit more things that we would see in our world, but I think probability cloud is the best way to think about it. So that is the 2px orbital, and then I haven't talked about how they get filled yet, but then you also have your 2py orbital, which'll go in this axis, but same idea, kind of a dumbbell shape in the y-direction, going in both along the y-axis, going in that direction and in that direction. Then, of course, so let me do this 2py, and then you also have your 2pz, and that goes in the z-direction up like that and then downwards like that. So when you keep adding electrons, the first-- so far, we've added four electrons. If you add a fifth electron, you would expect it to go into So even though this 2px orbital can fit two electrons, the first one goes there. The very next one won't go into that one. It actually wants to separate itself within the p orbital, so the very next electron that you add won't go into 2px, it'll go into 2py. And then the one after that won't go into 2py or 2px, it'll go into 2pz. They try to separate themselves. Then if you add another electron, if you add-- let's see, we've added one, two, three, four, five, six, seven. If you add an eighth electron, that will then go into the 2px orbital, so the eighth electron would go there, but it would have the opposite spin. So this is just a little bit of review with a little bit of visualization. Now, given what we just reviewed, let's think about what's happening with carbon. Carbon has six electrons. 1s orbital. Then 2s2, then 2p2, right? It only has two left, because it has a total of six electrons. Two go here, then there, then two are left to fill the p orbitals. If you go based on what we just drew and what we just talked about here, what you would expect for carbon-- let me just draw it out the way I did this. So you have your 1s orbital, your 2s orbital, and then you have your 2px orbital, your 2py orbital, and then you have your 2pz orbital. If you just go straight from the electron configuration, you would expect carbon, so the 1s orbital fills first, so that's our first electron, our second electron, our third electron. Then we go to our 2s orbital, That fills next, third electron, then fourth electron. Then you would expect maybe your fifth electron to go in the 2px."
    },
    {
        "Q": "At 14:00 methane has been drawn, but I see no 1s orbital drawn...is it there and just not mentioned, or did it get absorbed into the sp3's somehow?",
        "A": "I believe that he just did not put the 1s electrons in there since they play no role in the Hydrogens bonding to the Carbon.",
        "video_name": "u1eGSL6J6Fo",
        "timestamps": [
            840
        ],
        "3min_transcript": "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. have the hydrogens, so that's our carbon right there. Then you have your hydrogens. You have a hydrogen here. A hydrogen just has one electron in the 1s orbital, so the hydrogen has a 1s orbital. You have a hydrogen here that just has a 1s orbital. It has a hydrogen here, 1s orbital, hydrogen here, 1s orbital. So this is how the hydrogen orbital and the carbon orbitals get mixed. The hydrogens 1s orbital bonds with-- well, each of the hydrogen's 1s orbital bonds with each of the carbon's sp3 orbitals. Just so you get a little bit more notation, so when people talk about hybridized sp3 orbitals, all they're saying is, look, carbon doesn't bond. Once carbon-- this right here is a molecule of methane, right? This is CH4, or methane, and it doesn't bond like you would vanilla s and p orbitals. If you just went with straight vanilla s and p orbitals, the bonds would form. Maybe the hydrogen might be there and there, and if it had four hydrogens, maybe there and there, depending on how you want to think about it. But the reality is it doesn't look like that. It looks more like a tripod. It has a tetrahedral shape. The best way that that can be explained, I guess the shape of the structure, is if you have four equally-- four of the same types of orbital shapes, and those four types of orbital shapes are hybrids between s's and p's. One other piece of notation to know, sometimes people think it's a very fancy term, but when you have a bond between two molecules, where the orbitals are kind of pointing at each other, so you can imagine right here, this hydrogen orbital is pointing in that direction. This sp3 orbital is pointing that direction, and they're overlapping right around here."
    },
    {
        "Q": "1:33 what is an internal pressure?",
        "A": "I think there is an actual name for the internal pressure of a plant that allows it to stand upright. It s called Turgor Pressure.",
        "video_name": "zdvKhaQxvag",
        "timestamps": [
            93
        ],
        "3min_transcript": "we've talked a lot about cells in general but what I thought I would do in this video was focus on plant cells and in particular focus on the cell walls love plant cells so this right over here this is a drawing of a plant cell and my jump out at you immediately is instead of drawing it is this kind ever a roundish shape like that but we have drawn a lot of other cells I've drawn this is kind of a cubic structure rectangular prism and that's plant cells can have a structure like that and so the next question is what gives them that shape which allows them to to form the kind of cubic root angular prism and the answer is its the cell wall so this the cell cell wall so let's make sure we can already into ourselves properly in this picture so clearly if I was if I didn't have this cut out all I would be seeing is the outside all I would be seeing is the cell wall but we've cut it out and we can see the different layers we have the cell wall on the outside right below that right below that we have the cellular membrane or the plasma to this the cell cellular cellular membrane cellular membrane right under that and then under that the cell membrane is control is is containing in the cytoplasm inside of the cytoplasm we have all sorts of this big being that is taking up a lot of the volume inside this plant cell that the back you'll which we have described another in other in other videos vacuo as a combination of this internal pressure things like the back you all in just frankly the pressure from from all the fluid inside the cell pushing outwards plus the cell wall kinda holding it all in that's what gives plant their structure that's why plant is able to grow and be a plant is able to grow and be upright to that's my drawing over a plant active a plant in my room that i'm looking at right now in its able to in be upright and so you have the cell wall you the cellular membrane you have the other organelles at Suncorp last year we have our nuclear membrane urgency the shielding the inert nuclear membranes the DNA inside you the and OPA's make particular one kind containing that the rough ER containing the ribald or having the ribosomes on the membrane the smoothie are not having the ribosomes Golgi apparatus with us a little bit overview but our focus here is on the cell wall so let's go back to that so if we zoom in on this if we zoom in on the cell wall right over here we can look at we can look at this diagram and over here it might be a little bit surprising to you because when I've always imagined ace a a wall cell wall I imagine something like a brick wall something that's impenetrable but this drawing shows a something different and just to be clear what's going on here so this is our cellular membrane sorry routes or remembrance of right over here have my lipid by layer and then a right on top of that I have the cell wall but you see it isn't just ate"
    },
    {
        "Q": "At 2:06, how is it a probabilty of it happening?",
        "A": "A probability of it happening is the concentration of the molecules. If you have a high concentration, you will have a high probability. If you have a low concentration, then the probability is low.",
        "video_name": "TsXlTWgyItw",
        "timestamps": [
            126
        ],
        "3min_transcript": "Let's say we wanted to figure out the equilibrium constant for the reaction boron trifluoride in the gaseous plus 3 -- so for every mole of this, we're going to have 3 moles of H2O in the liquid state -- and that's in equilibrium. It's going forward and backwards with 3 moles of hydrofluoric acid, so it's in the aqueous state. It's been dissolved in the water. If it wasn't dissolved, if it was in the solid state, you would call this hydrogen fluoride. Once it's in water, you call it hydrofluoric acid, and we'll talk more about naming in the future, hopefully. Plus 1 mole of boric acid, also in the aqueous state. It's dissolved in the water. H3BO3 in the aqueous state. So what would the expression for the equilibrium constant So you might be tempted say, OK, that's easy enough, Sal. So the equilibrium constant, you just take the right-hand side. That's just the convention. There's symmetry here. I could've rewritten it either way, but let's just say you take the right-hand side and say, OK, this is dependent on the concentration of the hydrofluoric acid, he concentration of the HF, or the molarity of the HF, to the third power, times the concentration of the boric acid, H3BO3. And remember, this intuition of why you're taking this to the third power is what's the probability -- because in order for the reaction to go this way, you need to have 3 molecules of hydrofluoric acid being very close to 1 molecule of the boric acid. So if you watched the last video I just made about the intuition behind the equilibrium constant, the probability of this reaction happening or the probability of finding all of these molecules in the same place. Of course, you can adjust it with a constant and that's essentially what that does. But that's on the product side, or the reactant, depending on what direction you're viewing this equation, divided by the molarity of the boron trifluoride times -- and I'll do this in a different color-- the molarity of the h2O to the third power. And that's, of course, the H2O liquid. So there you go. We'll just figure this out. And my rebuttal to you is I want you to figure out the molarity of the water. What is the concentration of the water? Remember, the concentration is moles per volume, but in this case, what's happening? I'm putting some boron trifluoride gas"
    },
    {
        "Q": "at 3:18 whats a solvent. and how is it related to h20",
        "A": "Solvent is the liquid thing that is in abundant quantity. Notice how H2O is liquid (there s a small l written next to it, see?) and the others are aqueous. So, that means, water is everywhere. So it is in abundance. And thus, it is a solvent. I can t explain any better than that.",
        "video_name": "TsXlTWgyItw",
        "timestamps": [
            198
        ],
        "3min_transcript": "So you might be tempted say, OK, that's easy enough, Sal. So the equilibrium constant, you just take the right-hand side. That's just the convention. There's symmetry here. I could've rewritten it either way, but let's just say you take the right-hand side and say, OK, this is dependent on the concentration of the hydrofluoric acid, he concentration of the HF, or the molarity of the HF, to the third power, times the concentration of the boric acid, H3BO3. And remember, this intuition of why you're taking this to the third power is what's the probability -- because in order for the reaction to go this way, you need to have 3 molecules of hydrofluoric acid being very close to 1 molecule of the boric acid. So if you watched the last video I just made about the intuition behind the equilibrium constant, the probability of this reaction happening or the probability of finding all of these molecules in the same place. Of course, you can adjust it with a constant and that's essentially what that does. But that's on the product side, or the reactant, depending on what direction you're viewing this equation, divided by the molarity of the boron trifluoride times -- and I'll do this in a different color-- the molarity of the h2O to the third power. And that's, of course, the H2O liquid. So there you go. We'll just figure this out. And my rebuttal to you is I want you to figure out the molarity of the water. What is the concentration of the water? Remember, the concentration is moles per volume, but in this case, what's happening? I'm putting some boron trifluoride gas and it's creating these aqueous acids. These other molecules are dissolved completely in the water. So what's the solvent here? The solvent is H2O. This might be how the reaction happens, but pretty much, there's water everywhere. The water is in surplus. So if you were to really figure out the concentration of water, it's everywhere. I mean, you could say I mean, you could say everything but the boron trifluoride, but it's a very high number. And if you think about it from the probability point of view, if you say, OK, in order for this reaction to happen forward, I need to figure out the probability of finding a boron trifluoride atom or molecule -- actually, molecule-- in a certain volume, and it also needs 3 moles of water in that certain volume. But you say, hey, there's water everywhere. This is the solvent. There's water everywhere, so I really just need to worry about the concentration of the boron trifluoride."
    },
    {
        "Q": "what the heck does 1s^22s^22p^63s^1 in the video at 1:23 mean did i miss something because it makes no sense.",
        "A": "The superscripts are not exponents. They are the numbers of electrons in each shell or subshell. 1s\u00c2\u00b2 2s\u00c2\u00b22p\u00e2\u0081\u00b6 3s\u00c2\u00b9 means there are 2 electrons in the 1s shell, 2 in the 2s subshell, 6 in the 2p subshell, and 1 in the 3s subshell.",
        "video_name": "akm5H2JsccI",
        "timestamps": [
            83
        ],
        "3min_transcript": "Now that we've classified our elements into groups on the periodic table, let's see how to determine the number of valence electrons. And so for this video, we're only talking about the valence electrons for elements in the main groups. When we talk about the main groups, you're using the one through eight system for classifying groups. So one, two, three, four, five, six, seven, and eight. So we're going to ignore the other way to number the groups. And so therefore, we're going to ignore groups three through 12 for this video. And so if we're talking about the main groups, the valence electrons are the electrons in the outermost shell or the outermost energy level. And so let's see if we can figure out how many valence electrons sodium has. So for sodium, if I wanted to write an electron configuration for sodium-- I assume you already know how to do these-- so you would say it is 1S2, 2S2, 2P6. And that takes you all the way over here to neon. or the third energy level. And you have one more electron to worry about. And so that electron would go into a 3S orbital. So the full electron configuration is 1S2, 2S2, 2P6, and 3S1. When I want to figure out how many valence electrons sodium has, the number of valence electrons would be equal to the number of electrons in the outermost shell, the outermost energy level. For sodium, sodium has the first energy level, second energy level, and the third energy level. The outermost energy level would, of course, the third energy level. So if I see how many electrons sodium has in its outermost energy level, it's only one this time. So that means that sodium has one valence electron. And that's very convenient, because sodium is found in group one. And so we can say that for main groups, if you want to figure out how many valence electrons you have, it's just equal to the group number. So the group number is equal to the number And so that makes everything really easy. And so if I wanted to represent a neutral atom of sodium with its one valence electron, I could draw sodium here, and I could draw one valence electron next to sodium like that. All right. Let's go ahead and write the electron configuration for chlorine next. So here's chlorine over here. And so if I wanted to write the electron configuration for chlorine, it would be 1s2, 2s2, 2p6, and once again, that takes me all the way to neon. And so now, I'm over here in the third energy level, or the third period. I can see that I would fill 3s2-- so 3s3. And that puts me into my P orbitals. So how many electrons are in my P orbitals? One, two, three, four, five-- so I'm in the third energy level, I'm in P orbitals, and I have five electrons. And so that would be the electron configuration for chlorine."
    },
    {
        "Q": "At 9:38, when the tension is being found at the bottom of the circle, is the assumption made that the velocity is the same at the top and the bottom? The number 4 was used for both, and I thought the values would be different?",
        "A": "He s just showing you how the problem would be different if you have the velocity at the bottom instead of at the top. Consider it two different problems.",
        "video_name": "2lcaBPLLoLo",
        "timestamps": [
            578
        ],
        "3min_transcript": "Gravity's pointing away from the center, radially away from the center. That means tension still remains a positive force, but the force of gravity now, for this case down here, would have to be considered a negative centripetal force since it's directed away from the center of the circle. So, if we were to calculate the tension at the bottom of the path, the left hand side would still be v squared over r 'cause that's still the centripetal acceleration. The mass on the bottom would still be m 'cause that's the mass of the yo-yo going in a circle. But instead of T plus mg, we'd have T minus mg since gravity's pointing radially out of the center of the circle. And if we solve this expression for the tension in the string, we'd get that the tension equals, we'd have to multiply both sides by m, and then add mg to both sides. And we'd get that the tension's gonna equal m v squared over r, plus m g. This time, we add m g to this m v squared over r term, whereas over here, we had to subtract it. And that should make sense conceptually since before, up here, both tension and gravity to the total centripetal force, so neither one had to be as big as they might have been otherwise. But down here, not only is gravity not helping the tension, gravity's hurting the centripetal cause by pulling this mass out of the center of the circle, so the poor tension in this case not only has to equal the net centripetal force, it has to add up to more than the net centripetal force just to cancel off this negative effect from the force of gravity. And if we plugged in numbers, we'd see that the tension would end up being bigger. We'd actually get the same exact term here, except that instead of subtracting gravity, we have to add gravity to this net centripetal force expression. And we'd get that the tension would be 10.45 Newtons. So recapping, when solving centripetal force problems, we typically write the v squared over r on the left hand side as a positive acceleration, and by doing that, we've selected in toward the center of the circle as positive since that's the direction which means that all forces that are directed in toward the center of the circle also have to be positive. And you have to be careful because that means downward forces can count as a positive centripetal force as long as down corresponds to toward the center of the circle. And just because a force was positive during one portion of the trip, like gravity was at the top of this motion, that does not necessarily mean that that same force is gonna be positive at some other point during the motion."
    },
    {
        "Q": "at 2:50, why do we have to change degree Celsius to kelvin?",
        "A": "Because Kelvin is a measure of temperature in which 0 temperature corresponds to zero energy. C is a measure in which 0 is arbitrarily chosen. The ideal gas equation is therefore written in such a way (check the units) that it is expecting T to be in Kelvin.",
        "video_name": "erjMiErRgSQ",
        "timestamps": [
            170
        ],
        "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": "Hi, at time 8:21 in video, when doing final calculation, why is the calculation .082 divided by 303? shouldnt it be 4/(.082*303) as in 4 divided by (.082 multiplied by 303) ?",
        "A": "What he did is correct. 4/0.82/303 is equal to 4/(0.82*303) the same as 4/2/2 is equal to 4/(2*2)",
        "video_name": "erjMiErRgSQ",
        "timestamps": [
            501
        ],
        "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 6:14, why is there so many constant, i know in my chemistry class we only learned one. And what is the difference between all of them, is there different uses for them. And how do they come up with the constant reaction formula?",
        "A": "PV = nRT only has one constant.",
        "video_name": "erjMiErRgSQ",
        "timestamps": [
            374
        ],
        "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 8:21, why does Sal say \"4 divided by .082 divided by 303...\"? Wouldn't it be times 303? He used the multiplication dot when he wrote it out.",
        "A": "They both mean the same thing. You can divide 4 by 0.082*303 or you can divide 4 by 0.082 and then divide it by 303. You will get the same answer. Hope this helps :)",
        "video_name": "erjMiErRgSQ",
        "timestamps": [
            501
        ],
        "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 3:09, what does \"R\" mean in the equation PV = nRT?",
        "A": "The R is just a constant number that relates the other four properties, called the universal gas constant. The other four can change depending on the example or situation but R will always have the same value..",
        "video_name": "erjMiErRgSQ",
        "timestamps": [
            189
        ],
        "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 8:33, how did that work? why not divide 4 by ( 0.082 * 303 ), as in 4/(0.082*303) ?",
        "A": "What Sal did and what you ve suggested are mathematically equivalent. Sal effectively took (4/0.082)/303, which is the same as (4/0.082)*(1/303) which is the same as 4/(0.082*303).",
        "video_name": "erjMiErRgSQ",
        "timestamps": [
            513
        ],
        "3min_transcript": "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 You'd say, OK well one mole of H2 has a mass of what? The mass of one hydrogen is one atomic mass unit. The mass of two hydrogen when it's in its molecular form, is two atomic mass units. So a mole of it is going to be 2 grams. So in this case, we have 0.16 moles. So if we wanted to convert that to grams, this in the case of these hydrogen gas molecules would be 0.32 grams. And I just multiplied it by 2 because each mole is 2 grams. Anyway, I hope you found that there's a bunch more of these problems. Because I think the math is pretty straightforward here. The thing that always makes it daunting, I think, is the units and making sure you're using the right units. What is they are using meters cubed instead of liters, or kilopascals instead of atmospheres."
    },
    {
        "Q": "what did sal mean at 2:55 ? if the mountain in the image is very plane like a plane mirror without any up\nand downs then we will be able to see the image ?",
        "A": "Sal meant that mountain cause diffused reflection all around it and does not reflect specularly.",
        "video_name": "sd0BOnN6aNY",
        "timestamps": [
            175
        ],
        "3min_transcript": "And you can almost imagine that it bounces off at essentially the same angle, but in the other direction. So then it'll hit the surface, and then it'll bounce off, and it'll go just like that. And then we would call this the reflected ray, after it is kind of bounced off of the surface. Reflected ray. And you may have already noticed this if you've played around a lot with mirrors you would see-- and we're going to look at some images. So you can think about it a little better. Next time you're in front of the bathroom mirror you can think about this, and think about the angle of incidence and the angle of reflection. But they're actually equal. So let me define them right here. So if I were to just drop a straight line that is at a 90 degree, or that is perpendicular to the surface of the actual mirror right over here, we would define this, right here, as the angle of incidence. I'll just use theta. That's just a fancy letter to show that the angle at which and the vertical right there, that's the angle of incidence. And then the angle between that vertical and the blue ray right there, we call that the angle of reflection. And it's just a property of especially mirrors when you're having specular reflection. And you can see this for yourself at all the regular mirrors that you might experience is that the angle of incidence is equal to the angle of reflection. And actually we could see that in a couple of images over here. So let me show you some images of specular reflection, just to make it clear here. So you have some light from the sun hitting this mountain. And we're going to talk about diffuse reflection in a little bit, and that's what's happening. It's being reflected diffusely. That's why we don't see the actual image of the sun here. We just see the white. But then those white light rays, and they're actually are hitting the water. I'm going to try to match up parts of the mountain. So you have this part of the mountain. Let me do this in a better color. You have this part of the mountain up here, and the part of the reflection right over there. So what's happening right here is light is coming from that part of the mountain, hitting this part of the surface of the water. Let me see if I can draw this better. It's hitting this part of the surface of the water, and then it's getting reflected, specular reflection, to our eyes. And it's actually coming straight at us, but I'll draw it at a slight angle. And then it's just coming straight to our eyes like this. If our eye was-- Let's say our eye was here. It's actually coming straight out at us so I actually should just draw a vertical line, but hopefully this makes it clear. And what we just said, the angle of incidence is equal to the angle of reflection, so if you were to draw a vertical, and it might not be that obvious here, but this angle right over here-- Let me draw this a little darker color."
    },
    {
        "Q": "At 0:30, it is stated that particles move in rotation and curved paths but the kinetic molecular theory states that particles move only in straight lines. Error?",
        "A": "Definitely an error",
        "video_name": "eEJqaNaq9v8",
        "timestamps": [
            30
        ],
        "3min_transcript": "Voiceover: All bodies and systems possess a property called temperature. Most commonly temperature is used to refer to how hot or cold something is, but the real sciency definition of temperature is that it's a measure of the average kinetic energy of the particles in a system. I've got a system and I'm filling it with little individual particles. If we think about this microscopically each little particle in the system is moving in some way whether in rotation or in a straight line, or curving, or by kind of a combination of these means. All of these little particles are moving. The energy of motion is called the kinetic energy. All of these moving particles have kinetic energy. The faster those little particles are moving, the greater their kinetic energy. If each of those little particles in the system has greater kinetic energy, that means the system as a whole has a larger amount of total energy, and we would say that it has a greater temperature of the average kinetic energy of those particles. Because knowing the amount of energy in a system can be really useful in chemistry and in physics, we've developed temperature scales to help us quantify or measure the amount of this value, this value of energy. The three scales most widely used are the Kelvin scale, the Celsius scale, and the Fahrenheit scale. For all of these scales I'm going to draw a little thermometer, one for Kelvin. Then we have a thermometer for Celsius, and then another thermometer for the Fahrenheit scale. The two scales used most in the physical sciences are probably the Celsius scale and the Kelvin scale. As a point of comparison here on these thermometers, the freezing point of water occurs at zero degrees Celsius. That's where water freezes. Then the boiling point of water occurs at 100 degrees Celsius. So the boiling point of water occurs at 100 degrees Celsius. That's where water turns into steam. I'm going to write H20 here real quick just so we don't get confused that we're talking about the freezing and boiling point of water. Now when we use the Kelvin scale, we find that water's freezing point is 273.15 Kelvin. Then we find that water boils at 373.15 Kelvin. They differ fundamentally in the zero point. The Celsius and Kelvin scales differ in the zero points that they use, but between water's freezing point, and water's boiling point, we have a span of 100 temperature units for both scales."
    },
    {
        "Q": "At 5:23, when Mr. Khan says \"slide down the slope\", shouldn't the block go flying after the first hill?",
        "A": "It s a possibility. However, energy is still conserved, that is the initial PE + KE still equals the PE + KE at any place along the block s path, whatever the path truly is.",
        "video_name": "kw_4Loo1HR4",
        "timestamps": [
            323
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "At 5:50 Sal says that this problem would be extremely difficult using kinematics equations... I do not understand why it is difficult using kinematics. Vi = 0, a = 9.8, (y2-y1) = 10. Use Vi^2 = Vf^2 + 2a(y2-y1). Final velocity comes out to be 14 m/s. Gravity is the only force acting on the object and our displacement is simple. Why should I not use kinematics here?",
        "A": "You can, but most people find it easier to use CoE when it applies. Both will give you the same answer. Try it both ways and see what seems quicker and easier to you. Then see if you can think of a more complicated path that might be very difficult to solve with kinematics (like a curvy roller coaster, for example).",
        "video_name": "kw_4Loo1HR4",
        "timestamps": [
            350
        ],
        "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": "At minute 3:30 Sal plugs in the mass which is 1 Kg multiplied by (1/2) shouldn't it be .5 and if we divide both sides by .5 to get v^2 alone shouldn't 100/.5=50?",
        "A": "100/0.5 is not 50 Try it on your calculator 100 x 0.5 is 50",
        "video_name": "kw_4Loo1HR4",
        "timestamps": [
            210
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "At 1:56 if an object is in the ground it has 0 PE, so does an object on a cliff or mountain top also have 0 PE",
        "A": "Potential energy depends on where you re referential is, where you define your 0 to be. You can choose your 0 to be anywhere you want, as long as you stay true to your choice and don t change it in the middle of your calculations. If your 0 is on the ground, an object on the ground has 0 PE but the one on the cliff has not, because it has altitude relative to your 0 potential point.",
        "video_name": "kw_4Loo1HR4",
        "timestamps": [
            116
        ],
        "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": "at 3:00 he says that this formula only works for a point charge or for a spherical charge. My question is: does this formula work for a wire's cross-section.",
        "A": "it does but you have to apply a bit of calculus for it(or maybe Gauss law)",
        "video_name": "-LtvW5783zE",
        "timestamps": [
            180
        ],
        "3min_transcript": "The way we'll find a formula for the magnitude of the electric field is simply by inserting what we already know is the formula for the electric force. Coulomb's Law gives us the force between two charges, and we're just gonna put that right in here. Coulomb's Law says that the electric force between two charges is gonna be k, the electric constant, which is always nine times 10 to the ninth, multiplied by Q1, the first charge that's interacting, and that'd be this Q1 over here, multiplied by Q2, the other charge interacting, divided by the center to center distance between them squared, and then because we're finding electric field in here, we're dividing by Q2. Notice what happens here. Q2 is canceling, and we get that the magnitude of the electric field is gonna be equal to k, this electric constant, and I'll write that down over here so we know what it is. K is nine times 10 to the ninth, and it's got kinda weird units, but it makes sure that all the units come out okay when you multiply. And then, what do we still have up here? We've still got a Q1 divided by the center and you might be like, well, the other charge went away. We canceled it out. Centered between which two charges? Well, this could be to any point in space, really. So you imagine your test charge at any point you want. I could put it here, I can move it over to here. The r would just be the distance from the first charge, Q1, to wherever I wanna figure out what the electric field would be. But since this Q2 always divides out, we don't even need to talk about that. We can just figure out the electric field that's created by Q1 at any point in space, so this r is just the distance from the center of the charge creating the field to the point in space where you wanna determine the electric field. And now we've got it. This is a formula for the electric field created by a charge Q1. Technically, though, this is only true if this is a point charge. In other words, if it's really, really small compared to the other dimensions in the problem. Or, if this is spherically symmetric, then it doesn't matter. If you're outside of this charge and you've got a spherically symmetrical charge distribution, where all the charges are lumped on one side of this sphere, throughout, then this formula also works just as well when you're outside the sphere. And what's this formula saying? It's saying that the absolute value, or the magnitude of the electric field created at a point in space is equal to k, the electric constant, times the charge creating the field. This is important. This charge, Q1, is creating this electric field. And then you plug in the distance away from that charge that you wanna determine the electric field, r, you square it, and that'll tell you what the magnitude of the electric field is created by Q1 at any point in space around it. Now why are we being so careful, saying that this is just the magnitude? Here's why. Imagine we plugged in this charge as positive because the charge creating it is positive. You'd get a positive value for the electric field, and you might think, oh, that means positive. That means to the right. And in this case, it works out. It does go to the right at this point. But let's say you put those same calculations for a point over here, and you wanna determine what's the value of the electric field at this point? Well, if you plugged in k, it's a positive number, your Q is a positive number, r is gonna"
    },
    {
        "Q": "01:01 - 01:32 What does he mean by 'moles'?",
        "A": "A mole is just a number, like a dozen. But instead of 12, a mole is 6.02 * 10^23. You could use it to count anything, but since it is a big number is is mostly used to count very small things like atoms and molecules.",
        "video_name": "2f7YwCtHcgk",
        "timestamps": [
            61,
            92
        ],
        "3min_transcript": "In my humble opinion, the single most important biochemical reaction, especially to us, is cellular respiration. And the reason why I feel so strongly about that is because this is how we derive energy from what we eat, or from our fuel. Or if we want to be specific, from glucose. At the end of the day, most of what we eat, or at least carbohydrates, end up as glucose. In future videos I'll talk about how we derive energy from fats or proteins. But cellular respiration, let's us go from glucose to energy and some other byproducts. And to be a little bit more specific about it, let me write the chemical reaction right here. So the chemical formula for glucose, you're going to have six carbons, twelve hydrogens and six oxygens. So that's your glucose right there. So if you had one mole of glucose-- let me write that, 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."
    },
    {
        "Q": "At 3:40 you say that the energy produces 38 ATPs but I heard that it was 34, which one is correct?",
        "A": "4 ATP are generated before the electron transport chain. That, in addition to the 34 ATP, makes 38 ATP.",
        "video_name": "2f7YwCtHcgk",
        "timestamps": [
            220
        ],
        "3min_transcript": "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. you'll produce 38 ATPs. I was reading up a little bit before doing this video. And the reality is, depending on the efficiency of the cell in performing cellular respiration, it'll probably be more on the order of 29 to 30 ATPs. But there's a huge variation here and people are really still studying this idea. But this is all cellular respiration is. In the next few videos we're going to break it down into its kind of constituent parts. And I'm going to introduce them to you right now, just so you realize that these are parts of cellular respiration. The first stage is called glycolysis. Which literally means breaking up glucose. And just so you know, this part, the glyco for glucose and then lysis means to break up. When you saw hydrolysis, it means using water to break up a molecule. Glycolysis means we're going to be breaking up glucose."
    },
    {
        "Q": "at 1:54 and a few other places, Sal says a few \"moles\" of oxygen, hydrogen, etc. Bye \"moles\", does he mean molecules?",
        "A": "No, he means moles. A mole is a number, like a dozen, but much larger. 6.02*10^23. A mole of hydrogen atoms is 6.02*10^23 of those atoms.",
        "video_name": "2f7YwCtHcgk",
        "timestamps": [
            114
        ],
        "3min_transcript": "In my humble opinion, the single most important biochemical reaction, especially to us, is cellular respiration. And the reason why I feel so strongly about that is because this is how we derive energy from what we eat, or from our fuel. Or if we want to be specific, from glucose. At the end of the day, most of what we eat, or at least carbohydrates, end up as glucose. In future videos I'll talk about how we derive energy from fats or proteins. But cellular respiration, let's us go from glucose to energy and some other byproducts. And to be a little bit more specific about it, let me write the chemical reaction right here. So the chemical formula for glucose, you're going to have six carbons, twelve hydrogens and six oxygens. So that's your glucose right there. So if you had one mole of glucose-- let me write that, 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."
    },
    {
        "Q": "At 3:22 when Sal says that energy generates 38 ATPs isn't it 36 ATP?",
        "A": "A very efficient cell can produce 38 TOTAL ATP from one glucose molecule, but since glycolysis requires 2 ATP, the NET gain would be 36.",
        "video_name": "2f7YwCtHcgk",
        "timestamps": [
            202
        ],
        "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": "I assumed that Sal was saying 38 ATPs total at first because he was looking at the total number of ATPs not the net. However, in 10:30 , he says the net gain of ATP is 38. My books say 36. Am I understanding this wrong?",
        "A": "well the total ATPs produced in aerobic repiration should be 38... However, muscle cells & neurons produce only 36 molecules of ATP per glucose molecule. Its because the 2 molecules of NADH produced during glycolysis in muscle cells & neurons dont enter the ETC directly but through other carriers, which transfer the electrons and H+ to the cytochromes. Therefore, these two NADH molecules produce 2 molecules of ATP only, instead of the usual 3...",
        "video_name": "2f7YwCtHcgk",
        "timestamps": [
            630
        ],
        "3min_transcript": "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 And this part gets credit for producing the bulk of the ATPs. 34 ATPs. And this is also aerobic. It requires oxygen. So you can see, if you had no oxygen, if the cells weren't getting enough oxygen, you can produce a little bit of energy. But it's nowhere near as much as you can produce once you have the oxygen. And actually when you start running out of oxygen, this can't proceed forward, so what happens is some of these byproducts of glycolysis, instead of going into the Krebs cycle and the electron transport chain, where they need oxygen, instead they go through a side process called fermentation. For some organisms, this process of fermentation takes your byproducts of glycolysis and literally produces alcohol."
    },
    {
        "Q": "At 03:55 you mention there are 38 ATP molecules that are being released According to other sources it is 36 why is that ?",
        "A": "because they add the number of ATP molecules from the first and the second step of the cellular respiration. Since the first step is produces 2 ATP molecules and the second step produces 36 ATP molecules so when we add them together it is 38 ATP molecules.",
        "video_name": "2f7YwCtHcgk",
        "timestamps": [
            235
        ],
        "3min_transcript": "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. you'll produce 38 ATPs. I was reading up a little bit before doing this video. And the reality is, depending on the efficiency of the cell in performing cellular respiration, it'll probably be more on the order of 29 to 30 ATPs. But there's a huge variation here and people are really still studying this idea. But this is all cellular respiration is. In the next few videos we're going to break it down into its kind of constituent parts. And I'm going to introduce them to you right now, just so you realize that these are parts of cellular respiration. The first stage is called glycolysis. Which literally means breaking up glucose. And just so you know, this part, the glyco for glucose and then lysis means to break up. When you saw hydrolysis, it means using water to break up a molecule. Glycolysis means we're going to be breaking up glucose."
    },
    {
        "Q": "When Sal was talking about a net gain in ATP in the video about 9:25 in glycolysis, are the other processes in cellular respiration also have a net gain? Because Sal didn't write whether or not the 2 ATP in Krebs and the 34 in the transport chain were net gains. Thanks!",
        "A": "they are. In the ideal cell you have a net win of 38 ATPs. 2+2+34=38 It s a bit confusing if you look at the pictures of it. Sometimes it just says you ll win NADH or FADH2. But they are used to win ATP.",
        "video_name": "2f7YwCtHcgk",
        "timestamps": [
            565
        ],
        "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 08:48, why would you look for a resonance structure? Isn't the Oxygen in this case at a formal charge of 0 (6 - (4+2)). It doesn't make sense to me why you would push the electrons off the double bond and end up with a +O and -C as I thought nature disliked charges and prefers neutral atoms.",
        "A": "Yes, it does, but the contributor is still a minor contributor to the resonance hybrid.",
        "video_name": "UHZHkZ6_H5o",
        "timestamps": [
            528
        ],
        "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 1:59 how is the formal charge of oxygen negative one?",
        "A": "Formal charge = valence electrons - lone pair electrons - bonds 6 - 6 - 1 = -1",
        "video_name": "UHZHkZ6_H5o",
        "timestamps": [
            119
        ],
        "3min_transcript": "Voiceover: Let's look at a few of the patterns for drawing resonance structures, and the first pattern we're gonna look at, is a lone pair of electrons next to a pi bond. And so, here's a lone pair of electrons; I'm gonna highlight it in magenta, that lone pair of electrons is located on this carbon, let me go ahead and put this carbon in green, here. And I'm saying, there's a negative-one formal charge on that carbon in green, and so that carbon in green is also bonded to a hydrogen, so once again, you need to be very familiar with assigning formal charges. So we have a lone pair of electrons next to a pi bond, because over here, we have a double-bond between the carbon and the oxygen, one of those bonds is a sigma bond, and one of those bonds is a pi bond, so I'm just gonna say that these are the pi electrons. So our goal in drawing a resonance structure is to de-localize that negative-one formal charge, so spread out some electron density. And so, we could take the electrons in magenta, and move them into here, to form a double-bond, between the carbon in green and this carbon right here, and that'd be too many bonds to the carbon in yellow, onto this top oxygen here. So we go ahead, and draw in our brackets, and we put our double-headed resonance arrow, and we draw the other resonance structure, so we have our ring, like that, and then we have, now, a double-bond between those two carbons, and then this top oxygen here, now has only one bond to it. The oxygen used to have two lone pairs of electrons, now it has three, because it just picked up a pair of electrons from that pi bond. So let's go ahead, and follow the electrons. The electrons in magenta moved in here, to form our pi bond, like that, and the electrons in the pi bond, in blue, moved off, onto this oxygen, so I'm saying that they are those electrons. That gives the top oxygen a negative-one formal charge, and so we have our two resonance structures for the enalate anion. We know that both resonance structures contribute to the overall hybrid, and if you think about we have had a negative-one formal charge on the carbon in green, so that's a carb anion; and for the resonance structure on the right, we had a negative one formal charge on the oxygen, so that's an oxyanion. Oxygen is more electronegative than carbon, which means it's more likely to support a negative-one formal charge, and so the resonance structure on the right contributes more to the overall hybrid for an enalate anion. All right, let's do another pattern, a lone pair of electrons next to a positive charge, this time. So, let's look at nitromethane, and we could look at this lone pair of electrons here, on this oxygen, and that lone pair of electrons is next to a positive charge; this nitrogen has a plus one formal charge on it. And, so, let's think about drawing the resonance structure, so our goal is to de-localize charge, to spread charge out."
    },
    {
        "Q": "At 3:49, how come we start numbering at the left side for the second example, whereas Bromine is at 2?",
        "A": "Generally, you want the lowest numbers possible for your substituents. There are some exceptions, such as in alcohols, where you want the lowest number on the -OH group (even if it makes big numbers for any other substituents.)",
        "video_name": "nQ7QSV4JRSs",
        "timestamps": [
            229
        ],
        "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 4:28, why is it a oct-5-yn-4-ol but not 5-octyn-4-ol?",
        "A": "As it was explained to me, both are acceptable, but the first way is less confusing.",
        "video_name": "nQ7QSV4JRSs",
        "timestamps": [
            268
        ],
        "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 7:38 Sal says that the area of the piston = area of the base of the container but dosen't the base of container have walls on it's sides? So, the area of the base is not = to the area of piston!\nPlease explain me!",
        "A": "The area of the face of the piston is the same as the area of the base of the cylinder; the piston fits snuggly into the cylinder. I an not sure why you think the area of the walls of the cylinder are involved here.",
        "video_name": "obeGVTOZyfE",
        "timestamps": [
            458
        ],
        "3min_transcript": "system here did some work. So I'm claiming that it applied a force to this piston, and it applied that force to the piston for some distance. So let's figure out what that is, and if we can somehow relate it to other macro properties that we know reasonably well. Well we know the pressure and the volume, right? We know the pressure that's being exerted on the piston, at least at this point in time. Pressure is equal to force per area. Remember, this piston, you're just seeing it from the side, but it's a kind of a flat plate or a flat ceiling on top of this thing. And at what distance did it move it? You know I could blow it up a little bit. It moved it some-- I didn't draw it too big here-- some x, some distance x. So what is the force that it pushed it up? Well, the force, we know its pressure, the pressure's force per area. So if we want to know the force, we have to multiply pressure times area. If we multiply both sides of this times area, we get force. So we're essentially saying the area of this little ceiling to this container right there, you know, it could be, I could draw with some depth, but I think you It has some area. It's probably the same area as the base of the container. So we could say that the force being applied by our system-- let me do it in a new color-- the force is equal to our pressure of the system, times the area of the ceiling of our container of the piston. Now that's the force. Now what's the distance? The distance is-- I'll do it in blue-- it's this change right here. I didn't draw it too big, but that's that x. Now let's see if we can relate this somehow. Let me draw it a little bit bigger. And I'll try to draw in three dimensions. So let me draw the piston. What color did I do it in? I did it in that brown color. So our piston looks something-- I'll draw it as a elipse-- the piston looks like that. And it got pushed up. So it got pushed up some distance x. Let me see how good I can-- whoops. Let me copy and paste that same-- So the piston gets pushed up some distance x. Let me draw that. It got pushed up some distance x. And we're claiming that our-- oh sorry, this is the force. Sorry, let me be clear."
    },
    {
        "Q": "At 5:34 how does he get \"(1) in the equation.",
        "A": "He is showing that six \u00cf\u0080 electrons satisfy H\u00c3\u00bcckel s 4n+2 rule. 4n +2 = 6 4n = 6 \u00e2\u0080\u0093 2 = 4 n = 4/4 = 1 \u00e2\u0088\u00b4 Six \u00cf\u0080 electrons satisfy 4n +2 when n = 1",
        "video_name": "oDigu9YxXUg",
        "timestamps": [
            334
        ],
        "3min_transcript": "So we're going to start down here. So we're going to inscribe a hexagon. Let's see if we can put a hexagon in here. And so we have a six-sided figure here in our frost circle. The key point about a frost circle is everywhere your polygon intersects with your circle, that represents the energy level of a molecular orbital. And so this intersection right here, this intersection here, and then all the way around. And so we have our six molecular orbitals. And we have the relative energy levels of those six molecular orbitals. So let me go ahead and draw them over here. So we have three molecular orbitals which are above the center line. And those are higher in energy. And we know that those are called antibonding molecular orbitals. So these are antibonding molecular orbitals, which are the highest in energy. If we look down here, there are three molecular orbitals which are below the center line. So those are lower in energy. And if we had some molecular orbitals that were on the center line, those would be non-bonding molecular orbitals. We're going to go ahead and fill our molecular orbitals with our pi electrons. So go back over here. And remember that benzene has 6 pi electrons. And so filling molecular orbitals is analogous to electron configurations. You're going to fill the lowest molecular orbital first. And each orbital can hold two electrons, like electron configurations. And so we're going to go ahead and put two electrons into the lowest bonding molecular orbital. So I have four more pi electrons to worry about. And I go ahead and put those in. And I have filled the bonding molecular orbitals of benzene. So I have represented all 6 pi electrons. If I think about Huckel's rule, 4n plus 2, I have 6 pi electrons. So if n is equal to 1, Huckel's rule is satisfied. Because I would do 4 times 1, plus 2. And so 6 pi electrons follows Huckel's rule. If we look at the frost circle and we look at the molecular orbitals, we can understand Huckel's rule a little bit better visually. So if I think about these two electrons down here, you could think about that's where the two comes from in Huckel's rule. If think about these four electrons up here, that would be four electrons times our positive integer of 1. So 4 times 1, plus 2 gives us six pi electrons. And we have filled the bonding molecular orbitals of benzene, which confers the extra stability that we call aromaticity or aromatic stabilization. And so benzene is aromatic. It follows our different criteria. In the next few videos, we're going to look at several other examples of aromatic compounds and ions."
    },
    {
        "Q": "From 4:11 does hank mean the actual taxonomic ' class' when he mentions class ?",
        "A": "Yes, hank means classes",
        "video_name": "c7Yy9v8dH8s",
        "timestamps": [
            251
        ],
        "3min_transcript": "In humans, the notochord is reduced to the disks, the cartilage that we have beween our vertebra. Second, we have the nerve cord itself, called the dorsal hollow nerve cord, a tube made of nerve fibers that develops into the central nervous system. This is what makes chordates different from other animal phyla, which have solid ventral nerve cords, meaning they run along the front, or stomach, side. Third, all chordates have pharyngeal slits. In invertebrates like the lancelet here, they function as filters for feeding. In fish and other aquatic animals, they're heel slits, and in land-dwelling vertebrates like us, they disappear before we're born. But, that tissue develops into areas around our jaws, ears, and other structures in the head and neck. Finally, we can't forget our fourth synapomorphy, the post-anal tail. which is exactly what it sounds like. It helps propel aquatic animals through the water, makes our dog look happy when she wags it, and in humans, it shrinks during embryonic development, into what is known as the coccyx, or tailbone. It's right here. 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,"
    },
    {
        "Q": "What he calls a tetrad at 7:00, I thought (from lectures) was a bivalent - are these two names for the same thing, or is there a difference?",
        "A": "There both the same because the definition of bivalent is a pair of homologous chromosomes.",
        "video_name": "04gQ0bQu6xk",
        "timestamps": [
            420
        ],
        "3min_transcript": "A couple of things happen. The nuclear membrane begins to dissolve. This is very similar to prophase when we're looking at mitosis. So the nuclear envelope begins to dissolve. These things start to maybe migrate a little bit. So these characters are trying to go at different ends. And the DNA starts to bunch up into kind of its condensed form. So now I can draw it. So now I can start to draw it as proper. So this is the one from the father right over here. And this is the one from the mother. And I'm drawing, I'm overlapping on purpose because something very interesting happens especially in meiosis. So it's the mother right over here. Let me see. Let's now do the centromere in blue now. That's the centromere. These are the shorter ones from the mother. And actually, let me just do draw them on opposite sides just to show that they don't have to, the ones from the father aren't always on the left hand side. So this is the shorter one from the father. They couldn't be all on the left hand side but doesn't this all they have to be. And this is the shorter one from the mother. And I will draw this overlapping although they could have. Shorter one from the mother. And once again, each of these, this is a homologous pair, that's a homologous pair over there. Now, the DNA has been replicated so in each of the chromosomes in a homologous pair, you have two sister chromatids. And so, in this entire homologous pair, you have four chromatids. And so, this is sometimes called a tetrad. So let me just give ourselves some terminology. So this right over here is called a tetrad or often called a tetrad. Now, the reason why I drew this overlapping Let me label this. This is prophase I. You can get some genetic recombination, some homologous recombination. Once again, this is homologous pair. One chromosome from the father that I've gotten from the father. The species or the cell got it from its father's cell and one from the mother. And they're homologous. They might contain different base pairs, different actual DNA, but they code for the same genes. Over simplification, but in a similar place on each of these it might code for eye color or I don't know, personality. Nothing is that simple in how tall you get and it's not that simple in DNA but just to give you an idea of how it is. And the reason why I overlapped them like this is to show how the recombination can occur. So actually, let me zoom in. So this is the one from the father. Once again, it's on the condensed form. This is one chromosome made up of two sister chromatids"
    },
    {
        "Q": "At 7:40, why is the object exerting an equal force on the hand? I mean, after all it's the object that is getting accelerated, so the force on the object should be bigger than the force acting back on the hand, right? It seems to me like, if the forces were truly equal, nothing would move, not the hand, not the object...",
        "A": "Reaction pair forces are on different objects, not the same object. I push on the chair, the chair pushes on me. How many forces on the chair? Just one. So it accelerates. If you and the chair are floating in outer space, when you push on the chair it will also push on you and you will move in opposite directions. If you masses are equal, you will have equal but opposite acceleration.",
        "video_name": "By-ggTfeuJU",
        "timestamps": [
            460
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 6:30, wouldn't you have to throw an enormous object with A LOT of mass to accelerate backward enough to grab onto the space shuttle?",
        "A": "Yeah, that would be ideal, but you probably don t have a massive object on you when you re out in space. I assume it would require more energy to move around.",
        "video_name": "By-ggTfeuJU",
        "timestamps": [
            390
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 4:37, why is Sal saying that L goes in the same direction as current. My intuition tells me that the current is going through the L of the conductor and how can length have a direction?",
        "A": "You can consider L to be displacement rather than distance or length . We know displacement is a vector while distance and length are scalars. So L has a direction! :)",
        "video_name": "l3hw0twZSCc",
        "timestamps": [
            277
        ],
        "3min_transcript": "And then on this side of I1, what happens? Well, on this side, you can see the fingers come back around. So it pops out when it intersects with your video monitor. So on this side, the vectors-- this is the top of an arrow, coming out at you. Fair enough. So I1, by going in this direction, is generating a magnetic field that, at least where I2 is concerned, that magnetic field is going into the page. So what was our formula? And this all came from the first formula we learned about, the effect of a magnetic field soon. on a moving charge. But what was the formula of the net magnetic force on a current carrying wire? It was the force-- I'll do it in blue-- it's a vector, has a magnitude and direction-- is equal to the current. Well, in this case, we want to know the force on this Caused by this magnetic field, by magnetic field 1. So it will be equal to I2, the magnitude of this current, times L-- where L is-- because you can't just say, oh well, what is the effect on this wire? You have to know how much wire is under consideration. So let's say we have a length of wire. And then of course, if you know the length of wire and we knew its mass and we knew the force on it, we could figure out its acceleration in some directions. So let's say that this distance is L, and it's a vector. L goes in the same direction as the current. That's just the convention we're using. It makes things simple. So that's L. So the force on this wire, or at least the length L of this wire, is going to be equal to current 2 times L. We could call that even L2, just so that you know that it deals with wire 2. That's a vector quantity. 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."
    },
    {
        "Q": "Why do you use a c instead of an equal sign? Is this intentional or just a fault of drawing?(10:04)",
        "A": "It is an equal sign. Just the video has lower quality, so it looks like it is c.",
        "video_name": "7vHh1sfZ5KE",
        "timestamps": [
            604
        ],
        "3min_transcript": "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 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."
    },
    {
        "Q": "How can the voltage be equal at both the points (02:50)? In the previous video, Sal said that voltage is electric field times distance. Clearly, there is a difference in distance between the two points.",
        "A": "Yeah that it is quite confusing, the way to think about voltage in a circuit is work being done on charge . In circuits we are assuming that the work to move a charge through a conducting wire is negligible compared to the work getting the electrons through the resistors. Of course that isn t true, work is needed just to move the electrons through a wire even if it doesn t have any resistance but that amount of work is totally negligible.",
        "video_name": "7vHh1sfZ5KE",
        "timestamps": [
            170
        ],
        "3min_transcript": "And the higher the voltage, the more they really want to get to this positive terminal. So what's going to happen in this circuit? Actually, let me label everything. So let's call this R1, let's call this R2, let's call this R3. The first thing I want you to realize is that between elements that the voltage is always constant. And why is that? Well, we assume that this is a perfect conductor-- let's say this little segment right here, right? And so it's a perfect conductor. Well, let's look at it at this end. So you have all these electrons. This is a perfect conductor, so there's nothing stopping these electrons from just distributing themselves over this wire. Before you encounter an element in the circuit or device or whatever you want to call that, you can view this ideal conducting wire just from a schematic point of view as an extension of the negative terminal. And similarly, you can view this wire right here, this part of the wire, as an extension of And the reason why I want to say that is because it actually turns out that it doesn't matter if you measure the voltage here. So let's say if I take a measure of the voltage across those two terminals using what we call a voltmeter. And I'll later do a whole video on how voltmeters work, but remember, when we measure voltage, we have to measure it at two points. Because voltage is a potential difference. It's not some kind of absolute number. It's a difference between essentially how bad do electrons want to get from here to here. So if we measure the voltage between those two points, it would be the exact same thing as if we measured the voltage between these two points. Theoretically. As we know, no wires really have no resistivity. All wires have a little bit, but when we draw these schematics, we assume that the wires are perfect conductors and all the resistance takes place in the resistor. So that's the first thing I want you to realize, and it makes things very-- so, for example, everywhere along this wire, this part of the wire, the voltage is constant. Everywhere along this wire, the voltage is constant. get too messy. That's a big important realization when you later become an electrical engineer and have much harder problems to solve. Let me erase all of this. Let me erase all of that. Let me redraw that, because we can't have that gap there, because if there was that gap, current wouldn't flow. That's actually-- well, I'll draw later how you can draw a switch, but a switch is essentially a gap. It looks like a gap in the circuit that you can open or Because if you open it, no current will flow. If you close it, current will flow. OK, so you now know that the voltage between devices is constant. The other thing I want to convince you is that the current through this entire circuit is constant, and that applies to any circuit in series. Now, what do I mean by series? Series just means that everything in the circuit is after one another, right? If we take the convention and we say current flows in this direction, it'll hit this resistor, then the next"
    },
    {
        "Q": "at 1:12, sal says that the electrons are on the negative terminal... dont they repel??",
        "A": "yes, thats why current flows when a circuit is made",
        "video_name": "7vHh1sfZ5KE",
        "timestamps": [
            72
        ],
        "3min_transcript": "Let's make our circle a little bit more complicated now. So let's say I have a battery again, and let me do it in a different color just for variety. That's the positive terminal, that's the negative terminal. Let's say I have this perfect conductor, and let's say I have one resistor and I have another resistor. I don't know, just for fun, let's throw in a third resistor. And we know, of course, that the convention is that the current flows from positive to negative, that that's the flow of the current. And remember, current is just the charge that flows per unit of time or the speed of the charge flow. But we know, of course, that in reality what is happening, if there's any such thing as reality, is that we have a bunch of electrons here that, because of this voltage across the battery terminals, these electrons want to really badly And the higher the voltage, the more they really want to get to this positive terminal. So what's going to happen in this circuit? Actually, let me label everything. So let's call this R1, let's call this R2, let's call this R3. The first thing I want you to realize is that between elements that the voltage is always constant. And why is that? Well, we assume that this is a perfect conductor-- let's say this little segment right here, right? And so it's a perfect conductor. Well, let's look at it at this end. So you have all these electrons. This is a perfect conductor, so there's nothing stopping these electrons from just distributing themselves over this wire. Before you encounter an element in the circuit or device or whatever you want to call that, you can view this ideal conducting wire just from a schematic point of view as an extension of the negative terminal. And similarly, you can view this wire right here, this part of the wire, as an extension of And the reason why I want to say that is because it actually turns out that it doesn't matter if you measure the voltage here. So let's say if I take a measure of the voltage across those two terminals using what we call a voltmeter. And I'll later do a whole video on how voltmeters work, but remember, when we measure voltage, we have to measure it at two points. Because voltage is a potential difference. It's not some kind of absolute number. It's a difference between essentially how bad do electrons want to get from here to here. So if we measure the voltage between those two points, it would be the exact same thing as if we measured the voltage between these two points. Theoretically. As we know, no wires really have no resistivity. All wires have a little bit, but when we draw these schematics, we assume that the wires are perfect conductors and all the resistance takes place in the resistor. So that's the first thing I want you to realize, and it makes things very-- so, for example, everywhere along this wire, this part of the wire, the voltage is constant. Everywhere along this wire, the voltage is constant."
    },
    {
        "Q": "I might be wrong in my concepts......please correct me if I'm wrong....\n\nAt 1:55 isn't the compound supposed to be called \"2 butyl pentane\"(as the butyl group is linked to the 2nd carbon of the ring) instead of what sal tells us?",
        "A": "First you number the atoms in the ring. If there is only one group attached to the ring, that ring carbon is automatically number 1. Thus the names are butylcyclopentane and sec-butylcyclopentene.",
        "video_name": "TJUm860AjNw",
        "timestamps": [
            115
        ],
        "3min_transcript": "Let's see if we can get the molecular structure for butylcyclopentane. So you just break this up the way we've done it in the last several videos, the suffix is -ane, so it is an alkane, all single bonds. So single bonds. It's pentane, so we're dealing with five carbons on the base, or on the backbone. So this is five carbons and it's a cyclopentane, so it's five carbons in a ring. So its five-carbon ring is the backbone, and then we have a butyl group added to that five-carbon ring. Now, you might say, hey, Sal, how do I know which carbon to add it to? When you're dealing with a ring and you only have one group on the ring, it doesn't matter. Let me just show you what I mean. So let's draw the five-carbon ring. Let's draw the cyclopentane. So it'll just be a pentagon, so one, two, three, four, One, two, three, four, five. Now, it doesn't matter where I draw the butyl group. It's all symmetric around there. We just have a ring and it's connected to a butyl group at some point. It'll start to matter once we add more than one group. So we can just pick any of these carbons to add the butyl group to. Now, just as a review, the but- prefix, that refers to, remember, methyl, ethyl, propyl, or meth-, eth-, prop-, but-. This is four carbons. This is a four-carbon alkyl group. So let me just add it here. I could have added it to any of these carbons around this cyclopentane ring. So if I just add it right here, so I'm going to have four carbons. So one, two, three, four. That is the butyl part of this whole thing. And then let me just attach them up. So you might be tempted to just draw this right there. And actually, this would be right. This is butylcyclopentane. But a question might arise. 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."
    },
    {
        "Q": "At 4:23 when discussing naming this structurewould it be appropriate to call it a dibutyl cyclopentane",
        "A": "No.The common name is sec-butylcyclopentane. The IUPAC name is (1-methylpropyl)cyclopentane. A dibutylcyclopentane would have two separate butyl groups attached to the cyclopentane ring.",
        "video_name": "TJUm860AjNw",
        "timestamps": [
            263
        ],
        "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": "3:10 till 3:20 were a bit difficult to understand for me, can anyone please elaborate?",
        "A": "He is saying that there are two different ways of naming the side chains (alkyl groups) \u00e2\u0080\u0094 the common names and the official or IUPAC names. There are four different butyl groups, and their names in the two systems are Common IUPAC n-butyl butyl isobutyl 2-methylpropyl sec-butyl 1-methylpropyl tert-butyl 1,1-dimethylethyl",
        "video_name": "TJUm860AjNw",
        "timestamps": [
            190,
            200
        ],
        "3min_transcript": "One, two, three, four, five. Now, it doesn't matter where I draw the butyl group. It's all symmetric around there. We just have a ring and it's connected to a butyl group at some point. It'll start to matter once we add more than one group. So we can just pick any of these carbons to add the butyl group to. Now, just as a review, the but- prefix, that refers to, remember, methyl, ethyl, propyl, or meth-, eth-, prop-, but-. This is four carbons. This is a four-carbon alkyl group. So let me just add it here. I could have added it to any of these carbons around this cyclopentane ring. So if I just add it right here, so I'm going to have four carbons. So one, two, three, four. That is the butyl part of this whole thing. And then let me just attach them up. So you might be tempted to just draw this right there. And actually, this would be right. This is butylcyclopentane. But a question might arise. 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."
    },
    {
        "Q": "When Sal gives us the (1,1 methyl ethyl)cyclopentane towards the end of the video (12:45) he uses two one's and a di to describe the two methyls on the same carbon. Is it also correct to just say 1-dimethyl? If so, which would be more correct because I feel like 1,1-dimethyl is more referring to two groups of methyls (an ethyl) on each side of that carbon. Slightly confusing. =(",
        "A": "We use numbers to locate the substituents only when ther are many possibility for the positions of the substituents. In this case the name could be (dimethylethyl)cyclopentane because there is no other possibilty for the two methyl group to be attach on the ethyl substituent.",
        "video_name": "TJUm860AjNw",
        "timestamps": [
            765
        ],
        "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 think\nat 5:28 the vector denoted in blue should directed in downword direction",
        "A": "The string is pulling in the upward direction",
        "video_name": "_UrfHFEBIpU",
        "timestamps": [
            328
        ],
        "3min_transcript": "red point, is stationary. It's not accelerating in either the left/right directions and it's not accelerating in the up/down directions. So we know that the net forces in both the x and y dimensions must be 0. My second question to you is, what is going to be the offset? Because we know already that at this point right here, there's going to be a downward force, which is the force of gravity again. The weight of this whole thing. We can assume that the wires have no weight for simplicity. So we know that there's going to be a downward force here, this is the force of gravity, right? The whole weight of this entire object of weight plus wire is pulling down. So what is going to be the upward force here? Well let's look at each of the wires. This second wire, T2, or we could call it w2, I guess. It has no y components. It's not lifting up at all. So it's just pulling to the left. So all of the upward lifting, all of that's going to occur from this first wire, from T1. So we know that the y component of T1, so let's call-- so if we say that this vector here. Let me do it in a different color. Because I know when I draw these diagrams it starts to get confusing. Let me actually use the line tool. So I have this. Let me make a thicker line. So we have this vector here, which is T1. And we would need to figure out what that is. And then we have the other vector, which is its y component, and I'll draw that like here. This is its y component. And then of course, it has an x component too, and I'll do that in-- let's see. I'll do that in red. Once again, this is just breaking up a force into its component vectors like we've-- a vector force into its x and y components like we've been doing in the last several problems. And these are just trigonometry problems, right? We could actually now, visually see that this is T sub 1 x and this is T sub 1 sub y. Oh, and I forgot to give you an important property of this problem that you needed to know before solving it. Is that the angle that the first wire forms with the ceiling, this is 30 degrees. So if that is 30 degrees, we also know that this is a parallel line to this."
    },
    {
        "Q": "Could someone give me a hard question to this, i am not sure if i have got this or not, thanks! And how did at 9:28, did Mr. Khan get 200N, he said divide both sides by 1/2, that gives 50N, does it not? So what is happening here",
        "A": "It would have given 50 N if it were divided by 2. Diving by 1/2 is the same as multiplying by 2.",
        "video_name": "_UrfHFEBIpU",
        "timestamps": [
            568
        ],
        "3min_transcript": "So if we solve for T1 sub y we get T1 sine of 30 degrees is equal to T1 sub y. And what did we just say before we kind of dived into the math? We said all of the lifting on this point is being done by the y component of T1. Because T2 is not doing any lifting up or down, it's only pulling to the left. So the entire component that's keeping this object up, keeping it from falling is the y component of this tension vector. So that has to equal the force of gravity pulling down. This has to equal the force of gravity. That has to equal this or this point. So that's 100 Newtons. 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."
    },
    {
        "Q": "At 11:36 he said \"homologous\".Is it not homozygous?",
        "A": "Homozygous is when the two inherited alleles from both parents turn out to be both dominant or recessive at the same time, in the new organism.",
        "video_name": "DuArVnT1i-E",
        "timestamps": [
            696
        ],
        "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": "at 12:40 what is a genotype?",
        "A": "The genotype is the genetic makeup of a person, which phenotype is they physical makeup.",
        "video_name": "DuArVnT1i-E",
        "timestamps": [
            760
        ],
        "3min_transcript": "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 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."
    },
    {
        "Q": "What did Sal mean when he said noise at around 5:14?",
        "A": "He is referring to random changes that can occur during the budding process. The most likely change is a mutation of genes. Instead of an exact replica happening during the copying of the original, there is a error, and the end result is not identical to the original. An environmental event could also account for a change in the expression of the genes that are present, again resulting in a non-identical offspring.",
        "video_name": "DuArVnT1i-E",
        "timestamps": [
            314
        ],
        "3min_transcript": "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 But that would, as we already talked about, most of the time, we would have very little change, very little variation, and whatever variation does occur because of any kind of noise being introduced into this kind of budding process where I just replicate myself identically, most of the time it'll be negative. Most of the time, it'll break the organism. Now, when you have sexual reproduction, what happens? Well, you keep mixing and matching every possible combination of DNA in a kind of species pool of DNA. So let me make this a little bit more concrete for you. So let me erase this horrible drawing I just did. So we all have-- let me stick to humans because that's what we are. We have 23 pairs of chromosomes, and in each pair, we have one chromosome from our mother and one chromosome from our father."
    },
    {
        "Q": "Isn't the xy pair chromosomes a girl? And the xx is a man? In the picture at 8:55 he is a girl.",
        "A": "It actually depends on the species what type of chromosome pairs determine sexual characteristics; there are different sexual determination systems. For humans (and most mammals), the XY chromosomal pair denotes that you are biologically a male while the XX chromosomal pair denotes that you are biologically a female.",
        "video_name": "DuArVnT1i-E",
        "timestamps": [
            535
        ],
        "3min_transcript": "I have 23 from my mother. I have 23 from my father. Now, each of these chromosomes, and I made them right next to each other. So let me zoom in on one pair of these. So let's say we look at chromosome number-- I'll just call this chromosome number 3. So let me zoom in on chromosome number 3. I have one from my mother right here. Actually, maybe I'll do it this way. Remember, chromosome is just a big-- if you take the DNA, the DNA just keeps wrapping around, and it actually wraps around all these proteins, and it creates this structure, but it's just a big-- when you see it like that, you're like, oh, maybe the DNA-- no, but this could have millions of base pair, so maybe it'll look something like that. It's a densely wrapped version of-- well, it's a long string always the way it is, and we'll talk more about that, they draw it as densely packed like that. So let's say that's from my mother and that's from my father. Now, let's call this chromosome 3. They're both chromosome 3. And what the idea here is that I'm getting different traits from my father and from my mother. And I'm doing a gross oversimplification here, but this is really just to give you the idea of what's going on. This chromosome 3, maybe it contains this trait for hair color. And maybe my father had-- and I'll use my actual example. My father had very straight hair. So someplace on this chromosome, there's a gene for hair straightness. Let's say it's a little thing right there. And remember, that gene could be thousands of base pairs, but let's say this is hair straightness. So my father's version of that gene, he had the allele for I'll call it the allele straight for straight hair. Now, this other chromosome that my mother gave me, this essentially, and there are exceptions, but for the most part, it codes for the same genes, and that's why I put them next to each other. So this will also have the gene for hair straightness or curlyness, but my mom does happen to actually have curly hair. So she has the gene right there for curly hair. The version of the gene here is allele curly. The gene just says, look, this is the gene for whether or not your hair is curly. Each version of the gene is called an allele. Allele curly. Now, when I got both of these in my body or in my cells, and this is in every cell of my body, every cell of my body except for, and we'll talk in a few seconds about my germ"
    },
    {
        "Q": "At 18:07 when Sal talks about how a new combination of genes may or may not be passed on, what happens if it is not passed on, do an old combination of genes get passed on? or the genes from the other person get passed on?\nAlso can someone give me an exact definition for genotype and phenotype?",
        "A": "Genotypes are genes expressed as letters. (Ex. Tt) A phenotype is the appearance expressed in words. (Ex. tall)",
        "video_name": "DuArVnT1i-E",
        "timestamps": [
            1087
        ],
        "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": "Why do you throw e in there without explaining it? Please walk through the steps a little more deliberately. What is happening with e at 5:14?",
        "A": "It s assumed knowledge and it should be about the most complex algebra you need to know to get you through most of chemistry. lnK which is in that equation means the natural log of the variable K. To isolate K, we need to use the inverse function on both sides of the equation, which is e. e^(ln K) = K",
        "video_name": "U5-3wnY04gU",
        "timestamps": [
            314
        ],
        "3min_transcript": "for this balanced equation. So, let's write down our equation that relates delta-G zero to K. Delta-G zero is equal to negative RT, natural log of K. Delta-G zero is negative 33.0 kilojoules, so, let's write in here, negative 33.0, and let's turn that into joules, so times ten to the third, joules. This is equal to the negative, the gas constant is 8.314 joules over moles times K. So, we need to write over here, joules over moles of reaction. So, for this balanced equation, for this reaction, delta-G zero is equal to negative 33.0 kilojoules. So, we say kilojoules, or joules, over moles of reaction just to make our units work out, here. so, we write 298 K in here, Kelvin would cancel out, and then we have the natural log of K, our equilibrium constant, which is what we are trying to find. So, let's get out the calculator and we'll start with the value for delta-G zero which is negative 33.0 times 10 to the third. So, we're going to divide that by negative 8.314, and we'd also need to divide by 298. And so, we get 13.32. So, now we have 13.32, right, so our units cancel out here, and this is equal to the natural log of the equilibrium constant, K. So, how do we solve for K here? Well, we would take E to both sides. and E to the natural log of K on the right, this would cancel out and K would be equal to E to the 13.32, so let's do that. So, let's take E to the 13.32, and that's equal to, this would be 6.1, 6.1 times ten to the one, two, three, four, five. So, 6.1, 6.1 times ten to the fifth. And since we're dealing with gases, if you wanted to put in a KP here, you could. So, now we have an equilibrium constant, K, which is much greater than one. And we got this value from a negative value for delta-G zero. So, let's go back up to here, and we see that delta-G zero, right, is negative. So, when delta-G zero is less than zero,"
    },
    {
        "Q": "at about 6:20 hank mentions a tepane, what in the world is that? I am really curious now about that.",
        "A": "Pace or step.",
        "video_name": "cstic6WHr2E",
        "timestamps": [
            380
        ],
        "3min_transcript": "your sea anemones, your jellyfish, your corals, that have just one hole that serves as both mouth and anus. (laughing) Aren't you glad we're a little bit more complicated than that. It's worth noting that these animals have radial symmetry. All their junk kind of radiates out from a central point. That is their mouth hole/poo hole and that is a little bit more sophisticated than having no symmetry at all, like a sponge, but just barely. I mean, their anus and their mouth are the same thing. But more complex animals with the notable exception of the echinoderms, like starfish and sand dollars, exhibit bilateral symmetry. We have two-sided bodies that look the same on both sides. Something else we have in common is that we have an anus that is, get this, in a different place than our mouth. This separation is pretty key because it means that we, as animals, are basically built around a tube, a digestive tract, with a mouth at one end and an anus at the other. The process of forming this tract is called gastrulation and it's kind of a big deal. So, when we left our little blastula it was still just hanging out a little round hollow ball of cells. starts to form at a single point on the blastula. This place on the blastula that starts to invaginate or fold in on itself is called the blastopore. Now, for animals whose mouth and anus are the same thing, this is where the development stops, which is why they only have one hole for all their business, but in everything else the invagination continues until the indentation makes its way all the way through and opens on the other side creating what is essentially a hollow bead made of cells. Now we have a gastrula. Now two different things can happen at this point, depending on what kind of animal this is going to be. It can either be an animal whose mouth is the orifice that's formed by the blastopore, called a protostome or one whose anus is the structure that's created by the blastopore and that's called a deuterostome. So, guess which one you are? Write it down. I want to see your guesses. Chordates, that is to say, all vertebrates and a couple of our relatives like starfish are deuterostomes. Meaning that we were once just a butt hole and me, congratulations! And hopefully you're getting the idea here the formation of the digestive tract is the first thing that happens in the development of an animal and it happens to every living thing, whether it's going to be a tardigrade or a polar bear or a [tepane]. The miracle of life! Now so far, the little hollow bead of cells is basically two layers of tissue thick. An outer layer called the ectoderm and an inner layer called the endoderm and these are called your germ layers. For those organisms that stop developing at this point with that classy mouth/anus combo, they only get two germ layers. They're called diploblastic and they were born that way. It's totally okay, but for us more complex animals whose mouths are separate from our anuses, yes! We develop a third layer of tissue making us triploblasts. Here the ectoderm is going to end up being the animals skin and nerves and spinal cord and most of it's brain, while the endoderm"
    },
    {
        "Q": "At 9:06, how can P.E. be 5 Joules.\nWhen h=1/2d, P.E. should be equal to 5 units",
        "A": "Are the units Joules? If so, what s the problem?",
        "video_name": "vSsK7Rfa3yA",
        "timestamps": [
            546
        ],
        "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 3:51 , What do you mean by \"sp3 hybrid orbital gives 4 hybrid orbitals\" (Is this fixed) ?",
        "A": "in sp3 orbitals there are 4 orbitals as 1 of s orbital and 3 of p orbitals",
        "video_name": "BM-My1AheLw",
        "timestamps": [
            231
        ],
        "3min_transcript": "so that's like electron group geometry, you wanna think about the geometry of the entire molecule. I could think about drawing in those electrons, those bonding electrons, like that. So we have a wedge coming out at us in space, a dash going away from us in space, and then, these lines mean, \"in the plane of the page.\" And so, we can go ahead a draw in our hydrogens, and this is just one way to represent the methane molecule, which attempts to show the geometry of the entire molecule. So the arrangement of the atoms turns out to also be tetrahedral, so let's go ahead and write that. So, tetrahedral. And, let's see if we can see that four-sided figure, so a tetrahedron is a four-sided figure, so we can think about this being one face, and then let's go ahead and draw a second face. And if I draw a line back here, that gives us four faces to our tetrahedron. So our electron group geometry is tetrahedral, 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"
    },
    {
        "Q": "At, 8:40 - 8:42, there is repulsion between the lone pairs & the bond pairs but what about the lone pairs? Don't they repel themselves too?",
        "A": "Of course they do. And they repel one another the most.",
        "video_name": "BM-My1AheLw",
        "timestamps": [
            520,
            522
        ],
        "3min_transcript": "you have four SP three hybridized orbitals. So this oxygen is SP three hybridized, so I'll go ahead and write that in here, so oxygen is SP three hybridized. So we can draw that out, showing oxygen with its four SP three hybrid orbitals; so there's four of them. So I'm gonna go ahead and draw in all four. In terms of electrons, this orbital gets one, this orbital gets one, and these orbitals are going to get two, like that; so that takes care of oxygen's six valence electrons. When you're drawing in your hyrdogens, so let's go ahead and put in the hydrogen here, so, once again, each hydrogen with one electron, in a un-hybridized S orbital, like that. So in terms of overlap of bonds, here's one sigma bond, and here's another sigma bond; so that's our two sigma bonds for water. Once again, the arrangement of these electron pairs is tetrahedral, so VSEPR theory says the electrons repel, is tetrahedral, but that's not the geometry of the entire molecule, 'cause I was just thinking about electron groups, and these hybrid orbitals. The geometry of the molecule is different, so we'll go ahead and draw that over here. So we have our water molecule, and draw in our bonding electrons, and now let's put in our non-bonding electrons, like that, so we have a different situation than with ammonia. With ammonia, we had one lone pair of electrons repelling these bonding electrons up here; for water, we have two lone pairs of electrons repelling these bonding electrons, and so that's going to change the bond angle; it's going to short it even more than in the previous example. So the bond angle decreases, so this bond angle in here decreases to approximately 105 degrees, rounded up a little bit. So, thinking about the molecular geometry, or the shape of the water molecule, so this is, \"bent geometry,\" because you ignore the lone pairs of electrons, and that would just give you this oxygen here, and then this angle; so you could also call this, \"angular.\" So we have this bent molecular geometry, like that, or angular, and once again, for molecular geometry, ignore your lone pairs of electrons. So these are examples of three molecules, and the central atom in all three of these molecules is SP three hybridized, and so, this is one way to figure out your overall molecular geometry, and to think about bond angles, and to think about how those hybrid orbitals affect the structure of these molecules."
    },
    {
        "Q": "At 1:44 what is VSEPR theory?",
        "A": "VSEPR theory is the theory of hybridized orbitals. It is the theory that dictates unhybridized, sp1, sp2, and sp3 hybridizations. Basically the idea is that orbitals will hybridize to form bonds with equal energy.",
        "video_name": "BM-My1AheLw",
        "timestamps": [
            104
        ],
        "3min_transcript": "Voiceover: The concept of steric number is very useful, because it tells us the number of hybridized orbitals that we have. So to find the steric number, you add up the number of stigma bonds, or single-bonds, and to that, you add the number of lone pairs of electrons. So, let's go ahead and do it for methane. So, if I wanted to find the steric number, the steric number is equal to the number of sigma bonds, so I look around my carbon here, and I see one, two, three, and four sigma, or single-bonds. So I have four sigma bonds; I have zero lone pairs of electrons around that carbon, so four plus zero gives me a steric number of four. In the last video, we saw that SP three hybridized situation, we get four hybrid orbitals, and that's how many we need, the steric number tells us we need four hybridized orbitals, so we took one S orbital, and three P orbitals, and that gave us four, SP three hybrid orbitals, so this carbon must be SP three hybridized. So let's go ahead, and draw that in here. So this carbon is SP three hybridized, and in the last video, we also drew everything out, for that carbon, and we had one valence electron in each of those four, SP three hybrid orbitals, and then hydrogen had one valence electron, in an un-hybridized S orbital, so we drew in our hydrogens, and the one valence electron, like that. This head-on overlap; this is, of course, a sigma bond, so we talked about this in the last video. And so now that we have this picture of the methane molecule, we can think about these electron pairs, so these electron pairs are going to repel each other: like charges repel. And so, the idea of the VSEPR theory, tell us these electron pairs are going to repel, and try to get as far away from each other as they possibly can, in space. And this means that the arrangement of those electron pairs, ends up being tetrahedral. So let's go ahead and write that. So we have a tetrahedral arrangement of electron pairs around our carbon, like that. so that's like electron group geometry, you wanna think about the geometry of the entire molecule. I could think about drawing in those electrons, those bonding electrons, like that. So we have a wedge coming out at us in space, a dash going away from us in space, and then, these lines mean, \"in the plane of the page.\" And so, we can go ahead a draw in our hydrogens, and this is just one way to represent the methane molecule, which attempts to show the geometry of the entire molecule. So the arrangement of the atoms turns out to also be tetrahedral, so let's go ahead and write that. So, tetrahedral. And, let's see if we can see that four-sided figure, so a tetrahedron is a four-sided figure, so we can think about this being one face, and then let's go ahead and draw a second face. And if I draw a line back here, that gives us four faces to our tetrahedron. So our electron group geometry is tetrahedral,"
    },
    {
        "Q": "Around 7:45 Sal said that the radius of the observable universe is equal to 46 billion light years, does that mean the whole universe is 92 billion light years?",
        "A": "The diameter of the observable universe is 92 billion ly. We don t know how big the whole universe is. We have good reason to believe it is much, much bigger than the portion we can observe. Maybe infinite, we don t know.",
        "video_name": "06z7Q8TWPyU",
        "timestamps": [
            465
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 9:16 in the video he states that the light from the object that took 13.7 Billion years to reach us is now 46 Billion light years away. If it all began at the big bang 13.7 Billion years ago, how can anything be 46 Billion light years away?",
        "A": "Light years are a measure of distance not time",
        "video_name": "06z7Q8TWPyU",
        "timestamps": [
            556
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 5:54, why does he do a mol ratio of MnO4- over Fe2+ and not the other way around? I'm doing similar problems, and there's seemingly no rhyme or reason as to whether it should be 1/5 or 5/1. It varies from problem to problem.",
        "A": "That s because it doesn t matter which way he does it. He is trying to get the moles of Fe\u00c2\u00b2\u00e2\u0081\u00ba. If he had written 5/1 = x/0.000 4000, he would still have gotten x = 0.002 000 mol Fe\u00c2\u00b2\u00e2\u0081\u00ba.",
        "video_name": "EQJf8Gb8pg4",
        "timestamps": [
            354
        ],
        "3min_transcript": "we can see that we've used a certain volume of our solution. Let's say it took 20 milliliters. We used up 20 milliliters of our potassium permanganate solution to completely titrate our iron two plus. Our goal was to find the concentration of iron two plus. If we're going to find the concentration of iron two plus, we could figure out how many moles of permanganate were necessary to completely react with our iron two plus. We could figure out moles from molarity and volume. Let's get some more room down here. We know that molarity is equal to moles over liters. The molarity of permanganate is .02. We have .02 for the concentration of permanganate ions. Moles is what we're solving for. It took us 20 milliliters for our titration, which we move our decimal place one, two, three, So solve for moles. .02 times .02 is equal to .0004. So we have .0004. This is how many moles of permanganate were needed to completely react with all of the iron two plus that we originally had in our solution. It took .0004 moles of permanganate to completely react with our iron. All right. Next, we need to figure out how many moles of iron two plus that we originally started with. To do that, we need to use our balance redox reaction. We're going to look at the coefficients, because the coefficients tell us mole ratios. The coefficient in front of permanganate is a one. The coefficient in front of iron two plus is a five. If we're doing a mole ratio of permanganate to iron two plus, and iron would be a five. So we set up a proportion here. One over five is equal to ... Well, we need to keep permanganate in the numerator here. How many moles of permanganate were necessary to react with the iron two plus? That was .0004. So we have .0004 moles of permanganate. X would represent how many moles of iron two plus we originally started with. We could cross-multiply here to solve for x. Five times .0004 is equal to .002. X is equal to .002. X represents the moles of iron two plus that we originally had present. We're almost done, because our goal was to find the concentration of iron two plus cations. Now we have moles"
    },
    {
        "Q": "At 3:36 I don't understand, if there are excess MnO4-, then how can this be the end point? Wouldn't it mean that now there are extra ions which would contribute to an extra volume, and therefore become an inaccurate value to calculate the concentration?",
        "A": "The MnO4- has a purple color and the Fe2+ is colorless solution. If the Fe2+ solution is excess, it can be colorless unless the MnO4- is enough. If the color turns purple, that means the Fe2+ ion doesn t exist as Fe2+, they all turn into Fe3+.",
        "video_name": "EQJf8Gb8pg4",
        "timestamps": [
            216
        ],
        "3min_transcript": "so the oxidation state is plus two. Manganese is going from an oxidation state to plus two. That's a decrease or a reduction in the oxidation state. Therefore, manganese is being reduced in our redox reaction. Let's look at iron two plus. We have iron two plus as one of our reactants here. That means the oxidation state is plus two. For our products, we're making iron three plus, so an oxidation state of plus three. Iron is going from plus two to plus three. That's an increase in the oxidation state. Iron two plus is being oxidized in our redox reaction. As we drip in our potassium permanganate, we're forming our products over here. These ions are colorless in solution. As the permanganate reacts, this purple color disappears we should have a colorless solution. Let's say we've added a lot of our permanganate. Everything is colorless. But then we add one more drop, and a light purple color persists. Everything was clear, but then we add one drop of permanganate and then we get this light purple color. This indicates the endpoint of the titration. The reason why this is the endpoint is because our products are colorless. So if we get some purple color, that must mean we have some unreacted, a tiny excess of unreacted permanganate ions in our solution. That means we've completely reacted all the iron two plus that we originally had present. So we stop our titration at this point. We've reached the endpoint. We've used a certain volume of our potassium permanganate solution. Let's say we finished down here. we can see that we've used a certain volume of our solution. Let's say it took 20 milliliters. We used up 20 milliliters of our potassium permanganate solution to completely titrate our iron two plus. Our goal was to find the concentration of iron two plus. If we're going to find the concentration of iron two plus, we could figure out how many moles of permanganate were necessary to completely react with our iron two plus. We could figure out moles from molarity and volume. Let's get some more room down here. We know that molarity is equal to moles over liters. The molarity of permanganate is .02. We have .02 for the concentration of permanganate ions. Moles is what we're solving for. It took us 20 milliliters for our titration, which we move our decimal place one, two, three,"
    },
    {
        "Q": "How can he assume it is an acidic solution at minute 0:25?",
        "A": "The equation has H\u00e2\u0081\u00ba in it. That s an acid.",
        "video_name": "EQJf8Gb8pg4",
        "timestamps": [
            25
        ],
        "3min_transcript": "- [Voiceover] We've already seen how to do an acid-base titration. Now let's look at a redox titration. Let's say we have a solution containing iron two plus cations. We don't know the concentration of the iron two plus cations, but we can figure out the concentration by doing a redox titration. Let's say we have 10 milliliters of our solution, and let's say it's an acidic solution. You could have some sulfuric acid in there. In solution, we have iron two plus cations and a source of protons from our acid. To our iron two plus solution, we're going to add some potassium permanganate. In here, we're going to have some potassium permanganate, KMnO4. Let's say the concentration of our potassium permanganate is .02 molar. That's the concentration that we're starting with. Potassium permanganate is, of course, the source of permanganate anions, and MnO4 minus. Down here, we have a source of permanganate anions. We're going to drip in the potassium permanganate solution. When we do that, we're going to get a redox reaction. Here is the balanced redox reaction. If you're unsure about how to balance a redox reaction, make sure to watch the video on balancing redox reactions in acid. Let's look at some oxidation states really quickly so we can see that this is a redox reaction. For oxygen, it would be negative two. We have four oxygens, so negative two times four is negative eight. Our total has to add up to equal negative one. For manganese, we must have a plus seven, because plus seven and negative eight give us negative one. Manganese has an oxidation state of plus seven. Over here, for our products, we're going to make Mn two plus. so the oxidation state is plus two. Manganese is going from an oxidation state to plus two. That's a decrease or a reduction in the oxidation state. Therefore, manganese is being reduced in our redox reaction. Let's look at iron two plus. We have iron two plus as one of our reactants here. That means the oxidation state is plus two. For our products, we're making iron three plus, so an oxidation state of plus three. Iron is going from plus two to plus three. That's an increase in the oxidation state. Iron two plus is being oxidized in our redox reaction. As we drip in our potassium permanganate, we're forming our products over here. These ions are colorless in solution. As the permanganate reacts, this purple color disappears"
    },
    {
        "Q": "At 3:48, you said that lions and tigers can make ligers, but can animals of different species make hybrids of themselves?",
        "A": "Species are defined as being a distinct group that can breed within itself and produce viable, fertile offspring. Certain species (e.g. lions & tigers -> ligers, horses & donkeys -> mules) are close enough genetically that they can produce hybrids but these hybrids are sterile and not able to breed among themselves and produce offspring. Otherwise they would not represent a separate distinct species but rather a subpopulation or race.",
        "video_name": "Tmt4zrDK3dA",
        "timestamps": [
            228
        ],
        "3min_transcript": "but are actually closely related. And we'll talk about what it means to be closely related. And then we can see things that look very similar, that they have similar structures or they have similar behavior, like for example bats and birds, but they are actually all very, very distantly related. So we need a more exact definition for species than just things that look like each other, or just things that act like each other. And so the most typical definition for species are animals that can interbreed. And when we say interbreed, literally they can produce offspring with each other, and the offspring are fertile. Which means that the offspring can then further have babies, that they're not sterile, that they're capable of breeding with other animals and producing more offspring. You find a male lion and a female lioness and most of the time they will be able to have offspring, and those offspring can go and mate with other lions or lionesses, depending on their sex, and then they can have viable offspring. So it seems to work out pretty well for lions. Same thing is true of tigers. Now, it does turn out that if you get a male lion and a female tigress they can breed. They can breed and they can produce offspring. And their offspring-- which was made famous by Napoleon Dynamite, he was kind of fascinated by, these are kind of fascinating animals. Their offspring is called a liger. You get a male lion breeding with a female tiger you produce a liger, which is a hybrid, it's a cross between a lion and a tiger. And they're fascinating animals. They're actually larger than either lions or tigers. They are the largest cats that we know of. Or you can't say that lions and tigers are the same species, because even though they are able to interbreed, their offspring, for the most part, is not fertile, is not able to produce offspring. There have been one off stories about ligers being mated with either a lion or a tiger, but those are one off stories. In general, ligers can't interbreed. And in general, this combination isn't going to produce offspring that can keep interbreeding or that are fertile. So that's why we say that lions and tigers are different species. And that liger, we wouldn't even call it as a species at all. We would actually call it a hybrid between two species. Now the same thing is true-- and actually you might be asking yourself, well, this was a male lion and a female tigress, what if we went the other way around? What if we had a female lioness and a male tiger? In that case, you would produce something else called a tiglon or a tiglon, I actually don't"
    },
    {
        "Q": "At 3:13, Sal decides to integrate with a lower bound of zero and an upper bound of infinity. Could we also have integrated from infinity to zero, thereby summing all of the rings from an infinite distance away up to the point charge at a distance of zero?",
        "A": "Of course but you should then multiply the result by -1 as you reversed the limits of the integration.",
        "video_name": "TxwE4_dXo8s",
        "timestamps": [
            193
        ],
        "3min_transcript": "So at the end, we meticulously calculated what the y-component of the electric field generated by the ring is, at h units above the surface. So with that out of the way, let's see if we can sum up a bunch of rings going from radius infinity to radius zero and figure out the total y-component. Or essentially the total electric field, because we realize that all the x's cancel out anyway, the total electric field at that point, h units above the surface of the plane. So let me erase a lot of this just so I can free it up for some hard-core math. And this is pretty much all calculus at this point. So let me erase all of this. Watch the previous video if you forgot how it was derived. need a lot of space. There you go. OK, so let me redraw a little bit just so we never forget what we're doing here because that happens. So that's my plane that goes off in every direction. I have my point above the plane where we're trying to figure out the electric field. And we've come to the conclusion that the field is going to point upward, so we only care about the y-component. It's h units above the surface, and we're figuring out the electric field generated by a ring around this point of radius r. And what's the y-component of that electric field? We figured out it was this. So now what we're going to do is take the integral. So the total electric field from the plate is going to be the integral from-- that's a really ugly-looking integral-- So we're going to take a sum of all of the rings, starting with a radius of zero all the way to the ring that has a radius of infinity, because it's an infinite plane so we're figuring out the impact of the entire plane. So we're going to take the sum of every ring, so the field generated by every ring, and this is the field generated by each of the rings. Let me do it in a different color. This light blue is getting a little monotonous. Kh 2pi sigma r dr over h squared plus r squared to the 3/2. Now, let's simplify this a little bit. Let's take some constants out of it just so this looks like a slightly simpler equation. So this equals the integral from zero to-- So let's take"
    },
    {
        "Q": "At 5:10, Sal explains that when you dump (H^+) into the blood, the buffer system prevents the pH from going up by creating an equilibrium that bonds the H+ to the Bicarbonate. Why does that prevent the pH from going up? There is still a larger amount of H+ in the blood.",
        "A": "A solution gets more acidic (lower pH) when there are more free H+ ions in it. The bicarbonate takes these free H+ ions out of the solution. H+ that is bonded to something does not affect the pH.",
        "video_name": "gjKmQ501sAg",
        "timestamps": [
            310
        ],
        "3min_transcript": "But the topic of this video is why this is also useful for maintaining our blood pH in this range. Because these equilibrium reactions between carbon dioxide, carbonic acid, and bicarbonate this is a buffer system. This is a buffer, this is a buffer system. And the word \"buffer,\" in our everyday language, it refers to something that kind of smooths the impact of something, or it reduces the shock of something. And that's exactly what's happening here. Let's think about, remember, these are all equilibrium reactions, this is a weak acid, and you can even look at the different constituents of these molecules and account for them. You have one carbon here, one carbon here, one carbon there. You have one, two, three oxygens there. You have one, two, three oxygens there. One, two, three oxygens there. You have two hydrogens, two hydrogens, two hydrogens. But let's just think about what if you started dumping hydrogen ions in the blood. So, what if you were to dump hydrogen ions, Well, if you dump more hydrogen ions, if this right over here increases. Actually, let me put it this way, if you were to just dump hydrogen ions and if you didn't have this buffer system, then your pH would decrease. Your pH would go down, and if you do it enough, your pH, you would end up with acidosis. But lucky for us, we have this buffer system. And so if you increase your hydrogen ion concentrations, Le Chatelier's principle tells us, \"Hey, these equilibrium reactions are going to move to the left.\" So the more hydrogen ions you have sitting in the blood, the more likely they're gonna bump into the bicarbonate in just the right way to form carbonic acid. And the more carbonic acid that you have in the blood, well, it's the less likely that you're going to have the carbon dioxide reacting with the water to form more carbonic acid. So, as you add more hydrogen ions, by the bicarbonate. So this equilibrium, this set of equilibrium reactions is going to move to the left. So you're not going to have as big effect on pH. And similarly, if you dumped some base, let's say, you dumped some base in your blood right over here, well, instead of it just making your pH go up, and possibly give you alkalosis, well now, the base is going to sop up the hydrogen ions, and typically that would just make your pH go up, but if you have these things going down, well then, you have fewer of these to react and have the equilibrium reaction go to the left and so the reaction is going to move more and more to the right. And so this reaction, you're just gonna have more carbon dioxide being converted to carbonic acid being converted to bicarbonate. This whole thing is going to move to the right. And so it's going to be able, to some degree, replace the lost hydrogen ions. So this right over here's a buffer system."
    },
    {
        "Q": "I just wanted to thank you for this video. It was fantastic and it was very helpful. Just really quick question. This turtle doesn't have displacement because she is not changing position. She is just changing distance as seen at 3:33 but she doesn't move left or right? Because she is just moving forwards and backwards this is different than if she were moving left and backwards or right and forwards?",
        "A": "you can be displaced forward and backward just as well as left or right. any direction will do.",
        "video_name": "GtoamALPOP0",
        "timestamps": [
            213
        ],
        "3min_transcript": "If this turtle didn't go forward, down, and up, what did this turtle do? We'll start at t equals zero. We'll go up from there. And at t equals zero the value of this graph is three. And the value of this graph is representing the horizontal position, so the value of the graph is giving you the horizontal position. So at t equals zero, the turtle is at three meters. So let's put her over at three meters. She starts over here. Three meters, that's equal zero. Now what happens? So at t equals one second, same thing. We read our graph by going up, hit the graph, then we go left to figure out where we're at. Again turtle's still at three. At two seconds we come up, hit the graph. We come over to the left to figure out where we're at. This turtle's still at three, that's awkward. This turtle didn't even move. For the first two seconds this turtle's just sitting here. So a straight line a horizontal line on a position graph represents no motion whatsoever. This is awkward. Turtle was probably trying to figure out how to turn on her jet pack. 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."
    },
    {
        "Q": "5:27, where did you get the 8? I added from three and went down where it stopped and got more than 8.",
        "A": "I think he might have forgotten to count 0 when he was counting down from 3.",
        "video_name": "GtoamALPOP0",
        "timestamps": [
            327
        ],
        "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": "Why he made an assumption that it is moving horizontally not vertical? at 12:58",
        "A": "because objects can easily roll along the ground and maintain their speed, but it is not so easy to do that in the vertical direction. Gravity interferes.",
        "video_name": "GtoamALPOP0",
        "timestamps": [
            778
        ],
        "3min_transcript": "So at four seconds and negative five meters that's our .2. Alright so x two, that would be negative five, 'cause I'm just reading my graph, that .2. That's negative five. So I got negative five meters minus x one that's this. Don't make x one four. That's a time, that's not a position. So point one, the horizontal position was three. So positive three. Put the negative here 'cause the negative's in the formula. And then divide it by time two, that was four seconds. And minus t one was two seconds. And if you saw this thing, negative five and negative three and negative eight divided by two seconds. Oops can't figure out my units. Oh look at that. I got negative four meters per second. That was the instantaneous velocity at three seconds. Negative four meters per second. Negative because the turtle was going backwards. She turned on the reverse booster instead of the forward booster. Negative and four because look it going four meters every second. Made it eight meters and two seconds, that means she was going four meters per second on average. And since it's a straight line, that was the rate she was going at any moment. Beautiful. Alright that would've been if the follow up question is what is it at 2.4 seconds? Don't get concerned. Look it's the same everywhere. It'd be the same answer. Negative four meters per second. For this whole line. What else can we figure out one last thing. Let's say you were asked what's the instantaneous speed at a point? So I'm gonna write that as SINST instantaneous speed, or just s. 'Cause that's usually what we mean by speed. Equals average value, sorry, absolute value of the instantaneous velocity. So now here I've got to make an assumption. This is gonna get a little subtle. If all we're given is a horizontal position graph, This turtle could've gone back and forth, or the turtle could've been like flying upward, as she went back and forth. And if the horizontal location was the same the whole way, this would've looked exactly the same regardless of whether the turtle had any vertical motion at all. So we gotta be careful, 'cause the speed is the magnitude of the total velocity. This is the just the velocity in the x direction. So we're gonna make an assumption. I'm gonna assume this turtle was just moving horizontally. Instead of having the vertical motion. She's not ready for that yet. Alright so how do you get this? Speed is just the absolute value. The magnitude of the instantaneous velocity. And if this is the only component of velocity, then I can just figure this out pretty easy by saying that, oh I gotta give you time, makes no sense to say instantaneous speed. I gotta say instantaneous speed at a given moment. 'Cause the instantaneous speed here was zero."
    },
    {
        "Q": "At 12:50, doesn't the conjugate base create hydroxide ions in solution, adding to the total moles of OH-, causing [OH-] to be larger and pH to be lower? Or does acetate not affect the pH after the equivalence point?",
        "A": "You are correct! It will add to the total amount of OH-, increasing its molarity. But the change is so insignificant that the change is only in the slightest. I did the whole calculation for you and found out that the concentration changes only 9 (!!) places after the decimal. This will cause only a minuscule change in the pH. Also, the higher the concentration of OH-, the higher the pH.",
        "video_name": "WbDL7xN-Pn0",
        "timestamps": [
            770
        ],
        "3min_transcript": "So right about there, about 8.67. So, that's our equivalence point for a titration of a weak acid with a strong base for this particular example. Finally we're on to Part D, which asks us, what is the pH after the addition of 300.0 mL of a 0.0500 molar solution of sodium hydroxide? So, once again, we need to find the moles of hydroxide ions that we are adding. The concentration would be equal to 0.0500, so 0.0500 molar is our concentration of hydroxide ions. And that's equal to moles over liters. So, 300.0 mL would be 0.3000 liters. So, we have 0.3000 liters here. Multiply 0.0500 by 0.3000, and you get moles. So, 0.0500 times 0.300 moles of hydroxide ions. In Part C, we saw that we needed 0.0100 moles of hydroxide ions to completely neutralize the acid that we originally had present. So, we're going to use up... We're going to use up 0.0100 moles of hydroxide. That's how much was necessary to neutralize our acid. All right, so how many moles of hydroxide are left over after the neutralization? Well, that would just be 0.0050. So, 0.0050 mol of hydroxide ions are left over after all the acid has been neutralized. So, our goal is to find the pH. So, we could find the pOH if we found if we found the concentration of hydroxide. So, what is the concentration of hydroxide ions now after the neutralization has occurred? so it's 0.0050 divided by... What's our total volume? We started with 50.0, and we have now added 300.0 mL more, so 300.0 plus 50.0 is 350.0 mL. Or 0.3500 liters. So, what is our concentration of hydroxide ions? So, this is .005 divided by .35. So our concentration of hydroxide ions is 0.014. So, let's write 0.014 M here. Once we know that, we can calculate the pOH. So the pOH is the negative log of the concentration of hydroxide ions. So, it's the negative log of 0.014. So, we can do that on our calculator. Negative log of .014. And we get a pOH of 1.85."
    },
    {
        "Q": "at 4:36 why is it called The Bowman's Capsule?",
        "A": "A glomerulus is enclosed in the sac. Fluids from blood in the glomerulus are collected in the Bowman s capsule (i.e., glomerular filtrate) and further processed along the nephron to form urine. This process is known as ultrafiltration. The Bowman s capsule is named after Sir William Bowman, who identified it in 1842",
        "video_name": "UU366tJPovg",
        "timestamps": [
            276
        ],
        "3min_transcript": "And even though they do an amazing job, I'm not badmouthing your kidneys here, the way that they do it is frankly a little bit janky and inefficient. They start out by filtering a bunch of fluid and the stuff dissolved in the fluid out of your blood and then they basically reabsorb 99 percent of it back before sending that one percent on its way in the form of urine. Seriously 99 percent gets reabsorbed. On an average day your kidneys filter out about 180 liters of fluid from your blood but only 1.5 liters of that ends up getting peed out. So most of your excretory system isn't dedicated to excreting, it's dedicated to reabsorbing. But the system works, obviously. I'm still alive so we can't argue with that. Now it is time to get into the nitty-gritty details of how your kidneys do all this and it's pretty cool but there's lots of weird words so get ready. Your kidneys do all this work using a network of tiny filtering structures called nephrons, each one of your mango sized kidneys has about a million of them. If you were, don't do this, but if you were to unravel all of your nephrons and put them end to end they would stretch over 80 kilometers. we're just going to follow the flow from your heart to the toilet. Blood from the heart enters the kidneys through renal arteries and just so you know, whenever you hear the word renal, it means we're dealing with kidney stuff. As the blood enters it's forced into a system of tiny capillaries until it enters a tangle of porous capillaries called the glomerulus. This is the starting point for a single nephron. The pressure in the glomerulus is high enough that it squeezes some of the fluid out of the blood, about 20 percent of it, and into a cup-like sac called the bowman's capsule. The stuff that gets squeezed out is no longer blood, it is now called filtrate. It's made up of water, urea, some smaller ions and molecules like sodium, glucose and amino acids. The bigger stuff in your blood like the red blood cells and the larger proteins, they don't get filtered. Now the filtrate is ready to be processed from the bowman's capsule it flows into a twisted tube called the proximal convoluted tubule which means the tube near the beginning and that is all windy. Anyway, this is the first of two convoluted tubules in the nephron and these along with other tubules we're talking about are where the osmoregulation takes place. With all kinds of tricked out specialized pumps and other kinds of active and passive transport, they reabsorb water and dissolve materials to create whatever balance your body needs at the time. In the proximal tubule it's mainly organic solutes and the filtrate that are reabsorbed like glucose and amino acids and other important stuff that you want to hang onto. But it also helps to recapture some sodium and potassium and water that we're going to want later. From here the filtrate enters the loop of Henle, which is a long hairpin shaped tubule that passes through the two main layers of the kidney. The outermost layer is the renal cortex, that's where the glomerulus and the bowman's capsule and both convoluted tubules are and the layer beneath that is the renal medulla which is the center of the kidney. Cortex, by the way, is Latin for tree bark so whenever you see it in biology you know that it's the outside of something."
    },
    {
        "Q": "At 4:01, could you also call it a \"vectorial field?\"",
        "A": "Hadnt thought about it that way, but yes, I think you could call it a vectoral field Why do you ask? What are you thinking??",
        "video_name": "1E3Z_R5AHdg",
        "timestamps": [
            241
        ],
        "3min_transcript": "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 context in a second-- the average gravitational field at Earth's surface. And this is a little bit more of an abstract thing-- we'll talk about that in a second-- but it does help us think about how g is related to this scenario where I am not an object in free fall. A field, when you think of it in the physics context-- slightly more abstract notion when you start thinking about it in the mathematics context-- but in the physics context, a field is just something that associates a quantity with every point in space. So this is just a quantity with every point in space. And it can actually be a scalar quantity, in which case we call it a scalar field, and in which case it would just be a value. Or it could be a vector quantity, which would be a magnitude and a direction associated with every point in space. In which case you are dealing with a vector field. is, because at near Earth's surface, if you give me a mass-- so for example-- actually, I don't know what my mass is in kilograms. But if you're near Earth's surface and you give me a mass-- so let's say that mass right over there is 10 kilograms-- you can use g to figure out the actual force of gravity on that object at that point in space. So for example, if this has a mass of 10 kilograms, then we know-- and this right over here is the surface of the Earth, so that's the center of the Earth. So it actually associates a vector quantity whose magnitude,-- so its direction is towards the center of the Earth, and the magnitude of this vector quantity is going to be the mass times g. And you could take-- since we're already specifying the direction, we could say 9.81 meters per second squared towards the center of the Earth. And so in this situation, it would be 10 kilograms times 9.81 meters per second squared."
    },
    {
        "Q": "At 0:44, why is an object in free fall accelerating with a negative value since it is going towards the centre of the earth",
        "A": "Nature don t know you are positive or negative, you are free to choose sine conventions, either you choose negative or positive the important think is that you make your calculations right",
        "video_name": "1E3Z_R5AHdg",
        "timestamps": [
            44
        ],
        "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 0:26 he says \"methanol.\" I know this is a pretty elementary question, but why does methanol end in \"ol\"? As for a question actually pertaining to the video, does anyone have classic examples of molecules being used to teach oxidation states. (like practice problems?)",
        "A": "Methanol is a combination of the base methane plus an alcohol group, which tends to be simplified in nomenclature by ol . Thus, we have methanol.",
        "video_name": "CuGg-Tf8lPI",
        "timestamps": [
            26
        ],
        "3min_transcript": "- Both formal charge and oxidation states are ways of counting electrons, and they're both very useful concepts. Let's start with formal charge. So one definition for formal charge is the hypothetical charge that would result if all bonding electrons are shared equally. So let's go down to the dot structure on the left here, which is a dot structure for methanol, and let's assign a formal charge to carbon. We need to think about the bonding electrons or the electrons in those bonds around carbon, and we know that each bond consists of two electrons. So the bond between oxygen and carbon consists of two electrons. Let me go ahead and draw in those two electrons. Same for the bond between carbon and hydrogen, right? Each bond consists of two electrons, so I can go around and put in all of my bonding electrons. So if we want to assign a formal charge to carbon, we need to think about the number of valence electrons in the free atom or the number of valence electrons that carbon is supposed to have. four valence electrons, so I could put a four here, and from that four we're going to subtract the number of valence electrons in the bonded atom or the number of valence electrons that carbon has around it in our drawing. And since we're doing formal charge, we need to think about all those bonding electrons being shared equally. So we think about a covalent bond. So if we have two electrons and one bond, and those two electrons are shared equally, we could split them up. We could give one electron to oxygen and one electron to carbon in that bond. We go over here to this carbon hydrogen bond, and we could do the same thing. We have two electrons. We could split up those two electrons. We could give one to carbon and one to hydrogen, and we go all the way around, and we do the same thing over here. Split up those electrons and the same thing here. So how many valence electrons do we see around carbon now? So let me go ahead and highlight them. There's one, two, three, and four. around carbon in our drawing. So four minus four is equal to zero. So zero is the formal charge of carbon. So let me go ahead and highlight that here. So in this molecule the formal charge for carbon is zero. Now let's move on to oxidation states, right? So you could also call these oxidation numbers. So one definition for an oxidation state is the hypothetical charge that would result if all those bonding electrons are assigned to the more electronegative atom in the bond. So let's go to the dot structure on the right of methanol and let's assign an oxidation state to that carbon. We need to think about our bonding electrons again, so let's go ahead and put those in, all right? So we know that each bond consists of two electrons."
    },
    {
        "Q": "at @6:12 why didnt you change the final velocity vector into negative.",
        "A": "Because that is the magnitude it was coming down at, he was not trying to point out the direction.",
        "video_name": "sTp4cI9VyCU",
        "timestamps": [
            372
        ],
        "3min_transcript": "So the vertical component of our velocity is negative 29.06 times .03 in the downward direction. And the horizontal component of our velocity, we know, hadn't changed the entire time. That, we figured out, was 30 cosine of 80 degrees. So that over here, is 30 cosine of 80 degrees. I'll get the calculator out to calculate it. 30 cosine of 80 degrees, which is equal to 5.21. So this is 5.21 meters per second. These are both in meters per second. So what is the total velocity? Well, I can do the head to tails. So I can shift this guy over so that its tail is at the head of the blue vector. So it would look like that. The length of this-- the magnitude And then we could just use the Pythagorean theorem to figure out the magnitude of the total velocity upon impact. So the length of that-- we could just use the Pythagorean theorem. So the magnitude of our total velocity, that's this length right over here. The magnitude of our total velocity, our total final velocity I guess we can say, is going to be equal to-- well that's-- let me write it this way. The magnitude of our total velocity is going to be equal to square root-- this is just straight from the Pythagorean theorem-- of 5.21 squared plus 29.03 squared. And we get it as being the second-- the square root of 5.21 squared plus 29.03 squared This is equal to 29.49 meters per second. That is the magnitude of our final velocity, but we also need to figure out its direction. And so we need to figure out this angle. And now we're talking about an angle below the horizontal. Or, if you wanted to view it in kind of pure terms, it would be a negative angle-- or we could say an angle below the horizontal. So what is this angle right over here? So if we view it as a positive angle just in the traditional trigonometric way, we could say that the-- we could use any of the trig functions, we could even use tangent. Let's use tangent. We could say that the tangent of the angle, is equal to the opposite over the adjacent-- is equal to 29.03 over 5.21. Or that theta is equal to the inverse tangent, or the arctangent of 29.03 over 5.21."
    },
    {
        "Q": "At 10:17 Sal said that the reason why the car moves through the curve in a curved manner rather than going straight is because of friction. Shouldn't it be because of the change of vector velocity as Sal explained earlier in the satellite, yoyo and the the object travelling in a circular path in space examples?",
        "A": "the frictions is the cause of the change in vector velocity",
        "video_name": "vZOk8NnjILg",
        "timestamps": [
            617
        ],
        "3min_transcript": "Another example that you are probably somewhat familiar with or at least have heard about is if you have something in orbit around the planet So let's say this is Earth right here and you have some type of a satellite that is in orbit around Earth That satellite has some velocity at any given moment in time What's keeping it from not flying out into space and keeping it going in a circle is the force of gravity So in the example of a satellite or anything in the orbit even the moon in orbit around the Earth the thing that's keeping an orbit as opposed to flying out into space is a centripetal force of Earth's gravity Now another example, this is probably the most everyday example because we do it all the time If you imagine a car traveling around the racetrack Before I tell you the answer, I'll have you think about it It's circular. Let's view the racetrack from above If I have a car on a racetrack. I want you to pause it before I tell it to you because I think it's an interesting thing think about It seems like a very obvious thing that's happening We've all experienced; we've all taken turns in cars So we're looking at the top of a car. Tires When you see a car going at a constant speed so on the speedometer, it might say, 60 mph, 40 mph, whatever the constant speed but it's traveling in a circle so what is keeping--what is the centripetal force in that example? There's no obvious string being pulled on the car towards the center There is no some magical gravity pulling it towards the center of the circle There's obviously gravity pulling you down towards the ground but nothing pulling it to the side like this And I encourage you to pause it right now before I tell you the answer Assuming you now unpaused it and I will now tell you the answer The thing that's keeping it going in the circle is actually the force of friction It's actually the force between the resist movement to the side between the tires and the road And a good example of that is if you would remove the friction if you would make the car driving on oil or on ice or if you would shave the treads of the tire or something then the car would not be able to do this So it's actually the force of friction in this example I encourage you to think about that"
    },
    {
        "Q": "At 0:59, it was specified that the carbon turns into an sp2 hybridized carbon, meaning that all the substituents should be on the same plane, yet there are still wedges and dashes, indicating that the substituents of the carbon are not on the same plane. Can somebody explain this to me?",
        "A": "The substituents are still in the same plane. It is just that you are looking at the plane edge-on. This puts one bond closer to you (a wedge), one bond further away (a dashed line), and one that is equidistant at both ends (an ordinary line).",
        "video_name": "sDZDgctzbkI",
        "timestamps": [
            59
        ],
        "3min_transcript": "- [Narrator] In this video, we're going to look at the stereochemistry of the SN1 reaction. On the left is our alkyl halide, on the right is our nucleophile with a negative charge on the sulfur. We know that the first step of our SN1 mechanism should be loss of a leaving group. So if these electrons come off onto the bromine, we would form the bromide anion. And we're taking a bond away from the carbon in red. So the carbon in red should get a plus one formal charge. So let's draw the resulting carbocation here. So let me sketch that in. The carbon in red is this carbon. So that carbon should have a plus one formal charge. In the next step of our mechanism, our nucleophile will attack, alright? So the nucleophile attacks the electrophile and a bond will form between the sulfur and the carbon in red. But remember the geometry directly around that carbon in red, the carbons that are bonded to it, so this carbon in magenta, this carbon in magenta, and this carbon in magenta and so the nucleophile could attack from either side of that plane. At this point, I think it's really helpful to look at this reaction using the model set. So here's a screenshot from the video I'm going to show you in a second, and in that video I make bromine green, so here you can see this green bromine, this methyl group coming out at us in space is going to be red in the video. On the right side this ethyl group here will be yellow, and finally on the left side this propyl group will be gray. So here's our alkyl halide with our bromine going away from us, our methyl group coming out at us, our ethyl group on the right, and the propyl group on the left. So I'll just turn this a little bit so we get a different viewpoint, and we know that the first step is loss of our leaving group so I'm going to show these electrons coming off onto our bromine and leaving to form a carbocation. We need planar geometry around that central carbon. So here's another model which is more accurate. Now the nucleophile could attack from the left or from the right, and first let's look at what happens when the nucleophile attacks from the left. So we form a bond between the sulfur and the carbon, and let's go ahead and look at a model set of one of our products. So here's the product that results when the nucleophile attacks from the left side of the carbocation. Here's our carbocation again, and this time let's say the nucleophile approaches from the right side. So we're going to form a bond between this sulfur and this carbon. And let's make a model of the product that forms when the nucleophile attacks from the right. So here is that product. And then we hold up the carbocation so we can compare the two. Now let's compare this product with the product when the nucleophile attacked from the left side."
    },
    {
        "Q": "When Sal mentions that the helium core fuses into heavier elements at 6:02, why only up to iron 26? What about Uranium, gold, etc. ?\nSecondly what happens when A heavier element fuses with a light by one by somehow traveling from its shell or core to another?",
        "A": "It has to do with the strength of the Strong Nuclear force. The Strong Nuclear force has a limited distance that is works over. The size of the iron nucleus is the point where the Strong Nuclear force is weakened to the point where to make a larger atom it takes more energy to push the protons/neutrons together than comes out of the process. In systems called heavy ion colliders that they use them to produce heavy elements. So as long as there is enough energy in the collision they can fuse.",
        "video_name": "EdYyuUUY-nc",
        "timestamps": [
            362
        ],
        "3min_transcript": "as the Earth's orbit around the current sun. Or another way to view it is, where we are right now will be on the surface or near the surface or maybe even inside of that future sun. Or another way to put it, when the sun becomes a red giant, the Earth's going to be not even a speck out here. And it will be liquefied and vaporized at that point in time. So this is super, super huge. And we've even thought about it. Just for light to reach the current sun to our point in orbit, it takes eight minutes. So that's how big one of these stars are. To get from one side of the star to another side of the star, it'll take 16 minutes for light to travel, if it was traveling that diameter, and even slightly longer if it was to travel it in a circumference. So these are huge, huge, huge stars. And we'll talk about other stars in the future. They're even bigger than this when they become supergiants. But anyway, we have the hydrogen in the center-- sorry. Let me write this down. We have a helium core in the center. We're fusing faster and faster and faster. We're now a red giant. The core is getting hotter and hotter and hotter until it gets to the temperature for ignition of helium. So until it gets to 100 million Kelvin-- remember the ignition temperature for hydrogen was 10 million Kelvin. So now we're at 100 million Kelvin, factor of 10. And now, all of a sudden in the core, you actually start to have helium fusion. And we touched on this in the last video, but the helium is fusing into heavier elements. And some of those heavier elements, and predominately, it will be carbon and oxygen. And you may suspect this is how heavier and heavier elements form in the universe. They form, literally, due to fusion in the core of stars. 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,"
    },
    {
        "Q": "At 2:52, he mentioned that the star was starting to collapse. What exactly is meant by collapse?",
        "A": "i think he means that a while a red giant is expanding,there is also a white dwarf inside, which when the red giant sheds his outer layers, would reveal the white dwarf. but i m not sure.",
        "video_name": "EdYyuUUY-nc",
        "timestamps": [
            172
        ],
        "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": "Close to the end of a video, (12:21) it said that over many years, a white dwarf would eventually become burned out and turned into a black dwarf. Can you give me a time frame of how long that would take?",
        "A": "I have see estimates for around 8 to 11 billion years for a 0.5 to 1 solar mass white dwarf to cool to be a black dwarf.",
        "video_name": "EdYyuUUY-nc",
        "timestamps": [
            741
        ],
        "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": "At 1:40, he said our sun was even hotter than it was before because it is fusing faster. Is this causing global warming then?",
        "A": "no As compared to4.6 billion years ago it it hotter but global warming is caused by human activities in the mordern time(industrialisation). I would say that the solar output is stable since last 3 billion years and would be same for next 3-4 billion years",
        "video_name": "EdYyuUUY-nc",
        "timestamps": [
            100
        ],
        "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 7:48 would we be able to see a he flash from Earth?",
        "A": "Its not a literal flash. Its just a point in the Suns life cycle.",
        "video_name": "EdYyuUUY-nc",
        "timestamps": [
            468
        ],
        "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": "At 1:34, how is gas a fluid? I know it fills in the shape of what it is in, but how can it be a fluid? Also, at 6:30 what is that thing?t",
        "A": "a fluid is a substance which has the ability to flow. liquids can flow from a state of higher potential energy to lower potential energy. in gasses, the same thing applies. they are also considered as fluids because they flow to the edges of the container or from a hot surface to a cold surface. gasses are fluids because they have the ability to flow",
        "video_name": "Pn5YEMwQb4Y",
        "timestamps": [
            94,
            390
        ],
        "3min_transcript": "Let's learn a little bit about fluids. You probably have some notion of what a fluid is, but let's talk about it in the physics sense, or maybe even the chemistry sense, depending on in what context you're watching this video. So a fluid is anything that takes the shape of its container. For example, if I had a glass sphere, and let's say that I completely filled this glass sphere with water. I was going to say that we're in a zero gravity environment, but you really don't even need that. Let's say that every cubic centimeter or cubic meter of this glass sphere is filled with water. Let's say that it's not a glass, but a rubber sphere. If I were to change the shape of the sphere, but not really change the volume-- if I were to change the shape of the sphere where it looks like this now-- the water would just change its shape with the container. and in this case, I have green water. The same is also true if that was oxygen, or if that was just some gas. It would fill the container, and in this situation, it would also fill the newly shaped container. A fluid, in general, takes the shape of its container. And I just gave you two examples of fluids-- you have liquids, and you have gases. Those are two types of fluid: both of those things take the shape of the container. What's the difference between a liquid and a gas, then? A gas is compressible, which means that I could actually decrease the volume of this container and the gas will just become denser within the container. could squeeze that balloon a little bit. There's air in there, and at some point the pressure might get high enough to pop the balloon, but you can squeeze it. A liquid is incompressible. How do I know that a liquid is incompressible? Imagine the same balloon filled with water-- completely filled with water. If you squeezed on that balloon from every side-- let me pick a different color-- I have this balloon, and it was filled with water. If you squeezed on this balloon from every side, you would not be able to change the volume of this balloon. No matter what you do, you would not be able to change the volume of this balloon, no matter how much force or pressure you put from any side on it, while if this was filled with gas-- and magenta, blue in for gas-- you actually could decrease the volume by just increasing the pressure on all sides of the balloon. You can actually squeeze it, and make the entire volume smaller."
    },
    {
        "Q": "At 5:03 wouldn't you feel the G-force because 250 m/s > 9.8 m/s and couldn't you measure that?",
        "A": "you are comparing a velocity to an acceleration. It s not 9.8 m/s it s 9.8 m/s^2. That s acceleration. 250 m/s is a velocity.",
        "video_name": "3yaZ7lkQPUQ",
        "timestamps": [
            303
        ],
        "3min_transcript": "It would look like it's moving behind you or in this case, the way we're looking at it, to the right at 50 meters per second. Now, what would the plane look like? Well, the plane not only would it look like it's moving to the right at 250 meters per second, not only would it be just that 250 meters per second, but relative to you it'd look like it's going even faster 'cause you're going past it, you are going to the left from the stationary, from the ground's point of view at 50 meters per second. So the plane, to you, is gonna look like it's going 250 plus 50 meters per second. So the vector would look like this and so it would look like it's going to the right at 300, let me write that in that orange color, at 300 meters per second. What if we're talking about the plane's frame of reference? Why don't you pause this video and think about what the velocities would be of the plane, the car, and the ground from the plane's point of view. All right, now let's work through this together. So now, we're sitting in the plane and once again we shouldn't be flying the plane, we're letting someone else do that, we have our physics instruments out and we're trying to measure the velocities of these other things from my frame of reference. Well, the plane, first of all, is going to appear to be stationary and that might seem counterintuitive, but if you've ever sat in a plane, especially when there's no turbulence and the plane is already at altitude and it's not taking off or landing, oftentimes if you close your eyes you don't know if you are moving. In fact, if you close all of the windows, it feels like you are in a stationary object, that you might as well be in a house. So from the plane's point of view, you feel like, or from your point of view in the plane, Now the ground, however, looks like it's moving quite quickly. It'll look like it's moving past you at 250 meters per second. Whoops, try and draw a straight line. At 200... At 250. Sometimes my tools act funny. So, at 250 meters per second to the left. And the car, well it's moving to the left even faster. It's going to be moving to the left 50 meters per second faster than the ground is. So the car is gonna look not like it's just going 50 meters per second, it's gonna look like it's going 50 meters plus another 250 meters per second for a total of 300 meters per second to the left. So this gives you an appreciation"
    },
    {
        "Q": "Why does it take the same amount of time to cycle regardless of the amplitude? (that is to say, if the starting postion is A/2 it still takes the same amount time to go to -A/2 and return as it would if the starting postion were A)\n\nis it because the acceleration is less at A/2 than at A so the lower velocity means it takes longer to cycle? but why the same amount of time?\n\n(around 7:30 onwards)",
        "A": "Well from my understanding, the cycles both take the same amount of time because of its velocity, the one that is stretched more goes faster making it quicker, keeping a constant time, and the one that isnt stretched very far goes much slower, draging the time out even though it has less distance to cover. I m having a hard time explaining, it works well if you visualize it, imagine it and if you were to do it yourself.",
        "video_name": "oqBHBO8cqLI",
        "timestamps": [
            450
        ],
        "3min_transcript": "both sides of this equation times the inverse of the square root of k over m. And you get, t is equal to 2 pi times the square root-- and it's going to be the inverse of this, right? Of m over k. And there we have the period of this function. This is going to be equal to 2 pi times the square root of m over k. So if someone tells you, well I have a spring that I'm going to pull from some-- I'm going to stretch it, or compress it a little bit, then I let go-- what is the period? How long does it take for the spring to go back to its It'll keep doing that, as we have no friction, or no gravity, or any air resistance, or Air resistance really is just a form of friction. You could immediately-- if you memorize this formula, although you should know where it comes from-- you could It's 2 pi times m over k. That's how long it's going to take the spring to get back-- to complete one cycle. And then what about the frequency? If you wanted to know cycles per second, well that's just the inverse of the period, right? So if I wanted to know the frequency, that equals 1 over the period, right? Period is given in seconds per cycle. So frequency is cycles per second, and this is seconds per cycle. So frequency is just going to be 1 over this. Which is 1 over 2 pi times the square root of k over m. That's the frequency. But I have always had trouble memorizing this, and this. k over m, and m over k, and all of that. All you have to really memorize is this. And even that, you might even have an intuition as to why it's true. You can even go to the differential equations if you want to reprove it to yourself. Because if you have this, you really can answer any question The velocity of the mass, at any time, just by taking the Or the period, or the frequency of the function. As long as you know how to take the period and frequency of trig functions. You can watch my videos, and watch my trig videos, to get a refresher on that. One thing that's pretty interesting about this, is notice that the frequency and the period, right? This is the period of the function, that's how long it takes do one cycle. This is how many cycles it does in one second-- both of them are independent of A. So it doesn't matter, I could stretch it only a little bit, like there, and it'll take the same amount of time to go back, and come back like that, as it would if I stretch it a lot. It would just do that. If I stretched it just a little bit, the function would look like this. Make sure I do this right. I'm not doing that right. Edit, undo. If I just do it a little bit, the amplitude is going to be less, but the function is going to essentially do the"
    },
    {
        "Q": "1:25\nit said that that molecule's longest chain is 9-carbons\nbut i think that's wrong\nit's (i know this is the incorrect name, but i'm using it for an example) 2-propylheptane\n(emphasis on propyl)\npropyl means 3, but you wrote 4\nthe yellow line means it connects to the 9-carbons\nit's not the actual molecule\nso i believe the actual name is not 4-methylnonane, but 4-methyloctane!\nplease tell me if i am wrong or correct\ni'm just a bit confused...",
        "A": "The carbon atom that forms part of the long chain is counted as part of part of the alkane group but not the alkyl group. And remember that the lines are the bonds, the important things are the carbon atoms which are the end-points of the lines.",
        "video_name": "CFBKfgGTP98",
        "timestamps": [
            85
        ],
        "3min_transcript": "In the last video, we tried to draw a 2-Propylheptane. And we did our best attempt at drawing it, but it was pointed out that this wouldn't even be called a 2-Propylheptane to So you actually should never see something called 2-Propylheptane. Let me show you what I'm talking about. So when do you see something like this, you might immediately say-- and the way we drew it was actually correct, it just wouldn't be called 2-Propylheptane. So you say heptane. So that is a seven carbon alkane. No double bond, so one, two, three, four, five, six, seven. And then on the second carbon, so you have one, two, three, four, five, six, seven, we have a propyl group. Propyl, that is three carbons. So on the second carbon we have a propyl group. That's three carbons. So that is one, two, three. And so if someone gave you 2-Propylheptane, this would be what you would draw. But you wouldn't call this 2-Propylheptane. Because remember, if you're given the molecule you look for the longest chain and the longest chain here is not the It is not one, two, three, it is not this thing in magenta-- in this kind of mauve color. It's this chain where you start over here. If you start over here, one, two, three, four, five, six, seven, eight, nine, you actually get a longer chain. So this would actually be the backbone of this molecule right over there. That right over there would be the backbone, and so you would number it. You start numbering closest to the group that's attached, so one, two, three, four, five, six, seven, eight, nine. So you have nine carbons in your backbone, so we're dealing with nonane. We're dealing with nonane. And you have a methyl group: one carbon attached to the So this is going to be four methyl, this is our methyl group right here, 4-Methylnonane. So it was brought up, I think the user name is Minoctu, and they correctly corrected me, that there would never be such a thing as 2-Propylheptane. I just made that up. If someone were to, kind of, label this molecule they would call it 4-Methylnonane and ask you draw it. But either way, both of these would point you in the right direction. This would just be the incorrect name for it, because you'd be looking at-- if someone gave you this molecule and you named it this way, that would be incorrect. So I apologize for this. This is 4-Methylnonane. If you do heptane you're not finding the longest chain."
    },
    {
        "Q": "So, what would be the name of the actual neurotransmitter that is released as a response to the Glomus cell depolarizing? Would this be acetocholine or epinephrine? (6:24 in video)",
        "A": "Neurotransmitters actually known to be used by the glomus cells are : dopamine, noradrenaline, acetylcholine, substance P, vasoactive intestinal peptide and enkephalins.",
        "video_name": "cJXY3Cywrnc",
        "timestamps": [
            384
        ],
        "3min_transcript": "and these are the chemoreceptors that we're talking about. So these blue cells together make up a body of tissue, and that's where we get the term \"aortic body\" and \"carotid body.\" Now, on the carotid side, one interesting fact is that this body of tissue gets a lot of blood flow, in fact, some of the highest blood flow in the entire human body. It's about 2 liters per minute for 100 grams. And just to put that in perspective for the carotid body, imagine that you have a little 2-liter bottle of soda. I was thinking of something that would be about 2 liters, and soda came to mind. And you can imagine pouring this soda out over something that's about 100 grams-- maybe a tomato. That's about a 100-gram tomato. And if you could do this in one minute, if you could pour out this bottle in one minute, imagine how wet this tomato's going to get, how much profusion, in a sense, this tomato is going to get. That is how much profusion your carotid body gets. how much blood flow's going into that area. So let's now zoom in a little bit further. Let's say I have a capillary. And inside my capillary, I've got a little red blood cell here, floating around. And my red blood cell, of course, has some hemoglobin in it, which is a protein. And this protein has got some oxygen bound to it. I'm going to draw little blue oxygen molecules. And of course, there's some oxygen out here in the plasma itself as well. And if we're in our carotid body or aortic body, you might have these special little blue cells that I've been drawing, our peripheral chemoreceptor cells. And specifically they have a name. These things are called glomus cells. I had initially misstated it as a globus cell. But actually it's an M-- glomus. And these oxygen molecules-- these are oxygen molecules over here-- are going to diffuse down into the tissue and get into our glomus cell. It's going to look something like that. And if you have a lot of oxygen in the blood, of course, But if you don't have too much in here, then not too much is going to make its way into the cell. And that's actually the key point. Because what our cell is going to be able to do is start to detect low oxygen levels. Low oxygen levels in the glomus cell tells this cell that, actually, there are probably low levels in the blood. And when the levels are low, this cell is going to depolarize. Its membrane is going to depolarize. And what it has on the other side are little vesicles that are full of neurotransmitter. And so when these vesicles detect that, hey, there's a depolarization going on, these vesicles are going to dump their neurotransmitter out. And what you have waiting for them is this nice little neuron. So there's a nice little neuron waiting patiently for a signal, and that signal is going to come in the form of a neurotransmitter. So this is how the communication works. There's going to be a depolarization,"
    },
    {
        "Q": "At 09:43, Sal uses Coulomb's Law to calculate the \"force generated by the ring\". Coulomb's Law defines the electrostatic force that exists between two point charges, and not that between a thin ring and a point on it's axis. He is applying the Law to calculate the latter, which, obviously cannot be justified. Has he made a mistake?",
        "A": "No. He didn t make a mistake. It s like we assume a small amount of charge say dq and calculate its electric field at the axis of the coil then we tend to integrate the whole thing to obtain Electric field due to charge on entire coil.",
        "video_name": "prLfVucoxpw",
        "timestamps": [
            583
        ],
        "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": "At 0:10 why is the electric field constant? Doesn't it get weaker the further away you get from the electric field?",
        "A": "Not with infinite plates.",
        "video_name": "prLfVucoxpw",
        "timestamps": [
            10
        ],
        "3min_transcript": "In this video, we're going to study the electric field created by an infinite uniformly charged plate. And why are we going to do that? Well, one, because we'll learn that the electric field is constant, which is neat by itself, and then that's kind of an important thing to realize later when we talk about parallel charged plates and capacitors, because our physics book tells them that the field is constant, but they never really prove it. So we will prove it here, and the basis of all of that is to figure out what the electric charge of an infinitely charged plate is. So let's take a side view of the infinitely charged plate and get some intuition. Let's say that's the side view of the plate-- and let's say that this plate has a charge density of sigma. And what's charge density? It just says, well, that's coulombs per area. Charge density is equal to charge per area. That's all sigma is. So we're saying this has a uniform charge density. So before we break into what may be hard-core mathematics, might want to review some of the electrostatics from the physics playlist, and that'll probably be relatively easy for you. If you're watching this from the physics playlist and you haven't done the calculus playlist, you should not watch this video because you will find it overwhelming. But anyway, let's proceed. So let's say that once again this is my infinite so it goes off in every direction and it even comes out of the video, where this is a side view. Let's say I have a point charge up here Q. So let's think a little bit about if I have a point-- let's say I have an area here on my plate. Let's think a little bit about what the net effect of it is going to be on this point charge. Well, first of all, let's say that this point charge is at a height h above the field. Let me draw that. This is a height h, and let's say this is the point directly right here is r. So first of all, what is the distance between this part of our plate and our point charge? What is this distance that I'll draw in magenta? What is this distance? Well, the Pythagorean theorem. This is a right triangle, so it's the square root of this side squared plus this side squared. So this is going to be the square root of h squared plus r squared. So that's the distance between this area and our test charge. Now, let's get a little bit of intuition. So if this is a positive test charge and if this plate is positively charged, the force from just this area on the charge is going to be radially outward from this area, so it's going to be-- let me do it in another color because I don't want to-- it's going to go in that direction, right?"
    },
    {
        "Q": "At 7:45 and a little after, why are the oxygens positive? I thought when you have an oxygen-hydrogen bond the oxygen becomes partially negative?",
        "A": "The oxygen in this case is positive because it is sharing 2 of its non-valence electrons with that hydrogen. Its 2 valence electrons are already tied up in the two covalent bonds with the two carbons, and when it gives up some of its rights to the 2 non-valence electrons to the hydrogen, it reduces its negative charge (making it positive). :)",
        "video_name": "OpyTJbzA7Fk",
        "timestamps": [
            465
        ],
        "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": "At 3:43, Sal says that the nuclear membrane starts to disappear. Does this not break the law of conservation of mass? Where did it go? And how?",
        "A": "This just means it breaks down, not that it disappears into thin air.",
        "video_name": "TKGcfbyFXsw",
        "timestamps": [
            223
        ],
        "3min_transcript": "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 of where the cell is and its life cycle. It's amazing that this is happening. It's amazing that the structures, and what sometimes we consider to be a simple thing but it's actually incredibly complex thing. It kind of knows what to do even though it has no intelligence here. And a lot of what I'm talking about, the general overview of the process is well understood but scientists are still understanding exactly how do the different things happen when they should happen and by what mechanism and what's actually happening. Sometimes, molecular or atomic basis. But anyway, this first phase of mitosis, the nuclear envelope, the nuclear membrane starts to disappear. The centrosomes migrate to the opposite ends of the cell. And our DNA condenses into kind of the condensed form of the chromosomes. And we call this prophase. Prophase. Prophase of mitosis. In the next phase, let me draw my cell again. Drawing that same green color. In the next phase, your nuclear membrane is now gone. And the chromosomes start lining up in the middle of the cells. So you have the blue one right over here, the blue one that's one sister chromatid. Here's another sister chromatid and they're connected at the centromere, not to be confused with a centrosome. And then, here's the magenta, one of the magenta sister chromatids and another one. And once again, it's not magenta in real life, I'm just making it a magenta because it looks nice. That's the centromere right over there. Our centrosomes are at opposite ends of the cell at this point. At opposite ends of the cell."
    },
    {
        "Q": "At 4:48, Sal says that the chromosomes are not magenta in real life. What color are they in real life?",
        "A": "I believe it depends on the dye used on the cell before viewing it under the microscope. Though I am not fully certain.",
        "video_name": "TKGcfbyFXsw",
        "timestamps": [
            288
        ],
        "3min_transcript": "of where the cell is and its life cycle. It's amazing that this is happening. It's amazing that the structures, and what sometimes we consider to be a simple thing but it's actually incredibly complex thing. It kind of knows what to do even though it has no intelligence here. And a lot of what I'm talking about, the general overview of the process is well understood but scientists are still understanding exactly how do the different things happen when they should happen and by what mechanism and what's actually happening. Sometimes, molecular or atomic basis. But anyway, this first phase of mitosis, the nuclear envelope, the nuclear membrane starts to disappear. The centrosomes migrate to the opposite ends of the cell. And our DNA condenses into kind of the condensed form of the chromosomes. And we call this prophase. Prophase. Prophase of mitosis. In the next phase, let me draw my cell again. Drawing that same green color. In the next phase, your nuclear membrane is now gone. And the chromosomes start lining up in the middle of the cells. So you have the blue one right over here, the blue one that's one sister chromatid. Here's another sister chromatid and they're connected at the centromere, not to be confused with a centrosome. And then, here's the magenta, one of the magenta sister chromatids and another one. And once again, it's not magenta in real life, I'm just making it a magenta because it looks nice. That's the centromere right over there. Our centrosomes are at opposite ends of the cell at this point. At opposite ends of the cell. let me label this again. I labeled it in a previous video. That's centrosomes. Centrosomes where the two sister chromatids are connected. That's a centromere. Centromere. Now, you might heard of the word centrioles. Centrioles are actually, they help, they are exist inside the centrosomes. They're these two kind of cylindrical looking structures. Each of the centrosomes have two centrioles. But for the sake of this video, you can say, \"Well, the centrioles are just part of the centrosomes.\" But I'm just listing you all the words that involves centri something. Centrioles right over there and you have two of them per centrosome. So hopefully, that helps clarify some confusion. But these things line up. And a lot of what you're about to see, this orchestration, these things moving around in the cell, things lining up and soon things pulling apart, these are all coordinated with actually a fairly sophisticated mechanism of kind of a scaffold of this kind of ropes, these microtubules."
    },
    {
        "Q": "at 8:43, how can two chromosomes become four chromosomes?",
        "A": "The chromosomes (genetic material) duplicates in the previous cell phases (interphase/prophase) that occur before the splitting in mitosis",
        "video_name": "TKGcfbyFXsw",
        "timestamps": [
            523
        ],
        "3min_transcript": "What's going to happen next? What's going to happen next is ... Let me, I don't wanna draw it too big 'cause I wanna be able to fit it all in one page. What's going to happen next is that those microtubules are going to start pulling on each of the sister chromatids. Let me draw that. So you have this centrosome right over here. You have all these microtubules in your cell. You have the centrosome right over here, all those microtubules. And this one is going to be pulling, is going to be getting one of the blue chromatids to pull towards it or to move towards it. So let me draw that. So this is one blue chromatid right over here. aAd this one is going to be pulling the other blue chromatid towards it. And same thing for the magenta. And same thing for the magenta. And this one is being pulled that way. And just in case you're concerned about the, some of more of the words of vocabulary involved, the point it was this microtubules connect to what used to be sister chromatids, but now that they're apart, we now call them chromosomes. When they emerge, this is one chromosome and they have two sister chromatids. But now they're apart, we would actually now consider this each an independent chromosome. So now, you actually have four chromosomes over here. This point right over here, we call it kinetochore. And even exactly how that interphase works and exactly how things move is not fully understood. Some of this stuff is understood but some of this is still an area of research. So something even as basic as cell division, not so basic after all. So this right over here where you can start to see the DNA kind of migrating to their respective sides of the cell. We call this the anaphase. We call this the anaphase. and that is called telophase. Telophase. And in telophase, I'm gonna do my best shot to draw it. And you could already see I have started to draw that the cellular membrane is starting to pinch in kind of in preparation for cytokineses, in preparation for the cell splitting into two cells. So I'll do it a little bit more. Cytokineses is usually described as kinda being a separate process in mitosis although it's obviously they are, they kind of together help the cell fully turn into two cells. So now in telophase, so you have this, what used to be a sister chromatid, now we can call it a chromosome on its own. And we have this chromosome. And we have this chromosome right over there. We have to do it on both sides. So you're there and you have"
    },
    {
        "Q": "At 9:05 it is mentioned that if young boys have any precocious puberty it is pathological, but that may not be always the case in young girls. Is there any reason why is it like this?",
        "A": "Girls are more prone to environmental triggers for setting off puberty then boys are... Stress, changes in diet and eating habits, body weight and even sex itself can trigger puberty in women.",
        "video_name": "XhYQYVQq6K0",
        "timestamps": [
            545
        ],
        "3min_transcript": "If a girl comes in with just breast development but none of the body hair or body odor, we think, aha! she's been exposed to estrogens. And so we get a lot from the physical exam in terms of thinking what hormones to go after. okay, so even though the definitions of precocious is just any of these different physical developments happening too soon, the combination that you see is going to direct what yo uthing is going on wit hthem clinically. Exactly. Ok It will make us more or less worried and it will take us down one avenue or another, thinking about what could be wrong, what level of abnormality in the hormone system is at work. Ok. Can I ask you a question, just curious, so what are the most common diagnosis given to people coming in with their kids, that are concerned for precocious puberty? The common diagnosis, is \"variations are normal\". Aha. There is a condition called fancy doctor way of saying the adrenal glands producting their hormones, we see pubic hair, underarm hair, and acne. Uh... We also can see isolated breast development in little toddler girls that's not concerning. So that would just go away, or? Yes. Typically that begins between six months and two years and But even within the more full blown cases, where we're seeing breast and the pubic hair in a girl, or the enlagrement of the testes and the pubic hair in a boy, where's it's full on, full blown puberty, it may be idiopathic, or not pathological. We think that in girls, who develop fully, 95% of the time, even if they have the full blown puberty, it's idiopathic, or not pathological. So the condition is less common on boys, but when it occurs in boys, it's more worrisome. So it's less common for boys to have any signs of precocious puberty, but when it does happen you worry more. Exactly. Because it's more likeyl to be something bad. Exactly. So I'm guessing as an endocrinologist, once you've sort of ruled out the normal things, then you start th then you start thinking more about sort of what's going on in terms of all the different hormones that are controlling the se things, which maybe is something that we'll talk about in a different lecture exactly. is there any other sort of general things that you wanted to adress when we'll just talking about... I think the general advice I would give to a parent if see signs of puberty in their child, before 8 in a girl, before 9 in a boy, they should at least question their pediatrician"
    },
    {
        "Q": "5:20 why do we only divide 4.9 by 4 and not time?",
        "A": "4.9t is one term - the result of multiplying 4.9 and the change in time. If t is one, 4.9t=4.9. If t is two, 4.9t=9.8. If t is one (4.9t=4.9), then 4.9t/4=4.9/4=1.225. Dividing both by 4 would be 4.9/4/4, or 4.9t/16.",
        "video_name": "P7LKEkcNibo",
        "timestamps": [
            320
        ],
        "3min_transcript": "that we travel in the time up. So that'll give us our peak distance. Remember that distance-- or I should say displacement, in this situation-- displacement is equal to average velocity times change in time. The change in time that we care about is the time up. So that is our delta t over 2. Our total time divided by 2. This is our time up. And what's our average velocity? Well, the average velocity, if we assume constant acceleration, is your initial velocity plus your final velocity over 2. It's really just the mean of the two things. Well, we know what our initial velocity is. Our initial velocity is this thing over here. So this is this thing over here. Our final velocity-- remember, we're just talking about the first half of the time the ball is in the air-- We're talking when it gets to this peak point, right over here-- that's from two videos ago-- that peak point right over there. So our average velocity is just going to be this stuff divided by 2. So it's going to be 4.9 meters per second squared times delta t over 2. So this right here, this is our average velocity. Velocity average. So let's stick that back over here. So our maximum displacement is going to be our average velocity-- so that is 4.9 meters per second squared-- times delta t, all of that over 2. And then we multiply it again times the time up. So times delta t over 2 again. These are the same thing. And then we can simplify it. Our maximum displacement is equal to 4.9 meters per second squared times delta t squared, all of that over 4. And then we can just divide 4.9 divided by 4. 4.9 divided by 4 is-- let me just get the calculator out. I don't want to do that in my head, get this far and make a careless mistake. 4.9 divided by 4 is 1.225. So our maximum displacement is going to be 1.225 times our total time in the air squared, which is a pretty straightforward calculation."
    },
    {
        "Q": "At 3:14, why is delta-v equal to 0-initial velocity?",
        "A": "THat s how you find a change, you take final minus initial",
        "video_name": "P7LKEkcNibo",
        "timestamps": [
            194
        ],
        "3min_transcript": "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. that we travel in the time up. So that'll give us our peak distance. Remember that distance-- or I should say displacement, in this situation-- displacement is equal to average velocity times change in time. The change in time that we care about is the time up. So that is our delta t over 2. Our total time divided by 2. This is our time up. And what's our average velocity? Well, the average velocity, if we assume constant acceleration, is your initial velocity plus your final velocity over 2. It's really just the mean of the two things. Well, we know what our initial velocity is. Our initial velocity is this thing over here. So this is this thing over here. Our final velocity-- remember, we're just talking about the first half of the time the ball is in the air--"
    },
    {
        "Q": "At 5:16, why does Sal square delta t? To cancel out the units?",
        "A": "No..not to cancel the units, but to Simplify the calculation.",
        "video_name": "P7LKEkcNibo",
        "timestamps": [
            316
        ],
        "3min_transcript": "that we travel in the time up. So that'll give us our peak distance. Remember that distance-- or I should say displacement, in this situation-- displacement is equal to average velocity times change in time. The change in time that we care about is the time up. So that is our delta t over 2. Our total time divided by 2. This is our time up. And what's our average velocity? Well, the average velocity, if we assume constant acceleration, is your initial velocity plus your final velocity over 2. It's really just the mean of the two things. Well, we know what our initial velocity is. Our initial velocity is this thing over here. So this is this thing over here. Our final velocity-- remember, we're just talking about the first half of the time the ball is in the air-- We're talking when it gets to this peak point, right over here-- that's from two videos ago-- that peak point right over there. So our average velocity is just going to be this stuff divided by 2. So it's going to be 4.9 meters per second squared times delta t over 2. So this right here, this is our average velocity. Velocity average. So let's stick that back over here. So our maximum displacement is going to be our average velocity-- so that is 4.9 meters per second squared-- times delta t, all of that over 2. And then we multiply it again times the time up. So times delta t over 2 again. These are the same thing. And then we can simplify it. Our maximum displacement is equal to 4.9 meters per second squared times delta t squared, all of that over 4. And then we can just divide 4.9 divided by 4. 4.9 divided by 4 is-- let me just get the calculator out. I don't want to do that in my head, get this far and make a careless mistake. 4.9 divided by 4 is 1.225. So our maximum displacement is going to be 1.225 times our total time in the air squared, which is a pretty straightforward calculation."
    },
    {
        "Q": "At around 2:50, Sal divided the equation by (-1), but the right side of the equation remained the same, as he wrote it.\n\nIt matters for the direction of the velocity. Did it stay in the same direction or not?\n\nAnd if it did, why did he not put the velocity in the velocity in absolute value?\n\n\nThanks in advance for the helpers! :)",
        "A": "there was a negative symbol on both sides before he did that calculation",
        "video_name": "P7LKEkcNibo",
        "timestamps": [
            170
        ],
        "3min_transcript": "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. that we travel in the time up. So that'll give us our peak distance. Remember that distance-- or I should say displacement, in this situation-- displacement is equal to average velocity times change in time. The change in time that we care about is the time up. So that is our delta t over 2. Our total time divided by 2. This is our time up. And what's our average velocity? Well, the average velocity, if we assume constant acceleration, is your initial velocity plus your final velocity over 2. It's really just the mean of the two things. Well, we know what our initial velocity is. Our initial velocity is this thing over here. So this is this thing over here. Our final velocity-- remember, we're just talking about the first half of the time the ball is in the air--"
    },
    {
        "Q": "At 3:14, Sal starts writing the formula for displacement. Why is he using Time Up as the time instead of Total Time? Can't we use Total Time instead?",
        "A": "If he found the displacement over the total time, it would be zero. He is showing us how to find the peak height. The object reaches peak height when half of the total time has passed.",
        "video_name": "P7LKEkcNibo",
        "timestamps": [
            194
        ],
        "3min_transcript": "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. that we travel in the time up. So that'll give us our peak distance. Remember that distance-- or I should say displacement, in this situation-- displacement is equal to average velocity times change in time. The change in time that we care about is the time up. So that is our delta t over 2. Our total time divided by 2. This is our time up. And what's our average velocity? Well, the average velocity, if we assume constant acceleration, is your initial velocity plus your final velocity over 2. It's really just the mean of the two things. Well, we know what our initial velocity is. Our initial velocity is this thing over here. So this is this thing over here. Our final velocity-- remember, we're just talking about the first half of the time the ball is in the air--"
    },
    {
        "Q": "at 6:35, will this equation calculate the total displacement or the half of displacement? because it is only using half of the time.",
        "A": "I think it will calculate just how high the ball went in the air because if we calculated the whole displacement it would be 0 because it would land exactly from where we threw the ball,so it would be right at the place we started at, but i think we also calculated half the distance traveled",
        "video_name": "P7LKEkcNibo",
        "timestamps": [
            395
        ],
        "3min_transcript": "These are the same thing. And then we can simplify it. Our maximum displacement is equal to 4.9 meters per second squared times delta t squared, all of that over 4. And then we can just divide 4.9 divided by 4. 4.9 divided by 4 is-- let me just get the calculator out. I don't want to do that in my head, get this far and make a careless mistake. 4.9 divided by 4 is 1.225. So our maximum displacement is going to be 1.225 times our total time in the air squared, which is a pretty straightforward calculation. of how high we get displaced. Right when that ball is stationary, or has no net velocity, just for a moment, and starts decelerating downwards. So we can use that. If a ball is in the air for 5 seconds-- we can verify our computation from the last video-- our maximum displacement, 1.225, times 5 squared, which is 25, will give us 30.625. That's what we got in the last video. If the ball's in the air for, I don't know, 2.3 seconds-- so it's 1.225 times 2.3 squared-- then that means it went 6.48 meters in the air. So anyway, I just wanted to give you a simple expression that gives you the maximum displacement from the ground, assuming air resistance is negligible, as a function of the total time in the air. And it's a neat game to play."
    },
    {
        "Q": "At 3:05 when he multiplied the entire equation by a negative, shouldn't change of time be negative?",
        "A": "That depends on your definition, you can have negative time if you wanted to take X amount of seconds of something. Like how can I cut down the time it takes me to get to work, if it takes me 60 minutes to get to work taking one route then I take another route and it takes off 20 minutes of my time its 60-20 = 40 minutes. depends on the context of the expression.",
        "video_name": "P7LKEkcNibo",
        "timestamps": [
            185
        ],
        "3min_transcript": "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. that we travel in the time up. So that'll give us our peak distance. Remember that distance-- or I should say displacement, in this situation-- displacement is equal to average velocity times change in time. The change in time that we care about is the time up. So that is our delta t over 2. Our total time divided by 2. This is our time up. And what's our average velocity? Well, the average velocity, if we assume constant acceleration, is your initial velocity plus your final velocity over 2. It's really just the mean of the two things. Well, we know what our initial velocity is. Our initial velocity is this thing over here. So this is this thing over here. Our final velocity-- remember, we're just talking about the first half of the time the ball is in the air--"
    },
    {
        "Q": "At 4:13 Sal Khan says that the first part of the equation is the number of stars in the galaxy but my other source of information says that it is the rate that stars form. Which one is right?",
        "A": "I think that the average number of stars that are formed is a better value. Stars are born, and stars die. Using the average of stars currently alive at a given time makes more sense than just using the formation rate.",
        "video_name": "YYEgq1bweN4",
        "timestamps": [
            253
        ],
        "3min_transcript": "number of detectable civilizations in the Milky Way, in our galaxy. and once again, there could be civilizations, looking back in the star field right over here this star right over here, maybe it has a planet that is in the right place that has liquid water and maybe there is intelligent life on that planet, but they might not be detectable, because, they aren't technologically advanced enough that they are using electromagnetic radiation or maybe they just figured out some other way to communicate or maybe they are beyond using electromagnetic radiation, you know, radio waves and all the rest to communicate so we will never be able to detect them. We are talking about civilizations like ours that are, to some degree, using technology not too different than our own. That's what we mean by 'detectable', I like to start with just the total number of stars in our solar system. So let's just start with, I will call it, N*, so this is the number of stars in our galaxy, and our best guess, I said is this's going to be 100-400 billion stars. We don't even know how many there are, some of them are undetectable, and the center of our galaxy is just a big blur to us we don't even know what's on the other side of that, e can't even see all the stars that are packed into the center, so this is our best guess, 100-400 billion stars. Now obviously, there is going to be subset of those stars that even have planets. So, let's multiply the times that subset; so lets multiply times the frequency of having a planet. If you are a star, this is the percent chance or the frequency or so, I'll write this way \"fractions that have planets\", So, if this is a hundred billion, let's say I am making a guess here and we are learning more about this everyday, there are all these discoveries of 'exoplanets'- planets outside our solar systems, maybe this is one fourth. Then you could say, well that means there are 100 billion times one fourth, that means there are 25 billion stars that have planets around them. But that's still not enough to go to civilizations. We also need to think about planets, there could be a planet like Jupiter, and we don't know how life as we know it can survive on a planet like Jupiter or Neptune or Mercury. It has to have planets that are good for sustaining life. Preferably have a rocky core, liquid water on the outside, that's what we think are the ingredients we need for life, maybe we are just not being creative enough, that's what we know as life is being."
    },
    {
        "Q": "4:52 Sal said planets that are good for sustaining life preferably have a rocky core. Do he mean molten core?",
        "A": "Our core isn t molten, it is a solid ball of iron, nickel, and other assorted amounts of heavy metals. The surrounding area is liquid but the core isn t.",
        "video_name": "YYEgq1bweN4",
        "timestamps": [
            292
        ],
        "3min_transcript": "I like to start with just the total number of stars in our solar system. So let's just start with, I will call it, N*, so this is the number of stars in our galaxy, and our best guess, I said is this's going to be 100-400 billion stars. We don't even know how many there are, some of them are undetectable, and the center of our galaxy is just a big blur to us we don't even know what's on the other side of that, e can't even see all the stars that are packed into the center, so this is our best guess, 100-400 billion stars. Now obviously, there is going to be subset of those stars that even have planets. So, let's multiply the times that subset; so lets multiply times the frequency of having a planet. If you are a star, this is the percent chance or the frequency or so, I'll write this way \"fractions that have planets\", So, if this is a hundred billion, let's say I am making a guess here and we are learning more about this everyday, there are all these discoveries of 'exoplanets'- planets outside our solar systems, maybe this is one fourth. Then you could say, well that means there are 100 billion times one fourth, that means there are 25 billion stars that have planets around them. But that's still not enough to go to civilizations. We also need to think about planets, there could be a planet like Jupiter, and we don't know how life as we know it can survive on a planet like Jupiter or Neptune or Mercury. It has to have planets that are good for sustaining life. Preferably have a rocky core, liquid water on the outside, that's what we think are the ingredients we need for life, maybe we are just not being creative enough, that's what we know as life is being. So we don't necessarily know that they are going to have life, but they seem like they are just the right distance from the star, not too hot, not too cold. They have the right amount of gravity, water, all the other stuff, and we still don't know what this means, but this means average number, so given a number of solar systems with planets, what's the average number of planets capable of sustaining life, and once again, you don't know this answer. maybe it is 0.1, it's probably less than 1. Therefore any given solar system that has planets, the average number capable of sustaining life maybe its 0.1 maybe it's more than 1, I don't know. We don't know the exact answer here, but I will throw an answer in,"
    },
    {
        "Q": "At 3:38, why did Sal say \" really do 'EXCITE' the electrons in the chlorophyll.....\"",
        "A": "Because it means that they move faster. People often say excite rather than speed up.",
        "video_name": "-rsYk4eCKnA",
        "timestamps": [
            218
        ],
        "3min_transcript": "You have these fusion reactions in the sun 93 million miles away, and it's releasing these photons, and some small subset of those photons reach the surface of Earth. They make their way through clouds and whatever else. And then these plants and bacteria and algae are able to harness that somehow and turn them into sugars that we can then eat or maybe the cow eats them and we eat the cow if we're not vegetarians, and we can then use that for energy. Not that the cow is all carbohydrates, but this is essentially what is used as the fuel or the energy for all of the other important compounds that we eat. This is where we get all of our fuel. So this is fuel for animals. Or you know, if you eat a potato directly, you are directly getting your carbohydrates. But anyway, this is a very simple notion of photosynthesis, but it's not incorrect. I mean, if you had to know one thing about photosynthesis, this would be it. But let's delve a little bit deeper and try to get into the how this actually happens. I find it amazing that somehow photons of sunlight are used to create these sugar molecules or these carbohydrates. So let's delve a little bit deeper. So we can write the general equation for photosynthesis. Well, I've almost written it here. But I'll write it a little bit more scientifically specific. You start off with some carbon dioxide. You add to that some water, and you add to that-- instead of sunlight, I'm going to say photons because these are what really do excite the electrons in the chlorophyll that go down, and you'll see this process probably in this video, and we'll go in more detail in the next few videos. But that excited electron goes to a high energy state, and as it goes to a lower energy state, we're able to harness that energy to produce ATPs, and you'll see NADPHs, and those are used to produce carbohydrates. But we'll see that in a little bit. with these constituents, And then you end up with a carbohydrate. And a carbohydrate could be glucose, doesn't have to be glucose. So the general way we can write a carbohydrates is CH2O. And we'll put an n over here, that we could have n multiples of these, and normally, n will be at least three. In the case of glucose, n is 6. You have 6 carbons, 12 hydrogens and 6 oxygens. So this is a general term for carbohydrates, but you could have many multiples of that. You could have these long-chained carbohydrates, so you end up with a carbohydrate and then you end up with some oxygen. So this right here isn't so different than what I wrote up here in my first overview of how we always imagined photosynthesis in our heads. In order to make this equation balance-- let's see, I have n carbons so I need n carbons there. Let's see, I have two n hydrogens here."
    },
    {
        "Q": "At 2:51, why is Oxygen O2 and not just O. It is like this in my textbook as well and I do not know why.",
        "A": "oxygen is actually called oxygen oxide. this is the oxygen we breath",
        "video_name": "-rsYk4eCKnA",
        "timestamps": [
            171
        ],
        "3min_transcript": "as a living species. One, we need carbohydrates or we need sugars in order to fuel our bodies. You saw that in the cellular respiration videos. We generate all of our ATP by performing cellular respiration on glucose, which is essentially a byproduct, or a broken down carbohydrate. It's the simplest one for us to process in cellular respiration. And the second hugely important part is getting the oxygen. Once again, we need to breathe oxygen in order for us to break down glucose, in order to respire, in order to perform cellar respiration. So these two things are key for life, especially for life that breathes oxygen. So this process, other than the fact that it's interesting, that there are organisms around us, mostly You have these fusion reactions in the sun 93 million miles away, and it's releasing these photons, and some small subset of those photons reach the surface of Earth. They make their way through clouds and whatever else. And then these plants and bacteria and algae are able to harness that somehow and turn them into sugars that we can then eat or maybe the cow eats them and we eat the cow if we're not vegetarians, and we can then use that for energy. Not that the cow is all carbohydrates, but this is essentially what is used as the fuel or the energy for all of the other important compounds that we eat. This is where we get all of our fuel. So this is fuel for animals. Or you know, if you eat a potato directly, you are directly getting your carbohydrates. But anyway, this is a very simple notion of photosynthesis, but it's not incorrect. I mean, if you had to know one thing about photosynthesis, this would be it. But let's delve a little bit deeper and try to get into the how this actually happens. I find it amazing that somehow photons of sunlight are used to create these sugar molecules or these carbohydrates. So let's delve a little bit deeper. So we can write the general equation for photosynthesis. Well, I've almost written it here. But I'll write it a little bit more scientifically specific. You start off with some carbon dioxide. You add to that some water, and you add to that-- instead of sunlight, I'm going to say photons because these are what really do excite the electrons in the chlorophyll that go down, and you'll see this process probably in this video, and we'll go in more detail in the next few videos. But that excited electron goes to a high energy state, and as it goes to a lower energy state, we're able to harness that energy to produce ATPs, and you'll see NADPHs, and those are used to produce carbohydrates. But we'll see that in a little bit."
    },
    {
        "Q": "@7:42 Shouldn't he do (1.00029*ans)/(1.33*8.1)??",
        "A": "It would be the same thing. Apparantly his TI-85 can work by dividing a number, and then diving another number immediately after it, instead of putting parentheses around the items, but all in all, it would equate the same value.",
        "video_name": "10LuSfZZa3E",
        "timestamps": [
            462
        ],
        "3min_transcript": "And that's going to be equal to the index of refraction of water. So that's index of water is 1.33-- so let me do that in a different color. So that's going to be-- no, I wanted to do a different color. So that's going to be-- let me do it in this dark blue. So that's going to be 1.33 times sine of theta 2. And so, if we want to solve for sine of theta 2, you just divide both sides of this equation by 1.33. So let's do that. So I'll do it over here. So if you divide both sides by 1.33, we get 1.00029 times 7.92 over 8.1, and we're also going to divide by 1.33. So we're also dividing by 1.33. So let's figure what that is. So let's do that. Get the calculator out. So we have 1.00029 times 7.92. Well, actually I could even say times second answer, if we want this exact value. That was the last-- so I'm going to do that-- second answer. So that's the actual precise, not even rounding. And then we want to divide by 1.33, that's this right here. And then we want to divide by 8.1, and we get that. And that's going to be equal to the sine of theta 2. So that's going to be equal to the sine of theta 2. So let me write this down. So we have 0.735 is equal to the sine of theta 2. Now, we could take the inverse sine So we get theta 2 is equal to-- let's just take the inverse sine of this value. So I take the inverse sine of the value that we just had, so answer is just our last answer. And we have theta 2 being 47.3-- let's say rounded-- 47.34 degrees. So this is 47.34 degrees. So we were able to figure out what theta 2 is, 47.34 degrees. So now we just have to use a little bit of trigonometry to actually figure out this distance over here. Now what trig ratio involves-- so we know this angle. We want to figure out its opposite side, to that angle. And we know the adjacent side-- we know that this right here is 3. So what trig identity deals with opposite and adjacent? Well, tangent-- toa. Tangent is opposite over adjacent. So we know that the tangent of this angle right over here"
    },
    {
        "Q": "At 0:50 how'd do we have a large number of atoms?",
        "A": "Even a speck of a substance contains a huge number of atoms.",
        "video_name": "9REPnibO4IQ",
        "timestamps": [
            50
        ],
        "3min_transcript": "SAL: In the last video we saw all sorts of different types of isotopes of atoms experiencing radioactive decay and turning into other atoms or releasing different types But the question is, when does an atom or nucleus decide to decay? Let's say I have a bunch of, let's say these are all atoms. I have a bunch of atoms here. And let's say we're talking about the type of decay where an atom turns into another atom. So your proton number is going to change. Your atomic number is going to change. So it could either be beta decay, which would release electrons from the neutrons and turn them into protons. Or maybe positron emission turning protons into neutrons. But that's not what's relevant here. Let's say we have a collection of atoms. And normally when we have any small amount of any element, we really have huge amounts of atoms of that element. And we've talked about moles and, you know, one gram of carbon-12-- I'm sorry, 12 grams-- 12 grams of carbon-12 One mole of carbon-12. And what is one mole of carbon-12? That's 6.02 times 10 to the 23rd carbon-12 atoms. This is a ginormous number. This is more than we can, than my head can really grasp around how large of a number this is. And this is only when we have 12 grams. 12 grams is not a large mass. For example, one kilogram is about two pounds. So this is about, what? I want to say [? 1/50 ?] of a pound if I'm doing [? it. ?] But this is not a lot of mass right here. And pounds is obviously force. You get the idea. On Earth, well anywhere, mass is invariant. This is not a tremendous amount. So with that said, let's go back to the question of how do we know if one of these guys are going to decay in some way. And maybe not carbon-12, maybe we're talking about carbon-14 or something. How do we know that they're going to decay? They all have some probability of the decaying. At any given moment, for a certain type of element or a certain type of isotope of an element, there's some probability that one of them will decay. That, you know, maybe this guy will decay this second. And then nothing happens for a long time, a long time, and all of a sudden two more guys decay. And so, like everything in chemistry, and a lot of what we're starting to deal with in physics and quantum mechanics, everything is probabilistic. I mean, maybe if we really got in detail on the configurations of the nucleus, maybe we could get a little bit better in terms of our probabilities, but we don't know what's going on inside of the nucleus, so all we can do is ascribe some probabilities to something reacting. Now you could say, OK, what's the probability of any given molecule reacting in one second? Or you could define it that way. But we're used to dealing with things on the macro level, on dealing with, you know, huge amounts of atoms. So what we do is we come up with terms that help us get our head around this. And one of those terms is the term half-life."
    },
    {
        "Q": "8:10 - Ok, what is happening here? One equation is being subtracted from another without any substitutions or anything?\n\nI thought I knew my algebra but this has me confused.",
        "A": "When you have system of equation in two variables, then you want to cancel out some terms to solve for at least one variable. so he multiplied the equation and then subtracted the two in order to get rid of T2. You could substitute it as well. fell free to contact me for further clarification informjaka@gmail.com",
        "video_name": "zwDJ1wVr7Is",
        "timestamps": [
            490
        ],
        "3min_transcript": "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. Let's subtract this equation from this equation. So you can also view it as multiplying it by negative 1 and then adding the 2. So when you subtract this from this, these two terms cancel out because they're the same. And so then you're left with minus T2 from here. Minus this, minus 3 T2 is equal to 0 minus 20 square roots of 3. And so this becomes minus 4 T2 is equal to minus 20 square roots of 3. And then, divide both sides by minus 4 and you get T2 is equal to 5 square roots of 3 Newtons. So that's the tension in this wire. And now we can substitute and figure out T1. Let's use this formula right here because it looks suitably simple."
    },
    {
        "Q": "On 2:40, Sal says that oxygen and 2 hydrogen make up liquid water but i thought H2o makes up water who is right please help me!",
        "A": "H2O = water. This means that a molecule of water consists of 2 H (hydrogen) atoms and 1 molecule of O (oxygen).",
        "video_name": "Y3ATc9he254",
        "timestamps": [
            160
        ],
        "3min_transcript": "but if this ice were more dense, the solid water were more dense than the liquid water, well then the ice would sink. The ice would sink and collect at the bottom. The ice would collect at the bottom right over here. And then the water over here would freeze again. It would freeze again. But if that were more dense, then it would sink and collect at the bottom. And you keep going with this process on and on and on, and eventually the entire body of water, the entire lake would freeze solid. Would freeze... It would freeze solid. Now you can imagine this wouldn't be that good for the animals that are living in the water. If you imagine some fish in here. Those fish would then freeze. And most living things, there are a few simple organisms that can survive being frozen. But most living things would just die. And so this would not be a good environment for animals to live in, or for even biology to take place. it does not follow this pattern. When we're talking about water, when we're talking about water, when we go to the liquid state, when we go from liquid water to solid ice, to solid ice, we actually get, we actually get less dense. So this right over here is less dense. This is why ice floats. This is more dense, more dense. And this is less dense. And to think about why that is, it all goes back to the hydrogen bonding. So we've seen in previous videos. So, I'm gonna do it all in one color. Oxygen, hydrogen, hydrogen. Let's say this is the liquid state that I'm drawing right over here. This is liquid water. Liquid water. So then you have oxygen, and you have oxygen and hydrogen and hydrogen. And you have oxygen and hydrogen and hydrogen. We've already talked multiple times negative charges at the side away from the hydrogens. Partial negative, partial negative because oxygen is so electro-negative. And you have partial positive charges on the hydrogen ends. Partial positive charges at the hydrogens and these partial negatives and partial positives attract each other and this is called hydrogen bonding. Now the liquid state, you have enough energy. The temperature is just really average kinetic energy that these molecules are able to bounce around and flow past each other. These hydrogen bonds get broken and get reformed over and over again. These things flow past each other and also they have enough energy to kind of push even closer to each other than even the hydrogen bonds would dictate. Sometimes they go closer, sometimes they're further, sometimes they're pushing around. But as we get-- As we get cooler, as we get cooler and we lose heat, then they don't have the kinetic energy to kind of-- They get closer and bump up against each other and move right and flow right past each other and they form a lattice structure"
    },
    {
        "Q": "at 7:20 what does aq stand for?",
        "A": "aq stands for aqueous . that means dissolved in water or a solution in which water is the solvent. Hope this will help.............. ; )",
        "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": "if 110.mg of fluorine-18 is shipped at 10:00 A.M,how many milligrams of the radioisotope are still active when the sample arrives at the radiology laboratory at 5:20 P.M",
        "A": "It s half life is every 110 minutes, or 1 hour and fifty minutes. It will go through it s half life a total of ((7 1/3)x60)/110 times, or 4 times. You can divide 110 by 2 four times (55, 27.5, 13.75) and then 6.875 mg will be left.",
        "video_name": "dnYyMHSSb8M",
        "timestamps": [
            600,
            320
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 12:00, if the period is the time taken to complete one cycle then what does it mean the particle will move up and down in one second (frequency)?",
        "A": "The period is 1 second. That s not the frequency. The frequency is 1 per second.",
        "video_name": "tJW_a6JeXD8",
        "timestamps": [
            720
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "at 2:01, we suppose that the electrons go straight from one plate to the other, right?",
        "A": "The ions you mean. Positive and negative ions are the charge carriers here, not electrons. And no, here the system works differently, the positive ions move to the negative electrode and the negative ions move to the positive electrode. There they get discharged by gaining/losing electrons from/to the electrode. This is how electrolytes conduct electricity.",
        "video_name": "uUhBEufepWk",
        "timestamps": [
            121
        ],
        "3min_transcript": "- [Voiceover] Most solids offer some amount of resistance to the flow of current through them. This allows us to define things like the resistivity, or the conductivity, but the same is true for liquids. Consider this container full of a liquid. We can measure its resistivity. Now, if I took a battery, and I put one lead here, and one lead here, if there is a voltage and this liquid is able to conduct electricity, then this current should be able to flow through the intervening liquid over to the other side and then back up. Sometimes, this is done with AC current, otherwise, you might get electrolysis and then you get bubbles in here and that changes the liquid in some way. We want to measure the resistance and the resistivity of the liquid, not of some altered liquid. So, sometimes you use AC, but this is the general principal. Send in a voltage, a certain amount of current will flow. How can we use that, to determine the resistivity? Well, we know resistivity is equal to the resistance that we measure times the area divided by the length, I can imagine getting that. This length in here would just be this distance. There's my length, because my \"resistor,\" is this liquid in here. But, what's my area? So, this would be a bad experiment to do. If we want to measure the resistivity, what we really want, is something where we have a well defined area. Let me get rid of this. Imagine you had two plates. Take these two plates. You put them in the solution you want to measure the resistivity of, so, we put them in here. Stick them into there. They have a well defined area. We've got those. We can measure those if we want. We set them apart some known distance between them, L, and you hook them up to battery. So, take this one, hook it up to a known voltage, hook the other side up to the other plate, and if this solution, if this electrolytic solution in here can conduct electricity, current will flow from this side to the other side, and you can measure these quantities. You measure the area. You got that. How do we measure the resistance? Well, we know the voltage. We can have a known voltage of the battery up here, and you can stick ammeter in here to measure the current. If I stick an ammeter, ammeters measure the current. Now, I can just use Ohm's law, and I know that the resistance is just going to be the voltage divided by the current, and if plug all these values into here, I can get an experimental value for the resistivity of this liquid, sometimes it's called the electrolytic resistivity. Or, one over the electrolytic resistivity would be the electrolytic conductivity. So, this would be the electrolytic conductivity. So, this is an experimental way to do it. Honestly, you don't even have to go through all that much trouble. You can just take a solution. First, put a solution in here that has a known resistivity. That way, you can just do this: R equals Rho L over A."
    },
    {
        "Q": "what is glucose at 2:53?",
        "A": "Glucose is a sugar produced by plants in photosynthesis, it has a molecular formula of C6H12O6. In photosynthesis it is the result of a plant using water (H2O) and Carbon Dioxide (CO2) to make Glucose (C6H12O6) and Oxygen gas (O2). 6H2O + 6CO2 = C6H12O6 + 6O2",
        "video_name": "lzWUG4H5QBo",
        "timestamps": [
            173
        ],
        "3min_transcript": "And the beanstalk was helping him physically, but also was actually providing him with very precious oxygen. In fact, if the beanstalk didn't do that, he may not have even made it. And we also, we don't know for sure, but we think that perhaps some of this story may have taken place during the day. And in fact, we know that sunlight is quite important for this process. And we think that this process, the name that we give it for the beanstalk anyway, is photosynthesis. And so what is really happening-- we're actually going to kind of write it out here-- between Jack and the beanstalk, and really between all plants and animals? What is this process between them? We know that on the one hand, you have beanstalks doing photosynthesis, and on the other hand, you have folks like Jack doing cellular respiration. And there's this really kind of interesting symbiosis. 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."
    },
    {
        "Q": "AT 5:45 you said when the volume goes down pressure increases then why does bigger balloon bursts easily as compared to small ballon?",
        "A": "because bigger balloons have much more gas inside them which increases pressure significantly.",
        "video_name": "WScwPIPqZa0",
        "timestamps": [
            345
        ],
        "3min_transcript": "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. 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"
    },
    {
        "Q": "at 6:53 why would hydrogen get one electron ?",
        "A": "He explains it earlier on, at 1:14. It has plus one charge, which means that it lost an electron. Electrons have a -1 charge. Carbon is more electronegative than hydrogen, so it would take four electrons from the four hydrogen atoms. Each hydrogen atom gives an electron to the carbon, causing them to have 1 less electron, giving them 1+ for charge.",
        "video_name": "OE0MMIyMTNU",
        "timestamps": [
            413
        ],
        "3min_transcript": "to positive 4? Well, the way to increase your charge or your hypothetical charge is to lose electrons. Every time you lose an electron, this becomes less negative. And eventually, it'll become positive. So you have to lose 8 electrons. So I'll write plus 8 electrons right over here. You take these electrons, give it to this carbon. You're going to get to this side of this reaction. And the way I'm writing right now, these are called \"half reactions\" where I'm independently focusing on each of the elemental components of these reactions. So here, you have carbon. In this reaction, carbon-- in our hypothetical oxidation number world-- has lost 8 electrons. What do we call it when you are losing these hypothetical electrons? Well, we can remind ourselves OIL RIG-- oxidation is losing 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,"
    },
    {
        "Q": "At 6:15, how is carbon losing electrons therefore being oxidized if it has 8e- on the other side? Or is the 8e- just to balance the charges?",
        "A": "when something is oxidized it is losing electrons so the carbon in the reactant side lost 8 electrons to become the carbon on the products side. Now where will you put these 8 electrons? As i ve said they are lost or removed from the reactant so they are in product side. Hope this answered your question :)",
        "video_name": "OE0MMIyMTNU",
        "timestamps": [
            375
        ],
        "3min_transcript": "with the hydrogens-- in our hypothetical ionic bond world, oxygen is a good bit more electronegative. So we're assuming it's going to take the electrons from the hydrogens. So each of the hydrogens loses an electron, giving it an oxidation number of 1. I could even write it like this, if you like. An oxidation number of positive 1. The oxygen has gained 2 electrons. So that gives it an oxidation number of negative 2. So now that we've done that, let's think about who is getting oxidized and who is being reduced. So let's first focus on the carbon. The carbon starts off at an oxidation number of negative 4. The reaction takes place. And then, carbon now has an oxidation number of positive 4. to positive 4? Well, the way to increase your charge or your hypothetical charge is to lose electrons. Every time you lose an electron, this becomes less negative. And eventually, it'll become positive. So you have to lose 8 electrons. So I'll write plus 8 electrons right over here. You take these electrons, give it to this carbon. You're going to get to this side of this reaction. And the way I'm writing right now, these are called \"half reactions\" where I'm independently focusing on each of the elemental components of these reactions. So here, you have carbon. In this reaction, carbon-- in our hypothetical oxidation number world-- has lost 8 electrons. What do we call it when you are losing these hypothetical electrons? Well, we can remind ourselves OIL RIG-- oxidation is losing 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."
    },
    {
        "Q": "at 5:37 whats the differnce between a half reaction and a normal reaction",
        "A": "Take a normal reaction and now only consider the atoms being oxidized and the ones being reduced. Ignore all other atoms. Now split your equation into two parts: The reduction side: C(4-) ===> C(4+) + 8e- AND The oxidation side: 2 O2 + 8e- ===> 4 O(2-) The half equation is considering only the atoms being oxidized or reduced and where the electrons are flowing from and ending up. In this case carbon loses 8 electrons and oxygen gains 8 electrons. Hope that helps :)",
        "video_name": "OE0MMIyMTNU",
        "timestamps": [
            337
        ],
        "3min_transcript": "with the hydrogens-- in our hypothetical ionic bond world, oxygen is a good bit more electronegative. So we're assuming it's going to take the electrons from the hydrogens. So each of the hydrogens loses an electron, giving it an oxidation number of 1. I could even write it like this, if you like. An oxidation number of positive 1. The oxygen has gained 2 electrons. So that gives it an oxidation number of negative 2. So now that we've done that, let's think about who is getting oxidized and who is being reduced. So let's first focus on the carbon. The carbon starts off at an oxidation number of negative 4. The reaction takes place. And then, carbon now has an oxidation number of positive 4. to positive 4? Well, the way to increase your charge or your hypothetical charge is to lose electrons. Every time you lose an electron, this becomes less negative. And eventually, it'll become positive. So you have to lose 8 electrons. So I'll write plus 8 electrons right over here. You take these electrons, give it to this carbon. You're going to get to this side of this reaction. And the way I'm writing right now, these are called \"half reactions\" where I'm independently focusing on each of the elemental components of these reactions. So here, you have carbon. In this reaction, carbon-- in our hypothetical oxidation number world-- has lost 8 electrons. What do we call it when you are losing these hypothetical electrons? Well, we can remind ourselves OIL RIG-- oxidation is losing 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."
    },
    {
        "Q": "at 6:11 what does he mean by the lion goes ger? is it suppose to be a metaphor?",
        "A": "LEO (loss electrons is oxidation) goes GER (gaining electrons is reduction) it s just a way to remember whether a half equation is oxidation or reduction",
        "video_name": "OE0MMIyMTNU",
        "timestamps": [
            371
        ],
        "3min_transcript": "with the hydrogens-- in our hypothetical ionic bond world, oxygen is a good bit more electronegative. So we're assuming it's going to take the electrons from the hydrogens. So each of the hydrogens loses an electron, giving it an oxidation number of 1. I could even write it like this, if you like. An oxidation number of positive 1. The oxygen has gained 2 electrons. So that gives it an oxidation number of negative 2. So now that we've done that, let's think about who is getting oxidized and who is being reduced. So let's first focus on the carbon. The carbon starts off at an oxidation number of negative 4. The reaction takes place. And then, carbon now has an oxidation number of positive 4. to positive 4? Well, the way to increase your charge or your hypothetical charge is to lose electrons. Every time you lose an electron, this becomes less negative. And eventually, it'll become positive. So you have to lose 8 electrons. So I'll write plus 8 electrons right over here. You take these electrons, give it to this carbon. You're going to get to this side of this reaction. And the way I'm writing right now, these are called \"half reactions\" where I'm independently focusing on each of the elemental components of these reactions. So here, you have carbon. In this reaction, carbon-- in our hypothetical oxidation number world-- has lost 8 electrons. What do we call it when you are losing these hypothetical electrons? Well, we can remind ourselves OIL RIG-- oxidation is losing 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."
    },
    {
        "Q": "At 9:21, how is it that oxygen is reduced only by carbon and not carbon & hydrogen? On the right side of the equation is oxygen not receiving electrons from both hydrogen and carbon?",
        "A": "Hydrogen does not change its oxidation state, thus it cannot directly have been part of the oxidation/reduction process.",
        "video_name": "OE0MMIyMTNU",
        "timestamps": [
            561
        ],
        "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": "At 8:26 when he is saying that we wish to minimize the formal charge, why is doing so with one double bond preferred to doing so with two double bonds? Don't you get +1 for Sulfur, -1 for the left Oxygen and 0 for the right oxygen for the one double bond, whereas you get 0 for Sulfur and both oxygens for two double bonds? Wouldn't that be 'minimized' for Sulfur?",
        "A": "Did everyone see this pop up at 9:00 ? At 9:00, I drew one of the resonance structures for sulfur dioxide. While forming two double bonds would decrease the formal charge, both ways of representing sulfur dioxide are generally accepted.",
        "video_name": "3RDytvJYehY",
        "timestamps": [
            506
        ],
        "3min_transcript": "Here's two. And here's two more. So 18 minus 4 is 14 valence electrons left. All right, so we're going to start assigning some those leftover electrons to our terminal atoms, which are our oxygens. Oxygen's going to follow the octet rule. And so, therefore, each oxygen gets six more electrons to give each oxygen an octet. So we go ahead and do that. So I just represented 12 more valence electrons, right? Six on each oxygen. So 14 minus 12 is 2 valence electrons. So I have two valence electrons left over. And, remember, when you have leftover valence electrons, you go ahead and put them on your central atom. So we can go ahead and put those two valence electrons here on our central atom like that. Now, we're not quite done with our dot structure because sulfur doesn't have an octet. That's one way of thinking about it. You could also think about formal charge. Sulfur's formal charge is not minimized in this dot structure. 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"
    },
    {
        "Q": "At 3:00 he mentions that Boron doesn't need to follow the octet rule but it can. Does that mean that BF3 would have four resonance structures? The one that doesn't follow the octet rule AND the three that do?",
        "A": "It cannot because of the number of total Valence Electrons. Boron can follow the octet rule, but the total number of Valence Electrons prevents this. Boron normally wants 6-8 electrons.",
        "video_name": "3RDytvJYehY",
        "timestamps": [
            180
        ],
        "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": "At 12:04 it says bent, so is it 120 degrees or 104.5 degrees?",
        "A": "In this instance, the bond angle will be 120 degrees. If there was a lone pair on the central atom, then the bond angle would be reduced to 104.5 degrees. See the video titled VSEPR for 4 electron clouds for an example of this",
        "video_name": "3RDytvJYehY",
        "timestamps": [
            724
        ],
        "3min_transcript": "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. to be approximately 120 degrees, right? So let me go ahead and put this in here. So approximately 120 degrees, it's probably slightly less than that. But that is what we would predict the geometry of the electron cloud to be. So let's go back up and look at our rules here, our steps for predicting the shapes of molecules. So we've done step three, right? We have predicted the geometry of the electron clouds around our central atom. And now we go on to step four here. We're going to ignore any lone pairs on our central atom when we predict the geometry of the overall molecule. And so that now pertains to our example here. We're going to ignore that lone pair of electrons on the sulfur. And we're going to ignore this lone pair of electrons when we're talking about the shape of the molecule. So even though the electron clouds have a trigonal planar geometry, we say that the shape of the molecule has a bent or angular shape. And so if you look at that, if you just look at the atoms, you'll see this kind of bent or angular shape here. And so that's what we say is the shape of the molecule. So you could say bent, or you could say angular here. All right, so that's two examples of molecules with three electron clouds. And remember, it's not just the number of electron clouds. You have to ignore lone pairs of electrons to predict the final shape of the molecule."
    },
    {
        "Q": "From 8:12 on, could someone please simply define and discriminate between transcription and translation?",
        "A": "That is a great question! Transcription is when DNA transfers its genetic information to messenger RNA (mRNA), which carries this genetic information to a ribosome, where each codon (a set of three nucleotides) is translated to a specific protein (this part where the mRNA is in the ribosome is the translation step). Keep these in mind, and you ll be able to remember the difference! :) Note: Sal explains transcription around 8:06, and translation around 10:57.",
        "video_name": "6gUY5NoX1Lk",
        "timestamps": [
            492
        ],
        "3min_transcript": "it might define information for one gene, it could define a protein, this section right over here could be used to define another gene. And genes could be anywhere from several thousand base pairs long, all the way up into the millions. And as we'll see, the way that a gene is expressed, the way we get from the information for that section of DNA into a protein which is really how it's expressed, is through a related molecule to DNA, and that is RNA. Actually let me write this down. RNA. So RNA stands for ribonucleic acid. Ribonucleic acid, let me write that down. deoxyribonucleic acid, so the sugar backbone in RNA is a very similar molecule, well now it's got its oxy, it's not deoxyribonucleic acid, it's ribonucleic acid. The R, let me make it clear where the RNA come from, the R is right over there, then you have the nucleic, that's the n, and then it's a, acid. Same reason why we call the DNA nucleic acid. So you have this RNA. So what role does this play as we are trying to express the information in this DNA? especially if we're talking about cells with nucleii, the DNA sits there but that information has to for the most part get outside of the nucleus in order to be expressed. And one of the functions that RNA plays is to be that messenger, that messenger between a certain section of DNA and kind of what goes on outside of the nucleus, so that that can be translated into an actual protein. messenger RNA, is called transcription. Let me write that down. And what happens in transcription, let's go back to looking at one side of this DNA molecule. So let's say you have that right over there, let me copy and paste it. So there we go, actually I didn't wanna do that. I wanted the other side. So actually I think I'm on the wrong, let me go back here. And so let me copy and then let me paste. There we go. So let's say you have part of this DNA molecule, or you have 1/2 of it just like we did when we replicated it. But now we're not just trying to duplicate the DNA molecule,"
    },
    {
        "Q": "Do genes make DNA molecules or DNA molecules make up a gene (around 6:04)?",
        "A": "At around 6:29 , Sal mentions section of DNA . Genes are kind of like parts of DNA that code for the expression of specific traits. Hope this helps, and please let me know if I m wrong!",
        "video_name": "6gUY5NoX1Lk",
        "timestamps": [
            364
        ],
        "3min_transcript": "so thymine, adenine. Thymine, adenine. Guanine pairs with cytosine. And then cytosine pairs with guanine. So cytosine just like that. And so you can take half of each of this ladder, and then you can use it to construct the other half, and what you've essentially done is you've replicated the actual DNA. And this is actually a kind of conceptual level of how replication is done before a cell divides and replicates, and the entire cell duplicates itself. So that's replication. So the next thing you're probably thinking about, \"Okay, well it's nice to be able to replicate yourself \"but that's kind of useless if that information can't be \"used to define the organism in some way \"to express what's actually happening.\" And so let's think about how the genes in this DNA molecule are actually expressed. So I'll write this as \"expression\". because you hear sometimes the words DNA and chromosome and gene used somewhat interchangeably, and they are clearly related, but it's worth knowing what is what. So when you're talking about DNA you're talking literally about this molecule here that has this sugar phosphate base and it has the sequence of base pairs, it's got this double helix structure, and so this whole thing this could be a DNA molecule. Now when you have a DNA molecule and it's packaged together with other molecules and proteins and kind of given a broader structure, then you're talking about a chromosome. And when you're talking about a gene, you're talking about a section of DNA that's used to express a certain trait. Or actually used to code for a certain type of protein. So for example this could be, this whole thing could be a strand of DNA, but this part right over, let's say in orange I'll do it, it might define information for one gene, it could define a protein, this section right over here could be used to define another gene. And genes could be anywhere from several thousand base pairs long, all the way up into the millions. And as we'll see, the way that a gene is expressed, the way we get from the information for that section of DNA into a protein which is really how it's expressed, is through a related molecule to DNA, and that is RNA. Actually let me write this down. RNA. So RNA stands for ribonucleic acid. Ribonucleic acid, let me write that down."
    },
    {
        "Q": "At 10:48 does adenine pair with both uracil and thymine?",
        "A": "In a DNA molecule adenine pairs with thymine while in a RNA molecule adenine pairs with uracil. There are no uracil in DNA and no thymine in RNA.",
        "video_name": "6gUY5NoX1Lk",
        "timestamps": [
            648
        ],
        "3min_transcript": "a corresponding mRNA molecule. At least for that section of, at least for that gene. So this might be part of a gene Actually whoops, let me make sure I'm using the right tool. This might be part of a gene that is this section of our DNA molecule right over there. And so transcription is a very similar conceputal idea, where we're now going to construct a strand of RNA and specifically mRNA 'cause it's going to take that information outside of the nucleus. And so it's very similar except for when we're talking about RNA, adenine, instead of pairing with thymine, is now going to pair with uracil. So let me write this down, so now you're gonna have adenine pairs not with thymine but uracil. DNA has uracil instead of the thymine. But you're still going to have cytosine and guanine pairing. So for the RNA and in this case the mRNA that's going to leave the nucleus A is going to pair with U, so uracil, that's the base we're talking about, let me write it down, uracil. Thymine is still going to pair with adenine, just like that. Guanine is gonna pair with cytosine, and cytosine is going to pair with guanine. And so when you do that, now these two characters can detach, and now you have a single strand of RNA and in this case messenger RNA, that has all the information on that section of DNA. And so now that thing can leave the nucleus, go attach to a ribosome, and we'll talk more about that in future videos exactly how that's happened, and then this code can be used to actually code for proteins. Now how does that happen? And that process is called translation. and turning it into an amino acid sequence. Proteins are made up of sequences of amino acids. So translation. So let's take our mRNA or this little section of our mRNA, and actually let me draw it like this. Now let's see, I have it is U A C, so it's gonna be U A C then U U then A C G okay? And then we have an A, let me make sure I change it to the right color. We have an A there, and then we have this U U A, C G, alright, now let me put a C right over there, I'm just taking this and I'm writing it horizontally. I have a C here, not a G, it's a C. And then finally I have a G. And of course it'll keep going on and on and on. And what happens is each sequence of three,"
    },
    {
        "Q": "I was wondering why from 8:57 to 10:00, you multiplied -9.8 by two and took the square root of it all?",
        "A": "It was - -30 = 1/2* -9.8 t^2 If we multiply both side by 2 then the half in the RHS will cancel out and we will be left with- -30*2 = -9.8 t^2 minus minus cancels out and we get- (30*2)/9.8 = t^2 t = sq. rt. of [60/9.8] t = 2.47 s",
        "video_name": "jmSWImPs6fQ",
        "timestamps": [
            537,
            600
        ],
        "3min_transcript": "that means there is no vertical velocity to start with, I'm gonna have to plug this initial velocity in the y direction as zero. So that's the trick. Don't fall for it now you know how to deal with it. So we want to solve for displacement in the x direction, but how many variables we know in the y direction? I mean we know all of this. But we can't use this to solve directly for the displacement in the x direction. We need to use this to solve for the time because the time is gonna be the same for the x direction and the y direction. So I find the time I can plug back in over to there, because think about it, the time it takes for this trip is gonna be the time it takes for this trip. It doesn't matter whether I call it the x direction or y direction, time is the same for both directions. In other words, the time it takes for this displacement of negative 30 is gonna be the time it takes for this displacement of whatever this is that we're gonna find. So let's solve for the time. Now, how will we do that? Think about it. We know the displacement, we know the acceleration, we know the initial velocity, and we know the time. But we don't know the final velocity we don't want to know it. So let's use a formula that doesn't involve the final velocity and that would look like this. So if we use delta y equals v initial in the y direction times time plus one half acceleration in the y direction times time squared. Alright, now we can plug in values. My displacement in the y direction is negative 30. My initial velocity in the y direction is zero. This is where it would happen, this is where the mistake would happen, people just really want to plug that five in over here. But don't do it, it's a trap. So, zero times t is just zero so that whole term is zero. Plus one half, the acceleration is negative 9.8 meters per second squared. And then times t squared, alright, now I can solve for t. I'm gonna solve for t, and then I'd have to take the square root of both sides because it's t squared, and what would I get? I'd have to multiply both sides by two. and then I have to divide both sides by negative 9.8 meters per second squared, equals, notice if you would have forgotten this negative up here for negative 30, you come down here, this would be a positive up top. You'd have a negative on the bottom. You'd have to plug this in, you'd have to try to take the square root of a negative number. Your calculator would have been all like, \"I don't know what that means,\" and you're gonna be like, \"Er, am I stuck?\" So you'd start coming back here probably and be like, \"Let's just make stuff positive and see if that works.\" It would work because look at these negatives canceled but it's best to just know what you're talking about in the first place. So be careful: plug in your negatives and things will work out alright. So if you solve this you get that the time it took is 2.47 seconds. It's actually a long time. It might seem like you're falling for a long time sometimes when you're like jumping off of a table, jumping off of a trampoline, but it's usually like a fraction of a second. This is actually a long time, two and a half seconds"
    },
    {
        "Q": "At 8:19 why is the 1/2 there?",
        "A": "Because that s the formula",
        "video_name": "jmSWImPs6fQ",
        "timestamps": [
            499
        ],
        "3min_transcript": "was zero, there was no initial vertical velocity. This person was not launched vertically up or vertically down, this person was just launched straight horizontally, and so the initial velocity in the vertical direction is just zero. People don't like that. They're like \"hold on a minute.\" They're like, this person is gonna start gaining, alright, this person is gonna start gaining velocity right when they leave the cliff, this starts getting bigger and bigger and bigger in the downward direction. But that's after you leave the cliff. We're talking about right as you leave the cliff. That moment you left the cliff there was only horizontal velocity, which means you started with no initial vertical velocity. So this is the part people get confused by because this is not given to you explicitly in the problem. The problem won't say, \"Find the distance for a cliff diver \"assuming the initial velocity in the y direction was zero.\" Now, they're just gonna say, \"A cliff diver ran horizontally off of a cliff. \"Find this stuff.\" And you're just gonna have to know that okay, if I run off of a cliff horizontally that means there is no vertical velocity to start with, I'm gonna have to plug this initial velocity in the y direction as zero. So that's the trick. Don't fall for it now you know how to deal with it. So we want to solve for displacement in the x direction, but how many variables we know in the y direction? I mean we know all of this. But we can't use this to solve directly for the displacement in the x direction. We need to use this to solve for the time because the time is gonna be the same for the x direction and the y direction. So I find the time I can plug back in over to there, because think about it, the time it takes for this trip is gonna be the time it takes for this trip. It doesn't matter whether I call it the x direction or y direction, time is the same for both directions. In other words, the time it takes for this displacement of negative 30 is gonna be the time it takes for this displacement of whatever this is that we're gonna find. So let's solve for the time. Now, how will we do that? Think about it. We know the displacement, we know the acceleration, we know the initial velocity, and we know the time. But we don't know the final velocity we don't want to know it. So let's use a formula that doesn't involve the final velocity and that would look like this. So if we use delta y equals v initial in the y direction times time plus one half acceleration in the y direction times time squared. Alright, now we can plug in values. My displacement in the y direction is negative 30. My initial velocity in the y direction is zero. This is where it would happen, this is where the mistake would happen, people just really want to plug that five in over here. But don't do it, it's a trap. So, zero times t is just zero so that whole term is zero. Plus one half, the acceleration is negative 9.8 meters per second squared. And then times t squared, alright, now I can solve for t. I'm gonna solve for t, and then I'd have to take the square root of both sides because it's t squared, and what would I get? I'd have to multiply both sides by two."
    },
    {
        "Q": "at 5:53 do you not write 1,2 dimethylHept-2-ene because the Methyl at 1 is part of the chain?",
        "A": "Yes, the methyl at 1 is part of the chain so you wouldn t write 1,1 dimethylhex-1-ene. Although that name would indicate the molecule we are talking about, it wouldn t be listed in databases that way. 1,2-dimethylhept-2-ene would actually be written 3-methyloct-3-ene.",
        "video_name": "KWv5PaoHwPA",
        "timestamps": [
            353
        ],
        "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": "At 7:49 shouldn't Sal start from the place where the substituent is the closest i.e\nthe opposite direction",
        "A": "The numbering must include the alkenes as two consecutive numbers, although you write only the lower number in the name. In cyclohexene, the alkene carbons are automatically 1 and 2. C-1 becomes the one with the methyl group, and the numering must then go in the direction Sal used.",
        "video_name": "KWv5PaoHwPA",
        "timestamps": [
            469
        ],
        "3min_transcript": "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. cyclohexene would look just like this. Just like that. You don't have to specify where it is. It's just, one of these are going to be the double bond. Now when you have other constituents on it, by definition or I guess the proper naming mechanism, is one of the endpoints of the double bond will be the 1-carbon, and if any of those endpoints have something else on it, that will definitely be the 1-carbon. So these both are kind of the candidates for the 1-carbon, but this point right here also has this methyl group. We will start numbering there, one, and then you want to number in the direction of the other side of the double bond. One, two, three, four, five, six. So we have three methyl groups, one on one. So these are the-- let me circle the methyl groups. That's a methyl group right there. That's a methyl group right there. That's just one carbon. So we have three methyl groups, so this is going to So it is 1, 4, 6. We have three methyl groups, so it's trimethyl cyclohexene. 1, 4, 6-trimethylcyclohexene. That's what that is, hopefully you found that useful."
    },
    {
        "Q": "4:07 Can it also be named hept-2,4-diene?",
        "A": "Close. You have to keep the a of the ane ending when there is a multiplying prefix.. This makes it easier to pronounce. The name is hepta-2,4-diene.",
        "video_name": "KWv5PaoHwPA",
        "timestamps": [
            247
        ],
        "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": "At 6:00, Sal says it is \"2-methylhept-2ene\". Is it correct if I write \"2-methylheptene\" ?",
        "A": "Sal is correct on this one. Since there are two important structures that have to be pointed out (the methyl group and the double bond), you need to indicate these with two separate numbers. You can t just write the 2 once and say that it applies for both of the groups. Another acceptable name for 2-methylhept-2ene is: 2-methyl-2-heptane",
        "video_name": "KWv5PaoHwPA",
        "timestamps": [
            360
        ],
        "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": "At 4:20, Rishi says the LAIV can't cause you to get sick. If you had a really weak immune system, couldn't you actually get sick? Thanks in advance.",
        "A": "There are some groups of people (those with asthma, etc.) who do not qualify for the live vaccination. These people should get the killed vaccine.",
        "video_name": "wDghWK_Rr_E",
        "timestamps": [
            260
        ],
        "3min_transcript": "get the full protection from the vaccine. So maybe he was in that two-week time span right after he got the vaccination, and maybe he got sick then. And lastly, we have to remember the flu vaccine has three strains in it. So maybe he had a different strain. Maybe it was just one or the other flu virus strains that was out there that year. So who knows exactly why he got sick that year that he says he got sick. But that doesn't change the fact that the flu vaccine is still going to reduce his chances of getting sick in the future. Now, one thing, I'm going to stick with this gentleman here in the bottom left. He might say, well, a different strain? What the heck? If that's the case, then what's the point? I often hear that. He says, what's the point if you could have a flu illness from some other strain? And when people talk about this, I have to remind them that it's not perfect. So it's not an exact science. And what you have to remember is that it goes back to the seat belt issue. It does reduce your chances of getting sick from the flu, Nothing in life is going to reduce your chances to zero, and there's still a possibility that you could get sick. And so our best offer is that you can reduce the chance of getting sick. Let's move on to this gentleman in the middle then. And so he says to me, well, you know what? I think that the flu vaccine made me sick. So this is a little different. He's saying that the flu vaccine itself made him sick. And this is something a lot of people think. They think, well, maybe I was fine, and maybe it was only after getting the flu vaccine that I got sick. And this can be a very frustrating thing to feel. I mean, that's horrible that you were feeling fine, and then a vaccine came along and made you feel awful. But a couple of facts have to come to mind. So one fact, for example, is that we know that the flu vaccine cannot give you flu. So we know flu vaccine does not cause flu. That's a fact. And the reason I can say that with certainty Remember, there's the TIV and then there's the LAIV. The TIV is a dead vaccine, meaning the virus inside of it is dead. And the LAIV-- that's the other option-- this one is alive, but it's very weak. And so usually with this one you might get a runny nose at best. And so, if he says, I think that I got sick from the flu vaccine, I would say, well, I don't think that's possible from this one, right? If you got the injected version, then that's not possible at all. And if you got the live version, then you may have gotten some symptoms, but we wouldn't call that the flu. And I would also say that if that weak vaccine made you sick, can you only imagine how you would have felt if you had the wild virus, the one that circulates and we're trying to protect you from? What's much more likely for this gentleman is that he probably got sick from maybe this copycat virus, one of the other viruses that are circulating, probably around the time that he had the flu vaccine."
    },
    {
        "Q": "at 1:30, 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",
        "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": "If DMSO is more likely to take the left structure as Jay at 05:50 says, is it probable for it, used as the solvent in a reaction, to take the right, polar aprotic structure and increase the nucleophilicity? And are the polar aprotic solvents generally bulky, so that they, due to steric hindrance, generally do not solvate the anion of the nucleophile?",
        "A": "It never has either the structure on the right or the structure on the left. It always is a resonance hybrid of the two structures, with a partial negative charge on the O atom and a partial positive charge on the S. The major requirement for a polar aprotic solvent is that it have a large dipole moment and a large dielectric constant. It does not have to be bulky. Other examples of polar aprotic solvents are acetone, dimethylformamide, and acetonitrile.",
        "video_name": "My5SpT9E37c",
        "timestamps": [
            350
        ],
        "3min_transcript": "Next let's look at DMF. DMF is the short way of writing this one here. Again no hydrogen directly connected to an electronegative atom. This hydrogen is directly connected to this carbon and then this carbon would have three hydrogens on it and then this carbon would have three hydrogens on it. So DMF is a polar aprotic solvent. And finally let's look at this last one here. So the abbreviation would be HMPA. So let me write that down here. HMPA. Again no hydrogen is directly connected to an electronegative atom. Polar aprotic solvents favor an SN2 mechanism. So let's look at why. Down here I have an SN2 reaction. On the left we have this alkyl halide. Let's say we have sodium hydroxide. We could use DMSO as our solvent 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."
    },
    {
        "Q": "At 9:38, what are dopamine and serotonin?",
        "A": "Those are neurotransmitters, chemicals which let neurons communicate with each other.",
        "video_name": "TyZODv-UqvU",
        "timestamps": [
            578
        ],
        "3min_transcript": "The pump works against both of these conditions, collecting 3 positively charged sodium ions, and pushing them out into the positively charged sodium ion-rich environment. To get the energy to do this, the protein pump breaks up a molecule of ATP. ATP, adenosine tri-phopsphate, an adenosine molecule with 3 phosphate groups attached to it. So, when ATP connects with a protein pump, and enzyme breaks the covalent bond on one of those phophtaes in a bust of excitement and energy. The split releases enough energy to change the shape of the pump so that it opens outward and releases 3 sodium ions. This new shape also makes it a good fit for potassium ions that are outside the cell, so the pump lets 2 of those in. So, what you end up with is a nerve cell that is literally and metaphorically charged. It has all those sodium ions waiting outside with this intense desire to get inside of the cell, and when something triggers the nerve cell, it lets all of those in. And that gives the nerve cell a bunch of electric chemical energy, which you can then use to help you feel things There is still yet another way that stuff gets inside of cells, and this also requires energy, it's also a form of active transport. It's called vesicular transport, and the heavy lifting is done by vesicles, which are tiny sacs made of phohpholipids, just like the cell membrane. This kind of active transport is also called cytosis, from the Greek for cell action. When vesicles transport materials outside of a cell, it's called exocytosis, or outside cell action. A great example of this is going on in your brain right now, it's how your nerve cells release neurotransmitters. You've heard of neurotransmitters, they are very important in helping you feel different ways; they're like dopamine and serotonin. After neurotransmitters are synthesized and packaged into vesicles, they're transported until the vesicle reaches the membrane. When that happens, the 2 bilayers rearrange so that they fuse, and then the neurotransmitter spills out, and now I remember where I left my keys! Now, just play that process in reverse and you'll see how material gets inside the cell, and that's endocytosis. There are 3 different ways that this happens. My personal favorite is phagocytosis. that name itself means means devouring cell action. Check this out. So, this particle outside here is some kind of dangerous bacterium in your body, and this is a white blood cell. Chemical receptors on the blood cell membrane detect this punk invader and attach to it. Actually, reaching out around and engulfing it. Then the membrane forms a vesicle to carry it inside where it lays a total unholy beat down on it with enzymes and other cool weapons. Pinocytosis, or drinking action, is very similar to phagocytosis, except instead of surrounding whole particles, it just surrounds things that have already been dissolved. Here the membrane just folds in a little to form the beginning of a channel, and then pinches off to form a vesicle that holds the fluid. Most of your cells are doing this right now, because it's how our cells absorb nutrients. But what if a cell needs something that only occurs in very small concentrations? That's when cells use clusters of specialized receptor proteins in the membrane, that form a vesicle when receptors connect with the molecule"
    },
    {
        "Q": "At 8:26 in the breaking of the covalent bond, is the energy produced heat?",
        "A": "this energy can be sound, heat, electromagnetic, nuclear and mechanical energy",
        "video_name": "TyZODv-UqvU",
        "timestamps": [
            506
        ],
        "3min_transcript": "If our bodies were America, ATP would be credit cards. It's such an important form of information currency, that we're going to do an entire separate episode about it, which will be here. Ha, I was going in the wrong direction, but it will be here when we've done it. But for now, here's what you need to know. When a cell requires active transport it basically has to pay a fee, in the form of ATP, to a transport protein. A particularly important kind of frickin' sweet transport protein is called the sodium potassium pump. Most cells have them, but they're especially vital to cells that need lots of energy, like muscle cells and brain cells. (piano instrumental) Ah! (laughs) Biolography, (laughing) it's my favorite part of the show. The sodium potassium pump was discovered in the 1950s by a Danish medical doctor named Jens Christian Skou, who was studying how anesthetics work on membranes. He noticed that there was a protein in cell membranes and the way he got to know this pump was by studying the nerves of crabs, because crab nerves are huge compared to humans' nerves, and are easier to dissect and observe. But crabs are still small, so he need a lot of them. He struck a deal with a local fisherman, and over the years, studied approximately 25,000 crabs, each of which he boiled to study their fresh nerve fibers. He published his findings on the sodium potassium pump in 1957, and in the meantime became known for the distinct odor that filled the halls of the department of physiology at the university where he worked. 40 years after making his discovery, Sku was awarded the Nobel Prize in chemistry. And here's what he taught us. Turns out, these pumps work against 2 gradients at the same time. 1 is the concentration gradient, and the other is the electric chemical gradient. That's the difference in electrical charge on either side of a cell's membrane. So, then our cells that Sku was studying, like nerve cells in your brain, typically have a negative charge inside relative to the outside. The pump works against both of these conditions, collecting 3 positively charged sodium ions, and pushing them out into the positively charged sodium ion-rich environment. To get the energy to do this, the protein pump breaks up a molecule of ATP. ATP, adenosine tri-phopsphate, an adenosine molecule with 3 phosphate groups attached to it. So, when ATP connects with a protein pump, and enzyme breaks the covalent bond on one of those phophtaes in a bust of excitement and energy. The split releases enough energy to change the shape of the pump so that it opens outward and releases 3 sodium ions. This new shape also makes it a good fit for potassium ions that are outside the cell, so the pump lets 2 of those in. So, what you end up with is a nerve cell that is literally and metaphorically charged. It has all those sodium ions waiting outside with this intense desire to get inside of the cell, and when something triggers the nerve cell, it lets all of those in. And that gives the nerve cell a bunch of electric chemical energy, which you can then use to help you feel things"
    },
    {
        "Q": "At 0:52 why are there 2 hydrogens connected to Carbon and",
        "A": "In that structure you can only see 2 bonds to other carbons and there is not a charge indicated. With these structures you assume that the maximum possible bonds are there (4 for carbon), so there are two hydrogens bonded to that carbon.",
        "video_name": "qpP8D7yQV50",
        "timestamps": [
            52
        ],
        "3min_transcript": "- [Voiceover] I see a lot of mistakes when students draw resonance structures, and so I wanted to make a video on some of the more common mistakes that I've seen. So let's say we wanted to draw a resonance structure for this carbocation. Some students would take these electrons and move them down to here and say, all right, so on the right, now, I would have this, and this is my resonance structure. Let me highlight those electrons in blue here, so these electrons here move down to here. But this is incorrect, so let me write \"no\" here. So the resonance structure on the right, this is an incorrect resonance structure, why is this resonance structure not possible? Well, let's draw in the hydrogens on the carbons, and it will be much more obvious. So this carbon right here has one hydrogen on it, same with this carbon, and this carbon right here has two hydrogens on it, and the carbon with a plus one formal charge must have one hydrogen. So let's put in those hydrogens for the resonance structure on the right, and it should be obvious why this resonance structure Let's focus in on this carbon right here, the one I marked in red. How many bonds are there to that carbon? Well here's one bond, two, three, four, and five, that's five bonds to a carbon, that does not happen, you can't show carbon with five bonds, because that would be 10 electrons around this carbon, and carbon can never exceed an octet of electrons. Because of carbon's position on the periodic table, in the second period, there's four orbitals, and each orbital can hold a maximum of two electrons, which gives us four times two, which is eight. So carbon can never exceed an octet. There's another reason why this is wrong. If we go to this top carbon here, there's only three bonds around that carbon, so that carbon would have a plus one formal charge. So we added another formal charge, so this is incorrect, this is not a correct resonance structure. So what is the proper resonance structure to draw? Well, let's show that down here. You take your electrons, and you move them in the direction of the positive charge, of the positive one formal charge, and so let's show that. The electrons in, let me make them blue again, the electrons in blue move over to here, like that. And that moves the positive formal charge over to this carbon. If we draw in our hydrogens, it'll be clear why this is correct. So we put in a hydrogen here, we put in a hydrogen here, and we put in a hydrogen here. So let me draw in those three hydrogens on the right. Okay, now it's very obvious, let me point this out in red. It's obvious that this carbon here in red has a plus one formal charge, it has three bonds around it."
    },
    {
        "Q": "at 5:52 is magnesium actually originated in Magnesia?",
        "A": "Yes it did from the district of magnesia. It is a place in greece.",
        "video_name": "8Y4JSp5U82I",
        "timestamps": [
            352
        ],
        "3min_transcript": "So that might be magnetic north. And magnetic south, I don't know exactly where that is. But it can kind of move around a little bit. It's not in the same place. So it's a little bit off the axis of the geographic north pole and the south pole. And this is another slightly confusing thing. Magnetic north is the geographic location, where the north pole of a magnet will point to. But that would actually be the south pole, if you viewed the Earth as a magnet. So if the Earth was a big magnet, you would actually view that as a south pole of the magnet. And the geographic south pole is the north pole of the magnet. You could read more about that on Wikipedia, I know it's a little bit confusing. But in general, when most people refer to magnetic north, or the north pole, they're talking about the geographic north area. And the south pole is the geographic south area. But the reason why I make this distinction is because we know when we deal with magnets, just like electricity, or shortly-- is that opposite poles attract. So if this side of my magnet is attracted to Earth's north pole then Earth's north pole-- or Earth's magnetic north-- actually must be the south pole of that magnet. And vice versa. The south pole of my magnet here is going to be attracted to Earth's magnetic south. Which is actually the north pole of the magnet we call Earth. Anyway, I'll take Earth out of the equation because it gets a little bit confusing. And we'll just stick to bars because that tends to be a little bit more consistent. Let me erase this. There you go. I'll erase my Magnesia. I wonder if the element magnesium was first discovered in Magnesia, as well. Probably. And I actually looked up Milk of Magnesia, which is a laxative. And it was not discovered in Magnesia, but it has So I guess its roots could be in Magnesia if magnesium was discovered in Magnesia. Anyway, enough about Magnesia. Back to the magnets. So if this is a magnet, and let me draw another magnet. Actually, let me erase all of this. All right. So let me draw two more magnets. We know from experimentation when we were all kids, this is the north pole, this is the south pole. That the north pole is going to be attracted to the south pole of another magnet. And that if I were to flip this magnet around, it would actually repel north-- two north facing magnets would repel each other. And so we have this notion, just like we had in electrostatics, that a magnet generates a field."
    },
    {
        "Q": "At 10:10, how would you use light years to calculate diameter?",
        "A": "at center of universe we would use a light in any direction and then when it will reach the end of the universe that its radius and the twice is the diameter",
        "video_name": "5FEjrStgcF8",
        "timestamps": [
            610
        ],
        "3min_transcript": "and the Sun-- and you multiply that distance by 25,000. So if the Sun is right here, our nearest star will be in that same pixel. They'll actually be within-- you'd actually get a ton of stars within that one pixel, even though they're so far apart. And then this whole thing is 100,000 light years. It's 25,000 times the distance than the distance between the Sun and the nearest star. So we're talking about unimaginable, unfathomable distances, just for a galaxy. And now we're going to get our-- frankly, my brain is already well beyond anything that it can really process. At this point, it almost just becomes abstract thinking. It just becomes playing with numbers and mathematics. But to get a sense of the universe, itself, the observable universe-- and we have to be clear. Because we can only observe light Because that's how old the universe is. The observable universe is about 93 billion light years across. And the reason why it's larger than 13.7 billion is that the points in space that emitted light 13.7 billion years ago, those have been going away from us. So now they're on the order of 40 billion light years away. But this isn't about cosmology. This is just about scale and appreciating how huge the universe is. Just in the part of the universe that we can theoretically observe, you have to get-- and that we can observe, just because we're getting electromagnetic radiation from those parts of the universe-- you would have to multiply this number. So let me make this clear. 100,000 light years-- that's the diameter of the Milky Way. 1,000 would get you to 100 million light years. This is 100,000 times 1,000 is 100 million. You have to multiply by 1,000 again to get to 100 billion light years. And the universe, for all we know, might be much, much, much, much, larger. It might even be infinite. Who knows? But to get from just the diameter of the Milky Way to the observable universe, you have to multiply by a million. And already, this is an unfathomable distance. So in the whole scheme of things, not only are we pretty small, and not only are the things we build pretty small, and not only is our planet ultra small, and not only is our Sun ultra small, and our solar system ultra small, but our galaxy is really nothing compared to the vastness of the universe."
    },
    {
        "Q": "at 2:39-ish a pop up comes up stating the \"size\" of the earth, is that from pole to pole?",
        "A": "It is the diameter of the Earth which is acually long than from pole to pole",
        "video_name": "5FEjrStgcF8",
        "timestamps": [
            159
        ],
        "3min_transcript": "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. 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"
    },
    {
        "Q": "At 7:42, you mention how far apart stars are. What stars are the closest together, and how close are they?",
        "A": "The two closest stars every found are the stars in HM Cancri system. This system is made up of two white dwarfs co-orbiting each other with a period of about five minutes. Each stars is moving at about 500 kilometers per second around each other and have an average separation of about 80,000 kilometers from each other (a separation of around 7 earths).",
        "video_name": "5FEjrStgcF8",
        "timestamps": [
            462
        ],
        "3min_transcript": "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-- and the Sun-- and you multiply that distance by 25,000. So if the Sun is right here, our nearest star will be in that same pixel. They'll actually be within-- you'd actually get a ton of stars within that one pixel, even though they're so far apart. And then this whole thing is 100,000 light years. It's 25,000 times the distance than the distance between the Sun and the nearest star. So we're talking about unimaginable, unfathomable distances, just for a galaxy. And now we're going to get our-- frankly, my brain is already well beyond anything that it can really process. At this point, it almost just becomes abstract thinking. It just becomes playing with numbers and mathematics. But to get a sense of the universe, itself, the observable universe-- and we have to be clear. Because we can only observe light"
    },
    {
        "Q": "At about 5:49, what does one AU mean? What is an AU?",
        "A": "astronomical unit. Average distance between sun and earth.",
        "video_name": "5FEjrStgcF8",
        "timestamps": [
            349
        ],
        "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": "I thought astrophysicists have claimed that for some time after the creation of the universe, the particles that made up the young universe weren't ionized for some length of time. So wouldn't that mean that they couldn't transmit energy (light) across distances and that light from the early star formation couldn't be visible from our home planet?\n\nI probably worded this wrong but oh well.\n\ntime: ~9:00",
        "A": "Prior to about 380,000 years after the big bang the universe was too hot (above about 3000K) for electrons to be in atoms so the majority of matter had an electric charge so light could not travel freely. This is so early in the development of the universe that there was very little to see even if you could.",
        "video_name": "5FEjrStgcF8",
        "timestamps": [
            540
        ],
        "3min_transcript": "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-- and the Sun-- and you multiply that distance by 25,000. So if the Sun is right here, our nearest star will be in that same pixel. They'll actually be within-- you'd actually get a ton of stars within that one pixel, even though they're so far apart. And then this whole thing is 100,000 light years. It's 25,000 times the distance than the distance between the Sun and the nearest star. So we're talking about unimaginable, unfathomable distances, just for a galaxy. And now we're going to get our-- frankly, my brain is already well beyond anything that it can really process. At this point, it almost just becomes abstract thinking. It just becomes playing with numbers and mathematics. But to get a sense of the universe, itself, the observable universe-- and we have to be clear. Because we can only observe light Because that's how old the universe is. The observable universe is about 93 billion light years across. And the reason why it's larger than 13.7 billion is that the points in space that emitted light 13.7 billion years ago, those have been going away from us. So now they're on the order of 40 billion light years away. But this isn't about cosmology. This is just about scale and appreciating how huge the universe is. Just in the part of the universe that we can theoretically observe, you have to get-- and that we can observe, just because we're getting electromagnetic radiation from those parts of the universe-- you would have to multiply this number. So let me make this clear. 100,000 light years-- that's the diameter of the Milky Way."
    },
    {
        "Q": "At 4:28, Rishi writes pCO2 and pO2. What does p stand for?",
        "A": "partial pressure",
        "video_name": "QP8ImP6NCk8",
        "timestamps": [
            268
        ],
        "3min_transcript": "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. So this red blood cell goes off to do its delivery. Let's say, it's delivering some oxygen out here. And there is a tissue cell. And, of course, it doesn't know where it's going to go that day. But it's going to go wherever its blood flow takes it. So let's say, it takes a pass over to this thigh cell in your, let's say, upper thigh. So this thigh cell has been making CO2. And remember, sometimes we think of CO2 as being made only when the muscle has been working. But you could be napping. You could be doing whatever. And this CO2 is still being made because cellular respiration is always happening. So this red blood cell has moved into the capillary right by this thigh cell. So you've got a situation like this where now some of the CO2 is going to diffuse into the red blood cell like that. And what happens once it gets down there? Just so you get a closer view of what's going on. And we're in the thigh and the two big conditions in the thigh that we have to keep in mind. One is that you have a high amount of CO2 or partial pressure of CO2. And this is dissolved in the blood. And the other is that you have a low amount of oxygen, not too much oxygen in those tissues. So let's focus on that second point. If there's not too much oxygen in the tissues, and we know that the hemoglobin is kind of constantly bumping into oxygen molecules and binding them. And they fall off and new ones bind. So it's kind of a dynamic process. Now, when there's not too much oxygen around, these oxygen molecules are going to fall off as they always do in a dynamic situation. Except new ones are not going to bind. Because there's so little oxygen around in the area, that less and less oxygen is free and is available to bump into hemoglobin and bind to it."
    },
    {
        "Q": "At 4:05, how do you pronounce 'crastholatian' acid metabolism??",
        "A": "Crassulacean , after the Crassulaceae family of plants in which it was first discovered. This family includes orchids, bromeliads (of which pineapple is a member), purslane, agave, aloe and the succulent jade plant.",
        "video_name": "xp6Zj24h8uA",
        "timestamps": [
            245
        ],
        "3min_transcript": "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 But they both occur simultaneously, the light independent and light dependent, and only during the daytime. And if your stomata is closed, you need to perform photosynthesis, especially the Calvin cycle, you need CO2 (carbon dioxide). So how can you get around this? If I want to close my stomata during the day, but I need CO2 during the day, how can I solve this problem? And what desert plants, or many desert plants, have evolved to do, essentially does photosynthesis, but instead of fixing the carbon in outer cells and then pushing it in to inner cells and then performing the Calvin cycle, instead of outer and inner cells, they do it in the nighttime and in the daytime. So, in CAM plants - and these are called CAM plants because, It stands for crassulacean acid metabolism. And that's because it was first observed in that species of plant, the crassulacean plant. But these are just called, you could call it CAM photosynthesis or CAM plants. They're essentially a subset of C4 plants. But instead of performing C4 photosynthesis, kind of an outside cells and inside cells, they do it at the night time and the day. And what they do is, at night they keep their stomata open. And they perform, and they're able to fix - and everything occurs in the mesophyll cells and the CAM cells, in the CAM plants. So at night time, when they're not afraid of losing water - let's say this is a mesophyll cell right here - my stomata is open."
    },
    {
        "Q": "At 5:07 sal says that glass is a fluid,but how? in order to become one it should flow, right?",
        "A": "Glass is made of really long and hard to move molecules. Even though they want to move, its just too slow. Glass is not really a liquid, but also it isn t a solid, properly saying. Its state is called a supercooled liquid, or an amorphous solid. That happens when a liquid is cooled beyond its freezing point, but it doesn t freeze.",
        "video_name": "G4CgOF4ccXk",
        "timestamps": [
            307
        ],
        "3min_transcript": "Let me draw the cylinder in a more vibrant color so you can figure out the volume. So it equals this side, the left side of the cylinder, the input area times the length of the cylinder. That's the velocity of the fluid times the time that we're measuring, times the input velocity times time. That's the amount of volume that came in. If that volume came into the pipe-- once again, we learned several videos ago that the definition of a liquid is a fluid that's incompressible. It's not like no fluid could come out of the pipe and all of the fluid just gets squeezed. The same volume of fluid would have to come out of the pipe, Whatever comes into the pipe has to equal the volume coming out of the pipe. One assumption we're assuming in this fraction of time that we're dealing with is also that there's no friction in this liquid or in this fluid, that it actually is not turbulent and it's not viscous. A viscous fluid is really just something that has a lot of friction with itself and that it won't just naturally move without any resistance. When something is not viscous and has no resistance with itself and moves really without any turbulence, that's called laminar flow. That's just a good word to know about and it's the opposite of viscous flow. Different things have different viscosities, and we'll probably do more on that. Like syrup or peanut butter has a very, very high viscosity. Even glass actually is a fluid with a I think there's some kinds of compounds and magnetic fields that you could create that have perfect laminar flow, but this is kind of a perfect situation. In these circumstances, the volume in, because the fluid can't be compressed, it's incompressible, has to equal the volume out. What's the volume out over that period of time? Similarly, we could draw this bigger cylinder-- that's the area out-- and after T seconds, how much water has come out? Whatever water was here at the beginning of our time period will have come out and we can imagine the cylinder here. What is the width of the cylinder? What's going to be the velocity that the liquid is coming out on the right-hand side?"
    },
    {
        "Q": "At 2:59 when the nitrogen gets sp2 then the carbon with the negative charge gets sp3?.",
        "A": "In that particular resonance structure the carbon with the negative charge would be sp3, but the real structure has all the ring atoms sp2 hybridized with the \u00cf\u0080 electrons delocalized (spread over all of the ring atoms).",
        "video_name": "wvVdgGTrh-o",
        "timestamps": [
            179
        ],
        "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 2:40, I don't really understand where the numerator came from.",
        "A": "To figure total parallel resistance, one adds the reciprocal of all the resistors to get the reciprocal of the final resistance of the circuit. (i.e. one adds the conductances). The 16 is the least common denominator of the various fractions he is adding. He needs to convert each fraction to 16ths in order to add the numerators. For example, the 1/4 becomes 4/16, and so on.",
        "video_name": "3NcIK0s3IwU",
        "timestamps": [
            160
        ],
        "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 3:39 he begins to add up the resistance of the two resistors in series using ohms law by saying R1+R2 which gave him 3. my question is why is R3 on the vertical right hand side line not being added when it is also a resistor in series",
        "A": "R3 is not in series with (R1+R2), but rather in series with [(R1+R2) in parallel with R0]. So the total resistance would be (R1+R2)*R0/(R1+R2+R0) + R3 not just R1+R2+R3",
        "video_name": "3NcIK0s3IwU",
        "timestamps": [
            219
        ],
        "3min_transcript": "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? then we had a 1-ohm. So we had a 2-ohm and a 1-ohm in series, so those simplify to 3 ohms. Well, now this is getting really simple. So what do these two resistors simplify to? Well, 1 over their combined resistance is equal to 1/3 plus 1/3. 2/3. 1/R is equal to 2/3, so R is equal to 3/2, or we could say 1.5, right? So let's erase that and simplify our drawing. So this whole mess, the 3-ohm resistor in parallel with the other 3-ohm resistor is equal to one resistor with a 1.5 resistance."
    },
    {
        "Q": "at 6:55 the molecule which contains 2 lone pairs and 2 bond pairs\nnitrogen has valency of only 5\nin that 2 lone pair of electrons are there so there is only one e to be shared then how can it form two bonds\nis the one bond a dative bond?",
        "A": "Pretend that molecule had an -NH2 group on the end. Then pretend a very strong base came along and removed one of those hydrogens, and the electrons that were in that N-H bond stayed on nitrogen. What you get is nitrogen having 8 electrons still but only 4 of them are in bonds, the other 4 are in the 2 lone pairs.",
        "video_name": "5-MM39VCwc0",
        "timestamps": [
            415
        ],
        "3min_transcript": "for the formal charge. Alright, let me redraw that. So we have our nitrogen with four bonds to hydrogen and then nitrogen has a plus one formal charge. You should recognize this as being the ammonium ion from general chemistry. So this has a formal charge of plus one, so we have another pattern to think about here. So let's draw that in. We have one, two, three, four bonds and zero lone pairs of electrons. So when nitrogen has four bonds, four bonds and zero lone pairs, zero lone pairs of electrons, we've already seen the formal charge be equal to plus one. So let's look at some examples where nitrogen has a formal charge of plus one. So the example on the left, we can see there are four bonds and there are no lone pairs on that nitrogen, so that's a plus one formal charge. Here's one bond, two bonds, three bonds, and four bonds and no lone pairs, so a plus one formal charge on the nitrogen. Alright, finally, one more nitrogen to assign a formal charge to. So let's look at this one. Let's draw in the electrons in the bond. So here's two electrons and here's two electrons. What is the formal charge on nitrogen? Formal charge is equal to number of valence electrons nitrogen is supposed to have, which we know is five, and from that we subtract the number of valance electrons nitrogen actually has in our dot structure. So again we go over to here and we look at this bond and we give one electron to nitrogen and one electron to the other atom. And over here we give one electron to nitrogen and one electron to the other atom. And now we have two lone pairs of electrons on the nitrogen. So how many is that total? this would be one, two, three, four, five, and six. So five minus six gives us negative one. So a formal charge of negative one. Let me go ahead and redraw that. So I could draw it out here. So nitrogen with two lone pairs of electrons we just found has a formal charge of negative one. If I wanted to leave off the lone pairs of electrons I could do that, I could just write NH here and put a negative one formal charge, and because of this pattern, you should know there are two lone pairs of electrons on that nitrogen. Let me just clarify the pattern here. The pattern for a formal charge of negative one on nitrogen would be two bonds, here are the two bonds, and two lone pairs of electrons. So when nitrogen has two bonds and two lone pairs of electrons, nitrogen should have a formal charge of negative one. Let's look at some examples of that."
    },
    {
        "Q": "According to wikipedia: A parsec is the distance from the SUN (not from the EARTH) to an astronomical object that has a parallax angle of one arcsecond.\n\nSal said in 7:20, \"It's the distance that an object needs to be from EARTH in order for it to have a parallax angle of one arc second.\n\nYou could argue that it's approximately the same because 1 AU is miniscule compared to distance of a star from the earth. But still, the concept is different.",
        "A": "It s not just approximately the same, it s the same to an almost immeasurable degree. We are looking at objects that are light YEARS away. The distance between the sun and the earth is 8 light MINUTES. 8 minutes / 1 year = .0015%. The distance measurements themselves are nowhere near that precise.",
        "video_name": "6zV3JEjLoyE",
        "timestamps": [
            440
        ],
        "3min_transcript": "And we just use a little bit of trigonometry. The tangent of this angle, the tangent of 90 minus 1/3600, is going to be this distance in astronomical units divided Well, you divide anything by 1, it's just going to be that distance. So that's the distance right over there. So we get our calculator out. And we want to find the tangent of 90 minus 1 divided by 3,600. And we will get our distance in astronomical units, 206,264. We're going to say 265. So this distance over here is going to be equal to 206,265-- I'm just And if we want to convert that into light years, we just divide. So there are 63,115 astronomical units per light year. Let me actually write it down. I don't want to confuse you with the unit cancellation. So we're dealing with 206,265 astronomical units. And we want to multiply that times 1 light year is equal to 63,115 astronomical units. And we want this in the numerator and the denominator to cancel out. And so if you divide 206,265, this number up here, by 63,115, the number of astronomical units in a light year-- let me delete that right over there-- we get 3.2. Well, the way the math worked out here, it rounded to 3.27 light years. So I should just show it's approximate right over there. But that's where the parsec comes from. So hopefully now you just realize it is just a distance. But even more, you actually understand where it comes from. It's the distance that an object needs to be from Earth in order for it to have a parallax angle of one arc second. And that's where the word came from. Parallax arc second."
    },
    {
        "Q": "Hi all, I have a question. At 1:47, what does Sal mean when he says, \"Distance traveled as a function of theta.\"?? Thank you.",
        "A": "at 1:47 sal means that if theta is changed, the distance changes because distance depends upon theta",
        "video_name": "-h_x8TwC1ik",
        "timestamps": [
            107
        ],
        "3min_transcript": "We now know how long the object is going to be in the air, so we're ready to figure out how far it's going to travel. So we can just go back to kind of the core formula in all of really kinematics, all of kind of projectile motion or mechanical physical problems, and that's distance is equal to rate times time. Now, we're talking about the horizontal distance. So our distance is going to be equal to-- what's our rate in the horizontal direction? We care about horizontal distance traveled, so our rate needs to be the horizontal component of the velocity, or the magnitude of the horizontal component of the velocity. And we figured that out in the first video. That is s cosine of theta. So let's write that down right here. So our rate is s cosine of theta. And how long will we be traveling at Well, we'll be going at that speed as long as we are in the air. So how long are we in the air? Well, we figured that out in the last video. We're going to be in the air this long-- 2 s sine of theta divided by g. So the time is going to be 2 s sine of theta over g. So the total distance we're going to travel, pretty straightforward, rate times time. It's just the product of these two things. And we could put all of the constants out front, so it's a little bit clearer that it's a function of theta. So we can write that the distance traveled-- let me do that same green. The distance traveled as a function of theta is equal to-- I'll do that in this blue. This s times 2s divided by g is-- I'll do it in a neutral color actually. So 2s squared over g times cosine of theta times sine of theta. So now we have a general function. You give me an angle that I'm going to shoot something off at and you give me the magnitude of its velocity, and you give me the acceleration of gravity. I guess if we were on some other planet, who knows? And I will tell you exactly what the horizontal distance is."
    },
    {
        "Q": "At 0:30 what is a LCD?",
        "A": "Liquid Crystal Display. He mentions it in the vid",
        "video_name": "PSy6zQsk8z0",
        "timestamps": [
            30
        ],
        "3min_transcript": "- [Karl Voiceover] Okay, so here's our Vivitar ViviCam Camera, and we're gonna take it apart and show you how it works. We got the batteries here, and there's the on switch, and now that we have the camera on we're going to take a few photos. You can see the LCD display is showing what's going on there, and letting us see the pictures. Now the LCD is backlit with an LED, which is, LCD stands for liquid crystal display, and LED stands for light emitting diode. So, let's go ahead and get crackin', we'll take it apart. We'll take all the screws out that are holding the two halves of the camera together. So we'll remove the top part, and that's likely made out of ABS, and there's a electroplated bezel, that surrounds the lens of the camera, it's screwed in place. And that's also probably ABS, plated ABS. And now let's go and take a look at what the camera's latch looks like and what it's made out of. We've got some, looks like chrome-plated stamped steel, from the underside of the back side of the batteries, to the board, and so that's where the positive connection on the board is, and the negative connection is up at the top. So we're gonna remove the pin and we can take the latch out now. Now you can slide the latch back and forth, you can see how it moves. Again this is plated steel, you can see there's four little connections on there, we'll talk about those in a second, but let's remove this back piece. This is just a simple insert, and I think it's probably heat staked in place. What that means is really low-cost way to attach things. This metal piece is as well, you can see those four points there. They're heat staked in and it is magnetic, so it's plated steel. And so, now we're gonna remove these screws that are connecting the plastic bezel that surrounds the camera and connects the two sides. And so you can see, it's just an injection molded piece, but again it didn't have a plastic designation on it, so it's not 100%, could be polypropylene or maybe styrene. Probably styrene. So there's the on button and the off button, I should say, the on button and the shutter button, and they push down and trigger the membrane switches, which are mounted on our printed circuit board, or a piece of the printed circuit board that's connected. So now let's take out the screws that are holding the board in place. And we tried to remove the screws for our tripod mount, but one of them is stripped, so we're just going to leave it in place. And I think we'll just pop the back out there and we'll take a look at the two switch covers. These switches are injection molded and then they are painted, and then heat staked in place. And so you can see the silver colors from the paint, and the heat stake process is basically the plastic, just a little tab at the end of the plastic is melted against the outer part of the housing."
    },
    {
        "Q": "At 7:48, it says there's a decrease in Sn2 mechanism. I understand that, but if a strong nucleophile is put in a polar protic solvent, it becomes a weak nucleophile. Then, shouldn't Sn1 be favoured over Sn2 as a minor product ?",
        "A": "A polar protic solvent doesn t make a strong nucleophile into a weak nucleophile.It makes it into a weaker nucleophile. Ethoxide ion is a strong nucleophile in an aprotic solvent. It is weaker in a protic solvent, but it is still a pretty strong nucleophile even there.",
        "video_name": "vFSZ5PU0dIY",
        "timestamps": [
            468
        ],
        "3min_transcript": "We would have our ring and a double bond forms between carbon one and carbon six. So, that means a double bond forms in here. And then at carbon two, we still have a methyl group going away from us in space. So, let me draw that in like that. So, the electrons in red, hard to see, but if you think about these electrons in red back here, are gonna move in to form our double bond between what I've labeled as carbon one and carbon six. Let me label those again here. So, carbon one and carbon six. Again not IUPAC nomenclature just so we can think about our product compared to our starting material. So, this would be the major product of our reaction which is an E2 reaction. It would also be possible to get some products from an SN2 mechanism, but since heat is here, an elimination reaction is favored over a substitution. 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"
    },
    {
        "Q": "At 3:40 he says that the reaction would go by an Sn1 mechanism. Doesn't Sn1 only occurred in tertiary substrates? The one in the example is secondary",
        "A": "No, 2\u00c2\u00b0 substrates can react via SN1 or SN2, depending on the conditions. We have two competing processes. If the nucleophile attacks faster than the leaving group spontaneously leaves, the reaction is SN2. If the leaving group leaves before the nucleophile can successfully attack, we have SN1. In this case, the methanoic acid is a very weak nucleophile, so the rate of SN2 attack is slow. Br\u00e2\u0081\u00bb is a good leaving group. so the rate of the SN1 reaction predominates over that of SN2.",
        "video_name": "vFSZ5PU0dIY",
        "timestamps": [
            220
        ],
        "3min_transcript": "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 we have another secondary alkyl halide, so right now all four of these are possible until we look at our reagent which is sodium hydroxide, Na plus, OH minus, and we know that the hydroxide ion can function as a strong nucleophile or a strong base. So a strong nucleophile makes us think an SN2 reaction and not an SN1. The strong base makes us think about an E2 reaction and not an E1 reaction. Since we have heat, heat favors an elimination reaction over a substitution, so E2 should be the major reaction here. So, when we analyze our alkyl halide, the carbon bonded to the halogen is our alpha carbon and the carbons directly bonded to that would be our beta carbons. So, we have two beta carbons here and let me number this ring. I'm gonna say the alpha carbon is carbon one, I'm gonna go round clockwise, so that's one, two, three, carbon four, And next we're going to translate this to our chair confirmation over here. So, carbon one would be this carbon and then carbon two would be this one. This'll be carbon three, four, five and six. The bromine is coming out at us in space at carbon one which means it's going up. So, if I look at carbon one, we would have the bromine going up, which would be up axial. At carbon two, I have a methyl group going away from me in space, so that's going down, so at carbon two we must have a methyl group going down which makes it down axial. So, we care about carbon two. Let me highlight these again. So, we care about carbon two which is a beta carbon. We also care about carbon six which is another beta carbon. So, let's put in the hydrogens on those beta carbons. At carbon two, we would have a hydrogen that's up equatorial and at carbon six we would have"
    },
    {
        "Q": "around 9:30 in the video he said \"total distance\"... i thought we were finding the total displacement?",
        "A": "In this case, the distance and the displacement are the same, because the plane always moves in the same direction. But you are correct, if he was being super-careful, maybe he should have said displacement .",
        "video_name": "VYgSXBjEA8I",
        "timestamps": [
            570
        ],
        "3min_transcript": "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 physics class, but let's use it to actually figure out the acceleration that a pilot experiences when they're taking off of a Nimitz class carrier. So we have 2 times the acceleration times the distance, that's 80 meters, times 80 meters, is going to be equal to our final velocity squared. What's our final velocity? 72 meters per second. So 72 meters per second, squared, minus our initial velocity. So our initial velocity in this situation is just 0. So it's just going to be minus 0 squared, which is just going to be 0, so we don't even have to write it down. And so to solve for acceleration, to solve for acceleration, you just divide, so this is the same thing as 160 meters, well, let's just divide both sides by 2 times 80, so we get acceleration is equal to 72 m/s squared"
    },
    {
        "Q": "At 11:12, Sal uses the previous value stored in the calculator (72.222 m/s) as the final velocity. But if he had just used 72 m/s like he said during the video, then the answer we get would be slightly different, around an acceleration of 32.4 m/s^2. I know this is a very slight difference, but in the second case, the value rounds down to 32 m/s^2. Should we use the more precise value like Sal used or use the value that complies with the significant figures rule?",
        "A": "follow sig fig rules. But you don t round intermediate values - you only round the final answer (to the correct number of sig figs)",
        "video_name": "VYgSXBjEA8I",
        "timestamps": [
            672
        ],
        "3min_transcript": "physics class, but let's use it to actually figure out the acceleration that a pilot experiences when they're taking off of a Nimitz class carrier. So we have 2 times the acceleration times the distance, that's 80 meters, times 80 meters, is going to be equal to our final velocity squared. What's our final velocity? 72 meters per second. So 72 meters per second, squared, minus our initial velocity. So our initial velocity in this situation is just 0. So it's just going to be minus 0 squared, which is just going to be 0, so we don't even have to write it down. And so to solve for acceleration, to solve for acceleration, you just divide, so this is the same thing as 160 meters, well, let's just divide both sides by 2 times 80, so we get acceleration is equal to 72 m/s squared 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,"
    },
    {
        "Q": "at 4:24 where does the 2 come from in the denominator? I don't understand...it seems so random",
        "A": "The 2 is used because the average (mean) velocity is being found. The mean of a set is found by dividing the sum of the numbers in the set by the number of numbers. So the average mean speed is (V1+V2)/2, Hope this helped.",
        "video_name": "VYgSXBjEA8I",
        "timestamps": [
            264
        ],
        "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": "at 11:40, shouldn't Sal use 72 instead of 72.2222.... because it's only 2 sig figs? that would make 72^2 / 160 = 32.4, which rounded to 2 sig figs is 32, not 33.",
        "A": "Sal is often a bit careless about sig figs. The equation that you have written should have 2 sig figs in its answer, if that is the final answer. Remember not to discard extra sig figs until you report your final answer. You use them all during intermediate calculations.",
        "video_name": "VYgSXBjEA8I",
        "timestamps": [
            700
        ],
        "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 6:02, why does the hydrocarbon chain have no polarity?",
        "A": "Carbon and Hydrogen have very similar electronegativity, so the elements don t pull each other to a specific direction. Thus the hydrocarbon chain is not polar.",
        "video_name": "Pk4d9lY48GI",
        "timestamps": [
            362
        ],
        "3min_transcript": "And then in between, we have a carbon, and we call that the alpha carbon. We call that the alpha carbon. Alpha carbon, and that alpha carbon is bonded, it has a covalent bond to the amino group, covalent bond to the carboxyl group, and a covalent bond to a hydrogen. Now, from there, that's where you get the variation in the different amino acids, and actually, there's even some exceptions for how the nitrogen is, but for the most part, the variation between the amino acids is what this fourth covalent bond from the alpha carbon does. So you see in serine, you have this, what you could call it an alcohol. You could have an alcohol side chain. In valine right over here, you have a fairly pure hydrocarbon, hydrocarbon side chain. And so in general, we refer to these side chains as an R group, and it's these R groups and how they interact with their environment and the types of things they can do. And you can even see, just from these examples how these different sides chains might behave differently. This one has an alcohol side chain, and we know that oxygen is electronegative, it likes to hog electrons, it's amazing how much of chemistry or even biology you can deduce from just pure electronegativity. So, oxygen likes to hog electrons, so you're gonna have a partially-negative charge there. Hydrogen has a low electronegativity relative to oxygen, so it's gonna have its electrons hogged, so you're gonna have a partially positive charge, just like that, and so this has a polarity to it, and so it's going to be hydrophilic, it's going to, at least this part of the molecule is going to be able to be attracted and interact with water. And that's in comparison to what we have over here, this hydrocarbon side chain, this has no polarity over here, So this is going to be hydrophobic. And so when we start talking about the structures of proteins, and how the structures of proteins are influenced by its side chains, you could image that parts of proteins that have hydrophobic side chains, those are gonna wanna get onto the inside of the proteins if we're in an aqueous solution, while the ones that are more hydrophilic will wanna go onto the outside, and you might have some side chains that are all big and bulky, and so they might make it hard to tightly pack, and then you might have other side chains that are nice and small that make it very easy to pack, so these things really do help define the shape, and we're gonna talk about that a lot more when we talk about the structure. But how do these things actually connect? And we're gonna go into much more detail in another video, but if you have... If you have serine right over here, and then you have valine right over here, they connect through"
    },
    {
        "Q": "At 8:50\n\nWhy would Nitrogen grab a proton (H+) from the environment if this atom is very eletronegative?",
        "A": "The atom as a whole is not electronegative, if the carboxyl tail becomes negative by losing a H+ the amino head becomes positive by gaining a H+, to make a neutral molecule.",
        "video_name": "Pk4d9lY48GI",
        "timestamps": [
            530
        ],
        "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": "At 04:07, why is it that at the 3rd carbon it is not 3-dimethyl since you have two methyl groups coming off?",
        "A": "That second methyl has already been accounted for in the longest chain (it s the fourth carbon in the butane base), so you only need to name one methyl substituent at the 3 carbon.",
        "video_name": "peQsBg9P4ms",
        "timestamps": [
            247
        ],
        "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": "Hi, at 9:00 why does the #2 group move but not the #1 group?",
        "A": "He is rotating the whole molecule 120\u00c2\u00b0 about the axis joining the C and #1, so those two atoms stay in place while #2, #3, and #4 rotate about that axis.",
        "video_name": "peQsBg9P4ms",
        "timestamps": [
            540
        ],
        "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": "At around 4:53, why did Sal 4*56+3*16?",
        "A": "Each iron has a mass of 56 and each oxygen has a mass of 16. The total mass of Fe2O3 is calculated as 2*56+3*16. (You have a typo; should be 2*56 instead of 4*56)",
        "video_name": "SjQG3rKSZUQ",
        "timestamps": [
            293
        ],
        "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 7:00, do you always multiply that number by 2 or did he only do that because it was in front of Al? Would he have not multiplied it at all if it had nothing i front of the Al?",
        "A": "If you look at the reaction equation, there is a coefficient in front of Al. You multiply by the coefficient because that is the ratio in which these chemicals react. Of course, if the coefficient is 1 (in which case you don t write it just like you don t write a coefficient of 1 in algebra) there is no need to multiply.",
        "video_name": "SjQG3rKSZUQ",
        "timestamps": [
            420
        ],
        "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": "At about 7:12 sal says \"for every one molecule of this, we need one molecule of that.\" I thought Fe2O3 would have an ionic bond and therefore technically wouldn't be a molecule. Am I mistaken?",
        "A": "You are not mistaken, but we are often a little sloppy in terminology. We technically should be saying formula unit when speaking of Fe\u00e2\u0082\u0082O\u00e2\u0082\u0083 or any other ionic compound instead of molecule .",
        "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'm still confuse why use the ratio of 1:2 of FE2O3 and 2Al mole ratio , 85g FE2O2 is 0.53 moles right why the aluminum must be 2(0.53)mole?",
        "A": "So, the reaction asks for 1 unit of Fe2O3 for every 2 units of Al. Therefore, if you have 1*(0.53) moles of Fe2O3 then you need 2*(0.53) moles of Al",
        "video_name": "SjQG3rKSZUQ",
        "timestamps": [
            62
        ],
        "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": "At 5:18 is it okay to NOT use PEMDAS?",
        "A": "No, you should always compute numbers using the correct order of operations AKA PEMDAS.",
        "video_name": "SjQG3rKSZUQ",
        "timestamps": [
            318
        ],
        "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": "Why do we round the atomic mass numbers from the periodic table? Like for Iron at 4:43, we use 56, not 55.85",
        "A": "I understand that Iron has multiple isotopes and the atomic mass is an average of all those isotopes. 55.85 is the average mass of all Fe isotopes. Using 56 is simplifying a little bit...basically assuming you re using an Fe isotope with a mass of exactly 56 amu.",
        "video_name": "SjQG3rKSZUQ",
        "timestamps": [
            283
        ],
        "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 6:30 did he round that with the third number?",
        "A": "Yes, he did. 85 g Fe\u00e2\u0082\u0082O\u00e2\u0082\u0083 = 0.532 mol Fe\u00e2\u0082\u0082O\u00e2\u0082\u0083. Only 2 significant figures are justified, but this is an intermediate answer, so he should have carried an extra digit and then rounded off the final answer to 2 significant figures.",
        "video_name": "SjQG3rKSZUQ",
        "timestamps": [
            390
        ],
        "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": "At 4:53, Can you explain to me why you multiplied 2 x 56. I am not very clear...",
        "A": "you multiply 2X56 because the atomic mass of iron is approximately 56, times two because you have two atoms of Fe. Hope this helps! :)",
        "video_name": "SjQG3rKSZUQ",
        "timestamps": [
            293
        ],
        "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": "Things I am confused about:\n- At 5:20 you say that one molecule of iron three oxide is going to be 160 atomic mass units, then\n- At 5:41 you say that one mole of iron three oxide is going to have a mass of 160 grams.\nI am confused, at either 160 is recognized as grams or atomic mass unites?",
        "A": "A MOLECULE has mass of 160 amu. One MOLE of those molecules has mass of 160 grams. A mole is a number, and we picked it so that the grams to amu thing would work.",
        "video_name": "SjQG3rKSZUQ",
        "timestamps": [
            320,
            341
        ],
        "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 1:31 how did you predict the subscripts of the atoms, and how they would arrange themselves?",
        "A": "It s something that he just knows from past experience, and it is something that will come to you with time. When a metal is reacted with oxygen a metal oxide is formed. The subscripts are needed to make the formula work: Al^3+ O^2- The simplest way to make the charges cancel is 2x Al^3+ and 3x O^2- So the formula is Al2O3",
        "video_name": "SjQG3rKSZUQ",
        "timestamps": [
            91
        ],
        "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": "at 1:30 the iron oxide isFE2O3 why not FEO2?",
        "A": "Iron (Fe) in the complex will become a (Fe)3+ ion. The oxygens are each (O)2-. Therefore, in order to balance the charges and have a compound with zero net charge, you need 2(Fe)3+ and 3(O)2-",
        "video_name": "SjQG3rKSZUQ",
        "timestamps": [
            90
        ],
        "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": "At 5:36 Why does Sal add the atomic masses of O and Fe to get the atomic mass of Ironoxide",
        "A": "He added them to calculate the moles of Fe2O3 using the formula: moles = given mass/molar mass.",
        "video_name": "SjQG3rKSZUQ",
        "timestamps": [
            336
        ],
        "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": "At 11:27, in our class we say iron feroxide. Is it the same as iron three oxide as Sal says?",
        "A": "Do you mean ferric oxide ? If so, that is an older term for Iron (III) oxide. It is still widely used, but it is not the IUAPC approved term. Ferroxide is a brand name of a pigment, so that is not an acceptable scientific name. But, to answer your question, you should use the method demonstrated in the video with the Roman numerals. This is the officially sanctioned term and would be understood by any chemist.",
        "video_name": "SjQG3rKSZUQ",
        "timestamps": [
            687
        ],
        "3min_transcript": "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, Assuming we keep mixing it nicely and the whole reaction happens all the way. And we'll talk more about that in the future. And in that situation where we have more than 28.63 grams of aluminium, then this molecule will be the limiting reagent. Because we had more than enough of this, so this is what's going to limit the amount of this process from happening. If we have less than 28.63 grams of, I'll start saying aluminum, then the aluminum will be the limiting reagent, because then we wouldn't be able to use all the 85 grams of our iron molecule, or our iron three oxide molecule. Anyway, I don't want to confuse you in the end with that limiting reagents. In the next video, we'll do a whole problem devoted to limiting reagents."
    },
    {
        "Q": "at 3:36 he says Iron 3 Oxide, he means Iron 2 oxide right?",
        "A": "No, Fe\u00e2\u0082\u0082O\u00e2\u0082\u0083 is Iron (III) Oxide. The number after the metal ion references its oxidation state, not how many atoms are present per formula unit.",
        "video_name": "SjQG3rKSZUQ",
        "timestamps": [
            216
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "At 6:28 why asymmetric stretch is taking more energy than symmetric stretch ?",
        "A": "Here s my guess: Consider a CH\u00e2\u0082\u0082 group. In a symmetric stretch, the two H atoms are going in the same direction. The group dipole moment changes considerably because both bonds are going in and out at the same time. In an asymmetric stretch, they are going in opposite directions. One bond gets longer as the other gets shorter, so the change in dipole moment is much less. It takes energy to separate positive and negative charges from each other, so the symmetric vibration has a higher frequency (energy).",
        "video_name": "9GPuoukU8fM",
        "timestamps": [
            388
        ],
        "3min_transcript": "Let's compare it to butylamine. So over here, this is a primary amine. The nitrogen is bonded to one carbon, so we're talking about a primary amine now. And let's analyze the IR spectrum. So once again, we're gonna draw a line around 3,000, and we know that this in here is talking about the carbon-hydrogen bond stretch for an SP3 hybridized carbon. Alright, once again, let's look at just past that, right in the bond to hydrogen region, and we get two signals this time, right? So if we look over here, there are two signals. This signal, let's drop down, this is approximately 3,300, so we have one signal approximately 3,300. And then we have another signal. Let me go ahead and make that green here. So we've got another signal right here, which is a little bit higher in terms of the wave number. So we drop down, this signal is approximately 3,400. the nitrogen-hydrogen bond stretch. We get two signals, and we need to figure out what's going on here. Well, this has to do with symmetric and asymmetric stretching. So let's look at two generic amines here, and let's talk about what the difference is between symmetric and asymmetric stretching. If you have symmetric stretching, so these bonds are stretching in phase, if you will. You can think about the hydrogens stretching away from the nitrogen at the same time. So this is called symmetric stretching. This is symmetric stretching. And this one over here, let me go ahead and draw what's happening over here. So this time these two nitrogen-hydrogen bonds are stretching out of phase. So if that hydrogen is stretching this way, this hydrogen might be contracting here, so that's an asymmetric stretch. Let me go ahead and write that. So we're talking about an asymmetric stretch here. This is why we get these two different signals. It turns out it takes less energy to do the symmetric stretching. So if it takes less energy to do the symmetric stretching, this is the one that we find at a lower wave number. Remember, wave numbers correspond to energy. So it takes less energy to do a symmetric stretching, and so that's this signal. It takes a little more energy to do asymmetric stretch. And so that's this signal, right up here. So we get two different signals here for our primary amine. Two signals, right? And it's tempting to say, \"Oh, we get two signals \"because we have two nitrogen-hydrogen bonds. \"So here's a nitrogen-hydrogen bond, \"and here's a nitrogen-hydrogen bond.\" But that's not really what's happening. Some of the molecules are having a symmetric stretch, and some of the molecules are having an asymmetric stretch. And so that's why you see these two different signals. Once again, let's just really quickly compare these two different amines."
    },
    {
        "Q": "whats a solar mass as said at 1:28",
        "A": "One solar mass is not the mass, but the diameter of Sol(the Sun).",
        "video_name": "DxkkAHnqlpY",
        "timestamps": [
            88
        ],
        "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": "What would almost infinite be? (5:16)",
        "A": "There is no ALMOST infinite",
        "video_name": "DxkkAHnqlpY",
        "timestamps": [
            316
        ],
        "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": "Mentioned in 8:32, what are those stars in our solar system rapidly orbiting what is believed to be a black hole?",
        "A": "the star are called a high velocity star.",
        "video_name": "DxkkAHnqlpY",
        "timestamps": [
            512
        ],
        "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": "even at 8:15 the structure that is given\nis it benzophenone or dibenzophenone?",
        "A": "I wiki d it up It s benzophenone although some part of me really wants it to be called dibenzophenone :P",
        "video_name": "wD15pD5pCt4",
        "timestamps": [
            495
        ],
        "3min_transcript": "And I'll just name this systematically right here. And the more complicated things get, the more systematic people will want to name it. So if we have six carbons right here, and they're in a chain, so this is cyclohexane. You'd put the \"e\" there if this carbonyl group wasn't there. But since it is, we would call this cyclohexanone, So this right here tells us to name it cyclohexanone. And then in a ring like this, this would implicitly be the number one carbon. So if this is the number one carbon and we want to number in the direction so that the next groups have the lowest possible number, so we want to make this the two carbon. So this is 2,2-dichlorocyclohexanone. And I'll just show these to you because these tend to be referred to by their common names. So I just want to show them to you real fast. One is this molecule right here, where we have a methyl group on this side: CH3. And over here, we have a benzene ring. Now, that first super simple ketone that we saw, we called this acetone. And so the common name here is actually derived from acetone. Instead of calling it acetone, because it doesn't have just a methyl group here, this is called aceto-, and instead of acetone, it's acetophenone, because we have this phenyl group, that benzene ring right there. Acetophenone, which is a pretty common molecule, and you'll see it referred to this way. Now, the other one that you might see every now and then, looks like this, that has two benzene rings on it. It looks like that. And this is benzophenone. These last two I just really wanted to expose you to their But, in general, I think you have a decent idea at this point of how to name at least the simpler chains, either with the common names, for example, propyl, or methyl propyl ketone, or 2-pentanone. And these are the more typical or maybe the easier naming examples."
    },
    {
        "Q": "At 7:44 the structure that is drawn\nis it acetophenone or acetophenyl?",
        "A": "it s acetophenone. Phenyl is only used as a prefix--phenone is used as a suffix. (e.g. 2-phenylbutane)",
        "video_name": "wD15pD5pCt4",
        "timestamps": [
            464
        ],
        "3min_transcript": "And I'll just name this systematically right here. And the more complicated things get, the more systematic people will want to name it. So if we have six carbons right here, and they're in a chain, so this is cyclohexane. You'd put the \"e\" there if this carbonyl group wasn't there. But since it is, we would call this cyclohexanone, So this right here tells us to name it cyclohexanone. And then in a ring like this, this would implicitly be the number one carbon. So if this is the number one carbon and we want to number in the direction so that the next groups have the lowest possible number, so we want to make this the two carbon. So this is 2,2-dichlorocyclohexanone. And I'll just show these to you because these tend to be referred to by their common names. So I just want to show them to you real fast. One is this molecule right here, where we have a methyl group on this side: CH3. And over here, we have a benzene ring. Now, that first super simple ketone that we saw, we called this acetone. And so the common name here is actually derived from acetone. Instead of calling it acetone, because it doesn't have just a methyl group here, this is called aceto-, and instead of acetone, it's acetophenone, because we have this phenyl group, that benzene ring right there. Acetophenone, which is a pretty common molecule, and you'll see it referred to this way. Now, the other one that you might see every now and then, looks like this, that has two benzene rings on it. It looks like that. And this is benzophenone. These last two I just really wanted to expose you to their But, in general, I think you have a decent idea at this point of how to name at least the simpler chains, either with the common names, for example, propyl, or methyl propyl ketone, or 2-pentanone. And these are the more typical or maybe the easier naming examples."
    },
    {
        "Q": "When Sal first talks about the 2nd spaceship, he says \"at exactly the same velocity as her\", but then writes and says 1.5C. He loses me at 1:30. If the 2nd ship is traveling at 1.5C and she's at .5C, how do their relative positions stay the same?\nARE we dealing with SR here? It doesn't seem Newtonian to me.\nThank you",
        "A": "You are getting confused with the expressions, writing .5C means point five times the speed of light while writing 1.5C would mean one point five times the speed of light. The equation is not 1.5C its 1.5x10 to the power of 8 m/s which is half the speed of light (3x10 to the power of 8 m/s) So all the spaceships are traveling at the same speed, which is half of the speed of light. Hopefully this makes sense, have another look at the equation being written at 1:30 again.",
        "video_name": "F7BU1sXtul4",
        "timestamps": [
            90
        ],
        "3min_transcript": "- [Voiceover] In the last video, we started to construct a space-time diagram for my frame of reference and I'm just drifting through space and I'm assuming that I'm in an intertial frame of reference which means I'm moving at a constant velocity relative to all other frames of reference and we set up a situation where I emitted a photon right at time zero, so after one second, it would have moved 3 times 10 to the 8th meters. After two seconds, it would have moved 6 times 10 to the 8th meters, and then we had a, we added a little bit of flavor to our little scenario where right at time zero, a friend passes me up in her spaceship and she is traveling relative to me in the positive-x direction at half the speed of light, and so we plotted her path right at time zero. Her spaceship is right there. I could draw the spaceship. It's at the origin. Then after one, after one second, she would have traveled 1 1/2 times 10 to the 8th meters. After two seconds, she would have traveled 3 times 10 to the 8th meters, so this blue line in the last video was her path. Let's assume that we have actually a whole train of spaceships, all traveling in the positive-x direction at the same velocity as her spaceship. So this is her spaceship right over here, at time equals zero, she is exactly where I am, but let's say that 3 times 10 to the 8th meters in front of her, there's another spaceship, traveling at exactly 1 1/2 times 10 to the 8th meters per second, so they're traveling at the same relative velocity to me, but if you think about it, from each of their point of view, they would seem to be stationary, because the distance between them, in this x-direction is going to stay the same. So this person, at time zero, is going to be 3 times 10 to the 8th meters in the positive-x direction from me. Now, if I were to wait two seconds, they are going to be, so if I were to wait two seconds, they are going to be, they are going to be 6 times 10 to the 8th meters away from me, because they're going half and so I could draw their path and I'm going to do it in a slightly more muted, thinner color, so that we, what I'm essentially going to be doing here is I'm setting up gridlines for my friend's alternate frame of reference. I'm going to put that on top of my frame of reference and to be clear, I'm not assuming special relativity. I'm assuming a Newtonian world, classical mechanics. Just to get familiar with these ideas and to see where the Newtonian world is going to break down but let's say that's not the only ship, let's say there was another ship, that at time equals zero is 6 times 10 to the 8th meters away from me in the positive x-direction. Well, where are they going to be? Where are they going to be after two seconds? Well, after, let's see, after one second, they're going to be 1 1/2 times 10 to the 8th meters further. Then after two seconds, they're going to be 3 times 10 to the 8th meters further,"
    },
    {
        "Q": "At 4:10, there are bulges at the cup. Is there any relations with the convex meniscus?",
        "A": "Sort of. The bulge that Sal is talking about does have to do with cohesion, but in this situation the bulge is not truly a convex meniscus because it is formed by surface tension, through cohesion, rather than just cohesion.",
        "video_name": "_RTF0DAHBBM",
        "timestamps": [
            250
        ],
        "3min_transcript": "and that causes a phenomenon known as surface tension. So you have stronger, you have kind of a deeper, and this is still just hydrogen bonds, but since they're not being pulled in other directions by, upwards by the air, they're able to get a little bit more closely packed, a little bit tighter, and this we refer to as surface tension, surface tension. And you have probably observed surface tension many, many, many times in your life in the form of, say, a water droplet. A water droplet, it's able to have this roughly round shape because all the little water molecules on the surface of the water droplet, and here the surface might even be on the bottom of the water droplet. They are more attracted to each other than they are to the surrounding air, so they're able to form this type of a shape. You might've seen it if you go to a pond or a stream sometimes, so you see some still water. And let's say, let me do this in blue. So let's say that this is the surface You might have seen insects that are able to walk on the surface of the water. And I'm not doing a great job at drawing the insects. They don't look exactly like that. But they can walk on the surface of the water. You might've seen or you might've even tried to do something like put a paperclip on the water. And even though this thing is actually more dense than the water and you might expect it to sink, but because of the surface tension, which really forms something of a film on top of the water, the thing won't penetrate the surface, so the paperclip will float, unless you were to push on it a little bit and it allow it to puncture the surface, and then it would actually sink, which is what you would expect because it is actually denser. You'd even see this if you were to take a cup, if you were to take a cup and you were to fill it all the way up to the rim and then a little bit higher, it won't immediately overflow. It won't immediately overflow. If you're very careful, you'll see that you form a bulge here. And that bulges because those individual water molecules are more attracted to each other than they are to the surrounding air. Obviously if you keep pouring water, at some point, they're just gonna start overflowing because gravity's gonna take over there. Gravity's gonna overwhelm the surface tension. But this bulge will actually form. So surface tension, it is really due to the cohesion of the water. Remember, cohesion is when the molecules are attracted to each other. And it definitely, and especially because they're more attracted to each other than the surrounding air."
    },
    {
        "Q": "At 4:54, Sal says that 'the same average kinetic energy' . What is he referring to ?\n\nThanks in advance !",
        "A": "The wood and the metal have the same kinetic energy, or in other words they have the same temperature since temperature is just a mesure of the jiggling of atoms/molecules. Hope this helps ! If not, feel free to contact me",
        "video_name": "6f553BGaufI",
        "timestamps": [
            294
        ],
        "3min_transcript": "That is wood. And I have the metal surface, I'll do that in white. So I have the metal surface, right over here. And this metal surface, we already talked about, is going to feel colder. Let me draw the rest of my hand, actually. So the rest of my arm, you get the idea. So what's going on here. So let's just think about it at a microscopic level. So the wood, first of all, its surface is going to be uneven. So you're going to have atoms up here, but then you're going to have gaps, there's going to be air here. Let me actually scroll down a little bit. So it's going to be like this, so you're going to have gaps like that. And it also has internal gaps, like that. So this would be the wood, while the metal is much denser. So the metal, let me do the metal in that white color, the metal atoms are much more closely packed. It is much denser, the surface is smoother, it won't have any internal air pockets, it's not going to have any internal air pockets in it. And so what's going to happen? Well, we've always said, you're going to have a transfer of heat from the higher temperature system, or the higher temperature thing, to the lower temperature thing. And so, they're already going to have some kinetic energy, these things are going to have an average kinetic energy that's consistent with 70 degrees fahrenheit. So, let me just draw a couple of these arrows. Same thing over here, they're going to have the same average kinetic energy. So these things are all jostling around, bouncing around and pushing on each other with the electrostatic forces. But, my hand is warmer, my hand has a higher average kinetic energy. And so the atoms and molecules of my hand are going to bounce into the atoms and molecules of the wood, and they're going to transfer the kinetic energy. But we realize in the wood is, I'm making less contact. Because, first of all, the surface of the wood isn't smooth, so I'm making less contact. So this one over here might just bump into another air particle, it actually won't bump into a wood particle. But some of the wood particles will start to take some of the kinetic energy away from me. And I will sense that as being a little bit cool, so maybe that takes a little kinetic energy, that bumps into this guy. So the kinetic energy does get transferred down. But it's going to be transferred down a lot slower than what would happen in the metal. Because, one, I don't have as much surface contact between my hand and the wood, because of these gaps. I also have air pockets in the wood, like this."
    },
    {
        "Q": "At 1:00 how would the joules cancel out if we multiply them? We are supposed to divide it right? And where did the eV unit pop up from? If the unit of eV is (Joule)^2, then why is it written in the video that 1eV=1.6*10^-19 J?? I am not able to understand this part!",
        "A": "I don t think you have followed the units correctly. When you multiply by 1/x that s the same as dividing by x. So when he does: y J * 1 eV / z J, both J cancel out leaving just eV. He is converting an energy value from J to eV. 1 eV = 1.6 x 10^-19 J is the conversion factor between these.",
        "video_name": "nJ-PtF14EFw",
        "timestamps": [
            60
        ],
        "3min_transcript": "- [Voiceover] So in that last video, I showed you how to get this equation using a lot of Physics, and so it's actually not necessary to watch the previous video, you can just start with this video if you want. And E one, we said, was the energy associated with an electron, and the lowest energy level of hydrogen. And we're using the Bohr model. And we calculated the value for that energy to be equal to negative 2.17 times 10 to the negative 18 joules. And let's go ahead and convert that into electron volts, it just makes the numbers easier to work with. So one electron volt is equal to 1.6 times 10 to the negative 19 joules. So if I take negative 2.17 times 10 to the negative 18 joules, I know that for every one electron volts, right, one electron volt is equal to 1.6 times 10 to the negative 19 joules, and so I have a conversion factor here. And so, if I multiply these two together, the joules would cancel and give me electron volts as my units. you get negative 13.6 electron volts. So once again, that's the energy associated with an electron, the lowest energy level in hydrogen. And so I plug that back into my equation here, and so I can just rewrite it, so this means the energy at any energy level N is equal to E one, which is negative 13.6 electron volts, and we divide that by N squared, where N is an integer, so one, two, three, and so on. So the energy for the first energy level, right, we already know what it is, but let's go ahead and do it so you can see how to use this equation, is equal to negative 13.6 divided by, so we're saying the energy where N is equal to one, so whatever number you have here, you're gonna plug in here. So this would just be one squared, alright? Which is of course just one, and so this is negative 13.6 electron volts, so we already knew that one. so E two, this would just be negative 13.6, and now N is equal to two, so this would be two squared, and when you do that math you get negative 3.4 electron volts. And then let's do one more. So the energy for the third energy level is equal to negative 13.6, now N is equal to three, so this would be three squared, and this gives you negative 1.51 electron volts. So, we have the energies for three different energy levels. The energy for the first energy level is equal to negative 13.6. E two is equal to negative 3.4, and E three is equal to negative 1.51 electron volts. So energy is quantized using the Bohr models, you can't have a value of energy in between those energies. And also note that your energies are negative,"
    },
    {
        "Q": "Why can't we have a value of energy in between the integers which you mentioned at 2:59?",
        "A": "Because that s the way nature works. Energy is quantized. It was a surprising discovery, but it is true. Why it is that way, we don t know.",
        "video_name": "nJ-PtF14EFw",
        "timestamps": [
            179
        ],
        "3min_transcript": "you get negative 13.6 electron volts. So once again, that's the energy associated with an electron, the lowest energy level in hydrogen. And so I plug that back into my equation here, and so I can just rewrite it, so this means the energy at any energy level N is equal to E one, which is negative 13.6 electron volts, and we divide that by N squared, where N is an integer, so one, two, three, and so on. So the energy for the first energy level, right, we already know what it is, but let's go ahead and do it so you can see how to use this equation, is equal to negative 13.6 divided by, so we're saying the energy where N is equal to one, so whatever number you have here, you're gonna plug in here. So this would just be one squared, alright? Which is of course just one, and so this is negative 13.6 electron volts, so we already knew that one. so E two, this would just be negative 13.6, and now N is equal to two, so this would be two squared, and when you do that math you get negative 3.4 electron volts. And then let's do one more. So the energy for the third energy level is equal to negative 13.6, now N is equal to three, so this would be three squared, and this gives you negative 1.51 electron volts. So, we have the energies for three different energy levels. The energy for the first energy level is equal to negative 13.6. E two is equal to negative 3.4, and E three is equal to negative 1.51 electron volts. So energy is quantized using the Bohr models, you can't have a value of energy in between those energies. And also note that your energies are negative, because this is the one that's closest to zero, so E three is the highest energy level out of the three that we're talking about here. Alright, let's talk about the Bohr model of the hydrogen atom really fast. And so, over here on the left, alright, just to remind you, I already showed you how to get these different radii for the Bohr model, so this isn't drawn perfectly to scale. But if we assume that we have a positively charged nucleus, which I just marked in red here, so there's our positively charged nucleus. We know the electron orbits the nucleus in the Bohr model. So I'm gonna draw an electron here, so again, not drawn to scale, orbiting the nucleus. So the positively charged nucleus attracts the negatively charged electron. And I'm saying that electron is orbiting at R one, so that's this first radius right here. So R one is when N is equal to one,"
    },
    {
        "Q": "at 2:24, what is subscripts",
        "A": "It literally means written below, in this case it s the smaller letter. At 2:24, he means a smaller letter to distinguish different things. For example, if I m doing a physics problem and I have 2 velocities, of a car and of a train, I could use subscripts to tell them apart.",
        "video_name": "CFygKiTB-4A",
        "timestamps": [
            144
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 5:55, Sal did F*d, with F being 10 kg * 9.8. Why isn't F=0, because as he said, there is no net force because the elevator is going at a constant velocity?",
        "A": "The way it is going at constant velocity is it must have an upward force on it equal to its weight. Otherwise it would accelerate downward (fall)",
        "video_name": "3mier94pbnU",
        "timestamps": [
            355
        ],
        "3min_transcript": "slightly larger than the acceleration of gravity just to get the object moving. But let's say that the object is already at constant velocity. Let's say it's an elevator. And it is just going up with a constant velocity. And let's say the mass of the elevator is-- I don't know-- 10 kilograms. And it moves up with a constant velocity. It moves up 100 meters. So we know that the work done by whatever was pulling on this elevator, it probably was the tension in this wire that was pulling up on the elevator, but we know that the work done is the force necessary to pull up on it. Well that's just going to be the force of gravity. So we're assuming that the elevator's not accelerating, right? Because if the elevator was accelerating upwards, then the force applied to it would be more than the force of gravity. And if the elevator was accelerating downwards, or if would be less than the acceleration of gravity. But since the elevator is at a constant velocity moving up, we know that the force pulling upwards is completely equal to the force pulling downwards, right? No net force. Because gravity and this force are at the same level, so there's no change in velocity. I think I said that two times. So we know that this upward force is equal to the force of gravity. At least in magnitude in the opposite direction. So this is mg. So what's m? m is 10 kilograms. Times the acceleration of gravity. Let's say that's 9.8 meters per second squared. I'm not writing the units here, but we're all assuming kilograms and meters per second squared. And we're moving that for a distance of 100 meters. So how much work was put into this elevator, or into this object-- it doesn't have to be an elevator-- by whatever force that was essentially pushing up on it or pulling up on it? This would be 98 times 100. So it's 9,800 Newton meters or 9,800 joules. After we've moved up 100 meters, notice there's no change in velocity. So the question is, where did all that work get put into the object? And the answer here is, is that the work got transferred to something called potential energy. And potential energy is defined as-- well, gravitational potential energy. We'll work with other types of potential energy later with springs and things. Potential energy is defined as mass times the force of gravity times the height that the object is at. And why is this called potential energy? Because at this point, the energy-- work had to be put into the object to get it to this-- in the case of gravitational potential energy, work had to be put into the object to get it to this height. But the object now, it's not moving or anything, so it"
    },
    {
        "Q": "A little before 3:50, sal talks about pushing upwards with the acceleration of gravity while gravity is pulling downwards with the same acceleration, would the block even move then? Because if the block moves then there isnt any work done.",
        "A": "There is no net work done on the block and therefore we should not expect the kinetic energy of the block to increase, according to the work-kinetic energy theorem. But there is work done on the earth-block system, and therefore the potential energy of that system increases.",
        "video_name": "3mier94pbnU",
        "timestamps": [
            230
        ],
        "3min_transcript": "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. the acceleration of gravity. Mass times-- let's just call that gravity, right? 9.8 meters per second squared. And I were to apply this force for a distance of d upwards. Right? Or instead of d, let's say h. H for height. So in this case, the force times the distance is equal to-- well the force is mass times the acceleration of gravity, right? And remember, I'm pushing with the acceleration of gravity upwards, while the acceleration of gravity is pulling downwards. So the force is mass times gravity, and I'm applying that for a distance of h, right? d is h. So the force is this. This is the force. And then the distance I'm applying is going to be h. And what's interesting is-- I mean if you want to think of an exact situation, imagine an elevator that is already slightly larger than the acceleration of gravity just to get the object moving. But let's say that the object is already at constant velocity. Let's say it's an elevator. And it is just going up with a constant velocity. And let's say the mass of the elevator is-- I don't know-- 10 kilograms. And it moves up with a constant velocity. It moves up 100 meters. So we know that the work done by whatever was pulling on this elevator, it probably was the tension in this wire that was pulling up on the elevator, but we know that the work done is the force necessary to pull up on it. Well that's just going to be the force of gravity. So we're assuming that the elevator's not accelerating, right? Because if the elevator was accelerating upwards, then the force applied to it would be more than the force of gravity. And if the elevator was accelerating downwards, or if"
    },
    {
        "Q": "At 5:15, Sal said that there's no net force, so Fnet =0. But why he take 98 N as a force?",
        "A": "For Fnet to be zero the upward force must be equal to the weight (which is 98N). He is finding work done by this upward force.",
        "video_name": "3mier94pbnU",
        "timestamps": [
            315
        ],
        "3min_transcript": "the acceleration of gravity. Mass times-- let's just call that gravity, right? 9.8 meters per second squared. And I were to apply this force for a distance of d upwards. Right? Or instead of d, let's say h. H for height. So in this case, the force times the distance is equal to-- well the force is mass times the acceleration of gravity, right? And remember, I'm pushing with the acceleration of gravity upwards, while the acceleration of gravity is pulling downwards. So the force is mass times gravity, and I'm applying that for a distance of h, right? d is h. So the force is this. This is the force. And then the distance I'm applying is going to be h. And what's interesting is-- I mean if you want to think of an exact situation, imagine an elevator that is already slightly larger than the acceleration of gravity just to get the object moving. But let's say that the object is already at constant velocity. Let's say it's an elevator. And it is just going up with a constant velocity. And let's say the mass of the elevator is-- I don't know-- 10 kilograms. And it moves up with a constant velocity. It moves up 100 meters. So we know that the work done by whatever was pulling on this elevator, it probably was the tension in this wire that was pulling up on the elevator, but we know that the work done is the force necessary to pull up on it. Well that's just going to be the force of gravity. So we're assuming that the elevator's not accelerating, right? Because if the elevator was accelerating upwards, then the force applied to it would be more than the force of gravity. And if the elevator was accelerating downwards, or if would be less than the acceleration of gravity. But since the elevator is at a constant velocity moving up, we know that the force pulling upwards is completely equal to the force pulling downwards, right? No net force. Because gravity and this force are at the same level, so there's no change in velocity. I think I said that two times. So we know that this upward force is equal to the force of gravity. At least in magnitude in the opposite direction. So this is mg. So what's m? m is 10 kilograms. Times the acceleration of gravity. Let's say that's 9.8 meters per second squared. I'm not writing the units here, but we're all assuming kilograms and meters per second squared. And we're moving that for a distance of 100 meters. So how much work was put into this elevator, or into this object-- it doesn't have to be an elevator-- by whatever force that was essentially pushing up on it or pulling up on it?"
    },
    {
        "Q": "At 9:30, does the potential energy when it hits the ground get converted to kinetic energy? And would it have velocity 100 when it hits the ground?",
        "A": "There is no more PE when you reach the ground. It s all KE at that point. And the KE then gets converted to thermal energy",
        "video_name": "3mier94pbnU",
        "timestamps": [
            570
        ],
        "3min_transcript": "But it now has a lot of potential to do work. And what do I mean by potential to do work? Well after I move an object up 100 meters into the air, what's its potential to do work? Well, I could just let go of it and have no outside force other than gravity. The gravity will still be there. And because of gravity, the object will come down and be at a very, very fast velocity when it lands. And maybe I could apply this to some machine or something, and this thing could do work. And if that's a little confusing, let me give you an example. It all works together with our-- So let's say I have an object that is-- oh, I don't know-- a 1 kilogram object and we're on earth. And let's say that is 10 meters above the ground. So we know that its potential energy is equal to mass times So mass is 1. Let's just say gravitational acceleration is 10 meters per second squared. Times 10 meters per second squared. Times 10 meters, which is the height. So it's approximately equal to 100 Newton meters, which is the same thing is 100 joules. And what do we know about this? We know that it would take about 100-- or exactly-- 100 joules of work to get this object from the ground to this point up here. Now what we can do now is use our traditional kinematics formulas to figure out, well, if I just let this object go, how fast will it be when it hits the ground? And we could do that, but what I'll show you is even a faster way. And this is where all of the work and energy really becomes useful. We have something called the law of conservation of energy. It's that energy cannot be created or destroyed, it just And there's some minor caveats to that. But for our purposes, we'll just stick with that. So in the situation where I just take the object and I let go up here, up here it has a ton of potential energy. And by the time it's down here, it has no potential energy because the height becomes 0, right? So here, potential energy is equal to 100 and here, potential energy is equal to 0. And so the natural question is-- I just told you the law of conservation of energy, but if you look at this example, all the potential energy just disappeared. And it looks like I'm running out of time, but what I'll show you in the next video is that that potential energy gets converted into another type of energy. And I think you might be able to guess what type that is because this object is going to be moving really fast right before it hits the ground. I'll see you in the next video."
    },
    {
        "Q": "at 5:39 you said acceleration of gravity is 9.8 meter per second squared why is it not zero when the object is descending with constant velocity. It's not accelerating",
        "A": "If the object descends with constant velocity, it is not accelerating. But that s not how objects fall. They accelerate when they fall. That s why you can jump off a short step but you can t jump off a tall building.",
        "video_name": "3mier94pbnU",
        "timestamps": [
            339
        ],
        "3min_transcript": "slightly larger than the acceleration of gravity just to get the object moving. But let's say that the object is already at constant velocity. Let's say it's an elevator. And it is just going up with a constant velocity. And let's say the mass of the elevator is-- I don't know-- 10 kilograms. And it moves up with a constant velocity. It moves up 100 meters. So we know that the work done by whatever was pulling on this elevator, it probably was the tension in this wire that was pulling up on the elevator, but we know that the work done is the force necessary to pull up on it. Well that's just going to be the force of gravity. So we're assuming that the elevator's not accelerating, right? Because if the elevator was accelerating upwards, then the force applied to it would be more than the force of gravity. And if the elevator was accelerating downwards, or if would be less than the acceleration of gravity. But since the elevator is at a constant velocity moving up, we know that the force pulling upwards is completely equal to the force pulling downwards, right? No net force. Because gravity and this force are at the same level, so there's no change in velocity. I think I said that two times. So we know that this upward force is equal to the force of gravity. At least in magnitude in the opposite direction. So this is mg. So what's m? m is 10 kilograms. Times the acceleration of gravity. Let's say that's 9.8 meters per second squared. I'm not writing the units here, but we're all assuming kilograms and meters per second squared. And we're moving that for a distance of 100 meters. So how much work was put into this elevator, or into this object-- it doesn't have to be an elevator-- by whatever force that was essentially pushing up on it or pulling up on it? This would be 98 times 100. So it's 9,800 Newton meters or 9,800 joules. After we've moved up 100 meters, notice there's no change in velocity. So the question is, where did all that work get put into the object? And the answer here is, is that the work got transferred to something called potential energy. And potential energy is defined as-- well, gravitational potential energy. We'll work with other types of potential energy later with springs and things. Potential energy is defined as mass times the force of gravity times the height that the object is at. And why is this called potential energy? Because at this point, the energy-- work had to be put into the object to get it to this-- in the case of gravitational potential energy, work had to be put into the object to get it to this height. But the object now, it's not moving or anything, so it"
    },
    {
        "Q": "At 3:02, why exactly are we pushing the object up with a force mg? Wouldn't it cancel with the gravity?",
        "A": "If so happens then the body will stay at equilibrium rather we are pushing it with a bit greater force to overcome gravitational pull.",
        "video_name": "3mier94pbnU",
        "timestamps": [
            182
        ],
        "3min_transcript": "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. the acceleration of gravity. Mass times-- let's just call that gravity, right? 9.8 meters per second squared. And I were to apply this force for a distance of d upwards. Right? Or instead of d, let's say h. H for height. So in this case, the force times the distance is equal to-- well the force is mass times the acceleration of gravity, right? And remember, I'm pushing with the acceleration of gravity upwards, while the acceleration of gravity is pulling downwards. So the force is mass times gravity, and I'm applying that for a distance of h, right? d is h. So the force is this. This is the force. And then the distance I'm applying is going to be h. And what's interesting is-- I mean if you want to think of an exact situation, imagine an elevator that is already"
    },
    {
        "Q": "why is gravity negative at 5:52?",
        "A": "he said that the tension in the string is equal to mg..!! the net force in the vertical direction is equals zero..!! refer the video once more, carefully..!! gravity is not -ve !!",
        "video_name": "3mier94pbnU",
        "timestamps": [
            352
        ],
        "3min_transcript": "slightly larger than the acceleration of gravity just to get the object moving. But let's say that the object is already at constant velocity. Let's say it's an elevator. And it is just going up with a constant velocity. And let's say the mass of the elevator is-- I don't know-- 10 kilograms. And it moves up with a constant velocity. It moves up 100 meters. So we know that the work done by whatever was pulling on this elevator, it probably was the tension in this wire that was pulling up on the elevator, but we know that the work done is the force necessary to pull up on it. Well that's just going to be the force of gravity. So we're assuming that the elevator's not accelerating, right? Because if the elevator was accelerating upwards, then the force applied to it would be more than the force of gravity. And if the elevator was accelerating downwards, or if would be less than the acceleration of gravity. But since the elevator is at a constant velocity moving up, we know that the force pulling upwards is completely equal to the force pulling downwards, right? No net force. Because gravity and this force are at the same level, so there's no change in velocity. I think I said that two times. So we know that this upward force is equal to the force of gravity. At least in magnitude in the opposite direction. So this is mg. So what's m? m is 10 kilograms. Times the acceleration of gravity. Let's say that's 9.8 meters per second squared. I'm not writing the units here, but we're all assuming kilograms and meters per second squared. And we're moving that for a distance of 100 meters. So how much work was put into this elevator, or into this object-- it doesn't have to be an elevator-- by whatever force that was essentially pushing up on it or pulling up on it? This would be 98 times 100. So it's 9,800 Newton meters or 9,800 joules. After we've moved up 100 meters, notice there's no change in velocity. So the question is, where did all that work get put into the object? And the answer here is, is that the work got transferred to something called potential energy. And potential energy is defined as-- well, gravitational potential energy. We'll work with other types of potential energy later with springs and things. Potential energy is defined as mass times the force of gravity times the height that the object is at. And why is this called potential energy? Because at this point, the energy-- work had to be put into the object to get it to this-- in the case of gravitational potential energy, work had to be put into the object to get it to this height. But the object now, it's not moving or anything, so it"
    },
    {
        "Q": "If potential energy is the ability to do work (7:00), what work does the object do when it falls? Isn't this work (falling of the object) done by gravity?",
        "A": "Remember, gravity is an acceleration (9.81m/s^2); it s not a force. Therefore, 10kg of an object has potential to do work WITH GRAVITY, if it s located at proper position.",
        "video_name": "3mier94pbnU",
        "timestamps": [
            420
        ],
        "3min_transcript": "would be less than the acceleration of gravity. But since the elevator is at a constant velocity moving up, we know that the force pulling upwards is completely equal to the force pulling downwards, right? No net force. Because gravity and this force are at the same level, so there's no change in velocity. I think I said that two times. So we know that this upward force is equal to the force of gravity. At least in magnitude in the opposite direction. So this is mg. So what's m? m is 10 kilograms. Times the acceleration of gravity. Let's say that's 9.8 meters per second squared. I'm not writing the units here, but we're all assuming kilograms and meters per second squared. And we're moving that for a distance of 100 meters. So how much work was put into this elevator, or into this object-- it doesn't have to be an elevator-- by whatever force that was essentially pushing up on it or pulling up on it? This would be 98 times 100. So it's 9,800 Newton meters or 9,800 joules. After we've moved up 100 meters, notice there's no change in velocity. So the question is, where did all that work get put into the object? And the answer here is, is that the work got transferred to something called potential energy. And potential energy is defined as-- well, gravitational potential energy. We'll work with other types of potential energy later with springs and things. Potential energy is defined as mass times the force of gravity times the height that the object is at. And why is this called potential energy? Because at this point, the energy-- work had to be put into the object to get it to this-- in the case of gravitational potential energy, work had to be put into the object to get it to this height. But the object now, it's not moving or anything, so it But it now has a lot of potential to do work. And what do I mean by potential to do work? Well after I move an object up 100 meters into the air, what's its potential to do work? Well, I could just let go of it and have no outside force other than gravity. The gravity will still be there. And because of gravity, the object will come down and be at a very, very fast velocity when it lands. And maybe I could apply this to some machine or something, and this thing could do work. And if that's a little confusing, let me give you an example. It all works together with our-- So let's say I have an object that is-- oh, I don't know-- a 1 kilogram object and we're on earth. And let's say that is 10 meters above the ground. So we know that its potential energy is equal to mass times"
    },
    {
        "Q": "If you times 1/2 x 5 x 49 this is equal to 122.5 which is also 1.23x10 to the 23rd power... How does he get that answer at 1:41 - 1:58",
        "A": "Are you referring to his answer of 125 joules? He s using the approximation of 1/2 * 5 * 50 there. Also, 1.23 * 10 to the 23rd power is a huge number, I don t think that is what you meant to type.",
        "video_name": "3mier94pbnU",
        "timestamps": [
            101,
            118
        ],
        "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 5:40, why is force = 10*9.8? If force = m*a then shouldn't acceleration be -9.8 not 9.8, since it is downwards",
        "A": "You can define the signs of your direction however you want; it s arbitrary. If you want to make down be positive, you can. You just have to be consistent throughout the problem.",
        "video_name": "3mier94pbnU",
        "timestamps": [
            340
        ],
        "3min_transcript": "slightly larger than the acceleration of gravity just to get the object moving. But let's say that the object is already at constant velocity. Let's say it's an elevator. And it is just going up with a constant velocity. And let's say the mass of the elevator is-- I don't know-- 10 kilograms. And it moves up with a constant velocity. It moves up 100 meters. So we know that the work done by whatever was pulling on this elevator, it probably was the tension in this wire that was pulling up on the elevator, but we know that the work done is the force necessary to pull up on it. Well that's just going to be the force of gravity. So we're assuming that the elevator's not accelerating, right? Because if the elevator was accelerating upwards, then the force applied to it would be more than the force of gravity. And if the elevator was accelerating downwards, or if would be less than the acceleration of gravity. But since the elevator is at a constant velocity moving up, we know that the force pulling upwards is completely equal to the force pulling downwards, right? No net force. Because gravity and this force are at the same level, so there's no change in velocity. I think I said that two times. So we know that this upward force is equal to the force of gravity. At least in magnitude in the opposite direction. So this is mg. So what's m? m is 10 kilograms. Times the acceleration of gravity. Let's say that's 9.8 meters per second squared. I'm not writing the units here, but we're all assuming kilograms and meters per second squared. And we're moving that for a distance of 100 meters. So how much work was put into this elevator, or into this object-- it doesn't have to be an elevator-- by whatever force that was essentially pushing up on it or pulling up on it? This would be 98 times 100. So it's 9,800 Newton meters or 9,800 joules. After we've moved up 100 meters, notice there's no change in velocity. So the question is, where did all that work get put into the object? And the answer here is, is that the work got transferred to something called potential energy. And potential energy is defined as-- well, gravitational potential energy. We'll work with other types of potential energy later with springs and things. Potential energy is defined as mass times the force of gravity times the height that the object is at. And why is this called potential energy? Because at this point, the energy-- work had to be put into the object to get it to this-- in the case of gravitational potential energy, work had to be put into the object to get it to this height. But the object now, it's not moving or anything, so it"
    },
    {
        "Q": "At about 3:12 Sal says he's applying a force of mg upwards. Is this a net force? I fi t were just a force wouldn't it be counterbalanced by the force of gravity?",
        "A": "If he applies mg upward, and gravity pulls downward with a force of mg, then the net force is zero and the object will maintain its velocity.",
        "video_name": "3mier94pbnU",
        "timestamps": [
            192
        ],
        "3min_transcript": "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. the acceleration of gravity. Mass times-- let's just call that gravity, right? 9.8 meters per second squared. And I were to apply this force for a distance of d upwards. Right? Or instead of d, let's say h. H for height. So in this case, the force times the distance is equal to-- well the force is mass times the acceleration of gravity, right? And remember, I'm pushing with the acceleration of gravity upwards, while the acceleration of gravity is pulling downwards. So the force is mass times gravity, and I'm applying that for a distance of h, right? d is h. So the force is this. This is the force. And then the distance I'm applying is going to be h. And what's interesting is-- I mean if you want to think of an exact situation, imagine an elevator that is already"
    },
    {
        "Q": "At 7:38, Sal explains why the distance between us and the object emitting the photon has expanded due to the universe. Wouldn't the photon emitted from the object still be moving? It wouldn't have stayed in the same spot, right?",
        "A": "Right. It s own motion is bringing it closer, and that is offset to some degree by the expansion of space",
        "video_name": "6nVysrZQnOQ",
        "timestamps": [
            458
        ],
        "3min_transcript": "But I really just want to give you the idea of what's going on here. So let's just say, well, that photon says in about 10 million years, in roughly 10 million years, I should be right about at that coordinate. I should be about one third of the distance. But what happens over the course of those 10 million years? Well, over the course of those 10 million years, the universe has expanded some. The universe has expanded maybe a good deal. So let me draw the expanded universe. So after 10 million years, the universe might look like this. Actually it might even be bigger than that. Let me draw it like this. After 10 million years, the universe might have expanded a good bit. So this is 10 million years into the future. Still on a cosmological time scale, still almost at kind of the infancy of the universe because we're talking about 13.7 billion years. 10 million years go by. The universe has expanded. This coordinate, where we're sitting today at the present time, is now all the way over here. That coordinate where the photon was originally emitted is now going to be sitting right over here. And that photon, it said, OK, after 10 million light years, I'm going to get over there. And I'm approximating. I'm doing it in a very discrete way. But I really just want to give you the idea. So that coordinate, where the photon roughly gets in 10 million light years, is about right over here. The whole universe has expanded. All the coordinates have gotten further away from each other. Now, what just happened here? The universe has expanded. This distance that was 30 million light years now-- and I'm just making rough numbers here. I don't know the actual numbers here. Now, it is actually-- this is really just for the sake of giving you the idea of why-- well, giving you This distance now is no longer 30 million light years. Maybe it's 100 million. So this is now 100 million light years away from each other. The universe is expanding. These coordinates, the space is actually spreading out. You could imagine it's kind of a trampoline or the surface of a balloon. It's getting stretched thin. And so this coordinate where the light happens to be after 10 million years, it has been traveling for 10 million years, but it's gone a much larger distance. That distance now might be on the order of-- maybe it's on the order of 30 million light years. And the math isn't exact here. I haven't done the math to figure it out. So it's done 30 million light years. And actually I shouldn't even make it the same proportion. Because the distance it's gone and the distance it has to go, because of the stretching, it's not going to be completely linear. At least when I'm thinking about it in my head, it shouldn't be,"
    },
    {
        "Q": "in this video, at 13 minutes and 56 seconds (13:56) sal says that the distance between the two points is 46 billion light years. but isnt the universe 13.7 bilion years old, and wouldn't this imply that the distance between the two points is 13.7 light years???? i'm confused...",
        "A": "Some parts of the universe may be expanding faster than light.",
        "video_name": "6nVysrZQnOQ",
        "timestamps": [
            836
        ],
        "3min_transcript": "is a billion light years. So as you can see, the photon might getting frustrated. As it covers more and more distance, it looks back and says, wow, in only 50 million years, I've been able to cover 600 million light years. But it's frustrated because what it thought was it only had to cover 30 million light years in distance. That keeps stretching out because space itself is stretching. So the reality, just going to the original idea, this photon that is just reaching us, that's been traveling for-- let's say it's been traveling for 13.4 billion years. So it's reaching us is just now. So let me just fast forward 13.4 billion years from this point now to get to the present day. So if I draw the whole visible universe right over here, this point right over here is going to be-- where it was emitted from is right over there. We are sitting right over there. If I'm drawing the whole observable universe, the center actually should be where we are. Because we can observe an equal distance. If things aren't really strange, we can observe an equal distance in any direction. So actually maybe we should put us at the center. So if this was the entire observable universe, and the photon was emitted from here 13.4 billion years ago-- so 300,000 years after that initial Big Bang, and it's just getting to us, it is true that the photon has been traveling for 13.7 billion years. But what's kind of nutty about it is this object, since we've been expanding away from each other, this object is now, in our best estimates, this object is going to be about 46 billion light years away from us. This object is now 46 billion light years away from us. When we just use light to observe it, it looks like, just based on light years, hey, this light has been traveling 13.7 billion years to reach us. That's our only way of kind with light to kind of think about the distance. So maybe it's 13.4 or whatever-- I keep changing the decimal-- but 13.4 billion light years away. But the reality is if you had a ruler today, light year rulers, this space here has stretched so much that this is now 46 billion light years. And just to give you a hint of when we talk about the cosmic microwave background radiation, what will this point in space look like, this thing that's actually 46 billion light years away, but the photon only took 13.7 billion years to reach us? What will this look like? Well, when we say look like, it's based on the photons that are reaching us right now."
    },
    {
        "Q": "At 8:00, there's a point in space which emitted a photon 10mil. yrs ago and space is expanding between us so it takes longer to get to us. At 9:50,the photon isn\u00e2\u0080\u0099t travelling faster than the speed of light, even though it covered a greater distance.\n?\nIf space is expanding, then that photon must be travelling slower than the speed of light towards us, yet faster than the speed of light away from its origin. Ouch, my head! I realize this is getting into relativity, but can anyone expand on this",
        "A": "but the light will be red shifted because it s source is moving away.",
        "video_name": "6nVysrZQnOQ",
        "timestamps": [
            480,
            590
        ],
        "3min_transcript": "But I really just want to give you the idea of what's going on here. So let's just say, well, that photon says in about 10 million years, in roughly 10 million years, I should be right about at that coordinate. I should be about one third of the distance. But what happens over the course of those 10 million years? Well, over the course of those 10 million years, the universe has expanded some. The universe has expanded maybe a good deal. So let me draw the expanded universe. So after 10 million years, the universe might look like this. Actually it might even be bigger than that. Let me draw it like this. After 10 million years, the universe might have expanded a good bit. So this is 10 million years into the future. Still on a cosmological time scale, still almost at kind of the infancy of the universe because we're talking about 13.7 billion years. 10 million years go by. The universe has expanded. This coordinate, where we're sitting today at the present time, is now all the way over here. That coordinate where the photon was originally emitted is now going to be sitting right over here. And that photon, it said, OK, after 10 million light years, I'm going to get over there. And I'm approximating. I'm doing it in a very discrete way. But I really just want to give you the idea. So that coordinate, where the photon roughly gets in 10 million light years, is about right over here. The whole universe has expanded. All the coordinates have gotten further away from each other. Now, what just happened here? The universe has expanded. This distance that was 30 million light years now-- and I'm just making rough numbers here. I don't know the actual numbers here. Now, it is actually-- this is really just for the sake of giving you the idea of why-- well, giving you This distance now is no longer 30 million light years. Maybe it's 100 million. So this is now 100 million light years away from each other. The universe is expanding. These coordinates, the space is actually spreading out. You could imagine it's kind of a trampoline or the surface of a balloon. It's getting stretched thin. And so this coordinate where the light happens to be after 10 million years, it has been traveling for 10 million years, but it's gone a much larger distance. That distance now might be on the order of-- maybe it's on the order of 30 million light years. And the math isn't exact here. I haven't done the math to figure it out. So it's done 30 million light years. And actually I shouldn't even make it the same proportion. Because the distance it's gone and the distance it has to go, because of the stretching, it's not going to be completely linear. At least when I'm thinking about it in my head, it shouldn't be,"
    },
    {
        "Q": "At 16:09, is the \"white hot plasma\" the same thing as the \"cosmic background radiation\"?",
        "A": "The cosmic background radiation is the afterglow left over from the big bang. At first it was seen to be constant (the same everywhere). Now the COBE satellite shows that there are tiny fluctuations which are believed to be differences in density. This is what led to the forming of galaxies. But I m not sure what Sal means exactly by white hot plasma .",
        "video_name": "6nVysrZQnOQ",
        "timestamps": [
            969
        ],
        "3min_transcript": "This object is now 46 billion light years away from us. When we just use light to observe it, it looks like, just based on light years, hey, this light has been traveling 13.7 billion years to reach us. That's our only way of kind with light to kind of think about the distance. So maybe it's 13.4 or whatever-- I keep changing the decimal-- but 13.4 billion light years away. But the reality is if you had a ruler today, light year rulers, this space here has stretched so much that this is now 46 billion light years. And just to give you a hint of when we talk about the cosmic microwave background radiation, what will this point in space look like, this thing that's actually 46 billion light years away, but the photon only took 13.7 billion years to reach us? What will this look like? Well, when we say look like, it's based on the photons that are reaching us right now. So those photons are the photons being emitted from this primitive structure, from this white-hot haze of hydrogen plasma. So what we're going to see is this white-hot haze. So we're going to see this kind of white-hot plasma, white hot, undifferentiated not differentiated into proper stable atoms, much less stars and galaxies, but white hot. We're going to see this white-hot plasma. The reality today is that point in space that's 46 billion years from now, it's probably differentiated into stable atoms, and stars, and planets, and galaxies. And frankly, if that person, that person, if there is a civilization there right now and if they're sitting right there, and they're observing photons being emitted from our coordinate, from our point in space right now, they're not going to see us. They're going to see us 13.4 billion years ago. of our region of space when it really was just a white-hot plasma. And we're going to talk more about this in the next video. But think about it. Any photon that's coming from that period in time, so from any direction, that's been traveling for 13.4 billion years from any direction, it's going to come from that primitive state or it would have been emitted when the universe was in that primitive state, when it was just that white-hot plasma, this undifferentiated mass. And hopefully, that will give you a sense of where the cosmic microwave background radiation comes from."
    },
    {
        "Q": "At 9:07, Sal said there may be a black hole at the center of the galaxy, why would this be so?",
        "A": "We don t know why they are there, but it appears that most or maybe all galaxies have a super massive black hole in the center.",
        "video_name": "rcLnMe1ELPA",
        "timestamps": [
            547
        ],
        "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": "At 6:36, Sal mentions that our solar system is in the Orion Spur. Also, on a clear night, you sometimes see what looks like an arm/spur of the Milky Way. Which arm/spur is it?",
        "A": "That is the entire milky way. Every arm of it is in that same plane.",
        "video_name": "rcLnMe1ELPA",
        "timestamps": [
            396
        ],
        "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": "You state at 3:45 that our images of the galaxy are artist representations. What information do they have to draw from and what do they have to make educated guesses about in their representations? What do these artists use to generate these images?",
        "A": "The artists have to add the light and color. What our telescopes see is light waves of various frequencies. A lot of the light is not in the visible spectrum. The telescopes use electronic sensors to record the data. Someone has to decide what that data looks like. What wavelength should be what color? How bright is bright ? That s the artist.",
        "video_name": "rcLnMe1ELPA",
        "timestamps": [
            225
        ],
        "3min_transcript": "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 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."
    },
    {
        "Q": "In the picture at 12:07, there is a black space that forms a sort of horizontal line. Is that just empty space between stars? Or is it attributed to something else?",
        "A": "In these pictures, the whiter parts are usually gas clouds that reflect off the light, while the darkers areas are where dust clouds partially obscure the light of stars behind them.",
        "video_name": "rcLnMe1ELPA",
        "timestamps": [
            727
        ],
        "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 5:49, Sal mentions the Orion Spur. Where is that in the Milky Way?",
        "A": "We re in it.",
        "video_name": "rcLnMe1ELPA",
        "timestamps": [
            349
        ],
        "3min_transcript": "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 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."
    },
    {
        "Q": "At 9:03 Sal said that scientists think there is a super massive black hole in the middle. Would we one day get consumed by this black hole? Is our galaxy spinning into that black hole?",
        "A": "No, even a 4 million solar mass black hole is unable to get us at its distance. This black hole has not swallowed anything of significant size in millions of years.",
        "video_name": "rcLnMe1ELPA",
        "timestamps": [
            543
        ],
        "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": "At 7:40, he says that the boiling point for methane is around -164 degrees celsius. So, can methane be in a liquid form, if it is colder than -164 degrees celsius? And if so, is it present in a liquid form naturally anywhere?",
        "A": "Yes. Titan, Saturn s largest moon, has clouds, rain, rivers and lakes of liquid methane.",
        "video_name": "pBZ-RiT5nEE",
        "timestamps": [
            460
        ],
        "3min_transcript": "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 And if you do that, you can actually increase the boiling point of other hydrocarbons dramatically. And so even though London dispersion forces are the weakest, if you have larger molecules and you sum up all those extra forces, it can actually turn out to be rather significant when you're working with larger molecules. And so this is just a quick summary of some of the intermolecular forces to show you the application of electronegativity and how important it is."
    },
    {
        "Q": "2 questions.\n1. In the first example within the first 4:00, both carbons have the same number of substituents. So couldn't you add the proton and the OH to either one of the carbons?\n\n2. at 11:50, why do you not have to worry about sterochem? Just because you are ignoring it for the sake of teaching?",
        "A": "1. Correct. The alkene is symmetrical, so you can add the H and the OH to either end. 2. No. You don\u00c3\u00a8t have to worry about stereochemistry, because the product has no chiral centres.",
        "video_name": "dJhxphep_gY",
        "timestamps": [
            240,
            710
        ],
        "3min_transcript": "So what's left? In this acid base reaction, we took a proton away from H3O+, which leaves us H2O. So here we have H2O over here, so I'll go ahead and put lone pairs of electrons in on our water molecule. And we know that water can act as a nucleophile here. So this lone pair of electrons is going to be attracted to something that's positively charged. So nucleophilic attack on our carbocation. And this is technically at equilibrium as well, depending on the concentrations of your reactants. So let's go ahead and show that water molecule adding on to the carbon on the left. So the carbon on the right already had a hydrogen or proton added onto it, and the carbon on the left is going to have an oxygen now bonded to that carbon. Two hydrogens bonded to that oxygen, and there was a lone pair of electrons on that oxygen that did not participate in any kind of bonding. This gives this oxygen right here a plus 1 formal charge. And we're almost to our product. So we're almost there. We need one more acid base reaction to get rid of that proton on our oxygen. So water can function as a base this time. So water comes along, and this time it's going to act as a Bronsted Lowry base and accept a proton. So let's get those electrons in there. So this lone pair of electrons, let's say it takes that proton, leaving these electrons behind on my oxygen. Once again, I'll draw my equilibrium arrows here, acid base reaction. And I'm going to end up with an OH on the carbon on the left, and the carbon on the right there is a hydrogen, like that. So I added water. I ended up adding water across my double bond. And to be complete, this would regenerate my hydronium ion. I'd get H2O plus H+ would give me H3O+. And so there you go. So remember, a carbocation is present, so you have to think about Markovnikov addition. And since a carbocation is present, you have to think about possible rearrangements. So Markovnikov and rearrangements. Let's take a look at an example where you have a rearrangement here. So let's look at a reaction. So let's look at this as our starting alkene. And let's go ahead and think about the mechanisms. So we know H3O+ is going to be present. So H3O+ right here. So we're adding our alkene to a solution of water and sulfuric acid. And our first step in the mechanism, the pi electrons are going to function as a base and take a proton from our hydronium ion, leaving these electrons in here letting"
    },
    {
        "Q": "In the previous video it was said that temp is av KE by no of molecules but in this video at 3:20 it is said that temp is total KE by no of molecules. Which one is correct",
        "A": "average, he says total divded by n, which is number of molecules",
        "video_name": "HvYUKRMT0VI",
        "timestamps": [
            200
        ],
        "3min_transcript": "lot of different ways to measure temperature. We know that in Fahrenheit, what's freezing of water? It's 32 degrees Fahrenheit that's freezing, but that's also 0 degrees Celsius-- actually, that's how the Celsius scale was determined. They said, where does water freeze, and then where does water boil? 100 degrees for Celsius is boiling, and that's how they rated it. You could be colder than the freezing of water, and you'd have to go negative in that situation-- Fahrenheit, I'm actually not sure. I need to look that up in Wikipedia, or that might be something for you to do, and tell me how it came out. I think the boiling of water in Fahrenheit is 212 degrees, so it's a little arbitrary. I think Fahrenheit might be somehow related to human body temperature, but I'm just guessing. You can have different scales in this situation, and they were all kind of a bit arbitrary when they were designed. They were just to have some type of relative to scale-- hotter because they have a higher temperature then when things are freezing. You can't divide 100 by zero, but if something is 1 degree, is it necessarily the case that something that is 100 degrees Celsius is a hundred times hotter, or has a hundred times the kinetic energy? Actually, what we'll see is that no, it's actually not the case-- you don't have 100 times the kinetic energy, so this is a bit of an arbitrary scale. The actual interval is arbitrary-- you could pick the 1 degree as being one hundredth of the distance between zero and 100, but where you start-- at least in the Celsius scale-- is a bit arbitrary. They picked the freezing of water. Later on, people figured out that there is an absolute point to start at. And that absolute point to start at is the temperature at which a molecule or an atom has absolutely no kinetic energy. energy of the system, or the total kinetic energy of the system divided by the number of molecules. Or we could also say the average kinetic energy per molecule. The only way to really say that the temperature is zero-- and this is proportional, because the temperature scales are still a little bit arbitrary-- the only way to get to a temperature of zero should be when the kinetic energy of each and every molecule is zero, or the So they're not moving, they're not vibrating, they're not even blinking-- these molecules are stationary. The point at which that occurs is called absolute zero."
    },
    {
        "Q": "I've noticed this in a few videos but at 8:50 he draws a 'double head' arrow to show the movement of a electron from the Bromide ion to the carbocation but only one electron is moving so shouldn't it be a 'fish hook' arrow?",
        "A": "Yes, that s the convention. Most chemists consider this reaction to be a two-electron movement: Both \u00cf\u0080 electrons move to the H, and both of the electrons in the H-Br bond move on to the Br.",
        "video_name": "Z_GWBW_GVGA",
        "timestamps": [
            530
        ],
        "3min_transcript": "It has six protons, so it has a positive charge. This carbon right here has a positive charge. And another way to think about it is it was completely neutral and then it lost an electron. So now it will have a positive charge right over there. So this is what we are left after that step of the reaction. Oh, and of course, we can't forget. Bromine over here was neutral. It had seven valence electrons, and that's when bromine is neutral. But now it has eight, so now this will have a negative charge. This will have a negative charge, because it gained an electron. And in general, your total charge-- over here our total charge is zero, So? Our total charge will still be zero. We have a negative and we have a positive. They would cancel out, so our total charge is still zero. So what's likely to happen for the next step of our reaction? Well, we have this positive thing here. Maybe bromine just bumped, just the right way, to let go But now you have this guy who's negative and this guy who's positive. Maybe they'll be attracted to each other. Maybe they'll just bump into each other at the exact right way. And if they bump in the exact right way, maybe this guy can swipe the electron from the bromide ion, from this negative ion right here. And you might say, hey, isn't bromine more electronegative than carbon? Well, it might be, but this guy's electron rich. It's not just a regular bromine atom. This is a bromine plus an extra electron. So he's already hogged an electron, so he's electron rich. So in this situation, he's negative. He's positive. He can give this guy an electron. So if they bump in just the right way, this electron can be swiped by this carbon right over here. And this positive carbon, just so it gives you a little terminology, and we'll go over it in more detail in future videos, is actually called a carbocation. It's a positive ion of carbon. But anyway, if this electron gets swiped by this carbon, it will then form a bond. Because remember, this was the electron that was originally in a bond with this hydrogen. It's still going to be, you could imagine, paired up with this other purple or magenta electron right over there. So if that happens, then we're going to be left with-- so the next step is going to be-- so on this end of the molecule we have C, carbon, hydrogen, hydrogen. Then we have this orange hydrogen that we stole from the hydrogen bromide. Then we have this carbon right here. It has a hydrogen. You have the rest of the chain, CH2-CH2-CH3. And now, since this guy stole an electron, a bond will form with the bromine. Let me draw it. So he's going to steal this-- let me draw it this way. So a bond will form. He's stealing this electron, so now this electron is with"
    },
    {
        "Q": "At 5:23 Sal mentions continuing mountain chains between North America and Europe. What are some examples of this?",
        "A": "The Appalachian mountains in the US, the Scottish Highlands, and the Little Atlas Mountains in Morocco were all formed as parts of an ancient Pangean mountain range.",
        "video_name": "axB6uhEx628",
        "timestamps": [
            323
        ],
        "3min_transcript": "And the last time we had a supercontinent was Pangaea, about 250 million years ago. And now it's broken up into our current day geography. Now, I won't go into all of the detail why we believe that there was a Pangaea about 250 million years ago-- or, this diagram tells us, about 225 million years ago, give or take. But I'll go into some of the interesting evidence. On a very high level, you have a lot of rock commonalities between things that would have had to combine during Pangaea. And probably the most interesting thing is the fossil evidence. There are a whole bunch of fossils. And here are examples of it, from species that were around between 200 and 300 million years ago. And their fossils are found in a very specific place. This animal right here, cynognathus-- I hope I'm pronouncing that right-- cynognathus. This animal's fossils are only found here, and in this part of Africa. So not only does South America look like it fits very nicely into Africa. But the fossil evidence also makes it look like there was a nice clean band where this animal lived and where we find the fossils. So it really makes it seem like these were connected, at least when this animal lived, maybe on the order of 250 million years ago. This species right over here, its fossils are found in this area-- let me do it in a color that has more contrast-- in this area right over here. This plant, its fossils-- now, this starts to connect to a lot of dots between a lot of cont-- its fossils are found in this entire area, across South America, Africa, Antarctica, India, and Australia. And so not only does it look like the continents fit together in a puzzle piece, not only do we get it to a configuration like this if we essentially just rewind to the movement that we're seeing now-- but the fossil evidence also This animal right here, we find fossils on this nice stripe that goes from Africa through India, all the way to Antarctica. Now, this only gives us evidence of the Southern Hemisphere of Pangaea. But there is other evidence. We find kind of continuing mountain chains between North America and Europe. We find rock evidence, where just the way we see the fossils line up nicely. We see common rock that lines up nicely between South America and Africa and other continents that were at once connected. So all the evidence, as far as we can tell now, does make us think that there at one time was a Pangaea. And, for all we know, all the continents are going to keep moving. And maybe in a few hundred million years, we'll have another supercontinent. Who knows?"
    },
    {
        "Q": "At 7:04, David says that the force of tension is pointing towards the centre. But in previous videos, David said that the force of tension is only a PULL, not a PUSH. But if the force of tension is pointing towards the centre, wouldn't that be contradictory?",
        "A": "If the object that is moved around in a circle by a string, suddently lost the string, then the object would keep moving away from the center. This means that the string is pulling the object to stay in the circle. If this makes sense?",
        "video_name": "FfNgm-w9Krw",
        "timestamps": [
            424
        ],
        "3min_transcript": "for one of the directions. Which direction should we pick? Well, which force do we want to find? We want to find this force of tension, so even though I could if I wanted to use Newton's second law for this vertical direction, the tension doesn't even point that way, so I'm not gonna bother with that direction first. I'm gonna see if I can get by doing this in one step, so I'm gonna use this horizontal direction and that's gonna be the centripetal direction, i.e., into the circle. And when we're dealing with the centripetal force, we're gonna be dealing with the centripetal acceleration, so over here, when I use a and set that equal to the net force over mass, if I'm gonna use the centripetal force, I'm gonna have to use the centripetal acceleration. In other words, I'm gonna only plug forces that go into, radially into the circle here, and I'm gonna have the radial centripetal acceleration right here. And we know the formula for centripetal acceleration, that's v squared over r, so I'm gonna plug v squared over r into the left hand side. That's the thing that's new. When we used Newton's second law for just regular forces, but now, when you're using this law for the particular direction that is the centripetal direction, you're gonna replace a with v squared over r and then I set it equal to the net force in the centripetal direction over the mass. So what am I gonna plug in up here? What forces do I put up here? I mean, I've got normal force, tension, gravity. A common misconception is that people try to put them all into here. People put the gravitational force, the normal force, the tension, why not? But remember, if we've selected the centripetal direction, centripetal just means pointing toward the center of the circle, so I'm only going to plug in forces that are directed in toward the center of the circle, and that's not the normal force or the gravitational force. These forces do not point inward toward the center of the circle. The only force in this case that points toward the center of the circle is the tension force, and like we already said, that is the centripetal force. So over here, I'd have v squared over r, the only force acting as the centripetal force is the tension. Now, should that be positive or negative? Well, we're gonna treat inward as positive, so any forces that point inward are gonna be positive. Is it possible for a centripetal force to be negative? It is. If there was some force that pointed outward, if for some reason there was another string pulling on the ball outward, we would include that force in this calculation, and we would include it with a negative sign, so forces that are directed out of the circle, we're gonna count as negative and forces that are directed into the circle, we're gonna count as positive in here. And if they're not directed into or out, we're not gonna include them in this calculation at all. Now, you might object. You might say, wait a minute. There is a force out of the circle. This ball wants to go out of the circle. There should be a force this way. This is often referred to as the centrifugal force, and that doesn't really exist. So when people say that there's an outward force trying to direct this ball out of the circle,"
    },
    {
        "Q": "At 1:37, what is the angle between the plane containing two C-H bonds in CH4, and the plane containing the other two C-H bonds in the same molecule?",
        "A": "90 degrees is the angle",
        "video_name": "ka8Yt4bTODs",
        "timestamps": [
            97
        ],
        "3min_transcript": "Let's figure out the shape of the methane molecule using VSEPR theory. So the first thing that you do is draw a dot structure to show it the valence electrons. So for methane, carbon is in group four. So 4 valence electrons. Hydrogen is in group one, and I have four of them, so 1 times 4 is 4, plus 4 is 8 valence electrons that we need to show in our dot structure. Carbon goes in the center and carbon is bonded to 4 hydrogens, so I can go ahead and put my hydrogens in there like that. And this is a very simple dot structure. We've already shown all 8 of our valence electrons. Let me go ahead and highlight those here. 2, 4, six, and 8. So carbon has an octet and we are done. The next thing we need to do is count the number of electron clouds that surround our central atom. So remember, electron clouds are regions of electron density, So we can think about these bonding electrons here as being electron clouds. So that's one electron cloud. Here's another one down here. And then finally, here's another one. So we have four electron clouds surrounding our central atom. The next step is to predict the geometry of your electron clouds around your central atom. And so VSEPR theory tells us that those valence electrons are going to repel each other since they are negatively charged. And therefore, they're going to try to get as far away from each other as they can in space. And when you have four electron clouds, the electron clouds are farthest away from each other if they point towards the course of a tetrahedron, which is a four sided figure. So let me go ahead and draw the molecule here, draw the methane molecule. I'm going to attempt to show it in a tetrahedral geometry. And then I'll actually show you what a tetrahedron looks like here. So here's a quick sketch of what the molecule sort of looks like. And let me go ahead and draw tetrahedron over here so you can get a little bit better idea of the shape, all right? So four sided figure. And so there you go. Something like that. So you could think about the corners of your tetrahedron as being approximately where your hydrogens are 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."
    },
    {
        "Q": "At 4:15, why is the y-component 120 sin 30 degrees ?",
        "A": "The y- and x-component of the velocity is computed with trig. Go back to the 1-dimensional videos for a full explanation of how to use trig to break a vector into it s vertical and horizontal components. Sal does a good job of explaining how to break a vector into it s components, and why it is useful to do so in those earlier videos.",
        "video_name": "jl_gQ-eL3xo",
        "timestamps": [
            255
        ],
        "3min_transcript": "Its' a 30-degree angle to the horizontal. And there's a fence 350 feet away that's 30 feet high. It's roughly around there. That's 30. And what we need to do is figure out whether the ball can clear the fence. We figured out the last time when we used the unit vector notation that it doesn't clear the fence. But in this problem, or the second part of this problem, they said that there's a 5 meter per second wind gust to the right. So there's a wind gust of 5 meters per second right when I hit the ball. And you could go into the complications of how much does that accelerate the ball? Or what's the air resistance of the ball? I think for the simplicity of the problem, they're just saying that the x-component of the ball's velocity right after you hit it increases by 5 meters per second. I think that's their point. that we did it the last time, but we'll use a different notation. So we can write that equation that I had written before, that the position at any given time as a function of t is equal to the initial position-- that's an i right there-- plus the initial velocity. These are all vectors. Initial velocity times t plus the acceleration vector over 2t squared. So what's the initial position? And now we're going to use some of our new notation. The initial position when I hit the ball, its x-component is 0, right? It's almost like its coordinate, and they're not that different of a notation. And then the y-position is 4. Easy enough. Let me do it. So we can split it up into the x- and the y-components. The y-component is 120 sine of 30 degrees and then the x component is 120 cosine of 30 degrees. That's just the x-component after I hit it. But then they say there's this wind gust so it's going to be plus 5. I think that's their point when they say that there's this wind gust. They say that right when you hit it, for some reason in the x-direction, it accelerates a little bit by 5 meters per second. So the velocity vector. This notation actually is better, because it takes less space up, and you don't have all these i's and j's and pluses confusing everything. So the initial velocity vector, what's its x-component? It's 120 cosine of 30. Cosine of 30 is square root of 3/2 times 120 is 60 square roots of 3, and then you add 5 to it. Let me just solve it right now."
    },
    {
        "Q": "At 1:20, I believe there is a mistake. RBCs do not contain a nucleus.",
        "A": "That is not a RBC and you re correct, they shed their nuclei before they enter circulation. What he is drawing is endothelial cells lining the blood vessel. Hope this helped! :)",
        "video_name": "RQpBj8ebbNY",
        "timestamps": [
            80
        ],
        "3min_transcript": "Let's look at a blood vessel. A blood vessel is kind of like a tube. I think you'll agree with me. It's a tube through which blood travels. So here I'm drawing a tube, and I'm gonna ask you a question which is what makes up the walls of this tube? Now, for a blood vessel, what makes it up, is something called an endothelial cell. So, actually, the walls are made up of these kind of gooey endothelial cells that are stuck together tightly and that together form a tube through which the blood will travel. So here I'm drawing a bunch of cells tightly stuck together. They're stuck together tightly to prevent blood from coming out, of course. So this is more or less what it looks like. Each of these is a cell, and so, of course, each one has a nucleus, which I'll just quickly draw like that. Now let's delete that and let's change views So now here is the same blood vessel in cross section. And so these are the endothelial cells, which we're seeing in cross section, and maybe here we can draw the nucleus of this one. Maybe it's visible right there. So now we have blood moving through this blood vessel. Here are some red blood cells, and they're moving along, providing the body with oxygen. But a very important question to ask is what happens if this blood vessel gets damaged? So let's say that those two cells right there split open, and they break open. What's gonna happen if we don\u2019t fix this is that all of our blood is just going to rush out here, and we're gonna lose blood. So what is your body gonna do about this? Well, it's going to use a special player that we haven't talked about yet, which is the platelet. So I'm drawing a platelet here. It doesn't have a nucleus or anything. It's a tiny, little piece of a cell that your body uses to block up holes like this. So you have them floating around in your blood all the time. Here I'm drawing a bunch. And what happens is when you have a hole in your blood vessel, they're going to come together. They're going to stick together, and they're going to clog up this hole. And so, basically, they've built a little barrier so that we won't keep losing all our blood. So are you satisfied? Well, you shouldn\u2019t really be satisfied because there's a bit question here which is why are these platelets clumping here, and why aren't they clumping, you know, for example up here? Why don\u2019t they clump there or maybe even just in circulation? Why don\u2019t they clump up like this? Because if what they do is to clump up, how would they know to clump up here and not here? What's telling them? These are things that we don\u2019t want to happen"
    },
    {
        "Q": "AT 14:13\nIt is said, that the wavelength of that standing wave can be found using the formula\nWavelength (lambda) = 2* (Length of the tube)/ n , where n=1,2,3,.....\nBut, looking at the graph carefully, we can take n= no. of nodes...\nCan'nt we?",
        "A": "If we take n as no. of nodes, we just have to be careful! It works for this case, but not for all of them, ex. if you use n as number of nodes for standing waves on strings, it will not work out :)",
        "video_name": "BhQUW9s-R8M",
        "timestamps": [
            853
        ],
        "3min_transcript": "And so, how much of a wavelength is this? Let's try it out. Let's reference our one wavelength. So it starts at the bottom, and then it goes all the way up to the top, and then it goes all the way down to the bottom, but this one keeps going. This is more than a whole wavelength. Because that's just this part. That's one whole wavelength. Now I got to go all the way back up to the top. So this wave is actually one wavelength and a half. This amount is one extra half of a wavelength, so this was one wavelength and a half. So in this case, L, the total distance of the tube, that's not changing here. The total distance of the tube is L. This time, the wavelength in there is fitting, and one and 1/2 wavelengths fit in there. That's 3/2 of a wavelength. That means the lambda equals two L over three. So in this case, for lambda three, this is going to be called the third harmonic. This should be two L over three. And so, it keeps going. You can have the fourth harmonic, fifth harmonic, every time you add one more node in here, it's always got to be antinode on one end, antinode at the other. These are the possible wavelength, and if you wanted the possible, all of the possible ones, you can probably see the pattern here. Look: two L, and then just L, then two L over three, the next one turns out to be two L over four, and then two L over five, two L over six, and so, if you wanted to just write them all down, shoot ... lambda n equals -- this is all the possible wavelengths -- two L over n, where n equals one, two, three, four, and so on. And so, look at, if I had 'n equals one' in here, I'd have two L. That's the fundamental. You plug in n equals one, you get the fundamental. If I plug in n equals two, I get two L over two. That's just L. That's my second harmonic, because I'm plugging in n equals two. that's my third harmonic. This is telling me all the possible wavelengths that I'm getting for this standing wave. So that's open, open. In the next video, I'm going to show you how to handle open, closed tubes."
    },
    {
        "Q": "Aren't you supposed to write solute -> lowers FREEZING point at 3:17?",
        "A": "Its true, check the clarifications section.",
        "video_name": "z9LxdqYntlU",
        "timestamps": [
            197
        ],
        "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": "to be clear, when the solute is added, the boiling point is lower, and when the solute is added it creates a lower vapor pressure. What does vapor pressure have to do with the boiling point because that was mentioned at about 5:45",
        "A": "The standard boiling point of a substance is the temperature at which the vapour pressure of the substance is equal to the external pressure. The vapour pressure is decreased when a solute is added and thus, a higher temperature is required to bring the vapour pressure to be equal to the external pressure. This, we have a higher boiling point.",
        "video_name": "z9LxdqYntlU",
        "timestamps": [
            345
        ],
        "3min_transcript": "So at the surface, we said if I have a bunch of water molecules in the liquid state, we knew that although the average temperature might not be high enough for the water molecules to evaporate, that there's a distribution of kinetic energies. And some of these water molecules on the surface because the surface ones might be going fast enough to escape. And when they escape into vapor, then they create a vapor pressure above here. And if that vapor pressure is high enough, you can almost view them as linemen blocking the way for more molecules to kind of run behind them as they block all of the other ambient air pressure above them. So if there's enough of them and they have enough energy, they can start to push back or to push outward is the way I think about it, so that more guys can come in behind them. So I hope that lineman analogy doesn't completely lose you. Now, what happens if you were to introduce solute into it? 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."
    },
    {
        "Q": "at 6:04, sal says the solute lowers the vapour pressure but my question is if the solute particles occupy both the positions in surface as well as inside ,won't the solvent molecules (H2O) get dis-organised leading to more of the molecules escaping out of the system therby decreasing its boiling point ??",
        "A": "Within the solvent and at surface, the solute molecules become surrounded by layers of associated water molecules, or shells of water of solvation. Formation of these shells reduces number of solvent (water) molecules that have enough kinetic energy to escape as a vapor, decreasing vapor pressure and increasing boiling point.",
        "video_name": "z9LxdqYntlU",
        "timestamps": [
            364
        ],
        "3min_transcript": "So at the surface, we said if I have a bunch of water molecules in the liquid state, we knew that although the average temperature might not be high enough for the water molecules to evaporate, that there's a distribution of kinetic energies. And some of these water molecules on the surface because the surface ones might be going fast enough to escape. And when they escape into vapor, then they create a vapor pressure above here. And if that vapor pressure is high enough, you can almost view them as linemen blocking the way for more molecules to kind of run behind them as they block all of the other ambient air pressure above them. So if there's enough of them and they have enough energy, they can start to push back or to push outward is the way I think about it, so that more guys can come in behind them. So I hope that lineman analogy doesn't completely lose you. Now, what happens if you were to introduce solute into it? 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."
    },
    {
        "Q": "3:50 (ish) Why would the OCH3 donate the e- to the hydrogen, instead of the e- going directly to the chlorine? Doesn't going to the H before the Cl just make this unnecessarily convoluted?",
        "A": "Duh, wow. Sometimes I just miss the simplest things, I can t believe that went over my head haha thanks so much!",
        "video_name": "J0gXdEAaSiA",
        "timestamps": [
            230
        ],
        "3min_transcript": "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. methoxide base, then it can give away its electron to the rest of the molecule. It can give away this electron to the rest of the molecule. Now, carbon won't need the electron. Carbon doesn't want to have a negative charge. Maybe simultaneously that electron goes to that carbon right over there. But once again, this carbon doesn't want it. It already has four bonds, but what we see is we have this chloro here. This is a highly electronegative group. The chlorine is very electronegative, so the whole time, the chlorine was already tugging on this electron right here. Now, all of a sudden, it's all happening simultaneously, when an electron becomes available to this carbon, this carbon doesn't need this electron anymore. The chlorine already wanted it, so now the chlorine can take the electron. And just like that, in exactly one step, let's think about If all of these simultaneous reactions occurred, what are we left with? Let me redraw my two chlorobutanes. I'm going to have to change it now. So now, the chlorine has disappeared. So we clear it. The chlorine has now left. This chlorine is now up here. It had this electron right over there, which is right over there. The other electron it was paired with that was forming a bond with is now also on the chlorine. It becomes a chloride anion. Let me draw the rest of the valence electrons. One, two, three, four, five, six, and it has the seventh one right here. One, two, three, four, five, six, seven, eight, so it now"
    },
    {
        "Q": "At 6:40+ does the Chloride bond with the Sodium at any point?",
        "A": "no, remember that in solution the sodium and chloride would exist as ions anyways because of the ionic bonding",
        "video_name": "J0gXdEAaSiA",
        "timestamps": [
            400
        ],
        "3min_transcript": "This electron up here-- let me clear this part out as well. Now, this electron right here, this magenta electron, this is now given to this carbon. Let me draw it here. That magenta electron is now given to that carbon. And if we look at the other end, the one that it was paired with, that it was bonded with, it still is bonded with it, so that green electron is now still on that carbon and now they are bonded. Now, they're forming a double bond. This all of the sudden has become an alkene and now the methoxide took the hydrogen. Let me redraw the methoxide. OCH3, the oxygen had one, two, three, four, five. It had six. and then all of these six unpaired ones. Neutral oxygen has six. But now it gave one of them away to a hydrogen. It gave that green electron there to the hydrogen-- I'll make this hydrogen the same color-- to this hydrogen right over here, so now this hydrogen is now bonded with it. So what are we left with? We have a chloride anion, so this is chloride. We now have methanol. This was a strong base. Now it has become its conjugate acid. So now we have methanol, which is the same as the solvent, so it's now mixed in. And now we're left off with one, two, We still have four carbons, but now it's an alkene. We have a double bond, so we could call this but-2-ene, or sometimes called 2-butene. Let's think about what happened here. It all happened simultaneously. Both reactants were present. There was actually only one step so this is the rate-determining step. If we would try to name it, it would probably have a 2 in it someplace. And something got eliminated. The chloride, or I guess you could call it the chloro group, got eliminated. It was a leaving group in this situation. So this was eliminated, and this type of reaction where something is eliminated and both of the reactants are participating in the rate-determining step, and we only had one step here so that was the rate-determining step, is called an E2 reaction."
    },
    {
        "Q": "At 6:43, i'm not exactly sure what he means by \"faster than the speed of light\" because when you think about it, wouldn't the light travel that 1.5x10^8m with the ship and then travel the other 1.5x10^8m to be a total of 3.0x10^8m away from Sal in one second?",
        "A": "When you measure the speed of light when traveling through a vacuum it will always travel at 3 * 10^8 m/s regardless of the velocity of the source of the light when it was emitted. So if you have a rocket going 1.5 * 10^8 m/s and they fire a laser at you they see the light as traveling at 3 * 10^8 m/s and you see it traveling at 3 * 10^8 m/s as well not 4.5 * 10^8 m/s.",
        "video_name": "OIwp8m3S30c",
        "timestamps": [
            403
        ],
        "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": "7:57 i thought the OH takes place where the O was originally at , i mean where the C is before it",
        "A": "The O hasn t really moved, it s just drawn in a different position. It s still bonded to the same carbon as before.",
        "video_name": "rNJPNlgmhbk",
        "timestamps": [
            477
        ],
        "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 7:26, Sal describes the fluid part as \"Although we can't call it a liquid yet\". Aren't all fluids liquids, or is there a difference?",
        "A": "No, not all fluids are liquids, but all liquids are fluids. Gases are fluids. Sand can act like a fluid.",
        "video_name": "f2BWsPVN7c4",
        "timestamps": [
            446
        ],
        "3min_transcript": "is kind of in this magma, this deformable, somewhat fluid state, and depending on what depth you go into the mantle there are kind of different levels of fluidity. And then the core, the outer level layer of the core, the outer core is liquid, because the temperature is so high. The inner core is made up of the same things, and the temperature is even higher, but since the pressure is so high it's actually solid. So that's why the mantle, crust, and core differentiations don't tell you whether it's solid, whether it's magma, or whether it's really a liquid. It just really tells you what the makeup is. Now, to think about the makeup, and this is important for plate tectonics, because when we talk about these plates we're not talking about just the crust. We're talking about the outer, rigid layer. Let me just zoom in a little bit. Let's say we zoomed in right over there. So now we have the crust zoomed in. This right here is the crust. talking about the upper mantle. We haven't gotten too deep in the mantle right here. So that's why we call it the upper mantle. Now, right below the crust, the mantle is cool enough that it is also in real solid form. So this right here is solid mantle. And when we talk about the plates were actually talking about the outer solid layer. So that includes both the crust and the solid part of the mantle. And we call that the lithosphere. When people talk about plate tectonics, they shouldn't say crustal plates. They should call these lithospheric plates. And then below the lithosphere you have the least viscous part of the mantle, but the pressure isn't so large as what will happen when you go into the lower part of the mantle that the fluid can actually kind of move past each other, although still pretty viscous. This still a magma. So this is still kind of in its magma state. And this fluid part of the mantle, we can't quite call it a liquid yet, but over large periods of time it does have fluid properties. This, that essentially the lithosphere is kind of riding on top of, we call this the asthenosphere. So when we talk about the lithosphere and asthenosphere we're really talking about mechanical layers. The outer layer, the solid layers, the lithosphere sphere. The more fluid layer right below that is the asthenosphere. When we talk about the crust, mantle, and core, we are talking about chemical properties, what are the things actually made up of."
    },
    {
        "Q": "At around 1:30, Sal said that new land was forming. I don't get what he means by that. And also, he said that the land is pushing the two plates apart. What makes the land push the two plates apart? Is it the inside of the earth?",
        "A": "He means that when the plates get pushed apart magma comes up and cools creating more land also pushing the plates apart because of the growing land. The intense heat from the core moves the mantle moving the plates.",
        "video_name": "f2BWsPVN7c4",
        "timestamps": [
            90
        ],
        "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": "At 10:07, what does Kw stand for?",
        "A": "Kw= (1.0E-14) equilibrium constant for {H3O][OH-] Ka=(1.0E-7) [H3O] + C-base Kb=(1.0E-7) [OH-] + C-acid. These vales are important in determination of PH and POH of a solution",
        "video_name": "3Gm4nAAc3zc",
        "timestamps": [
            607
        ],
        "3min_transcript": "So we get Ka, our acidic equilibrium concentration, times Kb is equal to our hydrogen concentration times our hydroxide concentration. Remember, this is all in an aqueous solution. What do we know about this? What do we know about our hydrogen times our hydroxide concentration in an aqueous solution? For example, let me review just to make sure I'm jogging your memory properly. We could have H2O. It can autoionize into H plus. Plus OH minus. And this has an equilibrium. You just put the products. So the concentration of the hydrogen protons And you don't divide by this because it's the solvent. And we already figured out what this was. If we have just completely neutral water, this is 10 to the minus 7. And this is 10 to the minus 7. So this is equal to 10 to the minus 14. Now, these two things could change. I can add more hydrogen, I could add more hydroxide. And everything we've talked about so far, that's what we've been doing. That's what acids and bases do. They either increase this or they increase that. But the fact that this is an equilibrium constant means that, look, I don't care what you do to this. At the end of the day, this will adjust for your new reality of hydrogen protons. And this will always be a constant. As long as we're in an aqueous solution, a solution of water where water is a solvent at 25 degrees. I mean, in just pure water it's 10 to the minus 7. in an aqueous solution, the product is always going to be So that's the answer to this question. This is always going to be 10 to the minus 14. If you multiply hydrogen concentration times OH concentration. Now they won't each be 10 to the minus 7 anymore, because we're dealing with a weak acid or a weak base. So they're actually going to change these things. But when you multiply them, you're still going to get 10 to the minus 14. And let's just take the minus log of both sides of that. Let me erase all this stuff I did down here. I'll need the space. Let's say we take the minus logs of both sides of this equation. So you get the-- let me do a different color-- minus log, of course it's base 10, of Ka. Let me do it in the colors. Ka times Kb is going to be equal to"
    },
    {
        "Q": "At 5:28, how do you know that it can reach octet? Is there a proof or a way to determine? Does it have 8 valence electrons just because it is in the second period? Then what about other periods?",
        "A": "Experiments tell us this is how atoms behave",
        "video_name": "p7Fsb21B2Xg",
        "timestamps": [
            328
        ],
        "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": "At 3:59, I don't exactly understand what you mean when you say Cytosol is the \"Fluid between the organelles.\"",
        "A": "Hi, Dan! I get what you mean... So basically, suppose, that all the organelles are swimming in this swimming pool.... And instead of the water, there s something called Cytoplasm. Now, pretend that these organelles decide to leave it and get out of the pool. Now, because the organelles are out, the swimming pool changes its name to Cytosol. In a more realistic world, when all the cell organelles are taken out , the fluid that remains is called Cytosol. Hope that helps! TQ, Cookie!",
        "video_name": "6UqtgH_Zy1Y",
        "timestamps": [
            239
        ],
        "3min_transcript": "You have the DNA, you have the mRNA. It's all in here, this big jumble of chromatin inside the nucleus. How does it make its way outside of this double bilipid layer? And the way it makes its way out is through nuclear pores. So a nuclear pore is essentially a tunnel. And there are thousands of these. It's a tunnel through this bilipid layer. So the tunnel is made up of a bunch of proteins. So this right over here-- and this is kind of a cross section of it. But you could almost imagine it if you're thinking of it in three dimensions, you would imagine a tunnel. A protein-constructed-- a tunnel made out of proteins that goes through this double bilipid membrane. And so the mRNA can make its way out and get to a free ribosome, But this right over here is not the complete picture. Because when you translate a protein using a free ribosome, this is for proteins that are used inside the cell. So let me draw the entire cell right over here. This is the cell. This right over here is the cytosol of the cell. And you might be sometimes confused with the term cytosol and cytoplasm. Cytosol is all the fluid between the organelles. Cytoplasm is everything that's inside the cell. So it's the cytosol and the organelles and the stuff inside the organelles is the cytoplasm. So cytoplasm is everything inside of the cell. Cytosol is just the fluid that's between the organelles. is good for proteins used within the cell itself. The proteins can then float around the cytosol and used in whichever way is appropriate. But how do you get protein outside of the cell, or even inside the cellular membrane? Not within it, within the cell, but embedded in the cell membrane or outside of the cell itself. And we know that cells communicate in all sorts of different ways and they produce proteins for other cells or for use in the bloodstream, or whatever it might be. And that's what we're going to focus on in this video. So contiguous with this what's called a perinuclear space right over here, so the space between these two membranes-- So you have this perinuclear space between the inner and outer nuclear membrane. Let me just label that. That's the inner nuclear membrane. That's the outer nuclear membrane. You could continue this outer nuclear membrane, and you get into these kind of flaps and folds and bulges."
    },
    {
        "Q": "At 0:24, are proteins the same thing as amino acids if you were solving codons?",
        "A": "Proteins are amino acids. Protein is made up of long chains of amino acid.",
        "video_name": "6UqtgH_Zy1Y",
        "timestamps": [
            24
        ],
        "3min_transcript": "We've already talked about the process from going from DNA to messenger RNA. And we call that process transcription. And this occurs in the nucleus. And then that messenger RNA makes its way outside of the nucleus, and it attaches to a ribosome. And then it is translated into a protein. And so you could say that this part right over here, this is being facilitated by a ribosome. Or it's happening at a ribosome. With that high-level overview, I now want to think a little bit in more detail about how this actually happens, or the structure of things where this happens inside of a cell. And so I'm going to now draw the nucleus in a little bit more detail so that we can really see what's happening on its membrane. So this right over here is the nucleus. Actually, let me draw it like this. I'm going to draw it with two lines. Because it's actually a double bilipid membrane. So this is one bilipid layer right over here. And then this is another one right over here. And I'm obviously not drawing it to scale. I'm drawing it so you can get a sense of things. So each of these lines that I'm drawing, if I were to zoom in on this-- so if I were to zoom in on each of these lines, so let's zoom in. And if I got a box like that, you would see a bilipid layer. So a bilipid layer looks like this. You have the circle is a hydrophilic end and those lines are the fatty hydrophobic ends. So that's our bilipid layer. So that's each of these lines that I have drawn, each of them are a bilipid layer. So the question is, well, how does the mRNA-- obviously You have the DNA, you have the mRNA. It's all in here, this big jumble of chromatin inside the nucleus. How does it make its way outside of this double bilipid layer? And the way it makes its way out is through nuclear pores. So a nuclear pore is essentially a tunnel. And there are thousands of these. It's a tunnel through this bilipid layer. So the tunnel is made up of a bunch of proteins. So this right over here-- and this is kind of a cross section of it. But you could almost imagine it if you're thinking of it in three dimensions, you would imagine a tunnel. A protein-constructed-- a tunnel made out of proteins that goes through this double bilipid membrane. And so the mRNA can make its way out and get to a free ribosome,"
    },
    {
        "Q": "3:55 If cytoplasm is everything inside cell, can we say that a cell consist of cytoplasm and cell membrane?",
        "A": "no..cytoplasm is everything inside cell...everything such as cytosqueleton, cytoplasm, all organelles including nucleus, mitochondria, vesihecles and such... hope it helped",
        "video_name": "6UqtgH_Zy1Y",
        "timestamps": [
            235
        ],
        "3min_transcript": "You have the DNA, you have the mRNA. It's all in here, this big jumble of chromatin inside the nucleus. How does it make its way outside of this double bilipid layer? And the way it makes its way out is through nuclear pores. So a nuclear pore is essentially a tunnel. And there are thousands of these. It's a tunnel through this bilipid layer. So the tunnel is made up of a bunch of proteins. So this right over here-- and this is kind of a cross section of it. But you could almost imagine it if you're thinking of it in three dimensions, you would imagine a tunnel. A protein-constructed-- a tunnel made out of proteins that goes through this double bilipid membrane. And so the mRNA can make its way out and get to a free ribosome, But this right over here is not the complete picture. Because when you translate a protein using a free ribosome, this is for proteins that are used inside the cell. So let me draw the entire cell right over here. This is the cell. This right over here is the cytosol of the cell. And you might be sometimes confused with the term cytosol and cytoplasm. Cytosol is all the fluid between the organelles. Cytoplasm is everything that's inside the cell. So it's the cytosol and the organelles and the stuff inside the organelles is the cytoplasm. So cytoplasm is everything inside of the cell. Cytosol is just the fluid that's between the organelles. is good for proteins used within the cell itself. The proteins can then float around the cytosol and used in whichever way is appropriate. But how do you get protein outside of the cell, or even inside the cellular membrane? Not within it, within the cell, but embedded in the cell membrane or outside of the cell itself. And we know that cells communicate in all sorts of different ways and they produce proteins for other cells or for use in the bloodstream, or whatever it might be. And that's what we're going to focus on in this video. So contiguous with this what's called a perinuclear space right over here, so the space between these two membranes-- So you have this perinuclear space between the inner and outer nuclear membrane. Let me just label that. That's the inner nuclear membrane. That's the outer nuclear membrane. You could continue this outer nuclear membrane, and you get into these kind of flaps and folds and bulges."
    },
    {
        "Q": "At 0:28 what is the bronchai",
        "A": "the bronchi is the tubes that air goes through either the right or left lung. there are two types of bronchi, primary bronchi which are the first 2 tubes that creates a passageway to the lungs and the secondary bronchi which are the the tubes that branches of the primary bronchi. after the secondary bronchi, the bronchioles. the last one would be arterioles. hoped this helped",
        "video_name": "fLKOBQ6cZHA",
        "timestamps": [
            28
        ],
        "3min_transcript": ""
    },
    {
        "Q": "2:30 what is diffusion",
        "A": "Diffusion is the moment of a solute from a region of high concentration to one of a low concentration.",
        "video_name": "fLKOBQ6cZHA",
        "timestamps": [
            150
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 13:15 Sal says that \"They sop up 98.5% of oxygen \" what happens to the remaining 1.5% of oxygen ?",
        "A": "They pretty much stay wherever it was the other oxygen was sopped up from. If the oxygen was taken from the blood plasma, which is what I understood from the video, then the 1.5% will just remain in the blood plasma.",
        "video_name": "fLKOBQ6cZHA",
        "timestamps": [
            795
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 1:36 , Sal said that we take in 78% nitrogen , but i know that we cannot breath in nitrogen and we take it from plants because they are the only living organisms that can take in nitrogen ?",
        "A": "In fact we do dissolve a very tiny amount of nitrogen in our blood from the air we inhale. And the more the higher pressure around us. This is significant for divers who breathe under Deep Water. If a diver ascends to quickly nitrogen bubbles can be released in the bloodstream because the pressure diminishes too quickly to release the surplus of nitrogen throug normal breathing. This condition is leathal unless the diver very quickly is put in a pressuretank and is de-compressed slowly.",
        "video_name": "fLKOBQ6cZHA",
        "timestamps": [
            96
        ],
        "3min_transcript": ""
    },
    {
        "Q": "If red blood cells have DNA then why did he say they didn't. I don't understand. 14:26",
        "A": "they do have dna but they push it out to carry more hemoglobin, if you watch the entire video.",
        "video_name": "fLKOBQ6cZHA",
        "timestamps": [
            866
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 14:35, Sal said that red blood cells don't have DNA, so they can't reproduce. At 15:55, Sal also said that they don't live long. I wonder who who makes those RBCs for us.",
        "A": "At first, when produced in the bone marrow (most likely from the femur because it s the biggest bone in the body), the RBCs have nuclei. However, when they lose it the metabolism of the cell it s compromised and they live less then other cells.",
        "video_name": "fLKOBQ6cZHA",
        "timestamps": [
            875,
            955
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 10:27, Sal says that iron is the main component. Is that the reason why if you drink blood by accident it tastes like metal?",
        "A": "Yes, the hemoglobin molecule has a big-old Iron atom embedded in the complex protein.",
        "video_name": "fLKOBQ6cZHA",
        "timestamps": [
            627
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 3:21 why do you have to do all problems in kelvin units",
        "A": "The most common of the universal gas constants is in the units of ((L*atm)/(mol*K)), so in the instance of using that constant, you must be in kelvin. There are different constants for differing units, but Kelvin has the benefit of never being negative.",
        "video_name": "69V-60sga3M",
        "timestamps": [
            201
        ],
        "3min_transcript": "or force times one dimension of distance, which is joules, so this is in joules. So what the ideal gas constant essentially does is it converts -- we know that this is proportional to this, but it sets the exact constant of proportionality and it also makes sure we get the units right, so that's all it is. It just helps us translate from a world dealing with moles and Kelvins to a world of dealing with, well, in this case, atmospheres and liters, or bars and meters cubed, or kilopascals and meters cubed. But no matter what the units of the pressure and the volume are, the whole unit of pressure times volume is going to be joules. So it's just a translation between the two and we know that they're proportional. Now, with that said, let's do some more problems. So let's say we want to know how many grams of oxygen. So we want to know grams of O2. O2 in grams. that has a pressure of 12 atmospheres and the temperature is 10 degrees Celsius. Well, we've already broken out our ideal gas equation. Let's see, pressure is 12 atmospheres. So we can say 12. I'll keep the units there. 12 atmospheres times the volume. The volume should be in liters, so this is 300 milliliters, or 300 thousandths or 3 tenths. So that's 0.3 liters. 300 one thousandths of a liter is 0.3 liters. And that is equal to the number of moles we have of this, and that's what we need to figure out. If we know the number of moles, we know the number of grams. So this is equal to n times R. We're dealing with liters and atmospheres. We'll deal with this one. Times 0.082. And what's our temperature? It's 10 degrees Celsius. We always have to do everything in Kelvin. So it's 283 degrees Kelvin. So all we have to solve for is n. We just have to divide both sides of this equation by this right here, and so we have n is equal to 12 atmospheres -- I'm just swapping the sides -- times 0.3 liters divided by 0.082 times 283 degrees. Our answer will be in moles. And if you want to verify that, you can plug in all the units and use units for the ideal gas constant. The number of moles we're dealing with,"
    },
    {
        "Q": "At 3:23, why do you have to change the temperature into kelvin?",
        "A": "Because 0 on the Kelvin scale corresponds to 0 heat, but 0 on the celsius scale is arbitrary.",
        "video_name": "69V-60sga3M",
        "timestamps": [
            203
        ],
        "3min_transcript": "or force times one dimension of distance, which is joules, so this is in joules. So what the ideal gas constant essentially does is it converts -- we know that this is proportional to this, but it sets the exact constant of proportionality and it also makes sure we get the units right, so that's all it is. It just helps us translate from a world dealing with moles and Kelvins to a world of dealing with, well, in this case, atmospheres and liters, or bars and meters cubed, or kilopascals and meters cubed. But no matter what the units of the pressure and the volume are, the whole unit of pressure times volume is going to be joules. So it's just a translation between the two and we know that they're proportional. Now, with that said, let's do some more problems. So let's say we want to know how many grams of oxygen. So we want to know grams of O2. O2 in grams. that has a pressure of 12 atmospheres and the temperature is 10 degrees Celsius. Well, we've already broken out our ideal gas equation. Let's see, pressure is 12 atmospheres. So we can say 12. I'll keep the units there. 12 atmospheres times the volume. The volume should be in liters, so this is 300 milliliters, or 300 thousandths or 3 tenths. So that's 0.3 liters. 300 one thousandths of a liter is 0.3 liters. And that is equal to the number of moles we have of this, and that's what we need to figure out. If we know the number of moles, we know the number of grams. So this is equal to n times R. We're dealing with liters and atmospheres. We'll deal with this one. Times 0.082. And what's our temperature? It's 10 degrees Celsius. We always have to do everything in Kelvin. So it's 283 degrees Kelvin. So all we have to solve for is n. We just have to divide both sides of this equation by this right here, and so we have n is equal to 12 atmospheres -- I'm just swapping the sides -- times 0.3 liters divided by 0.082 times 283 degrees. Our answer will be in moles. And if you want to verify that, you can plug in all the units and use units for the ideal gas constant. The number of moles we're dealing with,"
    },
    {
        "Q": "at 3:00, why does \"n\" stand for moles instead of \"m\"?",
        "A": "n stands for number of molecules.",
        "video_name": "69V-60sga3M",
        "timestamps": [
            180
        ],
        "3min_transcript": "or force times one dimension of distance, which is joules, so this is in joules. So what the ideal gas constant essentially does is it converts -- we know that this is proportional to this, but it sets the exact constant of proportionality and it also makes sure we get the units right, so that's all it is. It just helps us translate from a world dealing with moles and Kelvins to a world of dealing with, well, in this case, atmospheres and liters, or bars and meters cubed, or kilopascals and meters cubed. But no matter what the units of the pressure and the volume are, the whole unit of pressure times volume is going to be joules. So it's just a translation between the two and we know that they're proportional. Now, with that said, let's do some more problems. So let's say we want to know how many grams of oxygen. So we want to know grams of O2. O2 in grams. that has a pressure of 12 atmospheres and the temperature is 10 degrees Celsius. Well, we've already broken out our ideal gas equation. Let's see, pressure is 12 atmospheres. So we can say 12. I'll keep the units there. 12 atmospheres times the volume. The volume should be in liters, so this is 300 milliliters, or 300 thousandths or 3 tenths. So that's 0.3 liters. 300 one thousandths of a liter is 0.3 liters. And that is equal to the number of moles we have of this, and that's what we need to figure out. If we know the number of moles, we know the number of grams. So this is equal to n times R. We're dealing with liters and atmospheres. We'll deal with this one. Times 0.082. And what's our temperature? It's 10 degrees Celsius. We always have to do everything in Kelvin. So it's 283 degrees Kelvin. So all we have to solve for is n. We just have to divide both sides of this equation by this right here, and so we have n is equal to 12 atmospheres -- I'm just swapping the sides -- times 0.3 liters divided by 0.082 times 283 degrees. Our answer will be in moles. And if you want to verify that, you can plug in all the units and use units for the ideal gas constant. The number of moles we're dealing with,"
    },
    {
        "Q": "At 3:35 he said that when our cells gets broken by damage we feel pain, but how about slight pain for example if you poke slightly with a toothpick to your hand? There are no visible damage done but it causes slight sensation of pain or it just kills less cells and releases less proteins who causes pain?",
        "A": "Dermic and hypodermic receptors, of which there are many, have a big impact on feeling slight damage/discomfort (e.g., toothpick example). There are also different kinds of nerves that carry different sensations (ex., tiny unmyelinated C fibers produce dull, burning ache while large myelinated axons convey quick, sharp pain) I d be happy to elaborate if you like.",
        "video_name": "D-oAsFIHqbY",
        "timestamps": [
            215
        ],
        "3min_transcript": "So this is the general idea behind a conformational change. So when heat is applied and also when pain applied via a particular molecule, you have a conformational change in the TrpV1 protein. So let's look at a diagram of a hand and go into this in a little bit more detail. OK, so here we have a hand. And as I mentioned before, we have cells located throughout the hand. And these cells are sensitive to temperature and to pain. And within these cells, there are TrpV1 receptors. So let's imagine that each one of these cells sends a little projection to a nerve that eventually reaches the brain. So these cells, whenever they are stimulated by either a change of temperature or the presence of some sort of painful stimulus-- So what can that be? So let's imagine that something pokes your hand. So let's imagine that we have a sharp object, and it pokes your hand. What happens is, the cell, when it gets poked, thousands of cells get broken up. So the cell gets broken up. And when it gets broken up, it releases all kinds of different molecules. And these molecules will travel around. So let's imagine it releases this little green molecule. It will travel around, and it will bind to one of the little TrpV1 receptors. And when it binds to a TrpV1 receptor, it causes the same conformational change that a change in temperature causes. And so that conformational change actually activates the cell, and the cell will send a signal to the brain. So this nerve over here actually contains three different types of fibers. So there are fast, medium, and slow fibers. So fast fibers are really, really fat in diameter. So we have these really big, fat fibers, and they have a lot myelin. So they are covered in myelin. And what myelin is, it's an insulator that basically allows the cell to conduct an action potential very quickly. So as an action potential or as a signal travels down the cell, if we have a lot of myelin surrounding the cell, the signal is able to travel really quickly. And another way that a signal is able to travel quickly is if the cell has a really big diameter. So a big diameter lowers the resistance. So you have less resistance and you have greater conductance because of the myelin. And these two things produce a very fast-- a cell that is able to produce-- send a signal pretty quickly."
    },
    {
        "Q": "At 5:45, Sal said that the initial velocity is 24.5 but earlier he said that the initial velocity is 0. i got mixed up. Can someone clarify it?",
        "A": "Earlier in the video he said final velocity is 0 m/s. I think this must be where you got confused.",
        "video_name": "IYS4Bd9F3LA",
        "timestamps": [
            345
        ],
        "3min_transcript": "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 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"
    },
    {
        "Q": "What makes the time up the same as the time down? (Around 1:15)",
        "A": "Acceleration remains constant throughout the flight so, when it has a velocity of zero (at the top), you can know it has to be at its halfway point during the flight. Hope that helps : )",
        "video_name": "IYS4Bd9F3LA",
        "timestamps": [
            75
        ],
        "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": "is the velocity calculated by Average on 6:25 because both the rise and fall of the object is taken into question?",
        "A": "No, Van. We are only dealing with the first part of the projectile-rising part, in which t=2.5s. This is because the falling part is essentially the reverse of rising part. We use average velocity because velocity is not constant, but acceleration is.",
        "video_name": "IYS4Bd9F3LA",
        "timestamps": [
            385
        ],
        "3min_transcript": "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 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"
    },
    {
        "Q": "At around 5:10, Sal found the initial velocity. Could you also use s=v_i*t+1/2a*t^2 and plug in 5 for t, -9.8 for a and 0 for s to find v_i? How does these two completely different methods get the same answer?",
        "A": "Approaches that are algebraically the same can have different forms",
        "video_name": "IYS4Bd9F3LA",
        "timestamps": [
            310
        ],
        "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": "At 6:50 Sal talks about the energy of the system being 0. Under what conditions would the internal energy change?",
        "A": "I had a look and could not see where he says the energy of the system is zero... However, the answer to your question is that internal energy = kinetic energy of the particles + potential energy of the particles. In an ideal gas, the potential energy changes are considered to be zero. Temperature is related to kinetic energy of the particles. So, a change in temp = change in KE = change in Internal energy",
        "video_name": "aAfBSJObd6Y",
        "timestamps": [
            410
        ],
        "3min_transcript": "Now, we've done a bunch of videos now. We said, OK, how much work was done on this system? Well, the work done on the system is the area under this curve. So some positive work was-- not done on the system, sorry. How much work was done by the system? We're moving in this direction. I should put the direction there. We're moving from left to right. The amount of work done by the system is pressure times volume. We've seen that multiple times. So you take this area of the curve, and you have the work done by the system from A to B. So let's call that work from A to B. Now, that's fair and everything, but what I want to think more about, is how much heat was transferred by my reservoir? Remember, we said, if this reservoir wasn't there, the temperature of my canister would have gone down as I expanded its volume, and as the pressure went down. So how much heat came into it? Well, let's go back to our basic internal energy formula. system minus the work done by the system Now, what is the change in internal energy in this scenario? Well, it was at a constant temperature the whole time, right? And since we're dealing with a very simple ideal gas, all of our internal energy is due to kinetic energy, which temperature is a measure of. So, temperature didn't change. Our average kinetic energy didn't change, which means our kinetic energy didn't change. So our internal energy did not change while we moved from left to right along this isotherm. So we could say our internal energy is zero. And that is equal to the heat added to the system minus the work done by the system. Right? So if you just-- we put the work done by the system on the other side, and then switch the sides, you get heat added to the system is equal to the work done by the system. And that makes sense. The system was doing some work this entire time, so it was essentially maybe some potential energy to these rocks. So it was giving energy away. It was giving energy outside of the system. So how did it maintain its internal energy? Well, someone had to give it some energy. And it was given that energy by this reservoir. So let's say, and the convention for doing this is to say, that it was given-- let me write this down. It was given some energy Q1. We just say, we just put this downward arrow to say that some energy went into the system here. Fair enough. Now let's take this state B and remove the reservoir, and completely isolate ourselves. So there's no way that heat can be transferred to and from our system. And let's keep removing some rocks. So if we keep removing some rocks, where do we get to? Let me go down here. So let's say we remove a bunch of more rocks."
    },
    {
        "Q": "Hi, at 10:02 Sal Khan says that catalysts lower activation energy but on some other websites and in books it is written that the catalysts do not lower the activation energy . What should I follow ?",
        "A": "Catalysts provide an alternate reaction pathway with a lower activation energy. The uncatalyzed pathway still has the same activation energy.",
        "video_name": "__zy-oOLPug",
        "timestamps": [
            602
        ],
        "3min_transcript": "you don't always have to add it, but if it doesn't happen spontaneously you're gonna have to add some energy to the system to get to this activated state. RIght, so this is when we were at this thing right here. We're there. So some energy has to be in the system, and this energy, the difference between the energy we're at when we were just hydrogen molecules and iodine molecules, and the energy we have to get to to get this activated state, this distance right here this is the activation energy. If we're able to get to somehow put enough energy in the system, then this thing will happen, they'll collide with enough energy and bonds will be broken and reformed. Activation energy. Sometimes it's written as Ea energy of activation, and in the future we'll maybe do reactions where we actually measure the activation energy. But the important thing is to conceptually understand that it's there, that things just don't spontaneously go from here to here. but you've probably heard of the word catalyst or something being catalyzed. And that's something, some other agent, some other thing in the reactions. So right now, so right now we're doing, we have H2 plus I2, yielding 2H hydrogen iodides. Now you could have a catalyst, and I'll just say plus C. And I actually don't know what a good catalyst would be for this reaction, and how a catalyst operates is, it can actually operate in many many different ways, so that's why I don't wanna do it in this video. But what a catalyst is, is something that doesn't change. It doesn't get consumed in the reaction. The catalyst was there before the reaction, the catalyst is there after the reaction. But what it does is it makes the reaction happen either faster, or it lowers the amount of energy for the reaction to happen. Which is kind of the same thing. So if you have a catalyst, then this activation energy And what it does is, it makes it, it might easily, it might be some molecule that allows some other transition state that has less of a potential energy so that you require less heat or less concentration of the molecules for them to bump into each other in the right direction, to get to that other state. So you require less energy. So given how we understand how these kinetics occur, these molecules interact with each other, what do you think are the things that will drive whether a reaction happens or not? I mean we already know that if we have a positive catalyst, there's something called a negative catalyst that will actually slow down a reaction. But if we have a positive catalyst, it lowers... Obviously it lowers the activation energy, so this makes reaction faster. More molecules are gonna bump into each other just right to be able to get over this hump 'cause the hump will be lower, when you have a catalyst."
    },
    {
        "Q": "@5:07 Could you explain the activated complex a little bit clearer? I was confused whether it is the bond between diatomic molecules or whether it was a bond formed between molecules already bonded.",
        "A": "It s more like a temporary phase (at a higher energy level) when the bond is first formed between the atoms of the reacting molecules, in this case an H atom of the H2 molecule and an I atom on an Iodine molecule. The activation complex is that moment when these two guys are trying to strengthen their bond and are JuSt breaking away from their original molecules to form HI . Hope I made sense to you",
        "video_name": "__zy-oOLPug",
        "timestamps": [
            307
        ],
        "3min_transcript": "The iodine might look something like this, it's a much bigger molecule. Where it's bonded together like this, it's also sharing some electrons in a covalent bond, everything's probabilistic. So in order for these two molecules to turn to this, somehow these bonds have to be broken, and new bonds have to be formed. And what has to happen is that these guys, there's a ton of these guys. I could draw a bunch of them. Or I could copy and paste. So there's a bunch of... There's a bunch of hydrogen molecules around, and some of these iodine gas molecules around. So what has to happen in order for us to get the hydrogen iodide is, they have to collide. And they have to collide in exactly the right way. So let's say this guy, wish I could show it. Let's say he's moving, this is neat, I'm just dragging and dropping. But he's moving. He has to hit this hydrogen molecule just right, And maybe just right, if he just happens to hit it then all of the sudden, let's say we get to this point right here, these electrons are gonna say \"Hey, you know, \"it's nice to be shared this way, \"we're in a stable configuration, \"we're filling the 1S shell, but look at this, \"there's this iodine that's close by \"and they really want me, \"they're much more electro-negative \"than me, the hydrogen\". So maybe they're kind of attracted here, they don't know whether they wanna be here between that hydrogen and this right here between that. And so they kind of enter this higher energy state. And similarly, you know these guys they say, \"Hey, wouldn't it be nicer, \"I don't have to be here, I could kind of go back home \"to my home atom if this guy comes in here\". Because then we're gonna have, then we're gonna have eight valence electrons and the same thing's happening here. And this complex right here, this kind of, right when the collision happens, this is actually a state, this is the high energy state of the transition state of the reaction, and this is called an activated complex. Sometimes you know I just drew it kind of visually, but you could draw it like this. So hydrogen has a covalent bond with another hydrogen, and then here comes along some iodine that has a covalent bond with some other iodine,, but all the sudden these guys like to bond as well. So they start forming, so there is kind of a you know, there's a little bit of an attraction on that side too. So this is another way of drawing the activation complex. But this is a high energy state, 'cause in order for the electrons, the way you can think of it, to kind of go from that bond to this bond, or this bond to that bond, or to go back, they have to enter into a higher energy state. A less stable energy state, than they were before. But they do that if there's enough energy, 'cause you can go from, so you're going from both of these things separate, let me just draw them separate,"
    },
    {
        "Q": "At around 8:00 he says that one has to add energy if its not spontaneous, I thought that even spontaneous reactions do have an activation energy so energy has to be added?",
        "A": "You re absolutely right that spontaneous reactions have activation energy -- what makes them spontaneous is that there s enough energy in the environment to overcome the activation energy. This is where conditions such as temperature and pressure can effect spontaneity (think delta H as it plays in to delta G, for example). Hope this helps!",
        "video_name": "__zy-oOLPug",
        "timestamps": [
            480
        ],
        "3min_transcript": "you have the hydrogen separate, plus the iodine separate. They go to this, which is a higher energy state. But if they can get to that higher energy state, if there's enough energy for the collision and they have enough kinetic energy when they hit in the right orientation, then, from this activated complex or this higher energy state, it will then go to the lowest energy state, and the lowest energy state is the hydrogen iodide. I wanna draw the iodide, and then the hydrogen. This is actually, this is actually a lower energy state than this. But in order to get here you have to go through a higher energy state. And I could do that with an energy diagram. So if we say that, let's say the X-axis is the progression of the reaction, we don't know how fast it's progressing, but this you can kind of view it as time on some dimension, and let's say this is the potential energy. I wanna draw thicker lines. See this is the potential energy. Right there, let me make this line thicker as well. So this is the potential energy. So initially, you are at this reality, and we can kind of view it as the combined potential energy, so this is where eventually, we start off here, and this is the H2 plus I2 And a lower potential energy is when we were in the hydrogen iodide. So this is the lower potential energy down here. Lower potential energy down here. This is the 2HI, right? But to get here, we have to enter this higher activation energy, where the electrons have to get, they have to have some energy to kind of be able to at least figure out what they wanna do with their lives. you don't always have to add it, but if it doesn't happen spontaneously you're gonna have to add some energy to the system to get to this activated state. RIght, so this is when we were at this thing right here. We're there. So some energy has to be in the system, and this energy, the difference between the energy we're at when we were just hydrogen molecules and iodine molecules, and the energy we have to get to to get this activated state, this distance right here this is the activation energy. If we're able to get to somehow put enough energy in the system, then this thing will happen, they'll collide with enough energy and bonds will be broken and reformed. Activation energy. Sometimes it's written as Ea energy of activation, and in the future we'll maybe do reactions where we actually measure the activation energy. But the important thing is to conceptually understand that it's there, that things just don't spontaneously go from here to here."
    },
    {
        "Q": "In 0:26 shouldn't it be one mole of hydrogen and one mole of iodine?",
        "A": "Quite correct. It should be 1 mol of H\u00e2\u0082\u0082 and 1 mol of I\u00e2\u0082\u0082.",
        "video_name": "__zy-oOLPug",
        "timestamps": [
            26
        ],
        "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": "At time 10:08 when he says 1 to 6 wouldn't be 1 to 12 since 6O2 is 12 oxygen?",
        "A": "He means 1 to 6 oxygen molecules",
        "video_name": "eQf_EAYGo-k",
        "timestamps": [
            608
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 0:42, why is oxygen in its molecular form?",
        "A": "This is a combustion reaction \u00e2\u0080\u0094 a reaction with oxygen \u00e2\u0080\u0094 and oxygen exists as O\u00e2\u0082\u0082 molecules.",
        "video_name": "eQf_EAYGo-k",
        "timestamps": [
            42
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 9:00 couldn't Sal have put down 0.14? does it matter how you round the number?",
        "A": "Actually, no he could not have written down .14. You have to look at the number with the least number of sig figs in the problem. Then you have to write your answer with that many sig figs and one more for uncertainty purposes. Since two was the least number of sig figs in the problem, Sal rounded his answer to three sig figs. Significant numbers do matter when you are doing stoichiometry problems. Never round midway.",
        "video_name": "eQf_EAYGo-k",
        "timestamps": [
            540
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 0:42, how does Sal know that the oxygen is in its molecular state? Is it the convention?",
        "A": "Yes that is the convention. If you see reacts with oxygen they mean O2.",
        "video_name": "eQf_EAYGo-k",
        "timestamps": [
            42
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 1:00, do you always have to start by balancing the equation?",
        "A": "Yes, it is necessary to balance the equation for any stoichiometry problem. Hope this helps :)",
        "video_name": "eQf_EAYGo-k",
        "timestamps": [
            60
        ],
        "3min_transcript": ""
    },
    {
        "Q": "To find the amount of C02 and H20 we used only glucose in that ratio at 13:00, but what about 02 molecule?",
        "A": "There is an excess of oxygen, so only all the glucose will be used up in the reaction. All of the moles of glucose will be used to make your products, carbon dioxide and water. There will be some oxygen left over after the reaction is complete because there are no more glucose for it to react with. Does this make sense? That s why we are only concerned with the LIMITING REACTANT or REAGENT.",
        "video_name": "eQf_EAYGo-k",
        "timestamps": [
            780
        ],
        "3min_transcript": ""
    },
    {
        "Q": "In 8:39 why does he divide 25 by 180?? soo confused right now",
        "A": "There is 25 grams of glucose The molar mass of glucose is approx 180 grams per mole If you divide like so: 25 g / 180 g/mol this will calculate how many moles of glucose there are. We always need to work in moles when using chemical equations. It s important to be comfortable doing this.",
        "video_name": "eQf_EAYGo-k",
        "timestamps": [
            519
        ],
        "3min_transcript": ""
    },
    {
        "Q": "0:25 if there is a vacuum everywhere, then wouldn't the water just go up?",
        "A": "between sun and earth, there is vacuum so why earth revolves in an elliptical orbit around the sun, by newton law of gravitation, every particle gravitate to other particles the force is the same but the acceleration is different bcoz of there difference in masses.",
        "video_name": "HnfBFeLunk4",
        "timestamps": [
            25
        ],
        "3min_transcript": "Before we move on, I just wanted to make sure that you understood that last point that I made at the end of that last video. We said that the pressure inputting into this, that we could view this cup with a hole in it as essentially a pipe, where the opening on the top of the cup is the input to the pipe, and this little mini-hole is the output to the pipe, and we said that this is a vacuum. Let's say this is vacuum all around. I know when I drew it last time, I closed it, but we have a vacuum everywhere. Since there's a vacuum everywhere, the pressure at this point P1 is equal to zero. The point I wanted to make is because we have a hole here, the pressure at that point at P2 is also equal to zero. You can almost view it as maybe the atmospheric pressure at that point, but since we're in a vacuum, that pressure is zero. That might have been a little confusing to you, because you said, well, wait, I thought at depth, if I had a point at that same height, that I would actually have a pressure at that point of rho gh. You do have an innate pressure in the liquid at that point of rho gh, and actually, that's what's causing the liquid to come out. But that's actually taken care of in the potential energy part of the equation. Let me rewrite Bernoulli's equation. The input pressure plus rho g h1 plus rho V1 squared over 2 is equal to the output pressure plus rho g h2 plus PV2 squared over 2. I think you understand that this term is pretty close to zero if the rate at which the surface moves is very slow if this surface area is much bigger than this hole. It's like if you poked a hole in Hoover Dam, that whole lake of the speed at which the water's coming out at the other end, so you could ignore this term. We also defined that the hole was at zero, so the height of h2 is zero. It simplified down to the input pressure, the pressure at the top of the pipe, or at the left side of the pipe, plus rho gh1. This isn't potential energy, but this was kind of the potential energy term when we derived Bernoulli's equation, and that equals the output pressure, or the pressure at the output of the hole, at the right side of the hole, plus the kinetic energy PV2. It's the kinetic energy term, because it doesn't actually doesn't add up completely to kinetic energy, because we manipulated it. I just wanted to really make the point that"
    },
    {
        "Q": "At 10:59 Sal says that the field lines correspond to the trajectory of the test charge but my textbook stated that this is a common misconception. So do they or do they not? If they don't how would the trajectory of the test charge compare to the field line?",
        "A": "The field lines are the force that would be exerted on a unit positive charge present in that field. Remember, it s force; not trajectory. If the charge is moving inside the electric field, the trajectory would depend on the force (and some fancy vector mechanics)",
        "video_name": "0YOGrTNgGhE",
        "timestamps": [
            659
        ],
        "3min_transcript": "a bunch more in between here. And that makes sense, right? Because as you get closer and closer to the source of the electric field, the charge gets stronger. Another way that you could have done this, and this would have actually more clearly shown the magnitude of the field at any point, is you could have-- you could say, OK, if that's my charge Q, you could say, well, really close, the field is strong. So at this point, the vector, the newtons per coulomb, is that strong, that strong, that strong, that strong. We're just taking sample points. You can't possibly draw them at every single point. So at that point, that's the vector. That's the electric field vector. But then if we go a little bit further out, the vector is going to be-- it falls off. This one should be shorter, then this one should be even shorter, right? You could pick any point and you could actually calculate 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"
    },
    {
        "Q": "is Sal saying that electric fields pervade the entire universe At 2:58? How do electric field lines extend infinite distances?",
        "A": "Electric fields do indeed pervade the universe, just like gravitational fields do.",
        "video_name": "0YOGrTNgGhE",
        "timestamps": [
            178
        ],
        "3min_transcript": "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. times the charge creating the field divided by the distance squared, the distance we are away from the charge. So essentially, we've defined-- if you give me a force and a point around this charge anywhere, I can now tell you the exact force. For example, if I told you that I have a minus 1 coulomb charge and the distance is equal to-- oh, I don't know. The distance is equal to let's say-- let's make it easy. Let's say 2 meters. So first of all, we can say, in general, what is the electric field 2 meters away from? So what is the electric field out here?"
    },
    {
        "Q": "At 4:48,how is the force of static friction is going to be 29.4 N? As we got this as a answer of budging force.",
        "A": "Budging force is the term Sal uses for the maximum possible force of static friction.",
        "video_name": "ZA_D4O6l1lo",
        "timestamps": [
            288
        ],
        "3min_transcript": "the direction is straight down towards the center of the earth The normal force, and that force is there because this block is not accelerating downwards So there must be some force that completely balances off the force of gravity And in this example, it is the normal force So it is acting 49 newtons upward and so these net out. And that's why this block does not accelerate upwards or downwards So what we have is the budge the magnitude of the budging force, needs to be equal to, over the magnitude of the normal force well this thing right over here is going to be 49 newtons Is equal to 0.60 Or we could say that the magnitude of the budging force is equal to 49 newtons times the coefficient of static fiction Or that's 49 newtons times 0.60 So the units here are still going to be in newtons So this 49 times .6 gives us 29.4 newtons This is equal to 29.4 newtons So that's the force that's started to overcome static friction which we are applying more than enough of so with a 100 newtons, we would just start to budge it and right when we are in just in that moment where that thing is just starting to move the net force-- so we have a 100 newtons going in that direction and the force of static friction is going to go in this direction-- maybe I could draw it down here to show it's coming from right over here The force of static friction is going to be 29.4 newtons that way and so right when I am just starting to budge this just when that little movement-- and then kinetic friction starts to matter, but just for that moment just for that moment I'll have a net force of 100 - 29.4 to the right, so I have a net force of 70.6 N for just a moment while I budge it So just exactly while I'm budging it While we're overcoming the static friction, we have a 70.6 N net force in the right direction And so just for that moment, you divide it by 5 kg mass So just for that moment, it will be accelerating at 14.12 m/s^2 So you'll have an acceleration of 14.1 m/s^2 to the right but that will just be for that absolute moment, because once I budge it"
    },
    {
        "Q": "At 5:40, Sal mentions sperm carry either an X or Y chromosome, determining gender. Are the X and Y chromosome not carried in every sperm?",
        "A": "Sperm will always carry one or the other unless they undergo nondisjunction in meiotic divisions. Remember that they are haploid, and as such can only carry one copy of each chromosome, including the sex chromosome.",
        "video_name": "-ROhfKyxgCo",
        "timestamps": [
            340
        ],
        "3min_transcript": "for her sex-determining chromosome. She's got two x's. That's what makes her your mom and not your dad. And then your dad has an x and a y-- I should do it in capital--and has a Y chromosome. And we can do a Punnett square. What are all the different combinations of offspring? Well,your mom could give this X chromosome, in conjunction with this X chromosome from your dad. This would produce a female. Your mom could give this other X chromosome with that X chromosome. That would be a female as well. Well,your mom's always going to be donating an X chromosome. And then your dad is going to donate either the X or the Y. So in this case,it'll be the Y chromosome. So these would be female,and those would be male. And it works out nicely that half are female and half are male. But a very interesting and somewhat ironic fact might pop out at you when you see this. what determines whether someone is or Who determines whether their offspring are male or female? Well,the mom always donates an X chromosome, so in no way does- what the haploid genetic makeup of the mom's eggs of the gamete from the female, in no way does that determine the gender of the offspring. It's all determined by whether--let me just draw a bunch of-- dad's got a lot of sperm,and they're all racing towards the egg. And some of them have an X chromosome in them and some of them have a Y chromosome in them. And obviously they have others. And obviously if this guy up here wins the race. Or maybe I should say this girl.If she wins the race, then the fertilized egg will develop into a female. If this sperm wins the race, then the fertilized egg will develop into a male. And the reason why I said it's ironic is throughout history, 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"
    },
    {
        "Q": "Would writing the angle of the second ball as -38 degrees be the same as when Sal says \" 38 degrees below the horizontal\" at 7:47 ?",
        "A": "It might be, if you specify that 0 is a horizontal line to the right. You need to be clear.",
        "video_name": "leudxqivIJI",
        "timestamps": [
            467
        ],
        "3min_transcript": "And maybe you're not familiar with arcsine yet because I don't think I actually have covered yet in the trig modules, although I will eventually. So we know it's just the inverse function of sine. So sine of theta is equal to 0.625. Then we know that theta is equal to the arcsine of 0.625. This is essentially saying, when you say arcsine, this says, tell me the angle whose sine is this number? That's what arcsine is. And we can take out Google because it actually happens that Google has a-- let's see. Google actually-- it's an automatic calculator. So you could type in arcsine on Google of 0.625. Although I think the answer they give you will be in radians. So I'll take that answer that will be in radians and I want to convert to degrees, so I multiply it times 180 over pi. And let's see what I get. So Google, you see, Google says 38.68 degrees. They multiplied the whole thing times 180 and then divided by pi, but that should be the same thing. So roughly 38.7 degrees is theta. Hope you understand that. You could pause it here if you don't, but let me just write that down. So it's 38 degrees. So theta is equal to 38.7 degrees. So then we're done. We figured out that ball B gets hit. This is ball B and it got hit by ball A. Ball A went off in that direction at a 30 degree angle, at a 30 degree angle at 2 meters per second. And now ball B goes at 38.-- or we could say roughly 39 degrees below the horizontal at a velocity of 3.2 meters per second. And does this intuitively make sense to you? Ball A had a mass of 10 kilograms while ball B had a mass of 5 kilograms. So it makes sense. So let's think about just the y direction. Ball A, we figured out, the y component of its velocity was 1 meter per second. And ball B's y component is 2 meters per second downwards. And does that makes sense? Well sure. Because their momentums have to add up to 0. There was no y component of the momentum before they hit each other. And in order for B to have the same momentum going downwards in the y direction as A going upwards, its velocity has to be essentially double, because its mass is half. And a similar logic, although the cosine-- it doesn't work out exactly like that. But a similar logic would mean that its overall velocity is going to be faster than the- than A's velocity. And so what was I just-- oh yeah."
    },
    {
        "Q": "At 4:13, If Br is so electronegative, why does it give it's electron to Fe in the first place?",
        "A": "it have 7 valence electrons and wants to fufill its orbitals with 8, the FeBr3 has an open spot so that both of the compounds have 8 and therefore are satisfied",
        "video_name": "K2tIixiXGOM",
        "timestamps": [
            253
        ],
        "3min_transcript": "with the right energies it can happen. So this electron-- let me do it in a different color that I haven't used yet, this green color-- let's say this electron right here gets nabbed by that iron. Then what do we have? Well then we have a situation, we have this bromine-- the blue one-- with one, two, three, four, five, six, seven valence electrons. We have the magenta bromine with one, two, three, four, five. Now it only has the sixth valence electron right here. The seventh got nabbed by the iron. So the iron has the seventh valence electron and then you have the rest of the molecule. So then you have your iron and it's attached, of course, to the three bromines. Just like that. And then our bonds, these guys were bonded. They still are bonded. And now these guys are bonded. This electron jumped over to the iron. And now we have another bond. But because this bromine lost an electron-- it was neutral, it lost an electron-- it now has a positive charge. And the iron, now that it gained this electron, now has a negative charge. So let's think about what's going to happen now. Now we're going to bring the benzene into the mix. So let me redraw the benzene. And we have this double bond, that double bond, and then just to make things clear, let me draw this double bond with the two electrons on either end. So we have the orange electron, you have your green electron right over there, and I'll draw the double bond as being green. Now let's think about this molecule right here. We have a bromine with a positive charge. Bromines are really, really, really, electronegative. positive charge, it really wants to grab an electron. And in the right circumstances, you can imagine where it really wants to grab that electron right there. So maybe, if there was just some way it could pull this electron. But the only way it could pull this electron is maybe if this-- because if it just took that electron than this bromine would have a positive charge, which isn't cool. So this bromine maybe would want to pull an electron. If this guy gets an electron, then this guy can get an electron. So you can imagine this thing as a whole really, really wants to grab an electron. It might be very good at doing it. So this is our strong electrophile. So what actually will happen in the bromination of this benzene ring-- let me draw some hydrogens here just to make things clear. We already have hydrogens on all of these carbons. Sometimes it's important to visualize this when we're doing electrophilic aromatic substitutions."
    },
    {
        "Q": "At 2:39, When do you know when to use negative for tension pulling up or negative for the force of gravity pulling down. I tried another pulley problem using the same concept pulling up 10kg at 2 m/s^2 and the answer was 2T - mg = ma (answer 60N) instead of 2T + mg = ma ( answer 40N upward). These both give two different answers so I am a bit confused.",
        "A": "You just have to decide, in each problem, which way do you want to be positive. If you decide up is positive, then it is always positive. If you decide down is positive, then it is always positive.",
        "video_name": "52wxpYnS64U",
        "timestamps": [
            159
        ],
        "3min_transcript": "per second, the acceleration of gravity. So the wire must be exerting some upward force on the object. And that is the force of tension. That is what's slowing-- that's what's moderating its acceleration from being 9.8 meters per second squared to being 4.13 meters per second squared. So essentially, what is the net force on this object? On just this object? Well the net force is-- and you can ignore what I said before about the net force in all the other places. But we know that the object is accelerating downwards. Well, we know it's 20 kilograms. So that's its mass. And we know that it's accelerating downwards at 4.13 meters per second squared. So the net force, 20 times-- see, times 20 is 82-- let's just say 83 Newtons. We know that the net force is 83 Newtons down. We also know that the tension force plus the force of gravity-- and what's the force of gravity? The force of gravity is just the weight of the object. So the force of tension, which goes up, plus the weight of -- the force of gravity is equal to the net force. And the way I set this up, tension's going to be a negative number. Just because I'm saying positive numbers are downwards, so a negative number would be upwards. So tension will be what is 83 minus 196? Minus 196 is equal to minus 113 Newtons. And the only reason why I got a negative number is because I used positive numbers for downwards. So minus 113 Newtons downwards, which is the same And so that is the tension in the rope. And you could have done the same thing on this side of the problem, although it would have been-- well, yeah. You could have done the exact same thing on this side of the problem. You would've said, well what would it have accelerated naturally if there wasn't some force of tension on this rope going backwards? And then you're saying, oh, well, we know it would have gone in this direction at some acceleration, but instead it's going in the other direction. So you use that. You figure out the net force, and then you say the tension plus all of these forces have to equal the net force. And then you should solve for the tension. And it would be the same tension. Now we will do a fun and somewhat simple, but maybe instructive problem. So I have a pie. This is the pie. This is parallel. And I have my hand. You can tell that my destiny was really to be a great"
    },
    {
        "Q": "09:00 You're placing the oxygen on the same side as Br but you're saying that it's inversion. How come?\nAlso, I really don't understand that triangle type bond, what's that?",
        "A": "See videos on stereochemistry. This video is way too advanced if you don t understand chirality or wedge and dash representations of bonds.",
        "video_name": "3LiyCxCTrqo",
        "timestamps": [
            540
        ],
        "3min_transcript": "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, is attached to only one alkyl group. This is a primary alkyl halide and that makes this a good SN2 reaction. The decreased steric hindrance allows the nucleophile to attack the electrophile."
    },
    {
        "Q": "At 16:24, it is said that work is done, but deltaU is zero because T does not change, which means Q-W is 0. If work is not zero, than Q must not be zero, which contradicts with a adiabatic process? What's the problem here?",
        "A": "I don t recall this process being described as adiabatic. Lots of heat comes from the reservoir to maintain T to offset the drop in T from work being done.",
        "video_name": "WLKEVfLFau4",
        "timestamps": [
            984
        ],
        "3min_transcript": "Alright. OK. So this is pressure, this is volume. Now. When we started off, before we blew away the wall, we had some pressure and some volume. So this is V1. And then we blew away the wall, and we got to-- Actually, let me do it a little bit differently. I want that to be just right there. Let me make it right there. So that is our V1. This is our original state that we're in. So state initial, or however we want it. That's our initial pressure. And then we blew away the wall, and our volume doubled, right? So we could call this 2V1. Our volume doubled, our pressure would have gone down, and we're here. That's our state 2. blew away the wall. Now, what we did was not a quasistatic process. I can't draw the path here, because right when I blew away the wall, all hell broke loose, and things like pressure and volume weren't well defined. Eventually it got back to an equilibrium where this filled the container, and nothing else was in flux. And we could go back to here, and we could say, OK, now the pressure and the volume is this. But we don't know what happened in between that. So if we wanted to figure out our Q/T, or the heat into the system, we learned in the last video, the heat added to the system is equal to the work done by the system. We'd be at a loss, because the work done by the system is the area under some curve, but there's no curve to speak of here, because our system wasn't defined while all the hell had broke loose. So what can we do? Well, remember, this is a state function. And this is a state function. And I showed that in the last video. So it shouldn't be dependent on how we got from there to there. with my words. This change in s, so s2 minus s1, should be independent of the process that got me from s1 to s2. So this is independent of whatever crazy path-- I mean, I could have taken some crazy, quasistatic path like that, right? So any path that goes from this s1 to this s2 will have the same heat going into the system, or should have the same-- let me take that-- Any system that goes from s1 to s2, regardless of its path, will have the same change in entropy, or their same change in s. Because their s was something here, and it's something different over here. And you just take the difference between the two. So what's a system that we know that can do that? Well, let's say that we did an isothermal. And we know that these are all the same isotherm, right?"
    },
    {
        "Q": "At around 2:00, Jay says that one oxygen fewer is hypochlorite. Why isn't perchlorate called HYPERchlorate? For example, you can have HYPOthermia and HYPERthermia.",
        "A": "Strangely enough, I believe in the past perchlorate was sometimes referred to as hyperchlorate but the name seems to have gone into disuse. Perhaps there was confusion between hypo and hyper so it was safer to drop the hyper.",
        "video_name": "DpnUrVXSLaQ",
        "timestamps": [
            120
        ],
        "3min_transcript": "- [Voiceover] When you take a general chemistry class, you often have to memorize some of the common polyatomic ions. So let's go through a list of some of the ones that you might see in your class. So we'll start off with Cation here, so a positively charged ion, NH four plus is called the Ammonium ion. And for Anions, there are many Anions that you should know. CH three COO minus is the Acetate ion. CN minus is the Cyanide ion. OH minus is the Hydroxide anion. MnO four minus is the Permanganate ion. And, when you get to NO three minus versus NO two minus, look at the endings. So NO three minus is Nitrate, so we have ate suffix, ate suffix here, which means more Oxygens. Versus the ite suffix, which means fewer Oxygens. So we can see that Nitrate has three Oxygens and Nitrite has two Oxygens. with some of the other polyatomic ions. For example, let's look at this next set here of four. And let's look at Chlorate. So Chlorate has three Oxygens. It's ClO three minus one. And Chlorite has fewer Oxygens, it has two Oxygens here, ClO two minus. So we have ate meaning more and ite meaning fewer here. What about Perchlorate? So here we have Chlorate, but we've added on a prefix this time and the prefix, per, means one more Oxygen. So Perchlorate means one more Oxygen than Chlorate. Chlorate had three Oxygens and for Perchlorate we add one on and we get four. So Perchlorate is ClO four minus. Next, let's look at Hypochlorite. So we talked about Chlorite up here, so here's Chlorite and then we put a prefix, hypo, in front of it. Hypo means one fewer, we take one away and now we have only one Oxygen. So that must be the Hypochlorite ion. We could have done this for a different halogen, here we're dealing with Chlorine, but let's say, instead of ClO three minus, let's do BrO three minus. ClO three minus was Chlorate, here we have Bromine instead of Chlorine, so this would be Bromate. So there's another polyatomic ion and we can do another example. So instead of ClO minus, which is Hypochlorite, we could have had BrO minus, which would therefore be Hypobromite. So this would be Hypobromite. Alright, let's look at our next set of polyatomic ions. Alright, so let's get some space down here. So we have SO four two minus, is called Sulfate."
    },
    {
        "Q": "4:30 how do we know we should use the second half-reaction?",
        "A": "To find a total cell potential we should take into account both oxidation and reduction reactions. It s not enough to know how much an element wants to be reduced, we also should check how strongly the other element wants to be oxidized, as electrons lost by one element are the electrons gained by another.",
        "video_name": "fYUwEAPejbY",
        "timestamps": [
            270
        ],
        "3min_transcript": "four hydroxide anions. So once again, this is a reduction reaction because those electrons are being incorporated into the molecule. And so you're left with the hydroxide right over here. And if you look at our metal air cell up here, where is that occurring? Well, you need oxygen, you need water and you need electrons. Well, if you look at the cathode right over here, you got your electrons coming in, you got your oxygen coming in, and water, we can assume, is available in this area. The electrolyte paste has water in it. The cathode is porous, it's a porous membrane-type substance. So they're all available over here. So you can assume that, that reaction could happen right over here in the cathode. So let me just write it, 02 in as a gas. So for every molecule of O2, you're gonna have two molecules of H2O in a liquid form. And then you got your electrons coming in on the wire, four hydroxide anions. Four hydroxide anions. And they're in a solution. They're part of that electrolyte paste that's in water. They're an aqueous solution, a water-based solution. And so that's going to happen right over there so the hydroxide anions, I can even say four hydroxide anions are going to be produced every time this reaction is happening. And they tell us the reduction potential. This has a positive potential, has a positive voltage. And notice they say the reduction potential at pH 11. So that's consistent with alkaline conditions because this paste might seep in through here a little bit and obviously you have all this hydroxide being produced. So you're gonna have a higher pH, a more basic pH. And so the fact that this is a positive voltage, so plus 0.34 volts means that the potential is going in this direction. going from left to right. Now they also said they give us the reduction potentials for the cathode, which we just talked about, this is the reduction potential for the cathode, and three possible metal anodes are given in the table below. So here there are three possible metal anodes. So we could have zinc, we could have sodium, we could have calcium, and so let's just go with zinc since it's the first one listed here. So if we assumed that the metal here was zinc, what's going to be going on? Well, you're going to have the zinc reacting with these hydroxides that are being produced over in the cathode, and then you're going to use that to produce zinc oxide water and electrons. So you're actually gonna have the reverse of this reaction. You're going to have, let me write the reverse. You're going to have zinc in the solid state. This whole anode is made out of metal zinc. And then for every molecule of that, you're gonna have two hydroxide anions that are dissolved"
    },
    {
        "Q": "At 7:04, Sal explained neutron stars by saying that it is just a ball of neutrons. How does this ball stay in place?",
        "A": "It dosn t stay in place . All stars - including neutron stars - orbit the super-massive black hole at the center of their galaxy. Furthermore, the galaxy itself is constantly in motion. So, the neutron star does not stay just sit there. It s moving.",
        "video_name": "qOwCpnQsDLM",
        "timestamps": [
            424
        ],
        "3min_transcript": "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 So we're talking about things being shot out at up to 10% of the speed of light. Now, that's 30,000 kilometers per second. That's almost circumnavigating the earth every second. So that's, I mean, this is unbelievably energetic events that we're talking about here. And so if the original star was-- and these are rough People don't have kind of a hard limit here. If the original star approximately 9 to 20 times the mass of the sun, then it will supernova. And the core will turn into what's called a neutron star. This is a neutron star, which you can imagine is just this dense ball. It's this dense ball of neutrons. 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,"
    },
    {
        "Q": "At 2:49, Sal mentioned that the electrons get captured into the nucleus. Well then, wouldn't the atom be annihilated?",
        "A": "Why would it be annihilated?",
        "video_name": "qOwCpnQsDLM",
        "timestamps": [
            169
        ],
        "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": "from 4:30 to 5:00\nsure there's loads of pressure on the core of the star, but why exactly does it suddenly go supernovae??\nis there some repulsion between the neutrons??",
        "A": "The collapse stops beyond the point when the neutrons are literally touching each other. The neutrons are actually smaller than normal neutrons because of this pressure. The forces trying to decompress the neutrons is incredible and they fuel the supernova.",
        "video_name": "qOwCpnQsDLM",
        "timestamps": [
            270,
            300
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "At 4:00, Sal says \"should.\" Can some cancer cells be so mutated that they do not produce MHC I at all?",
        "A": "Yes. Sometimes cancer cells are so mutated or too close to your actual cells that if your cyto cells attack them, they will also attack your healthy cells. That s why to be safe they don t attack them. They also don t recognize them so they don t kill them. That s why all people need help to fight off cancers since your body can t kill most cancers.",
        "video_name": "YdBXHm3edL8",
        "timestamps": [
            240
        ],
        "3min_transcript": "Once again this is what's occurring outside of cells. When we found stuff outside of cells, we engulf them and then we presented them on MHC two complexes. Now you're probably thinking well, I mean that's the outside of cells but there's MHC two, there's this helper T cells but we've also referred to cytotoxic T cells. What do those do? We've also, if there's MHC two, there's probably an MHC one complex. What does the MHC one complex do? We can recognize shady things that are happening outside of cells but don't shady things sometimes happen inside cells and how does our immune system respond to that? Actually as you can imagine all of those things will be answered in the rest of this video. Let's think about what happens when shady things start to happen inside the cell. For example it might not even be due to a virus could be the cell itself is gone awry. Let's say that this right over here is a cancer cell. It's had some mutations. It's starting to multiply like crazy. This is a cancer cell and a cancer cell because it had mutations are going to produce, it's going to produce some weird proteins. These cancer cells are going to produce some weird proteins. Every cell with a nucleus in your body and that's pretty much every cell except for red blood cells has MHC one complexes. The whole point of the MHC one complex is to bind two shady things that are produced inside of the cell, and then present them to the membrane. Even a malfunctioning cancer cell should be doing this. that are produced by the mutations inside of the nucleus and then it can present them. You could imagine what the appropriate immune response should be. These cancer cells should be killed and actually let me label this properly. That was MHC two, you're presenting an antigen that was found, those initially found outside of the cells engulfing and taken out. MHC one, it's binding to shady things inside the cell and then presenting it out. This thing should be killed. Now as you can imagine, what's going to kill it? Well that's where the cytotoxic T cell comes into the picture. The cytotoxic T cell is going to have, that's a receptor right there."
    },
    {
        "Q": "Around 6:23 when the Hydrogen breaks off was it in a hybridised sp2 orbital with the C? and when forming the double bond with the other C does it change to a p orbital?",
        "A": "Yes, at that point, the C atom is using an sp\u00c2\u00b2 orbital that is parallel to the vacant p orbital on the adjacent carbon. As the H leaves, the back lobe of the orbital becomes larger and the front lobe becomes smaller until, once the H atom is completely free, both lobes are the same size and you have a p orbital that contains two electrons. That p orbital overlaps with the adjacent empty p orbital to form a \u00cf\u0080 bond.",
        "video_name": "U9dGHwsewNk",
        "timestamps": [
            383
        ],
        "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": "Would the name of the new organic product formed at 8:39 be: 3-ethyl pentene?\nAlso, how would you indicate in the name where the double bond is?",
        "A": "You would name the compound 3-ethyl-pent-2-ene or 3-ethyl-2-pentene. The 2 indicating the location of the double bond.",
        "video_name": "U9dGHwsewNk",
        "timestamps": [
            519
        ],
        "3min_transcript": "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 off, and all the things like that. And we're going to see with E1, E2, SN1, and SN2, what kind of environments or reactants need to be there for each one of those to occur in different circumstances."
    },
    {
        "Q": "At 7:45 What if someone who gets blood on their hand washes their hands right after they got the blood on their hand? Would they still get ebola?",
        "A": "Not necessarily. The virus would need to enter the body through broken skin or mucous membranes, for example, if the blood on their hands got into a cut or came in contact with the eyes, nose, or mouth. Regular hand washing may remove the visible sign of blood, but the CDC recommends decontamination via an EPA-registered hospital disinfectant.",
        "video_name": "xYoQQCO15GE",
        "timestamps": [
            465
        ],
        "3min_transcript": "the world is kind of watching and intervening and helping to make sure that we get this under control. So, in addition to the supportive care the other big area that we kind of think about is how do you avoid transmitting the virus from one person to another. So let me just sketch in a second person here. So here I've drawn a second person and the question becomes how does Ebola get from the first person to the second person? How exactly does it move? And the quick answer is that it moves through bodily fluids. So bodily fluids include everything from blood to vomit and diarrhea or stool and these are kind of the big ones especially blood. Very, very infectious. But also stool and vomit because a lot of these patients are sick and bodily fluids can also be things like saliva and sweat. But again, these first few that I've kind of written out are the ones to really keep in mind, and blood in fact is an interesting one. and kind of illustrates the point where they showed that one drop. Just one drop of blood from a monkey that had Ebola was enough to carry with it a million little Ebola virus particles and that was thought to be enough to infect a million monkeys. So kind of the idea that one single drop could be enough to really decimate a population. So it kind of drives on the point that if there's a drop of blood, just a single drop or another bodily fluid that really gets over, let's say over into this person's hand and they can kind of go from there to if they touch their face. They can get entry into their body or through their nose or eyes or mouth or let's say they have a little cut. It could actually make its way into the blood through that cut. So, a few kind of interesting ways that bodily fluid can travel but really then it gets a ticket into the second person and the second person can get sick. with stool or vomit as well. Now another question that I think a lot of people ask is what about airborne? You know, can you breathe this stuff in? And that's a big question and the idea I think is literally is it kind of floating in the air? And the answer is no. You know, it's not really like other viruses like measles and some other ones where it's airborne. So let me just cross that out so it's very clear. It is not airborne. In fact, I think a lot of the confusion comes when you think about things like sneezes. So with a sneeze you can imagine like kind of you know, we've all seen those sneezes or had those sneezes where kind of a big droplet of goo or mucus kind of goes out, and you get this kind of shower, right, of droplets and this kind of right around where you're standing. Just right in your immediate vicinity and that's not gonna affect anyone that's kind of standing away from you, and you could imagine maybe in kind of a strange scenario. Let's say there's a person kind of hanging out right here"
    },
    {
        "Q": "At 6:00 what would 3.991 be rounded up to if it could only contain two significant figures? Would this result in 4.0? Or would it simply be 4?",
        "A": "4.0 since the number of significant digits are preferably preserved.",
        "video_name": "xHgPtFUbAeU",
        "timestamps": [
            360
        ],
        "3min_transcript": "Let's say we have another block, and this is the other block right over there. We have a, let's say we have an even more precise meter stick, which can measure to the nearest millimeter. And we get this to be 1.901 meters. So measuring to the nearest millimeter. And let's say those measurements were done a long time ago, and we don't have access to measure them any more, but someone says 'How tall is it if I were stack the blue block on the top of the red block - or the orange block, or whatever that color that is?\" So how high would this height be? Well, if you didn't care about significant figures or precision, you would just add them up. You'd add the 1.901 plus the 2.09. So let me add those up: so if you take 1.901 and add that to 2.09, 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,"
    },
    {
        "Q": "4:45 Sal said \"a mitochondria\" but the correct singular is \"a mitochondrion\", right?",
        "A": "Yes, that is correct.",
        "video_name": "J30zpvbmw7s",
        "timestamps": [
            285
        ],
        "3min_transcript": "from our NADH to the Oxygen, it would release a lot of energy but it would release so much energy that you wouldn't be able to capture most of it. You wouldn't be able to use it to actually do useful work, and so the process of Oxidative Phosphorylation is all about doing this at a series of steps and we do it by transferring these electrons from one electron acceptor to another electron acceptor, and every time we do that, we release some energy, and then that energy can be, in a more controlled way, be used to actually do work, and in this case, that work is pumping hydrogen protons across a membrane, and then that gradient that forms can actually be used to generate ATP, so let's talk through it a little bit more. So we're gonna go, these electrons, they're gonna be transferred, and I won't go into all of the details, this is to just give you a high-level overview of it. They're going to be transferred to different acceptors which then transfer it to another acceptor, so it might go to a Coenzyme, Cytochrome C, and it keeps going to different things, eventually getting to this state right over here, where those electrons can be accepted by the oxygen to actually form the water, and the process, every step of the way, energy is being released. Energy is being released, and this energy, as we will see in a second, is being used to pump hydrogen protons across a membrane, and we're gonna use that gradient to actually drive the production of ATP. So let's think about that a little bit more. So let's zoom in on, on a mitochondria. So this is mitochondria. Let's say that's our mitochondria, and let me draw the inner membrane and then, these folds in the inner membrane, the singular for them is crista. If we're talking about plurals, cristae. So we have these folds in the inner, So just to be clear, what's going on, this is the outer membrane, outer membrane. That is the inner membrane, inner membrane. The space between the outer and the inner membrane, the space right over here, that is the intermembrane space. Intermembrane, membrane space. And then the space inside the inner membrane, let me make that sure you can read that space properly, this space over here, this is the Matrix. This is the Matrix, and that is the location of our Citric acid Cycle or our Krebs Cycle, and I can symbolize that with this little cycle, we have a cycle going on here. And so that's where the bulk of the NADH is being produced. Now we also talked about some other coenzymes. In some books or classes, you might hear about FAD"
    },
    {
        "Q": "This isn't intuitive at all for me. If at 2:57 you do exactly what he said (index finger points in the direction of b and middle finger for a) with your right hand, i still get my thumb going into the screen, the same result as in the last video. How is he flexing his thumb to get it to point out of the page? I'm really confused.",
        "A": "I struggled with this too, but I think if you keep you index finger straight (as in fully extended and in line with your arm), then you should be able to figure it out.",
        "video_name": "o_puKe_lTKk",
        "timestamps": [
            177
        ],
        "3min_transcript": "Or when I drew it here, it would point into the page. So let's see what happens with b cross a, so I'm just switching the order. b cross a. Well, the magnitude is going to be the same thing, right? Because I'm still going to take the magnitude of b times the magnitude of a times the sine of the angle between them, which was pi over 6 radians and then times some unit vector n. But this is going to be the same. When I multiply scalar quantities, it doesn't matter what order I multiply them in, right? So this is still going to be 25, whatever my units might have been, times some vector n. And we still know that that vector n has to be perpendicular to both a and b, and now we have to figure out, well, is it, in being perpendicular, it can either kind of point into the page here or it could pop out of the page, or point out of the page. So which one is it? So what we do is we take our right hand. I'm actually using my right hand right now, although you can't see it, just to make sure I draw the right thing. So in this example, if I take my right hand, I take the index finger in the direction of b. I take my middle finger in the direction of a, so my middle figure is going to look something like that, right? And then I have two leftover fingers there. Then the thumb goes in the direction of the cross product, right? Because your thumb has a right angle right there. That's the right angle of your thumb. So in this example, that's the direction of a, this is the direction of b, and we're doing b cross a. That's why b gets your index finger. gets the second term, and the thumb gets the direction of the cross product. So in this example, the direction of the cross product is upwards. Or when we're drawing it in two dimensions right here, the cross product would actually pop out of the page for b cross a. So I'll draw it over. It would be the circle with the dot. Or if I were to draw it analogous to this, so this right here, that was a cross b. And then b cross a is the exact same magnitude, but it goes in the other direction. That's b cross a. It just flips in the opposite direction. And that's why you have to use your right hand, because you might know that, oh, something's going to pop in or out of the page, et cetera, et cetera, but you need to know your right hand to know whether it goes in or out of the page. Anyway, let's see if we can get a little bit more intuition of what this is all about because this is all about intuition."
    },
    {
        "Q": "What happens if you add to much force?\nFor example, from 4:28 on, what would happen to a 42N weight?",
        "A": "The weight would accelerate upwards until gravity brings it back down. It would jump , as you can try with any lever in real life.",
        "video_name": "DiBXxWBrV24",
        "timestamps": [
            268
        ],
        "3min_transcript": "This is getting monotonous. This is the output distance, from here to here. And let's figure out what the ratio has to be, for the ratio of the input distance to the output distance. Well, if we just divide both sides by 10, we get the distance input. It has to be 10 times the distance output, right? 100 divided by 10. So if the distance from the fulcrum to the weight is, I don't know, 5 meters, then the distance from where I'm applying the force to the fulcrum has to be 10 times that. It has to be 50 meters. So no matter what, the ratio of this length to this length has to be 10. And now what would happen? If I design this machine this way, I will be able to apply 10 newtons here, which is my maximum strength, 10 newtons downwards, and I will lift a 100 newton object. And now what's the trade off though? Nothing just pops out of thin air. The trade off is, is that I am going to have to push down for a much longer distance, for actually 10 times the distance as this object is going to move up. equal the work out. I can't through some magical machine-- and if you were able to invent one, you shouldn't watch this video and you should go build it and become a trillionaire-- but a machine can never generate work out of thin air. Or it can never generate energy out of thin air. That energy has to come from some place. Most machines actually you lose energy to friction or whatever else. But in this situation, if I'm putting in 10 newtons of force times some distance, whatever that quantity is of work, the work cannot change. The total work. It can go down if there is some friction in the system. So let's do another problem. And really they're all kind of the same formula. And then I'll move into a few other types of simple systems. I should use the line tool. We'll make this up on the fly. 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."
    },
    {
        "Q": "At 7:28, what does spontaneously mean?",
        "A": "With no additional energy / it will naturally just happen. This will also mean that delta G (change in free energy) is negative",
        "video_name": "g_snytB7iQ0",
        "timestamps": [
            448
        ],
        "3min_transcript": "indicating that a solid formed. This solid is called a precipitate. This solid spontaneously falls out of solution. This is the process of precipitation, which is the opposite of dissolution. In dissolution, we put a solid into water and we formed ions, right? In precipitation, the ions come together to form a solid, and that solid spontaneously falls out of solution. Silver chloride is our precipitate. We would still have some ions in solution. We would still have sodium cations and nitrate anions. In here, we would have sodium cations in solution and nitrate anions in solution. We could add them into here. We could say, NaNo3, aqueous, meaning those ions are present in solution. Let's look in more detail about what's happening We know that we had silver cations in solution. Here's our silver cation over here. We know that this ion is being surrounded by water molecules in the process of hydration, right? So these oxygens are partially negative right here. Since opposites attract, there's a force that's holding that ion in our solution, the forces of hydration. Same thing for the chloride anion, right? The partial positive charge on the hydrogen, opposite charges attract, right? So those water molecules are stabilizing the chloride anion in solution. But when we pour those two solutions together, we form our precipitate. We form silver chloride. We form this ionic crystal down here. Once again, opposite charges attract. So the positively-charged silver cation chloride anion here. Since we notice this solid to form, the electrostatic attractions of our ionic crystal must be stronger than the forces of hydration. This chloride anion would move into here, and then this silver cation would move into here. So the ions come out of solution and a precipitate spontaneously forms. We form our solid, silver chloride. This is one way to represent what's going on here. We could have also drawn out all of the ions, right? Instead of writing that, another way to represent what's happening would be to say, a solution of sodium chloride would be sodium cations in solution, so we write our sodium cations here, chloride anions, so Cl-. We added to that our solution of silver nitrate, which had silver ions in solution,"
    },
    {
        "Q": "At 3:20, would the equation NaCl (s) --H2O--> Na+ (aq) + Cl- (aq) the same as saying NaCl (s) --H2O--> NaCl (aq) ? Or are they structurally different in solution?",
        "A": "They mean the same thing, it is just that one is more specific than the other. NaCl (aq) requires knowing that this compound dissociates in aqueous solution. The other version spells that out explicitly.",
        "video_name": "g_snytB7iQ0",
        "timestamps": [
            200
        ],
        "3min_transcript": "is going to interact with the positive charge on the sodium. Right? Opposite charges attract. We have one water molecule here attracted to this sodium cation, and this water molecule would do the same, right? So partial negative charge attracted to the positive charge on the sodium. So the water molecules are going to pull off the sodium cations and eventually give you this situation over here. We have the partial negative oxygens, right? Partial negative oxygens are going to interact with the sodium cation, right? So the water is a dipole and the sodium cation is an ion. So we could call this an ion-dipole interaction. The water molecules break the ionic bonds, pull off the sodium cations, surround the sodium cation. We call this process hydration. where the ion is surrounded and stabilized by a shell of our solvent molecules. The same thing would happen with the chloride anions. Right up here, the chloride anion is negatively charged. But this time, the negative charge would be attracted to the positive part of our polar molecule, right? The oxygen is partially negative and this hydrogen here would be partially positive. Opposite charges attract, the positive charge is going to interact with the negative charge. Same for this molecule of water, partially positive hydrogen. Once again, this interaction is going to pull off that chloride anion and move it into solution. So we have our partial positive hydrogens interacting with our negatively-charged chloride anions here. Once again, we get ion-dipole interactions. The chloride anion is surrounded by our water molecules, The end result is, each sodium cation is surrounded by water molecules and each chloride anion is surrounded by water molecules. The sodium chloride has dissolved in water. We formed a solution, an aqueous solution, of sodium chloride. The way that you see that written, you would write, here we have solid sodium chloride, which we put into water, our solvent, and the water molecules surrounded our ions. Now we have sodium ions in solution, so we write an aq here for aqueous sodium ions, and we have chloride anions also in solution, so we write an aq here. So we have an aqueous solution of sodium chloride. The sodium chloride has dissolved in water. Now let's say we have one beaker that contains a solution of NaCl, so an aqueous solution of NaCl has sodium cations in solution"
    },
    {
        "Q": "When you say in 00:40ish \"the formula we've been speaking about\".. which video is this you are referring to? Thanks.",
        "A": "That formula is explained in the previous video, Putting it all together: Pressure, flow, and resistance.",
        "video_name": "fy_muPF0390",
        "timestamps": [
            40
        ],
        "3min_transcript": "Let's imagine that I'm here. And this is me at the age-- actually, I'll do this right now-- so let's say at the age of 31. And then I go into the future. So this is a line for the present. And then I go into the future. And this is me again. And this is now at the age of 71. So I've got 40 extra years. I've even picked up a cane to help me walk around. And finally, I go even further in the future, and I'm being very optimistic. And I'm going to go ahead and say this is at the age of 101. I live to be a centennial. And I'm sitting in a wheelchair, and I'm going to wave out at you. So that's me in a wheelchair at the age of 101. And we're going to apply that formula we've been talking about so much, delta P equals Q times R. So I said Q is cardiac output. And we've got R is resistance. So we've got this formula. And I go today, and I go in 40 years, and I go again when I'm 101. And today, they tell me my blood pressure is 120 over 80. And actually, I went not too long ago, and that's pretty much what they told me it was. And I go in the future, and in 40 years, they tell me it's actually gone up. My blood pressure is now 150 over 90. And in fact, I go again, when I'm 101, and they say it's 180 over, let's say, 105. So the blood pressure is rising, and that's basically what I'm told. And they say, well, you've got to make sure you eat well and exercise, and that should help your blood pressure. And I'm left wondering what the connection is between the two. So let's figure out what that connection is exactly. So my blood pressure, I just said, is 120 over 80. And if I want to figure out my mean arterial pressure, meaning the average pressure in my arteries, to give me a good guess as to what it's going to be. So I know that I spend about 1/3-- my mean arterial pressure is going to equal 1/3 times my systolic, because I know I spend about 1/3 of the time-- my heart does anyway-- spends 1/3 of its time beating. And it spends 2/3 of its time relaxing. And the relaxing pressure is the diastolic pressure, that's 80. And so that works out to about 95. And so that's how I came up with that 95 number. And that's also why it's not exactly 100, which is what you'd think an average would be between two numbers. But it's because we don't spend the exact same amount of time in systole as diastole. So then if I wanted to figure out here, it would be 1/3 times 150 plus 2/3 times 90. And that works out to 110. And if I wanted to do it at the age of 101, my mean arterial pressure would be 1/3 times 180 plus 2/3 times"
    },
    {
        "Q": "He only mentions \"BB\" as homozygous dominant (@11:10), but I am under the assumption that \"bb\" would be called homozygous recessive. Is this correct?",
        "A": "yes you right",
        "video_name": "eEUvRrhmcxM",
        "timestamps": [
            670
        ],
        "3min_transcript": "just said a brown-eyed gene, but what I should say is the brown-eyed version of the gene, which is the brown allele, or the blue-eyed version of the gene from my mom, which is the blue allele. Since the brown allele is dominant-- I wrote that up here --what's going to be expressed are brown eyes. Now, let's say I had it the other way. Let's say I got a blue-eyed allele from my dad and I get a brown-eyed allele for my mom. Same thing. The phenotype is going to be brown eyes. Now, what if I get a brown-eyed allele from both my mom and my dad? Let me see, I keep changing the shade of brown, but they're all supposed to be the same. So let's say I get two dominant brown-eyed alleles from my mom and my dad. Then what are you going to see? I'm still going to see brown eyes. So there's only one last combination because these are the only two types of alleles we might see in our population, although for most genes, there's more than two types. For example, there's blood types. There's four types of blood. But let's say that I get two blue, one blue allele from each of my parents, one from my dad, one from my mom. Then all of a sudden, this is a recessive trait, but there's nothing to dominate it. So, all of a sudden, the phenotype will be blue eyes. And I want to repeat again, this isn't necessarily how the alleles for eye color work, but it's a nice simplification to maybe understand how heredity works. There are some traits that can be studied in this simple way. But what I wanted to do here is to show you that many different genotypes-- so these are all different genotypes --they all coded for the same phenotype. didn't know exactly whether they were homozygous dominant-- this would be homozygous dominant --or whether they were heterozygotes. This is heterozygous right here. These two right here are heterozygotes. These are also sometimes called hybrids, but the word hybrid is kind of overloaded. It's used a lot, but in this context, it means that you got different versions of the allele for that gene. So let's think a little bit about what's actually happening when my mom and my dad reproduced. Well, let's think of a couple of different scenarios. Let's say that they're both hybrids. My dad has the brown-eyed dominant allele and he also"
    },
    {
        "Q": "At 3:15, Two oxygens that don't have double bond have 7 electrons? doesn't it need to have 6? so the formal charge is 1?",
        "A": "It has 7 electrons because of the one shared by Nitrogen. The formal charge is 1 because it has 7. Had it only 6, the formal charge would be zero.",
        "video_name": "bUCu7bPkZeI",
        "timestamps": [
            195
        ],
        "3min_transcript": "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, 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"
    },
    {
        "Q": "So at 6:59 when the germ cells go through Meiosis, to make the gametes, does that mean the germ cells are haploid?",
        "A": "first they are are diploid then after meiosis they become haploid",
        "video_name": "PvoigrzODdE",
        "timestamps": [
            419
        ],
        "3min_transcript": "These cells here eventually differentiate into the cells into the lungs, and obviously at this scale, the cells are way too small to even see. These cells differentiate into the cells of the heart. Now, I want to draw an important distinction here. Because most of the cells that I've just depicted here that are just a product of mitosis, these are your, I guess you could say these are your body cells, or these are your somatic cells. So these, all of these cells that I'm pointing out in your heart, your lung, your brain, these are somatic cells, or body cells. Somatic cells. And so you're probably wondering, well how do I eventually get these haploid number cells? How do I eventually get, if I'm talking about a male, how do I eventually get these haploid sex cells, these gametes, these sperm cells? I'm talking about a female, how do I eventually get these ova, these egg cells that have a haploid number? And the way that happens is some of these cells So they're going to differentiate into germ cells. In the case of, and they're going to differentiate when I say into germ cells, they're going to differentiate into your gonads. In the case of a female, the gonads are the ovaries. And in the case of a male, the gonads are the testes. The gonads are the testes. And the germ cells in the gonads or the cells that have differentiated into being part of the testes and ovaries, those germ cells. So we differentiate them from somatic cells. So there are germ cells there. Germ cells in your ovaries and testes. They, through the process of meiosis, they through the process of meiosis, 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,"
    },
    {
        "Q": "At 2:34 What does accreting mean?",
        "A": "accretion is the gradual gathering of matter.",
        "video_name": "VbNXh0GaLYo",
        "timestamps": [
            154
        ],
        "3min_transcript": "imagine is when you have the shockwave traveling out from a supernova, let's say you had a cloud of molecules, a cloud of gas, that before the shockwave came by just wasn't dense enough for gravity to take over, and for it to accrete, essentially, into a solar system. When the shockwave passes by it compresses all of this gas and all of this material and all of these molecules, so it now does have that critical density to form, to accrete into a star and a solar system. We think that's what's happened, and the reason why we feel pretty strongly that it must've been caused by a supernova is that the only way that the really heavy elements can form, or the only way we know that they can form is in kind of the heat of a supernova, and our uranium, the uranium that seems to be in our solar system on Earth, seems to have formed four and a half billion years ago, and we'll talk in a little bit more depth in future videos on exactly how people figure that out, but since the uranium seems about the same age as our solar system, it must've been formed at around the same time, and it must've been formed by a supernova, and it must be coming from a supernova, so a supernova shockwave must've passed through our part of the universe, and that's a good reason for gas to get compressed and begin to accrete. So you fast-forward a few million years. That gas would've accreted into something like this. It would've reached the critical temperature, critical density and pressure at the center for ignition to occur, for fusion to start to happen, for hydrogen to start fusing into helium, and this right here is our early sun. Around the sun you have all of the gases and particles and molecules that had enough angular velocity to not fall into the sun, to go into orbit around the sun. 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."
    },
    {
        "Q": "At 4:08 you say that the moon was a part of the earth ere it was hit by a proto planet named theia so if the earth and the moon is made up of same type of stuff why there is neither gravity nor life on the moon ?",
        "A": "The Moon does have gravity. It cannot support life because it isn t composed in the same way as the Earth. It is smaller, mostly composed of lighter elements like the Earth s crust, and lacks a significantly active core to generate any substantial magnetic field. As a result, it isn t able to maintain an atmosphere or liquid water. Thus life isn t really able to even get started on the Moon.",
        "video_name": "VbNXh0GaLYo",
        "timestamps": [
            248
        ],
        "3min_transcript": "four and a half billion years ago, and we'll talk in a little bit more depth in future videos on exactly how people figure that out, but since the uranium seems about the same age as our solar system, it must've been formed at around the same time, and it must've been formed by a supernova, and it must be coming from a supernova, so a supernova shockwave must've passed through our part of the universe, and that's a good reason for gas to get compressed and begin to accrete. So you fast-forward a few million years. That gas would've accreted into something like this. It would've reached the critical temperature, critical density and pressure at the center for ignition to occur, for fusion to start to happen, for hydrogen to start fusing into helium, and this right here is our early sun. Around the sun you have all of the gases and particles and molecules that had enough angular velocity to not fall into the sun, to go into orbit around the sun. 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"
    },
    {
        "Q": "at 2:35 sal says \"a few million years ago\" I Dont think thats right",
        "A": "2:42 corrects that",
        "video_name": "VbNXh0GaLYo",
        "timestamps": [
            155
        ],
        "3min_transcript": "imagine is when you have the shockwave traveling out from a supernova, let's say you had a cloud of molecules, a cloud of gas, that before the shockwave came by just wasn't dense enough for gravity to take over, and for it to accrete, essentially, into a solar system. When the shockwave passes by it compresses all of this gas and all of this material and all of these molecules, so it now does have that critical density to form, to accrete into a star and a solar system. We think that's what's happened, and the reason why we feel pretty strongly that it must've been caused by a supernova is that the only way that the really heavy elements can form, or the only way we know that they can form is in kind of the heat of a supernova, and our uranium, the uranium that seems to be in our solar system on Earth, seems to have formed four and a half billion years ago, and we'll talk in a little bit more depth in future videos on exactly how people figure that out, but since the uranium seems about the same age as our solar system, it must've been formed at around the same time, and it must've been formed by a supernova, and it must be coming from a supernova, so a supernova shockwave must've passed through our part of the universe, and that's a good reason for gas to get compressed and begin to accrete. So you fast-forward a few million years. That gas would've accreted into something like this. It would've reached the critical temperature, critical density and pressure at the center for ignition to occur, for fusion to start to happen, for hydrogen to start fusing into helium, and this right here is our early sun. Around the sun you have all of the gases and particles and molecules that had enough angular velocity to not fall into the sun, to go into orbit around the sun. 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."
    },
    {
        "Q": "At 6:20 should the empirical formula for water be HHO instead of H2O?",
        "A": "No H20 is the empirical formula on its own.",
        "video_name": "bmjg7lq4m4o",
        "timestamps": [
            380
        ],
        "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": "3:50 what's the meaningo of those double bonds ?",
        "A": "A double bond is where there are four electrons shared between two atoms. You will learn more about these in future videos.",
        "video_name": "bmjg7lq4m4o",
        "timestamps": [
            230
        ],
        "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 4:53, why do you have to draw a structural formula of a carbon with 3 single lines, and 3 other double lines bonding with each other?",
        "A": "Because that simply is what the structure of benzene is. Google it and you ll find many pictures that look just like that. You cannot give each carbon an octet of electrons without making those double bonds.",
        "video_name": "bmjg7lq4m4o",
        "timestamps": [
            293
        ],
        "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": "At 3:50, why is there a \"double bond\" on the hexagon of carbon atoms? What is a double bond? When should we use double bonds?",
        "A": "That molecule is benzene, it has the formula C6H6. The only way for each carbon to have 8 electrons around it is to form double bonds. Double bonds simply are just another bond between the atoms. As for when to use them the dot structure videos go over this. They re often needed to make elements follow the octet rule or to minimise formal charge.",
        "video_name": "bmjg7lq4m4o",
        "timestamps": [
            230
        ],
        "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 3:45 Sal draws the Benzene molecule, mentioning that every other atom is joined by a double bond, while the rest are single. What is the significance of this and how do you know what kind of bond (single, double, etc) it is?",
        "A": "Benzene (a carcinogen) has 2 possible formations. Going clockwise you can start with a double bond, then a single bond, etc. Or you can do the same thing but go anti-clockwise. the 2 structures are very similar. So instead of being one or the other it is actually somewhere between the 2. Each bond is not single or double but 1.5. The 1/2 is normally shown as a circle inside the carbon hexagon.",
        "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": "At 4:28, there is a structure of benzene shown. What I wanted to know is in the formula are the hydrogens connected to the carbons connected to each other?",
        "A": "Hydrogen is only capable of making a single bond, so all of the hydrogen atoms in benzene are bound only to a single atom of carbon.",
        "video_name": "bmjg7lq4m4o",
        "timestamps": [
            268
        ],
        "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 3:35, Sal draws a structural formula, but some covalent bonds have two lines. What do two lines mean in structural formulas?",
        "A": "It means a double bond. It s exactly how it sounds, 2 bonds between the two atoms instead of 1.",
        "video_name": "bmjg7lq4m4o",
        "timestamps": [
            215
        ],
        "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": "What happens to / with the positive electron from the battery? 3:51 \"gonna leave a net positive charge\"... Can you explain this? Thanks",
        "A": "the copper wire is neutral when no battery is attached when there is a + from the battery, Cu- will be attracted then the area will have less - , thus have a net + charge hope this solve your problem",
        "video_name": "ZRLXDiiUv8Q",
        "timestamps": [
            231
        ],
        "3min_transcript": "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. There's our little battery. This is the minus sign, this is the plus side. And we'll hook up a battery to this. What's going on in here? Inside this copper is a whole bunch of electrons that are associated with atoms. It's a neutral piece of metal. There's the same number of protons as there is electrons. But these electrons are a little bit loose. So if I put a plus over there, that's this situation right here, where a plus is attracting a minus. So an electron is gonna sort of wander over this way and go like that. And that's gonna leave a net positive charge in this region. So these electrons are all gonna start moving in this direction. And down at the end, travel in here, and it's gonna go in there and make up the difference. So if I had a net positive charge here from the electrons leaving and going to the left, this battery would fill those in. And I'm gonna get a net movement of charge, of negative charge, around in this direction, like this. The question is, how do I measure that? How do I measure or give a number to that amount of stuff that's going on? So we wanna quantify that, we wanna assign a number to the amount of current happening here. What we do is, in our heads, we put a boundary across here. So just make that up in your head. And it cuts all the way through the copper. And what we know, we're gonna stand right here. We're gonna keep our eye right on this boundary down in here. As we watch, what we're gonna do is, we're gonna count the number of electrons"
    },
    {
        "Q": "Why is it -q instead of +q in the copper wire at 5:03 because the net charge is positive since there will be more protons than electrons?",
        "A": "In metals, its the electrons that do the moving. Electrons have a negative charge, so they are labeled -q.",
        "video_name": "ZRLXDiiUv8Q",
        "timestamps": [
            303
        ],
        "3min_transcript": "There's our little battery. This is the minus sign, this is the plus side. And we'll hook up a battery to this. What's going on in here? Inside this copper is a whole bunch of electrons that are associated with atoms. It's a neutral piece of metal. There's the same number of protons as there is electrons. But these electrons are a little bit loose. So if I put a plus over there, that's this situation right here, where a plus is attracting a minus. So an electron is gonna sort of wander over this way and go like that. And that's gonna leave a net positive charge in this region. So these electrons are all gonna start moving in this direction. And down at the end, travel in here, and it's gonna go in there and make up the difference. So if I had a net positive charge here from the electrons leaving and going to the left, this battery would fill those in. And I'm gonna get a net movement of charge, of negative charge, around in this direction, like this. The question is, how do I measure that? How do I measure or give a number to that amount of stuff that's going on? So we wanna quantify that, we wanna assign a number to the amount of current happening here. What we do is, in our heads, we put a boundary across here. So just make that up in your head. And it cuts all the way through the copper. And what we know, we're gonna stand right here. We're gonna keep our eye right on this boundary down in here. As we watch, what we're gonna do is, we're gonna count the number of electrons and we're gonna have a stop watch and we're gonna time that. So we're gonna get, basically, this is charge, it's negative charge, and it's moving to the side. What we're gonna do is, at one little spot right here, we're just gonna count the number that go by in one second. So we're gonna get charge per second. It's gonna be a negative charge moving by. That's what we call current. It's the same as water flowing by in a river. That's the same idea. Now I'm gonna set up a different situation that also produces a current. And this time, we're gonna do it with water, water and salt. Let's build a tube of salt, of salt water, like this. We're gonna pretend this is some tube that's all full of water. I'm also gonna put a battery here. Let's put another battery."
    },
    {
        "Q": "At 6:15 in the video it discusses the difficulty of eating and digesting cellulose. Are cells in fruit not composed of cellulose, or does the plant do something magically different in the process of fruiting?",
        "A": "Fruit, being made of plant cells, definitely does contain cellulose. We can t digest it, but there are so many other good things in fruit (sugars, nutrients, etc.), that we eat it anyways. Cellulose is the fiber part that is advertised as being so healthy for your digestive system, by virtue of the fact that it can t be digested. Tree branches, on the other hand, are basically pure cellulose, with nothing else, so are noticeably less appetizing.",
        "video_name": "d9GkH4vpK3w",
        "timestamps": [
            375
        ],
        "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": "@8:33 when the compound reacts with H2 and Pt, since it goes from an alkyne all the way to an alkane, does the same reaction happen twice? syn addition twice with the reagents?",
        "A": "Yes, it is effectively the same reaction happening twice.",
        "video_name": "RdFfIEDxo18",
        "timestamps": [
            513
        ],
        "3min_transcript": "OK, so carbon triple bonded to another carbon, and we'll put a methyl group on each side like that. OK, so let's do a few different reactions with the same substrate here. So our first reaction will just be a normal hydrogenation with hydrogen gas, and let's use platinum as our catalyst. So this is not a poison catalyst. This is a normal catalyst. So what's going to happen is, first you're going to reduce the alkyne to an alkene. And then since there's no way of stopping it, it's going to reduce the alkene to an alkane. So this is going to reduce the alkyne all the way to an alkane. So if we go back up here to beginning, remember, we said that a poison catalyst will stop at the alkene, but if it's not a poison catalyst, down here. So this reaction is going to produce an alkane. Let's go ahead and draw the product. So we know that there are four carbons in my starting materials. There's going to be four carbons when I'm done here, so these two carbons in the center here are going to turn into CH2's. And then on either side, we still have our CH3's. So this is going to form butane as the product. All right, this time let's use a hydrogen gas, and let's use a Lindlar palladium here. This is our poisoned catalyst. So it's going to reduce our alkyne to an alkene, and then it's going to stop. And you have to think, what kind of alkene will you get? You will get a cis-alkene. So if we draw our two hydrogens adding on to the same sides, So our methyl groups will be going-- this and this would be our product, a cis-alkene. All right, let's do one more, same starting material. So this one right here, except this time we're going to add sodium, and we're going to use ammonia as our solvent. And remember, this will reduce our alkyne to an alkene, but it will form a trans-alkene as your product. So when you're drawing your product down here, you want to make sure that your two hydrogens are trans to each other. So they add on the mechanism, and then your two methyl groups would also be on the opposite side like that. So look very closely as to what you are reacting things with. Is it a normal hydrogenation reaction? Is it a hydrogenation reaction with a poison catalyst, which would form a cis-alkene? Or is it reduction with sodium and ammonia, which will give you a trans-alkene."
    },
    {
        "Q": "at 5:56 you get another Na with one electron, how do you know there are plenty left over after using Na with one electron already in the first step when u started off the reaction? Same with ammonia in the next step following 5:56?",
        "A": "There are billions and billions of Na atoms and NH\u00e2\u0082\u0083 molecules in the reaction mixture. The Na and ammonia are always in excess. The alkyne is the limiting reactant.",
        "video_name": "RdFfIEDxo18",
        "timestamps": [
            356,
            356
        ],
        "3min_transcript": "These electrons are going to repel, and they're going to want to try to be as far away from each other as they possibly can. So what's going to happen is, we have our two carbons right here. And let's say that these two electrons stay over here on this side. This one electron's going to go over to the opposite side. They're going to try to get as far away from each other as they possibly can. And same thing with these R group here, right? So this R group is going to try to get as far away from this R-prime group as it possibly can. So this trans conformation is the more stable one. So this is our negatively charged carbanion right here. So in the next step of the mechanism, we remember ammonia is present. So let's go ahead and draw an ammonia molecule floating around like that. So here is our ammonia molecule. And the carbanion is going to act as a base, and it's going to take a proton from the ammonia molecule. going to form a new bond with this proton, and these electrons are going to kick off onto the nitrogen. So let's go ahead and draw the results of that acid-base reaction. So now we have our two carbons, with an R group right here, R-prime right here. And now this carbon on the right is bonded to a proton. It bonded to a hydrogen like that. And then we still have our radical down here, so there is one electron on that carbon as well. All right, so the next step of our mechanism? Well, there's plenty of sodium present. So here's a sodium atom with one valence electron. The sodium is going to donate this electron to this carbon. So just use a half-headed arrow to show the movement of one electron. So if that sodium atom donates that one valence electron to that carbon, let's go ahead and draw the results of that. So we have two carbons double bonded, an R group over here, And this carbon had one electron around it. It just picked up one more from a sodium atom. So it's like that, which would give it a negative 1 formal charge. So this carbon has a negative 1 formal charge. So let's go ahead and draw that negative 1 formal charge. It's a carbanion. And once again, ammonia is floating around, so let's go ahead and draw ammonia right here, so NH3 like that. And the same thing is going to happen as did before, right? The negative charge is going to grab a proton. It's going to act as a base. And these electrons are going to kick off onto the nitrogen here. And so we protonate our carbanion, and we have completed our mechanism because now we have our two R groups across from each other. And we added on two hydrogens across from each other as well like that. So we formed a trans-alkene. All right, so that's the mechanism to form a trans-alkene. Let's look at a few examples."
    },
    {
        "Q": "At 9:50 in the video, 3-hexanol is explained to have a dipole-dipole moment between the O and the H, resulting in hydrogen bonding. My question is this: would there also be a dipole-dipole moment between the O and the C in 3-hexanol, and would that dipole-dipole moment be the same strength as that in 3-hexanone?",
        "A": "I agree there must be some polarization between the oxygen and the carbon in the alcohol, but I don t think it would be as strong as in the ketone. In the alcohol the oxygen is pulling electron density from both the hydrogen and the carbon, which is more electronegative than the hydrogen so the electron density shift is mostly away from hydrogen. In contrast, in the ketone the oxygen is pulling electron density exclusively from the carbon.",
        "video_name": "pILGRZ0nT4o",
        "timestamps": [
            590
        ],
        "3min_transcript": "So we have a dipole for this molecule, and we have the same dipole for this molecule of 3-hexanone down here. Partially negative oxygen, partially positive carbon. And since opposites attract, the partially negative oxygen is attracted to the partially positive carbon on the other molecule of 3-hexanone. And so, what intermolecular force is that? We have dipoles interacting with dipoles. So this would be a dipole-dipole interaction. So let me write that down here. So we're talk about a dipole-dipole interaction. Obviously, London dispersion forces would also be present, right? So if we think about this area over here, you could think about London dispersion forces. But dipole-dipole is a stronger intermolecular force compared to London dispersion forces. And therefore, the two molecules here of 3-hexanone are attracted to each other more than the two molecules of hexane. for these molecules to pull apart from each other. And that's why you see the higher temperature for the boiling point. 3-hexanone has a much higher boiling point than hexane. And that's because dipole-dipole interactions, right, are a stronger intermolecular force compared to London dispersion forces. And finally, we have 3-hexanol over here on the right, which also has six carbons. One, two, three, four, five, six. So we're still dealing with six carbons. If I draw in another molecule of 3-hexanol, let me do that up here. So we sketch in the six carbons, and then have our oxygen here, and then the hydrogen, like that. We know that there's opportunity for hydrogen bonding. Oxygen is more electronegative than hydrogen, so the oxygen is partially negative and the hydrogen is partially positive. The same setup over here on this other molecule of 3-hexanol. So partially negative oxygen, partially positive hydrogen. Let me draw that in. So we have a hydrogen bond right here. So there's opportunities for hydrogen bonding between two molecules of 3-hexanol. So let me use, let me use deep blue for that. So now we're talking about hydrogen bonding. And we know that hydrogen bonding, we know the hydrogen bonding is really just a stronger dipole-dipole interaction. So hydrogen bonding is our strongest intermolecular force. And so we have an increased attractive force holding these two molecules of 3-hexanol together. And so therefore, it takes even more energy for these molecules to pull apart from each other. And that's reflected in the higher boiling point for 3-hexanol, right? 3-hexanol has a higher boiling point than 3-hexanone and also more than hexane. So when you're trying to figure out boiling points, think about the intermolecular forces that are present between two molecules. And that will allow you to figure out"
    },
    {
        "Q": "around 0:34 sal shows a solar flare. what would be the effects if one of these were powerful enough to get to earth?",
        "A": "If a powerful solar eruption were to be aimed at earth, humans themselves would not be affected. But it would have a great affect on Earth s electrical grid. A shock like that could short out anything that isn t shielded and it could be years before the grid is fully restored. Such an event has happened in the past around the Civil war where telegraph systems stopped working.",
        "video_name": "jEeJkkMXt6c",
        "timestamps": [
            34
        ],
        "3min_transcript": ""
    },
    {
        "Q": "actually gravitational force is independent of mass right and in the video i saw tha sal said at4:57 as mass of the brick is larger force on the brick is also larger am i right in what i have asked",
        "A": "No, gravitational force is not independent of mass. That s quite a strange idea. Gravitational force is directly proportional to mass.",
        "video_name": "36Rym2q4H94",
        "timestamps": [
            297
        ],
        "3min_transcript": "this height and the center of the moon squared so they both have this exact expression on it So let's replace that expression, let's just call that the gravitational field on the moon define any number of the mass will tell you the weight of that object on the moon, or the gravitational force acting downard on that object on moon so this is the gravitational field on the moon it's called g sub m, and all of this is all of this quantity combined so we simplified that way, the force on the brick due to the moon is going to be equal to lower case g on the moon normally we will use this lower case g for the gravitational constant on earth with the gravitational field on earth or sometime the gravitational acceleration on earth but now it's refering to the moon so that's why the lower case subscript m it's equal to that times the mass of the brick for the case of the feather, the force on the feather is equal to all is equal to g sub m times the mass of the feather so now assuming that the mass of the brick is much greater than the mass of the feather, so let's going to assume, a reasonable thing to assume that the mass of the brick is greater than the mass of the feather, than the mass of the feather what's going to be their relative forces? Here you have the greater mass times the same quantity you have the smaller mass times the same quantity so the mass of the brick is greater than the mass of the feather completely reasonable to say that the force of gravity on the brick is going to be greater than the force of gravity on the feather so if you do all of this so every thing we've done to this point is correct, you might say hey it's gonna be more force due to gravity one the brick, and that's why the brick will be accelerated down that's being gravitational force on this brick but it also has greater mass and we remember the larger the mass is, the less acceleration it will experience for a given force So what really determine how quickly either of these things fall is their accelerations and let's figure out their accelerations So we knew that, I will do this on a neutral color we know that force is equal to mass time acceleration So if we wanna figure out the acceleration of the brick we could write it the other way the acceleration, if we divide both side by mass we get acceleration is equal to force divided by mass acceleration is vector quantity and force is also vector quantity and in this situation we will use, I'll use, we're not using any"
    },
    {
        "Q": "At about 9:00-9:20 Sal says Earth goes into whats called \"Snowball Earth\" and ices over. Is that like an Ice Age, and if not, how are Snowball Earth and an Ice Age different?",
        "A": "Snowball Earth is a period in earth s history where the whole surface of the earth was supposedly covered by ice. Snowball Earth periods are extreme ice ages. During most ice ages there was only glaciation in latitudes up to 40\u00c2\u00b0 or 50\u00c2\u00b0.",
        "video_name": "E1P79uFLCMc",
        "timestamps": [
            540,
            560
        ],
        "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": "I think there's a mistake at 7:00 minutes, how did he get 4. 75m/s?\nI've been calculating it over and over it it keeps appearing to be 3.7m/s.\nWhat's going on?",
        "A": "Try changing your calculator to degrees from radians, is that your problem?",
        "video_name": "_0nDUXO0k7o",
        "timestamps": [
            420
        ],
        "3min_transcript": "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 Remember if you're going to then go try to find out what one of these internal forces are, we neglected them because we treated this as a single mass. But you could ask the question, what is the size of this tension? Often that's like a part two because we might want to know what the tension is in this problem, if we do that now we can look at the 9 kg mass individually so I can say for just the 9 kg mass alone, what is the tension on it and what are the force? We can find the forces on it simply by saying the acceleration of the 9 kg mass is the net force on the 9 kg mass divided by the mass of the 9 kg mass. Now this is just for the 9 kg mass since I'm done treating this as a system. This trick of treating this two-mass system as a single object is just a way to quickly get the magnitude of the acceleration. Now that I have that and I want to find an internal force I'm looking at just this 9 kg box."
    },
    {
        "Q": "at 1:09 the term \"short\" is mentioned, and then again a few seconds later.\n\nplease may I know what is a short?",
        "A": "The term short in this context means to short-out; it basically occurs when a conductive element comes into contact with the circuit board, allowing voltage to bypass all resistors and capacitors. When this happens, this is bad. The voltage will rush through the element instead of the board with almost no resistance and destroy many, if not all, components. The solenoid safe-guards from this issue.",
        "video_name": "gFFvaLzhYew",
        "timestamps": [
            69
        ],
        "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": "at 9:40, would that tetra molecule be cis or trans?",
        "A": "That would be neither. It has the same group on all 4 positions.",
        "video_name": "AiGGaJfoQ1Y",
        "timestamps": [
            580
        ],
        "3min_transcript": "one more product for this reaction, and let's go back to our secondary carbocation over here. So, if it doesn't rearrange, you could actually take a proton from this secondary carbocation and form our last product. Let's think about the carbon in red right here. So we're looking at the carbons next door, so the carbon in red. Well, the carbon in magenta to the left doesn't have any proton, so we can't take a proton from that one, but the carbon to the right in magenta does. So if I squeeze in a hydrogen in here, finally, our base could come along and take this proton. And if that happen, then these electrons would move into here, so that gives us our final product. Just let me draw in our skeleton here, and then our double bond will form right... Actually, let me just redraw that double bond, so thing we're getting... We just redraw the whole thing. I think I have a little bit more space that I thought I did. our double bonded form here, and then we have our methyl groups coming off of that carbon. So the electrons in light blue moved into form our double bond. So we'd formed three products, three products from this E1 reaction. If we think about which one would be the major product, let's look at degrees of substitution. So let's go over to the alkene on the right. So when we think about the two carbons across our double bonds, there are one, two, three, four alkyl groups. So this is a tetrasubstituted alkene, so this should be the major product. This is the most stable alkene. Next, let's look at this one. So here's a carbon, and here's a carbon across our double bond. We can see this time we have only two alkyl groups. So this is a disubstituted alkene. And then finally, over here on the left, this would be a monosubstituted alkene. We have one alkyl group, so a monosubstituted alkene. And this one came from the secondary carbocation from no rearrangement, so this one is not gonna be a major product. Only a very small percentage of your products would be this monosubstituted alkene. Most of your products is gonna be your di and your tetrasubstituted alkene, So with your tetrasubstituted alkene being your major product since it is the most stable."
    },
    {
        "Q": "at 1:53 cant the compound be named as 3-Bromo-4,6-dimethylheptane if the functional group is given the first preference ?",
        "A": "It is not the functional group, but the rule of lowest numbers, that determines the numbering. In your name, the lowest number is 3. In the other name, the lowest number is 2. The lowest number wins. But the bromo wins in determining the order in which the substituents are listed in the name. So the compound really is 5-bromo-2,4-dimethylheptane.",
        "video_name": "aaZ-isZs4ko",
        "timestamps": [
            113
        ],
        "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 2:48 why does it lose both hydrogen protons.",
        "A": "Well..oxalic acid is a diprotic/dibasic acid, which means that it has two replaceable H atoms. Oxalic acid when it ionizes loses or liberates 2 hydrogen protons. It does this quite readily as the oxalate anion (acid with 2 H+ ions removed) is stabilized by -I effect and resonance.",
        "video_name": "XjFNmfLv9_Q",
        "timestamps": [
            168
        ],
        "3min_transcript": "We have that on the other carbon as well. This right here is what oxalic acid is. And it's an interesting acid, because it can actually donate This proton can be nabbed off, and this proton can also be contributed. And it's actually resident stabalized. If that doesn't mean anything to you, don't worry. You'll learn more about that in organic chemistry. But the important thing to realize here is that there's two protons to nab off of it. Now each molecule of sodium hydroxide-- remember when you put it in the water it really just dissolves, and you can really just think of it as hydroxide-- each molecule of hydroxide can nab one of the hydrogen protons. So for every one molecule of oxalic acid, you're going to need two hydroxides-- one to nab this hydrogen proton, and So let's write down the balanced equation that we're dealing with here. So we're going to start off with some oxalic acid. So that has two hydrogens-- so it's H2-- two carbons, and then four oxygens-- O4. It's dissolved in water, so it's an aqueous solution. And to that, we're going to add sodium hydroxide. Now I just told you that you're going to need two of the hydroxides to fully neutralize the oxalic acid. So you're going to need two of them. And this is also in our aqueous solution. And once the reaction happens, this guy will have lost both of the hydrogen protons, so let me draw that. So it will look like this. No more hydrogen, so it's C2O4. It'll have a negative 2 charge. And actually, you could imagine that it might be So this has a negative 2 charge. We could even write it there if you want-- 2 minus. And then you could have the sodiums over here. You have these two sodiums that have two plus. And this entire molecules becomes neutral. They are attracted to each other. They are still in an aqueous solution. And then, the hydroxide nabs the protons, and then you are left with just water. So plus 2 moles-- or 2 molecules depending on how we're viewing this-- plus two waters. I'll just use that same orange color. Plus two H2Os. One of the hydrogens in each of the water molecules are coming from the oxalic acid, and so two of these hydrogens in these two moles of the water are coming from one entire molecule of oxalic acid."
    },
    {
        "Q": "When writing the chemical name, does it matter the order of the elements, like are H2O and OH2 the same?\nI'm asking this because at 1:10, Sal wrote HO, but I've seen it mostly as OH.",
        "A": "technically doesn t matter but it is common to write the acid part on the left and the basic part of the right",
        "video_name": "XjFNmfLv9_Q",
        "timestamps": [
            70
        ],
        "3min_transcript": "I've taken this problem from Chapter 4 of the Chemistry & Chemical Reactivity book by Kotz, Treichel and Townsend, and I've done it with their permission. So let's do this example. A 1.034 gram sample of impure oxalic acid is dissolved in water and an acid-base indicator added. The sample requires 34.47 milliliters of 0.485 molar sodium hydroxide to reach the equivalence point. What is the mass of oxalic acid, and what is its mass percent in the sample? So before we even break into the math of this, let's just think about what's happening. We have some oxalic acid, which looks like this. It's really two carbolic acid groups joined together, if that means anything to you. Watch the organic chemistry play list if you want to learn more about that. So we have a double bond to one oxygen, and then another We have that on the other carbon as well. This right here is what oxalic acid is. And it's an interesting acid, because it can actually donate This proton can be nabbed off, and this proton can also be contributed. And it's actually resident stabalized. If that doesn't mean anything to you, don't worry. You'll learn more about that in organic chemistry. But the important thing to realize here is that there's two protons to nab off of it. Now each molecule of sodium hydroxide-- remember when you put it in the water it really just dissolves, and you can really just think of it as hydroxide-- each molecule of hydroxide can nab one of the hydrogen protons. So for every one molecule of oxalic acid, you're going to need two hydroxides-- one to nab this hydrogen proton, and So let's write down the balanced equation that we're dealing with here. So we're going to start off with some oxalic acid. So that has two hydrogens-- so it's H2-- two carbons, and then four oxygens-- O4. It's dissolved in water, so it's an aqueous solution. And to that, we're going to add sodium hydroxide. Now I just told you that you're going to need two of the hydroxides to fully neutralize the oxalic acid. So you're going to need two of them. And this is also in our aqueous solution. And once the reaction happens, this guy will have lost both of the hydrogen protons, so let me draw that. So it will look like this. No more hydrogen, so it's C2O4. It'll have a negative 2 charge. And actually, you could imagine that it might be"
    },
    {
        "Q": "at 2:30 how is one going to know it takes 2naoh to neutralize?",
        "A": "Oxalic acid has 2 Hydrogen protons which are formed when it disassociates in water. Thus, 2 OH- ions are required to neutralize them. Every NaOH molecule releases only 1 OH- ions. Thus, 2 NaOH are required to neutralize the oxalic acid. Thus, when there are two H+ ions which are formed on disassociation, two OH_ ions will be required to neutralize it. Hope this helps :)",
        "video_name": "XjFNmfLv9_Q",
        "timestamps": [
            150
        ],
        "3min_transcript": "I've taken this problem from Chapter 4 of the Chemistry & Chemical Reactivity book by Kotz, Treichel and Townsend, and I've done it with their permission. So let's do this example. A 1.034 gram sample of impure oxalic acid is dissolved in water and an acid-base indicator added. The sample requires 34.47 milliliters of 0.485 molar sodium hydroxide to reach the equivalence point. What is the mass of oxalic acid, and what is its mass percent in the sample? So before we even break into the math of this, let's just think about what's happening. We have some oxalic acid, which looks like this. It's really two carbolic acid groups joined together, if that means anything to you. Watch the organic chemistry play list if you want to learn more about that. So we have a double bond to one oxygen, and then another We have that on the other carbon as well. This right here is what oxalic acid is. And it's an interesting acid, because it can actually donate This proton can be nabbed off, and this proton can also be contributed. And it's actually resident stabalized. If that doesn't mean anything to you, don't worry. You'll learn more about that in organic chemistry. But the important thing to realize here is that there's two protons to nab off of it. Now each molecule of sodium hydroxide-- remember when you put it in the water it really just dissolves, and you can really just think of it as hydroxide-- each molecule of hydroxide can nab one of the hydrogen protons. So for every one molecule of oxalic acid, you're going to need two hydroxides-- one to nab this hydrogen proton, and So let's write down the balanced equation that we're dealing with here. So we're going to start off with some oxalic acid. So that has two hydrogens-- so it's H2-- two carbons, and then four oxygens-- O4. It's dissolved in water, so it's an aqueous solution. And to that, we're going to add sodium hydroxide. Now I just told you that you're going to need two of the hydroxides to fully neutralize the oxalic acid. So you're going to need two of them. And this is also in our aqueous solution. And once the reaction happens, this guy will have lost both of the hydrogen protons, so let me draw that. So it will look like this. No more hydrogen, so it's C2O4. It'll have a negative 2 charge. And actually, you could imagine that it might be"
    },
    {
        "Q": "At around 1:40 he is drawing the H atoms on the diagram and called them protons. Does he mean to say hydrons?",
        "A": "Hydrogen atoms are just protons with one electron.",
        "video_name": "XjFNmfLv9_Q",
        "timestamps": [
            100
        ],
        "3min_transcript": "I've taken this problem from Chapter 4 of the Chemistry & Chemical Reactivity book by Kotz, Treichel and Townsend, and I've done it with their permission. So let's do this example. A 1.034 gram sample of impure oxalic acid is dissolved in water and an acid-base indicator added. The sample requires 34.47 milliliters of 0.485 molar sodium hydroxide to reach the equivalence point. What is the mass of oxalic acid, and what is its mass percent in the sample? So before we even break into the math of this, let's just think about what's happening. We have some oxalic acid, which looks like this. It's really two carbolic acid groups joined together, if that means anything to you. Watch the organic chemistry play list if you want to learn more about that. So we have a double bond to one oxygen, and then another We have that on the other carbon as well. This right here is what oxalic acid is. And it's an interesting acid, because it can actually donate This proton can be nabbed off, and this proton can also be contributed. And it's actually resident stabalized. If that doesn't mean anything to you, don't worry. You'll learn more about that in organic chemistry. But the important thing to realize here is that there's two protons to nab off of it. Now each molecule of sodium hydroxide-- remember when you put it in the water it really just dissolves, and you can really just think of it as hydroxide-- each molecule of hydroxide can nab one of the hydrogen protons. So for every one molecule of oxalic acid, you're going to need two hydroxides-- one to nab this hydrogen proton, and So let's write down the balanced equation that we're dealing with here. So we're going to start off with some oxalic acid. So that has two hydrogens-- so it's H2-- two carbons, and then four oxygens-- O4. It's dissolved in water, so it's an aqueous solution. And to that, we're going to add sodium hydroxide. Now I just told you that you're going to need two of the hydroxides to fully neutralize the oxalic acid. So you're going to need two of them. And this is also in our aqueous solution. And once the reaction happens, this guy will have lost both of the hydrogen protons, so let me draw that. So it will look like this. No more hydrogen, so it's C2O4. It'll have a negative 2 charge. And actually, you could imagine that it might be"
    },
    {
        "Q": "AT 3:42 Sal said that the Water Molecule (H_2O) got a \"proton\" from the hydrogen ion .\n\"Can Protons be also added in a atom or molecule\", during a chemical reaction?\nI know that electrons do but can Protons tooo....?",
        "A": "When we say proton in chemical reactions it means a hydrogen cation H+, it doesn t mean one of the nuclei is gaining a proton.",
        "video_name": "Y4HzGldIAss",
        "timestamps": [
            222
        ],
        "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": "1:22 what is aqueous solution??",
        "A": "Solution whereby the solute is dissolved in water.",
        "video_name": "Y4HzGldIAss",
        "timestamps": [
            82
        ],
        "3min_transcript": "- [Voiceover] The first, I guess you could say, modern conception of an acid and base comes from this gentleman right over here, Svante Arrhenius, and he was actually the third recipient of the Nobel Prize in Chemistry in 1903. And his definition of acids, under his definition of acids and bases, an acid is something that increases the concentration, increases the concentration, concentration of Hydrogen protons, and we can say protons when put in an aqueous solution, when in aqueous, aqueous solution, and that's just a water solution. And then you can imagine what a base would be. You could think, oh maybe a base is something that decreases the protons and that's Or you could say, it decreases, or actually let me write this, it increases It increases the hydroxide concentration. when put in aqueous solution. When in aqueous, aqueous solution. So let's make that concrete. Let's look at some examples. So a strong Arrhenius acid, and actually, this would be a strong acid by other definitions as well, would be hydrochloric acid. Hydrochloric acid, you put it in a an aqueous solution. So that's the hydrogen. You have the chlorine. You put it in an aqueous solution. You put it in an aqueous solution, it will readily disassociate. This is a, this reaction occurs, strongly favors moving from the left to right. You're going to have the chlorine strip off the two electrons in the covalent bond with the hydrogen, 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,"
    },
    {
        "Q": "At 9:00, why isn't that molecule designated as \"cis\" or \"trans\"? Can't you have both an R/S designation and a cis/trans designation?",
        "A": "There are no double bonds and each of the 4 groups at the chiral centres are different. How could you determine cis/trans here? You can have R/S and cis/trans in cyclic molecules though.",
        "video_name": "kFpLDQfEg1E",
        "timestamps": [
            540
        ],
        "3min_transcript": "So that would be the lowest priority. So that would get a number four here. So the oxygen has the highest atomic number. So it's going to get a number one. And then we have a longer chain over here for this carbon on the right. So that's going to get a number two, we have a number three. And then the hydrogen would be a number four. So there's a little trick that I covered in an earlier video. So if you ignore the hydrogen, it looks like you're going around this way, it looks like you're going around counterclockwise. So it looks like it's S. But you have this lowest priority group is actually coming out at you. So remember the trick was, if it looks like it's S with those three, just reverse it. And so it must be R. It must have an absolute configuration of R at carbon three. So we'll go ahead and put in here a three R. And then we have to worry about the absolute configuration at carbon six. So at carbon six here, this is another chirality center. There's also a hydrogen attached to this carbon, going away from us like that. And let's think about the highest priority. Well, this chain over here on the left is going to get the highest priority. It has the most carbons, it has an oxygen over here. So it's definitely going to be highest priority There are more carbons this one over here on the right. So again, when you assign priority, this is going to get highest one. And then this is going to get a third up here. So this time, you're going around one, two, three, you're going around counterclockwise. But your lowest priority group, which is this hydrogen back here, is going away from you. So this actually is going to be S. So it's counterclockwise. So it's S. Three R, six S. And I went through those kind of fast. So you need to go back and watch some of the earlier videos on absolute configurations if that was a little bit too fast for you. All right. Let's look at cyclic alcohol. So ring systems. And then there's an O-H coming off of it like that. So six carbons without the O-H, we would call that cyclohexane. And since this is an alcohol, we would just change that to cyclohexanol. So that's very simple nomenclature. You don't really need a number. But you could write a one there. It's implied if you don't put in. So that would be cyclohexanol. What about something that has a ring system with two hydroxyl groups? So let's say we'll put in some stereochemistry, too. So let's say we have an O-H coming out at us. And let's say we have an O-H going away from us like that. So when you have a situation like this, when you have two alcohols in the same molecule, your prefixes be di. So this is actually a diol. And the nomenclature is based off the cyclohexane molecule. So you would write cyclohexane."
    },
    {
        "Q": "At 3:36, wouldn't the longest carbon chain start at the carbon in the top right (above what Jay designated as the first carbon) and continue down to what he designated at carbon 7? That would make a continuous 8 carbon chain",
        "A": "The OH group needs to be included in the main chain as it is the highest priority substituent in this molecule, even though there is a longer carbon chain.",
        "video_name": "kFpLDQfEg1E",
        "timestamps": [
            216
        ],
        "3min_transcript": "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? 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."
    },
    {
        "Q": "In 4:55, can't I write \"5-chloro-5-methyl-3-propyl-heptan-2-ol\" instead of 5-chloro-5-methyl-3-propyl-2-heptanol\"?\n\nMy chemistry teacher taught me that the first is correct. What's the difference between the names?",
        "A": "You can as long as you remove the dash between propyl and heptan. The preferred way to do it is to put the number in the middle like your first way. People don t always learn the new recommendations. It s also much easier to say 2-heptanol out loud than heptan-2-ol",
        "video_name": "kFpLDQfEg1E",
        "timestamps": [
            295
        ],
        "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": "At 1: 26, in my textbook, it shows that the name should be propan-1-ol (or propanol) and similarly the name of 2-propanol should also be propan-2-ol.\nAlso, at 4:48, according to the rules in my textbook, the name should be 5-chloro-5-methyl-3-propylheptan-3-ol.\nI am very confused now because I don't know if should put -ol at first or at last.\nPlease help me!",
        "A": "Yeah ! you are correct, in my textbook too it is written as propan -2-ol. And the textbook contains the correct nomenclature",
        "video_name": "kFpLDQfEg1E",
        "timestamps": [
            288
        ],
        "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": "I thought water was already a conductor. 8:20",
        "A": "It s actually not the water itself that is the conductor. All water forms hydronium (H3O+) and hydroxyl (OH-) ions, which conduct the electricity. In pure water the concentrations of these ions are very low (about (about 10^-7 mols/liter of each), so you need to add impurities (such as salts or acids) to increase the number of ions and the conductivity.",
        "video_name": "Rw_pDVbnfQk",
        "timestamps": [
            500
        ],
        "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": "At 1:37ish, wouldn't the carbons be connected slightly differently? More like a single uniform \"diamond\" shape, rather than a spread out tree.",
        "A": "actually, it depends on the cooling rate. atleast what i learned when working with metals was when they were melted and cooling, if cooled slowly, they would form their perfect crystaline shapes but if quenched (dipped in water for a quicker cool) they would create many irregular crystal patterns, this irregularity is what would cause quenched metals to bend easier than slowly cooled ones. if theres a big difference in strength of a metal and its ability to bend, same metals different cooling rate.",
        "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 6:35, isn't HF held together with hydrogen bonding, not dipole-dipole forces?",
        "A": "If we re getting technical hydrogen bonding is a strong type of dipole-dipole interaction.",
        "video_name": "Rw_pDVbnfQk",
        "timestamps": [
            395
        ],
        "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": "At 4:20, Sal says that when we cut something, we are breaking atomic bonds. But I thought we were breaking molecular bonds. Or are they the same ?",
        "A": "What he means is that we are breaking bonds between atoms. We don t have knives or scissors sharp enough to cut atoms in pieces.",
        "video_name": "Rw_pDVbnfQk",
        "timestamps": [
            260
        ],
        "3min_transcript": "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 Chlorine has seven electrons and it's dying to get a new. So sodium essentially donates its electron to chlorine, and then the chlorine becomes negative, the sodium becomes positive, and they want to be near each other, right? So you have a positive sodium ion and a negative chlorine ion, and the structure of this is going to look something like this, where they're all -- so let me do the sodium in green. So you have a bunch of sodium ions that are positive, and then you have a bunch of chlorine ions that are maybe -- this isn't the exact way that they actually are, but I think you get the idea, that one atom is positive and one atom is negative, so they really, really want to be close to each other. And so this is a pretty strong bond, and it has very-- not a very high boiling point. It can have a pretty high boiling point, and this type of structure is actually quite brittle. So if you take some dry table salt, not dissolved in water, and you slam it with a hammer, you'll see that you'll get, a big slice of it. It'll just fall off, right? Because you're essentially just cutting it along one of these lines really fast. That's the interesting thing. Whenever you do something on a macroscale, like cut sth. you really fundamentally are breaking atomic bonds. So the strength of the atomic bonds really do tell you about how hard or strong something is. Now, the metallic crystal we've talked a lot about. Metals, they like to get rid of their electrons, or not get rid of them, they like to share them. So what happens is, let's say in the case of iron, you have a bunch of iron atoms. This is all iron. And their electrons are allowed to roam free in the neighborhood. These are all the electrons. They're allowed to roam free. And because of this, it forms this sea of electrons that are negative, and that makes it a very good conductor of electricity."
    },
    {
        "Q": "At 1:46 Sal says that dinosaurs may have been smarter than us, but they couldn't of been smarter than us because if they were, they could stopped the asteroid... Right? Or, they were smart enough, but they didn't have what they needed to stop it? What do you think, because that's confusing me. (We'll probably never know)",
        "A": "They were never aware of an asteroid. Even if they had tech, it would have struck unexpectedly. Like today, last year we almost got hit with an asteroid 50 feet across. We didnt know until it got close.",
        "video_name": "T5DGZIsfK-0",
        "timestamps": [
            106
        ],
        "3min_transcript": "I've talked a bunch about the Drake equation, or our own version of the Drake equation that starts with the number of stars in the galaxy, but I haven't given it a shot yet. I haven't tried my own attempt at thinking about how many detectable civilizations there are. So let's actually do that here. So let's just assume that there are 100 billion stars. 100 billion stars. So that's my first term right over there. Let's say that 1/4 will develop planets. And let's say of the solar systems that develop planets, on average let's say that they develop an average of 0.1 planets capable of sustaining life. Or really, that you'll have one planet for every 10 of these solar systems with planets. That's just my assumption there. I don't know if that's right. Now let's multiply that times the fraction of these planets capable of sustaining And I don't know what that is, but I hinted in previous videos that life is one of those things that it seems like if you have all the right ingredients, it's so robust. And you have life it these underwater volcanoes, you have bacteria that can process all sorts of weird things. So let's say that probability is pretty high. Let's say that is 50%, or half of the plans that are capable of getting life actually do have life. I would guess that that might even be higher. But once again, just a guess. Now we have to think about of the life, what fraction becomes intelligent? What becomes intelligent over some point in the history? Well, I'll say it's a tenth. A tenth of all-- maybe if the asteroids didn't kill the dinosaurs, it wouldn't have happened on Earth. Who knows? Or maybe we'd just have some very intelligent dinosaurs We don't know. And maybe there's other forms of intelligent life even on our own planet that we haven't fully appreciated. Dolphins are a good candidate. Some people believe that octopuses, there's a theory that they could develop eventually the ability to kind one day, if their brains mature, and all of the rest, make tools the same way primitive primates eventually were able to have larger brain sizes and actually manipulate things to make tools. I don't want to get into all of that. So there's a 1 in 10 chance that you get intelligent life, and then assuming that intelligent life shows up, what fraction is going to become detectable? I don't know. I don't know whether dolphins will ever communicate via radio or not. So let's just say that is-- I don't know. Let's say that is another 1 in 10 chance, or I'll say 0.1. And then we have to multiply it times the detectable life of the civilization on average. Once again, huge assumptions being here, but the detectable life of a civilization, let me just put it at 10,000 years."
    },
    {
        "Q": "At 2:47-2:51 where did \"Milkomeda\" come from?",
        "A": "To be fair, Sal didn t make it up, it is the already made name for the future galaxy.",
        "video_name": "QXYbGZ3T3_k",
        "timestamps": [
            167,
            171
        ],
        "3min_transcript": "so this video is actually spanning billions of years but when you actually speed up time like that you'll see that it really gives you the sense of the actual dynamics of these interactions. The other thing that I want to talk about before I start the video is to make you realise that when we talk about galaxies colliding it doesn't mean that the stars are colliding. In fact there are going to be very few stars that actually collide the probability of a star-star collision is very low and that's because we learned when we learned about interstellar scale that there's mostly free space in between stars the closest star to us is 4.2 light years away and that's roughly 30 million times the diameter of the sun so you have a lot more free space than star space or even solar system space. So lets start up this animation it's pretty amazing and what you'll see here so these are just the... obviously... so one rotation is actually 250 million years give or take. But now you see these stars right here are starting to get attracted to this core and then some of the stuff in that core was attracted to those stars and they get pulled away that was the first pass of these 2 galaxies. Some stuff is just being thrown off into intergalactic space and you might worry that might happen to the earth and there's some probability that it would happen to the earth but it really wouldn't affect what happens within those star's solar system this is happening so slow, you wouldn't feel like some type of acceleration or something. And then this is the second pass so they passed one pass and once again this is occuring over hundreds of millions or billions of years and on the second pass they finally are able to merge. And all of these interactions are just through the gravity over interstellar or you could call it intergalactic distances. You can see they merge into what could be called as \"Milkomeda\" or maybe the Andromedy Way I don't know whatever you want to call it but even though they've merged, a lot of the stuff has still been thrown off into intergalactic space but this is a pretty amazing animation to me but it's also pretty neat how a super computer can do all of the computations to figure out what every particle which is really a star or cluster of stars or group of stars is actually doing to actually give us a sense of the actual dynamics here, but this is pretty neat look at that! Every little dot is whole groups of stars, thousands of stars potentially."
    },
    {
        "Q": "At 2:11; what is the difference between the Hydrogen atom and Deuterium??",
        "A": "Deuterium is a hydrogen atom with one neutron. Protium has no neutrons in the atom. Tritium has 2 neutrons.",
        "video_name": "I-Or4bUAIfo",
        "timestamps": [
            131
        ],
        "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 5:35 we know oxygen has 6 valence electrons it is making a single covalent bond with sulfur so it must be contributing one of its electrons in bonding which leaves 5 lone electrons with it doesn't it then why is he drawing 6 lone electrons with it?",
        "A": "Both of the electrons in the S-O bond are coming from the sulfur.",
        "video_name": "dNPs-cr_6Bk",
        "timestamps": [
            335
        ],
        "3min_transcript": "And once again you ignore hydrogen so it's between sulfur and oxygen. And if you look at the periodic table, you'll see that oxygen is higher in group six than sulfur is. Therefore oxygen is more electronegative. And so therefore, we're going to put sulfur at the center. So we're going to put sulfur right here. Once again, look at the rules from the previous video if that didn't make quite sense to you here. So we have sulfur attached to four oxygens. And I'll go ahead and put my four oxygens in there like that. And then I have two hydrogens. And by experience, you are talking about an acid here. You're going to put your hydrogens on oxygens. And so we're going to go ahead and put our hydrogens here. And let's see how many valence electrons we've used up a drawing this skeleton here. So we have two, four, six, eight, ten, and twelve. So we've used up 12 valence electrons so 32 minus 12 gives us 20 valence electrons left to worry about. to assign some of those left over electrons to some of the terminal atoms. But again, we're not going to assign those electrons to hydrogen because hydrogen's already surrounded by two electrons. And so we're going to try to assign some electrons to oxygen. And oxygen's going to follow the octet rule. So let's examine, let's say the top oxygen here. And we could see the top oxygen is surrounded by two electrons already right there in green. And so if we're going to give it an octet it needs six more. So we have one, two, three, four, five, six. Same thing for this oxygen down here, it needs an octet. So we go ahead and give it six more electrons like that. Right so, we also have these other oxygens over here to worry about. So let me go ahead and use green again. So let's say this oxygen over here on the left, the one bonded to this hydrogen here. Here's two electrons and here's another two for four. So for that option to have an octet, it needs four more. on this oxygen. And then we're going to do the same thing for this oxygen as well. So let's see, how many electrons did we just represent there? Well we had six on the top oxygen, six in the bottom oxygen. That's 12. And then we had four on the left and four more on the right. So that's eight. So 12 plus 8 is 20. So we've now represented all of the valence electrons that we needed to show. And let's think about this as possibly being the final dot structure. So we have an octet around sulfur, an octet around oxygen, and hydrogen's fine. So you might think that we are done here. However, let's go ahead and assign some formal charges. And let's see what that does. So let me go ahead and draw in some electrons here. So we know that each bond consists of two electrons. I'm going to go ahead and make them red here like that. And let's assign a formal charge to the top oxygen here."
    },
    {
        "Q": "At about 5:30, the K_a of HCl is discussed. Since HCl is a strong acid, so the reaction is irreversible, and K_a is an equilibrium constant, how is this possible? I thought it was only possible to measure the K_a of a weak acid.",
        "A": "Is not completely irreversible though very likely to favor porducts. That s why the KB value is so tiny Strong Acid has a very weak base",
        "video_name": "DGMs81-Rp1o",
        "timestamps": [
            330
        ],
        "3min_transcript": "This is equal to 1.0 times 10 to the negative 14 which is our value for Kw. When you add reactions together to get a net reaction you multiply the equilibrium constants to get the equilibrium constant for the net reaction which in this case is Kw for the auto-ionization of water. Let's go ahead and go in even more detail here. Ka, that's your products over your reactants. That'd be H3O plus, the concentration of H3O plus times the concentration of ammonia. Let's go ahead and do that. The concentration of H3O plus times the concentration of ammonia, and that's all over the concentration of ammonium. This is Ka. This is all over the concentration of ammonium. This represents, let me go ahead and highlight this here. Next let's think about Kb. Over here is Kb, that'd be the concentration of your products, so NH4 plus times OH minus. Let's go ahead and do that. Let's put this in parenthesis here. We have the concentration of ammonium, NH4 plus, times the concentration of OH minus. That's over the concentration of NH3. That's over the concentration of NH3 here. What do we get? The ammonium would cancel out. That cancels here. Then the NH3 cancels out. We're left with H3O plus times OH minus which we know is equal to 1.0 times 10 to the negative 14. Just another way to think about this. This can be important, relating Ka and Kb to Kw. If you know one you can figure out the other. Let's think about HCl. The conjugate base to HCl would be Cl minus, the chloride anion here. Let's think about what this equation means. HCl is a strong acid which means a very high value for Ka. An extremely, extremely high value for Ka. What does that say about Kb for the conjugate base? The conjugate base here is the chloride anion. If Ka is very large then Kb must be very small for this to be equal to Kw. Kb is extremely small here, so a very small value for Kb. This mathematically describes what we talked about earlier the stronger the acid the weaker the conjugate base. HCl is a very strong acid, so it has a very, very high value for Ka. And the conjugate base is the chloride anion,"
    },
    {
        "Q": "At about 0:45 you mention Heat of Fusion. Don't you mean Heat of Vaporization?",
        "A": "We re looking at liquid becoming gaseous, in other words vaporization, so I think he meant to say heat of vaporization.",
        "video_name": "hA5jddDYcyg",
        "timestamps": [
            45
        ],
        "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": "At 9:27, why do they want to evaporate at a high KE?",
        "A": "High KE means they are moving fast. The ones that are moving fast are the ones that might be able to escape the surface and be carried away.",
        "video_name": "hA5jddDYcyg",
        "timestamps": [
            567
        ],
        "3min_transcript": "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 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"
    },
    {
        "Q": "At 9:11, you reference the Ka value of Ammonium (taken from another video). If solving this problem requires this Ka value, would the value normally be displayed as part of the question?",
        "A": "The Ka will either be given to you, or it will be possible to calculate it based on what else the equation gives you",
        "video_name": "kWucfgOkCIQ",
        "timestamps": [
            551
        ],
        "3min_transcript": "does react with water, but the chloride anion does not react appreciably with water. So you don't have to worry about this being present, but the NH four plus will affect the pH, which our goal, of course, is to figure out the pH. Alright, so let's get some more room and let's write in our initial concentration of NH four plus, which is 0.05. So the initial concentration of NH four plus is 0.05. And, once again, for these problems, these weak acid equilibrium problems, we assume that we're starting with zero for the concentration of our products and our change, we're going to lose a certain concentration of ammonium. A certain concentration of ammonium is going to react with water. So we're gonna lose x. And then whatever we lose for ammonium turns into ammonia. So if we lose x for the concentration of ammonium, for NH three, for ammonia. And therefore, that's also the same concentration of NH three O plus. So at equilibrium, we're gonna have 0.05 minus x for the concentration of NH four plus. For the concentration on H three O plus, we have x. And for the concentration of NH thee, we also have x. Next, this is functioning as an acid, alright? NH four plus is donating a proton to water, so for the equilibrium expression, we have to write Ka here. And Ka is equal to concentration of products over reactants, so we have concentration of H three O plus, which is x times concentration of NH three, which is x all over the concentration of NH four plus, which is 0.05 minus x. So 0.05 minus x. Ka for NH four plus, Ka for ammonium, and we found it to be 5.6 times ten to the negative ten. So this is equal to... This is equal to, x times x is x squared and 0.05 minus x, if x is a very small concentration, 0.05 minus x is approximately the same as 0.05. So we write 0.05 in here, to make our lives easier for the calculation and we can get out the calculator and solve for x. 5.6 times ten to the negative ten. Alright, we're gonna multiply that by 0.05 and then we're gonna take the square root of our answer and solve for x. X is equal to 5.3 times ten to the negative six. So let's go ahead and write that down. X is equal to 5.3 times ten to the negative six"
    },
    {
        "Q": "At 9:20, you said that you can leave out the \"X\" because is \"X\" would have a small amount. How do you know that ? I saw that the result was small, but I just don't understand when you can leave out the\"X\"",
        "A": "Most instructors say that x is very small if it is less than 5 % of the initial concentration of the acid or base. A quick test for this is if the initial concentration is greater than or equal to 400 \u00c3\u0097 K. If [Acid]\u00e2\u0082\u0080/K \u00e2\u0089\u00a5 400, x is small enough to ignore. At 9:20, [Acid]\u00e2\u0082\u0080/K = 0.0500/5.6\u00c3\u009710\u00e2\u0081\u00bb\u00c2\u00b9\u00e2\u0081\u00b0 = 8.9 \u00c3\u0097 10\u00e2\u0081\u00b7. This is much larger than 400, so you can safely ignore x in the denominator of the equation.",
        "video_name": "kWucfgOkCIQ",
        "timestamps": [
            560
        ],
        "3min_transcript": "does react with water, but the chloride anion does not react appreciably with water. So you don't have to worry about this being present, but the NH four plus will affect the pH, which our goal, of course, is to figure out the pH. Alright, so let's get some more room and let's write in our initial concentration of NH four plus, which is 0.05. So the initial concentration of NH four plus is 0.05. And, once again, for these problems, these weak acid equilibrium problems, we assume that we're starting with zero for the concentration of our products and our change, we're going to lose a certain concentration of ammonium. A certain concentration of ammonium is going to react with water. So we're gonna lose x. And then whatever we lose for ammonium turns into ammonia. So if we lose x for the concentration of ammonium, for NH three, for ammonia. And therefore, that's also the same concentration of NH three O plus. So at equilibrium, we're gonna have 0.05 minus x for the concentration of NH four plus. For the concentration on H three O plus, we have x. And for the concentration of NH thee, we also have x. Next, this is functioning as an acid, alright? NH four plus is donating a proton to water, so for the equilibrium expression, we have to write Ka here. And Ka is equal to concentration of products over reactants, so we have concentration of H three O plus, which is x times concentration of NH three, which is x all over the concentration of NH four plus, which is 0.05 minus x. So 0.05 minus x. Ka for NH four plus, Ka for ammonium, and we found it to be 5.6 times ten to the negative ten. So this is equal to... This is equal to, x times x is x squared and 0.05 minus x, if x is a very small concentration, 0.05 minus x is approximately the same as 0.05. So we write 0.05 in here, to make our lives easier for the calculation and we can get out the calculator and solve for x. 5.6 times ten to the negative ten. Alright, we're gonna multiply that by 0.05 and then we're gonna take the square root of our answer and solve for x. X is equal to 5.3 times ten to the negative six. So let's go ahead and write that down. X is equal to 5.3 times ten to the negative six"
    },
    {
        "Q": "How can we predict that a neutral molecule such as ethanol or water molecule would act as a base as shown at 10:26?",
        "A": "Any molecule that contains an atom with a lone pair of electrons, such as the O in ethanol or water, can accept a proton from an acid. For example. H2O: + H-Cl --> [H2O-H]+ + Cl-. The [H2O-H]+ is usually written as H3O+. Since the water is accepting a proton from the HCl, it is behaving as a Br\u00c3\u00b8nsted-Lowry base.",
        "video_name": "l-g2xEV-z7o",
        "timestamps": [
            626
        ],
        "3min_transcript": "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. it is the conjugate base to water, but water is not a great asset, and we know that from the pKa value here, so water is not great at donating a proton, which means that the hydroxide anion is not that stable, and since the hydroxide anion is not that stable, it's not a great leaving group. So let's go ahead and take off this arrow here, because the first step is not loss of a leaving group, the first step is a proton transfer. We have a strong acid here, sulfuric acid, and the alcohol will act as a base and take a proton from sulfuric acid. And that would form water as your leaving group, and water is a much better leaving group than the hydroxide anion, and again, we know that by pKa values. Water is the conjugate base to the hydronium ion, H3O+, which is much better at donating a proton, the pKa value is much, much lower. And that means that water is stable, when you are doing an E1 mechanism with an alcohol is to protonate the OH group. So here's our alcohol, and the carbon bonded to the OH is our alpha carbon, and then these carbons next to the alpha carbon would all be beta carbons. We just saw the first step is a proton transfer, a lone pair of electrons on the oxygen take a proton from sulfuric acid, so we transfer a proton, and let's go ahead and draw in what we would have now, so there'd be a plus-one formal charge on the oxygen, so let's highlight our electrons in magenta, these electrons took this proton to form this bond, and now we have water as a leaving group, let me just fix this hydrogen here really fast, and these electrons can come off onto our oxygen, so that gives us water as our leaving group, and let me go ahead and draw in the water molecule here, and let me highlight electrons,"
    },
    {
        "Q": "at 1:41, why does the alcohol group act like a base, when oxygen does not like to be positively charged?",
        "A": "Bases remove protons, nucleophiles form bonds. Hyroxide removes the proton and so is a base not a nucleophile. If it formed a bond and stayed, then it would be a nucleophile. In step 1 it is a nuc, step 2 it is a base. Overall it is a base I belive, I could be wrong though",
        "video_name": "l-g2xEV-z7o",
        "timestamps": [
            101
        ],
        "3min_transcript": "- [Instructor] Let's look at the mechanism for an E1 elimination reaction, and we'll start with our substrate, so on the left. Let's say we're dealing with alkyl halide. So the carbon that's bonded to our halogen would be the alpha carbon, and the carbon next to that carbon would be the beta carbon, so we need a beta hydrogen for this reaction. The first step of an E1 elimination mechanism is loss of our leaving group, so loss of leaving group, let me just write that in here really quickly, and in this case, the electrons would come off onto our leaving group in the first step of the mechanism. So we're taking a bond away from this carbon, the one that I've circled in red here, so that carbon is going from being sp3 hybridized to being sp2 hybridized. So now we have a carbocation, and we know that carbocations, sp2 hybridized carbons have planar geometry around them, so I've attempted to show the planar geometry So that's the first step, loss of the leaving group to form a carbocation. In the second step, our base comes along and takes this proton, which leaves these electrons behind, and those electrons move in to form our alkene, so this is the second step of the mechanism, which is the base takes, or abstracts, a proton, so base takes a proton to form our alkene. And let me go ahead and highlight those electrons, so these electrons here in magenta moved in to form our double bond, and we form our product, we form our alkene. So the first step of the mechanism, the loss of the leaving group, this turns out to be the rate determining step, so this is the slowest step of the mechanism. So if you're writing a rate law, the rate of this reaction would be equal to the rate constant k times the concentration of your substrate, so that's what studies have shown, 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,"
    },
    {
        "Q": "4:11 I don't understand.... 20mph per second? Can someone please help?",
        "A": "If you are in your car and you go from 0 mph to 20 mph, and you do it in 1 second, then your acceleration was 20 mph per second.",
        "video_name": "FOkQszg1-j8",
        "timestamps": [
            251
        ],
        "3min_transcript": ""
    },
    {
        "Q": "5:50 If there were no atmospheric pressure it would turn automatically into a gas?",
        "A": "Yes (but it will also depend on temperature), because there is nothing holding things together. In fact, if you had liquid water in space it would freeze because it is cold and evaporate because there is no pressure at the same time.",
        "video_name": "tvO0358YUYM",
        "timestamps": [
            350
        ],
        "3min_transcript": "You see that the units work out. This kelvin is going to cancel out with that kelvin in the denominator. This gram in the numerator will cancel out with that grams. And it makes sense. Specific heat is the amount of energy per mass per degree that is required to push it that 1 degree. So here we're doing 58 degrees, 1,000 grams, you just The units cancel out. So you have kelvin canceling out with kelvin. You have grams canceling out with grams. And we are left with-- take out the calculator, put it on the side here. So we have 58.29 times 1,000-- times one, two, three-- times 2.44 is equal to-- and we only have three 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"
    },
    {
        "Q": "what does he mean by sending an axon to the optic nerve at 9:42 in the video? How do you send an axon? Does he mean send a message to an axon in the optic nerve",
        "A": "All the axons of ganglion cells compose the optic nerve. send would be just a figure of speech.",
        "video_name": "CqN-XIPhMpo",
        "timestamps": [
            582
        ],
        "3min_transcript": "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. so there are on-center and there are off-center bipolar cells. So the on-center bipolar cells normally are being turned off when this rod cell is turned off. But as we mentioned, due to the phototransduction cascade, the rods turn off, which actually turns on the bipolar cell. So basically, on-center bipolar cells get turned on with light and get turned off when there's no light. So that's how they get their names. So when the bipolar cell gets turned on, it activates a retinal ganglion cell, which then sends an axon to the optic nerve, and then into the brain. And so that process is known as the phototransduction cascade, and it basically allows your brain to recognize that there's light entering the eyeball."
    },
    {
        "Q": "At 11:30, why do you divide mole on both sides?",
        "A": "Review basic algebra - solving simple equations.",
        "video_name": "VqAa_cmZ7OY",
        "timestamps": [
            690
        ],
        "3min_transcript": "And I'll take out my calculator. And I have 0.162 divided by 0.03 is equal to 5.4. And actually more significant is, we could really say it's 5.40 since we have at least three significant digits in both situations. So 5.40 is our proportionality constant. And you would actually divide by 1 in both cases. We just want the number here. But if you wanted the units, you'd want to divide by that 1 centimeters as well. Now we can use this to figure out the exact answer to our problem without having to eyeball it like we just did. We know that for potassium permanganate at 540 nanometers, the absorbance is going to be equal to 5.4 The units of this proportionality constant right here is liters per centimeter mole. And you'll see it'll just cancel out with the distance which is in centimeters, or the length, and the molarity which is in moles per liter. And it just gives us a dimension list, absorbance. So times-- in our example the length is 1 centimeter-- times 1 centimeter, times the concentration. Now in our example they told us the absorbance was 0.539. That's going to be equal to 5.4 liters per centimeter mole, times 1 centimeter, times our concentration. over there. And then we can just divide both sides by 5.4 liters per mole. So let's do that. Let's divide both sides by 5.4 liters per mole, and what do we have? So on the right-hand side, all of this business is going to We're just going to have this concentration left over. So our concentration is equal to-- let's figure out what this number is. So we have 0.539 divided by 5.4 gives us-- so we only have-- well this is actually 5.40. So we actually have three significant digits. So we could say 0.0998."
    },
    {
        "Q": "At 9:03, what does the epsilon mean and what does it refer to?",
        "A": "It means permittivity.",
        "video_name": "VqAa_cmZ7OY",
        "timestamps": [
            543
        ],
        "3min_transcript": "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. take the first one, we get-- the absorbance was 0.162. That's going to be equal to this constant of proportionality times 1 centimeter. That's how wide the vial was. Times-- now what is the concentration? Well when the absorbance was 0.162, our concentration was 0.03 times 0.-- actually, I'll write all the significant digits there-- 0.0300. So if we want to solve for this epsilon, we can just divide both sides of this equation by 0.0300. So you divide both sides by 0.0300 and what do we get? These cancel out, this is just a 1. And so you get epsilon is equal to-- let's figure out And I'll take out my calculator. And I have 0.162 divided by 0.03 is equal to 5.4. And actually more significant is, we could really say it's 5.40 since we have at least three significant digits in both situations. So 5.40 is our proportionality constant. And you would actually divide by 1 in both cases. We just want the number here. But if you wanted the units, you'd want to divide by that 1 centimeters as well. Now we can use this to figure out the exact answer to our problem without having to eyeball it like we just did. We know that for potassium permanganate at 540 nanometers, the absorbance is going to be equal to 5.4"
    },
    {
        "Q": "At 4:57, at the bottom left of the virgo supercluster image, doesn't it say 250 million light years or am I mistaken?",
        "A": "oh yeah",
        "video_name": "JiE_kNk3ucI",
        "timestamps": [
            297
        ],
        "3min_transcript": "This is 4 million light years. Four light years is just the distance from us to the Alpha Centauri. So that was nothing. That would only take that a Voyager 1 80,000 years to get. This is 4 million light years. So 4 million times the distance to the nearest star. But even this is-- I mean I'm starting to stumble on my words because there's really no words to describe it-- even this is small on an intergalactic scale. Because when you zoom out more, you can see our Local Group. Our local group is right over here. And this right over here is the Virgo Super Cluster. And each dot here is at least one galaxy. But it might be more than one galaxy. And the diameter here is 150 million light years. 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."
    },
    {
        "Q": "At 8:31 you talk about the universe Expanding. Does that mean the Sun will one Day be out of range for us to even see it? Please answer my question if you understanding, because the Universe is just mind blowing :)",
        "A": "No, it doesn t, because gravity keeps earth near the sun. the effect of the expansion of space is only meaningful over intergalactic distances.",
        "video_name": "JiE_kNk3ucI",
        "timestamps": [
            511
        ],
        "3min_transcript": "to the beginning of the actual universe. And the reason why it's the visible universe is there might have been something a little bit further out. Maybe it's light hasn't reached us yet or maybe the universe itself, and we'll talk more about this, it's expanding so fast that the light will never, ever reach us. So it's actually a huge question mark on how big the actual universe is. And then some people might say, well, does it even matter? Because this by itself is a huge distance. And I want to make it clear, you might say, OK, if this light over here, if this is coming from 13 billion light years away, or if this is 13 billion light years away, then you could say, hey, so everything that we can observe or that we can even observe the past of, the radius is about 26 billion light years. But even there, we have to be careful because remember the universe is expanding. When this light was emitted-- and I'll do a whole video on this because the geometry of it 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"
    },
    {
        "Q": "8:35 We used to be close but it took the light 13B years to catch up with us. Does this mean the galaxies are moving apart almost at the speed of light? Or some even faster?",
        "A": "even faster.",
        "video_name": "JiE_kNk3ucI",
        "timestamps": [
            515
        ],
        "3min_transcript": "to the beginning of the actual universe. And the reason why it's the visible universe is there might have been something a little bit further out. Maybe it's light hasn't reached us yet or maybe the universe itself, and we'll talk more about this, it's expanding so fast that the light will never, ever reach us. So it's actually a huge question mark on how big the actual universe is. And then some people might say, well, does it even matter? Because this by itself is a huge distance. And I want to make it clear, you might say, OK, if this light over here, if this is coming from 13 billion light years away, or if this is 13 billion light years away, then you could say, hey, so everything that we can observe or that we can even observe the past of, the radius is about 26 billion light years. But even there, we have to be careful because remember the universe is expanding. When this light was emitted-- and I'll do a whole video on this because the geometry of it 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"
    },
    {
        "Q": "in 8:35 sal said that the universe would keep expanding, if so, in a large amount of time, would we move farther and farther away from the sun or the other planets or galaxies?",
        "A": "The expansion of space is only measurable over distances that are much, much greater than even the distance to nearby galaxies. The effect between the earth and the sun is very tiny, and in any event, the sun s gravity offsets the expansion and pulls the earth back even as the expansion of space minutely stretches it away",
        "video_name": "JiE_kNk3ucI",
        "timestamps": [
            515
        ],
        "3min_transcript": "to the beginning of the actual universe. And the reason why it's the visible universe is there might have been something a little bit further out. Maybe it's light hasn't reached us yet or maybe the universe itself, and we'll talk more about this, it's expanding so fast that the light will never, ever reach us. So it's actually a huge question mark on how big the actual universe is. And then some people might say, well, does it even matter? Because this by itself is a huge distance. And I want to make it clear, you might say, OK, if this light over here, if this is coming from 13 billion light years away, or if this is 13 billion light years away, then you could say, hey, so everything that we can observe or that we can even observe the past of, the radius is about 26 billion light years. But even there, we have to be careful because remember the universe is expanding. When this light was emitted-- and I'll do a whole video on this because the geometry of it 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"
    },
    {
        "Q": "At 12:30, do the neurotransmitters bind to receptors on sodium ion channels or on the post synaptic membrane?",
        "A": "Receptors are sometimes on sodium ion channels, and the ion channels are in the membrane, and sometimes the receptors are in the membrane but connected with the sodium channels with another molecules. I think they will explain us in some other video, probably next one :)",
        "video_name": "Tbq-KZaXiL4",
        "timestamps": [
            750
        ],
        "3min_transcript": "neurotransmitters. And when the calcium channels-- they're voltage gated-- when it becomes a little more positive, they open calcium floods in and what the calcium does is, it bonds to these proteins that have docked these vesciles. So these little vesicles, they're docked to the presynpatic membrane or to this axon terminal membrane These proteins are actually called SNARE proteins. It's an acronym, but it's also a good word because they've literally snared the vesicles to this membrane. So that's what these proteins are. And when these calcium ions flood in, they bond to these proteins, they attach to these proteins, and they change the confirmation of the proteins just enough that these proteins bring these vesicles closer to the membrane and also kind of pull apart the two membranes so that the 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"
    },
    {
        "Q": "At 3:50, there is just one product drawn. Shouldn't there be one more? The methylgroup is also an ortho- and para-directing activator, which means that there must be an isomer where -SO3H is next to the -CH3.",
        "A": "Jay is only showing the major product, which is based on the hydroxyl being a stronger activator.",
        "video_name": "iF-f2-KSw6E",
        "timestamps": [
            230
        ],
        "3min_transcript": "And first I look at the OH group, which I know is an ortho para director, right. So this is an ortho para director here. And I'd look at where that would be on my ring. Well, once again this would be the ortho position. And again by symmetry, the same ortho position over there. The para position is once again taken up by this methyl group here. When I think about the methyl group-- so let me just use a different color for that one. The methyl group is also an ortho para director. So what's ortho and para to the methyl group. Well this would be ortho, right. These two spots would be ortho, again identical because of symmetry. The para spot is taken up by this OH here. And so now we have a case where we have two ortho para directors directing to different spots. And so the way to figure out which one wins is to think about the activating strength of these two ortho para directors. So the strongest activator is going to be the directing group. So this is a strong activator. And the methyl group we saw is a weak activator from some of the earlier videos here. So the spot that's going to get the SO3H is this carbon, which again by symmetry would be this one right here. So we can go ahead and draw the product of this sulfonation So let me go ahead and sketch in my benzene ring. Let me do another one here. That one wasn't very good. And we have our ring. We have our OH. We have our methyl group. And since the OH is the strongest activator, we're going to go ahead and put SO3H ortho to the OH. And that is our final product. Let's do one more example. So let's see what happens with this reaction here. So once again I can identify this as being a halogenation reaction where And let's go ahead and analyze the substituents once again. So I have a methyl group again, which I know is an ortho para director. So I'll go ahead and mark the spots that are ortho and para to that methyl group. So this would be the ortho position. This would be the ortho position. And then this time the para position is free, so we could possibly put the chlorine there. If I look at what else is on my ring, right-- so I have another substituent here, which is a chlorine. We know that halogens are also ortho para directors because of the lone pair of electrons that are on the chlorine there. So an ortho para director for the chlorine. So what's ortho to this chlorine here. Well this spot is ortho. So is this spot and so is this spot. And so we have a couple of different possible products here. And let's go ahead and start drawing them here. So if I think about the product--"
    },
    {
        "Q": "ok so on 5:38 he talks about hydrophobic stuff. So is oil hydrophobic?",
        "A": "Yes. The reason being that oil cannot be mixed into water, instead is seprerates and that is the meaning of hydrophobic",
        "video_name": "QymONNa5C6s",
        "timestamps": [
            338
        ],
        "3min_transcript": "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 of the eccentrics, a man named Henry Cavendish. He communicated with his female servants only via notes, and added a staircase to the back of his house to avoid contact with his housekeeper. Some believe he may have suffered from a form of autism, but just about everyone will admit that he was a scientific genius. He's best remembered as the first person to recognize hydrogen gas as a distinct substance and to determine the composition of water. In the 1700s, most people thought that water itself was an element, but Cavendish observed that hydrogen, which he called inflammable air, reacted with oxygen, known then by the awesome name, dephlogisticated air, to form water. Cavendish didn't totally understand what he'd discovered here, in part because he didn't believe in chemical compounds. He explained his experiments with hydrogen in terms of a fire-like element called phlagiston. Nevertheless, his experiments were groundbreaking. Like his work in determining the specific gravity basically the comparative density of hydrogen"
    },
    {
        "Q": "why is the water blue at 3:39",
        "A": "Blue food coloring/dye was mixed with the water in order to make it more visible on the kitchen counter (since water is clear it is difficult to see).",
        "video_name": "QymONNa5C6s",
        "timestamps": [
            219
        ],
        "3min_transcript": "This is actually the way that it appears. It is v-shaped. Because this big old oxygen atom is a little bit more greedy for electrons, it has a slight negative charge; whereas, this area here with the hydrogen atoms has a slight positive charge. Thanks to this polarity, all water molecules are attracted to one another; so much so that they actually stick together and these are called hydrogen bonds. We talked about them last time, but essentially what happens is that the positive pole around those hydrogen atoms bonds to the negative pole around the oxygen atoms of a different water molecule. It's a weak bond, but look, they're bonding! Seriously, I cannot overstate the importance of this hydrogen bond. When your teacher asks you, \"What's important about water?\" start out with the hydrogen bonds and you should put it in all caps and maybe some sparkles around it. One of the cool properties that results from these hydrogen bonds is a high cohesion for water which results in high surface tension. Cohesion is the attraction between two like things, like attraction between one molecule of water and another molecule of water. Water has the highest cohesion of any nonmetalic liquid. 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"
    },
    {
        "Q": "at 13:50 what happens to the H on the Oxygen that attacks the electrophilic carbon (leading to a cycle product)?",
        "A": "During the workup , the reaction mixture is probably treated with dilute sodium carbonate solution. The hydroxide ions would neutralize the H from the oxonium ion.",
        "video_name": "8-ccnvn9DxI",
        "timestamps": [
            830
        ],
        "3min_transcript": "reacting it with ethylene glycol, and, once again, we use Toluenesulfonic acid, as our catalyst. And this one's a little bit different, because we can see we have a diol, as one of our reactants; up here, we just had butanol, only one OH, but this one has two on it. So, trying to figure out the product here, sometimes it helps just to run through the mechanism really quickly, and so the Toluenesulfonic acid is going to help us to protonate our carbon EEL, and then we have our nucleophile attack, so one of these OHs is going to attack here. And so, without going through all the steps in the mechanism again, that was obviously a pretty complicated mechanism, I'll jump to one of the later steps of the mechanism, where we have already lost water, so minus H two O, so we've already gotten past the dehydration step. And then that would give us this as our intermediate, so there is actually gonna be a plus one formal charge on this oxygen. And then we have these two carbons over here, so let's go ahead, and color-coordinate some of our atoms once again. So, this oxygen has already bonded, we've already lost water, so that oxygen is this oxygen, right here. And then, we still have another OH on this molecule, and that's this one over here, like that. So when we get to this step, we're actually gonna get an intra-molecular, nucleophilic attack. So, these electrons are going to attack this carbon, and kick these electrons off, onto this oxygen. And so, when you think about the final product, you're actually gonna get a cyclic product here, a cyclic acetone. So, we would have our four carbons, and then we would have this oxygen, and then two carbons, and then this oxygen, and they're both bonded to this carbon right here. And so, once again, let's highlight some of those carbons: so this carbon right here, and this carbon right here, or this carbon, and this carbon, and, in our final product, like that. so a little bit trickier than the previous reactions. So in the next video, we'll see a use of cyclic acetals as a protecting group."
    },
    {
        "Q": "At 8:43. why does ethanol attack the carbonyl carbon and not the oxygen as oxygen possesses the positive charge (3 bonds) while the carbonyl doesn't? Also wouldn't there be less steric hindrance if it attacked the oxygen?",
        "A": "That would make the oxygen violate the octet rule",
        "video_name": "8-ccnvn9DxI",
        "timestamps": [
            523
        ],
        "3min_transcript": "then that would kick these electrons off onto the oxygen, and then we would have water. So, this is the dehydration portion, so we're gonna form water. So let me go ahead, and mark this as being the next step, right? So, in the next step, when those electrons kick in there, so this would be step five, we're going to lose H two O, so the dehydration step. And we would be left with, once again, our ring, and, this time, a double bond to this oxygen, with an ethyl coming off of that oxygen like this. So let's go ahead and make sure we still have a lone pair of electrons on this oxygen, and a plus one formal charge, and the electrons in green, so these electrons in here, moved in here to give us our double bond once again. And, once again, we have a plus one formal charge on the oxygen, so if you drew a resonance structure for this, as being very electrophilic. So, once again, we're going to get a nucleophile attacking our electrophile in the next step, so this would be step six. So, step six would be a nucleophilic attack. So, in step six, a nucleophile comes along, once again, ethanol is our nucleophile, so here is ethanol, so let's go ahead and show ethanol right here, with lone pairs of electrons. And one of these lone pairs of electrons, of course, would attack our electrophile, so nucleophile attacks electrophile, and that would push these electrons in here off onto this oxygen. So let's go ahead, and draw what we have next. Alright, so we now have an oxygen, with still a hydrogen on it, and ethyl right here, a lone pair of electrons, a plus one formal charge on this oxygen. So, let's highlight those electrons: so, in magenta here, these electrons formed a bond, And then over here on the right, we have an oxygen, with an ethyl group, and now there are two lone pairs of electrons on this oxygen. So, we are almost there, right, last step. So, step seven would be a deprotonation step. So in step seven here, all we have to do is take that proton off, and we would form our acetal product. So, once again, we could have a molecule of ethanol come along, and function as a base, and so, a lone pair of electrons take this proton, leaving these electrons behind, on the oxygen, and then finally we are able to draw our acetal products. So we would have, let's go ahead and make this a little bit more angled, so on the left, we would have our oxygen, with an ethyl, and then this carbon is also bonded to another oxygen, with an ethyl coming off of it like that. And so, let's go ahead and show those final electrons here, on our oxygen like this, and, once again, highlight these electrons,"
    },
    {
        "Q": "At 1:50, how was Sal able to infer that acceleration = gravity = -10m/s^2? Was there some kind of formula or rule he used or is that a constant of some sort?",
        "A": "That s the acceleration due to gravity on the surface of the earth. We measure it.",
        "video_name": "15zliAL4llE",
        "timestamps": [
            110
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 4:24 the equation that was used was d=v*t . Could we have used the formula vf^2=vi^2+2ad",
        "A": "Yes, They ll both give the same answer as long as you know the value of a",
        "video_name": "15zliAL4llE",
        "timestamps": [
            264
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 5:06, Sal gives us the equation delta V=Vf-Vi. Isn't it necessary for (vf-vi) to be divided by time to find the average velocity?",
        "A": "No, it isn t. The change in velocity divided by the change in time gives you the acceleration.",
        "video_name": "15zliAL4llE",
        "timestamps": [
            306
        ],
        "3min_transcript": ""
    },
    {
        "Q": "how come distance be -500m at 9:08 as told in the video??",
        "A": "at 9:08 the distance is -500m because the average velocity is -50m/s the time is 10s and if they both are multiplied the answer would be -500m if you watch the video again you will find how will it come.",
        "video_name": "15zliAL4llE",
        "timestamps": [
            548
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 6:15, how did you get 10? Did you find the square root of 100 to get 10?",
        "A": "When you look at the equation, you ll realize that the change in velocity = acceleration X time. We know that the change in velocity (final velocity - initial velocity) is -100 because -100 - 0 = -100. We also know that acceleration is -10 m/s^2 because it s gravity. Then this gives -100 = -10 X t. It s clear now that t=10. Hope that helped!",
        "video_name": "15zliAL4llE",
        "timestamps": [
            375
        ],
        "3min_transcript": ""
    },
    {
        "Q": "Sal talks about 7:07 into the video that since momentum is conserved, the speed must be adjusted to the new weight. What happens in an inelastic collision where the two objects stop dead in their tracks? Any mass times 0 velocity always equals zero, so how can momentum be conserved in an inelastic collision?",
        "A": "Let s say (to make it easy) that the colliding objects have equal mass and equal speed. Note that equal speed does not mean equal velocity - the objects are headed toward each other, so one of them has velocity exactly equal to the negative of the other one. Before collision, total momentum is: m* v1 + m*v2 = m*(v1 -v2) = 0! So you see now where this is going, right? After the collision, they are stuck together sitting there, so momentum is 0. 0 before, 0 after means momentum was conserved.",
        "video_name": "XFhntPxow0U",
        "timestamps": [
            427
        ],
        "3min_transcript": "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. vehicle called a car truck, its momentum will have to be the same as the car and the truck's momentum when they were separate. So what do we know about this car truck object? Well we know its new mass. The car truck object, it will be the combined mass of the two. So it's 1,000 kilograms plus 2,000 kilograms. So it's 3,000 kilograms. And now we can use that information to figure out its velocity. How? Well, its momentum-- this 3,000 kilogram object's momentum-- has to be the same as the momentum of the two objects before the collision. So it still has to be 9,000 kilogram meters per second. So once again, mass times velocity. So mass is 3,000 times the new velocity. So we could call that, I don't know, new velocity, v sub n. That will equal 9,000."
    },
    {
        "Q": "you said we need the mass of the moving object, but what happens if I am not given the mass of the car? like said at 5:30",
        "A": "To calculate momentum, mass is an essential ingredient. However, you might be given the mass indirectly so to say. E.g. One of the cars can move with 40N with an acceleration with 4m/s. This automatically tells you that the mass is 10kg. Thus either way, you must have the mass somehow. Or else it is impossible to calculate the momentum.",
        "video_name": "XFhntPxow0U",
        "timestamps": [
            330
        ],
        "3min_transcript": "they use for impulse. But another way of viewing impulse is force times change in time. Well that's the same thing as change in momentum over change in time times change in time. Right? Because this is just the same thing as force. And that's just change in momentum, so that's impulse as well. And the unit of impulse is the joule. And we'll go more into the joule when we do work in all of that. And if this confuses you, don't worry about it too much. The main thing about momentum is that you realize it's mass times velocity. And since force is change in momentum per unit of time, if you don't have any external forces on a system or, on say, on a set of objects, their combined, or their net momentum won't change. And that comes from Newton's Laws. The only way you can get a combined change in momentum is if you have some type of net force acting on the system. 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"
    },
    {
        "Q": "At 4:54, why can't we assume switching the positions of CH3 and Br? Since they're all single bonds, wouldn't they rotate freely anyways?",
        "A": "No, that s the point. No matter how you try you ll find it s impossible to rotate them in such a way that the 2 molecules can fit on top of one another with each atom in the same place. If you cannot visualise it, a molecular model kit will be able to prove it.",
        "video_name": "tk-SNvCPLCE",
        "timestamps": [
            294
        ],
        "3min_transcript": "should say usually, are carbons, especially when we're dealing in organic chemistry, but they could be phosphoruses or sulfurs, but usually are carbons bonded to four different groups. And I want to emphasize groups, not just four different atoms. And to kind of highlight a molecule that contains a chiral atom or chiral carbon, we can just think of one. So let's say that I have a carbon right here, and I'm going to set this up so this is actually a chiral atom, that the carbon specific is a chiral atom, but it's partly a chiral molecule. And then we'll see examples that one or both of these are true. Let's say it's bonded to a methyl group. Let's say there's a bromine over here. Let's say behind it, there is a hydrogen, and then above it, we have a fluorine. Now if I were to take the mirror image of this thing right here, we have your carbon in the center-- I want to do it in that same blue. You have the carbon in the center and then you have the fluorine above the carbon. You have your bromine now going in this direction. You have this methyl group. It's still popping out of the page, but it's now going to the right instead of to the left, So CH3. And then you have the hydrogen still in the back. These are mirror images, if you view this as kind of the mirror and you can see on both sides of the mirror. Now, why is this chiral? Well, it's a little bit of a visualization challenge, but no matter how you try to rotate this thing right here, you will never make it exactly like this thing. You might try to rotate it around like that and try to So let's try to do that. If we try to get the methyl group over there, what's going to happen to the other groups? Well, then the hydrogen group is going-- or the hydrogen, I The hydrogen atom is going to move there and the bromine is going to move there. So this would be superimposable if this was a hydrogen and this was a bromine, but it's not. You can imagine, the hydrogen and bromine are switched. And you could flip it and do whatever else you want or try to rotate it in any direction, but you're not going to be able to superimpose them. So this molecule right here is a chiral molecule, and this carbon is a chiral center, so this carbon is a chiral carbon, sometimes called an asymmetric carbon or a chiral center. Sometimes you'll hear something called a stereocenter. A stereocenter is a more general term for any point in"
    },
    {
        "Q": "I actually cannot understand why will the molecule be not superimposble (at 5:36).",
        "A": "He is arguing, correctly, that the images are NOT superimposable because the carbon is chiral.",
        "video_name": "tk-SNvCPLCE",
        "timestamps": [
            336
        ],
        "3min_transcript": "Let's say there's a bromine over here. Let's say behind it, there is a hydrogen, and then above it, we have a fluorine. Now if I were to take the mirror image of this thing right here, we have your carbon in the center-- I want to do it in that same blue. You have the carbon in the center and then you have the fluorine above the carbon. You have your bromine now going in this direction. You have this methyl group. It's still popping out of the page, but it's now going to the right instead of to the left, So CH3. And then you have the hydrogen still in the back. These are mirror images, if you view this as kind of the mirror and you can see on both sides of the mirror. Now, why is this chiral? Well, it's a little bit of a visualization challenge, but no matter how you try to rotate this thing right here, you will never make it exactly like this thing. You might try to rotate it around like that and try to So let's try to do that. If we try to get the methyl group over there, what's going to happen to the other groups? Well, then the hydrogen group is going-- or the hydrogen, I The hydrogen atom is going to move there and the bromine is going to move there. So this would be superimposable if this was a hydrogen and this was a bromine, but it's not. You can imagine, the hydrogen and bromine are switched. And you could flip it and do whatever else you want or try to rotate it in any direction, but you're not going to be able to superimpose them. So this molecule right here is a chiral molecule, and this carbon is a chiral center, so this carbon is a chiral carbon, sometimes called an asymmetric carbon or a chiral center. Sometimes you'll hear something called a stereocenter. A stereocenter is a more general term for any point in groups that it is joined to. But all of these, especially when you're in kind of in introductory organic chemistry class, tends to be a carbon bonded to four different groups. And I want to to stress that it's not four different atoms. You could have had a methyl group here and a propyl group here, and the carbon would still be bonded directly to a carbon in either case, but that would still be a chiral carbon, and this would still actually be a chiral molecule. In the next video, we'll do a bunch of examples. We'll look at molecules, try to identify the chiral carbons, and then try to figure out whether the molecule itself is--"
    },
    {
        "Q": "At 6:00 Sal says that an uniform electric field can be formed. But,as i have understood, it is uniform only along the plate not perpendicular to it and thus we cant construct one even with an infinitely long charged plate.(the way it is shown in the diagram i.e. perpendicular to the plate as the field is proportional to 1/d where d is the distance from the wire.\nIs this correct?",
        "A": "It is close to uniform as long as the distance between the plates is much smaller than the plates. Also, it is not uniform near the edges.",
        "video_name": "elJUghWSVh4",
        "timestamps": [
            360
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 2:58 Sal says that the upward force we would need to apply to push a certain object up is equal to the force of gravity. But wouldn't tht mean tht the net force on the object equals 0 and it would not move upwards?\nPlease help? Sorry if I'm missing something here.",
        "A": "Sal is saying that after getting the object to accelerate a little then we have to apply the same force as gravity. As we have already caused the object to accelerate we then only have to keep the object moving to go up. And only by applying the same and opposite force as gravity, we can do that.",
        "video_name": "elJUghWSVh4",
        "timestamps": [
            178
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 7:23,velocity at time t=0 should be 2 m/s right why is it 1 m/s?",
        "A": "You probably mixed the concept of velocity and change in velocity (or acceleration). At t=0, the velocity David assume is 1 m/s, while the acceleration is 2 m/s^2.",
        "video_name": "DD58B2siDv0",
        "timestamps": [
            443
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "at 8:05, how do we know that the velocity at 0(zero) second was 1?",
        "A": "It is just an assumption.",
        "video_name": "DD58B2siDv0",
        "timestamps": [
            485
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "At 6:10, shouldn't the 'd' of 'Dicyclobutyl' be in small case, i.e. 'dicyclobutyl' ?",
        "A": "The D would be capitalized at the beginning of a sentence, but not anywhere else.",
        "video_name": "ygXkdSKXQoA",
        "timestamps": [
            370
        ],
        "3min_transcript": "we're trying to minimize the numbers here. We might want to go clockwise so we hit two right over here, three, four, five, six, seven. So in this situation where do we hit interesting things? We hit interesting things at the one, at the two, or where do we have groups attached to the chain? One, two, four, six, and seven. So that's one option. We'll have groups attached at one, two, four, six, and seven if we start right over there, and if we were to go clockwise. Our other option, is to start, and let me erase those, at this other cyclobutyl group here on the left. Let me erase these so I don't mess up the diagram too much. The other option is to start here, make this the one, number this as one, to hit the cyclopropyl, to hit something else as soon as possible. So two, three, four, five, six, seven. So now where we are getting interesting things? We have groups at the one carbon, the two carbon, the four carbon, the five carbon, and at the seven carbon. So actually this second numbering is preferable. Both of them have something at the one, the two, the four, and the seven, but the second one has something at a five while the first one had something at a six. So we would actually want to do this numbering right over here, and this is the numbering that we have now listed right over here. Let's now write what the name of this molecule actually is. So once again, we start first in alphabetical order. So we're going to start with the cyclobutyl groups. And we have two of them, one at the one carbon, So we could say 1 comma 4. And since there are two cyclobutyl groups we would say dicyclobutyl. 1, 4 dicyclobutyl. And then what comes next in alphabetical order is the cyclopropyl. That comes before isobutyl. \"C\" comes before \"I.\" That's at the two carbon. So then we can say 2 cyclopropyl. And now we can get to the two isobutyls. So there's an isobutyl at the five carbon and at the seven carbon. So we can say 5, 7, and since there's two of them we'd say di isobutyl. 5, 7 di isobutyl."
    },
    {
        "Q": "At 5:40 how would we number it if the numbers were\n1)1,2,4,6,7\n2)1,2,4,5,8\nBecause ive heard we always take the lowest sum, but what if the sum were equal?",
        "A": "5 is lower than 6, the sum doesn t actually matter.",
        "video_name": "ygXkdSKXQoA",
        "timestamps": [
            340
        ],
        "3min_transcript": "we're trying to minimize the numbers here. We might want to go clockwise so we hit two right over here, three, four, five, six, seven. So in this situation where do we hit interesting things? We hit interesting things at the one, at the two, or where do we have groups attached to the chain? One, two, four, six, and seven. So that's one option. We'll have groups attached at one, two, four, six, and seven if we start right over there, and if we were to go clockwise. Our other option, is to start, and let me erase those, at this other cyclobutyl group here on the left. Let me erase these so I don't mess up the diagram too much. The other option is to start here, make this the one, number this as one, to hit the cyclopropyl, to hit something else as soon as possible. So two, three, four, five, six, seven. So now where we are getting interesting things? We have groups at the one carbon, the two carbon, the four carbon, the five carbon, and at the seven carbon. So actually this second numbering is preferable. Both of them have something at the one, the two, the four, and the seven, but the second one has something at a five while the first one had something at a six. So we would actually want to do this numbering right over here, and this is the numbering that we have now listed right over here. Let's now write what the name of this molecule actually is. So once again, we start first in alphabetical order. So we're going to start with the cyclobutyl groups. And we have two of them, one at the one carbon, So we could say 1 comma 4. And since there are two cyclobutyl groups we would say dicyclobutyl. 1, 4 dicyclobutyl. And then what comes next in alphabetical order is the cyclopropyl. That comes before isobutyl. \"C\" comes before \"I.\" That's at the two carbon. So then we can say 2 cyclopropyl. And now we can get to the two isobutyls. So there's an isobutyl at the five carbon and at the seven carbon. So we can say 5, 7, and since there's two of them we'd say di isobutyl. 5, 7 di isobutyl."
    },
    {
        "Q": "At 9:42 Sal draws a circle and a dot in the centre to say that the vector is pointing out of the page, though what do you draw/signal to say that the vector pointing away from you? Thank you to whoever can answer this.",
        "A": "You draw a circle with an x in it. The idea is that the one coming toward you looks like the head of an arrow coming at you and the one going away from you looks like the feathers of an arrow going away.",
        "video_name": "s38l6nmTrvM",
        "timestamps": [
            582
        ],
        "3min_transcript": "The magnitude of our force vector times sine of theta, that gave us the component of the force vector that is perpendicular to the arm. And we just multiply that times the magnitude of r, and we got the magnitude of the torque vector, which was 15. We can leave out the newton meters for now. 15, and then its direction is this vector that we specified by n. We can call it the normal vector. And what do we know about this vector? It's perpendicular to both r-- this is r, right-- and it's perpendicular to F. And the only way that I can visualize in our three-dimensional universe, a vector that's perpendicular to both this and this is if it pops in or out of this page, right? Because both of these vectors are in the plane that are defined by our video. So if I'm a vector that is perpendicular to your screen, whatever you're watching this on, then it's going to be perpendicular to both of these vectors. into the page? We use the right hand rule, right? In the right hand rule, we take-- r is our index finger, F is our middle finger, and whichever direction our thumb points in tells us whether or not we are-- the direction of the cross product. So let's draw it. Let me see if I can do a good job right here. So if that is my index finger, and you could imagine your hand sitting on top of this screen. So that's my index finger representing r, and this is my Remember, it only works with your right hand. If you do your left hand, it's going to be the opposite. And then my middle finger is going to go in the direction you to draw this. So if I were to draw it-- let me draw my nails just so you know what this is. So this is the nail on my index finger. This is the nail on my middle finger. And so in this situation, where is my thumb going to be? My thumb is going to be popping out. I wish I could-- that's the nail of my thumb. Hopefully, that makes some sense, right? That's the palm of my hand. That's the other side of my-- and I could keep drawing, but hopefully, that makes some sense. This is my index finger. This is the middle finger. My thumb is pointing out of the page, so that tells us that the torque is actually pointing out of the page. So the direction of this unit vector n is going to be out of the page, and we could signify that by a circle with a dot. And I'm almost at my time limit, and so there you have it: the cross product as it is applied to torque. See you in the next video."
    },
    {
        "Q": "@ 1:35 how do we predict when something will dissociate and when it will not? Why doesn't the Carbon or any other part of the sorbate dissociate?",
        "A": "This sort of thing will come to you with time doing practice problems. In general though just remember that water is not strong enough to break C-C or C-H bonds.",
        "video_name": "jzcB3faNdq0",
        "timestamps": [
            95
        ],
        "3min_transcript": "- [Voiceover] Potassium sorbate, and they give us its formula right over here, has a molar mass of 150 grams per mole. They put this decimal here to show us that these are actually three significant figures. Even the zero is a significant digit here. Is commonly added to diet soft drinks as a preservative. A stock solution of potassium sorbate, dissolvent's an aqueous solution here, of known concentration must be prepared. A student titrates 45 millileters of the stock solution with one point two five molar hydrochloric acid using both an indicator and a pH meter. The value of K A for sorbic acid is one point seven times ten to the negative fifth. All right, so let's tackle this piece by piece. Write the net ionic equation for the reaction between potassium sorbate and hydrochloric acid. All right, so first off, let's write the ionic equation, and then we'll, I'll write the net ionic equation, and hopefully you'll see the difference. So ionic, ionic equation. 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."
    },
    {
        "Q": "What is coulom force at 6:09?",
        "A": "A Coulomb force is the attractive force between positive and negative charges.",
        "video_name": "q--2WP8wXtk",
        "timestamps": [
            369
        ],
        "3min_transcript": "there the electrons in that bond could spend some of their time on this atom and some of their time on this atom right over here. And so when you have a covalent bond like this, you can then find the distance between the 2 nuclei and take half of that and call that call that the atomic radius. So these are all different ways of thinking about it. Now, with that out of the way, let's think about what the trends for atomic size or atomic radii would be in the periodic table. So the first thing to think about is what do you think will be the trend for atomic radii as we move through a period. So let's say we're in the fourth period and we were to go from potassium to krypton. What do you think is going to be the trend here? And if you want to think about the extremes, how do you think potassium is going to compare to krypton in terms of atomic radius. I encourage you to pause this video and think about that on your own. the outermost electrons are going to be in your fourth shell. Here, you're filling out 4S1, 4S2. Then you start back filling into the 3D subshell and then you start filling again in 4P1 and so forth. You start filling out the P subshell. So as you go from potassium to krypton, you're filling out that outermost fourth shell. Now what's going on there? Well, when you're at potassium, you have 19- 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19. You have 19 protons and you have 19 electrons. Well I'll just draw those. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, but you only have 1 electron in that outermost, in that fourth shell, so let's just say that's that electron at a moment, just for visually. It doesn't necessarily have to be there but just to visualize that. So that 1 electron right over there, 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,"
    },
    {
        "Q": "At 6:38 Sal said \" 2s and 2p\". But aren't those 4s and 4p orbitals?",
        "A": "Yes, they are 4s and 4p orbitals",
        "video_name": "q--2WP8wXtk",
        "timestamps": [
            398
        ],
        "3min_transcript": "the outermost electrons are going to be in your fourth shell. Here, you're filling out 4S1, 4S2. Then you start back filling into the 3D subshell and then you start filling again in 4P1 and so forth. You start filling out the P subshell. So as you go from potassium to krypton, you're filling out that outermost fourth shell. Now what's going on there? Well, when you're at potassium, you have 19- 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19. You have 19 protons and you have 19 electrons. Well I'll just draw those. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, but you only have 1 electron in that outermost, in that fourth shell, so let's just say that's that electron at a moment, just for visually. It doesn't necessarily have to be there but just to visualize that. So that 1 electron right over there, 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,"
    },
    {
        "Q": "At 3:38,sal says atooms are joined by covalent bonds ......so my first question is what are covalent bonds and how can we find radius if atom is not joined to anyone??",
        "A": "Covalent bonds are bonds between two atoms sharing electrons.",
        "video_name": "q--2WP8wXtk",
        "timestamps": [
            218
        ],
        "3min_transcript": "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 there the electrons in that bond could spend some of their time on this atom and some of their time on this atom right over here. And so when you have a covalent bond like this, you can then find the distance between the 2 nuclei and take half of that and call that call that the atomic radius. So these are all different ways of thinking about it. Now, with that out of the way, let's think about what the trends for atomic size or atomic radii would be in the periodic table. So the first thing to think about is what do you think will be the trend for atomic radii as we move through a period. So let's say we're in the fourth period and we were to go from potassium to krypton. What do you think is going to be the trend here? And if you want to think about the extremes, how do you think potassium is going to compare to krypton in terms of atomic radius. I encourage you to pause this video and think about that on your own."
    },
    {
        "Q": "At 7:15, do you have to write the CH2 as carbon bonded with 2 hydrogen atoms? Can't you just write it as CH2?",
        "A": "The formula with two separate H atoms is a structural formula. The formula with a CH\u00e2\u0082\u0082 group is a condensed structural formula. You decide which one you want to use for your own purposes.",
        "video_name": "pMoA65Dj-zk",
        "timestamps": [
            435
        ],
        "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 1:14 isn't the metals \"magic number\" 18, as they want to fill their d-orbital.",
        "A": "Yes, but that isn t something that should come up in organic chemistry as it almost always deals with C H N O (and a few others here and there...)",
        "video_name": "pMoA65Dj-zk",
        "timestamps": [
            74
        ],
        "3min_transcript": "The one thing that probably causes some of the most pain in chemistry, and in organic chemistry, in particular, is just the notation and the nomenclature or the naming that we use. And what I want to do here in this video and really the next few videos is to just make sure we have a firm grounding in the notation and in the nomenclature or how we name things, and then everything else will hopefully not be too difficult. So just to start off, and this is really a little bit of review of regular chemistry, if I just have a chain of carbons, and organic chemistry is dealing with chains of carbons. Let me just draw a one-carbon chain, so it's really kind of ridiculous to call it a chain, but if we have one carbon over here and it has four valence electrons, it wants to get to eight. That's the magic number we learned in just regular chemistry. For all molecules, that's the stable valence structure, I guess you could say it. A good partner to bond with is hydrogen. So it has four valence electrons and then hydrogen electron with each other and then they both look pretty happy. I said eight's the magic number for everybody except for hydrogen and helium. Both of them are happy because they're only trying to fill their 1s orbital, so the magic number for those two guys is two. So all of the hydrogens now feel like they have two electrons. The carbon feels like it has eight. Now, there's several ways to write this. You could write it just like this and you can see the electrons explicitly, or you can draw little lines here. So I could also write this exact molecule, which is methane, and we'll talk a little bit more about why it's called methane later in this video. I can write this exact structure like this: a carbon bonded to four hydrogens. And the way that I've written these bonds right here you could imagine that each of these bonds consists of two 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"
    },
    {
        "Q": "why is hydrogen the best partner as explained by sal in 0:56?\nwhy can't other atoms be used?",
        "A": "other atoms can be used but the main problem is that we have to draw other electrons of the atom also. For eg:in CO2 we have to draw the rest 4 electrons of oxygen & also number compounds of carbon-hydrogen is more than 1000.",
        "video_name": "pMoA65Dj-zk",
        "timestamps": [
            56
        ],
        "3min_transcript": "The one thing that probably causes some of the most pain in chemistry, and in organic chemistry, in particular, is just the notation and the nomenclature or the naming that we use. And what I want to do here in this video and really the next few videos is to just make sure we have a firm grounding in the notation and in the nomenclature or how we name things, and then everything else will hopefully not be too difficult. So just to start off, and this is really a little bit of review of regular chemistry, if I just have a chain of carbons, and organic chemistry is dealing with chains of carbons. Let me just draw a one-carbon chain, so it's really kind of ridiculous to call it a chain, but if we have one carbon over here and it has four valence electrons, it wants to get to eight. That's the magic number we learned in just regular chemistry. For all molecules, that's the stable valence structure, I guess you could say it. A good partner to bond with is hydrogen. So it has four valence electrons and then hydrogen electron with each other and then they both look pretty happy. I said eight's the magic number for everybody except for hydrogen and helium. Both of them are happy because they're only trying to fill their 1s orbital, so the magic number for those two guys is two. So all of the hydrogens now feel like they have two electrons. The carbon feels like it has eight. Now, there's several ways to write this. You could write it just like this and you can see the electrons explicitly, or you can draw little lines here. So I could also write this exact molecule, which is methane, and we'll talk a little bit more about why it's called methane later in this video. I can write this exact structure like this: a carbon bonded to four hydrogens. And the way that I've written these bonds right here you could imagine that each of these bonds consists of two 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"
    },
    {
        "Q": "at 6:30, why is the final velocity -5m/s? Wouldn't it be 0m/s because the rocket is not moving at all, it is sitting on the ground? I understand that negatives are used when the rocket is going down, but the final velocity would be when the velocity at the very end of the experiment, when the rocket lands and is sitting stationary on the ground.",
        "A": "The equation of motion for a projectile does not include the effect of collisions, so the final velocity is the velocity the projectile has the moment it reaches a surface. After that point, the projectile EOM no longer holds since there are additional forces acting on it.",
        "video_name": "ZZ39o1rAZWY",
        "timestamps": [
            390
        ],
        "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 11:41, why is the average velocity in the horizontal direction is 5 square roots of 3 metres per second? I know Sal said it is because it doesn't change, but why does it not change?",
        "A": "Gravity only affects the velocity in the vertical direction, and since we are assuming that there is no air resistance, there is nothing to change the horizontal velocity.",
        "video_name": "ZZ39o1rAZWY",
        "timestamps": [
            701
        ],
        "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": "How can you prove mathematically that the initial velocity will be the same as the final velocity in this situation? ( 5:37 ) Can I calculate this somehow from v = at or s = 1/2(at^2)?",
        "A": "There are several approaches. Some have to do with potential and kinetic energy. And it s not v = at, it s (v-u) = at, the formula most appropriate to that is v^2=u^2 + 2as, when the displacement (vertical) is zero then v^2=u^2 and since you know that the object will be moving in the opposite direction (vertically) when it lands than it was when it was launched then it is necessary that v = -u.",
        "video_name": "ZZ39o1rAZWY",
        "timestamps": [
            337
        ],
        "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": "7:08 shouldn't the 10 be positive since it is actually 5-(-5) = 5+5",
        "A": "At 7:08 the equation he just finished was -5 -5 = (-5) + (-5) = -10",
        "video_name": "ZZ39o1rAZWY",
        "timestamps": [
            428
        ],
        "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": "8:46 If the answer to acceleration or velocity is \"0\", is it necessary to include units?",
        "A": "It s better if you do. Best ask your teacher if you really feel like being lazy :)",
        "video_name": "d-_eqgj5-K8",
        "timestamps": [
            526
        ],
        "3min_transcript": "to be the distance traveled. And lucky for us, this is just going to be a triangle, and we know how to figure out the area for triangle. So the area of a triangle is equal to one half times base times height. Which hopefully makes sense to you, because if you just multiply base times height, you get the area for the entire rectangle, and the triangle is exactly half of that. So the distance traveled in this situation, or I should say the displacement, just because we want to make sure we're focused on vectors. The displacement here is going to be-- or I should say the magnitude of the displacement, maybe, which is the same thing as the distance, is going to be one half times the base, which is five seconds, times the height, which is five meters per second. Times five meters. Five meters per second. The seconds cancel out with the seconds. And we're left with one half times five times five meters. So it's one half times 25, which is equal to 12.5 meters. And so there's an interesting thing here, well one, there's a couple of interesting things. Hopefully you'll realize that if you're plotting velocity versus time, the area under the curve, given a certain amount of time, tells you how far you have traveled. The other interesting thing is that the slope of the curve tells you your acceleration. What's the slope over here? Well, It's completely flat. And that's because the velocity isn't changing. So in this situation, we have a constant acceleration. The magnitude of that acceleration is exactly zero. Our velocity is not changing. Here we have an acceleration of one meter per second squared, and that's why the slope of this line right over here is one. The other interesting thing, is, if even if you have constant acceleration, you could still figure out the distance We were able to figure out there we were able to get 12.5 meters. The last thing I want to introduce you to-- actually, let me just do it until next video, and I'll introduce you to the idea of average velocity. Now that we feel comfortable with the idea, that the distance you traveled is the area under the velocity versus time curve."
    },
    {
        "Q": "At 7:18, the displacement is taken to be the area under the slope i.e., a triangle. why isn't the other side of the slope considered, tat would form a rectangle?",
        "A": "If you go at a constant velocity of 10 m/s, the graph of that is a straight horizontal line at 10 m/s, right? If you go that velocity for 10 s, how far did you go? 100 m, right? You got that by doing 10 m/s x 10 s = 100 m, which is the area UNDER that horizontal line. Now, did we need to add the area above the straight line? No, why would we do that, right? It doesn t have to do with anything. The same idea holds when the line is tilted. The area above the line has no signficance. It s infinite, too.",
        "video_name": "d-_eqgj5-K8",
        "timestamps": [
            438
        ],
        "3min_transcript": "we have a constant slope. So we have just a line here. We don't have a curve. Now what I want to do is think about a situation. Let's say that we accelerate it one meter per second squared. And we do it for-- so the change in time is going to be five seconds. And my question to you is how far have we traveled? Which is a slightly more interesting question than what we've been asking so far. So we start off with an initial velocity of zero. And then for five seconds we accelerate it one meter per second squared. So one, two, three, four, five. So this is where we go. This is where we are. So after five seconds, we know our velocity. Our velocity is now five meters per second. But how far have we traveled? So we could think about it a little bit visually. We could say, look, we could try to draw rectangles over here. of one meter per second. So if I say one meter per second times the second, that'll give me a little bit of distance. And then the next one I have a little bit more of distance, calculated the same way. I could keep drawing these rectangles here, but then you're like, wait, those rectangles are missing, because I wasn't for the whole second, I wasn't only going one meter per second. I kept accelerating. So I actually, I should maybe split up the rectangles. I could split up the rectangles even more. So maybe I go every half second. So on this half-second I was going at this velocity. And I go that velocity for a half-second. Velocity times the time would give me the displacement. And I do it for the next half second. Same exact idea here. Gives me the displacement. So on and so forth. But I think what you see as you're getting-- is the more accurate-- the smaller the rectangles, you try to make here, the closer you're going to get to the area under this curve. And just like the situation here. to be the distance traveled. And lucky for us, this is just going to be a triangle, and we know how to figure out the area for triangle. So the area of a triangle is equal to one half times base times height. Which hopefully makes sense to you, because if you just multiply base times height, you get the area for the entire rectangle, and the triangle is exactly half of that. So the distance traveled in this situation, or I should say the displacement, just because we want to make sure we're focused on vectors. The displacement here is going to be-- or I should say the magnitude of the displacement, maybe, which is the same thing as the distance, is going to be one half times the base, which is five seconds, times the height, which is five meters per second. Times five meters."
    },
    {
        "Q": "At 9:58, Sal says that He fuses to C and O.Why does He not get fused into other elements like Lithium,Boron or Nitrogen etc.?",
        "A": "It does, but lithium, boron, and nitrogen actually have a lower fusion temperature than helium so they fuse instantly.",
        "video_name": "kJSOqlcFpJw",
        "timestamps": [
            598
        ],
        "3min_transcript": "And the temperature here keeps going up. So we said that the first ignition, the first fusion, occurs at around 10 million Kelvin. This thing will keep heating up until it gets to 100 million Kelvin. And now I'm talking about a star that's about as massive as the sun. Some stars will never even be massive enough to condense the core so that its temperature reaches But let's just talk about the case in which it does. So eventually, you'll get to a point-- so we're still sitting in the red giant phase, so we're this huge star over here. We have this helium core. And that helium core keeps getting condensed and condensed and condensed. And then we have a shell of hydrogen that keeps fusing into helium around it. So this is our hydrogen shell. Hydrogen fusion is occurring in this yellow shell over here that's causing the radius of the star to get bigger and bigger, to expand. But when the temperature get sufficiently hot-- and now I think you're going to get a sense of how heavier and heavier elements form in the universe, and all of the heavy elements that you see around us, including the ones that are in you, were formed it this way from, initially, hydrogen-- when it gets hot enough at 100 million Kelvin, in this core, because of such enormous pressures, then the helium itself will start to fuse. So then we're going to have a core in here where the helium itself will start to fuse. And now we're talking about a situation. You have helium, and you had hydrogen. And all sorts of combinations will form. But in general, the helium is mainly going to fuse into carbon and oxygen. And it'll form into other things. And it becomes much more complicated. But let me just show you a periodic table. I didn't have this in the last one. I had somehow lost it. But we see hydrogen here has one proton. It actually has no neutrons. It was getting fused in the main sequence into helium, two protons, two neutrons. You need four of these to get one of those. Because this actually has an atomic mass of 4 if we're talking about helium-4. And then the helium, once we get to 100 million Kelvin, can start being fused. If you get roughly three of them-- and there's all of these other things that are coming and leaving the reactions-- you can get to a carbon. You get four of them, four of them at least as the starting raw material. You get to an oxygen. So we're starting to fuse heavier and heavier elements. So what happens here is this helium is fusing into carbon and oxygen. So you start building a carbon and oxygen core. So I'm going to leave you there. I realize I'm already past my self-imposed limit of 10 minutes. But what I want you to think about is what is likely to happen. What is likely to happen here if this star will never"
    },
    {
        "Q": "at 1:12 why is hydrogen fusing into helium?",
        "A": "beryllium is also an element ,heavier than hydrogen or helium but lighter than oxygen or carbon",
        "video_name": "kJSOqlcFpJw",
        "timestamps": [
            72
        ],
        "3min_transcript": "In the last video, we had a large cloud of hydrogen atoms eventually condensing into a high pressure, high mass, I guess you could say, ball of hydrogen atoms. And when the pressure and the temperature got high enough-- and so this is what we saw the last video-- when the pressure and temperature got high enough, we were able to get the hydrogen protons, the hydrogen nucleuses close enough to each other, or hydrogen nuclei close enough to each other, for the strong force to take over and fusion to happen and release energy. And then that real energy begins to offset the actual gravitational force. So the whole star-- what's now a star-- does not collapse on itself. And once we're there, we're now in the main sequence of a star. What I want to do in this video is to take off from that starting point and think about what happens in the star next. So in the main sequence, we have the core of the star. So this is the core-- star's core. And it's releasing just a ton of energy. And that energy is what keeps the core from imploding. It's kind of the outward force to offset the gravitational force that wants to implode everything, that wants to crush everything. And so you have the core of a star, a star like the sun, and that energy then heats up all of the other gas on the outside of the core to create that really bright object that we see as a star, or in our case, in our sun's case, the sun. Now, as the hydrogen is fusing into helium, you could imagine that more and more helium is forming in the core. So I'll do the helium as green. So more, more, and more helium forms in the core. It'll especially form-- the closer you get to the center, the higher the pressures will be, In fact, the bigger the mass of the star, the more the pressure, the faster the fusion occurs. And so you have this helium building up inside of the core as this hydrogen in the core gets fused. Now what's going to happen there? Helium is a more dense atom. It's packing more mass in a smaller space. So as more and more of this hydrogen here turns into helium, what you're going to have is the core itself is going to shrink. So let me draw a smaller core here. So the core itself is going to shrink. And now it has a lot more helium in it. And let's just take it to the extreme point where it's all helium, where it's depleted. But it's much denser. That same amount of mass that was in this sphere"
    },
    {
        "Q": "In 9:26, How does the Hydrogen heat up? Is it hot in space? Is the star heating it up? If the star is heating it up, whats it's source?",
        "A": "As gravity pulls in the atoms, they are confined to a smaller and smaller volume. When you compress a gas, it heats up.",
        "video_name": "kJSOqlcFpJw",
        "timestamps": [
            566
        ],
        "3min_transcript": "wavelength than this thing over here. This thing, the core, was not burning as furiously as this thing over here. But that energy was being dissipated over a smaller volume. So this has a higher surface temperature. This over here, the core is burning more-- sorry, the core is no longer burning. The core is now helium that's not burning. It's getting denser and denser as the helium packs in on itself. But the hydrogen fusion over here is occurring more intensely. It's occurring in a hotter way. But the surface here is less hot because it's just a larger surface area. So it doesn't make-- the increased heat is more than mitigated by how large the star has become. Now, this is going to keep happening. And this core is keep-- the pressures keep intensifying because more and more helium is getting And the temperature here keeps going up. So we said that the first ignition, the first fusion, occurs at around 10 million Kelvin. This thing will keep heating up until it gets to 100 million Kelvin. And now I'm talking about a star that's about as massive as the sun. Some stars will never even be massive enough to condense the core so that its temperature reaches But let's just talk about the case in which it does. So eventually, you'll get to a point-- so we're still sitting in the red giant phase, so we're this huge star over here. We have this helium core. And that helium core keeps getting condensed and condensed and condensed. And then we have a shell of hydrogen that keeps fusing into helium around it. So this is our hydrogen shell. Hydrogen fusion is occurring in this yellow shell over here that's causing the radius of the star to get bigger and bigger, to expand. But when the temperature get sufficiently hot-- and now I think you're going to get a sense of how heavier and heavier elements form in the universe, and all of the heavy elements that you see around us, including the ones that are in you, were formed it this way from, initially, hydrogen-- when it gets hot enough at 100 million Kelvin, in this core, because of such enormous pressures, then the helium itself will start to fuse. So then we're going to have a core in here where the helium itself will start to fuse. And now we're talking about a situation. You have helium, and you had hydrogen. And all sorts of combinations will form. But in general, the helium is mainly going to fuse into carbon and oxygen. And it'll form into other things. And it becomes much more complicated."
    },
    {
        "Q": "At 0:50, it is said that the compund breaks into individual ions, when dissolved in water. But, if this happens, they will no longer be compunds. How are they be able to retain the characterictics of the initial compund?",
        "A": "When they dissolve, they become a solution of the compound. It is still the same compound, but it is now dissolved.",
        "video_name": "BgTpPM9BMuU",
        "timestamps": [
            50
        ],
        "3min_transcript": "- [Instructor] What we have here is a molecular equation describing the reaction of some sodium chloride dissolved in water plus some silver nitrate, also dissolved in the water. They're going to react to form sodium nitrate, still dissolved in water, plus solid silver chloride and if you were to look at each of these compounds in their crystalline or solid form before they're dissolved in water, they each look like this. But once you get dissolved in water, and that's what this aqueous form tells us, it tells us that each of these compounds are going to get dissolved in water, they're no longer going to be in that crystalline form, crystalline form. Instead, you're going to have the individual ions disassociating. So for example, in the case of sodium chloride, the sodium is going to disassociate in the water. Sodium is a positive ion, or cation, and so it's going to be attracted to the partially negative oxygen end. That's what makes it such a good solvent. Now, the chloride anions, similarly, are going to dissolve in water 'cause they're going to be attracted to the partially positive hydrogen ends of the water molecules and the same thing is gonna be true of the silver nitrate. Silver ... The silver ion, once it's disassociated, is going to be positive and the nitrate is a negative. It is an anion. Now, in order to appreciate this and write an equation that better conveys the disassociation of the ions, we could instead write the equation like this. This makes it a little bit clearer that look, the sodium and the chloride aren't going to be necessarily together anymore. The sodium is going to dissolve in the water, like we have here. The chloride is gonna dissolve in the water. The silver ions are going to dissolve in the water So this makes it a little bit clearer and similarly on this end with the sodium nitrate stays dissolved so we can write it like this with the individual ions disassociated. But the silver chloride is in solid form. You can think of it as precipitating out of the solution. This does not have a high solubility, so it's not going to get dissolved in the water and so we still have it in solid form. Now you might say, well which of these is better? Well it just depends what you are trying to go for. This form up here, which we see more typically, this is just a standard molecular equation. Molecular ... Molecular equation. It's in balanced form. We always wanna have our equations balanced. This right over here is known as a complete ionic equation. The complete's there because we've put in all of the ions and we're going to compare it to a net ionic equation"
    },
    {
        "Q": "At 5:31, why can't Carbon have +2 charge?",
        "A": "In theory, carbon can have +2 charge, but for only like, i don t know, one millionth of a nanosecond? Carbocations are highly unstable (most of them!) and react quickly. The bigger the charge, the less stable they are.",
        "video_name": "7p2qfyqiXHc",
        "timestamps": [
            331
        ],
        "3min_transcript": "this carbon will have trigonal planar geometry around it, and again, that's important when you do your organic chemistry mechanics, so carbocations are extremely important to understand. Let's look at some other examples of carbocations and analyze them a little bit too. So let's start with the carbocation on the far left. The carbon with the plus one formal charge is this one, in the center here, and what is this carbon in red bonded to? Well, the carbon in red is bonded to a CH3 group up here, which we call a methyl group in organic chemistry, the carbon in red is bonded to another CH3 group here, and another CH3 group here. So the carbon in red already has three single bonds with zero loan pairs of electrons, and so the carbon in red is a plus one formal charge. Let's look at this carbocation right here, let's highlight the carbon with the plus one formal charge, it's this one, so this carbon in red is bonded to so we only have two bonds here, we only have two bonds at this point, but we know in order for that carbon in red to have a plus one formal charge, we need three bonds, like the example on the left, in the example on the left we have three bonds here to that carbon, and so where is the last bond? The last bond, of course, must be to a hydrogen, so we draw it in here like that, so the carbon in red is bonded to a hydrogen. Usually you leave off your hydrogens when you make these drawings, but it's important to understand what's actually there. Move on to the last example, this time the positive one formal charge is on this carbon in red, and that carbon in red is directly bonded to one other carbon, so that's one bond, but we know we need a total of three bonds, so the carbon in red must have two more bonds, and those two other bonds must be to hydrogen, so we draw in, there's one bond to hydrogen, and there's another bond to hydrogen, Let's do another formal charge, let's assign formal charge to another carbon. Let's put in our electrons in our bonds, so we put those in, and our goal is find the formal charge on carbon, and so 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 that carbon actually has in our drawing. So we divide up these electrons here in these bonds, and this time, carbon has a loan pair of electrons on it, so how many electrons are around carbon in our drawing? This time, there's one, two, three, and then two more from this loan pair, so four and five, so four minus five gives us a formal charge of negative one, so carbon is supposed to have four valence electrons, here it has five, so it's like it's gained an extra electron, which gives it a negative one formal charge."
    },
    {
        "Q": "At 3:35, does he mean percentage for that one drop, or for the whole plasma part?",
        "A": "Those percentages are applicable for both the drop and the entirety of the plasma.",
        "video_name": "5MOn8X-tyFw",
        "timestamps": [
            215
        ],
        "3min_transcript": "Let me write it out here. Centrifugation. And the machine is called a centrifuge. So it's basically going to spin really quickly, let's say, in one direction or the other. And as a result, what happens is that the blood starts separating out. And the heavy parts of blood kind of go to the tip of the tube. And the less dense part of blood actually rises towards the lid. So after you've centrifuged-- let's say you've actually gone through this process, and you centrifuge the blood. Now you have the same tube, but I'm going to show you kind of an after picture. So let's say this was before I actually spun the tube, and now I've got an after. This is my after picture. So after I spin the tube, what does it look like? Let me draw the tube. And the biggest key difference here is that instead of having one similar looking homogeneous liquid, like we had before, now it actually You've got three different layers, in fact. I'm going to draw all three layers for you. So this is the first layer. And this is the most impressive layer. The largest volume of our blood is going to be in this top layer. So remember, this is the least dense, right? It's not very dense, and that's why it stayed near the lid. And it's actually going to make up about 55% of our total volume. And we call it plasma. So if you've ever heard that word plasma, now you know what it means. So if I was to take a drop of this stuff-- let's say I took a little drop of this plasma, and I wanted to take a good hard look at what was in my drop-- 90% of plasma is going to be nothing more than water. So that's interesting, right, because the major part of blood is plasma, and the major part of plasma is water. So now you're seeing why it is that we always say, well, make sure you drink a lot of water. Make sure you're hydrated. And in fact, that's true for the rest of your body as well. But I want to stress that it's true for blood as well. So that leaves the rest, right? We've got 90%, we have to get to 100%. So what is 8% of this plasma made up of? It's protein. And let me give you some examples of this protein. So one would be, for example, albumin. And albumin, if you're not familiar with it, it's an important protein in your plasma that keeps the liquid from kind of leaking away out of the blood vessels. Another important protein, the antibody. And this, I'm sure you've heard of, but antibodies are basically involved in your immune system, making sure that you stay nice and healthy and don't get sick with infections. And another part of the protein, another type of protein, to kind of keep in mind, would be fibrinogen. And this is one important protein involved in clotting."
    },
    {
        "Q": "what does Sal mean by culum of water at 6:20 ...is the liquid in the test tube not mercury ?",
        "A": "I think this is a mistake because at @6:50 he writes the density of mercury 13600kg/m-3",
        "video_name": "i6gz9VFyYks",
        "timestamps": [
            380
        ],
        "3min_transcript": "how to do all the math, and all of a sudden you go, what is specific gravity? All specific gravity is, is the ratio of how dense that substance is to water. All that means is that mercury is 13.6 times as dense as water. Hopefully, after the last video-- because I told you to-- you should have memorized the density of water. It's 1,000 kilograms per meter cubed, so the density of mercury-- let's write that down, and that's the rho, or little p, depending on how you want to do it-- is going to be equal to 13.6 times the density of water, or times Let's go back to the problem. What we want to know is how high this column of mercury is. We know that the pressure-- let's consider this point right here, which is essentially the base of this column of mercury. What we're saying is the pressure on the base of this column of mercury right here, or the pressure at this point down, has to be the same thing as the pressure up, because the mercury isn't moving-- we're in a static state. We learned several videos ago that the pressure in is equal to the pressure out on a liquid system. Essentially, I have one atmosphere pushing down here on the outside of the surface, so I must have one atmosphere pushing up here. The pressure pushing up at this point right here-- we again, and just imagine where the pressure is hitting-- is one atmosphere, so the pressure down right here must be one atmosphere. What's creating the pressure down right there? It's essentially this column of water, or it's this formula, which we learned in the last video. What we now know is that the density of the mercury, times the height of the column of water, times the acceleration of gravity on Earth-- which is where we are-- has to equal one atmosphere, because it has to offset the atmosphere that's pushing on the outside and pushing up here. The density of mercury is this: 13.6 thousand, so 13,600 kilogram meters per meter cubed. That's the density times the height-- we don't know what"
    },
    {
        "Q": "At around 9:37 can some one please explain how 1N is 1kgm^2/s? I thought that force=mass*acceleration leading to the unit kgm/s^2. Thank you in advance!",
        "A": "A Newton is a kg*m/s^2 (mass * acceleratin) A Joule is a kg*m^2/s^2 (Force * distance)",
        "video_name": "i6gz9VFyYks",
        "timestamps": [
            577
        ],
        "3min_transcript": "13,600 kilograms per meter cubed times 9.8 meters per second squared. Make sure you always have the units right-- that's the hardest thing about these problems, just to know that an atmosphere is 103,000 pascals, which is also the same as newtons per meter squared. Let's just do the math, so let me type this in-- 103,000 We were dealing with newtons, so height is equal to 0.77 meters. And you should see that the units actually work, because we have a meters cubed in the denominator up here, we have a meters cubed in the denominator down here, and then we have kilogram meters per second squared here. We have newtons up here, but what's a newton? A newton is a kilogram meter squared per second, so when you divide you have kilogram meters squared per second squared, and here you have kilogram meter per second squared. When you do all the division of the units, all you're left with is meters, so we have 0.77 meters, or roughly 77 centimeters-- is how high this column of mercury is. And you can make a barometer out of it-- you can say, let me make a little notch on this test tube, and that represents one atmosphere. different parts of the globe are. Anyway, I've run out of time. See you in the next video."
    },
    {
        "Q": "At 2:43, if there is vacuum in the tube, shouldn't mercury climb up the whole tube to compensate for the difference in pressure?",
        "A": "it is usual to think that, if there is a vacuum then things will naturally get sucked into that space. But actually, that only happens if there is enough pressure pushing it. The atmospheric pressure is wha pushes it and that is only big enough to get the mercury so far. The weight of mercury is also pushing against the atmospheric pressure hope that makes sense :)",
        "video_name": "i6gz9VFyYks",
        "timestamps": [
            163
        ],
        "3min_transcript": "We are actually on Earth, or actually in Paris, France, at sea level, because that's what an atmosphere is defined as-- the atmospheric pressure. Essentially, the way you could think about it-- the weight of all of the air above us is pushing down on the surface of this bowl at one atmosphere. An atmosphere is just the pressure of all of the air above you at sea level in Paris, France. And in the bowl, I have mercury. is actually going to go up this column a little bit. We're going to do the math as far as-- one, we'll see why it's going up, and then we'll do the math to figure out how high up does it go. Say the mercury goes up some distance-- this is all still mercury. And this is actually how a barometer works; this is something that measures pressure. Over here at this part, above the mercury, but still within our little test tube, we have a vacuum-- there is no air. Vacuum is one of my favorite words, because it has two u's in a row. We have this set up, and so my question to you is-- how high is this column of mercury going to go? thing is going up to begin with. We have all this pressure from all of the air above us-- I know it's a little un-intuitive for us, because we're used to all of that pressure on our shoulders all of the time, so we don't really imagine it, but there is literally the weight of the atmosphere above us. That's going to be pushing down on the surface of the mercury on the outside of the test tube. Since there's no pressure here, the mercury is going to go upwards here. This state that I've drawn is a static state-- we have assumed that all the motion has stopped. So let's try to solve this problem. Oh, and there are a couple of things we have to know before we do this problem. It's mercury, and we know the specific gravity-- I'm using terminology, because a lot of these problems, the hardest part is the terminology-- of mercury is 13.6."
    },
    {
        "Q": "At 2:16, Sal calls the water molecule a tetrahedron. What exactly is a tetrahedron?",
        "A": "if i remember corectly, a tetrahedron has 6 edges and 4 vertices",
        "video_name": "6G1evL7ELwE",
        "timestamps": [
            136
        ],
        "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": "at 2:58, why is their a bubble around... i din't realy understand.",
        "A": "The bubble is supposed to illustrate that the electrons in a water molecule (and any other molecule) aren t actually stationary in one place (like we usually draw them, with the dots and lines), but are constantly buzzing around.",
        "video_name": "6G1evL7ELwE",
        "timestamps": [
            178
        ],
        "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": "@2:50 why did he split the cube/pentacyclo ring the way that he did?",
        "A": "Because of the symmetry of the molecule, there are many ways in which he could have done it. The point is that he had to draw the largest ring possible (8 C atoms) that contained two bridgehead carbons. The particular one he chose was probably the easiest to draw.",
        "video_name": "ayKHmN90ncc",
        "timestamps": [
            170
        ],
        "3min_transcript": "on the left, I'm going to start cutting bonds. And let's see how many cuts it takes to get to an open chain alkane. For example, I could start by cutting right here. So we'll say that's our first cut. And then our second cut, we could make a cut right back here like that. So we could make that my second cut here on my cubane. And then for my third cut, I'm going to go for this one right up here. So we'll take care of that one. So that's three cuts so far. And then, if I just go ahead and take out of this one, this bond right here, and then this bond right here, so that's cuts four and five. I now get an open chain alkane. So it took five cuts for us to do that. So there are five rings in cubane, so it's pentacyclo. So that's just not immediately obvious, to me anyway, as to why there are five rings in cubane. to start the IUPAC name here, so pentacyclo, meaning five rings. And then we start our brackets, just like we did in the video on bicyclic nomenclature. And to finish naming cubane, we're going to pretend like it is a bicyclic compound. And the first thing we do is identify our bridgehead carbons. So the carbons that are common to both of the two rings here. So hopefully it's obvious those are two rings and those are the bridgehead carbons that connect those two rings. When you number a bicyclic compound, you start at one of the bridgehead carbons and then you go the longest path first. So I'm going to start at this carbon, and I'm going to go the longest path, which would be up here. So this would be number 2, this would be number 3, carbon number 4, carbon number 5, carbon number 6, which takes me to the other bridgehead carbon. And then you name your next longest path. and then make this one back here carbon 8. So those are my eight carbons of cubane. And so, once again, I can continue to pretend like it's a bicyclic molecule. And the next thing I would do is I would name the number of carbons in my longest path. So the number of carbons in my longest path would be this one, so there'd be 1, 2, 3, 4 carbons. Remember, you exclude the bridgehead carbons when you're doing this, so we're going to start with a 4 right here, like that. Next, you do the number of carbons in your second longest path. So we can see my second longest path would be this one right here. And there are two carbons in my second longest past. So I go ahead and put a 2 over here like that. And then finally, it's the number of carbons between the bridgehead carbons, which in this example, of course, there are no carbons between my two bridgehead carbons. So I would put a 0 here like that. But of course, cubane is not a bicyclic compound,"
    },
    {
        "Q": "At 0:06 Sal refers to the neuron as a cell. However, the neuron has bits that are cells, like the Schwanne cells. Doesn't this make the neuron a tissue? Or is it a cell with add-ons?",
        "A": "It s more like a cell with add-ons. The Schwann cells are helper cells/glial cells that specialize in helping a certain type of cell. These cells are just as important as any other cell except not so well known.",
        "video_name": "ob5U8zPbAX4",
        "timestamps": [
            6
        ],
        "3min_transcript": "We could have a debate about what the most interesting cell in the human body is, but I think easily the neuron would make the top five, and it's not just because the cell itself is interesting. The fact that it essentially makes up our brain and our nervous system and is responsible for the thoughts and our feelings and maybe for all of our sentience, I think, would easily make it the top one or two cells. So what I want to do is first to show you what a neuron looks like. And, of course, this is kind of the perfect example. This isn't what all neurons look like. And then we're going to talk a little bit about how it performs its function, which is essentially communication, essentially transmitting signals across its length, depending on the signals it receives. So if I were to draw a neuron-- let me pick a better color. So let's say I have a neuron. It looks something like this. So in the middle you have your soma and then from the soma-- let me draw the nucleus. This is a nucleus, just like any cell's nucleus. then the neuron has these little things sticking out from it that keep branching off. Maybe they look something like this. I don't want to spend too much time just drawing the neuron, but you've probably seen drawings like this before. And these branches off of the soma of the neuron, off of its body, these are called dendrites. They can keep splitting off like that. I want to do a fairly reasonable drawing so I'll spend a little time doing that. So these right here, these are dendrites. And these tend to be-- and nothing is always the case in biology. Sometimes different parts of different cells perform other functions, but these tend to be where the neuron receives its signal. And we'll talk more about what it means to receive and 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."
    },
    {
        "Q": "What is the myelin sheath and what does it do? Mentioned at 4:42.",
        "A": "The myelin sheath helps keep the electrical signals insulated and protected; it is the same concept as rubber on a power cord.",
        "video_name": "ob5U8zPbAX4",
        "timestamps": [
            282
        ],
        "3min_transcript": "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. 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"
    },
    {
        "Q": "At 5:25, it is mentioned that the effects of the impulses are combined. Does that mean that the multiple impulses are transformed into one or does it mean that the effects of each impulse are measured which will then decide whether or not access is granted to the axon. Or, are you saying the impulse effects are measured and then another signal is created and then travels down the axon?",
        "A": "Thanks, you really explained it well :)",
        "video_name": "ob5U8zPbAX4",
        "timestamps": [
            325
        ],
        "3min_transcript": "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. 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"
    },
    {
        "Q": "At 7:40, why is it quasi-static? What does quasi mean?",
        "A": "that means almost such that you get almost static",
        "video_name": "lKq-10ysDb4",
        "timestamps": [
            460
        ],
        "3min_transcript": "small amount of time. And so you wouldn't have thrown that system into this, you know, havoc that I did this last time. Of course, we haven't moved all the way here yet. But what we have done is, we would have moved from that point maybe to this other point right here that's just a little bit closer to there. I've just removed a little bit of the weight. So my pressure went down just a little bit. And my volume went up a just a little bit. Temperature probably went down. And the key here is I'm trying to do it in such small increments that as I do it, my system is pretty much super close to equilibrium. I'm just doing it just slow enough that at every step it achieves equilibrium almost immediately. Or it's almost in equilibrium the whole time I'm doing it. And then I do it again, and do it again. And I'll just draw my drawings a little less neat, just for the sake of time. infinitely small mass. And now my little piston will move just a little bit higher. And I have, remember I have one less sand up here than I had over here. And then my volume in my gas increases a little bit. My pressure goes down a little bit. And I've moved to this point here. What I'm doing here is I'm setting up what's called a quasi-static process. And the reason why it's called that is because it's almost static. It's almost in equilibrium the whole time. Every time I move a grain of sand I'm just moving a little bit closer. And obviously even a grain of sand, the reality is if I were to do this in real life, even a grain of sand on a small scale is going to reek a little bit of havoc on my system. This piston is going to go up a little bit. So say, let me just do even a smaller grain of sand, and do equilibrium. So you can imagine this is kind of a theoretical thing. If I did an infinitely small grains of sand, and did it just slow enough so that it's just gently moved from this point to this point. But we like to think of it theoretically, because it allows us to describe a path. Because remember, why am I being so careful here? Why am I so careful to make sure that the state, the system is in equilibrium the whole time when I get from there to there? Because our macrostates, our macro variables like pressure, volume, and temperature, our only defined when we're in equilibrium. So if I do this process super slowly, in super small increments, it allows me to keep my pressure and volume and actually my temperature of macrostates at any point in time. So I could actually plot a path. So if I keep doing it small, small, small, I could actually plot a path to say, how did I get from state 1 to state 2 on this on this PV diagram."
    },
    {
        "Q": "At 1:20, how come the force due to gravity is mg? In free fall particle videos earlier, I remember it being just g (-9.81 m/s^2)?",
        "A": "You remember wrong. g is acceleration due to gravity, not force due to gravity. (It is also force per unit of mass.) To find the force of gravity on an object you have to multiply m by g.",
        "video_name": "TC23wD34C7k",
        "timestamps": [
            80
        ],
        "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 11:59 He said that when \u00ce\u00b8=0 all forces will act perpendicularly to the plane. Then does that not mean that \u00ce\u00b8=90? Furthermore if \u00ce\u00b8=0, then how do you explain the magnitude of the normal force since mg(cos\u00ce\u00b8) = 0 [cos\u00ce\u00b8=0, mg(0)=0], the normal force should be the same as the weight of the object when \u00ce\u00b8 is zero, could someone please clear my doubts. Thank you",
        "A": "Theta is the angle between the plane and the ground. When that angle is 0, the force of the weight is perpendicular to the plane. Look at the drawing.",
        "video_name": "TC23wD34C7k",
        "timestamps": [
            719
        ],
        "3min_transcript": "over the magnitude of the force due to gravity-- which is the magnitude of mg-- that is going to be equal to what? This is the opposite side to the angle. So the blue stuff is the opposite side, or at least its length, is the opposite side of the angle. And then right over here this magnitude of mg, that is the hypotenuse. So you have the opposite over the hypotenuse. Opposite over hypotenuse. Sine of an angle is opposite over hypotenuse. So this is going to be equal to the sine of theta. This is equal to the sine of theta. Or you multiply both sides times the magnitude of the force due to gravity and you get the component of the force due to gravity that is parallel to the ramp is going to be the force due to gravity total times Times sine of theta. And hopefully you should see where this came from. Because if you ever have to derive this again 30 years after you took a physics class, you should be able to do it. But if you know this right here, and this right here, we can all of a sudden start breaking down the forces into things that are useful to us. Because we could say, hey, look, this isn't moving down into this plane. So maybe there's some normal force that's completely netting it out in this example. And maybe if there's nothing to keep it up, and there's no friction, maybe this thing will start accelerating due to the parallel force. And we'll think a lot more about that. And if you ever forget these, think about them intuitively. You don't have to go through this whole parallel line and transversal and all of that. If this angle went down to 0, then we'll be talking about essentially a flat surface. And if this angle goes down to 0, then all of the force should be acting perpendicular to the surface of the plane. So if this going to 0, if the perpendicular force should be the same thing as the total gravitational force. And that's why it's cosine of theta. Because cosine of 0 right now is 1. And so these would equal each other. And if this is equal to 0, then the parallel component of gravity should go to 0. Because gravity will only be acting downwards, and once again, if sine of theta is 0. So the force of gravity that is parallel will go to 0. So if you ever forget, just do that little intuitive thought process and you'll remember which one is sine and which one is cosine."
    },
    {
        "Q": "Exactly...At 2:37. Mr. Khan literates that the notation that he will be using is unconventional. Then what is a conventional notation?",
        "A": "conventional notation is scientific notation",
        "video_name": "TC23wD34C7k",
        "timestamps": [
            157
        ],
        "3min_transcript": "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. to show something that is parallel to the surface. So this is the component of force due to gravity that's perpendicular, component of force that is parallel. So let's see if we can use a little bit a geometry and trigonometry, given that this wedge is at a theta degree incline relative to the horizontal. If you were to measure this angle right over here, you would get theta. So in future videos we'll make it more concrete, like 30 degrees or 45 degrees or whatever. But let's just keep in general. If this is theta, let's figure out what these components of the gravitational force are going to be. Well, we can break out our geometry over here. This, I'm assuming is a right angle. And so if this is a right angle, we know that the sum of the angles in a triangle add up to 180. So if this angle, and this 90 degrees-- right angle says 90 degrees-- add up to 180, then that"
    },
    {
        "Q": "At 2:36, I tried searching up the conventional notation for vector magnitudes but came away a little confused, so does anyone know what it is?",
        "A": "Unsure what you mean exactly by conventional notation, since mathematicians and physicists can t agree, but the most COMMON notation is as follows: Math: Vector Q is located at <3,4,5> Meaning 3 units in the x-direction, 4 units in the y-direction, and 5 units in the z-direction. Physics: Vector Q is located at 3i+4j+5k, where i is the x-component of Vector Q (Qsubx), j is the y-component (Qsuby), and k is the z component (Qsubz). That s how it s usually written.",
        "video_name": "TC23wD34C7k",
        "timestamps": [
            156
        ],
        "3min_transcript": "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. to show something that is parallel to the surface. So this is the component of force due to gravity that's perpendicular, component of force that is parallel. So let's see if we can use a little bit a geometry and trigonometry, given that this wedge is at a theta degree incline relative to the horizontal. If you were to measure this angle right over here, you would get theta. So in future videos we'll make it more concrete, like 30 degrees or 45 degrees or whatever. But let's just keep in general. If this is theta, let's figure out what these components of the gravitational force are going to be. Well, we can break out our geometry over here. This, I'm assuming is a right angle. And so if this is a right angle, we know that the sum of the angles in a triangle add up to 180. So if this angle, and this 90 degrees-- right angle says 90 degrees-- add up to 180, then that"
    },
    {
        "Q": "at 7:30, how can we write V+ as Vin.. isn't Vin = V+ - V-??",
        "A": "Hello Karthikeyan, In this case the non-inverting input terminal (V+) is connected directly to the voltage source (Vin). You are also correct that the op-amp amplifies the difference between the input terminals: Vout = Gain * (V+ - V-) Yes these circuits can be tricky. Different things have the same names. You will need to look at the context to figure out the meaning. Regards, APD",
        "video_name": "_Ut-nQ535iE",
        "timestamps": [
            450
        ],
        "3min_transcript": "Now what else do I know? Let's look at this resistor chain here. This resistor chain actually looks a lot like a voltage divider, and it's actually a very good voltage divider. Remember we said this current here, what is this current here? It's zero. I can use the voltage divider expression that I know. In that case, I know that V minus, this is the voltage divider equation, equals V out Times the bottom resistor remember this? R2 over R1 plus R2, so the voltage divider expression says that when you have a stack of resistors like this, with the voltage on the top and ground on the bottom, 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."
    },
    {
        "Q": "at 10:54, you said ''because x is <<1\", for me to get that I had to calculate it first, so how did you know that it is much smaller than 1 without any calculations?Since when x<<1 we make assumption, that means subtracting/adding it does not make any significant difference, so from that I also wanna know if I would be wrong if I DO NOT make assumptions instead I just work it out the way it is.",
        "A": "You can confidently ignore the value of x when the Kb value is of order 10^(-5) or smaller. The reason is that small Ka would mean small dissociation. Also, if you do not ignore x in the equation, you will arrive at almost the same answer (there may be a difference but it is insignificant). But, you may notice that if you do not ignore x, you will arrive at a quadratic equation which increases our effort to solve for the answer(and its going to get you the same answer). Hope this helps!",
        "video_name": "223KLPnJCBI",
        "timestamps": [
            654
        ],
        "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": "Forgive my ignorance, but when he speaks of a step-up transformer @ 1:35, what exactly happens to the voltage? I assume it isn't making free energy...?",
        "A": "True, Though the voltage is increased in step up transformer, the current is decreased to make Power= Voltage*Current consant.",
        "video_name": "xuQcB-oo-4U",
        "timestamps": [
            95
        ],
        "3min_transcript": "- [Karl] Today, we're gonna take apart an alarm clock radio and we're gonna see what's inside it and how it works. There's basically four systems that we're going to evaluate. There's the power system. There's the alarm clock or the clock itself. And the structure of the device in the interface. Then we're also gonna take a look at the radio. The first thing is let's take a look at the power system. I've already cut apart this plug here. You can see the prongs. That's where the power comes from. We've got the two wires here. The two wires connect to what's called the transformer. In the transformer, there are three key components. We've got a primary coil, a secondary coil, and an iron core. The primary coil is wound around a certain number of times. The secondary coil is wound around fewer times in this particular transformer. That means it's a step-down transformer. What this transformer does is it converts AC down to nine volt AC. Because the components in the alarm clock need lower voltage. It steps the power down. The way it does it is this coil induces a current flow in this coil, and the iron core helps that to happen. Because this coil has fewer turns, it's a step-down transformer which means that the voltage is less coming out of this part. If there were more turns, it would be a step-up transformer. The iron core, again, facilitates that process. It's called electromagnetic induction. This coil induces the current flow in this coil. Anyway, the power will travel through the cable here, the wire, and it comes to the alarm clock. Let's take a look at the housing, first of all. It's fairly low cost housing. It's made out of injection-molded plastic. Let's see where that power comes in and where it goes. is they basically just only used one fastener, one separate fastener. This is a screw. The more screws you have in things, the easier they are oftentimes to put together and take apart, but they are also more expensive. Every screw requires either a robot or a person to assemble it and it's an expensive cost adder. The more screws you can take out, the more cost you can reduce. All these fasteners in this are actually molded into the body panels. There's a pin or a tab there, so we can pull this top part off. This part right here is the front plate or the front vessel. It's made out of a tinted acrylic and it's injection-molded. There's two parts of the mold that come together and molded plastic is injected, and then this comes out. The reason it's injection-molded is that it creates very precise part. You can get a nice, clean finish."
    },
    {
        "Q": "At about 1:00 the transformer is for what?",
        "A": "the transformer lowers the voltage so that the alarm clock radio doesn t over heat",
        "video_name": "xuQcB-oo-4U",
        "timestamps": [
            60
        ],
        "3min_transcript": "- [Karl] Today, we're gonna take apart an alarm clock radio and we're gonna see what's inside it and how it works. There's basically four systems that we're going to evaluate. There's the power system. There's the alarm clock or the clock itself. And the structure of the device in the interface. Then we're also gonna take a look at the radio. The first thing is let's take a look at the power system. I've already cut apart this plug here. You can see the prongs. That's where the power comes from. We've got the two wires here. The two wires connect to what's called the transformer. In the transformer, there are three key components. We've got a primary coil, a secondary coil, and an iron core. The primary coil is wound around a certain number of times. The secondary coil is wound around fewer times in this particular transformer. That means it's a step-down transformer. What this transformer does is it converts AC down to nine volt AC. Because the components in the alarm clock need lower voltage. It steps the power down. The way it does it is this coil induces a current flow in this coil, and the iron core helps that to happen. Because this coil has fewer turns, it's a step-down transformer which means that the voltage is less coming out of this part. If there were more turns, it would be a step-up transformer. The iron core, again, facilitates that process. It's called electromagnetic induction. This coil induces the current flow in this coil. Anyway, the power will travel through the cable here, the wire, and it comes to the alarm clock. Let's take a look at the housing, first of all. It's fairly low cost housing. It's made out of injection-molded plastic. Let's see where that power comes in and where it goes. is they basically just only used one fastener, one separate fastener. This is a screw. The more screws you have in things, the easier they are oftentimes to put together and take apart, but they are also more expensive. Every screw requires either a robot or a person to assemble it and it's an expensive cost adder. The more screws you can take out, the more cost you can reduce. All these fasteners in this are actually molded into the body panels. There's a pin or a tab there, so we can pull this top part off. This part right here is the front plate or the front vessel. It's made out of a tinted acrylic and it's injection-molded. There's two parts of the mold that come together and molded plastic is injected, and then this comes out. The reason it's injection-molded is that it creates very precise part. You can get a nice, clean finish."
    },
    {
        "Q": "@ 9:39, he said that the copper wire works as an antenna . why does that work?",
        "A": "Because as the electricity goes through it will make contact with the satellite. Hope that helps. :)",
        "video_name": "xuQcB-oo-4U",
        "timestamps": [
            579
        ],
        "3min_transcript": "We have a jumper here which helps to, you can use jumpers to alter the functionality of the seven-segment display. This one may be programmed or maybe set up to function in a certain way so that this jumper allows you to transfer power to the different part of the display. Then we have these little white spots here, and what that is is the back of this module, this seven-segment module, has these little pieces of plastic that stick through and there's a hot plate that basically pushes on those pins that stick through and melts them, and it holds the plate against the printed circuit board. This is just a cheap way or an inexpensive way to fasten things together. It works pretty well. That's the clock portion. The buttons, let's talk a little bit about the interface. The buttons, when you push down on the buttons here, they trigger these pins and you can see the pins flex pretty good right here. The pins are connected by these little standoffs. Because the standoffs are really thin, they can flex. 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"
    },
    {
        "Q": "@8:39 how does he know that the equilibrium lies to the left? How do you know if you have more reactants than products? I'm having difficulty understanding whether to point the arrow to the left or to the right. Thanks",
        "A": "If Ka > 1, the position of equilibrium lies to the right. If Ka < 1, the position of equilibrium lies to the left.",
        "video_name": "BeHOvYchtBg",
        "timestamps": [
            519
        ],
        "3min_transcript": "so negative one charge on the oxygen. Let me show those electrons. These electrons in green move off onto the oxygen right here, giving it a negative charge. We're also gonna form a hydronium. All right, so H3O plus, so let me go ahead and draw in hydronium. So plus one formal charge on the oxygen and let's show those electrons in red. All right, so this electron pair picks up the acidic proton. Forming this bond that we get H3O plus. So another way to write this acid base reaction would be just to write acetic acid, CH3, COOH plus H2O gives us the acetate anion, CH3COO minus plus H3O plus. Now acetic acid is a weak acid and weak acids don't donate protons very well. So acetic acid is gonna stay mostly protonated. you're gonna have a relatively high concentration of your reactants here. When we write the equilibrium expression, write KA is equal to the concentration of your product so CH3COO minus times the concentration of H3O plus, all over the concentration of acetic acid because we leave water out. So all over the concentration of acetic acid. All right, the equilibrium lies to the left because acetic acid is not good at donating this proton. So we're going to get a very large number for the denominator, for this concentration so this is a very large number and a very small number for the numerator. All right, so this is a very small number. So if you think about what that does to the KA, all right, a very small number divided by an ionization constant much less than one. All right, so this value is going to be much less than one and that's how we recognize, that's one way to recognize a weak acid. Look at the KA value."
    },
    {
        "Q": "At 7:47, I know it sounds silly but could acetic acid have acted as a base, and ACCEPTED a proton from water? For example, either the double-bonded oxygen or single bonded oxygen could have accepted the proton. Thanks",
        "A": "It can happen, but that reaction is not very favourable in water as water is a much better base (which means it s much more likely to accept H+) If you added a strong acid to pure acetic acid it could happen forming CH3COOH2^+",
        "video_name": "BeHOvYchtBg",
        "timestamps": [
            467
        ],
        "3min_transcript": "competing base strength. All right, so here we have Bronsted-Lowry. Base water is acting as a Bronsted-Lowry base and accepting a proton. And over here if you think about the reverse reaction, the chloride anion would be trying to pick up a proton from hydronium for the reverse reaction here but since HCL is so good at donating protons, that means that the chloride anion is not very good at accepting them. So the stronger the acid, the weaker the conjugate base. Water is a much stronger base than the chloride anion. Finally let's look at acetic acids. Acetic acid is going to be our Bronsted-Lowry acid and this is going to be the acidic proton. Water is gonna function as a Bronsted-Lowry base and a lone pair of electrons in the auction is going to take this acidic proton, leaving these electrons behind on the oxygen. So let's go ahead and draw our products. We would form the acetate anions. so negative one charge on the oxygen. Let me show those electrons. These electrons in green move off onto the oxygen right here, giving it a negative charge. We're also gonna form a hydronium. All right, so H3O plus, so let me go ahead and draw in hydronium. So plus one formal charge on the oxygen and let's show those electrons in red. All right, so this electron pair picks up the acidic proton. Forming this bond that we get H3O plus. So another way to write this acid base reaction would be just to write acetic acid, CH3, COOH plus H2O gives us the acetate anion, CH3COO minus plus H3O plus. Now acetic acid is a weak acid and weak acids don't donate protons very well. So acetic acid is gonna stay mostly protonated. you're gonna have a relatively high concentration of your reactants here. When we write the equilibrium expression, write KA is equal to the concentration of your product so CH3COO minus times the concentration of H3O plus, all over the concentration of acetic acid because we leave water out. So all over the concentration of acetic acid. All right, the equilibrium lies to the left because acetic acid is not good at donating this proton. So we're going to get a very large number for the denominator, for this concentration so this is a very large number and a very small number for the numerator. All right, so this is a very small number. So if you think about what that does to the KA, all right, a very small number divided by"
    },
    {
        "Q": "At 1:26 why does the Cl ion have a negative charge when it has a full octet?",
        "A": "A chlorine atom has 7 valence electrons. If it has 8 (a full octet) then it has to have gained 1 electron which is why it has a -1 charge.",
        "video_name": "BeHOvYchtBg",
        "timestamps": [
            86
        ],
        "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": "At 2:40 why we do 2^x =4 Then X=? ,what does X means here?",
        "A": "To be more specific, x is the order of the reaction in NO (which is the exponent for [NO] in the rate law).",
        "video_name": "Ad0aaYixFJg",
        "timestamps": [
            160
        ],
        "3min_transcript": "concentration of nitric oxide affects the rate of our reaction. We're going to look at experiments one and two here. The reason why we chose those two experiments is because the concentration of hydrogen is constant in those two experiments. The concentration of hydrogen is point zero zero two molar in both. If we look at what we did to the concentration of nitric oxide, we went from a concentration of point zero zero five to a concentration of point zero one zero. We increased the concentration of nitric oxide by a factor of two. We doubled the concentration. What happened to the initial rate of reaction? Well the rate went from one point two five times 10 to the negative five to five times 10 to the negative five. The rate increased by a factor of four. We increased the rate by a factor of four. you could just use a calculator and say five times 10 to the negative five and if you divide that by one point two five times 10 to the negative five, this would be four over one, or four. This rate is four times this rate up here. Now we know enough to figure out the order for nitric oxide. Remember from the previous video, what we did is we said two to the X is equal to four. Over here, two to the X is equal to four. Obviously X is equal to two, two squared is equal to four. So we can go ahead and put that in for our rate law. Now we know our rate is equal to K times the concentration of nitric oxide this would be to the second power. So the reaction is second order in nitric oxide. So this time we want to choose two experiments where the concentration of nitric oxide is constant. That would be experiment two and three where we can see the concentration of nitric oxide has not changed. It's point zero one molar for both of those experiments. But the concentration of hydrogen has changed. It goes from point zero zero two to point zero zero four. So we've increased the concentration of hydrogen by a factor of 2 and what happened to the rate of reaction? Well it went from five times 10 to the negative five to one times 10 to the negative four so we've doubled the rate. The rate has increased by a factor of two. Sometimes the exponents bother students. How is this doubling the rate? Well, once again, if you can't do that in your head,"
    },
    {
        "Q": "At 5:38, by Time rate of change, does he mean the speed the electrons are moving, and doesn't this relate closely with Voltage?",
        "A": "Time rate of change is always of something. Like: time rate of change of voltage, or time rate of change of position. If the time rate of change of voltage is 5 volts/second, then in the capacitor equation (i = C dv/dt) you would substitute 5 for dv/dt.",
        "video_name": "l-h72j2-X0o",
        "timestamps": [
            338
        ],
        "3min_transcript": "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. that we have in our minds that we're gonna try to build in the real world but we'll come close. We'll come very close. We now have a wonderful set of equations: V equals iR, i equals C, dv, dt. v equals L, di, dt. These are gonna be like poetry for you pretty soon and these ideal equations will produce all kinds of really cool circuits for us."
    },
    {
        "Q": "At 13:20, I am a little confused on how you went from 0 ice to 0 degrees water ?",
        "A": "For each kg of ice, you need a certain amount of energy to melt it to water. During the melting, the temperature stays the same. The amount of energy you need is given by L, the latent heat of fusion, which is in units joules per kilogram.",
        "video_name": "zz4KbvF_X-0",
        "timestamps": [
            800
        ],
        "3min_transcript": "Which is this right here, that distance right here. I forgot to figure out how much energy to turn that 100 degree water into 100 degree vapor. So that's key. So I really should have done that up here before I calculated the vapor. But I'll do it down here. So to do 100 degree water to 100 degree vapor. That's this step right here, this is the phase change. I multiply the heat of vaporization which is 2,257 joules per gram times 200 grams. And this is equal to 451,400. I'll do it in that blue color. for our sample of 200 grams. This piece right here was 83,000 joules. This piece right here was 3,780 joules. So to know the total amount of energy the total amount of heat that we had to put in the system to go from minus 10 degree ice all the way to 110 degree vapor, we just add up all of the energies which we had to do in all of these steps. Let's see. And I'll do them in order this time. So to go from minus 10 degree ice to zero degree ice. Of course we have 200 grams of it. It was 4,100. Plus the 67,000. So plus 67,110. Plus 83,000. That's to go from 0 degree water to 100 degree water. Plus 83,560. So we're at 154,000 right now, and just to get to 100 degree water. And then we need to turn that 100 degree water into 100 degree vapor. So you add the 451,000. So, plus 451,400 is equal to 606. And then finally, we're at 100 degree vapor, and we want to convert that to 110 degree vapor. So it's another 3,700 joules. So plus 3,780 is equal to 609,950 joules. So this whole thing when we're dealing with 200 grams"
    },
    {
        "Q": "At 9:55 Sal only crossed out the I's to make the equation simpler. why is that and why didn't he cross out the R's?",
        "A": "This is because I(current) is the same throughout the circuit. so Sal cancelled the I s. But the resistance keeps on changing after going through each of the resistor. so he simply cannot cancel out the R s.",
        "video_name": "ZrMw7P6P2Gw",
        "timestamps": [
            595
        ],
        "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": "13:00\n\nIs there a name for this little period where the temperature stays constant although we are adding or removing a certain amount of heat?",
        "A": "Heat of vaporazation or the heat of fusion.",
        "video_name": "pKvo0XWZtjo",
        "timestamps": [
            780
        ],
        "3min_transcript": "add heat my temperature will go up. Temperature is average kinetic energy. Let's say I'm in the solid state here. And I'll do the solid state in purple. No I already was using purple. I'll use magenta. So as I add heat, my temperature will go up. Heat is a form of energy. And when I add it to these molecules, as I did in this example, what did it do? It made them vibrate more. Or it made them have higher kinetic energy, or higher average kinetic engery, and that's what temperature is a measure of; average kinetic energy. So as I add heat in the solid phase, my average kinetic energy will go up. And let me write this down. This is in the solid phase, or the solid state of matter. Now something very interesting happens. 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"
    },
    {
        "Q": "In 0:05 he says that there are only 4 truth is there are 5 states of matter.They are Solid,Liquid,Gas,plasma and Bose- Einstine compound.",
        "A": "I learned there were five as well.",
        "video_name": "pKvo0XWZtjo",
        "timestamps": [
            5
        ],
        "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": "Are the polar bonds forming between the H2O molecules @ 4:06 hydrogen bonds? If not what are hydrogen bonds?",
        "A": "Yes, in a hydrogen bond hydrogen is bound to a highly electronegative atom, such as nitrogen, oxygen or fluorine.",
        "video_name": "pKvo0XWZtjo",
        "timestamps": [
            246
        ],
        "3min_transcript": "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. When we talk about the whole state of the whole matter, we actually think about how the molecules are interacting with Not just how the atoms are interacting with each other within a molecule. I just drew one oxygen, let me copy and paste that. But I could do multiple oxygens. And let's say that that hydrogen is going to want to be near this oxygen. Because this has partial negative charge, this has a partial positive charge. And then I could do another one right there. And then maybe we'll have, and just to make the point clear, you have two hydrogens here, maybe an oxygen wants to hang out there. So maybe you have an oxygen that wants to be here because it's got its partial negative here. And it's connected to two hydrogens right there that have their partial positives. But you can kind of see a lattice structure. Let me draw these bonds, these polar bonds that start forming These bonds, they're called polar bonds because the molecules themselves are polar. And you can see it forms this lattice structure. And if each of these molecules don't have a lot of kinetic energy. Or we could say the average kinetic energy of this matter is fairly low. And what do we know is average kinetic energy? Well, that's temperature. Then this lattice structure will be solid. These molecules will not move relative to each other. I could draw a gazillion more, but I think you get the point that we're forming this kind of fixed structure. And while we're in the solid state, as we add kinetic energy, as we add heat, what it does to molecules is, it just makes them vibrate around a little bit. If I was a cartoonist, they way you'd draw a vibration is to put quotation marks there."
    },
    {
        "Q": "At round 14:02, Sal says that water takes much more time to vaporize as compared to the time period that ice takes to melt.\nWhy does this happen?",
        "A": "In order to vaporize water, you have to give a lot of energy to each of the molecules so that they have enough energy to move far away from each other. To melt ice into water, you also have give the molecules energy, but you only have to give enough to break the bonds between the water molecules, and that doesn t require nearly as much energy as it does to actually pull the molecules away from each other.",
        "video_name": "pKvo0XWZtjo",
        "timestamps": [
            842
        ],
        "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 4:07 when the ring breaks do the electrons in magenta need to rotate in order to form a bond with the other halogen atom(anti addition)?",
        "A": "Nothing needs to rotate, the magenta electrons are not forming a new bond they are going to the Br that was in the ring. The electrons in the new C-Br bond comes from the bromide anion. It would help if Jay coloured these in.",
        "video_name": "Yiy84xYQ3es",
        "timestamps": [
            247
        ],
        "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": "At 8:50 in the video the mechanism is described as happening two more times and the result is a trialkylborane molecule. I can't think of where the other two O's have come from. I don't understand why B, with an octet around it has a negative charge.",
        "A": "From another two peroxides. Formal charge = valence electrons - lone pair electrons - bonds 3 - 4 = -1",
        "video_name": "00qYQahwuSQ",
        "timestamps": [
            530
        ],
        "3min_transcript": "So this is the weird part. Alright so these electrons right here are actually going to form a bond with the oxygen. At the same time these electrons are gonna come off on to this oxygen. So, we form hydroxide. Let me go ahead and show that first. That's maybe the easier part to understand of what's going on here. So we have three long pairs of electrons on the oxygen, negative one formal charge, we're gonna make these electrons green. These electrons in green here come off on to the oxygen. So the oxygen oxygen bond is weak. So, it's relatively easy to break this bond, and let me highlight the other electrons here. Well, these electrons in magenta formed this bond between the oxygen and the boron, and now we get our alkyl group migration and I'll show these electrons in the blue. So the electrons in blue are forming a bond between this carbon and this oxygen. It's pretty hard to see. Let me go ahead and draw what we have and then we'll highlight some electrons here. So now we have our oxygen bonded to this group and our boron has two alkyl groups still bonded to it like that. So let's follow those electrons. The electrons in magenta represent the bond between the oxygen and the boron, and the electrons in blue, right, would be these electrons right here. So it's this carbon is now bonded to this oxygen. So the migration of the alkyl group removes the formal charge on the boron and it breaks the weak oxygen oxygen bond and kicks off hydroxide. Alright so now this process happens two more times. Alright cause we have these two other alkyl groups, and so when that happens two more times we get, let me go ahead and draw it in, a trialkylborane. our group coming off of that and then over here we would have our oxygen and then we have our group coming off of that and then also down here. Alright, our next step is the hydroxide anion functions as a nucleophile. So the hydroxide attacks the boron. Once again our boron has an empty orbital. So nucleophilic attack let's draw the results of our nucleophilic attack. Now we have OH bonded to the boron and the boron still has all of these groups around it. So let me sketch all of those in, and you can see why this mechanism is getting rather cumbersome at this point. Right, drawing in all these groups is a little bit annoying here but we still have a negative one. I should say we have a negative one formal charge and a boron and we still have some lone pairs of electrons on this oxygen and I'm putting those in because in the next step to get rid of the negative one formal charge these electrons come off"
    },
    {
        "Q": "How is Chlorine less electronegative than oxygen? 6:40",
        "A": "Sal explains this in the video on trends in the periodic system (chemistry video no. 10). Simply put, chlorine (compared to oxygen) has an additional shell of valence electrons (and a lot more electrons overall) that repels the electrons of other atoms or molecules. Watch the video or look at Wikipedia for a better explanation!",
        "video_name": "8qfzpJvsp04",
        "timestamps": [
            400
        ],
        "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": "9:15 Hydrogen Flourine is HF not H-FL, right?",
        "A": "Yes Hydrogen Flourine is HF.",
        "video_name": "8qfzpJvsp04",
        "timestamps": [
            555
        ],
        "3min_transcript": "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 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,"
    },
    {
        "Q": "At 7:27 you said the electron from the hydrogen spends most of its time with the chlorine, but why doesn't the chlorine just take the electron permanently, like an ionic bond?",
        "A": "hydrogen bonding can also only form if it s with the elements FON. Flourine Oxygen and Nitrogen. and it s also covalently bonded.",
        "video_name": "8qfzpJvsp04",
        "timestamps": [
            447
        ],
        "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": "At 5:19, how doe he say that the lone pair will move up angularly away from the Central Atom O ?",
        "A": "The electron pairs in water point towards the corners of a tetrahedron. The bonding pairs are in the plane of the paper. One lone pair is coming angularly out of the paper, and the other lone pair is pointing angularly behind the paper.",
        "video_name": "q3g3jsmCOEQ",
        "timestamps": [
            319
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 8:07 i did not quite understand why the tetrachloromethane molecule will have 0 D??",
        "A": "Same case as in CO2. Since the power of Cl in C(CL)4 attracting the electron to themselves have the same magnitude individually, they cancel each other. Hope this helps.",
        "video_name": "q3g3jsmCOEQ",
        "timestamps": [
            487
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 8:00 it is said that we would expect CCl_4 not to be polar, but aren't the individual C-Cl bonds polar making the molecule polar?",
        "A": "Yes, the individual bonds are all polar, but they cancel each other out, so the molecule as a whole is nonpolar.",
        "video_name": "q3g3jsmCOEQ",
        "timestamps": [
            480
        ],
        "3min_transcript": ""
    },
    {
        "Q": "1. at 12:05 , once we have the moles quantity. cant we use PV=nRT to figure out the pressure for each?\n\n2. For PV= nRT dont we always have to use Liters as the volume?\n\n3. Would the total number of moles always equall 100 moles when we add them up coming from % ?\n\nThanks for helping. and you are changing the world with your videos, not just how people are learning things.",
        "A": "for number 1,yes you can , you just have to replace n with the number of moles of each molecule",
        "video_name": "d4bqNf37mBY",
        "timestamps": [
            725
        ],
        "3min_transcript": "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. So, the partial pressure of oxygen, let me write that as pressure due to oxygen, due to O2. It's going to be 20% of the total pressure. 20% times, let me write 56.746 kilopascals. I just took this pressure measurement. If I wanted atmospheres, times 0.56 atmospheres. So the partial pressure of oxygen -- well, I already have the 0.56 written there. So times 0.2 is equal to 0.112 atmospheres. It's just 20% of this. How did I get 20%? We have a total of 100 mole molecules in our balloon. 20 moles of them are oxygen. So 20% are oxygen. So 20% of the pressure is going to be due to oxygen. If I took 20% times the 56,000. 0.2 times 56, you get roughly 11.2 kilopascals. I'm just multiplying 20% by any of these numbers. And the numbers are going to change depending on the units. So you do the same process. What is the partial pressure due to nitrogen? Well even though 2/3 of the mass is nitrogen, only 50% of the actual particles are nitrogen. So 50% of the pressures is due to the nitrogen particles. Remember, you have to convert everything to moles. Because we only care about the number of particles. So if you want to know the partial pressure due to the nitrogen molecules, it's 50% of these. So it's 28,373 pascals."
    },
    {
        "Q": "I know this might be a simple math question but dont you have to divide the entire side of the equation by 4 instead of just the 100 @ 8:52 ?",
        "A": "this might help: since there are no variables or anything complicated, the entire side is just multiplying numbers. because of the order of operations (BEDMAS, or BEMDAS, or PEMDAS, or however you learned it), it doesn t matter whether you divide the whole thing by at the end, or divide one of the numbers by 4 in the middle. you re just dividing by 4! try it on a calculator, you get the same result :)",
        "video_name": "d4bqNf37mBY",
        "timestamps": [
            532
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "At 4:28, why did he change oxygen from 16 to 32?",
        "A": "The relative atomic mass of one oxygen atom (O) is 16, while the relative atomic mass of one oxygen molecule (O2) is twice that at 32. It depends on if you are talking about the atom or the molecule.",
        "video_name": "d4bqNf37mBY",
        "timestamps": [
            268
        ],
        "3min_transcript": "So let's do it in grams. Because when we talk about molecular mass it's always in grams. It doesn't have to be. But it makes it a lot simpler to convert between atomic mass units and mass in our world. So this is 2/3 of 2100, that's 1400 grams of N2. Now what's the molar mass of this nitrogen molecule? Well we know that the atomic mass of nitrogen is 14. So this molecule has two nitrogens. So its atomic mass is 28. So one of these molecules will have a mass of 28 atomic mass units. Or one mole of N2 would have a mass of 28 grams. So one mole is 28 grams. We have 1400 grams -- or we say grams per mole, if we want to keep our units right. by 28 grams per mole we should get the number of moles. So 1400 divided by 28 is equal to 50. That worked out nicely. So we have 50 moles of N2. We could write that right there. All right. Let's do oxygen next. So we do the same process over again. 30% is oxygen. So let's do oxygen down here, O2. So we take 30%. Remember, these percentages I gave you, these are the percentages of the total mass, not the percentage of the moles. So we have to figure out what the moles are. So 30.48% of 2100 grams is equal to about 640. So this is equal to 640 grams. And then what is the mass of one mole of the oxygen gas molecule? The atomic mass of one oxygen atom is 16. You can look it up on the periodic table, although you should probably be pretty familiar with it by now. So the atomic mass of this molecule is 32 atomic mass units. So one mole of O2 is going to be 32 grams. We have 640 grams. So how many moles do we have? 640 divided by 32 is equal to 20. We have 20 moles of oxygen. Let me write that down. We have 20 moles. Now we just have to figure out the hydrogen."
    },
    {
        "Q": "At 7:52 how did you get 3 different R equations? And how can you be sure that you are choosing the right one, is it that you choose which ever one matches your units ?",
        "A": "You re right - you always want to choose the R equation that matches the units you re using (or, you have to convert your units to match your R equation!).",
        "video_name": "d4bqNf37mBY",
        "timestamps": [
            472
        ],
        "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": "At 3:43 you say that the star will enlarge its radius like a red giant, but not get to the same size as a sun-sized star becoming a red giant. Does it also a experience a color change toward red, and if so what color was it in the first place? Does the creation of heavier elements in the core affect the color as well?",
        "A": "No, as the stars surface gets farther from the core, it gets cooler. Color is entirely related to temperature the cooler surface glows red instead of yellow.",
        "video_name": "UhIwMAhZpCo",
        "timestamps": [
            223
        ],
        "3min_transcript": "But anyway, let's think about what happens. And so far, just the pattern of what happens, it's going to happen faster because we have more pressure, more gravity, more temperature. But it's going to happen in pretty much the same way as what we saw with a star the mass of the sun. Eventually that helium-- sorry, that hydrogen is going to fuse into a helium core that's going to have a hydrogen shell around it. It's going to have a hydrogen shell around it, hydrogen fusion shell around it. And then you have the rest of the star around that. So let me label it. This right here is our helium core. And more and more helium is going to be built up as this hydrogen in this shell fuses. And in a star the size of our sun or the mass of our sun, this is when it starts to become a red giant. Because this core is getting denser and denser and denser as more and more helium is produced. And as it gets denser and denser and denser, being put on the hydrogen, on this hydrogen shell out here, where we have fusion still happening. And so that's going to release more outward energy to push out the radius of the actual star. So the general process, and we're going to see this as the star gets more and more massive, is we're going to have heavier and heavier elements forming in the core. Those heavier and heavier elements, as the star gets denser and denser, will eventually ignite, kind of supporting the core. But because the core itself is getting denser and denser and denser, material is getting pushed further and further out with more and more energy. Although if the star is massive enough, it's not going to be able to be pushed out as far as you will have in kind of a red giant, with kind of a sun-like star. But let's just think about how this pattern is going to continue. So eventually, that helium, once it gets dense enough, it's going to ignite and it's going to fuse into carbon. And you're going to have a carbon core forming. So that is carbon core. That's a carbon core. Around that, you have a helium core. you have a shell of helium fusion-- that's helium, not hydrogen-- turning into carbon, making that carbon core denser and hotter. And then around that, you have hydrogen fusion. Have to be very careful. You have hydrogen fusion. And then around that, you have the rest of the star. And so this process is just going to keep continuing. Eventually that carbon is going to start fusing. And you're going to have heavier and heavier elements form the core. And so this is a depiction off of Wikipedia of a fairly mature massive star. And you keep forming these shells of heavier and heavier elements, and cores of heavier and heavier elements until eventually, you get to iron. And in particular, we're talking about iron 56. Iron with an atomic mass of 56. Here on this periodic table that 26 is its atomic number. It's how many protons it has. 56, you kind of view it as a count of the protons"
    },
    {
        "Q": "4:00-4:30 are there others stars that have different elements or have we discovered all of the elements ?",
        "A": "we have discovered all the elements that naturally exist, and even if we hadn t there would have to be stars as big as the milky way to fuse elements beyond iron.",
        "video_name": "UhIwMAhZpCo",
        "timestamps": [
            240,
            270
        ],
        "3min_transcript": "But anyway, let's think about what happens. And so far, just the pattern of what happens, it's going to happen faster because we have more pressure, more gravity, more temperature. But it's going to happen in pretty much the same way as what we saw with a star the mass of the sun. Eventually that helium-- sorry, that hydrogen is going to fuse into a helium core that's going to have a hydrogen shell around it. It's going to have a hydrogen shell around it, hydrogen fusion shell around it. And then you have the rest of the star around that. So let me label it. This right here is our helium core. And more and more helium is going to be built up as this hydrogen in this shell fuses. And in a star the size of our sun or the mass of our sun, this is when it starts to become a red giant. Because this core is getting denser and denser and denser as more and more helium is produced. And as it gets denser and denser and denser, being put on the hydrogen, on this hydrogen shell out here, where we have fusion still happening. And so that's going to release more outward energy to push out the radius of the actual star. So the general process, and we're going to see this as the star gets more and more massive, is we're going to have heavier and heavier elements forming in the core. Those heavier and heavier elements, as the star gets denser and denser, will eventually ignite, kind of supporting the core. But because the core itself is getting denser and denser and denser, material is getting pushed further and further out with more and more energy. Although if the star is massive enough, it's not going to be able to be pushed out as far as you will have in kind of a red giant, with kind of a sun-like star. But let's just think about how this pattern is going to continue. So eventually, that helium, once it gets dense enough, it's going to ignite and it's going to fuse into carbon. And you're going to have a carbon core forming. So that is carbon core. That's a carbon core. Around that, you have a helium core. you have a shell of helium fusion-- that's helium, not hydrogen-- turning into carbon, making that carbon core denser and hotter. And then around that, you have hydrogen fusion. Have to be very careful. You have hydrogen fusion. And then around that, you have the rest of the star. And so this process is just going to keep continuing. Eventually that carbon is going to start fusing. And you're going to have heavier and heavier elements form the core. And so this is a depiction off of Wikipedia of a fairly mature massive star. And you keep forming these shells of heavier and heavier elements, and cores of heavier and heavier elements until eventually, you get to iron. And in particular, we're talking about iron 56. Iron with an atomic mass of 56. Here on this periodic table that 26 is its atomic number. It's how many protons it has. 56, you kind of view it as a count of the protons"
    },
    {
        "Q": "At 2:10 how do you know it is cosine and not tangent or sine?",
        "A": "Think about the problem, if your moving something forward (applied force is greater then friction), and were to use sin0\u00c2\u00b0 then your work F(d)(0)=0 and that doesn t make sense because there is some type of work being done BC the object is moving",
        "video_name": "udgMh3Y-dTk",
        "timestamps": [
            130
        ],
        "3min_transcript": "I'm going to show you some examples of how to solve problems involving work. Imagine a 4 kilogram trashcan. The trashcan is disgusting. So someone ties a string to it and pulls on the string with a force of 50 newtons. The force of kinetic friction on the trashcan while it slides is 30 newtons. The trash can slides across the ground for a distance of 10 meters. Let's try to find the work done by each force on the trash can as it slides across the ground. To find the work done by each force, we should recall the formula definition of work. Work equals Fd cosine theta, where theta is the angle between the force doing the work and the direction the trashcan is moving. There are four forces involved here-- tension, the normal force, the gravitational force, and the force of kinetic friction. In finding the work done for all of these forces, the displacement is going to be 10 meters. But the value of the force and the angle between that force and the displacement is going to differ for each of the forces. we'll plug in the size of the tension, which is 50 newtons. The displacement is 10 meters. And since the tension force is pointed in the same direction as the displacement, the angle between the force of tension and the displacement is 0 degrees. And since cosine of 0 is 1, the work done by the tension force is 500 joules. To find the work done by friction, we'll plug in the size of the force of friction, which is 30 newtons. The displacement is still 10 meters. And since the force of friction points in the opposite direction as the displacement, the angle between the force of friction and the displacement is 180 degrees. Since cosine of 180 is negative 1, the work done by the force of friction is negative 300 joules. Now let's figure out the work done by the gravitational force. The force of gravity is mg. So the force of gravity is 4 kilograms times 9.8 meters per second squared, which is 39.2 newtons. 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"
    },
    {
        "Q": "At 7:00, Sal says that astronauts can't tell whether they're in free fall near an object with a gravitational pull, or in deep space without any noticeable gravitational forces. How is this possible? Won't they feel the difference in acceleration?",
        "A": "How can they feel the acceleration of free fall? The astronauts on the ISS are in free fall. What do you think they feel?",
        "video_name": "oIZV-ixRTcY",
        "timestamps": [
            420
        ],
        "3min_transcript": "If they were to just slow themselves down, if they were to just brake relative to the Earth, and if they were to just put their brakes on right over there, they would all just plummet to the Earth. So there's nothing special about going 300 or 400 miles up into space, that all of a sudden gravity disappears. The influence of gravity, actually on some level, it just keeps going. You can't, it might become unnoticeably small at some point, but definitely for only a couple of hundred miles up in the air, there is definitely gravity there. It's just they're in orbit, they're going fast enough. So if they just keep falling, they're never going to hit the Earth. And if you want to simulate gravity, and this is actually how NASA does simulate gravity, is that they will put people in a plane, and they call it the vomit rocket because it's known to make people sick, and they'll make them go in a projectile motion. So if this is the ground, in a projectile path or in a parabolic path I should say, so the plane will take off, and it or in a parabolic path. And so anyone who's sitting in that plane will experience free fall. So if you've ever been in, if you've ever right when you jump off of a or if you've ever bungee jumped or skydived or even the feeling when a roller coaster is going right over the top, and it's pulling you down, and your stomach feels a little ill, that feeling of free fall, that's the exact same feeling that these astronauts feel because they're in a constant state of free fall. But that is an indistinguishable feeling from, if you were just in deep space and you weren't anywhere close any noticeable mass, that is an identical feeling that you would feel to having no gravitational force around you. So hopefully that clarifies things a little bit. To someone who's just sitting in the space shuttle, and if they had no windows, there's no way of them knowing whether they are close to a massive object and they're just in free fall around it, they're in orbit, or whether they're and they really are in a state of or in a place where there's very little gravity."
    },
    {
        "Q": "At 0:01 what is \"level 4\" multiplication? Does Sal mean grade 4?",
        "A": "no its just to show your going in a more complicated type of multplication",
        "video_name": "_k3aWF6_b4w",
        "timestamps": [
            1
        ],
        "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": "at 1:03 where did he get a and b?",
        "A": "For Ax^2 +Bx +C a*b = AC a + b = B He set the equality, then factored the resultants to obtain candidates for a and b. This takes experience and trial and error.",
        "video_name": "dstNU7It-Ro",
        "timestamps": [
            63
        ],
        "3min_transcript": "Simplify the rational expression and state the domain. Once again, we have a trinomial over a trinomial. To see if we can simplify them, we need to factor both of them. That's also going to help us figure out the domain. The domain is essentially figuring out all of the valid x's that we can put into this expression and not get something that's undefined. Let's factor the numerator and the denominator. So let's start with the numerator there, and since we have a 2 out front, factoring by grouping will probably be the best way to go, so let's just rewrite it here. I'm just working on the numerator right now. 2x squared plus 13x plus 20. We need to find two numbers, a and b, that if I multiply them, a times b, needs to be equal to-- let me write it over here on the right. a times b needs to be equal to 2 times 20, so it has to be equal to positive 40. And then a plus b has to be equal to 13. are 5 and 8, right? 5 times 8 is 40. 5 plus 8 is 13. We can break this 13x into a 5x and an 8x, and so we can rewrite this as 2x squared. It'll break up the 13x into-- I'm going to write the 8x first. I'm going to write 8x plus 5x. The reason why I wrote the 8x first is because the 8 shares common factors with the 2, so maybe we can factor out a 2x here. It'll simplify it a little bit. 5 shares factors with the 20, so let's see where this goes. We finally have a plus 20 here, and now we can group them. That's the whole point of factoring by grouping. You group these first two characters right here. Let's factor out a 2x, so this would become 2x times-- well, 2x squared divided by 2x is just going to be x. 8x divided by 2x is going to be plus 4. And if we factor out a 5, what do we get? We get plus 5 times x plus 4. 5x divided by 5 is x, 20 divided by 5 is 4. We have an x plus 4 in both cases, so we can factor that out. We have x plus 4 times two terms. We can undistribute it. This thing over here will be x plus four times-- let me do it in that same color-- 2x plus 5. And we've factored this numerator expression right there. Now, let's do the same thing with the denominator I'll do that in a different-- I don't want to run out of colors. So the denominator is right over here, let's do the same exercise with it. We have 2x squared plus 17x plus 30."
    },
    {
        "Q": "why we need the slope? he mention it in min 2:34",
        "A": "Since the line of reflection is not horizontal or vertical, he noted that it has a slope of 1. We use iy-intercepts and slopes to graph lines.",
        "video_name": "3aDV3L8aZtY",
        "timestamps": [
            154
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 2:30 he takes the derivative of e^(-2x^2) with respect to (-2x^2). Why doesn't the power rule apply here?",
        "A": "The power rule applies when the base is a variable and the power is a number. This is an exponential, the base is a number and the variable is in the power. They are very different functions, and therefore have different derivatives.",
        "video_name": "MUQfl385Yug",
        "timestamps": [
            150
        ],
        "3min_transcript": "Let's say that f of x is equal to x times e to the negative two x squared, and we want to find any critical numbers for f. I encourage you to pause this video and think about, can you find any critical numbers of f. I'm assuming you've given a go at it. Let's just remind ourselves what a critical number is. We would say c is a critical number of f, if and only if. I'll write if with two f's, short for if and only if, f prime of c is equal to zero or f prime of c is undefined. If we look for the critical numbers for f we want to figure out all the places where the derivative of this with 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,"
    },
    {
        "Q": "At 5:15 why is the answer a positive and a negative? e.g. + or - 1/2",
        "A": "It s because you are taking the square root of (1/4). When you take the square root you ll have two solutions ( + or - 1/2 in this case ) since when you square each one you ll get (1/4).",
        "video_name": "MUQfl385Yug",
        "timestamps": [
            315
        ],
        "3min_transcript": "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 Then what x values is this true at? We just take the plus or minus square root of both sides and you get x is equal to plus or minus one half. Negative one half squared is one fourth, positive one half squared is one fourth. If x equals plus or minus one half f prime, or the derivative, is equal to zero. Let me write it this way. f prime of one half is equal to zero, and you can verify that right over here. And f prime of negative one half is equal to zero. If someone asks what are the critical numbers here, they are one half and negative one half."
    },
    {
        "Q": "At about 2:29, Sal is saying that 3a(a^5)(a^2) is 3(a^1*a^7). How? I am probably just confused on this topic, though.",
        "A": "a = a^1 3\u00e2\u0080\u00a2a^1\u00e2\u0080\u00a2a^5\u00e2\u0080\u00a2a^2 = 3\u00e2\u0080\u00a2a^(1 + 5 + 2) = 3\u00e2\u0080\u00a2a^8",
        "video_name": "-TpiL4J_yUA",
        "timestamps": [
            149
        ],
        "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": "At 1:28 why is he making those markings? He draws a line and then crosses it out... as to say, what? If he wants to make it easier for himself to follow along, shouldn't he be crossing out the 2 digit and 3 digit instead?",
        "A": "He circles the numbers to see which numbers he s multiplying. He s just trying to help us understand by giving us a visual. :)",
        "video_name": "TqRReFvbpXA",
        "timestamps": [
            88
        ],
        "3min_transcript": "Let's multiply 7 times 253 and see what we get. So just like in the last example, what I like to do is I like to rewrite the largest number first. So that's 253. And then write the smaller number below it and align the place value, the 7. It only has a ones place, so I'll put the 7 right over here below the ones place in 253. And then put the multiplication symbol right over here. So you could read this as 253 times 7, which we know is the same thing as 7 times 253. And now we are ready to compute. And there are many ways of doing this, but this one you could call the standard way. So what I do is I start with my 7. And I multiply it times each of the numbers up here, and I carry appropriately. So first I start with 7 times 3. Well, 7 times 3 we know is 21. Let me write that down. 7 times 3 is equal to 21. You could do this part in your head, but I just want to make it clear where I'm getting these numbers from. is I would write the 1 into 21 down here, but then carry the 2 to the tens place. Now I want to figure out what 7 times 5 is. We know from our multiplication tables that 7 times 5 is equal to 35. Now, we can't just somehow put the 35 down here. We still have to deal with this 2 that we carried. So we compute 7 times 5 is 35, but then we also add that 2. So it's 35 plus 2 is 37. Now, we write the 7 right over here in the tens place and carry the 3. Now we need to compute what 7 times 2 is. We know that 7 times 2 is 14 from our multiplication tables. We can't just put a 14 down here. We have this 3 to add. So 7 times 2 is 14, plus 3 is 17. because 2 is the last number that we had to deal with. And so we have our answer. 7 times 253 is 1,771."
    },
    {
        "Q": "At 4:33, how would you find the area of an inscribed polygons?",
        "A": "I would break it into triangles, each with a vertex at the center of the circle, then use trigonometry (or special triangles) to solve for the sides and area.",
        "video_name": "LrxZMdQ6tiw",
        "timestamps": [
            273
        ],
        "3min_transcript": "So we can use that information to figure out what the other angles are. Because these two base angles-- it's an isosceles triangle. The two legs are the same. So our two base angles, this angle is going to be congruent to that angle. If we could call that y right over there. So you have y plus y, which is 2y, plus 60 degrees is going to be equal to 180. Because the interior angles of any triangle-- they add up to 180. And so subtract 60 from both sides. You get 2y is equal to 120. Divide both sides by 2. You get y is equal to 60 degrees. Now, this is interesting. I could have done this with any of these triangles. All of these triangles are 60-60-60 triangles, which tells us-- and we've proven this earlier on when we first started studying equilateral triangles-- we know that all of the angles of a triangle are 60 degrees, then we're dealing with an equilateral triangle, which means that all the sides have the same length. So if this is 2 square roots of 3, then so is this. And this is also 2 square roots of 3. So pretty much all of these green lines are 2 square roots of 3. And we already knew, because it's a regular hexagon, that each side of the hexagon itself is also 2 square roots of 3. So now we can essentially use that information to figure out-- actually, we don't even have to figure this part out. I'll show you in a second-- to figure out the area of any one of these triangles. And then we can just multiply by 6. So let's focus on this triangle right over here and think about how we can find its area. We know that length of DC is 2 square roots of 3. We can drop an altitude over here. We can drop an altitude just like that. And then if we drop an altitude, we know that this is an equilateral triangle. And we can show very easily that these two triangles are symmetric. These are both 90-degree angles. We know that these two are 60-degree angles already. And then if you look at each of these two independent triangles, you'd have to just say, well, So this has to be 30 degrees. This has to be 30 degrees. All the angles are the same. They also share a side in common. So these two are congruent triangles. So if we want to find the area of this little slice of the pie right over here, we can just find the area of this slice, or this sub-slice, and then multiply by 2. Or we could just find this area and multiply by 12 for the entire hexagon. So how do we figure out the area of this thing? Well, this is going to be half of this base length, so this length right over here. Let me call this point H. DH is going to be the square root of 3. And hopefully we've already recognized that this is a 30-60-90 triangle. Let me draw it over here. So this is a 30-60-90 triangle. We know that this length over here is square root of 3. And we already actually did calculate that this is 2 square roots of 3. Although we don't really need it. What we really need to figure out is this altitude height. And from 30-60-90 triangles, we know"
    },
    {
        "Q": "At 2:10, how ar the triangles congruent by SSS? I thought SSS was a similarity postulate and the triangles are congruent through the SAS Postulate.",
        "A": "SSS is both a congruency and a similarity postulate. We cannot use SAS because we haven t proven that all the central angles are congruent.",
        "video_name": "LrxZMdQ6tiw",
        "timestamps": [
            130
        ],
        "3min_transcript": "We're told that ABCDEF is a regular hexagon. And this regular part-- hexagon obviously tells us that we're dealing with six sides. And you could just count that. You didn't have to be told it's a hexagon. But the regular part lets us know that all of the sides, all six sides, have the same length and all of the interior angles have the same measure. Fair enough. And then they give us the length of one of the sides. And since this is a regular hexagon, they're actually giving us the length of all the sides. They say it's 2 square roots of 3. So this side right over here is 2 square roots of 3. This side over here is 2 square roots of 3. And I could just go around the hexagon. Every one of their sides is 2 square roots of 3. They want us to find the area of this hexagon. Find the area of ABCDEF. And the best way to find the area, especially of regular polygons, is try to split it up into triangles. And hexagons are a bit of a special case. Maybe in future videos, we'll think about the more general case of any polygon. But with a hexagon, what you could think about is if we take this point right over here. And let's call this point G. And let's say it's the center of the hexagon. I'm talking about a point. It can't be equidistant from everything over here, because this isn't a circle. But we could say it's equidistant from all of the vertices, so that GD is the same thing as GC is the same thing as GB, which is the same thing as GA, which is the same thing as GF, which is the same thing as GE. So let me draw some of those that I just talked about. So that is GE. There's GD. There's GC. All of these lengths are going to be the same. So there's a point G which we can call the center of this polygon. And we know that this length is equal to that length, which is equal to that length, which is equal to that length, which is equal to that length, which is equal to that length. We also know that if we go all the way around the circle like that, we've gone 360 degrees. And we know that these triangles are all going to be congruent to each other. And there's multiple ways that we could show it. But the easiest way is, look, they have two sides. All of them have this side and this side And they all have this third common side of 2 square roots of 3. So all of them, by side-side-side, they are all congruent. What that tells us is, if they're all congruent, then this angle, this interior angle right over here, is going to be the same for all six of these triangles over here. And let me call that x. That's angle x. That's x. That's x. That's x. That's x. And if you add them all up, we've gone around the circle. We've gone 360 degrees. And we have six of these x's. So you get 6x is equal to 360 degrees. You divide both sides by 6. You get x is equal to 60 degrees. All of these are equal to 60 degrees. Now there's something interesting. We know that these triangles-- for example, triangle GBC-- and we could do that for any of these six triangles. It looks kind of like a Trivial Pursuit piece. We know that they're definitely isosceles triangles,"
    },
    {
        "Q": "At around 1:40, couldn't you take the (positive and negative) square roots of both sides?",
        "A": "You could. (and then x = -2 and x = 2) Good question!",
        "video_name": "bml74_PsfwA",
        "timestamps": [
            100
        ],
        "3min_transcript": "Solve and eliminate any extraneous solutions. And what they mean by extraneous solutions are, in the course of solving this rational equation right here, we might get some solutions that if we actually put it into the original problem would give us undefined expressions. And so those solutions are extraneous solutions. They actually don't apply. You actually want to throw them out. And so let's look at this equation. We have x squared over x plus 2 is equal to 4 over x plus 2. So right from the get-go, we don't know if this is going to necessarily be a solution to this equation. But we know, just looking at this, that if x is equal to negative 2, then this denominator and this denominator are going to be 0. And you're dividing by 0. It would be undefined. So we can, right from the get-go, exclude x is equal to negative 2. So x cannot be equal to negative 2. That would make either of these expressions undefined, on either side of the equation. So with that out of the way, let's try to solve it. So as a first step, we want to get the x plus 2 out of the denominator. So let's multiply both sides by x plus 2. And we can assume that x plus 2 isn't 0. So it's going to be defined. x plus 2 divided by x plus 2 is just 1. And so our equation has simplified to x squared is equal to 4. And you could probably do this in your head, but I want to do it properly. So you can write this. You could subtract 4 from both sides. Do it in kind of the proper quadratic equation form. So x squared minus 4 is equal to 0. I just subtracted 4 from both sides over here. And so you could factor this. This is a difference of squares. You get x plus 2 times x minus 2 is equal to 0. And then if this is equal to 0, if the product of two things are equal to 0, that means either one or both of them are equal to 0. So this tells us that x plus 2 is equal to 0 or x minus 2 is equal to 0. If you subtract 2 from both sides of this equation right If you add 2 to both sides of this equation right over here, you get x is equal to 2. And we're saying that either of these would make this last expression 0. Now, we know that we need to exclude one of them. We know that x cannot be equal to negative 2. So x equals negative 2 is an extraneous solution. It's not really a solution for-- it is a solution for this, once we got rid of the rational expressions. But it's not a solution for this original problem up here, because it would make the expressions undefined. It would cause you to divide by 0. So the only solution here is x is equal to 2. And you can check it yourself. If you do 2 squared, you get 4, over 2 plus 2 is 4. And that should be equal to 4 over 2 plus 2, over 4, which it definitely does. 1 is equal to 1."
    },
    {
        "Q": "At 2:33 Sal says it will make sense. I never really got past the 2 to the 0 power. I don't understand how he took two, put a zero above it, and then turned it into a one?! How does that work?",
        "A": "What he showed helped me understand why it does that, so now let me try to explain it for you: He changed the way he did the exponents to multiplying 1 times how many numbers (the number that the exponent is) to that one. So when it is 2\u00e2\u0081\u00b0 = 1 because there aren t any 2s to multiply by. So something as big as 1,000,000\u00e2\u0081\u00b0 = 1 Now let s do it regular: When you have 4\u00e2\u0081\u00b6 = 1 \u00c3\u0097 4 \u00c3\u0097 4 \u00c3\u0097 4 \u00c3\u0097 4 \u00c3\u0097 4 \u00c3\u0097 4 = 4096 Well, that is a bit too big of a number so let s do 3\u00c2\u00b2 = 1 \u00c3\u0097 3 \u00c3\u0097 3 = 9 Ask me to clarify anything.",
        "video_name": "dAvosUEUH6I",
        "timestamps": [
            153
        ],
        "3min_transcript": ""
    },
    {
        "Q": "is there other basic rigid motions other than reflect,translate, and rotate?\n\nas said in 1:07",
        "A": "Those three translations are the three basic geometric translations besides dilation.",
        "video_name": "EDlZAyhWxhk",
        "timestamps": [
            67
        ],
        "3min_transcript": "So we have another situation where we want to see whether these two figures are congruent. And the way we're going to test that is by trying to transform this figure by translating it, rotating it, and reflecting it. So the first thing that might-- let me translate it. So if these two things are congruent, looks like point E and the point that I'm touching right now, those would correspond to each other. Let me try to rotate it a little bit. So let me rotate-- whoops, I don't want to rotate there, I want to rotate around point E since I already have those on top of each other. So just like that. And it looks like now, if I reflect it across that line, I'm going to be-- oh no, I'm not there. You see, this is tricky. See, when I reflect it, this point, this point, this point, and this point seem to be in the exact same place, but point C does not correspond with that point right over there. So these two polygons are not congruent. And this is why it's important to do this, to make sure the rotations work out. So these two are not congruent. through translations, rotations, and reflections. So are these polygons congruent? No."
    },
    {
        "Q": "At 1:09 Sal says that he doesn't like using FOIL. FOIL is really easy and is much less confusing then the way Sal did the distributive property twice even though you get the same answer. Why does Sal not like FOIL?",
        "A": "FOIL won t help you if you have to expand a product that isn t two binomials multiplied together; for example, two trinomials multiplied together. It s usually better to understand what you re doing instead of relying on mnemonics. For example: (a + b + c) * (d + e + f) = ad + ae + af + bd + be + bf + cd + ce + cf",
        "video_name": "JKvmAexeMgY",
        "timestamps": [
            69
        ],
        "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 around 2:00-2:10, why is it when x=1 and when x=3? Why not something else?",
        "A": "You can pick any x you like. 0 and 1 are frequently picked because 0*a=0 and 1*a=a. Since the multiplication is easy, you can solve the problem quicker.",
        "video_name": "7QMoNY6FzvM",
        "timestamps": [
            120,
            130
        ],
        "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": "1:47\nHow do you determine whether the parabola faces upward \"U\" or downward?\nSOS",
        "A": "Clarifiction to the prior response you were given: You need to look at the coefficient of the X^2 terms or the coefficient in front of (X+b)^2. The coefficient of whatever is being squared determines the direction of the parabola. If the number is positive, then the parabola opens upward. If the coefficient is negative, the parabola opens downward.",
        "video_name": "7QMoNY6FzvM",
        "timestamps": [
            107
        ],
        "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": "At 5:00, doesn't the \"function\" Sal draws fail the vertical line test? Why does he still call it a function?",
        "A": "That s called a piecewise function - they re defined like this: f(x) = { x + 1 for x > 1 x - 1 for x < 1 x for x = 1 }",
        "video_name": "8VgmBe3ulb8",
        "timestamps": [
            300
        ],
        "3min_transcript": "from the beginning-- because this is really the definition of an even function-- is when you look at this, you're like hey, what does this mean? f of x is equal to f of negative x. And all it does mean is this. It means that if I were to take f of 2, f of 2 is 4. So let me show you with a particular case. f of 2 is equal to f of negative 2. And this particular case for f of x is equal to x squared, they are both equal to 4. So really, this is just another way of saying that the function can be reflected, or the left side of the function is the reflection of the right side of the function across the vertical axis, across the y-axis. Now just to make sure we have a decent understanding here, let me draw a few more even functions. And I'm going to draw some fairly wacky things just so you really kind of learn to visually recognize them. So let's say a function like this, it does something like that. And then on this side, it does the same thing. It's the reflection, so it jumps up here, then it goes like this, then it goes like this. I'm trying to draw it so it's the mirror image of each other. This is an even function. You take what's going on on the right hand side of this function and you literally just reflect it over the y-axis, and you get the left hand side of the function. And you could see that even this holds. If I take some value-- let's say that this value right here is, I don't know, 3. And let's say that f of 3 over here is equal to, let's say, that that is 5. So this is 5. We see that f of negative 3 is also going to be equal to 5. And I can draw, let me just draw one more to really make sure. I'll do the axis in that same green color. Let me do one more like this. And you could have maybe some type of trigonometric looking function that looks like this, that looks like that. And it keeps going in either direction. So something like this would also be even. So all of these are even functions. Now, you are probably thinking, well, what is an odd function? And let me draw an odd function for you. So let me draw the axis once again. x-axis, y-axis, or the f of x-axis. And to show you an odd function, I'll give you a particular odd function, maybe the most famous of the odd functions. This is probably the most famous of the even functions. And it is f of x-- although there are probably"
    },
    {
        "Q": "at 4:07, when he draws the funny function graph, isn't it true that it isn't a function because it doesn't pass the vertical line test? It appears that the line would go through the open circle and the wing-shaped line on the left side.",
        "A": "Perhaps he just didn t draw it clearly. I think he meant to draw an open circle as you suggested. If that was not his intention, then you are correct, it isn t a function.",
        "video_name": "8VgmBe3ulb8",
        "timestamps": [
            247
        ],
        "3min_transcript": "And this, hopefully, or maybe makes complete sense to you. You're like, well, Sal, obviously if I just reflect this function over the y-axis, that's going to be the case. Whatever function value I get at the positive value of a number, I'm going to get the same function value at the negative value. And this is what kind of leads us to the formal definition. If a function is even-- or I could say a function is even if and only of-- so it's even. And don't get confused between the term even function and the term even number. They're completely different kind of ideas. So there's not, at least an obvious connection that I know of, between even functions and even numbers or odd functions and odd numbers. So you're an even function if and only if, f of x is equal to f of negative x. from the beginning-- because this is really the definition of an even function-- is when you look at this, you're like hey, what does this mean? f of x is equal to f of negative x. And all it does mean is this. It means that if I were to take f of 2, f of 2 is 4. So let me show you with a particular case. f of 2 is equal to f of negative 2. And this particular case for f of x is equal to x squared, they are both equal to 4. So really, this is just another way of saying that the function can be reflected, or the left side of the function is the reflection of the right side of the function across the vertical axis, across the y-axis. Now just to make sure we have a decent understanding here, let me draw a few more even functions. And I'm going to draw some fairly wacky things just so you really kind of learn to visually recognize them. So let's say a function like this, it does something like that. And then on this side, it does the same thing. It's the reflection, so it jumps up here, then it goes like this, then it goes like this. I'm trying to draw it so it's the mirror image of each other. This is an even function. You take what's going on on the right hand side of this function and you literally just reflect it over the y-axis, and you get the left hand side of the function. And you could see that even this holds. If I take some value-- let's say that this value right here is, I don't know, 3. And let's say that f of 3 over here is equal to, let's say, that that is 5. So this is 5. We see that f of negative 3 is also going to be equal to 5."
    },
    {
        "Q": "AT 9:39, why is there a gap between the reflection about the y-axis and the reflection about the x-axis?\nAt 6:58, there was no gap.",
        "A": "It s just a different equation he s graphing (I m not sure what the equation is; if anyone knows I d love to hear). It does look rather strange as a function, but it s still classified as an odd function.",
        "video_name": "8VgmBe3ulb8",
        "timestamps": [
            579,
            418
        ],
        "3min_transcript": "we figured out f of 2 is 8. 2 to the third power is 8. We know that f of negative 2 is negative 8. Negative 2 to the third power is negative 8. So you have the negative of negative 8, negatives cancel out, and it works out. So in general, you have an odd function. So here's the definition. You are dealing with an odd function if and only if f of x for all the x's that are defined on that function, or for which that function is defined, if f of x is equal to the negative of f of negative x. Or you'll sometimes see it the other way if you multiply both sides of this equation by negative 1, you would get negative f of x is equal to f of negative x. And sometimes you'll see it where it's swapped around where they'll say f of negative x is equal to-- let me write that careful-- I just swapped these two sides. So let me just draw you some more odd functions. So I'll do these visually. So let me draw that a little bit cleaner. So if you have maybe the function does something wacky like this on the right hand side. If it was even, you would reflect it there. But we want to have an odd function, so we're going to reflect it again. So the rest of the function is going to look like this. So what I've drawn in the non-dotted lines, this right here is an odd function. And you could even look at the definition. If you take some value, a, and then you take f of a, which would put you up here. This right here would be f of a. If you take the negative value of that, if you took negative a here, f of negative a is going to be down here. the same distance from the horizontal axis. It's not completely clear the way I drew it just now. So it's maybe going to be like right over here. So this right over here is going to be f of negative a, which is the same distance from the origin as f of a, it's just the negative. I didn't completely draw it to scale. Let me draw one more of these odd functions. I think you might get the point. Actually, I'll draw a very simple odd function, just to show you that it doesn't always have to be something crazy. So a very simple odd function would be y is equal to x, something like this. Whoops. y is equal going through the origin. You reflect what's on the right onto to the left. You get that. And then you reflect it down, you get all of this stuff in the third quadrant. So this is also an odd function. Now, I want to leave you with a few things that are not odd functions and that sometimes might"
    },
    {
        "Q": "at 0:10, he says rational #s are different from natural numbers. What exactly ARE rational numbers?",
        "A": "natural numbers are the numbers you learned for counting: 1, 2, 3 ... whole numbers are the numbers of apples you might have: 0, 1, 2, 3 ... integers are positive or negative numbers that are not fractions: ... -3, -2, -1, 0, 1, 2, 3 ... rational numbers are any number that can be expressed as a fraction of integers.",
        "video_name": "i1i2_9wg6N8",
        "timestamps": [
            10
        ],
        "3min_transcript": "- [Voiceover] What I'd like to do in this video is order these six numbers from least to greatest. So the least of them being on the left hand side, and the greatest on the right. And I encourage you to pause this video, and see if you can do it on your own, before we work through it together. So assuming you've had a go at it, so let's do it together, and to help us there, let's plot these numbers on a number line, and I have a number line up here, so there you go, there is a handy number line. And let's just take them one by one. So the first number here, we have 7/3. So let's see if we can express that in a different way, if we can write that as a mixed number. So 7/3, how many wholes are here? And the whole is going to be 3/3. So this is going to be 3/3, plus another 3/3, is going to get us to 6/3. And so you're going to have one more third left. So this is 7/3. Three plus three, plus one is seven. So this is 3/3 is one whole, three thirds is one whole, so this is two So seven thirds, same thing as one, two, and you see, between consecutive integers, we have three spaces, so we are essentially marking off thirds, so two and 1/3, is going to be 1/3 of the way between two and three, so it's going to be right over there, so that is 7/3. Then we have negative 5/2, so same logic. Negative, let me do that in that green color. So, I can do it over here. Negative five over two, well that's the same thing as the negative of 5/2, and 5/2 is going to be 2/2 plus another 2/2, plus 1/2 so this is two and 1/2, this is one, this is one, and that's 1/2. So this is going to be one, plus one, plus 1/2, two and 1/2, we have our negative out there, so it's negative two and 1/2. and the negative two and 1/2 is going to be halfway between negative two, and negative three, so it's going to be right over there. So that is negative 5/2. Then we have zero, not too difficult. It's actually labeled on our number line for us. Then we have negative two. Negative two, once again, on our number line for us. Two, two steps, two whole numbers to the left of zero. So negative two is going to put us right over there. Then we have negative 12/4 so it might jump out at you immediately, 12 divided by four is three, so this is going to be the same thing as negative three. So this is negative 12/4 or if you want to use the type of logic that we used in these first two numbers, you could say negative 12/4 is the same thing as the negative of 12/4 which is 4/4 plus another 4/4"
    },
    {
        "Q": "At 0:21, what does Sal mean by \" arbrutary\"?",
        "A": "Any old An arbitrary angle measure is a random angle or just any old angle",
        "video_name": "0gzSreH8nUI",
        "timestamps": [
            21
        ],
        "3min_transcript": "Thought I would do some more example problems involving triangles. And so this first one, it says the measure of the largest angle in a triangle is 4 times the measure of the second largest angle. The smallest angle is 10 degrees. What are the measures of all the angles? Well, we know one of them. We know it's 10 degrees. Let's draw an arbitrary triangle right over here. So let's say that is our triangle. We know that the smallest angle is going to be 10 degrees. And I'll just say, let's just assume that this right over here is the measure of the smallest angle. It's 10 degrees. Now let's call the second largest angle-- let's call that x. So the second largest angle, let's call that x. So this is going to be x. And then the first sentence, they say the measure of the largest angle in a triangle is 4 times the measure of the second largest angle. So the second largest angle is x. 4 times that measure is going to be 4x. So the largest angle is going to be 4x. of the angles inside of a triangle is that they add up to 180 degrees. So we know that 4x plus x plus 10 degrees is going to be equal to 180 degrees. It's going to be equal to 180. And 4x plus x, that just gives us 5x. And then we have 5x plus 10 is equal to 180 degrees. Subtract 10 from both sides. You get 5x is equal to 170. And so x is equal to 170/5. And let's see, it'll go into it-- what is that, 34 times? Let me verify this. So 5 goes into-- yeah, it should be 34 times because it's going to go into it twice as many times as 10 would go into it. 10 would go into 170 17 times. 5 would go into 170 34 times. So we could verify it. Go into 170. 5 goes into 17 three times. 3 times 5 is 15. 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."
    },
    {
        "Q": "Would 3 quadrupled by shown as a radical with a little four in the \"notch\", as a cube root is described at 4:32?",
        "A": "4th roots would have a little 4 in the notch of the radical symbol. 5th roots would have a 5 in that position. 6th roots would have a 6 in that position. See the pattern?",
        "video_name": "87_qIofPwhg",
        "timestamps": [
            272
        ],
        "3min_transcript": "Well the volume is going to be two, times two, times two, which is two to the third power or two cubed. This is two cubed. That's why they use the word cubed because this would be the volume of a cube where each of its sides have length two and this of course is going to be equal to eight. But what if we went the other way around? What if we started with the cube? What if we started with this volume? What if we started with a cube's volume and let's say the volume here is eight cubic units, so volume is equal to eight and we wanted to find the lengths of the sides. So we wanted to figure out what X is cause that's X, that's X, and that's X. It's a cube so all the dimensions have the same length. Well there's two ways that we could express this. We could say that X times X times X or we could use the cube root symbol, which is a radical with a little three in the right place. Or we could write that X is equal to, it's going to look very similar to the square root. This would be the square root of eight, but to make it clear, they were talking about the cube root of eight, we would write a little three over there. In theory for square root, you could put a little two over here, but that'd be redundant. If there's no number here, people just assume that it's the square root. But if you're figuring out the cube root or sometimes you say the third root, well then you have to say, well you have to put this little three right over here in this little notch in the radical symbol right over here. And so this is saying X is going to be some number that if I cube it, I get eight. So with that out of the way, let's do some examples. Let's say that I have... Let's say that I want to calculate the cube root of 27. Well if say that this is going to be equal to X, this is equivalent to saying that X to the third or that 27 is equal to X to the third power. So what is X going to be? Well X times X times X is equal to 27, well the number I can think of is three, so we would say that X, let me scroll down a little bit, X is equal to three. Now let me ask you a question. Can we write something like... Can we pick a new color? The cube root of, let me write negative 64. I already talked about that if we're talking the square root, it's fairly typical that hey you put a negative number in there at least until we learn about imaginary numbers, we don't know what to do with it. But can we do something with this? Well if I cube something, can I get a negative number? Sure. So if I say this is equal to X,"
    },
    {
        "Q": "At 5:02 Sal say's that he is going to calculate the CUBE root of 27.\nI didn't understasnd how it is calculated?",
        "A": "So what Sal is saying is he is breaking it down into 3 * 3 = 9 9 * 3 = 27 So, 3 * 3 * 3 = 27. Therefore, 3 is the cube root of 27. Just like cube root of y = x * x * x",
        "video_name": "87_qIofPwhg",
        "timestamps": [
            302
        ],
        "3min_transcript": "Well the volume is going to be two, times two, times two, which is two to the third power or two cubed. This is two cubed. That's why they use the word cubed because this would be the volume of a cube where each of its sides have length two and this of course is going to be equal to eight. But what if we went the other way around? What if we started with the cube? What if we started with this volume? What if we started with a cube's volume and let's say the volume here is eight cubic units, so volume is equal to eight and we wanted to find the lengths of the sides. So we wanted to figure out what X is cause that's X, that's X, and that's X. It's a cube so all the dimensions have the same length. Well there's two ways that we could express this. We could say that X times X times X or we could use the cube root symbol, which is a radical with a little three in the right place. Or we could write that X is equal to, it's going to look very similar to the square root. This would be the square root of eight, but to make it clear, they were talking about the cube root of eight, we would write a little three over there. In theory for square root, you could put a little two over here, but that'd be redundant. If there's no number here, people just assume that it's the square root. But if you're figuring out the cube root or sometimes you say the third root, well then you have to say, well you have to put this little three right over here in this little notch in the radical symbol right over here. And so this is saying X is going to be some number that if I cube it, I get eight. So with that out of the way, let's do some examples. Let's say that I have... Let's say that I want to calculate the cube root of 27. Well if say that this is going to be equal to X, this is equivalent to saying that X to the third or that 27 is equal to X to the third power. So what is X going to be? Well X times X times X is equal to 27, well the number I can think of is three, so we would say that X, let me scroll down a little bit, X is equal to three. Now let me ask you a question. Can we write something like... Can we pick a new color? The cube root of, let me write negative 64. I already talked about that if we're talking the square root, it's fairly typical that hey you put a negative number in there at least until we learn about imaginary numbers, we don't know what to do with it. But can we do something with this? Well if I cube something, can I get a negative number? Sure. So if I say this is equal to X,"
    },
    {
        "Q": "at 0:31 why when you square 49 the result is 49?",
        "A": "He is squaring the result of \u00e2\u0088\u009a49, which is 7. 7 squared is 49, so (\u00e2\u0088\u009a49)^2 = 49.",
        "video_name": "87_qIofPwhg",
        "timestamps": [
            31
        ],
        "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 0:58, what are Lucas numbers??",
        "A": "Lucas numbers are defined in pretty much the same way as the Fibonacci numbers except that the first Lucas number is 2 instead of 1. So, for the Lucas sequence, you get: 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, ....",
        "video_name": "14-NdQwKz9w",
        "timestamps": [
            58
        ],
        "3min_transcript": "So say you're me and you're in math class and you're trying to ignore the teacher and doodle Fibonacci spirals, while simultaneously trying to fend off the local greenery. Only, you become interested in something that the teacher says by accident, and so you draw too many squares to start with. So you cross some out, but cross out too many, and then the teacher gets back on track, and the moment is over. So oh well, might as well try and just do the spiral from here. So you make a three by three square. And here's a four by four, and then seven, and then 11 This works because then you've got a spiral of squares. So you write down the numbers, 1, 3, 4, 7, 11, 18. It's kind of like the Fibonacci series, because 1 plus 3 is 4, 3 plus 4 is 7, and so on. Or maybe it starts at 2 plus 1, or negative 1 plus 2. Either way, it's a perfectly good series. And it's got another similarity with the Fibonacci series. The ratios of consecutive numbers also approach Phi. So, a lot of plants have Fibonacci numbers of spirals, but to understand how they do it, we can learn from the exceptions. This pine cone that has seven spirals one way, and eleven the other, might be showing Lucas numbers. And since Fibonacci numbers and Lucas numbers are related, maybe that explains it. by always growing new parts a Phi-th of a circle around. What angle would give Lucas numbers? In this pine cone, each new pine cone-y thing is about 100 degrees around from the last. We're going to need a Lucas angle-a-tron. It's easy to get a 90 degree angle-a-tron, and if I take a third of a third of that, that's a ninth of 90, which is another 10 degrees. Now you can use it to get spiral patterns like what you see on the Lucas number plants. It's an easy way to grow Lucas spirals, if plants have an internal angle-a-tron. Thing is, 100 is pretty far from 137.5. If plants were somehow measuring angle, you'd think the anomalous ones would show angles close to a Phi-th of a circle. Not jump all the way to 100. Maybe I'd believe different species use different angles, but two pine cones from the same tree? Two spirals on the same cauliflower? And that's not the only exception. A lot of plants don't grow spirally at all. Like this thing. With leaves growing opposite from each other. And some plants have alternating leaves, 180 degrees from each other, which is far from both Phi and Lucas angles. have a fundamentally different growth pattern, and are in a different class of plant or something, but wouldn't it be even better if there were one simple reason for all of these things? These variations are a good clue that maybe these plants get this angle, and Fibonacci numbers, as a consequence of some other process and not just because it mathematically optimizes sunlight exposure if the sun is right overhead, which it pretty much never is, and if the plan are perfectly facing straight up, which they aren't. So how do they do it? Well, you could try observing them. That would be like science. If you zoom in on the tip of a plant, the growing part, there's this part called the meristem. That's where new plant bits form. The biggest plant bits where the first to form off the meristem, and the little ones around the center are newer. As the plants grow, they get pushed away from the meristem, but they all started there. The important part is that a science observer would see the plant bits pushing away, not just from the meristem, but from each other. A couple physicists once tried this thing where they dropped drops of a magnetized liquid in a dish of oil."
    },
    {
        "Q": "The age of vinod and sanjeev are in the ratio of 5:7. Ten year later the ratio of their age will be 7:9. Find the present age.",
        "A": "Let v be vinod age now and s be sajeev age now. 5c= 7v so c = 7/5 v 10 years from now, 7(c+10) = 9(v+10) so 7c + 70 = 9v + 90 or 7c = 9v +20 Substitute into c, so 7(7/5v) = 9v + 20 multiply whole equation by 5 49v = 45v + 100 subtract 45v 4v = 100, v = 25 and c = 7/5(25) = 35 Ratio of 25:35 reduces to 5:7 Ten years, v will be 35 and c will be 45 so ratio of 35:45 or 7:9",
        "video_name": "W-5liMGKgHA",
        "timestamps": [
            307,
            429
        ],
        "3min_transcript": "And on the right hand side, I can distribute this 3. So 3 times 2 is 6. 3 times y is 3y. 6 plus 3y. And then it's always nice to get all of our constants on one side of the equation, all of our variables on the other side of the equation. So we have a 3y over here. We have more y's on the right hand side than the left hand So let's get rid of the y's on the left hand side. You could do it either way, but you'd end up with negative numbers. So let's subtract a y from each side. And we are left with, on the left hand side, 18. And on the right hand side you have 6 plus 3 y's. Take away one of those y's. You're going to be left with 2 y's. Now we can get rid of the constant term here. So we will subtract 6 from both sides. 18 minus 6 is 12. The whole reason why we subtracted 6 from the right was to get rid of this, 6 minus 6 is 0, so you have 12 Two times the number of years it will take is 12, and you could probably solve this in your head. But if we just want a one-coefficient year, we would divide by 2 on the right. Whatever we do to one side of an equation, we have to do it on the other side. Otherwise, the equation will not still be an equation. So we're left with y is equal to 6, or y is equal to 6. So going back to the question, how many years will it take for Arman to be three times as old as Diya? Well, it's going to take six years. Now, I want you to verify this. Think about it. Is this actually true? Well, in six years, how old is Arman going to be? He's going to be 18 plus 6. We now know that this thing is 6. So in six years, Arman is going to be 18 plus 6, which is 24 years old. How old is Diya going to be? Well, she's going to be 2 plus 6, which is 8 years old. as 8. In 6 years-- Arman is 24, Diya is 8-- Arman is three times as old as Diya, and we are done."
    },
    {
        "Q": "Okay, quite possibly my concept of integration is wrong but shouldn't increasing the number of trapeziums bring the result closer to the value obtained by integrating f(x)? in this case integrating it brings the value to 7.4, which is close enough to the value obtained at 8:26 , but if i increase the no. of trapeziums to 10, the value of delx becomes 0.5, so the total becomes 3.26. it becomes less accurate.",
        "A": "If you increase the number of trapeziums to 10, then \u00ce\u0094x = 0.5, and you got that the area is equal to: A = 0.5/2 [ f(1) + 2f(1.5) + 2f(2) + 2f(2.5) + 2f(3) + 2f(3.5) + 2f(4) + 2f(4.5) + 2f(5) + 2f(5.5) + f(6) ] A = 0.25 [ 0 + 2\u00e2\u0088\u009a0.5 + 2\u00e2\u0088\u009a1 + 2\u00e2\u0088\u009a1.5 + 2\u00e2\u0088\u009a2 + 2\u00e2\u0088\u009a2.5 + 2\u00e2\u0088\u009a3 + 2\u00e2\u0088\u009a3.5 + 2\u00e2\u0088\u009a4 + 2\u00e2\u0088\u009a4.5 + \u00e2\u0088\u009a5 ] A \u00e2\u0089\u0085 7.3847 Which is a much closer result.",
        "video_name": "1p0NHR5w0Lc",
        "timestamps": [
            506
        ],
        "3min_transcript": "f of 1, let's just remind ourselves what our original function was. Our original function was the square root of x minus 1. So f of 1 is the square root of 1 minus 1, so that is just going to be 0. This expression right over here is going to be 2 times the square root of 2 minus 1. The square root of 2 minus 1 is just 1, so this is just going to be 2. Actually, let me do it in that same-- well, I'm now using the purple for a different purpose than just the first trapezoid. Hopefully, you realized that. I was just sticking with that pen color. Then f of 3. 3 minus 1 is 2-- square root of 2. So the function evaluated at 3 is the square root of 2. So this is going to be 2 times the square root of 2. Then the function evaluated at 4. When you evaluate it at 4, you get the square root of 3. So this is going to be 2 times the square root of 3. And then you get 2 times the square root of 4-- 5 minus 1 2 times the square root of 4 is just four. And then finally, you get f of 6 is square root of 6 minus 1, is the square root of 5. And I think we're now ready to evaluate. So let me get my handy TI-85 out and calculate this. So it's going to be-- well I'm just going to calculate-- well, I'll just multiply. So 0.5 times open parentheses-- well, it's a 0. I'll just write it, just so you know what I'm doing. 0 plus 2 plus-- whoops. Lost my calculator. Plus 2 times the square root of 2 plus 2 times the square root of 3 plus 4-- I'm almost done-- plus the square root of 5-- so let me write that-- gives me-- and I'll just round it-- 7.26. So the area is approximately equal to 7.26 under the curve y is equal to the square root of x minus 1 between x equals 1 and x equals 6. And we did this using trapezoids."
    },
    {
        "Q": "On the third example, at about 7:25, Sal says that the length marked 2 and the segment parallel to it are equal. He says both are length 2. He says \"We know that these are both length 2 [because] these are all 90 degree angles..\" -- I don't understand how that lets us assume they are equal.",
        "A": "Actually, you are right. He can t assume they are equal. BUT, the reason they wouldn t be equal is because the length of either the vertical purple side or the white side (or both) would have been changed. In that case, though, the green side would have changed exactly in the same amount that the other two (white an vertical purple) would have changed, except that if they got shorter, it would have gotten longer, and vice versa.",
        "video_name": "vWXMDIazHjA",
        "timestamps": [
            445
        ],
        "3min_transcript": "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 I am going to shift and put it right over there. Then let me keep using different colors, and then this side right over here I am going to shift it and put it right up here. Then finally Iam going to have this side right over here, I can shift it and put it right over there and I think you see what is going on right now. Now all of these sides combined are going to be the same as this side kind of building, even you know this thing was not a rectangle,its its perimeter is going to be a little bit interesting. All we have to think about is this 2 right over here, now lets think about all of these sides that is going up and down. So this side i can shift it all the way to the right and go right over here. Let me make it clear all inside goes all the way to the end, right that it is the exact same all insde. Now this white side I can shift all the way to the right over there, then this green side I can shift right over there and then I have, and then I can shift, and then i can shift this. so I have not, I have not done anything yet, let me be clear I have not done anything yet with that and that I have not shift them over and let me take this side right over here and shift it over. So let me take this entire thing and shift over there and shift it over there. So before I count these two pieces right over here and we know that each have length 2 this 90 degrees angle, so this has link to and this has link to. Before I count these two pieces, I shifted everything else so I was able to form a rectangle. So at least counting everything else I have 7 plus 6, so lets see 7 plus 6 all of these combined are also going to be 7, plus 7, and all of these characters combined are all also going to be 6, plus 6, and then finally I have this 2, right here that I have not counted before, this 2, plus this 2, plus this 2. And then we have our perimeter, so what is this giving us,"
    },
    {
        "Q": "At 3:02 when Sal mentions altitude, what does he mean exactly?",
        "A": "Altitude is a geographic term used to describe the height of land.",
        "video_name": "vWXMDIazHjA",
        "timestamps": [
            182
        ],
        "3min_transcript": "it is going to be equal to the perimeter of the 5 triangles is equal to perimeter of 5 outer triangles. Just call them 5 triangles like this minus their basis, right, if i take the perimeter of all of these sides If i added up the part that should not be part of the perimeter of the star would be this part,that part, that part,that part, that part and that part. those are not the part, those are not the part of the perimeter of the star so should be the perimeter of the 5 triangles minus the links of their bases links of their 5 bases. So what is the perimeter of the 5 triangles? well, the perimeter of each of them is 30, perimeter of 5 of them is going to be 5 times 30 which is 150, now we want to subtract out the links of their 5 bases right over here. So this inner pentagon has a perimeter 50, that is the sum of the 5 bases. So that right over here is 50, so the perimeter of the star is going to be 150 minus 50, or or 100. All we need is to get the perimeter of all triangles, subtracted out these bases which was the perimeter of the inner pentagon and we are done. Now lets do the next problem. What is the area of this this quadrilateral, something that has 4 sides of ABCD? And this is a little bit we have not seen a figure quite like this just yet, it on the right hand side looks like a rectangle, and on the left hand side looks like a triangle and this is actually trapezoid, but we can actually as you could imagine the way we figure out the area of several triangles splitting it up into pieces we can recognise. And the most obvious thing to do here is started A and just drop a rock 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."
    },
    {
        "Q": "Sal @ 18:00 c2= 1/3 (x2 - 2x1) you forgot the 2 from the equation above. love you Sal. you are the greatest.",
        "A": "Sal corrected this error at the end of the video.",
        "video_name": "Qm_OS-8COwU",
        "timestamps": [
            1080
        ],
        "3min_transcript": "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. this first equation that says c1 plus 0 is equal to x1, so c1 is equal to x1. So that one just gets us there. So c1 is equal to x1. So you give me any point in R2-- these are just two real numbers-- and I can just perform this operation, and I'll tell you what weights to apply to a and b to get to If you say, OK, what combination of a and b can get me to the point-- let's say I want to get to the point-- let me go back up here. Oh, it's way up there. Let's say I'm looking to get to the point 2, 2. So x1 is 2. Let me write it down here. Say I'm trying to get to the point the vector 2, 2. What combinations of a and b can be there? Well, I know that c1 is equal to x1, so that's equal to 2, and c2 is equal to 1/3 times 2 minus 2."
    },
    {
        "Q": "At 2:21, I noticed the following: 3a+ -2c. This is a bit of a silly question but when I was in middle school we were taught to write the above as 3a+(-2c) as a parenthesis was thought to be needed between the +,- signs. Is the way I was taught to write it wrong/redundant?",
        "A": "Either putting parenthesis around the negative number or writing it in the form of 3a-2c is preferred for clarity. In school students are often taught to only do one operation at a time in each math step. However you may choose to combine multiple operations in a single step for speed purposes. Not following these type of conventions just increases the chance of confusion and mistakes but does not invalidate the math. To me it is similar to the difference between formal and informal writing.",
        "video_name": "Qm_OS-8COwU",
        "timestamps": [
            141
        ],
        "3min_transcript": "One term you are going to hear a lot of in these videos, and in linear algebra in general, is the idea of a linear combination. And all a linear combination of vectors are, they're just a Let me show you what that means. So let's say I have a couple of vectors, v1, v2, and it goes all the way to vn. And they're all in, you know, it can be in R2 or Rn. Let's say that they're all in Rn. They're in some dimension of real space, I guess you could call it, but the idea is fairly simple. A linear combination of these vectors means you just add up the vectors. It's some combination of a sum of the vectors, so v1 plus v2 plus all the way to vn, but you scale them by arbitrary constants. So you scale them by c1, c2, all the way to cn, where member of the real numbers. That's all a linear combination is. Let me show you a concrete example of linear Let me make the vector. Let me define the vector a to be equal to-- and these are all bolded. These purple, these are all bolded, just because those are vectors, but sometimes it's kind of onerous to keep bolding things. So let's just say I define the vector a to be equal to 1, 2. And I define the vector b to be equal to 0, 3. What is the linear combination of a and b? Well, it could be any constant times a plus any constant times b. So it could be 0 times a plus-- well, it could be 0 times a plus 0 times b, which, of course, would be what? That would be 0 times 0, that would be 0, 0. That would be the 0 vector, but this is a completely valid linear combination. big bold 0 like that. I could do 3 times a. I'm just picking these numbers at random. 3 times a plus-- let me do a negative number just for fun. So I'm going to do plus minus 2 times b. What is that equal to? Let's figure it out. Let me write it out. It's 3 minus 2 times 0, so minus 0, and it's 3 times 2 is 6. 6 minus 2 times 3, so minus 6, so it's the vector 3, 0. This is a linear combination of a and b. I can keep putting in a bunch of random real numbers here and here, and I'll just get a bunch of different linear combinations of my vectors a and b. If I had a third vector here, if I had vector c, and maybe"
    },
    {
        "Q": "At 3:37, he says that (if it's continuous) Psi xy = Psi yx . Is there a video where he proves that?",
        "A": "It s proven in all Calc III texts, but most likely in the section that he talks about partial derivatives, it s proven.",
        "video_name": "a7wYAtMjORQ",
        "timestamps": [
            217
        ],
        "3min_transcript": "So that's one situation to consider. What happens when we take the partial, with respect to x, and then y? So with respect to x, you hold y constant to get just the partial, with respect to x. Ignore the y there. And then you hold the x constant, and you take the partial, with respect to y. So what's the difference between that and if we were to switch the order? So what happens if we were to-- I'll do it in a different color-- if we had psi, and we were to take the partial, with respect to y, first, and then we were to take the partial, with respect to x? So just the notation, just so you're comfortable with it, that would be-- so partial x, partial y. And this is the operator. And it might be a little confusing that here, between these two notations, even though they're the same thing, the order is mixed. That's just because it's just a different way of This says, OK, partial first, with respect to x, then y. This views it more as the operator, so we took the partial of x first, and then we took y, like you're But anyway, so this can also be written as the partial of y, with respect to x-- sorry, the partial of y, and then we took the partial of that with respect to x. Now, I'm going to tell you right now, that if each of the first partials are continuous-- and most of the functions we've dealt with in a normal domain, as long as there aren't any discontinuities, or holes, or something strange in the function definition, they usually are continuous. And especially in a first-year calculus or differential course, we're probably going to be dealing with continuous functions in soon. our domain. If both of these functions are continuous, if both of the first partials are continuous, then these two are going to be equal to each other. So psi of xy is going to be equal to psi of yx. Now, we can use this knowledge, which is the chain knowledge to now solve a certain class of differential equations, first order differential equations, called exact equations. And what does an exact equation look like? An exact equation looks like this. The color picking's the hard part. So let's say this is my differential equation. I have some function of x and y. So I don't know, it could be x squared times cosine of y or something. I don't know, it could be any function of x and y. Plus some function of x and y, we'll call that n, times dy, dx is equal to 0. This is-- well, I don't know if it's an exact equation yet, but if you saw something of this form, your first impulse should be, oh-- well, actually, your very first impulse is, is this separable? And you should try to play around with the algebra a little bit to see if it's separable, because that's always the most straightforward way. If it's not separable, but you can still put it in this form,"
    },
    {
        "Q": "At 3:00, where are the sample sets s1, s2, etc coming from. Sal is just picking out numbers without describing the process.",
        "A": "The sets s1, s2 and s3 are outcomes that could be obtained by throwing the unfair dice defined by the distribution in yellow, but they are just examples and could be totally different.",
        "video_name": "JNm3M9cqWyc",
        "timestamps": [
            180
        ],
        "3min_transcript": "And let's say it's very likely to get a 6 like that. So that's my probability distribution function. If I were to draw a mean-- this the symmetric, so maybe the mean would be something like that. The mean would be halfway. So that would be my mean right there. The standard deviation maybe would look-- it would be that far and that far above and below the mean. But that's my discrete probability distribution function. Now what I'm going to do here, instead of just taking samples of this random variable that's described by this probability distribution function, I'm going to take samples of it. But I'm going to average the samples and then look at those samples and see the frequency of the averages that I get. And when I say average, I mean the mean. Let me define something. Let's say my sample size-- and I could put any number here. But let's say first off we try a sample size of n is equal to 4. And what that means is I'm going to take four samples from this. So let's say the first time I take four samples-- Let's say I get another 1. And let's say I get a 3. And I get a 6. So that right there is my first sample of sample size 4. I know the terminology can get confusing. Because this is the sample that's made up of four samples. But then when we talk about the sample mean and the sampling distribution of the sample mean, which we're going to talk more and more about over the next few videos, normally the sample refers to the set of samples from your distribution. And the sample size tells you how many you actually took from your distribution. But the terminology can be very confusing, because you could easily view one of these as a sample. But we're taking four samples from here. We have a sample size of four. And what I'm going to do is I'm going to average them. So let's say the mean-- I want to be very careful when I say average. The mean of this first sample of size 4 is what? 1 plus 1 is 2. 2 plus 3 is 5. 5 plus 6 is 11. 11 divided by 4 is 2.75. Let me do another one. My second sample of size 4, let's say that I get a 3, a 4. Let's say I get another 3. And let's say I get a 1. I just didn't happen to get a 6 that time. And notice I can't get a 2 or a 5. It's impossible for this distribution. The chance of getting a 2 or 5 is 0. So I can't have any 2s or 5s over here. So for the second sample of sample size 4, my second sample mean is going to be 3 plus 4 is 7. 7 plus 3 is 10 plus 1 is 11. 11 divided by 4, once again, is 2.75. Let me do one more, because I really want to make it clear what we're doing here. Actually, we're going to do a gazillion more. But let me just do one more in detail. So let's say my third sample of sample size 4--"
    },
    {
        "Q": "At 2:47, he starts using numbers like two and eight that didn't fit with what he was doing a minute before. Are those numbers suppose to be there to help solve the equation or just random numbers he pulled out?",
        "A": "They are just numbers he chose to plug in to the equation so we could see that the property was true.",
        "video_name": "PupNgv49_WY",
        "timestamps": [
            167
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 3:07 Why did he multiply instead of adding and where did he get the 16 to add to the 240?",
        "A": "He multiplied because when you are adding two logarithms of the same base (in this video, it was base 2) you can rewrite the logarithm as the product of the two numbers that are on the inside. log(8) + log(32) is the same thing as log(8 *32) 8*32 = 256, so we get log(8*32) = log(256) (I didn t write in the base on the logarithm, which would have been base 2.)",
        "video_name": "PupNgv49_WY",
        "timestamps": [
            187
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 2:18 and on, what criteria is used to choose the limiting numbers 1, 2, 3/2, 5/3, 8/5 etc.?",
        "A": "Its fibonacci. 1,1,2,3,5,8,13,21 1/1 2/1 3/2 and so on. Its late, but someone will probably like this.",
        "video_name": "lOIP_Z_-0Hs",
        "timestamps": [
            138
        ],
        "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": "At 4:41, in the third line, Sal writes f(x+h) without the denominator of h that it had in line 2. Why isn't this a mistake?\nIf I did the same thing with numbers, it would be as if I rewrote (3 + 5)/2 as 3 + 5/2. That IS a mistake. What's different here?",
        "A": "He factored it out, which is not a mistake. This would be like doing the following: (6+10)/5 = 2 \u00e2\u0088\u0099 [(3+5)/5] This is valid because: (ab)/c = a \u00e2\u0088\u0099 (b/c)",
        "video_name": "L5ErlC0COxI",
        "timestamps": [
            281
        ],
        "3min_transcript": "I just added and subtracted the same thing, but now this thing can be manipulated in interesting algebraic ways to get us to what we all love about the product rule. And at any point you get inspired, I encourage you to pause this video. Well to keep going, let's just keep exploring this expression. So all of this is going to be equal to, it's all going to be equal to the limit as H approaches zero. So the first thing I'm gonna do is I'm gonna look at, I'm gonna look at this part, this part of the expression. And in particular, let's see, I am going to factor out an F of X plus H. So if you factor out an F of X plus H, this part right over here is going to be F of X plus H, F of X plus H, times you're going to be left with G of X plus H. G of X plus H, that's that there, minus G of X, minus G of X, oops, I forgot the parentheses. Oops, it's a different color. I got a new software program and it's making it hard for me to change colors. My apologies, this is not a straightforward proof and the least I could do is change colors more smoothly. Alright, (laughing) G of X plus H minus G of X, that's that one right over there, and then all of that over this H. All of that over H. So that's this part here and then this part over here this part over here, and actually it's still over H, so let me actually circle it like this. So this part over here I can write as. actually here let me, let me factor out a G of X here. So plus G of X plus G of X times this F of X plus H. Times F of X plus H minus this F of X. Minus that F of X. All of that over H. All of that over H. Now we know from our limit properties, the limit of all of this business, well that's just going to be the same thing as the limit of this as H approaches zero plus the limit of this as H approaches zero. And then the limit of the product is going to be the same thing as the product of the limits. So if I used both of those limit properties, I can rewrite this whole thing as the limit, let me give myself some real estate, the limit as H approaches zero of F of X plus H,"
    },
    {
        "Q": "So if we can claim that the the limit as h\u00e2\u0086\u00920 of f(x+h) is f(x), as was stated in the video at 7:30, Why can't you evaluate it as f(x+h)g(x+h)-f(x)g(x)/h = f(x)g(x+h)-f(x)g(x)/h = f(x)((g(x+h)-g(x))/h), which would be equal to f(x)g'(x). This result is clearly wrong, but I can't see where exactly I've made a mistake.",
        "A": "You need to put the entirety of the first expression in parentheses as it all must be divided by h",
        "video_name": "L5ErlC0COxI",
        "timestamps": [
            450
        ],
        "3min_transcript": "actually here let me, let me factor out a G of X here. So plus G of X plus G of X times this F of X plus H. Times F of X plus H minus this F of X. Minus that F of X. All of that over H. All of that over H. Now we know from our limit properties, the limit of all of this business, well that's just going to be the same thing as the limit of this as H approaches zero plus the limit of this as H approaches zero. And then the limit of the product is going to be the same thing as the product of the limits. So if I used both of those limit properties, I can rewrite this whole thing as the limit, let me give myself some real estate, the limit as H approaches zero of F of X plus H, times the limit as H approaches zero, of all of this business, G of X plus H minus G of X, minus G of X, all of that over H, I think you might see where this is going. Very exciting. Plus, plus the limit, let me write that a little bit more clearly. Plus the limit as H approaches zero of G of X, our nice brown colored G of X, times, now that we have our product here, the limit, the limit as H approaches zero of F of X plus H. Of F of X plus H minus F of X, all of that, all of that over H. And let me put the parentheses where they're appropriate. So that, that, that, that. And all I did here, the limit, the limit of this sum, that's gonna be the sum of the limits, that's gonna be the limit of this plus the limit of that, and then the limit of the products is gonna be the same thing as the product of the limits. So I just used those limit properties here. But now let's evaluate them. What's the limit, and I'll do them in different colors, what's this thing right over here? The limit is H approaches zero of F of X plus H. Well that's just going to be F of X. Now, this is the exciting part, what is this? The limit is H approaches zero of G of X plus H minus G of X over H. that's the definition of our derivative. That's the derivative of G. So this is going to be, this is going to be the derivative of G of X, which is going to be G prime of X."
    },
    {
        "Q": "At 6:00 , how come there are no restrictions on x-3y? How can we tell if there will be restrictions on a certain part or not?",
        "A": "If a factor disappears from the denominator, then its restriction needs to be stated. Otherwise there d be no way of knowing that it had ever existed in the original problem. If a factor is still left in the denominator of the simplified fraction, then its restriction does not need to be stated because it is still visible and we all know that division by zero is not allowed.",
        "video_name": "e7vA_S7abSY",
        "timestamps": [
            360
        ],
        "3min_transcript": "I'm gonna get four y squared and if I add them, I get four y? It looks like two y would do the trick. So it seems like we can rewrite the numerator. This is going to be. So let me draw a little line here to make it clear that this is, this is going to be equal to five times x plus two y times, I could say this is x plus two y squared or I could just say x plus two y times x plus two y. Once again, two y times two y is four y squared. Two y plus two y is four y. And that's all going to be over, that is all going to be over x minus three y times x plus two y. And so now, I have a common factor, x plus two y in both the numerator and the denominator, well that's just going to be one if we assume that x plus two y does not equal zero. And that's actually an important constraint because once we cancel this out, you lose that information. If you want this to be algebraically equivalent, we could say that x plus two y cannot be equal to zero or another way you could say it is that x cannot be equal to, cannot be equal to negative two y. I just subtracted two y from both sides there. And so what you're left with, and we can redistribute this five if we wanna write it out in expanded form. We could rewrite it as, the numerator would be five x. Let me write it over here. Five x plus 10 y. And the denominator is x minus three y. But once again, if we want it to be algebraically equivalent, we would have to say x cannot be equal to, x cannot be equal to negative two y. And now this is algebraically equivalent and you can argue that it's a little bit simpler."
    },
    {
        "Q": "at 0:45 Sal says that 5^2/3 is equal to 25/9. Shouldn't it be 5^2/3^2 or (5/3)^2 is equal to 25/9?",
        "A": "Well, he didn t say that, actually. What he said was that 5/3 squared was equal to 25/9 What he wrote as he segued to the next example was subtly different, and it is true that he should have put his Segue into Park long enough to put the 5/3 into parentheses to emphasize that he was squaring the whole fraction.",
        "video_name": "bIFdW0NZ9W4",
        "timestamps": [
            45
        ],
        "3min_transcript": "Fractional exponents can be a little daunting at first, so it never hurts to do as many examples as possible. So let's do a few. What if we had 25/9, and we wanted to raise it to the 1/2 power? So we're essentially just saying, well, what is the principal square root of 25/9? So what number times itself is going to be 25/9? Well, we know 5 times 5 is 25, and 3 times 3 is 9. So why don't we just go with 5/3? Because notice, if you have 5/3 times 5/3, that is going to be 25/9. Or another way of saying this, that 5/3 squared is equal to 25/9. So 25/9 to the 1/2 is going to be equal to 5/3. Now let's escalate things a little bit. Let's take a really hairy one. Let's raise 81/256 to the negative 1/4 power. So what's going on here? This negative-- the first thing I always like to do is I want to get rid of this negative in the exponent. So let me just take the reciprocal of this and raise it to the positive. So I could just say that this is equal to 256/81 to the 1/4 power. And so now I can say, well, what number times itself times itself times itself is going to be equal to 256, and what number times itself times itself times itself-- did I say that four times? Well, what number, if I take four of them and multiply, do I get 81? And one way to think about it, this is going to be the same thing-- and we'll talk about this in more depth later on when we talk about exponent properties. to the 1/4 over 81 to the 1/4. You, in fact, saw it over here. This over here was the same thing as the square root of 25 over the square root of 9. Or 25 to the 1/2 over 9 to the 1/2. So we're just doing that over here. So one, we still have to think about what number this is. And this is a little bit of, there's no easy way to do this. You kind of have to just play around a little bit to come up with it. But 4 might jump out at you if you recognize that 16 times 16 is 256. We know that 4 to the fourth power, or you're about to know this, is 4 times 4 times 4 times 4. And 4 times 4 is 16, times 4 is 64, times 4 is equal to 256. So 4 to the fourth is 256, or we could say 4 is equal to 256 to the 1/4 power. Fair enough? Now what about 81? Well, 3 might jump out at you. We know that 3 to the fourth power is equal to 3 times 3 times 3 times 3, which is equal to 81."
    },
    {
        "Q": "at 1:25 isn't that the transitive property of equality?",
        "A": "It can be taken that way. If boys=tall and Bill=tall, that s like saying x=tall and y=tall. Therefore, you can substitute x for tall and say x=y. If you write the equation as boys =tall and tall=bill, then it s like saying x=y and y=z, therefore x=z, that is more the transitive property. This really depends on your perspective, and as I debated today, it can be taken either way.",
        "video_name": "GluohfOedQE",
        "timestamps": [
            85
        ],
        "3min_transcript": "All right, we're doing the California Standards released questions in geometry now. And here's the first question. It says, which of the following best describes deductive reasoning? And I'm not a huge fan of when they ask essentially definitional questions in math class. But we'll do it and hopefully it will help you understand what deductive reasoning is. Although, I do think that deductive reasoning itself is probably more natural than the definition they'll give here. Well, actually before I even look at the definitions, let me tell you what it is. And then we can see which of these definitions matches it. Deductive reasoning is, if I give you a bunch of statements and then from those statements you deduce, or you come to some conclusion that you know must be true. Like if, I said that all boys are tall. If I told you all boys are tall. And if I told you that Bill is a boy. Bill is a boy. So if you say, OK, if these two statements are true, what can you deduce? Well I say well, Bill is a boy and all boys are tall. Then Bill must be tall. You deduced this last statement from these two other statements that you knew were true. And this one has to be true if those two are true. So Bill must be tall. That is deductive reasoning. Deductive reasoning. Bill must be tall, not Bill must be deductive. So anyway, you have some statements and you deduce other statements that must be true given those. And you often hear another type of reasoning, inductive reasoning. And that's when you're given a couple of examples and you generalize. Well, I don't want to get too complicated here, because this is a question on deductive reasoning. But generalizations often aren't a good thing. But if you see a couple of examples and you see a pattern of broader generalization. That's inductive reasoning. But that's not what they're asking us about this. Let's see if we can find the definition of deductive reasoning in the California Standards language. Use logic to draw conclusions based on accepted statements. Yeah, well actually that sounds about right. That's what we did here. We used logic to draw conclusions based on accepted statements, which were those two. So I'm going to go with A, so far. Accepting the meaning of a term without definition. Well, I don't even know how one can do that. How do you accept the meaning of something without it having being defined? Let's see, so it's not B. I don't think anything is really B. C, defining mathematical terms to correspond with physical objects. No, that's not really anything related to deductive reasoning either. D, inferring a general truth by examining a number of specific examples."
    },
    {
        "Q": "At 0:57, why did Sal draw those lines over the z? Please click the time to see what I'm talking 'bout",
        "A": "Some people draw a line in their z s because it helps other people tell them apart from 2 s, since sloppy z s and 2 s can look like each other.",
        "video_name": "cTveNRjWQYo",
        "timestamps": [
            57
        ],
        "3min_transcript": "Beatrice is ordering pizzas for a party. Each pizza is cut into 12 pieces. Beatrice wants to have enough pizza so that each person can have 4 pieces. And 16 people will be at the party. A lot of information in this problem. If Beatrice buys 7 pizzas, how many extra pieces will there be after each person eats 4 pieces? So to do this, let's think about how many total pieces she will have. And then let's think about how many total pieces will be eaten by the 16 people. And then we'll figure how much that she had extra. So first, let's think about how many total pieces. And I encourage you to pause the video and try to figure it out yourself. She ordered 7 pizzas. How many total pieces are there going to be? Well, you have 7 pizzas. 7-- that's the number of pizzas. you multiply by 12 to get the total number of pieces. So this is number of pieces per pizza. And 7 times 12 is equal to 84 total pieces. So that's how many pieces she's got. Now let's think about how many get eaten by the 16 people. So you have 16 people. 16-- that's the number of people. And this little symbol, that's shorthand for number. 16 people, and they each eat 4 pieces. So 16 people times 4 pieces per person. So number of pieces per person. So that's 64-- not 84-- 64 pieces are eaten by the 16 people. So how much does she have left over? Well, if you start with 84 pieces and 64 get eaten, 84 minus 64 is 20. We deserve a drum roll now. You have 20 extra pieces after everyone eats their share."
    },
    {
        "Q": "3:50 wouldn't length 'a' equal length 'b' then?",
        "A": "You can t make that assumption. As you shift to a different point on the circle, the length of a and b will vary. If the point was very close to the x-axis, then b would be very short compared to a . If the point was close to the y-axis, then b would be very long compared to a . Hope this makes sense.",
        "video_name": "1m9p9iubMLU",
        "timestamps": [
            230
        ],
        "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": "my doubts are in the right triangle in the video.At 3:46, why is the length of the opp side b? and at 4:08, why is the adj side equal to a? what does it have to do with the coordinates of the intersection point?",
        "A": "(a,b) in the video is the x,y coordinate of the line from the origin to the perimeter point on the unit circle.",
        "video_name": "1m9p9iubMLU",
        "timestamps": [
            226,
            248
        ],
        "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 1:47 .... how is he sure that \" arctan( tan ( g^-1(x) - 3pi/2 ) ) \" will give out ( g^-1(x) - 3pi/2 ) ?\ni mean .... how is he sure , that ( g^-1(x) - 3pi/2 ) is restricted to the domain of tan ?",
        "A": "Tangent isn t actually a completely invertible function since it outputs the same value from different inputs. So the fact that the question asks for the inverse function implies it s looking for the simplest inverse function. Since it s just looking for the simplest inverse function, we consider arctangent as if it perfectly cancels tangent.",
        "video_name": "QGfdhqbilY8",
        "timestamps": [
            107
        ],
        "3min_transcript": "Voiceover:We're told given g of x is equal to ten of x minus three pi over two plus six, find the g inverse of x. They want us to type that in here and then they also want us to figure out what is the domain of g inverse, the domain of g inverse of x. I've got my little scratch pad here to try to work that through. Let's figure out what g inverse of x is. This is g of x, so g inverse of x. Essentially, let me just read this is g of x right over here, g of x is equal to tangent of x minus three pi over two plus six. G inverse of x, I essentially can swap the \u2026 I can replace the x with the g inverse of x and replace to g of x with an x and then solve for g inverse of x. I could write that x is equal to tangent of g inverse of x minus three pi over two plus six. I actually encourage you to pause this video and try to work through this out or work it out on your own. Let's subtract six from both sides to at least get rid of this six here so I'm left with x minus six is equal to the tangent of g inverse of x minus three pi over two. Now let's take the inverse tangent of both sides of this equation so the inverse tangent on the left hand side is the inverse tangent of x minus six and on the right hand side the inverse tangent of tangent. If we restrict the domain in the proper way and we'll talk about that in a little bit is just going to be what the input into the tangent function is. If you restrict the domain in the right way, inverse tangent of the tangent of something, Once again if we restrict the domain, if we restrict what the possible values of theta are in the right way. Let's just assume that we're doing that and so the inverse tangent of the tan, of this is going to be just this stuff right over here. It's just going to be that, it's going to be g inverse of x minus three pi over two. Now we're in the [home] stretch to solve for g inverse of x we could just add three pi over two to both sides so we get and actually let me just swap both sides. We get g inverse of x is equal to the inverse tangent of x minus six and then we're adding three pi over two to both sides so this side is now on this side so plus three pi over two. Let me actually type that and I'm going to see if I can remember it"
    },
    {
        "Q": "At 0:04 it says two sets of parallel sides. Isn't that the same thing as two sets of parallel lines. Sides on a parallelogram are lines. Even if the question says so, you can also say sets of sides. Not to be critical, just thought I'd point it out.",
        "A": "Yes, two sets of parallel sides are the same things as lines. It s just another way of saying it. Just like how you might call a sphere, a ball. There might be many different ways to say that, like two pairs of parallel lines.",
        "video_name": "1pHhMX0_4Bw",
        "timestamps": [
            4
        ],
        "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 3:39, why is it that we minus the dy/dx from the other side? Because it is being multiplied by the (2x-2y) on the left wouldn't we have to divide the other side by dy/dx?",
        "A": "... but then we would have an answer that was for (dy/dx)^-1 = dx/dy ... which is not what we want!",
        "video_name": "9uxvm-USEYE",
        "timestamps": [
            219
        ],
        "3min_transcript": "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 We are left with 1 minus 2x plus 2y. So let me write it that way. Or we could write this as-- so negative, negative 2y is just a positive 2y. And then we have minus 2x. And then we add that 1, plus 1. And now to solve for dy dx, we just have to divide both sides by 2y minus 2x minus 1. And we are left with-- we deserve a little bit of a drum roll at this point. As you can see, the hardest part was really the algebra to solve for dy dx. We get the derivative of y with respect to x is equal to 2y minus 2x plus 1 over 2y minus 2x minus 1."
    },
    {
        "Q": "At 2:17, Sal distributes the (2x-2y) to the -(dy/dx). I understand that the answer originally is (-2x+2y)(dy/dx), and that he's just rearranging things when he writes it as (2y-2x)(dy/dx). But I don't understand how he gets (-2x+2y)(dy/dx) in the first place! I would have written (2x-2y)(-(dy/dx)). Is he simply applying the negative sign of the (dy/dx) to the (2x-2y)? That's what it looks like, but I don't understand why that is being done. Why can't the negative stay with the (dy/dx)?",
        "A": "He is multiplying (2x - 2y)(-(dy/dx)) by one, but by a special form of one which is (-1)(-1). Now he has (-1)(-1)(2x - 2y)(-(dy/dx)). Since we can multiply in any order let s shift things to: (-1)(2x -2y)(-1)(-(dy/dx)). If you multiply the first two terms together, and the 3rd and 4th term together you get: (2y - 2x)(dy/dx).",
        "video_name": "9uxvm-USEYE",
        "timestamps": [
            137
        ],
        "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 1:25ish. Angle BGC and DGC would also be \" Adjacent\", correct?",
        "A": "Close. Adjacent angles share a ray, but neither is interior to the other. Each of those angles would be interior to angle BGD. However, the angles you mention are adjacent to each other.",
        "video_name": "vAlazPPFlyY",
        "timestamps": [
            85
        ],
        "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": "At 7:27 what about \"x\" does not equal 3?\nx=3 will make the function undefined too right?\nHelp me! I'm a little bit confused.",
        "A": "Values that make both the numerator and denominator zero are undefined points (like x = -3 in this example). Values that make only the denominator zero are vertical asymptotes (like x = 3 in this example). Yes, both points are undefined for this function, but in different ways.",
        "video_name": "P0ZgqB44Do4",
        "timestamps": [
            447
        ],
        "3min_transcript": "but by itself it does not make a vertical asymptote. Let's just think about this denominator right over here so we can factor it out. Actually let's factor out the numerator and the denominator. We can rewrite this as F of X is equal to the numerator is clearly every term is divisible by three so let's factor out three. It's going to be three times X squared minus six X minus 27. All of that over the denominator each term is divisible by six. Six times X squared minus 9 and let's see if we can factor the numerators and denominators out further. This is going to be F of X is equal to three times let's see, two numbers, their product is negative 27, their sum is negative six. Negative nine and three seem to work. You could have X minus nine times X plus three. Just factor the numerator over the denominator. This would be X minus three times X plus three. When does the denominator equal zero? The denominator equals zero when X is equal to positive three or X is equal to negative three. Now I encourage you to pause this video for a second. Think about are both of these vertical asymptotes? Well you might realize that the numerator also equals zero when X is equal to negative three. What we can do is actually simplify this a little bit and then it becomes a little bit clear where our vertical asymptotes are. We could say that F of X, we could essentially divide the numerator and denominator by X plus three and we just have to key, if we want the function to be identical, we have to keep the [caveat] that the function itself is not defined when X is equal to negative three. We have to remember that but that will simplify the expression. This exact same function is going to be if we divide the numerator and denominator by X plus three, it's going to be three times X minus nine over six times X minus three for X does not equal negative three. Notice, this is an identical definition to our original function and I have to put this qualifier right over here for X does not equal negative three because our original function is undefined at X equals negative three. X equals negative three is not a part of the domain of our original function. If we take X plus three out of the numerator and the denominator, we have to remember that. If we just put this right over here, this wouldn't be the same function because this without the qualifier is defined for X equals negative three"
    },
    {
        "Q": "5:34 So is codomain the same as range?",
        "A": "No, range is a subspace (or subset) of codomain. Range is the specific mapping from the elements in set X to elements in set Y, where codomain is every element in set Y. To concretize: Let f map X --> Y Where X = { 0, 1} Where Y = { 2, 3, 4, 5, 6...}. Let f(0) = 2 & f(1) = 3. Clearly, the range of X = [ 1, 2} where as the codomain of X = { 1, 2 ,3, 4, 5, 6..}.",
        "video_name": "BQMyeQOLvpg",
        "timestamps": [
            334
        ],
        "3min_transcript": "This statement you've probably never seen before, but I like it because it shows the mapping or the association more, while this association I think that you're putting an x into a little meat grinder or some machine that's going to ground up the x or square the x, or do whatever it needs to do to the x. This notation to me implies the actual mapping. You give me an x, and I'm going to associate another number in real numbers called x squared. So it's going to be just another point. And just as a little bit of terminology, and I think you've seen this terminology before, the set that you are mapping from is called the domain and it's part of the function definition. I, as the function creator, have to tell you that every valid input here has to be a set of real numbers. Now the set that I'm mapping to is called the codomain. Sal, when I learned all of this function stuff in algebra II or whenever you first learned it, we never used this codomain word. We have this idea of range, I learned the word range when I was in 9th or 10th grade. How does this codomain relate to range? And it's a very subtle notation. So the codomain is a set that you're mapping to, and in this example this is the codomain. In this example, the real numbers are the domain and the codomain. So the question is how does the range relate to this? So the codomain is the set that can be possibly mapped to. You're not necessarily mapping to every point in the codomain. I'm just saying that this function is generally mapping from members of this set to that set. 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."
    },
    {
        "Q": "why is 0 not in the range? it is a real number and the product of any number and zero would be zero, so I don\u00c2\u00b4t get why Sal says in 13:36 that it is not a member of the range?!",
        "A": "Elements of the range are not numbers, but triples of numbers. So asking whether or not 0 is in the range makes no sense. Yes it is true that we can make the third coordinate 0, but can we make the first coordinate 5, the second coordinate 1, and the third coordinate 0 all at the same time? The answer is no.",
        "video_name": "BQMyeQOLvpg",
        "timestamps": [
            816
        ],
        "3min_transcript": "Let's take our h of-- let me use my other notation-- let's say that I said h, and I wanted to find the mapping from the point in R2, let's say the point 2 comma 3. And then my function tells me that this will map to the point in R3. I add the two terms, the 2 plus 3, so it's 5. I'd find the difference between x2 and x1-- so 3 minus 2 is 1-- and then I multiply the two, 6. So clearly this will be in the range, this is a member of the range. So for example the point 2, 3, which might be right there, will be mapped to the three dimensional point, it's kind but I think you get the idea, would be mapped to the three dimensional point 5, 1, 6. So this is definitely a member of the range. Now my question to you, if I have some point in R3, let's say that this is the point 5, 1, 0. Is this point a member of the range? It's definitely a member of the codomain, it's in R3. It's definitely in here, and this by definition is the codomain. But is this in our range? 5 has to be the sum of two numbers, the 1 has to be the difference of two numbers, and then the 0 would have to be the product of two numbers. And clearly we know 5 is the sum, and 1 is the difference, we're dealing with 2 and 3, and there's no way you can get the product of those numbers to be equal to 0. So the range would be the subset of all of these points in R3, so there'd be a ton of points that aren't in the range, and there'll be a smaller subset of R3 that is in the range. Now I want to introduce you to one more piece of terminology when it comes to functions. These functions up here, this function that mapped from points in R2 to R, so its codomain was R. This function up here is probably the most common function you see in mathematics, this is also mapping to R. These functions that map to R are called scalar value or real value, depending on how you want to think about it. But if they map to a one dimensional space, we call them a scalar valued function, or a real valued function."
    },
    {
        "Q": "What does he mean at 12:42?",
        "A": "The function h maps an ordered pair onto an ordered triple. He is observing that the result of the function is an ordered triple of real numbers, an element of the range R^3.",
        "video_name": "BQMyeQOLvpg",
        "timestamps": [
            762
        ],
        "3min_transcript": "And notice I'm going from a space that has two dimensions to a space it has three dimensions, or three But I can always associate some point with x1, x2 with some point in my R3 there. A slightly trickier question here is, what is the range? Can I always associate every point-- maybe this wasn't the best example because it's not simple enough -- but can I associate every point in R3-- so this is my codomain, my domain was R2, and my function goes from R2 to R3, so that's h. And so my range, as you could see, it's not like every coordinate you can express in this way in some way. Let me give you an example. Let's take our h of-- let me use my other notation-- let's say that I said h, and I wanted to find the mapping from the point in R2, let's say the point 2 comma 3. And then my function tells me that this will map to the point in R3. I add the two terms, the 2 plus 3, so it's 5. I'd find the difference between x2 and x1-- so 3 minus 2 is 1-- and then I multiply the two, 6. So clearly this will be in the range, this is a member of the range. So for example the point 2, 3, which might be right there, will be mapped to the three dimensional point, it's kind but I think you get the idea, would be mapped to the three dimensional point 5, 1, 6. So this is definitely a member of the range. Now my question to you, if I have some point in R3, let's say that this is the point 5, 1, 0. Is this point a member of the range? It's definitely a member of the codomain, it's in R3. It's definitely in here, and this by definition is the codomain. But is this in our range? 5 has to be the sum of two numbers, the 1 has to be the difference of two numbers, and then the 0 would have to be the product of two numbers. And clearly we know 5 is the sum, and 1 is the difference, we're dealing with 2 and 3, and there's no way you can get the product of those numbers to be equal to 0."
    },
    {
        "Q": "At 2:30, why does a number to the negative power a decimal?",
        "A": "When you have a negative exponent, you re essentially dividing the current number, like 5 squared, by the base, or 5. If you keep doing that, you get 1, 0.2, 0.04, and so on. I hope this helped you!",
        "video_name": "6phoVfGKKec",
        "timestamps": [
            150
        ],
        "3min_transcript": "Express 0.0000000003457 in scientific notation. So let's just remind ourselves what it means to be in scientific notation. Scientific notation will be some number times some power of 10 where this number right here-- let me write it this way. It's going to be greater than or equal to 1, and it's going to be less than 10. So over here, what we want to put here is what that leading number is going to be. And in general, you're going to look for the first non-zero digit. And this is the number that you're going to want to start off with. This is the only number you're going to want to put ahead of or I guess to the left of the decimal point. So we could write 3.457, and it's going to be multiplied by 10 to something. Now let's think about what we're going to have to multiply it by. To go from 3.457 to this very, very small number, to move the decimal to the left a bunch. You have to add a bunch of zeroes to the left of the 3. You have to keep moving the decimal over to the left. To do that, we're essentially making the number much much, much smaller. So we're not going to multiply it by a positive exponent of 10. We're going to multiply it times a negative exponent of 10. The equivalent is you're dividing by a positive exponent of 10. And so the best way to think about it, when you move an exponent one to the left, you're dividing by 10, which is equivalent to multiplying by 10 to the negative 1 power. Let me give you example here. So if I have 1 times 10 is clearly just equal to 10. 1 times 10 to the negative 1, that's equal to 1 times 1/10, which is equal to 1/10. 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?"
    },
    {
        "Q": "In 9:15 - he simplified 0.1 into 1/10, but it could have been into 10/100 as well... If you do it like that, you have root10 / root100 = root10 / 10\n\nSeems correct to me but has different answer.... why not do it this way?",
        "A": "You can do it either way, but depending on what the problem is, it might be easier to do it your way or to do it Sal s way. For instance, if you had 0.2 , then that would simplify to 1/5 Sal s way; but your way, it would simplify to 20/200 , which would turn into root20 / root200 , and that s a little more messy. So, sometimes your way is easier, and sometimes Sal s is. But you can do it whichever way you prefer; they both give correct answers.",
        "video_name": "BpBh8gvMifs",
        "timestamps": [
            555
        ],
        "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": "at 2:35 shouldn't 2/(square root of) 6 be the other way around",
        "A": "Sal had 2*2*2*3 inside his radical. 2*2=4, 2*3=6. So you have 4*6 inside a radical. The sqrt of 4 is a rational number ( a number that can be expressed as a fraction a/b where b doesn t = 0) so we can go ahead and work with it. We remove the sqrt of four from the radical, leaving us with 2 (the sqrt of 4) times the sqrt of 6.",
        "video_name": "BpBh8gvMifs",
        "timestamps": [
            155
        ],
        "3min_transcript": "square root of 2 times 2 times 2 times 3. That's the same thing as 24. Well, we see here, we have one perfect square right there. So we could rewrite this. This is the same thing as the square root of 2 times 2 times the square root of 2 times 3. Now this is clearly 2. This is the square root of 4. The square root of 4 is 2. And then this we can't simplify anymore. We don't see two numbers multiplied by itself here. So this is going to be times the square root of 6. Or we could even right this as the square root of 2 times the square root of 3. Now I said I would talk about whether things are rational or not. This is rational. This part A can be expressed as the ratio of 2 integers. This is rational. This is irrational. I'm not going to prove it in this video. But anything that is the product of irrational numbers. And the square root of any prime number is irrational. I'm not proving it here. This is the square root of 2 times the square root of 3. That's what the square root of 6 is. And that's what makes this irrational. I cannot express this as any type of fraction. I can't express this as some integer over some other integer like I did there. And I'm not proving it here. I'm just giving you a little bit of practice. And a quicker way to do this. You could say, hey, 4 goes into this. 4 is a perfect square. Let me take a 4 out. This is 4 times 6. The square root of 4 is 2, leave the 6 in, and you would have gotten the 2 square roots of 6. Which you will get the hang of it eventually, but I want to do it systematically first. 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."
    },
    {
        "Q": "What does corresponding angles mean? [8:16] and other parts of the video. Thanks!",
        "A": "Corresponding angles are created where a transversal crosses other (usually parallel) lines. The corresponding angles are the ones at the same location at each intersection.",
        "video_name": "TErJ-Yr67BI",
        "timestamps": [
            496
        ],
        "3min_transcript": "that could be parallel, if the alternate interior angles are And we see that they are. These two are kind of candidate alternate interior angles, and they are congruent. So AB must be parallel to CD. Actually, I'll just draw one arrow. AB is parallel to CD by alternate interior angles congruent of parallel lines. I'm just writing in some shorthand. Forgive the cryptic nature of it. I'm saying it out. And so we can then do the exact same-- we've just shown that these two sides are parallel. We could then do the exact same logic to show that these two sides are parallel. And I won't necessarily write it all out, but it's the exact same proof to show that these two. So first of all, we know that this angle Actually, let me write it out. So we know that angle AEC is congruent to angle DEB. They are vertical angles. And that was our reason up here, as well. And then we see the triangle AEC must be congruent to triangle DEB by side-angle-side. So then we have triangle AEC must be congruent to triangle DEB by SAS congruency. Then we know that corresponding angles must be congruent. So for example, angle CAE must be congruent to angle BDE. of congruent triangles. So CAE-- let me do this in a new color-- must be congruent to BDE. And now we have a transversal. The alternate interior angles are congruent. So the two lines that the transversal is intersecting must be parallel. So this must be parallel to that. So then we have AC must be parallel to be BD by alternate interior angles. And we're done. We've just proven that if the diagonals bisect each other, if we start that as a given, then we end at a point where we say, hey, the opposite sides of this quadrilateral must be parallel, or that ABCD is a parallelogram."
    },
    {
        "Q": "At 16:03, why are you multiplying 1/4sin(t) by 2sintcost? Wasn't it ...+sin2t?",
        "A": "At 15:23 Sal reminded you of the trigonometric identity: sin(2\u00ce\u00b8) = 2\u00c2\u00b7sin(\u00ce\u00b8)\u00c2\u00b7cos(\u00ce\u00b8). And even when he s writing it at 16:03, he clearly says this is just a trig identity .",
        "video_name": "IW4Reburjpc",
        "timestamps": [
            963
        ],
        "3min_transcript": "And we're going to evaluate this whole thing at t, so you get 1/2 sine squared of t minus 1/2 the sine of 0 squared, which is just 0, so that's just minus 0. So far, everything that we have written simplifies to-- let me multiply it all out. So I have 1/2-- let me just pick a good color-- 1/2t sine of t-- I'm just multiplying those out-- plus 1/4 sine of t sine of 2t. And then over here I have minus 1/2 sine squared t times I just took the minus cosine t and multiplied it through here and I got that. Now, this is a valid answer, but I suspect that we can simplify this more, maybe using some more trigonometric identities. And this guy right there looks ripe to simplify. And we know that the sine of 2t-- another trig identity you'll find in the inside cover of any of your books-- is 2 times the sine of t times the cosine of t. So if you substitute that there, what does our whole expression equal? You get this first term. Let me scroll down a little bit. You get 1/2t times the sine of t plus 1/4 sine of t times this thing in here, so times 2 sine of t cosine of t. And then finally I have this minus 1/2 sine squared t cosine of t. No one ever said this was going to be easy, but hopefully it's instructive on some level. At least it shows you that you didn't memorize your trig identities for nothing. So let me rewrite the whole thing, or let me just rewrite this part. So this is equal to 1/4. Now, I have-- well let me see, 1/4 times 2. 1/4 times 2 is 1/2. And then sine squared of t, right? This sine times this sine is sine squared of t cosine of t. And then this one over here is minus 1/2 sine squared of t cosine of t. And luckily for us, or lucky for us, these cancel out. And, of course, we had this guy out in the front. We had this 1/2t sine t out in front. Now, this guy cancels with this guy, and all we're left"
    },
    {
        "Q": "at 10:00 if we substitute sin\u00cf\u0084.cos\u00cf\u0084 equals to sin2\u00cf\u0084/2 . answer turns out to be different at end. isn't it ? Terms donot cancel out each other in end",
        "A": "You do get the same answer. Try again, and make sure that you evaluate it a 0 as well as a t. The integral of \u00c2\u00bdsin(2\u00cf\u0084) becomes -\u00c2\u00bccos(2\u00cf\u0084). That is evaluated at t and at 0, so we are left with [-cos(2(t))/4] - [-cos(2(0))/4] = \u00c2\u00bc - \u00c2\u00bccos(2t). The cos(2t) can be expanded into cos\u00c2\u00b2(t) - sin\u00c2\u00b2(t), so we are left with \u00c2\u00bc + \u00c2\u00bcsin\u00c2\u00b2(t) - \u00c2\u00bccos\u00c2\u00b2(t). All this must be multiplied by the -cos(t) out front, and you are left with \u00c2\u00bccos\u00c2\u00b3(t) - \u00c2\u00bccos(t)sin\u00c2\u00b2(t) - \u00c2\u00bccos(t). You will find that that will cancel with the other integral.",
        "video_name": "IW4Reburjpc",
        "timestamps": [
            600
        ],
        "3min_transcript": "to the cosine of tau, just the derivative of sine. Or we could write that du is equal to the cosine of tau d tau. We'll undo the substitution before we evaluate the endpoints here. But this was a little bit more of a conundrum. I don't know how to take the antiderivative of cosine squared of tau. It's not obvious what that is. So to do this, we're going to break out some more trigonometric identities. And in a video I just recorded, it might not be the last video in the playlist, I showed that the cosine squared of tau-- I'm just using tau as an example-- is equal to 1/2 And once again, this is just a trig identity that you'll find really in the inside cover of probably your calculus book. So we can make this substitution here, make this substitution right there, and then let's see what our integrals become. So the first one over here, let me just write it here. We get sine of t times the integral from 0 to t of this thing here. Let me just take the 1/2 out, to keep things simple. So I'll put the 1/2 out here. That's this 1/2. So 1 plus cosine of 2 tau and all of that is d tau. And then we have this integral right here, minus cosine of t times the integral from-- let me be very clear. This is tau is equal to 0 to tau is equal to t. And then this thing right here, I did some u subsitution. If u is equal to sine of t, then this becomes u. And we showed that du is equal to cosine-- sorry, u is equal to sine of tau. And then we showed that du is equal to cosine tau d tau, so this thing right here is equal to du. So it's u du, and let's see if we can do anything useful now. So this integral right here, the antiderivative of this is"
    },
    {
        "Q": "At 6:20, what does Sal mean by the solution satisfying the diff eq for all x's?",
        "A": "In this case, it basically means writing the equation in such as a way that you have no differentials remaining and no excess variables. If the differential was in the xy-plane, then x- and y-variables are the intend. In the xyz-space, only x-, y-, and z-variables. Suitable substitutions (like here with m) are allowed.",
        "video_name": "zid7J4EhZN8",
        "timestamps": [
            380
        ],
        "3min_transcript": "is equal to some coefficient times x, plus some other constant value. Well, in order for a constant value to be equal to a coefficient times x plus some other constant value the coefficient on X must be equal to zero. Another way to think about it is, this should be, you could rewrite the left-hand side here as zero x plus m. So you see, you kind of match the coefficients. So zero must be equal to 3m minus 2, and m is equal to 3b, m is equal to 3b minus 5. m is equal to 3b minus 5. So let's use that knowledge, that information, to solve for m and b. So we could use this first one. So 3m minus 2 must be equal to zero. So let's write that. 3m minus 2 is equal to zero, or 3m is equal to 2, or m is equal to 2/3. So we figured out what m is. because we know that, we know that m is equal to, is equal to 3b minus 5. m is 2/3, so we get 2/3 is equal to 3b minus 5. We could add 5 to both sides, which is the same thing as adding 15/3 to both sides. Is that, did I do that right? Yeah, adding 5 to both sides is the same thing as adding 15/3 to both sides. So let's do that. 15/3 plus 15/3, these cancel out. That's just 5 right over there. On the left-hand side we have 17/3 is equal to 3b, or if you divide both sides by 3 you get b is equal to 17, b is equal to 17/9, and we're done. We just found a particular solution The solution is y is equal to 2/3x plus 17/9. And I encourage you, after watching this video, to verify that this particular solution indeed does satisfy this differential equation for all x's. For all x's."
    },
    {
        "Q": "At 1:04 is altitude the proper term?",
        "A": "YES! According the the definition An altitude is the line segment which join vertex and its opposite sides (known as base) perpendicular to it. And interesting thing about it that it is the the shortest path to reach opposite side from the vertex. A STRAIGHT PATH IS THE SHORTEST PATH!",
        "video_name": "APNkWrD-U1k",
        "timestamps": [
            64
        ],
        "3min_transcript": "I will now do a proof of the law of sines. So, let's see, let me draw an arbitrary triangle. That's one side right there. And then I've got another side here. I'll try to make it look a little strange so you realize it can apply to any triangle. And let's say we know the following information. We know this angle -- well, actually, I'm not going to say what we know or don't know, but the law of sines is just a relationship between different angles and different sides. Let's say that this angle right here is alpha. This side here is A. The length here is A. Let's say that this side here is beta, and that the length here is B. Beta is just B with a long end there. So let's see if we can find a relationship that connects A and B, and alpha and beta. So what can we do? And hopefully that relationship we find will be the law of sines. So let me draw an altitude here. I think that's the proper term. If I just draw a line from this side coming straight down, and it's going to be perpendicular to this bottom side, which I haven't labeled, but I'll probably, if I have to label it, probably label it C, because that's A and B. And this is going to be a 90 degree angle. I don't know the length of that. I don't know anything about it. All I know is I went from this vertex and I dropped a line that's perpendicular to this other side. So what can we do with this line? Well let me just say that it has length x. The length of this line is x. Can we find a relationship between A, the length of this line x, and beta? Well, sure. Let's see. Let me find an appropriate color. That's, I think, a good color. So what's the relationship? If we look at this angle right here, beta, x is opposite to it and A is the hypotenuse, if we look at this right triangle right here, right? So what deals with opposite and hypotenuse? Whenever we do trigonometry, we should always just right soh cah toa at the top of the page. Soh cah toa. So what deals with opposite of hypotenuse? Soh, and you should probably guess that, because I'm proving the law of sines. So the sine of beta is equal to the opposite over the hypotenuse. It's equal to this opposite, which is x, over the hypotenuse, which is A, in this case. And if we wanted to solve for x, and I'll just do that, because it'll be convenient later, we can multiply both sides of this equation by A and you get A sine of beta is equal to x. Fair enough. That got us someplace."
    },
    {
        "Q": "At 3:10 you multiply by Y' but why is the reason for this? If the product rule is just f'(x)g(x)+f(x)g'(x) which would mean from my understanding (1)(y^2) + (x)(2y)",
        "A": "It s the notation that is confusing. I had a hard time with this as well. If you mentally substitute write in dy/dx here rather than multiply by y I bet you would feel more comfortable",
        "video_name": "ZtI94pI4Uzc",
        "timestamps": [
            190
        ],
        "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": "4:49 where did the +1 come from?",
        "A": "That is basic factoring: 2x + 6x\u00c2\u00b2 Factors to 2x(1+3x) The same thing is happening here, only with slightly more complicated expressions.",
        "video_name": "ZtI94pI4Uzc",
        "timestamps": [
            289
        ],
        "3min_transcript": "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 Now let's get all of our y primes on one side. So let's add y prime to both sides. So let's add-- and just to be clear, I'm adding one y prime to both sides. So let's add a y prime to both sides. And let's subtract this business from both sides. So let's subtract y squared e to the xy squared subtracting from both sides. So we're going to subtract y squared e to the xy squared. And we are left with 2xye to the xy squared plus 1 times y prime. We had this many y primes and then we add another 1y prime so we have this many plus 1 y primes. That's going to be equal to-- well, I purposely added y prime to both sides and so we are left with 1 to the xy squared. And now we just have to divide both sides by this. And we're left with the derivative of y with respect to x is equal to this, which I will just copy and paste. Actually, let me just rewrite it. Scroll down a little bit. It's equal to 1 minus y squared e to the xy squared over this business. Let me get some more space. 2xye to the xy squared plus 1. And we're done. It was kind of crazy, but fundamentally no different than what we've been doing in the last few examples."
    },
    {
        "Q": "at 4:50 why does he get \"+1)y'\" and not just 2y'? it seems like there is already a y' on that side so now that he added another there should be two right?",
        "A": "He just skipped one step, when he added y he got 2xy(e^xy^2)*y +y Now you can t just add this and get 2y but they both have a y so you can do this y (2xy*e^xy^2+1)",
        "video_name": "ZtI94pI4Uzc",
        "timestamps": [
            290
        ],
        "3min_transcript": "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 Now let's get all of our y primes on one side. So let's add y prime to both sides. So let's add-- and just to be clear, I'm adding one y prime to both sides. So let's add a y prime to both sides. And let's subtract this business from both sides. So let's subtract y squared e to the xy squared subtracting from both sides. So we're going to subtract y squared e to the xy squared. And we are left with 2xye to the xy squared plus 1 times y prime. We had this many y primes and then we add another 1y prime so we have this many plus 1 y primes. That's going to be equal to-- well, I purposely added y prime to both sides and so we are left with 1 to the xy squared. And now we just have to divide both sides by this. And we're left with the derivative of y with respect to x is equal to this, which I will just copy and paste. Actually, let me just rewrite it. Scroll down a little bit. It's equal to 1 minus y squared e to the xy squared over this business. Let me get some more space. 2xye to the xy squared plus 1. And we're done. It was kind of crazy, but fundamentally no different than what we've been doing in the last few examples."
    },
    {
        "Q": "I don't understand how the equation represents a surface (after 7:00). Can anyone recommend any sites that might show an animation to help me visualize it?",
        "A": "This is fairly easy. Just imagine now that the only variables we have are m and b. Then you can come up with an equation that looks like this: z=x^2-2*x-2*y+2*x*y+y^2 where x is m, y is b and z is SL error. If you plot that (copy paste it into google search) you ll end up with a surface (3 variables). Here it is drawn as a parabola (z=x^2+y^2), however it resembles only some similarity with the latter. To check it just copy paste provided equations into google search and it will plot it for you.",
        "video_name": "f6OnoxctvUk",
        "timestamps": [
            420
        ],
        "3min_transcript": "And then we have plus m squared times n times the mean of the x squared values. And then we have-- almost there, home stretch-- we have this over here which is plus 2mb times n times the mean of the x values. And then, finally, we have plus nb squared. So really, in the last two to three videos, all we've done is we simplified the expression for the sum of the squared differences from the those n points to this line, y equals mx plus b. So we're finished with the hard core algebra stage. The next stage, we actually want to optimize this. this expression right over here. We want to find the m and the b values that minimize it. And to help visualize it, we're going to start breaking into a little bit of three-dimensional calculus here. But hopefully it won't be too daunting. If you've done any partial derivatives, it won't be difficult. This is a surface. If you view that you have the x and y data points, everything here is a constant except for the m's and the b's. We're assuming that we have the x's and y's. So we can figure out the mean of the squared values of y, the mean of the xy product, the mean of the y's, the mean of the x squareds. We assume that those are all actual numbers. So this expression right here, it's actually going to be a surface in three dimensions. So you can imagine, this right here, that is the m-axis. And then, you could imagine the vertical axis to be the squared error. This is the squared error of the line axis. So for any combination of m and b, if you're in the mb plane, you pick some combination of m and b. You put it into this expression for the squared It'll give you a point. If you do that for all of the combinations of m's and b's, you're going to get a surface. And the surface is going to look something like this. I'm going to try my best to draw it. It's going to look like this. You could almost imagine it as a kind of a bowl."
    },
    {
        "Q": "In 6:33 when Sal says that pi is an irrational number, if someone does find out that pi terminates or repeats does that make pi a rational number?",
        "A": "Pi cannot be rational. However, if someone can prove that it terminates or repeats, then yes, it would be rational",
        "video_name": "qfQv8GzyjB4",
        "timestamps": [
            393
        ],
        "3min_transcript": "will cancel out. And we just have to figure out what 34,028 minus 340 is. So let's just figure this out. 8 is larger than 0, so we won't have to do any 2 is less than 4. So we will have to do some regrouping, but we can't borrow yet because we have a 0 over there. And 0 is less than 3, so we have to do some regrouping there or some borrowing. So let's borrow from the 4 first. So if we borrow from the 4, this becomes a 3 and then this becomes a 10. And then the 2 can now borrow from the 10. This becomes a 9 and this becomes a 12. And now we can do the subtraction. 8 minus 0 is 8. 12 minus 4 is 8. 9 minus 3 is 6. 3 minus nothing is 3. 3 minus nothing is 3. So 9,900x is equal to 33,688. So we get 33,688. Now, if we want to solve for x, we just divide both sides by 9,900. Divide the left by 9,900. Divide the right by 9,900. And then, what are we left with? We're left with x is equal to 33,688 over 9,900. Now what's the big deal about this? Well, x was this number. x was this number that we started off with, this number that just kept on repeating. And by doing a little bit of algebraic manipulation and subtracting one multiple of it from another, we're able to express that same exact x as a fraction. Now this isn't in simplest terms. I mean they're both definitely divisible by 2 and it looks like by 4. So you could put this in lowest common form, but we All we care about is the fact that we were able to represent x, we were able to represent this number, as a fraction. As the ratio of two integers. So the number is also rational. It is also rational. And this technique we did, it doesn't only apply to this number. Any time you have a number that has repeating digits, you could do this. So in general, repeating digits are rational. The ones that are irrational are the ones that never, ever, ever repeat, like pi. And so the other things, I think it's pretty obvious, this isn't an integer. The integers are the whole numbers that we're dealing with. So this is someplace in between the integers. It's not a natural number or a whole number, which depending on the context are viewed as subsets of integers. So it's definitely none of those. So it is real and it is rational. That's all we can say about it."
    },
    {
        "Q": "At 6:31 Sal said that Pi never repeats. however, doesn't Pi always repeat?",
        "A": "What he meant was pi doesn t have a pattern. You never see the same set of numbers twice. Unlike a repeating decimal such as 0.3838383838.... and so on.",
        "video_name": "qfQv8GzyjB4",
        "timestamps": [
            391
        ],
        "3min_transcript": "will cancel out. And we just have to figure out what 34,028 minus 340 is. So let's just figure this out. 8 is larger than 0, so we won't have to do any 2 is less than 4. So we will have to do some regrouping, but we can't borrow yet because we have a 0 over there. And 0 is less than 3, so we have to do some regrouping there or some borrowing. So let's borrow from the 4 first. So if we borrow from the 4, this becomes a 3 and then this becomes a 10. And then the 2 can now borrow from the 10. This becomes a 9 and this becomes a 12. And now we can do the subtraction. 8 minus 0 is 8. 12 minus 4 is 8. 9 minus 3 is 6. 3 minus nothing is 3. 3 minus nothing is 3. So 9,900x is equal to 33,688. So we get 33,688. Now, if we want to solve for x, we just divide both sides by 9,900. Divide the left by 9,900. Divide the right by 9,900. And then, what are we left with? We're left with x is equal to 33,688 over 9,900. Now what's the big deal about this? Well, x was this number. x was this number that we started off with, this number that just kept on repeating. And by doing a little bit of algebraic manipulation and subtracting one multiple of it from another, we're able to express that same exact x as a fraction. Now this isn't in simplest terms. I mean they're both definitely divisible by 2 and it looks like by 4. So you could put this in lowest common form, but we All we care about is the fact that we were able to represent x, we were able to represent this number, as a fraction. As the ratio of two integers. So the number is also rational. It is also rational. And this technique we did, it doesn't only apply to this number. Any time you have a number that has repeating digits, you could do this. So in general, repeating digits are rational. The ones that are irrational are the ones that never, ever, ever repeat, like pi. And so the other things, I think it's pretty obvious, this isn't an integer. The integers are the whole numbers that we're dealing with. So this is someplace in between the integers. It's not a natural number or a whole number, which depending on the context are viewed as subsets of integers. So it's definitely none of those. So it is real and it is rational. That's all we can say about it."
    },
    {
        "Q": "isn't a 360 degree a cricle i'm confused and whats the difference between supplementary and complemtary at 3:14",
        "A": "A supplementary is a pair of angles with 180 degrees total while complementary is a pair of angles that equal 90degrees.",
        "video_name": "zrqzG6xKa1A",
        "timestamps": [
            194
        ],
        "3min_transcript": "x plus y is equal to 180 degrees. And why does that make sense? Because look, if we add up x plus y we have gone halfway around the circle. So that's 180 degrees, right? So this is part of the way, and this is the rest of the way. So x plus y are going to equal 180 degrees. So hopefully we have learned that. And then let me switch colors for the sake of variety. Let me use my line tool. If I have-- let's see, I'm going to draw perpendicular lines. If I have that line, and then I have that line. And they are perpendicular. And then I have another line. Let's say it goes like that. And then I say that this is angle x. This is angle x. And this is angle y. Well, I said this line and this line are perpendicular, right? So that means that they intersect at a 90 degree angle. So we know that this whole thing is 90 degrees. And so what do we know about x plus y? Well, x plus y is going to equal 90 degrees. Or we could say that x and y are complementary. And I always get confused between supplementary and complementary. You just got to memorize it. I don't know if there's any-- let's see, is there any easy way? 180, supplementary. You could say that 180-- 100 starts with an O, which supplementary does not start with. So there. There's your mnemonic. And 90 starts with an N, and complementary does That's your other mnemonic. Complementary. I don't know if I'm spelling it right. Who cares? Let's move on. So let's learn some more stuff about angles. And what I'm going to do is I'm going to give you an arsenal, and then once you have that arsenal you can just tackle these beastly problems that I'm going to throw at you. So just take these for granted right now, and then in a few videos, probably, we're going to tackle some beastly problems. And you know, I'm using variables here. And if you're not familiar with variables you can put numbers here. If x was 30 degrees, then y is going to be 60 degrees. Or in this case, if x is, I don't know, 45 degrees, then y is going to be 135 degrees. That other way. Let me draw another property of angles of intersecting lines. So if I have two angles, two lines that intersect like this."
    },
    {
        "Q": "I'm confused about the proof at 7:20. Exactly afterwards when he did 180-x+y=180? Why did he do that?",
        "A": "He proved earlier that z = 180 - x , so then he could use this knowledge to replace the meaning of z: (z) + y = 180 (180 - x) + y = 180 The items in parentheses above are the same thing: we know this because it was proven earlier. This is how math proofs happen: you reuse the equations you ve already solved to help rewrite the equation you re trying to solve in a different way. That way, you reduce it into what you know is a fact.",
        "video_name": "zrqzG6xKa1A",
        "timestamps": [
            440
        ],
        "3min_transcript": "Well what do we know about angle x and angle z? It may not be obvious to you because I've drawn it slightly different, but I'll give you a small hint with an appropriately interesting color. So what angle is this whole thing right here? Well I'm just going along a line, right? That's halfway around a circle. So what angle is that? Well that's 180 degrees. So what does x plus z equal? Well, x plus z is going to equal that larger angle. x plus purple z is going to equal-- I think I'll switch to the blue; maybe it's taking too much time for me to switch-- is equal to 180 degrees. Or x and z are supplementary. So what do we know about z? Well z is equal to 180 minus x. Because x plus z is 180. Fine. Now, what's the relationship between z and y? Well, z and y are also supplementary. Because look, if I drew this angle here. Look at this big angle. Well once again I'm still going halfway around the circle. But now I'm using this line right here. So that's 180 degrees. So we know that angle z plus angle y is also equal to 180 degrees. Or, I don't want to keep writing it, but z and y are also supplementary. But we just figured out that z is 180 minus x. Right? So let's just substitute that back in here. So we get 180 minus x plus y is equal to 180 degrees. Why don't we subtract 180 degrees from both sides of this equation. That cancels out, and we get minus x plus y is equal to 0. And then add x to both sides of this equation, and we get y is equal to x. That was a very long way of showing you something that is fairly simple-- that opposite angles are equal to each other. So x is equal to y. And if you've played around with this, if you just drew a bunch of straight lines and they intersected at different angles, I think when you eyeball it it would make sense."
    },
    {
        "Q": "At 01:26, why does he put y/x? Would it also work the other way round?",
        "A": "He does that because that s how you find the answer to the problem, and no it does not work the other way around.",
        "video_name": "Iqws-qzyZwc",
        "timestamps": [
            86
        ],
        "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 13:41, when Sal says that the slope is undefined, does that just mean that there is no slope for that particular line at all?",
        "A": "Not really. Remember, the slope of a line is a fraction: change in y / change in x. If the slope is undefined, it is telling you that the change in x = 0 (denominator = 0). We can t divide by 0, which makes the slope undefined. If you say the line has no slope, this can be confused with a slope = 0. In this case, you have no change in Y (numerator = 0). And, you will have a horizontal line. So, to avoid confusion, it is better to say the slope = 0 or the slope - undefined. Hope this helps.",
        "video_name": "Iqws-qzyZwc",
        "timestamps": [
            821
        ],
        "3min_transcript": "These are interesting. Let's do the line e right here. Change in y over change in x. So our change in y, when we go from this point to this point-- I'll just count it out. It's one, two, three, four, five, six, seven, eight. It's 8. Or you could even take this y-coordinate 2 minus negative 6 will give you that distance, 8. What's the change in y? Well the y-value here is-- oh sorry what's the change in x? The x-value here is 4. The x-value there is 4. X does not change. So it's 8/0. Well, we don't know. 8/0 is undefined. So in this situation the slope is undefined. When you have a vertical line, you say your slope is undefined. Because you're dividing by 0. But that tells you that you're dealing probably with a vertical line. Now finally let's just do this one. This seems like a pretty straight up vanilla slope You have that point right there, which is the point 3, 1. So this is line f. You have the point 3, 1. Then over here you have the point negative 6, negative 2. So our slope would be equal to change in y. I'll take this as our ending point, just so you can go in different directions. So our change in y-- now we're going to go down in that direction. So it's negative 2 minus 1. That's what this distance is right here. Negative 2 minus 1, which is equal to negative 3. Notice we went down 3. And then what is going to be our change in x? Well, we're going to go back that amount. What is that amount? Well, that is going to be negative 6, that's our end point, minus 3. That gives us that distance, which is negative 9. Which is the same thing as if we go forward 9, we're going to go up 3. All equivalent. And we see these cancel out and you get a slope of 1/3. Positive 1/3. It's an upward sloping line. Every time we run 3, we rise 1. Anyway, hopefully that was a good review of slope for you."
    },
    {
        "Q": "What is the difference between 80:1 and 1:80? He said it can go the other way around.",
        "A": "Both scenarios gives you something that is 80 times as big as another object/line. So changing the order doesn t change much. However, you need to be specific to which object the 1 and the 80 belongs to. Building a miniature ship with a 80:1 ratio is a bit confusing, because the miniature ship is prob 80 times smaller instead of 80 times larger than the real one.",
        "video_name": "byjmR7JBXKc",
        "timestamps": [
            4801,
            140
        ],
        "3min_transcript": ""
    },
    {
        "Q": "How did sal in 3:55 get 12.5 from 128,000? -dazed and confused",
        "A": "Well, at 3:50 Sal got 12.8 square meters (m^2), in your question you said 12.5 (maybe a typo?), but the conversion is a simple conversion from square centimeters (cm^2) to square meters (m^2).",
        "video_name": "byjmR7JBXKc",
        "timestamps": [
            235
        ],
        "3min_transcript": "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, And 320 centimeters, well, that's 3.2 meters. And you would say, OK, 4 times 3.2, that is 12.8 square meters. But either way, the area of the living room in the real world in meters squared, or square meters, is 12.8."
    },
    {
        "Q": "How many more proportion can we make from 2:4::3:6\nPlz answer. My answer is total 8 including ratio given above.",
        "A": "Since a ratio is a fraction, we can create a infinite number of proportions from one ratio. Take any ratio, multiply its 2 parts by the same number and you will get an equivalent ratio. Since the 2 ratios are equal, you have a proportion. The number you select to multiply with can be any number. Since there are an infinite set of numbers, you can create an infinite set of ratios.",
        "video_name": "qYjiVWwefto",
        "timestamps": [
            124,
            186
        ],
        "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": "why in 3:55 the ecuation is equal to 1?",
        "A": "try to compare with this simple numeric example 6 / 3 = 2 and then take 2 to other side dividing in the denominator and we have 6 / (3 * 2) = 1",
        "video_name": "6YRGEsQWZzY",
        "timestamps": [
            235
        ],
        "3min_transcript": "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, Now let's see, let's subtract 2v squared from both sides of this. And we will be left with 2xv v prime is equal to 1 plus-- let's see, we're subtracting 2v squared from both sides. So we're just left with a 1 plus v squared here, right? 3v squared minus 2v squared is just v squared. And let's see, we want it to be separable, so let's put all the v's on the left hand side. So we get 2xv v prime divided by 1 plus v squared is equal to 1. And let's divide both sides by x. So we get the x's on the other side. So then we get 2v-- and I'll now switch back Instead of v prime, I'll write dv dx. 2v times the derivative of v with respect to x divided by 1 plus v squared is equal to-- I'm dividing both sides by x, notice I didn't write the x on this side-- so that is equal to 1 over x. And then, if we just multiply both sides of this times dx, we've separated the two variables and we can integrate So let's do that. Let's go up here. I'll switch to a different color, so you know I'm working on a different column now. So multiply both sides by dx. I get 2v over 1 plus v squared dv is equal to 1 over x dx. And now we can just integrate both sides of this equation. This is a separable equation in terms of v and x. And what's the integral of this? At first, you might think, oh boy, this is complicated. This is difficult, maybe some type of trig function."
    },
    {
        "Q": "At 3:49, why did he divide 2xvv' by the entire right hand side as opposed to subtracting v^2 from both sides?",
        "A": "When solving a separable equation you don t want to have a term you are adding to the dy/dx (or in this case dv/dx) term. The problem is that it will make it harder to separate the variables but lets try: 2x dv/dx - v^2 = 1 2 dv/dx - v^2/x = 1/x 2 dv - ((v^2/x) dx) = 1/x dx So by creating an additional term on the left side of the equation you have a mix of terms and so you can t integrate.",
        "video_name": "6YRGEsQWZzY",
        "timestamps": [
            229
        ],
        "3min_transcript": "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, Now let's see, let's subtract 2v squared from both sides of this. And we will be left with 2xv v prime is equal to 1 plus-- let's see, we're subtracting 2v squared from both sides. So we're just left with a 1 plus v squared here, right? 3v squared minus 2v squared is just v squared. And let's see, we want it to be separable, so let's put all the v's on the left hand side. So we get 2xv v prime divided by 1 plus v squared is equal to 1. And let's divide both sides by x. So we get the x's on the other side. So then we get 2v-- and I'll now switch back Instead of v prime, I'll write dv dx. 2v times the derivative of v with respect to x divided by 1 plus v squared is equal to-- I'm dividing both sides by x, notice I didn't write the x on this side-- so that is equal to 1 over x. And then, if we just multiply both sides of this times dx, we've separated the two variables and we can integrate So let's do that. Let's go up here. I'll switch to a different color, so you know I'm working on a different column now. So multiply both sides by dx. I get 2v over 1 plus v squared dv is equal to 1 over x dx. And now we can just integrate both sides of this equation. This is a separable equation in terms of v and x. And what's the integral of this? At first, you might think, oh boy, this is complicated. This is difficult, maybe some type of trig function."
    },
    {
        "Q": "Hi sal, at \"7:49\" you subtract the cx^3 to the left side of the equation. Wouldn't you have to divide that expression to the left side? I was confused.",
        "A": "He brought the entire CX^3 term to the left side. Whereas in dividing, he would have to have left either the C or X^3 term on the right side. 1 + 2 = 3 -> 1 + 2 - 3 = 0 hope this helps!",
        "video_name": "6YRGEsQWZzY",
        "timestamps": [
            469
        ],
        "3min_transcript": "So that's how I knew that this was the antiderivative. And if you don't believe me, use the chain rule and take the derivative of this, and you'll get this. And hopefully, it will make a little bit more sense. But anyway, that's the left hand side, and then that equals-- Well, this one's easy. That's the natural log, the absolute value of x. We could say, plus c, but just so that we can simplify it a little bit, an arbitrary constant c, we can really just write that as the natural log of the absolute value of some constant c. I mean, this is still some arbitrary constant c. So we can rewrite this whole equation as the natural log of 1 plus v squared is equal to-- when you add natural logs, you can essentially just multiply the two numbers that you're taking the natural log of-- the natural log of, we could say, the absolute value of cx. And so the natural log of this is equal to the natural log of this. So we could say that 1 plus v squared is equal to cx. So we know v is equal to y over x, so let's do that. So we get 1 plus y over x squared is equal to cx. Let me scroll this down a little bit. Let's multiply both sides of the equation times x squared. We could rewrite this as y squared over x squared. So we multiply both sides times x squared, you get x squared plus y squared is equal to cx to the third. And we're essentially done. If we want to put all of the variable terms on left hand side, we could say that this is equal to x squared plus y squared minus cx to the third is equal to 0. And this implicitly defined function, or curve, or however you want to call it, is the solution to our original homogeneous first order differential equation. I will see you in the next video. And now, we're actually going to do something. We're going to start embarking on higher order And actually, these are more useful, and in some ways, easier to do than the homogeneous and the exact equations that we've been doing so far. See you in the next video."
    },
    {
        "Q": "At 6:56 couldn't he have taken everything as a power of e. to cancel the natural log. Or would it have become e^(ln(x) + C) which is the same as e^(ln(x))*e^c which is cx.\n\nSo it would be the same solution, so, nvm. Just figured that in the fly. Was my logic correct?",
        "A": "Your logic is correct, I regualary use the approach you suggested in my Differential Equations class and the teachers does it too. I think Sal just took a different approach",
        "video_name": "6YRGEsQWZzY",
        "timestamps": [
            416
        ],
        "3min_transcript": "reverse chain rule. We have a function here, 1 plus v squared, an expression here. And we have its derivative sitting right there. So the antiderivative of this, and you can make a substitution if you like. You could say u is equal to 1 plus v squared, then du is equal to 2v dv. And then, well, you would end up saying that the antiderivative is just the natural log of u. Or, in this case, the antiderivative of this is just the natural log of 1 plus v squared. We don't even have to write an absolute value there. Because that's always going to be a positive value. So the natural log of 1 plus v squared. And I hope I didn't confuse you. That's how I think about it. I say, if I have an expression, and I have its derivative multiplied there, then I can just take the antiderivative of the whole expression. And I don't have to worry about what's inside of it. So if this was a 1 over an x, or 1 over u, it's just the So that's how I knew that this was the antiderivative. And if you don't believe me, use the chain rule and take the derivative of this, and you'll get this. And hopefully, it will make a little bit more sense. But anyway, that's the left hand side, and then that equals-- Well, this one's easy. That's the natural log, the absolute value of x. We could say, plus c, but just so that we can simplify it a little bit, an arbitrary constant c, we can really just write that as the natural log of the absolute value of some constant c. I mean, this is still some arbitrary constant c. So we can rewrite this whole equation as the natural log of 1 plus v squared is equal to-- when you add natural logs, you can essentially just multiply the two numbers that you're taking the natural log of-- the natural log of, we could say, the absolute value of cx. And so the natural log of this is equal to the natural log of this. So we could say that 1 plus v squared is equal to cx. So we know v is equal to y over x, so let's do that. So we get 1 plus y over x squared is equal to cx. Let me scroll this down a little bit. Let's multiply both sides of the equation times x squared. We could rewrite this as y squared over x squared. So we multiply both sides times x squared, you get x squared plus y squared is equal to cx to the third. And we're essentially done. If we want to put all of the variable terms on left hand side, we could say that this is equal to x squared plus y squared minus cx to the third is equal to 0. And this implicitly defined function, or curve, or however you want to call it, is the solution to our original homogeneous first order differential equation."
    },
    {
        "Q": "At 3:33, how did Sal get 0.1 from the second equation?",
        "A": "The -0.1 is part of the function. It was given as part of the problem definition. Sal didn t create it.",
        "video_name": "GA_yxxeFYBU",
        "timestamps": [
            213
        ],
        "3min_transcript": "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. but let's go the other way around. Here we have a, we have a sequence defined recursively, and I want to create a function that defines a sequence explicitly. Let's think about this. One way to think about it, this sequence, when n is equal to one it starts at 9.6, and then every term is the previous term minus 0.1. The second term is gonna be the previous term minus 0.1, so it's gonna be 9.5. Then you're gonna go to 9.4. Then you're gonna go to 9.3. We could keep going on and on and on. If we want, we could make a little table here, and we could say this is n, this is h of n, and you see when n is equal to one, h of n is 9.6. When n is equal to two, we're now in this case over here, it's gonna h of two minus one, so it's gonna be h of one minus 0.1. which is going to be 9.5. When h is three, it's gonna be h of two, h of two minus 0.1, minus 0.1. H of two is right over here. You subtract a tenth you're gonna get 9.4, exactly what we saw over here. Let's see if we can pause the video now and define this... Create a function that constructs or defines this arithmetic sequence explicitly. Here it was recursively. We wanna define it explicitly. So let's just call it, I don't know, let's just call it f of n. We can say look, it's gonna be 9.6, but we're gonna subtract, we're gonna subtract 0.1 a certain number of times depending on what term we're talking about. We're gonna subtract 0.1, but how many times are we gonna subtract it as a function of n? If we're talking about the first term we subtract zero times."
    },
    {
        "Q": "At 4:40 when Sal solves the integral and gets 2*ln|x-1|, can I just forget the absolute value of x-1 because of the logarithmic rules? it's the same like ln(x-1)^2 so my argument is always positive and therefore always valid for my ln. Or did I overlook something?",
        "A": "Since you took the square out , you must put a modulus there as negative log isn t defined. But if the square is there, its always positive so no worries.",
        "video_name": "5j81gyHn9i0",
        "timestamps": [
            280
        ],
        "3min_transcript": "but it's actually much more useful for finding the integral. Negative 1/2 plus two over x minus one, d x. Now, how do we evaluate this? Well, the antiderivative of negative 1/2 is pretty straight forward. That's just going to be negative 1/2 x, plus the entire derivative of two over x minus one. You might reduce here ahead. The derivative of x minus one is just one so you could say that the derivative is sitting there. We can essentially use substitution in our heads and say, \"Okay, let's just take the entire derivative, \"we could say with respect to x minus one which will be \"the natural log of the absolute value of x minus one.\" If all of that sounds really confusing, I'll let you do the u-substitution. If I were just trying to evaluate I could see, okay, the derivative of x minus one is just one so I could say u is equal to x minus one, and then d u is going to be equal to d x. This is going to be, we can rewrite in terms of u as two, I'll just take the constant out two times the integral of one over u, d u, which we know as two times the natural log of the absolute value of u plus c. In this case, we know that u is x minus one. This is equal to two times the natural log of x minus one plus c. That's what we're going to have right over here. So plus two times the natural log of the absolute value of x minus one plus c. The plus c doesn't just come from this one, this general overtaking the integral of the whole thing. we take the derivative and the constant will go away. Let me just put the plus c right over there, and we are done."
    },
    {
        "Q": "Just before 2:31 you got 2*2*x*x*x**y and I did not understand how you got that because dont they factor out or something?",
        "A": "That 2*2*x*x*x*y, or 4x^3y, is the greatest common factor of the two polynomials, and he factors it out of both of them.",
        "video_name": "_sIuZHYrdWM",
        "timestamps": [
            151
        ],
        "3min_transcript": "of its basic constituents. Now let's do the same thing for 8-- I'll color code it-- 8x to the third. Let me do it in similar colors. So in this situation we have 8x to the 1/3 y. So the prime factorization of 8 is 2 times 2 times 2. It's 2 times 2 times 2. Prime, or I should say the factorization of x to the third, or the expansion of it, is just times x times x times x. x multiplied by itself three times. And then we are multiplying everything by a y here, times y. So what factors are common to both of these? And we want to include as many of them as possible to find this greatest common factor. So we have two 2's here, three 2's here. So we only have two 2's in common in both of them. So we only have three x's in common. Three x's and three x's. And we have a y here and a y here. So y is common to both expressions. So the greatest common factor here is going to be 2 times 2. So it's going to be 2 times 2 times x times x times x times Or 4x to the third y. So this is what we want to factor out. So that means we can write this thing as-- if we factor out a 4x to the third y, where essentially we have to divide each of these by 4x to the third. We're factoring it out. So let me rewrite this. So this is 4x to the fourth y plus 8x to the third y. And we're going to divide each of these by 4x to the third y. If we were to multiply this out, we would distribute this 4x to the third y on each terms. And then it would cancel with the denominator. You would have the same thing in the numerator and denominator. And then you would get this expression over here. So hopefully this makes sense that these are the exact same expression. But when you write it this way, then it becomes pretty clear that this is 4x to the third y. And then you just simplify each of these expressions. 4 cancels with 4. x to the fourth divided by x to the third is x. y divided by y is just 1. So you have x plus 8 divided by 4 is 2. x to the third divided by x to the third is 1. Y divided by y is 1. So x plus 2. Another way to see what's left over when you factor it out is if you were to take out the common factor. So we took out this and this. What was left over in 4x to the fourth y when we took this stuff out? When we undistributed it? Well, the only thing that was left was this x right over here. Let me do that in another color."
    },
    {
        "Q": "say you have 6 peanut cookies and you have 6 friends two of your friends cannot eat nuts. So the ratio is 6:2 (thats how you write a ratio) Now say you baked six more cookies to give to 12 of your friends. For every 6 peanut cookies you have two friends. So what is the ratio if you have 12 cookies?",
        "A": "Solving this problem is easier than it seems. The way I do it is that I turn the ratio into a fraction multiply the numerator and denominator by x (in your case,2) then turn it back into a fraction. 6:2 = 6/2 6 X 2 = 12 -------------------- 2 X 2 =4 12/4 = 12:4 By observing the diagram, we can tell the ratio is now 12:4. Hope This Helps!",
        "video_name": "bIKmw0aTmYc",
        "timestamps": [
            362
        ],
        "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": "at 2:05 how is 6/9 the same thing as 2/3?",
        "A": "6/9 simplifies into 2/3",
        "video_name": "bIKmw0aTmYc",
        "timestamps": [
            125
        ],
        "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": "kate and nora each have a sum of money. the ratio of the amount of money Kate has to that of Nora is 3:5. After Nora gives $150 to Kate, the ratio of the money is 7:9. Find the sum of money Kate had initially?",
        "A": "I think it goes like this Kate to Nora = 3:5 Kate to Total sum = 3:8 Kate+150 to Nora = 7:9 Kate+150 to Total sum = 7: 16 We want Kate initial sum to total moeny. 3/8 and 7/16 have different denominators and we need them the same. 3*2/8*2 = 6/16 Total sum = 150/(1/16) = $2400. Kates initial sum = 2400*(6/16) = $900",
        "video_name": "bIKmw0aTmYc",
        "timestamps": [
            185,
            429
        ],
        "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 about 1:09 shouldn't he have added fraction form?",
        "A": "I think it is covered in the next video.",
        "video_name": "bIKmw0aTmYc",
        "timestamps": [
            69
        ],
        "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": "Why can\u00e2\u0080\u0099t there be another way to write a ratio, and can there be a negative ratio like -4:10",
        "A": "Hello random stranger!! There are 3 different ways to write a ratio. Ex: 2:4 2/4 2 to 4 Hope that helps random stranger! XD",
        "video_name": "bIKmw0aTmYc",
        "timestamps": [
            250
        ],
        "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": "Ratio of boys to girls is 3:4 if 252 students attends how many are boys",
        "A": "SInce ratios are the same thing as fractions, if the ratio of the boys to girls is 3:4, that means 3/7 children are boys. Therefor, you would have to find 1/7 of 252 (or 36). Then, since 3/7 children are boys, you multiply 1/7x3, or 36+36+36=108. Finally, your answer is 108 boys.",
        "video_name": "bIKmw0aTmYc",
        "timestamps": [
            184
        ],
        "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 9:46 how is it that you can compare the matrices\n[a] and [b]\n[c] ___ [d]\nto the fractions a/c=b/d ??? I thought matrices were used to notate vectors, not ratios....",
        "A": "We are looking at the situation where the determinant is 0. So ad - bc =0 and thus ad = bc ad/cd = bc/cd a/c = b/d",
        "video_name": "UqyN7-tRS00",
        "timestamps": [
            586
        ],
        "3min_transcript": "And not only would they intersect, they would intersect in an infinite number of places. But still you would have no unique solution. You'd have no one solution to this equation. It would be true at all values of x and y. So you can kind of view it when you apply the matrices to this problem. The matrix is singular, if the two lines that are being represented are either parallel, or they are the exact same line. They're parallel and not intersecting at all. Or they are the exact same line, and they intersect at an infinite number of points. And so it kind of makes sense that the A inverse wasn't defined. So let's think about this in the context of the linear combinations of vectors. That's not what I wanted to use to erase it. So when we think of this problem in terms of linear That this is the same thing as the vector ac times x plus the vector bd times y, is equal to the vector ef. So let's think about it a little bit. We're saying, is there some combination of the vector ac and the vector bd that equals the vector ef. But we just said that if we have no inverse here, we know that because the determinant is 0. And if the determinant is 0, then we know in this situation that a/c must equal b/d. So a/c is equal to b/d. So what does that tell us? Well let me draw it. And maybe numbers would be more helpful here. But I think you'll get the intuition. I'll just assume both sectors are in the first quadrant. Let me draw. The vector ac. Let's say that this is a. Let me do it in a different color. So I'm gonna draw the vector ac. So if this is a, and this is c, then the vector ac looks like that. Let me draw it. I want to make this neat. The vector ac is like that. And then we have the arrow. And what would the vector bd look like? Well the vector bd-- And I could draw it arbitrarily someplace. But we're assuming that there's no derivative-- sorry, no determinant. Have I been saying derivative the whole time?"
    },
    {
        "Q": "At 0:53, he writes: zero to the second power, is 1*0*0. Why would this equation be 1*0*0? Since any number times one is that number, what would be the difference between 1*0*0 and 0*0? And where did the one come in anyway? Wouldn't zero to the second power just be written as 0*0? Maybe i'm just missing something. Thanks!\n\nLorenzo, age 9",
        "A": "I m not exactly sure why he put a one there either. I think it s out of habit from other videos he s done, when taking the powers of non-zero numbers. I think one example is proving that any non-zero number is 1, like this: 5^2 = 1*5*5 5^1 = 1*5 5^0 = 1 His putting a one in front of the equation probably came from that. Perhaps someone might use that to prove that 0^0 is also one if that s his opinion.",
        "video_name": "PwDnpb_ZJvc",
        "timestamps": [
            53
        ],
        "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 2:15 why so many zero and why is he talking about stuff we havent even learned yet",
        "A": "Different schools go at different paces and introduce things in different orders. Now expand that idea to all the schools in your country and then expand it again to all the schools in the world. It s not possible to make a website that exactly matches every school and every teacher.",
        "video_name": "PwDnpb_ZJvc",
        "timestamps": [
            135
        ],
        "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": "Wait... at 0:43 Sal says that 0 to the first power is 0*1, but I thought it was 0*0, can anyone explain why? Thanks!",
        "A": "Ah, 0^1 = 0 and 0^0 is undefined. Does that help?",
        "video_name": "PwDnpb_ZJvc",
        "timestamps": [
            43
        ],
        "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 1:46 Sal says that this property extends to negative exponents as well. But 0^-1 power would be 1/0^1 = 1/0 which is undefined right?",
        "A": "Yes you are correct. By definition, a negative exponent implies a reciprocal. That is, any number, x^(-1) is equivalent to 1/x.",
        "video_name": "PwDnpb_ZJvc",
        "timestamps": [
            106
        ],
        "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": "Why can he switch natural logarithm with the limit at 6:24 ? He explained it but it didn't really convince me. My textbook doesn't seem to mention anything about that kind of limit rule.",
        "A": "Think about how we approach analyzing composite functions. For example, if I am looking at the lim as x --> \u00e2\u0088\u009e for sin(1/x), analyzing the sin part of the function doesn t really matter until I determine the behavior of 1/x as x --> \u00e2\u0088\u009e. Therefore, I can rewrite the limit as sin (lim as x --> \u00e2\u0088\u009e of 1/x) and this will not change the answer. Try it and see for yourself.",
        "video_name": "3nQejB-XPoY",
        "timestamps": [
            384
        ],
        "3min_transcript": "So let's make that substitution. So all of this is going to be equal to the limit as -- now we've gotten rid of our delta x. We're going to say the limit as an approaches infinity of the natural log -- I'll go back to that mauve color -- the natural log of 1 plus -- now, I said that instead of delta x over x, I made the substitution that that is equal to 1 over n. So that's 1 plus 1 over n. And then what's 1 over delta x? Well delta x is equal to x over n, so 1 over delta x is going to be the inverse of this. It's going to be n over x. And then we can re-write this expression right here -- let me re-write it again. This is equal to the limit as n approaches infinity of the natural log of 1 plus 1 over n. I could say this is to the n, and then all of this to the 1 over x. Once again, this is just an exponent property. If I raise something to the n and then to the 1 over x, I could just multiply the exponent to get to the n over x. So these two statements are equivalent. But now we can use logarithm properties to say hey, if this is the exponent, I can just stick it out in front of the coefficient right here. I could put it out right there. And just remember, this was all of the derivative with respect to x of the natural log of x. So what is that equal to? We could put this 1 out of x in the front here. In fact, that 1 out of x term, it has nothing to do with n. It's kind of a constant term when you think of it in terms of n. So we can actually put it all the way out here. We could put it either place. So we could say 1 over x times all that stuff in mauve. The limit as n approaches infinity of the natural log The natural log of all of that stuff. Or, just to make the point clear, we can re-write this part -- let me make a salmon color -- equal to 1 over x times the natural log of the limit as n approaches infinity. I'm just switching places here, because obviously what we care is what happens to this term as it approaches infinity, of 1 plus 1 over n to the n. Well what is -- this should look a little familiar to you on some of the first videos where we talked about e -- this is one of the definitions of e. e is defined. I'm still not using this at all. I'm just stating that the definition of e, e is equal to the limit as n approaches infinity of 1 plus 1 over n to the n."
    },
    {
        "Q": "At 3:25, why did Sal use the substitution. I don't get how he just defines delta(x) / x equal to 1/n.\nDid I miss something from the previous video? (I just started watching from this video)\n\nPlease help.",
        "A": "Because n hasn t appeared anywhere in any equations we are free to use it how we like. we re just making the equation look different to make it easier to recognize what s going on. Sorry if my answer doesn t help.",
        "video_name": "3nQejB-XPoY",
        "timestamps": [
            205
        ],
        "3min_transcript": "natural log of x plus delta x minus the natural log of x. All of that over delta x. Now we can just use the property of logarithms. If I have the log of a minus the log of b, that's the same thing as a log of a over b. So let me re-write it that way. So this is going to be equal to the limit as delta x approaches 0. I could take this 1 over delta x right here. 1 over delta x times the natural log of x plus delta x divided by this x. Just doing the logarithm properties right there. Then I can re-write this -- first of all, when I have this coefficient in front of a logarithm I can make this the exponent. And then I can simplify this in here. 0 of the natural log -- let me do this in a new color. Let me do it in a completely new color. The natural log of -- the inside here I'll just divide everything by x. So x divided by x is 1. Then plus delta x over x. Then I had this 1 over delta x sitting out here, and I can make that the exponent. That's just an exponent rule right there, or a logarithm property. 1 over delta x. Now I'm going to make a substitution. Remember, all of this, this was all just from my definition of a derivative. This was all equal to the derivative of the natural log of x. I have still yet to in any way use this. And I won't use that until I actually show it to you. I've become very defensive about these claims of circularity. They're my fault because that shows that I wasn't clear to be more clear this time. So let's see if we can simplify this into terms that we recognize. Let's make the substitution so that we can get e in maybe terms that we recognize. Let's make the substitute delta x over x is equal to 1 over n. If we multiply -- this is the same thing. This is the equivalent to substitution. If we multiply both sides of this by x, as saying that delta x is equal to x over n. These are equivalent statements. I just multiplied both sides by x here. Now if we take the limit as n approaches infinity of this term right here, that's equivalent -- that's completely equivalent to taking the limit as delta x approaches 0. If we're defining delta x to be this thing, and we take the limit as its denominator approaches 0, we're going"
    },
    {
        "Q": "at 2:54, why did sal write 8? wasn't it 7?",
        "A": "No it is f(7) = 8, so at x = 7, y reaches it maximum height of 8. Range has to do with possible ys.",
        "video_name": "sXP7VhU1gYE",
        "timestamps": [
            174
        ],
        "3min_transcript": "is greater than or equal to negative 6. Or we could say negative 6 is less than or equal to x, which is less than or equal to 7. If x satisfies this condition right over here, the function is defined. So that's its domain. So let's check our answer. Let's do a few more of these. The function f of x is graphed. What is its domain? Well, exact similar argument. This function is not defined for x is negative 9, negative 8, all the way down or all the way up I should say to negative 1. At negative 1, it starts getting defined. f of negative 1 is negative 5. So it's defined for negative 1 is less than or equal to x. And it's defined all the way up to x equals 7, including x equals 7. So this right over here, negative 1 is less than or equal to x is less than or equal to 7, the function is defined for any x that satisfies this double inequality right over here. Let's do a few more. What is its range? So now, we're not thinking about the x's for which this function is defined. We're thinking about the set of y values. Where do all of the y values fall into? Well, let's see. The lowest possible y value or the lowest possible value of f of x that we get here looks like it's 0. The function never goes below 0. So f of x-- so 0 is less than or equal to f of x. It does equal 0 right over here. f of negative 4 is 0. And then the highest y value or the highest value that f of x obtains in this function definition is 8. f of 7 is 8. It never gets above 8, but it does equal 8 right over here when x is equal to 7. So 0 is less than f of x, which is less than or equal to 8. So that's its range. Let's do a few more. This is kind of fun. The function f of x is graphed. So once again, this function is defined for negative 2. Negative 2 is less than or equal to x, which is less than or equal to 5. If you give me an x anywhere in between negative 2 and 5, I can look at this graph to see where the function is defined. f of negative 2 is negative 4. f of negative 1 is negative 3. So on and so forth, and I can even pick the values in between these integers. So negative 2 is less than or equal to x, which is less than or equal to 5."
    },
    {
        "Q": "At 2:13, he says that the formula is clean, what does he mean by that?",
        "A": "When he says that the formula of the volume of the cone is clean, he means that the cone is EXACTLY 1/3 of the volume of a cylinder. As you can see in the video, the volume of a cylinder is h(pi)*r^2, and the area of a cylinder is exactly 1/3 of that, which is known as 1/3*h(pi)*r^2. So when he says clean, he means that the volume of a cylinder is exactly 1/3 the volume of a cone.",
        "video_name": "hC6zx9WAiC4",
        "timestamps": [
            133
        ],
        "3min_transcript": "Let's think a little bit about the volume of a cone. So a cone would have a circular base, or I guess depends on how you want to draw it. If you think of like a conical hat of some kind, it would have a circle as a base. And it would come to some point. So it looks something like that. You could consider this to be a cone, just like that. Or you could make it upside down if you're thinking of an ice cream cone. So it might look something like that. That's the top of it. And then it comes down like this. This also is those disposable cups of water you might see at some water coolers. And the important things that we need to think about when we want to know what the volume of a cone is we definitely want to know the radius of the base. So that's the radius of the base. Or here is the radius of the top part. You definitely want to know that radius. And you want to know the height of the cone. 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,"
    },
    {
        "Q": "Why is f(x,y)=p(x,y)i+q(x,y)j a function of 't' as Sal mentions around @15:28",
        "A": "Well, f(x,y) is a function of x and y. But if we re following the path defined by r(t) = x(t)i + y(t)j then our x and y are themselves just functions of t, so f(x,y) becomes f(x(t),y(t)), in short it depends completely on t (though indirectly through x(t) and y(t)). I.e. f(x(t),y(t)) becomes f(t).",
        "video_name": "t3cJYNdQLYg",
        "timestamps": [
            928
        ],
        "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": "5:48\nhow does variable summation applies to 1",
        "A": "There are a few ways to think about it. One is, when i is 1, then 1 is still 1. When i is 2, then 1 is still 1 right? When i is some number k, then 1 is still 1. When i is N, 1 is still one. So when you sum along the index i, 1 is simply added to itself again and again N times. So the answer is N. The other way to think about it, which is essentially the same is this. Since i is never 0, we can rewrite 1 as i/i. Then Sum(1)=Sum(i/i)=1/1+2/2+3/3+...+(n-2)/(n-2)+(n-1)/(n-1)+n/n=1+1+1+...+1+1+1=N.",
        "video_name": "sRVGcYGjUk8",
        "timestamps": [
            348
        ],
        "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": "Around 3:30 Sal references the Calculus playlist--I'm not even CLOSE to that playlist yet. Am I watching these videos too soon? It seems like the Statistics playlist is showing up really early on my practice map and I may not have the skills to successfully accomplish the unit. Do you think this could be true? I did okay up through standard deviation, but z-scores, empirical rule and some references are throwing me off!",
        "A": "This is like a side tour, sightseeing in a cool neighborhood. You don t need to move into the calculus house to work in statistics. For example, I think the formula for the Standard Deviation of a uniform distribution is (b-a)/sqrt(12). I wanted to know, Why 12? I asked Doctor Math and he (Doctor Anthony) gave me an explanation that I (frankly) didn t understand, but trust. I don t need to know where the 12 came from to use the formula, but I find it comforting to know that someone knows.",
        "video_name": "sRVGcYGjUk8",
        "timestamps": [
            210
        ],
        "3min_transcript": "This is the same thing as x sub i squared minus-- this is your little algebra going on here. So when you square it-- I mean, we could multiply it out. We could write it. x sub i minus mu times x sub i minus mu. So we have x sub i times x sub i, that's x sub i squared. Then you have x sub i times minus mu. And then you have minus mu times x sub i. So when you add those two together, you get minus 2x sub i mu, because you have it twice. x sub i times mu, that's 1 minus x sub i mu. And then you have another one, minus mu x sub i. When you add them together, you get minus 2x sub i mu. I know it's confusing with me saying sub i and all of that. But it's really no different than when Just the variables look a little bit more complicated. And then the last term is minus mu times minus mu, which is plus mu squared. Fair enough. Let me switch colors just to keep it interesting. Let me cordon that off. The sum of this is the same thing as the sum of-- because if you think about it, we're going to take each x sub i. For each of the numbers in our population, we're going to perform this thing. And we're going to sum it up. But if you think about it, this is the same thing as-- if you're not familiar with sigma notation this is a good thing to know in general, just a little bit of intuition. That this is the same thing as-- I'll do it here to have space. The sum from i is equal to 1 to n of the first term, x sub i squared minus-- and actually, we can bring out the constant terms. is the thing that has the i-th term. So in this case, it's x sub i. So x sub 1, x sub 2. So that's the thing that you have to leave on the right hand side of the sigma notation. And if you've done the calculus playlists already, sigma notation is really like a discrete integral on some level. Because in an integral, you're summing up a bunch of things and you're multiplying them times dx, which is a really small interval. But here you're just taking a sum. And we showed in the calculus playlist that an integral actually is this infinite sum of infinitely small things, but I don't want to digress too much. But this was just a long way of saying that the sum from i equals 1 to n of the second term is the same thing as minus 2 times mu of the sum from i is equal to 1 to n of x sub i."
    },
    {
        "Q": "At 2:05 Sal says that the line he just drew was called a transversal, because it transversed across the other two lines. But what does it mean to transverse?",
        "A": "To transverse means to intercept.",
        "video_name": "H-E5rlpCVu4",
        "timestamps": [
            125
        ],
        "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 5:00, if we were given the measure of the yellow angle in the second diagram, could we find out what the measure of the green angle was? If so, how?",
        "A": "No, you would not. Since the two angles exist on different lines, and the lines are not parallel, then there is no way to know for certain if the angles would compliment each other or not.",
        "video_name": "H-E5rlpCVu4",
        "timestamps": [
            300
        ],
        "3min_transcript": "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 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."
    },
    {
        "Q": "Hi friends :-) I have a question for you-\n1> at 6:27 \"Sal\" proved us that alternate interior angles are equal , right? so, here is my question ?\nwe know that :- b=c,f=g\nnow b and c are vertical angles and we also know that b = f (corresponding angle) so why can't we say that c = f ,right? and why not alternate exterior angles?",
        "A": "Ok, so you said b=c,f=g. Which says b=c and f=g. If you look back at Sals blackboard on the video there is no comma, it simply says b=c=f=g which means they are ALL equal, so yes , c does equal f and c equals g, and b equals g etc. etc.. As for alternate exterior angles, the same rule applies. a=d=e=h. They are all equal.",
        "video_name": "H-E5rlpCVu4",
        "timestamps": [
            387
        ],
        "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": "At 3:11 Sal says that lim kf(x) = K*limf(x)\nx-->c x-->c\nshouldn't it be equal to lim k * limf(x)\nx-->c x-->c",
        "A": "At 3:11 if one takes lim k, one will just get K because it is not a function, so one could put it as a function as g(x)=K, but then it would always return K and it would always approach K no matter where you get your limit. so instead of taking lim K, you just use K because K is the limit of K. This is because it is a constant, not a variable. so for example, 2f(2) where f(x)=2x will get you 2(2*2) which is 2*4 which is 8. Hope that made sense.",
        "video_name": "lSwsAFgWqR8",
        "timestamps": [
            191
        ],
        "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 2:44 How do I solve this question?\n\n(3r^3-19r^2+27r+4) / (r-4)",
        "A": "3r^2+31r-87-344/r-4",
        "video_name": "UquFdMg6Z_U",
        "timestamps": [
            164
        ],
        "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": "It took me a minute to figure this out, but I wanted to check if I am right. At 4:30, the +1 -2/3x can not be simplified because it can also be stated as + 1x^0 - 2/3x^-1, which makes the x terms different. Is this correct?",
        "A": "You are correct. 1 is a constant term (x^0) and is not a like term with (2/3)x^(-1).",
        "video_name": "UquFdMg6Z_U",
        "timestamps": [
            270
        ],
        "3min_transcript": "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. here, but x to the negative 1 is the exact same thing as 1 over x. And so let's just write our answer completely simplified. So it's going to be 3x to the third minus 1/2 x plus 1-- because this thing right here is just 1-- and then minus 2 times 1 in the numerator over 3 times x in the denominator. And we are done. Or we could write this. Depending on what you consider more simplified, this last term right here could also be written in minus 2/3 x to the negative 1. But if you don't want a negative exponent, you could write it like that."
    },
    {
        "Q": "1:47 why did you add -4x on that side? Not as in addition but you rewrote it as a subtraction sentence using -4x . -8 doesn't have a coefficient so why did Sal do that? I thought it was X minus X or X plus X. Really Confused Here !",
        "A": "In order to graph an equation, it is easiest if it is in y=mx+b form. Therefore you want y alone on the left. Since the 4x is hanging out with the 2y, by subtracting it to both sides it helps get y all by itself.",
        "video_name": "V6Xynlqc_tc",
        "timestamps": [
            107
        ],
        "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": "At 2:50 Sal says that the point at beginning of the interval is a relative maximum point. How do we know that to be true when the points on both sides of the point in question have to have a y-coordinate equal to or less than it, but then everything to the left of the point Sal was referring to was undefined?",
        "A": "Well, that means that there is nothing to the left so it only needs to worry about the existing or defined parts of the graph.",
        "video_name": "zXyQI4lD4wI",
        "timestamps": [
            170
        ],
        "3min_transcript": "They would be the bottom of your valleys. So that's a relative minimum point. This right over here is a relative minimum point, even if there are other parts of the function that are lower. Now there's also an edge case for both relative maxima and relative minima, and that's where the graph is flat. So if you have parts of your function where it's just constant, these points would actually be both. For example, if this is our x-axis right over here, that's our x-axis, if this is our y-axis right over there, and if this is x equals c, if you construct an open interval around c, you notice that the value of our function at c, f of c, is at least as large as the values of the function around it. And it is also at least as small as the values of the function around it, so this point would also be considered But that's an edge case that you won't encounter as often. So with that primer out of the way, let's identify the relative extrema. So first the relative maximum points. Well that's a top of a hill right over there, this is the top of a hill. You might be tempted to look at that point and that point, but notice, at this point right over here, if you go to the right, you have values that are higher than it. So it's really not at the top of a hill. And right over if you go to the left, you have values that are higher than it, so it's also not the top of a hill. And what about the relative minimum points? Well this one right over here is a relative minimum point. This one right over here is a relative minimum point. And this one over here is a relative minimum point. Now let's do an example dealing with absolute extrema. So here we're told to mark the absolute maximum and the absolute minimum points So once again, pause this video and see if you can have a go at this. So you have an absolute maximum point at let's say x equals c if and only if, so I'll write iff for if and only if, f of c is greater than or equal to f of x for all the x's in the domain of the function. And you have an absolute minimum at x equals c if and only if, iff, f of c is less than or equal to f of x for all the x's over the domain. So another way to think about it is, absolute maximum point is the high point. So over here, that is the absolute maximum point. And then the absolute minimum point is interesting because in this case, it would be actually one of, it would happen at one of the endpoints of our domain."
    },
    {
        "Q": "At 02:52, Sal says that the point (3,-8) would be a relative maximum point but how is that possible? The function is only till -8. How can we assume that the function will have the greatest value considering the points around it? I hope I made my question clear.",
        "A": "Sal told us at the beginning of the video that the domain was closed, that is, it included the end points. The domain is [-8,6]. On this particular graph, if we start at x=-8, and move towards the right on the x axis, the next immediate f(x) is less than it was at x=-8. Because we are on a closed interval, that makes the point (-8, 3) a relative maximum. (Make sure you put your x coordinate first when referring to a point on a graph\u00f0\u009f\u0098\u008a.)",
        "video_name": "zXyQI4lD4wI",
        "timestamps": [
            172
        ],
        "3min_transcript": "They would be the bottom of your valleys. So that's a relative minimum point. This right over here is a relative minimum point, even if there are other parts of the function that are lower. Now there's also an edge case for both relative maxima and relative minima, and that's where the graph is flat. So if you have parts of your function where it's just constant, these points would actually be both. For example, if this is our x-axis right over here, that's our x-axis, if this is our y-axis right over there, and if this is x equals c, if you construct an open interval around c, you notice that the value of our function at c, f of c, is at least as large as the values of the function around it. And it is also at least as small as the values of the function around it, so this point would also be considered But that's an edge case that you won't encounter as often. So with that primer out of the way, let's identify the relative extrema. So first the relative maximum points. Well that's a top of a hill right over there, this is the top of a hill. You might be tempted to look at that point and that point, but notice, at this point right over here, if you go to the right, you have values that are higher than it. So it's really not at the top of a hill. And right over if you go to the left, you have values that are higher than it, so it's also not the top of a hill. And what about the relative minimum points? Well this one right over here is a relative minimum point. This one right over here is a relative minimum point. And this one over here is a relative minimum point. Now let's do an example dealing with absolute extrema. So here we're told to mark the absolute maximum and the absolute minimum points So once again, pause this video and see if you can have a go at this. So you have an absolute maximum point at let's say x equals c if and only if, so I'll write iff for if and only if, f of c is greater than or equal to f of x for all the x's in the domain of the function. And you have an absolute minimum at x equals c if and only if, iff, f of c is less than or equal to f of x for all the x's over the domain. So another way to think about it is, absolute maximum point is the high point. So over here, that is the absolute maximum point. And then the absolute minimum point is interesting because in this case, it would be actually one of, it would happen at one of the endpoints of our domain."
    },
    {
        "Q": "At 4:06, wouldn't 6/13 plus 6/13 be equal to 12/13? How does it add to zero?",
        "A": "The term in the original equation was MINUS 6/13. Sal then added POSITIVE 6/13 to both sides. That s where the zero came from. When combining terms (or numbers) in algebra, you always have to take note of the sign as well as the numerical amount.",
        "video_name": "XD-FDGdWnR8",
        "timestamps": [
            246
        ],
        "3min_transcript": "1/3 plus 4/3 is indeed equal to 5/3. Let's do another one of these. So let's say that we have the equation K minus eight is equal to 11.8. So once again I wanna solve for K. I wanna have just a K on the left-hand side. I don't want this subtracting this eight right over here. So in order to get rid of this eight, let's add eight And of course, if I do it on the left-hand side, I have to do it on the right-hand side as well. So we're gonna add eight to both sides. The left-hand side, you are substracting eight and then you're adding eight. That's just going to cancel out, and you're just going to be left with K. And on the right-hand side, 11.8 plus eight. Well, 11 plus eight is 19, so it's going to be 19.8. And we're done, and once again, what's neat about equations, 19.8 minus eight is 11.8. Let's do another one, this is too much fun. Alright, so let's say that I had 5/13 is equal to T minus 6/13. Alright, this is interesting 'cause now I have my variable on the right-hand side. But let's just leave it there. Let's just see if we can solve for T by getting rid of everything else on the right-hand side. And like we've done in the past, if I'm subtracting 6/13, so why don't I just add it? Why don't I just add 6/13? I can't just do that on the right-hand side. Then the two sides won't be equal anymore, so I gotta do it on the left-hand side if I wanna hold the equality. So what happens? On the left-hand side I have, let me give myself a little bit more space, I have 5/13 plus 6/13, plus 6/13 are equal to, Well, I was subtracting 6/13, now I add 6/13. Those are just going to add to zero. 6/13 minus 6/13 is just zero, so you're left with T. So T is equal to this. If I have 5/13 and I add to that 6/13, well I'm gonna have 11/13. So this is going to be 11/13 is equal to T, or I could write that the other way around. I could write T is equal to 11/13."
    },
    {
        "Q": "at 4:45 is Sal writing (e^ln2)^u or (e)^ln2*u? I mean e^ln2 equals 2 and 2 is raised to the power u...so why should it be the latter and not the former?",
        "A": "if you remember your logarithmic rules, you will know that log(x^y) = y*log(x)",
        "video_name": "C5Lbjbyr1t4",
        "timestamps": [
            285
        ],
        "3min_transcript": "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 product of those exponents. So this is equal to, let me switch colors, this is equal to the integral of e, to the u, e to the, let me write it this way. e to the natural log of 2 times u. I'm just multiplying these two exponents. I raise something to something, then raise it again, we know from our exponent rules, it's just a product of those two exponents. du. Now, this is just a constant factor, right here. This could be, you know, this could just be some number. We could use a calculator to figure out what this is. We could set this equal to a. But we know in general that the integral, this is pretty straightforward, we've now put it in this form. The antiderivative of e to the au, du, is just 1 over a e to the au. This comes from this definition up here, and of course plus If we take the derivative of this, we take the derivative of the inside, which is just going to be a. We multiply that times the one over a, it cancels out, and we're just left with e to the au. So this definitely works out. So the antiderivative of this thing right here is going to be equal to 1 over our a, it's going to be 1 over our constant term, 1 over the natural log of 2 times our whole expression, e e. And I'm going to do something. This is just some number times u, so I can write it as u times some number. And I'm just doing that to put in a form that might help us simplify it a little bit. So it's e to the u times the natural log of 2, right? All I did, is I swapped this order. I could have written this as e to the natural log of 2 times u. If this an a, a times u is the same thing as u times a. Plus c. So this is our answer, but we have to kind of reverse"
    },
    {
        "Q": "At 6:01 we have,\n\n= exp [u*ln(2)] / ln(2) + C\n\nWhy can't we substitute u=ln(x) now and use ln(x)ln(2)=ln(x+2) ? i.e.,\n\n= exp [ln(x+2)] / ln(2) + C\n= (x+2) / ln(2) + C\nCan anyone see what I'm doing wrong?",
        "A": "You ve confused the property a little bit; ln(x)*ln(2) is not equal to ln(x+2). Rather, ln(x) + ln(2) = ln(2*x), and ln(x)*ln(2) = ln(2^(ln(x))). Hope that helps.",
        "video_name": "C5Lbjbyr1t4",
        "timestamps": [
            361
        ],
        "3min_transcript": "product of those exponents. So this is equal to, let me switch colors, this is equal to the integral of e, to the u, e to the, let me write it this way. e to the natural log of 2 times u. I'm just multiplying these two exponents. I raise something to something, then raise it again, we know from our exponent rules, it's just a product of those two exponents. du. Now, this is just a constant factor, right here. This could be, you know, this could just be some number. We could use a calculator to figure out what this is. We could set this equal to a. But we know in general that the integral, this is pretty straightforward, we've now put it in this form. The antiderivative of e to the au, du, is just 1 over a e to the au. This comes from this definition up here, and of course plus If we take the derivative of this, we take the derivative of the inside, which is just going to be a. We multiply that times the one over a, it cancels out, and we're just left with e to the au. So this definitely works out. So the antiderivative of this thing right here is going to be equal to 1 over our a, it's going to be 1 over our constant term, 1 over the natural log of 2 times our whole expression, e e. And I'm going to do something. This is just some number times u, so I can write it as u times some number. And I'm just doing that to put in a form that might help us simplify it a little bit. So it's e to the u times the natural log of 2, right? All I did, is I swapped this order. I could have written this as e to the natural log of 2 times u. If this an a, a times u is the same thing as u times a. Plus c. So this is our answer, but we have to kind of reverse the antiderivative with respect to x. But before I do that, let's see if I can simplify this a little bit. What is, if I have, just from our natural log properties, or logarithms, a times the natural log of b. We know this is the same thing as the natural log of b to the a. Let me draw a line here. Right? That this becomes the exponent on whatever we're taking the natural log of. So u, let me write this here, u times the natural log of 2, is the same thing as the natural log of 2 to the u. So we can rewrite our antiderivative as being equal to 1 over the natural log of 2, that's just that part here, times e to the, this can be rewritten based on this logarithm property, as the natural log of 2 to the u, and of course we still have our plus c there. Now, what is e raised to the natural log of 2 to the u? The natural log of 2 to the u is the power that you have to"
    },
    {
        "Q": "At 5:00, why does the integral of e^(au) become 1/a * e^(au)? Where did the 1/a come from? Thanks!",
        "A": "The chain rule, I presume would be the answer, since the expression uses a function of a function ( the thing you wrote up). Hope that helps :)",
        "video_name": "C5Lbjbyr1t4",
        "timestamps": [
            300
        ],
        "3min_transcript": "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 product of those exponents. So this is equal to, let me switch colors, this is equal to the integral of e, to the u, e to the, let me write it this way. e to the natural log of 2 times u. I'm just multiplying these two exponents. I raise something to something, then raise it again, we know from our exponent rules, it's just a product of those two exponents. du. Now, this is just a constant factor, right here. This could be, you know, this could just be some number. We could use a calculator to figure out what this is. We could set this equal to a. But we know in general that the integral, this is pretty straightforward, we've now put it in this form. The antiderivative of e to the au, du, is just 1 over a e to the au. This comes from this definition up here, and of course plus If we take the derivative of this, we take the derivative of the inside, which is just going to be a. We multiply that times the one over a, it cancels out, and we're just left with e to the au. So this definitely works out. So the antiderivative of this thing right here is going to be equal to 1 over our a, it's going to be 1 over our constant term, 1 over the natural log of 2 times our whole expression, e e. And I'm going to do something. This is just some number times u, so I can write it as u times some number. And I'm just doing that to put in a form that might help us simplify it a little bit. So it's e to the u times the natural log of 2, right? All I did, is I swapped this order. I could have written this as e to the natural log of 2 times u. If this an a, a times u is the same thing as u times a. Plus c. So this is our answer, but we have to kind of reverse"
    },
    {
        "Q": "at 0:42,what does reciprical mean?",
        "A": "The reciprocal of a fraction is simply the fraction flipped over. For example, the reciprocal of 6/9 is 9/6.",
        "video_name": "yb7lVnY_VCY",
        "timestamps": [
            42
        ],
        "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": "1:15 what does he mean",
        "A": "He means that if the weekend was a square split into 20 parts, one of those parts would be spent studying for a single subject (math, science, english, history, etc)",
        "video_name": "yb7lVnY_VCY",
        "timestamps": [
            75
        ],
        "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": "At 2:07, why do you have to do:\n(4y^2 -6y) + (10y - 15)\ninstead of\n(4y^2 +10y) + (-6y - 15) <-- it's clearly not the same thing right?\n\nThanks",
        "A": "Either way works as long as you can factor it, as both expressions are equivalent. One factors to 2y(2y-3) + 5(2y-3) = (2y+5)(2y-3) The other equals 2y(2y+5) - 3(2y +5) = (2y+5)(2y-3) Hope this helps!",
        "video_name": "u1SAo2GiX8A",
        "timestamps": [
            127
        ],
        "3min_transcript": "We're asked to factor 4y squared plus 4y, minus 15. And whenever you have an expression like this, where you have a non-one coefficient on the y squared, or on the second degree term-- it could have been an x squared-- the best way to do this is by grouping. And to factor by grouping we need to look for two numbers whose product is equal to 4 times negative 15. So we're looking for two numbers whose product-- let's call those a and b-- is going to be equal to 4 times negative 15, or negative 60. And the sum of those two numbers, a plus b, needs to be equal to this 4 right there. So let's think about all the factors of negative 60, or 60. And we're looking for ones that are essentially 4 apart, because the numbers are going to be of different signs, because their product is negative, so when you take two numbers of different signs and you sum them, you kind of view it as the difference of their absolute values. If that confuses you, don't worry about it. But this tells you that the numbers, since they're going to be of different size, their absolute values are going to 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."
    },
    {
        "Q": "(9:40 p.m.) Please help me with this problem. I've been out of school for years and I'm preparing the the GMAT. x^2 - 4x + 4 > or = 0",
        "A": "Factor it as (x-2)^2, and then you find that it s 0 when x is 2. So you test on either side of 2 to find what the sign will be. If you plug in 1 it s positive, and when you plug in 3 it s positive, so your solutions are all real numbers.",
        "video_name": "u1SAo2GiX8A",
        "timestamps": [
            580
        ],
        "3min_transcript": ""
    },
    {
        "Q": "In 4:21 how did he get the m<bec =60 degrees ?",
        "A": "Triangle BEC is equilateral which means that all angles are equal. All angles add up to 180 degrees. 180 divided by 3 equals 60 degrees.",
        "video_name": "3n0LvI99-KM",
        "timestamps": [
            261
        ],
        "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": "At 2:41 Khan didt divide -8 by 10 he just made it into -8/10 then divided it by 2 which was -4/5 doest he need to divide both sides by 10 because its an equation?",
        "A": "At about 2:05, Sal has 10x = -8 To solve an equation, you need to get to x = a number. x must be by itself. He multiplies both sides by 1/10 (which is the same as dividing by 10 to create x = -8/10 or x = -4/5. This is a solution. He has x by itself. If he now multiplies by 10, he would have 10x = -8 again. Hope this helps.",
        "video_name": "Zn-GbH2S0Dk",
        "timestamps": [
            161
        ],
        "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": "In 4:29, he puts (-1/3) -3x= 11 (-1/3) . What I'm confused about, is he taught us in I believe Linear Equations level 2, that you should do it the other way around. Example: (-3/1) -3x=11 (-3/1) Am I making sense lol? I mean I understand that if you do it out, it won't work but still its a bit confusing since it contradicts what I learned earlier.",
        "A": "You make sense, except you probably confused yourself Sal doesn t take the recipricol of 1/2 he multiples 1/2 Hope this helps!",
        "video_name": "Zn-GbH2S0Dk",
        "timestamps": [
            269
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "I didn't understand why the y intercept is -6. Can someone please help? (at 1:55)",
        "A": "If x = 0, you get y = 0\u00c2\u00b2 + 0 - 6 = -6",
        "video_name": "xdiBjypYFRQ",
        "timestamps": [
            115
        ],
        "3min_transcript": "Welcome to the presentation on quadratic inequalities. Before we get to quadratic inequalities, let's just start graphing some functions and interpret them and then we'll slowly move to the inequalities. Let's say I had f of x is equal to x squared plus x minus 6. Well, if we wanted to figure out where this function intersects the x-axis or the roots of it, we learned in our factoring quadratics that we could just set f of x is equal to 0, right? Because f of x equals 0 when you're intersecting the x-axis. So you would say x squared plus x minus 6 is equal to 0. And you just factor this quadratic. x plus 3 times x minus 2 equals 0. And you would learn that the roots of this quadratic How would we visualize this? Well let's draw this quadratic function. Those are my very uneven lines. So the roots are x is equal to negative 3. So this is, right here, x is at minus 3y0 -- by definition one of the roots is where f of x is equal to 0. So the y, or the f of x axis here is 0. The coordinate is 0. And this point here is 2 comma 0. Once again, this is the x-axis, and this is the f of x-axis. We also know that the y intercept is minus 6. This isn't the vertex, this is the y intercept. And that the graph is going to look something like this -- not general idea if you've ever seen a clean parabola. It looks like that with x minus 3 here, and x is 2 here. Pretty straightforward. We figured out the roots, we figured out what it looks like. Now what if we, instead of wanting to know where f of x is equal to 0, which is these two points, what if we wanted to know where f of x is greater than 0? What x values make f of x greater than 0? Or another way of saying it, what values make the statement true? x squared plus x minus 6 is greater than 0, Right, this is just f of x. Well if we look at the graph, when is f of x greater than 0? Well this is the f of x axis, and when are we Well f of x is greater than 0 here -- let me draw that another color -- is greater than 0 here, right? Because it's above the x-axis."
    },
    {
        "Q": "At 5:29, how does he go from x^2+3x-28 to (x+7)(x-4)??",
        "A": "By factoring.",
        "video_name": "xdiBjypYFRQ",
        "timestamps": [
            329
        ],
        "3min_transcript": "So just visually looking at it, what x values make this true? Well, this is true whenever x is less than minus 3, right, or whenever x is greater than 2. Because when x is greater than 2, f of x is greater than 0, and when x is less than negative 3, f of x is greater than 0. So we would say the solution to this quadratic inequality, and we pretty much solved this visually, is x is less than minus 3, or x is greater than 2. And you could test it out. You could try out the number minus 4, and you should get f of x being greater than 0. You could try it out here. Or you could try the number 3 and make sure that this works. 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."
    },
    {
        "Q": "Why do you have to change the sign > to <?? It looks something like you change both sides equations...but instead of doing that,you change the signs? Why not use = instead of </>?(at 5:13)",
        "A": "I use a simple example whenever I can. Say 10>0. If you change the sign on 10, then -10<0. The same exact change happens if there are multiple numbers, as long as you change all of the signs. BTW, if you have numbers on both sides of inequality, say, 10>9, you just change signs on both sides and flip the arrow. -10<-9. Hope this helps.",
        "video_name": "xdiBjypYFRQ",
        "timestamps": [
            313
        ],
        "3min_transcript": "So just visually looking at it, what x values make this true? Well, this is true whenever x is less than minus 3, right, or whenever x is greater than 2. Because when x is greater than 2, f of x is greater than 0, and when x is less than negative 3, f of x is greater than 0. So we would say the solution to this quadratic inequality, and we pretty much solved this visually, is x is less than minus 3, or x is greater than 2. And you could test it out. You could try out the number minus 4, and you should get f of x being greater than 0. You could try it out here. Or you could try the number 3 and make sure that this works. 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."
    },
    {
        "Q": "Hi,\n\nwhen sal took any vector from the original subset V and multiplied it with a vector from the \"orthogonal complement\" subspace at 10:05 , he found that their dot product was equal to 0.\nHow can we be sure that this is the case? I mean, the orthogonal subspace is supposed to have the vectors that are orthogonal to a specific subspace of vectors, not to any in V. I hope I made my question clear...",
        "A": "thats the definition of an orthogonal complement... the z axis is the orthogonal complement of the xy plane in a 3 dimensional space. obviously, any vector on the z axis will be orthogonal to vectors in the xy plane, even though the xy plane is a 2 dimensional subspace.",
        "video_name": "zlI8mx8Hc8o",
        "timestamps": [
            605
        ],
        "3min_transcript": "Let me draw rn again. Let me draw all of rn like that. Now, we have the orthogonal complement. Let me just draw that first. So v perp And then you have the orthogonal complement of the orthogonal complement which could be this set right here. Right? This is v perp. I haven't even drawn the subspace v. All I've shown is, I have some subspace here, which I happen to call v perp. And then I have the orthogonal complement of that subspace. So this means that anything in rn can be represented as the sum of a vector that's here and a vector that's here. So, if I say that w-- let me do it in purple. If I say the vector w-- let me write it this way. The vector v can be represented as the sum of the orthogonal complement of v or v perp And x is a member of its orthogonal complement. Notice, all I'm saying, I could have called this set s. And then this would have been s and its orthogonal complement. And we learned that anything in rn could be represented as the sum of something in a subspace and the subspace's orthogonal complement. So it doesn't matter that v is somehow related to this. It can be represented as a sum of a vector here plus a vector there. Fair enough. Now, what happens if I dot v with w? I'm doing the exact same argument that I did before. Well, if you take anything that's a member of our orthogonal complement, that's going to give us 0. What else is that going to be equal to? If we write v in this way, v dot w is the same thing as this thing dot w. So w plus x dot-- and this is going to be equal to w dot w plus x dot w. And what's x dot w? x is in the orthogonal complement of your orthogonal complement. And w is in the orthogonal complement. So if you take the dot product, you're going to get 0. They're orthogonal to each other. So this is just equal to w dot w or the length of w squared. And since since has to equal 0, we just have a bunch of equals here, that tells us that once again the vector w has to be equal to 0. So that tells us v is equal to w plus x."
    },
    {
        "Q": "At 4:04 why is the x negative?",
        "A": "The x is negative because after going down 4 for the change in y (negative 4), you have to go six to the left to meet up with the line again, (negative six). Going up is positive, going right is positive, left and down are negative.",
        "video_name": "R948Tsyq4vA",
        "timestamps": [
            244
        ],
        "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": "At about 4:10. Why do the negatives cancel out. Because wouldn't it be -4/6??",
        "A": "Review your sign rules. A negative divided by a negative = a positive. So, -4 / (-6) = +4/6",
        "video_name": "R948Tsyq4vA",
        "timestamps": [
            250
        ],
        "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": "at 3:38 how does he get negative change in y? I get why the x is cause he's moving in the opposite direction but the change in y seems the same as the first example at 2:10. why is it different?",
        "A": "In the first example, he was going from a lower point to a higher point, so there was a positive change in y. In the second example, he is going from a higher point to a lower point, so the y is going down. There is a negative change in y. Another way to think about it is this: The y value is going from 1 at the starting point to -3 at the ending point. You get from 1 to -3 by subtracting 4, so the change in y is -4.",
        "video_name": "R948Tsyq4vA",
        "timestamps": [
            218,
            130
        ],
        "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": "At 2:20 Sal says that the slope is 2/3. I am kinda confused about how you would draw that line using the information found out? Can someone please help me?",
        "A": "Okay, using the information Sal mentioned at 2:33 in the video you can t draw a line on a graph. This is because the equation of any linear line on a Cartesian plane can be delfined by the formula y=mx+b. Since we know that m, or the slope equals 2/3 we can add it in the equation. Without any point known on the line, though, we can t solve the equation because any point s coordinates could be plugged into the equation along with the slope to solve for b. Therefore the line is undefined. Hope this helps!",
        "video_name": "R948Tsyq4vA",
        "timestamps": [
            140
        ],
        "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 0:55 when you are talking about the starting point, what if a certain graph/problem does not give you a certain starting point. Would you start at (0,0)",
        "A": "A problem would not be worded like that, it would be more clear.",
        "video_name": "R948Tsyq4vA",
        "timestamps": [
            55
        ],
        "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": "2:28 How the heck did he get a NEGATIVE 32?! What made it negative???? That concept has never been explained to me before!",
        "A": "because he multiply 8 * (-4) :) if you have 8(3x - 4), youll have 8 *(3x) + 8*(-4) and this is 24x + (-32), which is basically 24x - 32 :) in this case you have 8(3/8 x - 4 ) so you are left with 3x - 32",
        "video_name": "PL9UYj2awDc",
        "timestamps": [
            148
        ],
        "3min_transcript": "We have the equation 3/4x plus 2 is equal to 3/8x minus 4. Now, we could just, right from the get go, solve this the way we solved everything else, group the x terms, maybe on the left-hand side, group the constant terms on the right-hand side. But adding and subtracting fractions are messy. So what I'm going to do, right from the start of this video, is to multiply both sides of this equation by some number so I can get rid of the fractions. And the best number to do it by-- what number is the smallest number that if I multiply both of these fractions by it, they won't be fractions anymore, they'll be whole numbers? That smallest number is going to be 8. I'm going to multiply 8 times both sides of this equation. You say, hey, Sal, how did you get 8? And I got 8 because I said, well, what's the least common multiple of 4 and 8? Well, the smallest number that is divisible by 4 and 8 is 8. get rid of the fractions. And so let's see what happens. So 8 times 3/4, that's the same thing as 8 times 3 over 4. Let me do it on the side over here. That's the same thing as 8 times 3 over 4, which is equal to 8 divided by 4 is just 2. So it's 2 times 3, which is 6. So the left-hand side becomes 8 times 3/4x is 6x. And then 8 times 2 is 16. You have to remember, when you multiply both sides, or a side, of an equation by a number, you multiply every term by that number. So you have to distribute the 8. So the left-hand side is 6x plus 16 is going to be equal to-- 8 times 3/8, that's pretty easy, the 8's cancel out and you're just left with 3x. And then 8 times negative 4 is negative 32. And now we've cleaned up the equation a good bit. Now the next thing, let's try to get all the x terms on the left-hand side, and all the constant terms on the right. Let's subtract 3x from both sides to do it. That's the best way I can think of of getting rid of the 3x from the right. The left-hand side of this equation, 6x minus 3x is 3x. 6 minus 3 is 3. And then you have a plus 16 is equal to-- 3x minus 3x, that's the whole point of subtracting 3x, is so they cancel out. So those guys cancel out, and we're just left with a negative 32. Now, let's get rid of the 16 from the left-hand side. So to get rid of it, we're going to subtract 16 from both sides of this equation. Subtract 16 from both sides. The left-hand side of the equation just becomes-- you have this 3x here; these 16's cancel out, you don't have to write anything-- is equal to negative 32 minus 16 is negative 48. So we have 3x is equal to negative 48."
    },
    {
        "Q": "at 2:06 how do you go from 1 to 3",
        "A": "I think Sal is trying to show you that the slope remains constant across the entire line. He started with 2/1 Then, he picked another 2 points and the slope was 6/3. You need to recognize that these are = values. If reduce / divide 6/3, you get 2/1. Same slope. He could have gone up 10 and to the right 5. he would still be on the line. Slope would be: 10/5, which still = 2/1. Hope this helps. Hope that helps.",
        "video_name": "MeU-KzdCBps",
        "timestamps": [
            126
        ],
        "3min_transcript": "- [Voiceover] As we start to graph lines, we might notice that they're differences between lines. For example, this pink or this magenta line here, it looks steeper than this blue line. And what we'll see is this notion of steepness, how steep a line is, how quickly does it increase or how quickly does it decrease, is a really useful idea in mathematics. So ideally, we'd be able to assign a number to each of these lines or to any lines that describes how steep it is, how quickly does it increase or decrease? So what's a reasonable way to do that? What's a reasonable way to assign a number to these lines that describe their steepness? Well one way to think about it, could say well, how much does a line increase in the vertical direction for a given increase in the horizontal direction? So let's write this down. So let's say if we an increase increase, in vertical, in vertical, for a given increase a given increase in horizontal. So, how can this give us a value? Well let's look at that magenta line again. Now let's just start at an arbituary point in that magenta line. But I'll start at a point where it's going to be easy for me to figure out what point we're at. So if we were to start right here, and if I were to increase in the horizontal direction by one. So I move one to the right. To get back on the line, how much do I have to increase in the vertical direction? Well I have to increase in the vertical direction by two. So at least for this magenta line, it looks like our increase in vertical is two, whenever we have an increase in one in the horizontal direction. Let's see, does that still work if I were to direction by one, if I were increase in the horizontal direction... So let's increase by three. So now, I've gone plus three in the horizontal direction, then to get back on the line, how much do I have to increase in the vertical direction? I have to increase by one, two, three, four, five, six I have to increase by six. So plus six. So when I increase by three in the horizontal direction, I increase by six in the vertical. We were just saying, hey, let's just measure how much to we increase in vertical for a given increase in the horizontal? Well two over one is just two and that's the same thing as six over three. So no matter where I start on this line, no matter where I start on this line, if I take and if I increase in the horizontal direction by a given amount, I'm going to increase twice as much twice as much in the vertical direction. Twice as much in the vertical direction."
    },
    {
        "Q": "After looking at 5:08, does that mean the slope of a graph basically is a rate of change in a table and graph? In a nutshell, the rate of change and the slope of a graph is the same, right?",
        "A": "Yes you are correct. Adding on to that, it can really be applied to Physics. For example, if you found the slope for a velocity over time graph, you get the velocity. See, isnt that cool? like how everything is connected in this universe.",
        "video_name": "MeU-KzdCBps",
        "timestamps": [
            308
        ],
        "3min_transcript": "divided by increase in horizontal, this is what mathematicians use to describe the steepness of lines. And this is called the slope. So this is called the slope of a line. And you're probably familiar with the notion of the word slope being used for a ski slope, and that's because a ski slope has a certain inclination. It could have a steep slope or a shallow slope. So slope is a measure for how steep something is. And the convention is, is we measure the increase in vertical for a given in increase in horizontal. So six two over one is equal to six over three is equal to two, this is equal to the slope of this magenta line. So let me write this down. So this slope right over here, the slope of that line, is going to be equal to two. And one way to interpret that, for whatever amount you increase in the horizontal direction, you're going to increase twice as much in the vertical direction. What would be the slope of the blue line? Well, let me rewrite another way that you'll typically see the definition of slope. And this is just the convention that mathematicians have defined for slope but it's a valuable one. What is are is our change in vertical for a given change in horizontal? And I'll introduce a new notation for you. So, change in vertical, and in this coordinate, the vertical is our Y coordinate. divided by our change in horizontal. And X is our horizontal coordinate in this coordinate plane right over here. So wait, you said change in but then you drew this triangle. Well this is the Greek letter delta. This is the Greek letter delta. And it's a math symbol used to represent change in. So that's delta, delta. And it literally means, change in Y, change in Y, change in X. So if we want to find the slope of the blue line, we just have to say, well how much does Y change for a given change in X? So, the slope of the blue line. So let's see, let me do it this way. Let's just start at some point here. And let's say my X changes by two so my delta X is equal to positive two. What's my delta Y going to be? What's going to be my change in Y? Well, if I go by the right by two, to get back on the line, I'll have to increase my Y by two. So my change in Y is also going to be plus two. So the slope of this blue line, the slope of the blue line, which is change in Y over change in X. We just saw that when our change in X is positive two, our change in Y is also positive two. So our slope is two divided by two, which is equal to one."
    },
    {
        "Q": "2:00 What if the line is not strait so the slope cannot only have one number?",
        "A": "Good question, you re getting a bit ahead of yourself though. There are other types of equations that will describe curved lines.",
        "video_name": "MeU-KzdCBps",
        "timestamps": [
            120
        ],
        "3min_transcript": "- [Voiceover] As we start to graph lines, we might notice that they're differences between lines. For example, this pink or this magenta line here, it looks steeper than this blue line. And what we'll see is this notion of steepness, how steep a line is, how quickly does it increase or how quickly does it decrease, is a really useful idea in mathematics. So ideally, we'd be able to assign a number to each of these lines or to any lines that describes how steep it is, how quickly does it increase or decrease? So what's a reasonable way to do that? What's a reasonable way to assign a number to these lines that describe their steepness? Well one way to think about it, could say well, how much does a line increase in the vertical direction for a given increase in the horizontal direction? So let's write this down. So let's say if we an increase increase, in vertical, in vertical, for a given increase a given increase in horizontal. So, how can this give us a value? Well let's look at that magenta line again. Now let's just start at an arbituary point in that magenta line. But I'll start at a point where it's going to be easy for me to figure out what point we're at. So if we were to start right here, and if I were to increase in the horizontal direction by one. So I move one to the right. To get back on the line, how much do I have to increase in the vertical direction? Well I have to increase in the vertical direction by two. So at least for this magenta line, it looks like our increase in vertical is two, whenever we have an increase in one in the horizontal direction. Let's see, does that still work if I were to direction by one, if I were increase in the horizontal direction... So let's increase by three. So now, I've gone plus three in the horizontal direction, then to get back on the line, how much do I have to increase in the vertical direction? I have to increase by one, two, three, four, five, six I have to increase by six. So plus six. So when I increase by three in the horizontal direction, I increase by six in the vertical. We were just saying, hey, let's just measure how much to we increase in vertical for a given increase in the horizontal? Well two over one is just two and that's the same thing as six over three. So no matter where I start on this line, no matter where I start on this line, if I take and if I increase in the horizontal direction by a given amount, I'm going to increase twice as much twice as much in the vertical direction. Twice as much in the vertical direction."
    },
    {
        "Q": "What does Sal mean at 3:30 when he says \"Let's see if we can break down 115 any further\"? What method does he use, and how does it work?",
        "A": "He is literally trying to factorize or break down 115 as much as possible.The method he has showed you is prime factorization . In this you find 2 numbers that when multiplied give you a certain result. you keep doing this until you only have prime numbers.It can also be called a factor tree.",
        "video_name": "O64YFlX1_aI",
        "timestamps": [
            210
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 1:44 how is it a ray without a arrow on the other end?",
        "A": "Think of it with everyday things. Take for example sun rays. they go in one direction until they hit an object (us, the atmosphere or a mirror ect.). Do they have arrows? As long as one line looks like it is going to continue on and on, they it should be considered a ray.",
        "video_name": "DkZnevdbf0A",
        "timestamps": [
            104
        ],
        "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 5:28 , I dont understand why Sal put +(-14). it is so confusing.",
        "A": "He meant 6 plus negative 14, which is 6 minus 14. Hopefully this helped you.",
        "video_name": "-4bTgmmWI9k",
        "timestamps": [
            328
        ],
        "3min_transcript": "Our goal is four right over here. So this is one, two, three, four, get to positive four. That's positive four right there. We're starting at negative two... Let me do this in a different color. We're starting at negative two. We're saying four is the same thing as negative two, negative two is right over here, negative two plus some amount. And it's clear we're gonna be moving to the right by, let's see, we're gonna move to the right by one, two, three, four, five, six. So we moved to the right by six. So we added six. So negative two plus six is equal to four. This is fascinating. Actually, let's just do several more of these, I can't stop. (laughs) Alright. So let's say we wanted to figure out, Six plus blank is equal to negative eight. Like always, try to pause the video and figure out what this blank is going to be. Let me throw my number line back here. So my number line. And one way to think about it is I am starting at six. So it's five, this is six right over here. And I'm gonna add something to get to negative eight. To get to negative eight this is negative five, negative six, negative seven, negative eight. I want to get right over here, I want to get to negative eight. So what do I have to do to get from six to negative eight? To go from six to negative eight. We're clear I have to go to the left on the number line. And how much do I have to go to the left? Well, let's count it. I have to go one, two, three, four, five, six, seven, eight, nine, So going 14 to the left, you could say that I just subtracted 14. And if we phrase it as six plus what is equal to negative eight? Well, six plus negative 14. If this said six minus something I could have just said six minus 14, but if it's saying six plus what it's going to be six plus negative 14."
    },
    {
        "Q": "what is a checker board pattern? Sal mentions it at 0:20",
        "A": "A checkerboard pattern is when every other square is black or white both in the horizontal and vertical directions. With numbers instead of colors, it could look like this: 1 0 1 0 1 0 1 0 1",
        "video_name": "u00I3MCrspU",
        "timestamps": [
            20
        ],
        "3min_transcript": "As a hint, I will take the determinant of another 3 by 3 matrix. But it's the exact same process for the 3 by 3 matrix that you're trying to find the determinant of. So here is matrix A. Here, it's these digits. This is a 3 by 3 matrix. And now let's evaluate its determinant. So what we have to remember is a checkerboard pattern when we think of 3 by 3 matrices: positive, negative, positive. So first we're going to take positive 1 times 4. So we could just write plus 4 times 4, the determinant of 4 submatrix. And when you say, what's the submatrix? Well, get rid of the column for that digit, and the row, and then the submatrix is what's left over. So we'll take the determinant of its submatrix. So it's 5, 3, 0, 0. Then we move on to the second item in this row, in this top row. But the checkerboard pattern says we're going to take the negative of it. So it's going to be negative of negative 1-- times the determinant of its submatrix. You get rid of this row, and this column. You're left with 4, 3, negative 2, 0. And then finally, you have positive again. Positive times 1. This 1 right over here. Let me put the positive in that same blue color. So positive 1, or plus 1 or positive 1 times 1. Really the negative is where it got a little confusing on this middle term. But positive 1 times 1 times the determinant of its submatrix. So it's submatrix is this right over here. You get rid of the row, get rid of the column 4, 5, negative 2, 0. So the determinant right over here is going to be 5 times 0 minus 3 times 0. And all of that is going to be multiplied times 4. Well this is going to be 0 minus 0. So this is all just a 0. So 4 times 0 is just a 0. So this all simplifies to 0. Now let's do this term. We get negative negative 1. So that's positive 1. So let me just make these positive. Positive 1, or we could just write plus. Let me just write it here. So positive 1 times 4 times 0 is 0. So 4 times 0 minus 3 times negative 2. 3 times negative 2 is negative 6. So you have 4-- oh, sorry, you have 0 minus negative 6, which is positive 6. Positive 6 times 1 is just 6. So you have plus 6. And then finally you have this last determinant."
    },
    {
        "Q": "I don't understand the intersect stuff @2:52. Is there a video that talks about this? Or could someone explain it for me? Thx :)",
        "A": "What don t you understand about it",
        "video_name": "VTlvg4wJ1X0",
        "timestamps": [
            172
        ],
        "3min_transcript": "If something-squared is equal to four, that means that the something, that means that this something right over here, is going to be equal to the positive square root of four or the negative square root of four. Or it's gonna be equal to positive or negative two. And so we could write that x plus three could be equal to positive two or x plus three could be equal to negative two. Notice, if x plus three was positive two, two-squared is equal to four. If x plus three was negative two, negative two-squared is equal to four. So, either of these would satisfy our equation. So, if x plus three is equal to two, we could just subtract three from both sides to solve for x and we're left with x is equal to negative one. Or, over here we could subtract three from both sides to solve for x. So, or, x is equal to negative two minus three So, those are the two possible solutions and you can verify that. Take these x-values, substitute it back in, and then you can see when you substitute it back in if you substitute x equals negative one, then x plus three is equal to two, two-squared is four, minus four is zero. And when x is equal to negative five, negative five plus three is negative two, squared is positive four, minus four is also equal to zero. So, these are the two possible x-values that satisfy the equation. Now let's do another one that's presented to us in a slightly different way. So, we are told that f of x is equal to x minus two squared minus nine. And then we're asked at what x-values does the graph of y equals f of x intersect the x-axis. So, if I'm just generally talking about some graph, so I'm not necessarily gonna draw that y equals f of x. So if I'm just, so that's our y-axis, this is our x-axis. And so if I just have the graph of some function. 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."
    },
    {
        "Q": "4:14, Shouldn't it be the first n+1 squares since we're starting at 0 or do most people start at 1 so it's the first n squares? If the latter, why is Sal starting at 0?",
        "A": "I think he started at 0 to eliminate the D term (in the previous video). Our Sigma in this starts at 0, also, so he had to start at 0 to evaluate the formula.",
        "video_name": "MkGXR8umLco",
        "timestamps": [
            254
        ],
        "3min_transcript": "is equal to negative 9. And now if we add both sides, on the left-hand side, we have-- let's see. 24A minus 18A is 6A-- these cancel-- is equal to 11 minus 9 is 2. Divide both sides by 6. We get A is equal to 2/6, which is the same thing as 1/3. And so now we can substitute back to solve for B. So let's see. We have 6 times 1/3. Our A is 1/3. Plus 2B is equal to 3. 6 times 1/3 is 2. 2 plus 2B is equal to 3. Subtract 2 from both sides. 2B is equal to 1. Divide both sides by 2. So A is 1/3. B is 1/2. Now we just have to solve for C. So we can go back to this original equation right over here. So we have 1/3 plus 1/2 plus C is equal to 1. Actually, let me do that over here so I have some space. So I have some space, let me do it right over here. So I have 1/3 plus 1/2 plus C is equal to 1. Now we find ourselves a common denominator. So let's see. A common multiple of 3, 2, and-- I guess you could say 1-- this is 1 over 1-- is going to be 6. So I can rewrite this as 2/6 plus 3/6 plus C is equal to 6/6. 1 is the same thing as 6/6. So this is 5/6 plus C is equal to 6/6. Subtract 5/6 from both sides. We get C is equal to 1/6. And there we are. We deserve a drum roll now. We've figured out a formula for the sum of the first n squares. So we can rewrite this. This formula is now going to be-- A is 1/3. So it's 1/3 n to the third power plus 1/2 n squared-- let me make sure it doesn't look like that n is a-- 1/2 n squared plus 1/6 n. So this is a pretty handy formula. If now you wanted to find 0 squared plus 1 squared plus 2 squared plus 3 squared, all the way to 100 squared, instead of squaring 100 numbers and then adding them together, you could figure out 1/3 of 100 to the third power plus 1/2 times 100 squared plus 1/6 times 100."
    },
    {
        "Q": "Okay at 2:13 on question 4 he says Z is 1 but then he says that Z is over 4. I'm really confused about that how can he say one thing but it mean another?",
        "A": "The equation is 1/4 * Z So to solve you do: 1/4 * Z/1 (remember Z/1 is still equal to Z) Then you multiply across, first the two numerators (1*Z which equals 1Z or Z) then the two denominators (4*1 which equals 4) And your final answer is Z/4",
        "video_name": "aoXUWSwiDzE",
        "timestamps": [
            133
        ],
        "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": "@ 4:45, Sal skips from question 6 to question 10.\nCould he possibly either do a re upload where he does all the questions, or maybe someone provide the answers to the rest of the questions so that those who are working directly from the sheet can see if their workings are correct.",
        "A": "He is not trying to answer all the questions. The whole point of these videos is to help you understand the concepts. The only reason you could think these videos are not doing their jobs is if, even after you watch the video, you are confused on the topic. If that is true, you might want to try talking to a teacher or parent to help you delve further into the topic. After all, they probably know different ways to help you learn. Not everyone is a visual learner and learns through videos.",
        "video_name": "aoXUWSwiDzE",
        "timestamps": [
            285
        ],
        "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": "From 15:32 to 15:43, Sal mentions his wife being a doctor, and if people don't see the decimal point they'll overdose on medication. What does he mean by that?",
        "A": "If you are suppose to give a patient 3.2 mg of some medicine And, instead you lose the decimal point and give them 32mg of medicine. You will have given your patient 10 times the amount of medicine they should have. Depending on the medicine, this can have drastic consequences.",
        "video_name": "trdbaV4TaAo",
        "timestamps": [
            932,
            943
        ],
        "3min_transcript": "the number-- what if we had a 7-- let me think of it this way. Let's say we had a 7, 3 over there. So what would we do? Well, we'd want to go to the first digit right here because this is kind of the largest power of 10 that could go into this thing right here. So if we wanted to represent that thing, let me do another decimal that's like that one. So let's say I did 0.0000516 and I wanted to represent this in scientific notation. I'd go to the first non-digit 0-- the first non-zero digit, not non-digit 0, which is there. And I'm like, OK, what's the largest power of 10 that will fit into that? So I'll go 1, 2, 3, 4, 5. So it's going to be equal to 5.16. So I take 5 there, then everything else is going to be behind the decimal point. So this is going to be the largest power of 10 that fits into this first non-zero number. So it's 1, 2, 3, 4, 5. So 10 to the minus 5 power. Let me do another example. So the point I wanted to make is you just go to the first-- if you're starting at the left, the first non-zero number. That's what you get your power from. That's where i got my 10 to the minus 5 because I counted 1, 2, 3, 4, 5. You got to count that number just like we did over here. And then, everything else will be behind the decimal. Let me do another example. Let's say I had 0.-- and my wife always point out that I have to write a 0 in front of my decimal points because she's a doctor. And if people don't see the decimal point, someone might overdose on some medication. So let's write it her way, 0.0000000008192. Clearly, this is a super cumbersome number to write. And you might forget about a 0 or add too many 0's, which could be costly if you're doing some important scientific research. 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,"
    },
    {
        "Q": "In 6:22, how does it have 23 zeros? 602200000000000000000000 (phew) only has 21 zeros.",
        "A": "He was saying 23 digits after the 6.",
        "video_name": "trdbaV4TaAo",
        "timestamps": [
            382
        ],
        "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": "In the video, at 11:10, Sal said that scientific notation can be used to express any number in an easier way. Does that imply fractions and integers too? Or is it only natural numbers?",
        "A": "I m really not sure. I think so, but I don t think that includes repeating decimals. I m somewhat new to scientific notation too.",
        "video_name": "trdbaV4TaAo",
        "timestamps": [
            670
        ],
        "3min_transcript": "And I wanted to write it in scientific notation. So I guess the best way to think about it is, it's 7,345. So how can I represent a thousand? Well, I wrote it over here, 10 to the third is 1,000. So we know that 10 to the third is equal to 1,000. So that's essentially the largest power of 10 that I can fit into this. This is seven 1,000's. So if this is seven 1,000's, and then it's 0.3 1,00's, then it's 0.4 1,000's-- I don't know if that helps you, we can write this as 7.345 times 10 to the third because it's going to be seven 1,000's plus 0.3 1,000's. What's 0.3 times 1,000? 0.3 times 1,000 is 300. What's 0.04 times 1,000? That's 40. What's 0.05 times 1,000? That's a 5. So 7.345 times 1,000 is equal to 7,345. So if I took 7.345 times 1,000. The way I do it is I ignore the 0's. I essentially multiply 1 times that guy up there. So I get 7, 3, 4, 5. Then I had three 0's here, so I put those on the end. And then I have three decimal places. So 1, 2, 3. Put the decimal right there. And there you have it, 7.345 times 1,000 is indeed 7,345. Let's do a couple of them. Let's say we wanted to write the number 6 in scientific notation. Obviously, there's no need to write in scientific notation. But how would you do it? Well, what's the largest power of 10 that fits into 6? Well, the largest power of 10 that fits into 6 is just 1. So we could write it as something times 10 to the 0. This is just 1, right? That's just 1. 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."
    },
    {
        "Q": "at 8:04 Avogadro's number is written as 6.022 x 10^23 but shouldn't it be 6.022 x 10^20 because there is 20 zeroes after the 6022?",
        "A": "Everything is fine in the video, notice that there is a dot between 6 and 0. Of course 6.022x 10^23 = 6022x 10^20, but 6022x 10^20 is not a scientific notation, because scientific notation is a number that is at least 1 and less than 10 multiplied by a power of 10.",
        "video_name": "trdbaV4TaAo",
        "timestamps": [
            484
        ],
        "3min_transcript": "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. as some multiple of this guy? Because if we multiplied this guy by 6-- if we multiply 6 times 10 to the 23rd, what do we get? Well, we're just going to have a 6 with twenty three 0's. We're going to have a 6, and then you're going to have twenty three 0's. You're going to have twenty three 0's like that. Because all I did, if you take 6 times this. You know how to multiply. You'd have the 6 times this 1. And then all the 6 times the 0's will all be 0. So you'll have 6 followed by twenty three 0's. So that's pretty useful. But still, we're not getting quite to this number. I mean, this had some 2's in there. So how could we do it a little bit better? Well, what if we wrote it as a decimal? This number right here is identical to this number if these 2's were 0's. But if we want to put those 2's there, what can we do? We could put some decimals here. We could say that this is the same thing as 6.022 times 10 And now, this number is identical to this number, but it's a much easier way to write it. And you could verify it, if you like. It will take you a long time. Maybe we should do it with a smaller number first. But if you multiply 6.022 times 10 to the 23rd, and you write it all out, you will get this number right there. You will get Avogadro's number. Avogadro's number. And although this is complicated or it looks a little bit unintuitive to you at first, this was just a number written out. This has a multiplication and then a 10 to a power. You might say, hey, that's not so simple. But it really is. Because you immediately know how many 0's there are. And it's obviously a much shorter way to write this number. Let's do a couple of more. I started with Avogadro's number because it really shows you the need for a scientific notation. So you don't have to write things like that over and over again. So let's do a couple of other numbers. And we'll just write them in scientific notation."
    },
    {
        "Q": "At 2:00 Sal says Avogadro's number has \" 6 followed by the 23 digits or the 6022 followed by 20 zeros\". Then at 7:15 we multiply 6*10^23, that is 6 followed by 23 digits. At 8:01 we do 6.022*10^23. My question is why does he multiply it by the power of 23 ? Isn't the answer now have 6 and 26 digits now?",
        "A": "In both cases, the leading digit before the decimal point is just the 6 . In 6.022 , the other 3 values are to the right of the decimal point. The 10^23 moves the decimal point from where it is currently located in the number. Another way to compare them... You originally stated 6022 followed by 20 zeroes. In this situation, the decimal point is at the end of 6022. To get to 6.022 we only moved the decimal 3 places. So, we need 20+3 = 23 as our exponent. Hope this makes sense.",
        "video_name": "trdbaV4TaAo",
        "timestamps": [
            120,
            435,
            481
        ],
        "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": "So, for clarification, scientific notation is an easier way to write 12,000,000,000,000. So how do you know where to put the decimal place in, say, 6,510,000,000,000? Is there a way to know? I'm sorry if its in the video, I'm only at 3:46.",
        "A": "The scientific notation formal format is x*10^y, where 1 <= x < 10 and y is an integer. In other words, when writing in scientific notation, the number you multiply the power of ten must be between 1 (included) and 10. Here are the numbers you suggested as an example: 12,000,000,000,000 = 1.2*10^13 6,510,000,000,000 = 6.51*10^12",
        "video_name": "trdbaV4TaAo",
        "timestamps": [
            226
        ],
        "3min_transcript": "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. 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"
    },
    {
        "Q": "I don't get problem 13 at 8:31\nCan anyone help me?",
        "A": "For the 2 triangles to be similar the corresponding angles all have to have congruent measurements. For obvious reasons, angle DBE is the same on both tri. For the other 2 corresponding angles to be proven to be congruent, the 2 sides AC and DE must be parallel to eachother. This is true because the transversal of line AB would make the 2 corresponding angles of the triangle congruent ecause of the corresponding angles of the transversal are these angles. This is true for the other transversal as well.",
        "video_name": "bWTtHKSEcdI",
        "timestamps": [
            511
        ],
        "3min_transcript": "Whatever angle this us, let's call this x. Let's call this angle y and this angle y. We know that x plus 2y is equal to 180, or that 2y is equal to 180 minus x, Or y is equal to 90 minus x over 2. Now if this is x, And let's call these z and z, So we know that x plus 2z is equal to 180. All the angles in a triangle have to add up to 180. Subtract x from both sides, you get 2z is equal to 180 minus x. Divide by 2, you get z is equal to 90 minus x over 2. So z and y are going to be the same angles. So all the angles are the same, so we're dealing with similar triangles. So choice D was definitely correct. 13. Which of the following facts would be sufficient to prove that triangles ABC, that's the big triangle, and triangle DBE, so that's a small one, are similar? So we have to prove that all of their angles are similar. I cannot even look at the choices and I can guess where So we want to prove that those are similar. So first of all, they share the same angle. Angle ABC, this angle, is the same as angle DBE. So they share that same angle. So we got one angle down. Now let's think about it. If we knew that this angle is equal to that angle and that angle is equal to that angle, we'd be done. And the best way to come to that conclusion is if they told us that this and this are parallel. I'm guessing that's where they're going. Now I might have gone on a completely wrong tangent. Because it those two are parallel, then these two lines So that angle and that angle would be corresponding angles, so they would be congruent, and then that angle and that angle would be corresponding angles so they'd also be congruent. So if they told us that these are parallel, we're done. These are definitely similar triangles. And sure enough, choice C, they tell us that AC and DE These are parallel, that's a transversal, this is a corresponding angle of that, so they're congruent. This is a corresponding angle to this, congruent, so all of the angles are congruent. So we have a similar triangle. Problem 14. OK."
    },
    {
        "Q": "weren't all the sides on the paralellogram on 10:34 the same?",
        "A": "There are four paralellograms: Rhombus, Rhomboid, Rectangle, and a Square. The square is the only paralellogram must have four equal sides. While a Rhombus can have 4 equal lengths, for this problem, it is not a given that this is the case. We are looking for proof only that triangles inside the paralellogram are congruent. The answer to your question is the angles could be the same, either way, you should be able to get the right answer even if they were not.",
        "video_name": "bWTtHKSEcdI",
        "timestamps": [
            634
        ],
        "3min_transcript": "So that angle and that angle would be corresponding angles, so they would be congruent, and then that angle and that angle would be corresponding angles so they'd also be congruent. So if they told us that these are parallel, we're done. These are definitely similar triangles. And sure enough, choice C, they tell us that AC and DE These are parallel, that's a transversal, this is a corresponding angle of that, so they're congruent. This is a corresponding angle to this, congruent, so all of the angles are congruent. So we have a similar triangle. Problem 14. OK. Fair enough. Parallelogram: that tells us that the opposites sides are parallel. That's parallel to that, and then this is parallel to that. And all of the choices got clipped at the bottom, but I'll copy them over. Maybe I'll copy them above the question. Well, let me see what I can do. I think that's good enough. A little unconventional. OK. Parallelogram is shown below. They say which pair of triangles can be established to be congruent to prove that angle DAB is congruent to angle BCD? So DAB is this. Let me do it in another color. DAB is that angle, is congruent to BCD. the same angle measure. OK, and what do we have to show? They say what pair of triangles can be established to be congruent to prove that. OK, if these are both part of two different congruent triangles and they are the corresponding angles, then we know that they're congruent and we'd be done. So let's see what they say. Triangle ADC and BCD. BCD has this angle in it. BCD does help us because it has this angle in it, but triangle ADC does not have this angle in it, right? Triangle ADC has this the smaller angle in it. ADC doesn't involve this whole thing, so that's not going to help us. Triangle AED, once again, does not involve this larger angle, does not involve the angle DAB. It only involves the little smaller angle, so that's not going to help us. Triangle DAB."
    },
    {
        "Q": "At 5:40, Sal says that if you have two sides and an angle, you can figure everything else out. The problem here is that SSA Congruence is NOT a valid congruence. SSA Congruence includes two sides and an angle. This is because you most likely end up with two solutions for the third side. Does he mean that if you are given two sides and an angle and the angle is between the sides, then you could find everything else?",
        "A": "I think when you have a SSA triangle you would just list both possible answers. Like if you were asked to find side C of an SSA triangle you say C=x or C=y",
        "video_name": "VjmFKle7xIw",
        "timestamps": [
            340
        ],
        "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 2:50, how did you get 1/4?",
        "A": "1/2 divided by 2 = 1/4 he got the 1/2 because sin 30 = 1/2",
        "video_name": "VjmFKle7xIw",
        "timestamps": [
            170
        ],
        "3min_transcript": "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, 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."
    },
    {
        "Q": "At around 4:30, why do you need to take the reciprocal of both sides to solve the law of sines?",
        "A": "The goal was to isolate the variable. There are several ways of accomplishing this, but since the variable was in the denominator, taking the reciprocal of both sides seemed a useful choice.",
        "video_name": "VjmFKle7xIw",
        "timestamps": [
            270
        ],
        "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 12:16 why does Sal evaluate one of the Sin(t)'s at Pi all of a sudden (instead of Pi/2)? I got for that problem:\n2Pi + 2 and sal has 4 + 2Pi. Is this a mistake?",
        "A": "The second expression with a sine was sin(2t), which, evaluated at pi/2, is sin(2pi/2)=sin(pi)=0",
        "video_name": "wyTjyQMVvc4",
        "timestamps": [
            736
        ],
        "3min_transcript": "all of that dt. Now this should be reasonably straight forward to get the antiderivative of. Let's just take it. The antiderivative of this is antiderivative of cosine of t; that's a sine of t. The derivative of sine is cosine. So this is going to be 4 sine of t-- the scalars don't affect anything --and then, well let me just distribute this 4. So this is 4 times 1 which is 4 minus 4 cosine of 2t. So the antiderivative of 4 is 4t-- plus 4t --and then the antiderivative of minus 4 cosine of u00b5 t? Let's see it's going to be sine of 2t. The derivative of sine of 2t is 2 cosine of 2t. We're going to have to have a minus sign there, and put a 2 What's the derivative of minus 2 sine of t? Take the derivative of the inside 2 times minus 2 is minus 4. And the derivative of sine of 2t with respect to 2t is cosine of 2t. So there we go; we've figured out our antiderivative. Now we evaluate it from 0 the pi over 2. And what do we get? We get 4 sine-- let me write this down, for I don't want to skip too many --sine of pi over 2 plus 4 times pi over 2-- that's just 2 pi minus 2 sine of 2 times pi over 2 sine of pie, and then all of that minus all this evaluated at 0. That's actually pretty straightforward because sine of 0 is 0. 4 times 0 is 0, and sine of 2 times 0, that's also 0. So everything with the 0's work out nicely. And then what do we have here? Sine of pi over 2-- in my head, I think sine of 90 degrees; And then sine of pi is 0, that's 180 degrees. So this whole thing cancels out. So we're left with 4 plus 2 pi. So just like that we were able to figure out the area of this first curvy wall here, and frankly, that's the hardest part. Now let's figure out the area of this curve. And actually you're going to find out that these other curves as they go along the axes are much, much, much easier, but we're going to have to find different parametrizations for this. So if we take this curve right here, let's do a parametrization for that. Actually, you know what? Let me continue this in the next video because I realize I've been running a little longer. I'll do the next two walls and then we'll sum them all up."
    },
    {
        "Q": "Wouldn't it be 1/2 +pi^2/8? if not, what happened to the +1 at 10:23 ?could someone please help, because I don't understand where that 1 went.",
        "A": "The 1 was multiplied by the sin(t)*cos(t) function inside the integral. The square root of negative sine squared plus cosine squared is one.",
        "video_name": "uXjQ8yc9Pdg",
        "timestamps": [
            623
        ],
        "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": "At 1:37 why would it be mm squared?",
        "A": "when you do an area it is always in mm squared or square mm (or whatever measurement it is in. This is because it would take that many 1mm x 1mm squares to fill up the area of the circle. don t forget a circle is a 2D shape, i may not have used candy for an example on area but just pretend it s the circle bit on one side of the candy lol",
        "video_name": "ZyOhRgnFmIY",
        "timestamps": [
            97
        ],
        "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": "At 2:47\nDo you get the same anwser if you do 3.14 times R and just add the second power?",
        "A": "Not quite. Because of PEMDAS, you square R, then multiply that by pi, or 3.14. Squaring 3.14r would get a completely different answer.",
        "video_name": "ZyOhRgnFmIY",
        "timestamps": [
            167
        ],
        "3min_transcript": "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. Now, we are using the calculator's internal representation of pi which is going to be more precise than what I had in the last one. And you can 201.06 (to the nearest hundred) So, more precise is 201.06 square millimeters. So, this is closer to the actual answer, because the calculator's representation is more precise than this very rough approximation of what pi is."
    },
    {
        "Q": "At 1:30, why does Sal say that the answer is still squared, after already squaring the radius?",
        "A": "the units are squared, not re-squaring the numbers. Areas are always in square units, Volumes are in cubed units.",
        "video_name": "ZyOhRgnFmIY",
        "timestamps": [
            90
        ],
        "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": "At 1:53, why after you have squared 8 (64) do you still leave it squared?",
        "A": "Because the equation to find the area of a circle is \u00cf\u0080 r2. Therefore to find the area you have to have the radius squared and then you multiply it by \u00cf\u0080. So when you square 8 you get 64 and to find the area you multiply it by \u00cf\u0080 ( \u00cf\u0080 r2 ). He leaves 64 that way because its r2 ( 8*8 ). \u00cf\u0080*64=Area. Hope you got your question answered !!",
        "video_name": "ZyOhRgnFmIY",
        "timestamps": [
            113
        ],
        "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": "@ 1:18 how did he get 64 mm from 8 mm",
        "A": "he found the square root of the radius of the circle",
        "video_name": "ZyOhRgnFmIY",
        "timestamps": [
            78
        ],
        "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": "2:59 doesnt he mean to say millimeters sq?",
        "A": "You can say square millimeters or millimeters squared. Either is correct.",
        "video_name": "ZyOhRgnFmIY",
        "timestamps": [
            179
        ],
        "3min_transcript": "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. Now, we are using the calculator's internal representation of pi which is going to be more precise than what I had in the last one. And you can 201.06 (to the nearest hundred) So, more precise is 201.06 square millimeters. So, this is closer to the actual answer, because the calculator's representation is more precise than this very rough approximation of what pi is."
    },
    {
        "Q": "Starting from 0:45, why do we do \"radius^2\", or to be more specific, why \"^2\" to come to the conclusion of the area? any explanation of this somewhere?",
        "A": "It comes from a higher branch of mathematics called calculus, which is really good at deriving formulae for areas of shapes. When you try it for a circle, the r^2 pops out of the formula.",
        "video_name": "ZyOhRgnFmIY",
        "timestamps": [
            45
        ],
        "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": "at 1:35 how does prime factorization work?",
        "A": "prime factorization is making a number as small as it can by only using prime numbers",
        "video_name": "DKh16Th8x6o",
        "timestamps": [
            95
        ],
        "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": "When you're drawing a vector with given points (for example, (1,2) for vector v @ 12:57), how do you know what direction the vector is going in? Sal keeps drawing the vector going up but why can't it go down?",
        "A": "One Can Draw A Vector In Two Cases. 1)A Point Through Which The Vector Passes And Its Inclination(Angle)With Respect To Any Of The Co-ordinate Axes Are Known Or 2)Minimum Of Two Points Are Known Through Which The Vector Passes. One Can Draw A Vector Through A Single Point In Arbitrary Direction Only.",
        "video_name": "r4bH66vYjss",
        "timestamps": [
            777
        ],
        "3min_transcript": "we scale our vectors. When we multiply it times some scalar factor. So let me pick new vectors. Those have gotten monotonous. Let me define vector v. v for vector. Let's say that it is equal to 1, 2. So if I just wanted to draw vector v in standard position, I would just go 1 to the horizontal and then 2 to the vertical. That's the vector in standard position. If I wanted to do it in a non standard position, I could do it right here. 1 to the right up 2, just like that. Equally valid way of drawing vector v. Equally valid way of doing it. Now what happens if I multiply vector v. What if I have, I don't know, what if I have 2 times v? 2 times my vector v is now going to be equal to 2 times 2, and then 2 times 2 which is 4. Now what does 2 times vector v look like? Well let me just start from an arbitrary position. Let me just start right over here. So I'm going to go 2 to the right, 1, 2. And I go up 4. 1, 2, 3, 4. So this is what 2 times vector v looks like. This is 2 times my vector v. And if you look at it, it's pointing in the exact same direction but now it's twice as long. And that makes sense because we scaled it by a factor of 2. When you multiply it by a scalar, or you're not changing its direction. Its direction is the exact same thing as it was before. You're just scaling it by that amount. And I could draw this anywhere. I could have drawn it right here. I could have drawn 2v right on top of v. Then you would have seen it, I don't want to cover it. You would have seen that it goes, it's exactly, in this case when I draw it in standard It's along the same line, it's just twice as far. it's just twice as long but they have the exact same direction. Now what happens if I were to multiply minus 4 times our vector v? Well then that will be equal to minus 4 times 1, which is minus 4. And then minus 4 times 2, which is minus 8. So this is on my new vector. Minus 4, minus 8. This is minus 4 times our vector v. So let's just start at some arbitrary point. Let's just do it in standard position. So you go to the right 4. Or you go to the left 4. So so you go to the left 4, 1, 2, 3, 4. And then down 8. Looks like that. So this new vector is going to look like this. Let me try and draw a relatively straight line. There you go. So this is minus 4 times our vector v. I'll draw a little arrow on it to make sure you know it's a vector."
    },
    {
        "Q": "At 19:40, it is said that \"2x -y -z +3x\" must be equal to 0 in order for b to be valid.\nIsn't it that we have then:\nif b is valid => 5x -y -z = 0\n(i.e. not yet \"<=>\". Necessary condition only).\nThen b is in the plan of equation 5x -y -z = 0, in other words:\n\"the plan of equation 5x -y -z = 0\" is included in C(A). Finally, as C(A) is a plan (dim C(A) = 2), then C(A) = \"the plan of equation 5x -y -z = 0\". (now we have \"<=>\" . b is valid <=> 5x -y -z = 0 )",
        "A": "He states Ax = b at 14:00 ish, and asks what are all the possible b s, which form C(A). So if you find out that Ax forms a plane, that equality gives you equivalence .. Ax forms a plane <=> All the b s form a plane, because LHS = RHS (or in other words the validity of b is given). At least that s how I d interpret it.",
        "video_name": "EGNlXtjYABw",
        "timestamps": [
            1180
        ],
        "3min_transcript": "So let me from the get go try to zero out this third row. And the best way to zero out this third row is to just replace the third row. So the first row-- well, I won't even write the first row. The second row is 0, 1, minus 2, minus 1, and 2x minus y. I'm not even going to worry about the first row right now. But let's replace the third row, just in our attempt to go into reduced row echelon form. Let's replace it with the second row minus the third row. So you get 2x minus y minus z plus 3x. I just took this minus this. So minus z plus 3x. So 0 minus 0 is 0. 1 minus 1 is 0. Minus 2 minus minus 2 is 0, and that's also 0. So we're only going to have a valid solution to Ax equals b What happens if he's not equal to zero? Then we're going to have a bunch of zeroes equaling some number, which tells us that there's no solution. So if I pick a b where this guy does not equal zero, then I'll have no solution. If this guy equals 5, if I pick x, y, and z's such as that this expression is equal to 5, then Ax equal to b will have no solution, because it will have 0 is equal to 5. So this has to equal 0. So 2x minus y minus z plus 3x must be equal to 0 in order for b to be valid, in order for b to be a member of the column space of A, in order for it to be a valid vector that Ax can become, or the product A times x can become for some x. So what does this equal to? If we add the 2x plus the 3x, I get 5x minus y minus z is figured out the basis vectors. We said oh, you know what? The basis vectors, they have to be in the column space themselves by definition. So let me find a normal vector to them both by taking the cross product. I did that, and I said the cross product times any valid vector in our space minus one of the basis vectors has to be equal to zero, and then I got this equation. Or we could have done it the other way. We could've actually literally solved this equation setting our b equal to this. We said what b's will give us a valid solution? And our only valid solution will be obtained when this guy has to be equal to zero, because the rest of his row became zero. And when we set that equal to zero, we got the exact same equation. So, hopefully, you found this to be mildly satisfying, because we were able to tackle the same problem from two"
    },
    {
        "Q": "I'm so confused.. at about 11:00 Sal decided to factor out 27\u00c2\u00b712x^2 - 4x^6 = 0 to 4x^2(27\u00c2\u00b73-x^4) = 0 .. When I was doing it on my own I multiplied the constants and got 324x^2-4x^6 = 0, factoring out to 4x^2(324-x^4) = 0... Is this an example of where BEDMAS is really important?",
        "A": "You forgot to divide the 324 in your final equation by 4. :-)",
        "video_name": "zC_dTaEY2AY",
        "timestamps": [
            660
        ],
        "3min_transcript": "But if the derivative is equal to 0, the second derivative is equal to 0, you cannot assume that is an inflection point. So what we're going to do is, we're going to find all of the point at which this is true, and then see if we actually do have a sign change in the second derivative of that point, and only if you have a sign change, then you can say it's an inflection point. So let's see if we can do that. So just because a second derivative is 0, that by itself does not tell you it's an inflection point. It has to have a second derivative of 0, and when you go above or below that x, the second derivative has to actually change signs. Only then. So we can say, if f prime changes signs around x, then we can say that x is an inflection. And if it's changing signs around x, then it's definitely 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."
    },
    {
        "Q": "I think at 19:00 the way this notation is formed is as x --> 3(neg. side)\nThe arrow meaning approaching 3 from the negative side.",
        "A": "Hay Bennet! Join us as a life-long learner! We can use guys like you in KA. It s been a year since you ve asked this question, & I haven t seen you around since! Miss you around! :)",
        "video_name": "zC_dTaEY2AY",
        "timestamps": [
            1140
        ],
        "3min_transcript": "a little bit less than 3. If x is a little bit less than 3, if it's like 2.9999, this number is going to be less than 81, so this is also going to be positive. And of course, the denominator is always positive. So as x is less than 3, is approaching from the left, we are concave upwards. This thing's going to be a positive. Then f prime prime is greater than 0. We are upwards, concave upwards. When x is just larger than 3, what's going to happen? Well, this first term is still going to be positive. But if x is just larger than 3, x to the fourth is going to be just larger than 81, and so this second term is going to be negative in that situation. Let me do it ina new color. It's going to be negative when x is larger than 3. Because this is going to be larger than 81. So if this is negative and this is positive, then the whole thing is going to be negative, because this denominator is still going to be positive. going to be concave downwards. One last one. What happens when x is just a greater than minus 3? So just being greater than minus 3, that's like minus 2.99999. So when you take minus 2.99 square it, you're going to get a positive number, so this is going to be positive. And if you take minus 2.99 to the fourth, that's going to be a little bit less than 81, right? Because 2.99 to the fourth is a little bit less than 81, so this is still going to be positive. So you have a positive times a positive divided by a positive, so you're going to be concave upwards, because your second derivative is going to be greater than 0. Concave upwards. And then finally, when x is just, just less than negative 3, remember, when I write this down, I don't mean for all x's larger than negative 3,or all x's smaller than negative 3. There's actually no, well, I can't think of the notation from the left, but what happens if we just go to minus 3.11? Or 3.01, I guess is a better one, or 3.1? Well, this term right here is going to be positive. But if we take minus 3.1 to the fourth, that's going to be larger than positive 81, right? The sign will become positive, it'll be larger than 81, so this'll become negative. So in that case as well, we'll have a positive times a negative divided by a positive, so then our second derivative is going to be negative. And so we're going to be downwards. So I think we're ready to plot. so first of all, is x plus or minus 3 inflection points? As we approach x is equal to 3 from the left, we are concave upwards, and then as we cross 3, the second derivative is 0. The second derivative's 0, I lost it up here. The second derivative is 0. And then, as we go to the right of 3, we become concave downwards."
    },
    {
        "Q": "At 3:41, I thought that you put dots over the top of repeating decimals. Can you do both?",
        "A": "Some students learn to write repeating decimals by putting dots over the repeating numbers, but in this case Sal puts a line over the repeating numbers. However If you want to use Sal s method you can but if you were taught to put dots, its better you do that and yes, you CAN use both lines or dots for your answer. Hope that helped !!",
        "video_name": "Y2-tz27nKoQ",
        "timestamps": [
            221
        ],
        "3min_transcript": "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 but we've got to bring down another 0. So you have 27, which goes into 100 \u2013 (We've already done this.) \u20133 times. So you see something happening here. It's 0.703703. And we're just going to keep repeating 703. This is going to be equal to 0.703703703703 \u2013 on and on and on forever. So the notation for representing a repeating decimal like this is to write the numbers that repeat \u2013 in this case 7, 0, and 3 \u2013 and then you put a line over all of to indicate that they repeat. So you put a line over the 7, the 0, and the 3, which means that the 703 will keep repeating on and on and on. So let's actually input it into the answer box now. So it's 0.703703. the first six digits of the decimal in your answer. And they don't tell us to round or approximate \u2013 because, obviously, if they said to round to that smallest, sixth decimal place, then you would round up because the next digit is a 7. But they don't ask us to round. They just say, \"Include only the first six digits of the decimal in your answer.\" So that should do the trick. And it did."
    },
    {
        "Q": "At 1:50 he says that [sq rt]of 3x3x13 = [sq rt] of 3x3x[sq rt] of 13......Why does x13 = [sq rt] of 13? What did I miss?",
        "A": "Basically Sal is saying sqrt(3*3*13). Both 3s and the 13 are under the square root symbol. What he is doing is taking out the sqrt(13) , so that sqrt(3*3*13)=sqrt(3*3)*sqrt(13) I hope this helps you.",
        "video_name": "cw3mp8oNASk",
        "timestamps": [
            110
        ],
        "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": "0:42, 3 doesn't go into 11 3 times. or am i missing something?",
        "A": "You are missing something. Since 117 (if I add digits, they =9, so I know that it is divisible by three and 9). When he says 3 goes into 11 3 times, he then multiplies and subtracts (11-9) and drops down the remainder of 2 and divides 27 by 3",
        "video_name": "cw3mp8oNASk",
        "timestamps": [
            42
        ],
        "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": "Aight a question, Should i watch this after watching the exponent properties playlist?\nBecause im going in order and this one goes first, but sal says at 1:38, and we know this by our exponent properties.",
        "A": "then by using Sal s advise watch the exponents video.",
        "video_name": "cw3mp8oNASk",
        "timestamps": [
            98
        ],
        "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": "Ask a question...I don't know why at 2:04 he crosses out the 3 ?!",
        "A": "As he explained in the video, the square root of 3*3 is the same thing as saying the square root of 9. The square root of 9 is 3, so he simplified the square root of 3*3 that he had written down into just 3.",
        "video_name": "cw3mp8oNASk",
        "timestamps": [
            124
        ],
        "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 it is explained that 5 sqrt 3*3 (also 5 sqrt 9) equals 15. Sal crosses out one of the 3's and writes a little 3 above the radical. Could someone explain this to me or lead me to a video explaining it?",
        "A": "He is just using the basic definition of a square root. Square roots reverse the process of squaring a number. 3^2 = 9 So, sqrt(9) = 3 If you don t understand this concept, you need to go back to the videos that introduce square roots. Use the search bar and search for introduction to square roots to find the video.",
        "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": "Which video (and where) explains why you can add up the digits of a number to see if it's divisible by 3 like at 0:25 - 0:36?",
        "A": "go to pre- algabra and in the factors and multiples section you will find divisablity tests at the top of the list and it explains the rule for 3 in the first video",
        "video_name": "cw3mp8oNASk",
        "timestamps": [
            25,
            36
        ],
        "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 1:48 sal splits the square root.why?",
        "A": "He does show to show you that the sqrt(3*3)=3 so that can be taking out of the entire radical.",
        "video_name": "cw3mp8oNASk",
        "timestamps": [
            108
        ],
        "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 0:09, Sal said that 117 is not a perfect square. What does that mean?",
        "A": "It means that a any number times itself won t equal 117",
        "video_name": "cw3mp8oNASk",
        "timestamps": [
            9
        ],
        "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 hte first why did you pick x how do you know that was x 2:35",
        "A": "He did not know that it was x (which isn t a number anyway). He just picked x as a variable or as a placeholder for an unknown value. He could have put in T, Delta or Cow, and the results would be the same in the end. The variables would just have wonky names, so he picked x, which sounds sensible.",
        "video_name": "1uWZNW5PF-s",
        "timestamps": [
            155
        ],
        "3min_transcript": "simplified form if you divide both this height and this width So that's why they're saying greatest common divisor 1. And then they say, find the perimeter of the rectangle. So let's see what we can do here. And I encourage you to pause it and try to do it on your own before I bumble my way through this problem. So let's start in the beginning. Let's start at this square right over here, the center square. And they did tell us that they're all squares. So let's say that that square right over here has a length x and a height x. It's an x by x square. So let me write it. So this is an x, and that is an x. So this is an x by x square right over there. And then you have this square right over here. And we don't know its measurements. So let's say that this square right over here is y by y. So it has y width, and it also has y height. Well, this is an x plus y by an x plus y square, because the width of these two squares combined made the width of this larger square. So what I'm going to do is-- actually, this might be an easier way to write it. Since these are all squares, I'm going to write the dimension of that square inside the square. So this is going to be an x by x square. Kind of a non-conventional notation, but it'll help us keep things a little bit neat. This is going to be a y by y square. So I'm not saying the area is y. I'm saying it's y by y. This over here is going to be an x plus y times-- an x plus y is going to be each of its dimensions. So it's going to be x plus y height and x plus y width. Then this one over here-- well, if this dimension is x plus y and this dimension right over here is x, then this whole side or any of the sides of the square So x plus x plus y is 2x plus y. You can imagine that I'm just labeling the left side of each of these squares. The left side of this square has length y. Left side of this one, x. This one has x plus y. And then this is 2x plus y. And then we can go do this one up here. Well, if this distance right over here is 2x plus y and this distance right over here is x plus y, you add them together to get the entire dimension of one side of the square. So it's going to be 3x plus 2y. I've just added the 2x plus the x and the y plus the y to get 3x plus 2y is the length of one dimension or one side of this square. And they're all the same. Now let's go to this next square. Well, if this length is 3x plus 2y and this length is 2x plus y, then this entire length right over here"
    },
    {
        "Q": "At 6:13, Sal got 2 differant answers for 2 sides of a square. How did he get the 2 answers?",
        "A": "To get the dimension on the left side of the large rectangle, he added the lengths of a side of each square on the left. To get the dimension on the right side of the large rectangle, he added the lengths of a side of each square on the right. Since we know opposite sides of a rectangle have equal length, we can set up an equation (13x+7y=8x+9y) to solve for the ratio of x to y (x=2/5*y).",
        "video_name": "1uWZNW5PF-s",
        "timestamps": [
            373
        ],
        "3min_transcript": "5x plus 3y is going to be that entire length right over there. And we can also go to this side right over here where we have this length-- let me do that same color. This length is 3x plus 2y. This is x plus y. And this is y. So if you add 3x plus 2y plus x plus y plus y, you get 4x plus-- what is that-- 4y, right? 2y, 3y, 4y. And then we can express this character's dimensions in terms of x and y because this is going to be 5x plus 3y. Then you're going to have 2x plus y. And then you're going to have x. So you add the x's together. 5x plus 2x is 7x, plus x is 8x. And then you add the y's together, 3y plus y, and then you don't have a y there. So that's going to be plus 4y. And then finally, we have this square right over here. Its dimensions are going to be the y plus the 4x plus 4y. So that's 4x plus 5y. And then if we think about the dimensions of this actual rectangle over here, if we think about its height right over there, that's going to be 5x plus 3y plus 8x plus 4y. So 5 plus 8 is 13. So it's 13x plus 3 plus 4 is 7y. So that's its height. But we can also think about its height by going on the other side of it. And maybe this will give us some useful constraints because this is going to have to be the same length as this over here. And so if we add 4x plus 4x, we get 8x. So these are going to have to be equal to each other, so that's an interesting constraint. So we have 13x plus 7y is going to have to equal 8x plus 9y. And we can simplify this. If you subtract 8x from both sides, you get 5x. And if you subtract 7y from both sides, you get 5x is equal to 2y. Or you could say x is equal to 2/5 y. In order for these to show up as integers, we have to pick integers here. But let's see if we have any other interesting constraints if we look at the bottom and the top of this, if this gives us any more information. So if we add 5x plus 3y plus 3x plus 2y plus 4x plus 4y,"
    },
    {
        "Q": "At 3:56 isn't it negative 3/2",
        "A": "No: -6/-4 = 3/2 Those minus signs cancel He plugs -2 into 6(x+1)/(x-2) = -6/-4 = 3/2 looks correct",
        "video_name": "oUgDaEwMbiU",
        "timestamps": [
            236
        ],
        "3min_transcript": "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? except that the function is not defined at x equals negative 2. So that's why we have to put that constraint here if we wanted this to be the same thing as our original function. But if we wanted to re-engineer the function so it is continuous at that point then we just have to set f of x equal to whatever this expression would have been when x is equal to negative 2. So let's think about that. Let's think about that. So 6 times negative 2 plus 1 over negative 2 minus 2 is equal to-- this is 6 times negative 1. So it's negative 6 over negative 4, which is equal to 3/2. So if we redefine f of x, if we say f of x is equal to 6x For x not equal positive or negative 2, and it's equal to 3/2 for x equals negative 2. Now this function is going to be the exact same thing as this right over here. This f of x, this new one. This new definition-- this extended definition of our original one-- is now equivalent to this expression, is equal to 6 times x plus 1 over x minus 2. But just to answer their question, what value should be assigned to f of negative 2 to make f of x continuous to that point? Well f of x should be-- or f of negative 2 should be 3/2."
    },
    {
        "Q": "At 1:10, I don't get why do you start multiplying (-3x -2). It's getting me confused.",
        "A": "If you have a variable, like X in a fraction, you usually want to get that variable isolated. So in order to get X by itself you have to multiply both sides by the denominator. When that happens, the side with the fraction has its denominator cancelled out, but what you do to one side, you must do to the other. So that is why he had to multiply both sides by (-3x-2). Hope this helped!!",
        "video_name": "PPvd4X3Wv5I",
        "timestamps": [
            70
        ],
        "3min_transcript": "So we have 14x plus 4 over negative 3x minus 2 is equal to 8. And I'll give you a few moments to see if you can tackle it on your own. So this equation right here, at first it doesn't look like a straightforward linear equation. We have one expression on top of another expression. But as we'll see, we can simplify this to turn it into a linear equation. So the first thing that I want to do is, I don't like this negative 3x plus 2 sitting here in the denominator, it makes me stressed. So I want to multiply both sides of this equation times negative 3x minus 2. What does that do for us? Well, on the left-hand side, you have this negative 3x minus 2, it's going to be over negative 3x minus 2, they will cancel out. And so you're left with, on the left-hand side, your 14x plus 4. have to multiply 8 times negative 3x minus 2. So you are left with-- well, 8 times negative 3x is negative 24x, and then 8 times negative 2 is negative 16. And there you have it. We have simplified this to just a traditional linear equation. We've got variables on both sides, so we can just keep simplifying. So the first thing I want to do, let's just say we want to put all of our x terms on the left-hand side. So I want to get rid of this negative 24x right over here. So the best way to do that, I'm going to add 24x to the right-hand side. I can't just do it to the right-hand side, I have to also do it to the left-hand side. And so I am left with, on the left-hand side, 14x plus 24x is 38x. And then I have the plus 4-- Is equal to, well, negative 24x plus 24x. are left with just the negative 16. Now we just have to get rid of this 4 here. Let's subtract that 4 from both sides. And we are left with-- and this is the home stretch now-- we are left with 38x is equal to negative 16 minus 4 is negative 20. And so we can divide both sides of this equation by 38. And we are left with x is equal to negative 20 over 38, which can be simplified further. Both the numerator the denominator is divisible by 2. So let's divide the numerator and denominator by 2, and we get negative 10 over 19. x is equal to negative 10 over 19, and we are done. I encourage you to validate this for yourself. It's a little bit of a hairy number right over here,"
    },
    {
        "Q": "At 4:30, the distance formula is mentioned. what is the distance formula? And why is it the same thing as the Pythagorean Theorem? I thought they were two completely different formulas.",
        "A": "The distance formula is a formula you can use to find the shortest distance between any 2 points on the coordinate plane. You are correct that the distance formula and Pythagorean theorem are 2 different things but the distance formula is derived from the Pythagorean theorem. The distance formula is: d = \u00e2\u0088\u009a[(({x_1} - {x_2})^(2)) + (({y_1} - {y_2})^(2))]",
        "video_name": "iATjsfAX8yc",
        "timestamps": [
            270
        ],
        "3min_transcript": ""
    },
    {
        "Q": "7:10-end you ended up calculating the area of the circle at the start and not the rate of change (I think). r was a function of t, and if you think about the problem just generally thinking about it, the rate of change should not be linear b/c the formula for area is quadratic. In fact, now that I am looking back, I think that this is because you solved for circumference rather than area.",
        "A": "A = pi\u00e2\u0080\u00a2r^2 d/dt(A) = d/dt(pi\u00e2\u0080\u00a2r^2) d/dt(A) = pi\u00e2\u0080\u00a2d/dt(r^2) d/dt(A) = 2\u00e2\u0080\u00a2pi\u00e2\u0080\u00a2r\u00e2\u0080\u00a2d/dt(r) dA/dt = 2\u00e2\u0080\u00a2pi\u00e2\u0080\u00a2r\u00e2\u0080\u00a2dr/dt dr/dt = 1 r(0) = 3 dA/dt = 2\u00e2\u0080\u00a2pi\u00e2\u0080\u00a23\u00e2\u0080\u00a21 dA/dt = 6\u00e2\u0080\u00a2pi C = 2\u00e2\u0080\u00a2pi\u00e2\u0080\u00a2r d/dt(C) = d/dt(2\u00e2\u0080\u00a2pi\u00e2\u0080\u00a2r) d/dt(C) = 2\u00e2\u0080\u00a2pi\u00e2\u0080\u00a2d/dt(r) dC/dt = 2\u00e2\u0080\u00a2pi\u00e2\u0080\u00a2dr/dt dr/dt = 1 r(0) = 3 dC/dt = 2\u00e2\u0080\u00a23\u00e2\u0080\u00a21 dC/dt = 6",
        "video_name": "kQF9pOqmS0U",
        "timestamps": [
            430
        ],
        "3min_transcript": "So the rate at which r of t changes with respect to time? Well, we could just write that as dr dt. These are equivalent expressions. And of course, we have our pi out front. And I just want to emphasize this is just the chain rule right over here. The derivative of something squared with respect to time is going to be the derivative of the something squared with respect to the something. So that's 2 times the something, times the derivative of that something with respect to time. I can't emphasize enough. What we did right over here, this is the chain rule. That is the chain rule. So we're left with pi times this is equal to the derivative of our area with respect to time. Now let me rewrite all this again just so it cleans up a little bit. So we have the derivative of our area with respect to time is equal to pi times-- actually let me put that 2 out front. Is equal to 2 times pi times-- I can now We know that r is a function of t. So I'll just write 2 pi times r times dr dt. Actually, let me make the r in blue. 2 pi r dr dt. Now what do we know? We know what r is. We know that r, at this moment right in time, is 3 centimeters. Right now r is 3 centimeters. We know dr dt right now is 1 centimeter per second. We know this is 1 centimeter per second. So what's da dt going to be equal to? Well, it's going to be equal to-- do that same green-- 2 pi times 3 times 3 times 1 times-- that's purple-- times 1 centimeter per second. And let's make sure we get the units right. So it's going to be centimeters. That's too dark of a color. It's going to be square centimeters, centimeters times centimeters, square centimeter per second, which is the exact units we need for a change in area. So we have da dt is equal to this. da, the rate at which area is changing with respect to time, is equal to 6 pi. So it's going to be a little bit over 18 centimeters squared per second. Right at that moment. Yep, 3 times 2 pi. So 6 pi centimeters squared per second is how fast the area is changing. And we are done."
    },
    {
        "Q": "At 4:04: Why can you rewrite d/Dt [pi*r^2] as pi*d/dt [r(t)?",
        "A": "Because pi is a constant, and you can do that with constants when you are taking derivatives.",
        "video_name": "kQF9pOqmS0U",
        "timestamps": [
            244
        ],
        "3min_transcript": "Well, they say at what rate is the area of the circle growing? So we need to figure out at what rate is the area of the circle-- where a is the area of the circle-- at what rate is this growing? This is what we need to figure out. So what might be useful here is if we can come up with a relationship between the area of the circle and the radius of the circle and maybe take the derivative with respect to time. And we'll have to use a little bit of the chain rule So what is the relationship at any given point in time between the area of the circle and the radius of the circle? Well, this is elementary geometry. The area of a circle is going to be equal to pi times the radius of the circle squared. Now what we want to do is figure out the rate at which the area is changing with respect to time. So why don't we take the derivative of both sides of this with respect to time? And let me give myself a little more real estate. Actually, let me just rewrite what I just had. So pi r squared. I'm going to take the derivative of both sides of this with respect to time. So the derivative with respect to time. I'm not taking the derivative with respect to r, I'm taking the derivative with respect to time. So on the left-hand side right over here, I'm going to have the derivative of our area. Actually, let me just write it in that green color. I'm going to have the derivative of our area with respect to time on the left-hand side. And on the right-hand side, what do I have? Well, if I'm taking the derivative of a constant times something, I can take the constant out. So let me just do that. Pi times the derivative with respect to time of r squared. And to make it a little bit clearer what I'm about to do, why I'm using the chain rule, we're assuming that r is a function of time. If r wasn't a function of time then area wouldn't be a function of time. So instead of just writing r, let me make it explicit I'll write r of t. So it's r of t, which we're squaring. And we want to find the derivative of this with respect to time. And here we just have to apply the chain rule. We're taking the derivative of something squared with respect to that something. So the derivative of that something squared with respect to the something is going to be 2 times that something to the first power. Let me make it clear. This is the derivative of r of t squared with respect to r of t. The derivative of something squared with respect to that something. If it was a derivative of x squared with respect to x, we'd have 2x. If it was the derivative of r of t squared with respect to r of t, it's 2r of t. But this doesn't get us just the derivative with respect to time. This is just the derivative with respect to r of t. The derivative at which this changes with respect to time, we have to multiply this times the rate at which r of t"
    },
    {
        "Q": "at 1:36 why does sal multiply a/b with n/n and m/n with b/b",
        "A": "I think he did it so he would have the same denominators for both fractions.",
        "video_name": "HKUJkMQsGkM",
        "timestamps": [
            96
        ],
        "3min_transcript": "What I want to do in this video is think about whether the product or sums of rational numbers are definitely going to be rational. So let's just first think about the product of rational numbers. So if I have one rational number and-- actually, let me instead of writing out the word rational, let me just represent it as a ratio of two integers. So I have one rational number right over there. I can represent it as a/b. And I'm going to multiply it times another rational number, and I can represent that as a ratio of two integers, m and n. And so what is this product going to be? Well, the numerator, I'm going to have am. I'm going to have a times m. And in the denominator, I'm going to have b times n. Well a is an integer, m is an integer. So you have an integer in the numerator. And b is an integer and n is an integer. So you have an integer in the denominator. So now the product is a ratio of two integers right over here, so the product is also rational. So this thing is also rational. you're going to end up with a rational number. Let's see if the same thing is true for the sum of two rational numbers. So let's say my first rational number is a/b, or can be represented as a/b, and my second rational number can be represented as m/n. Well, how would I add these two? Well, I can find a common denominator, and the easiest one is b times n. So let me multiply this fraction. We multiply this one times n in the numerator and n in the denominator. And let me multiply this one times b in the numerator and b in the denominator. Now we've written them so they have a common denominator of bn. And so this is going to be equal to an plus bm, So b times n, we've just talked about. This is definitely going to be an integer right over here. And then what do we have up here? Well, we have a times n, which is an integer. b times m is another integer. The sum of two integers is going to be an integer. So you have an integer over in an integer. You have the ratio of two integers. So the sum of two rational numbers is going to give you another. So this one right over here was rational, and this one is right over here is rational. So you take the product of two rational numbers, You take the sum of two rational numbers, you get a rational number."
    },
    {
        "Q": "At 0:55 could b^1 also be b^0?",
        "A": "Only if b = 1.",
        "video_name": "X6zD3SoN3iY",
        "timestamps": [
            55
        ],
        "3min_transcript": "Simplify 25 a to the third and a to the third is being raised to the third power, times b squared and all of that over 5 a squared, b times b squared So we can do this in multiple ways, simplify different parts. What I want to do is simplify this part right over here. a to the third power, and we're raising that to the third power. So this is going to be from the power property of exponents, or the power rule this is going to be the same thing as a to the 3 times 3 power So this over here (let me scroll up a little bit) is going to be equal to a the 3 times 3 power, or a to the ninth power. We could also simplify this b times b squared over here. This b times b squared, that is the same thing as b to the first power remember, b is just b to the first power. So it's b to the first power times b to the second power. So b to the first times b to the second power is just equal to b to the one plus two power, which is equal to b to the third power and then last thing we could simplify, just right off the bat just looking at this: or we could say it's going to give us 5 over 1 if you view it as dividing the numerator and the denominator both by 5. So what does our expression simiply to? We have 5a to the ninth, and then we still have this b squared here, b squared. All of that over a squared times b to the third power... times b to the third power. Now, we can use the quotient property of exponents. You have an a to the ninth Let me use a slightly different color. We have an a to the ninth. over a squared. What's that going to simplify to? Well, that's going to simplify to be the same thing, a to the ninth over a squared, the same thing as a to the nine minus two, which is equal to a to the seventh power. and this will get a little bit interesting here So that simplifies too. So b squared over b to the third is equal to b to the two minus three power, which is equal to b to the negative one power. And we'll leave it alone like that right now. So this whole expression simplifies to It simplifies to: 5 times a to the seventh power (because this simplifies to a to the seventh) a to the seventh, times (the bs right here simplify too) b to the negative one. We could leave it like that, you know, that's pretty simple but we may not want a negative exponent there we just have to remeber that b to the negative one power is the same thing as one over b. Now if we remember that, then we can rewrite this entire expression as, the numerator will have a five and will have a to the seventh 5 a to the seventh. And then the denominator will have the b. So we're multiplying this times one over b."
    },
    {
        "Q": "At 5:34, Sal says that the notation for the length/magnitude of a vector is notated by using double lines around the vector, like this.\n||a||\nCan somebody explain why we use this \"double absolute value\" notation to signify the magnitude of a vector?",
        "A": "Well, maybe just as a convention sort of .Someone kept that symbol and maybe we are using it.",
        "video_name": "WNuIhXo39_k",
        "timestamps": [
            334
        ],
        "3min_transcript": ""
    },
    {
        "Q": "when he means the length of vector.. does he mean the magnitude of the vector. is it another word for magnitude or is it completely different things. please explain... thanks (4:44 min)",
        "A": "i think it s exactly the same thing",
        "video_name": "WNuIhXo39_k",
        "timestamps": [
            284
        ],
        "3min_transcript": ""
    },
    {
        "Q": "1:58 Why is 3+5i a complex \"number\"?\nIt consists of 2 numbers... Real and Imaginary.\nCouldn't it be called \"complex numbers\"?",
        "A": "Z is the complex number, comprised of a Real Part (5) and an Imaginary Part (3i).",
        "video_name": "SP-YJe7Vldo",
        "timestamps": [
            118
        ],
        "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": "at 3:18 why do we choose y-axis as imaginary and x-axis as real part ,is there any proof for it??",
        "A": "It s just a natural representation. The key point here is that the imaginary axis and the real axis must be perpendicular to each other. Imagine rotating your graph by 90 degrees, now the imaginary axis is the horizontal one.",
        "video_name": "SP-YJe7Vldo",
        "timestamps": [
            198
        ],
        "3min_transcript": "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. For example, z right over here which is five plus three i, the real part is five so we would go one, two, three, four, five. That's five right over there. The imaginary part is three. One, two, three, and so on the complex plane, on the complex plane we would visualize that number right over here. This right over here is how we would visualize z on the complex plane. It's five, positive five in the real direction, positive three in the imaginary direction. We could plot other complex numbers. Let's say we have the complex number a which is equal to let's say it's negative two plus i. Where would I plot that? Well, the real part is negative two, negative two, and the imaginary part is going to be you could imagine this as plus one i"
    },
    {
        "Q": "As discussed in 0:57, you divide the numerator and the denominator by the same number. However, for 30/45 and 54/81, do they both have to be divided by 5?",
        "A": "No. You see, you could factor out 5 ONLY from 30/45, since both 54 and 81 are not divisible by 5( 54/5=10.8, and 81/5=16.2). You then get 6/9. 3 is divisible by both 6 and 9.You get 2/3. This is can no longer be factored. Now, for 54/81, you see automatically (if you know you re times tables) that 54 and 81 are divisible by 9. You get 6/9.Like 30/45, you see that the numerator and denominator are divisible by 3.You get 2/3. 54/81 and 30/45 are equivalent. I hope this helped.",
        "video_name": "Io9i1JkKgN4",
        "timestamps": [
            57
        ],
        "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": "which formula is used here Square root at 6:00",
        "A": "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 3:59, why is Sal using 3 coordinates? Its only the supposed to be x and y correct?",
        "A": "its a triangle so he has to use three points",
        "video_name": "GiGLhXFBtRg",
        "timestamps": [
            239
        ],
        "3min_transcript": "or around the center of mass. But anyway, the point of this video is not to focus on physics and throwing iron triangles. The point here is I want to show you a neat property of medians. And the property is that if you pick any median, the distance from the centroid to the midpoint of the opposite side-- so this distance-- is going to be half of this distance. So if this distance right here is a, then this distance right here is 2a. Or another way to think about it is this distance is 2/3 of the length of the entire median, and this distance right here is 1/3 of the length of the entire median. And let's just prove it for ourselves just so you don't have to take things on faith. And to do that, I'll draw an arbitrary triangle. I'll do a two-dimensional triangle, and I'll do it in three dimensions because at least in my mind, it makes the math a little bit easier. In general, whenever you have an n-dimensional figure it makes the math a little bit easier. The actual tetrahedron problem that we did, you could actually embed it in four dimensions and it would make the math easier. It's just much harder to visualize, so I didn't do it that way. But let's just have an arbitrary triangle. And let's say it has a vertex and there, a vertex there, and a vertex there. So I'm not making any assumptions about the triangle. I'm not saying it's isosceles, or equilateral or anything. It's just an arbitrary triangle. And so let's say this coordinate right over here is-- I'll call this the x-axis. So, this is the x-axis, the y-axis, and the z-axis. I know some of y'all are used to swapping these two axes, but it doesn't make a difference. So let's call this coordinate right here a, 0, 0. So it's a along the x-axis. Let's call this coordinate 0, b, 0. And let's call this coordinate up here, 0, 0, c. And if you connect the points, you're going to have a triangle just like that. 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"
    },
    {
        "Q": "What is tangent? At 0:15.",
        "A": "A tangent is a line on the outside of a circle or curve that only touches at one point.",
        "video_name": "ZiqHJwzv_HI",
        "timestamps": [
            15
        ],
        "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": "I don't understand the end behaviors @4:25",
        "A": "End behavior is what the function looks like when it reaches really high or really low values of x.",
        "video_name": "tZKzaF28sOk",
        "timestamps": [
            265
        ],
        "3min_transcript": "So when we talk about end behavior, we're talking about the idea of what is this function? What does this polynomial do as x becomes really, really, really, really positive and as x becomes really, really, really, really negative? And kind of fully recognizing that some weird things might be happening in the middle. But we just want to think about what happens at extreme values of x. Now obviously for the second degree polynomial nothing really weird happens in the middle. But for a third degree polynomial, we sort of see that some interesting things can start happening in the middle. But the end behavior for third degree polynomial is that if a is greater than 0-- we're starting really small, really low values-- and as a becomes positive, we get to really high values. If a is less than 0 we have the opposite. And these are kind of the two prototypes for polynomials. Because from there we can start thinking about any degree polynomial. So let's just think about the situation of a fourth degree polynomial. plus bx to the third plus cx squared plus dx plus-- I don't want to write e because e has other meanings in mathematics. I'll say plus-- I'm really running out of letters here. I'll just use f, although this isn't the function f. This is just a constant f right over here. So let's just think about what this might look like. Let's think about its end behavior, and we could think about it relative to a second degree polynomial. So its end behavior, if x is really, really, really, really negative, x to the fourth is still going to be positive. And if a is greater than 0 when x is really, really, really negative, we're going to have really, really positive values, just like a second degree. And when x is really positive, same thing. x to the fourth is going to be positive, times a is still going to be positive. So its end behavior is going to look Now, it might do-- in fact it probably will do some funky stuff in between. It might do something that looks kind of like that in between. But we care about the end behavior. I guess you could call the stuff that I've dotted lined in the middle, this is called the non-end behavior, the middle behavior. This will obviously be different than a second degree polynomial. But what happens at the ends will be the same. And the reason why, when you square something, or you raise something to the fourth power, you raise anything to any even power for a very large-- as long as a is greater than 0, for very large positive values, you're going to get positive values. And for very large negative values, you're going to get very large positive values. You take a negative number, raise it to the fourth power, or the second power, you're going to get a positive value. Likewise, if a is less than 0, you're going to have very similar end behavior to this case. For a polynomial where the highest degree"
    },
    {
        "Q": "At 0:33 what does \"zen\" mean",
        "A": "The voice recognition software got a bit confused. I m pretty sure Sal says and we essentially picked no y s then .",
        "video_name": "_hrN4rVCOfI",
        "timestamps": [
            33
        ],
        "3min_transcript": "Voiceover:What I want to do in this video is hopefully give more intuition as to why the binomial theorem or the binomial formula involves combinatorics. Let's just think about what this expansion would be. I'm just using a particular example that's pretty simple, x plus y to the third power which is x plus y, times x plus y, times x plus y. We already know what the terms look like, we could pick an x from each of these. If we pick exactly an x from each of these that's going to be our x to the third term, and we essentially picked no y [zen] because we picked an x from each of this when we multiplied out to get this term. How many ways are there to construct that? How many ways are there, if we're picking from three things. We're picking from three things, how many ways can we choose exactly zero y's of not picking a y? There's only one way to do that, there's only one possible way of doing that. By not picking a y from each of them The coefficient right over here is one x to the third. There's only one way, when you take this product out, only one of the terms initially is going to have an x to the third in it. Now, what about the next term? So plus. Now we're going to pick from three buckets but we want to pick one y. Another way of thinking about it you have three people or we have three people, this person, this person, and this person. One of them you're going to choose to be I guess be your friend that's another way of saying which of these are you going to pick the y from? This is the exact same thing as a combinatorics problem, from three things you are choosing one of them to be y. Of course three choose one, this is equal to three. If you're picking one y that means you're picking two x's, so it's x squared times y to the one and we could keep doing that logic, so plus. Now we're going to think about the situation where we are picking two y's out of three things. It's x to the first, y squared. We're starting with three things and we're going to pick two to two of this y's so it could be this y and this y to construct the product y. It could be y times y times that x or it could be y times y, times that x or it could be y times x, times y. This is three choose two, out of three things what are the different ways, what are the combinations of picking two things. If you have three friends how many combinations can you put two of them in your two seater car where you don't care about who sits in which seat. You're just thinking about how many different combinations can you pick from that and once again this evaluates to three"
    },
    {
        "Q": "at 3:04 couldnt he just added 5 to 35 and add 20?",
        "A": "Well, he could have, but doing that is basically doing exactly what he did do. He probably chose to do it how he did because if you aren t working with time then you would have to subtract, and he want s us to build good habits in problem solving.",
        "video_name": "UhMM68fq9FA",
        "timestamps": [
            184
        ],
        "3min_transcript": "We know that there are 60 minutes in an hour. So 1:59 is 1 minute before 2:00 PM. So I'm not drawing it completely it scale, but this right over here is 1:59 PM. So we use 1 minute to get to 2:00 PM. And then, let's see, we have to get up to 96 minutes. So then you get another 60 minutes to go to 3:00 PM. So, so far, we've used a cumulative 61. So this is 1 minute. So this is how long the test has been going on since the beginning. So 1 minute by 2:00 PM, 61 minutes by 3:00 PM. If we go all the way to 4:00 PM, that's going to be another 60. That'd be 121 minutes, which is longer than the actual test. So we know that the entire test is 96 minutes. So we're going to get to some point right over here, some point like this. And what we need to figure out is, is what is left over? What is this distance right over here? Or what is this time, this difference in time right over here? And so to figure that out, we had to figure out the time at which 96 minutes have passed since the beginning of the exam. So at 3 o'clock, 61 minutes have passed. And so if we, if we say well how many more minutes have to go on for the end of the test? It's going to be 96 minus 61. So the amount of time that elapses past 3 minutes is 96 minus 61, which is 35 minutes. So this right over here is going to be 3:35. And so now our question is, how much time is there between 3:35 and 4:00 PM? Well, once again, 60 minutes in an hour. 4:00 PM is essentially the 60th minute. So she has a total of 25 minutes remaining after the test before she has to get to volleyball practice. Just as a review. One minute between 1:59 and 2:00, then another 60. That gets us to 61 total minutes. Then another 35 minutes gets us to 96 minutes have passed. And that also gets us to 3:35, which means we have 25 minutes until 4:00."
    },
    {
        "Q": "so at 2:32 it says that all four base angles are congruent,but is the angle at the top congruent to them also?",
        "A": "Not necessarily, Because the sides of the rhombus do not necessarily have to be equal to one of the diagonals. However, if you have a rhombus with angles 60 and 120 degrees, then the shorter diagonal will split the rhombus into two equilateral triangles and so the vertical angle of the triangle will be also equal to the other two base angles (ie, all the angles will be 60 degrees).",
        "video_name": "_QTFeOvPcbY",
        "timestamps": [
            152
        ],
        "3min_transcript": "Now, let me draw one of its diagonals. And the way I drew it right here is kind of a diamond. One of its diagonals will be right along the horizontal, right like that. Now, this triangle on the top and the triangle on the bottom both share this side, so that side is obviously going to be the same length for both of these triangles. And then the other two sides of the triangles are also the same thing. They're sides of the actual rhombus. So all three sides of this top triangle and this bottom triangle are the same. So this top triangle and this bottom triangle are congruent. They are congruent triangles. If you go back to your ninth grade geometry, you'd use the side-side-side theorem to prove that. Three sides are congruent, then the triangles themselves are congruent. But that also means that all the angles in the triangle are congruent. So the angle that is opposite this side, this shared side right over here will be congruent to the corresponding angle in the other triangle, the angle So it would be the same thing as that. Now, both of these triangles are also isosceles triangles, so their base angles are going to be the same. So that's one base angle, that's the other base angle, right? This is an upside down isosceles triangle, this is a right side up one. And so if these two are the same, then these are also going to be the same. They're going to be the same to each other, because this is an isosceles triangle. And they're also going to be the same as these other two characters down here, because these are congruent triangles. Now, if we take an altitude, and actually, I didn't even have to talk about that, since actually, I don't think that'll be relevant when we actually want to prove what we want to prove. If we take an altitude from each of these vertices down to this side right over here. So an altitude by definition is going to be perpendicular down here. Now, an isosceles triangle is perfectly symmetrical. If you drop an altitude from the-- I or the unique vertex in an isosceles triangle-- you will split it into two symmetric right triangles. Two right triangles that are essentially the mirror images of each other. You will also bisect the opposite side. This altitude is, in fact, a median of the triangle. Now we could do it on the other side. The same exact thing is going to happen. We are bisecting this side over here. This is a right angle. And so essentially the combination of these two altitudes is really just a diagonal of this rhombus. And it's at a right angle to the other diagonal of the rhombus. And it bisects that other diagonal of the rhombus. And we could make the exact same argument over here. You could think of an isosceles triangle over here. This is an altitude of it."
    },
    {
        "Q": "At 3:05, why is it just answer 17 but at 3:31, 5 is 25?",
        "A": "He is squaring each number. So at 3:05 he squares the squared root of 17, the square root of 17x the square root of 17 equals 17. The square root of 17 is a number slightly bigger than 4, because 4x4 equals 16, so this is just a little bit more than that. At 3:31 he square 5. 5x5=25 The concept is that if you square each number you can compare the numbers without the radical signs........",
        "video_name": "KibTbfkoPTs",
        "timestamps": [
            185,
            211
        ],
        "3min_transcript": "Because their squares are not going to be irrational numbers. It's going to be much easier to compare, and then we can order them. Because if we order the squares, then they'll tell us what happens if we order their square roots. What am I talking about? Well, I'm just gonna square each of these. So if I take this to the second power, this is going to be four square roots of two, times four square roots of two. You can change the order of multiplication. That's four times four times the square root of two times the square root of two. Now, four times four is 16. Square root of two times square root of two, well, that's just going to be two. So it's gonna be 16 times two which is equal to 32. Now what about two square roots of three? Well, same idea. Let's square it, let's square it. And i'll do this one a little bit faster. So if we square two square roots of three, times square root of three squared. So it's going to be two squared times the square root of three squared. Well, two squared is going to be four. Square root of three squared is going to be three. So this is going to be equal to 12. That's this thing squared. If this step seems a little bit confusing, if you have the product of two things raised to a power, that's the same thing as raising each of them to that power, and then taking the product. And you can actually see, I worked it out here, why that actually makes sense. Notice when I just changed the order of multiplication you had four times four, or four squared, times square root of two squared, which is going to be two. So let's keep doing that. So what is this value squared? It's gonna be three squared, which is nine, times square root of two squared, which is two. What's the square root of 17 squared? That's just going to be seventeen. Do that in blue. This is just going to be 17. What is three square roots of three squared? It's gonna be three squared, which is nine, times square root of three squared. The square root of three times the square root of three is three. So it's gonna be nine times three, or 27. And what is five squared? This is pretty straightforward. That's going to be 25. So let's order them from least to greatest. Which of them, when I square it, gives me the smallest value? Compare 32 to 12 to 18 to 17 to 27 to 25. 12 is the smallest value. So if their square is the smallest, and these are all positive numbers, then this is going to be the smallest value out of all of them. Let me write that first. Two square roots of three. So I've covered that one."
    },
    {
        "Q": "It states at 4:56 that zero is a positive number. I thought all positive numbers have an opposite. If they do then what is zero's opposite?",
        "A": "Yes, all numbers, either positive or negative, should have the opposite, except 0.",
        "video_name": "XHHYA2Ug9lk",
        "timestamps": [
            296
        ],
        "3min_transcript": "this whole thing is going to be a negative. But let's keep on doing, we said, look if all of them are negative, then this thing would be negative, but that's because I had three numbers here. What if I had four numbers here? What if I had times, What if I had times d here? And if I told you all of these numbers were negative? Let's think about it, if negative, negative, negative, negative. And I can do the multiplication in any order, but I'll just go left to right. A times b, negative times a negative. That would yield a positive. Now if you multiply that product times c, positive times a negative, positive times a negative, positive times a negative that would give you a negative. And then you multiply this negative times this negative, so this whole product, a, b, c, is going to be negative. But then we multiply it times a negative. Well a negative times a negative is going to be a positive. So this whole thing is going to be a positive. And so you're probably seeing a pattern here. if you're multiplying a bunch of numbers, and if you have an odd number of negatives, odd number of negatives being multiplied, and or divided. I just did multiplication here, but this would have also been true if these were all division symbols. If these were all division symbols, we would've been able to say the exact same thing. If you have an odd number of negative numbers in your product or in your quotient, well then you're going to have a negative, you're going to have a negative value for the entire expression. If you have an even, if you have an even number of negative, well then the whole thing is going to be, the whole thing is going to be positive. And so you can view these as generalizations of what we just saw here. Positive times a positive. Zero is actually an even number, so this would be positive. Negative times a negative. Well that's an even number of negatives. That's this case again, so you're going to be positive. Either of these cases, you have one negative. You have one negative in either of these cases, so it's going to be this case, odd number of negatives. So that's going to be a negative. And so, we can use this knowledge to start dealing with negative numbers and exponents. So if I were to say, I would have a to the, let me throw out a wild number here, A to the 101st power, And we know that a is less than zero. What is this going to be? Well this is taking 101 A's and multiplying them together. You have an odd number, an odd number of negatives being multiplied together? Well this whole then is going, this whole thing is going to be less than zero. If we knew what a was, we could calculate it somehow, but we know that this thing is going to be negative. Let's do something, so, and we could do it other ways. We could say something like this."
    },
    {
        "Q": "At 3:28, Sal says that he can do multiplication in any order, but thought you had to go from left to right.",
        "A": "You can do multiplication in any order because of the commutative property of multiplication, which means that you ll get the same answer no matter what order you put it in.",
        "video_name": "XHHYA2Ug9lk",
        "timestamps": [
            208
        ],
        "3min_transcript": "and now I'm going to write the dot for times. A times b times c, a times b times c. Now if I told you that these were all positive numbers, then you say, \"OK a times b times c is going \"to be positive.\" \"A times b would be positive, and that times c is positive.\" Now what would happen if I were to tell you that they were all negative numbers. What if a, b, and c were all negative? Well if there were all negative, let me write it that way. Let me actually write, let me write, a, b, and c, a, b, and c, they're all going to be negative. So if that's the case, what is this product going to be equal to? Well you're going to have a negative here, times a negative. So a negative times a negative, a times b, if you do that first, and we can when we multiply these numbers. That's going to give you a positive. But then you're going to multiply that times c. You're going to multiply that times c, which is a negative. So you're going to have a positive times a negative, which is going to be a negative. So this one, if a, b, and c are all less than zero, then the product, a, b, c is going to be less than zero as well. This whole thing is going to be negative. Now, if I did something else. If I said, \"There's other ways \"that I can make the product negative.\" If a is, let's say that a, actually let me just write it this way. Let's say that a is positive, b is negative. B is negative, and c is positive. And c is positive. Well here, positive times a negative, if you do this first. Positive times a negative is going to give you a negative. And then a negative times a positive, different signs, is going to give you a negative. this whole thing is going to be a negative. But let's keep on doing, we said, look if all of them are negative, then this thing would be negative, but that's because I had three numbers here. What if I had four numbers here? What if I had times, What if I had times d here? And if I told you all of these numbers were negative? Let's think about it, if negative, negative, negative, negative. And I can do the multiplication in any order, but I'll just go left to right. A times b, negative times a negative. That would yield a positive. Now if you multiply that product times c, positive times a negative, positive times a negative, positive times a negative that would give you a negative. And then you multiply this negative times this negative, so this whole product, a, b, c, is going to be negative. But then we multiply it times a negative. Well a negative times a negative is going to be a positive. So this whole thing is going to be a positive. And so you're probably seeing a pattern here."
    },
    {
        "Q": "@0:28 Why is it 2cm?? Really confused",
        "A": "That is because 2 is the height. When you look at it, two cm is the line coming up that makes the figure 3D, so that is why it is the height. Also, the length and the height make the base and they intersect before the height comes up at the edge.",
        "video_name": "feNWZEln6Nc",
        "timestamps": [
            28
        ],
        "3min_transcript": "- [Voiceover] Let's see if we can figure out the volume of this figure over here. They've given us some of the dimensions. We see this side over here is two centimeters, this is seven centimeters, this is 12 centimeters, this is five centimeters, this is three centimeters. And so like always, pause this video and see if you can figure it out. Well there's a bunch of ways to do this, but the way I'd like to do it is just to break it up into two rectangular prisms. So what I'm gonna do is, in fact most of the reasonable ways to do this would be to break it up into two rectangular prisms, and the ones that jump out at me is one prism like this that is three centimeters wide, five centimeters high, and then it is seven centimeters long, or seven centimeters deep. So this one right over here. And if this part right over here was transparent you would see it look just like this. You would see it look just like this. And so this one once again, it is three centimeters wide, So this distance right over here is going to be the same as this distance right over here. So seven centimeters long. So the width times the length times the height is five centimeters. Gets us to, let's see. Three times seven is 21, times five is equal to, 20 times five is 100, one times five is five. So it's going to be 105. We can say 105 cubic centimeters, cause you have centimeters times centimeters times centimeters. So this blue part right over here, this blue rectangular prism, has a volume of 105 cubic centimeters. So now we can separately figure out the volume of what I'm now highlighting in this magenta color. What I'm highlighting in this magenta color. If this was transparent, you would see this part back over here and right over here. Well, we know its height is two centimeters, we know that this dimension right over here, I guess you could say its depth, we could call it that, is seven centimeters. But what is this right over here? If we want to consider this, maybe it's length, or maybe it's width, depending on what we want to call it. Well, let's see, this whole thing is 12 centimeters, from here to here is 12 centimeters, and we know that from here to here is three centimeters, so this piece right over here must be nine centimeters. So that must be nine centimeters, is this distance right over here. So the volume of this magenta part is going to be nine centimeters times seven centimers times the height, times two centimeters. Which is going to get us, let's see, nine times seven is 63, 63 times two is equal to,"
    },
    {
        "Q": "At about 1:35, you mention that 5 and 12 are relatively prime. What exactly does that mean?\n\nThe more I think about this, the more puzzled I become. I guess I'm having trouble with how prime-ness can have different degrees. The one solution that I've come up with isn't exactly satisfying: that the two numbers together 'may as well be'/ are 'as good as' prime, since they only have one as a common factor and all primes have 1 as a common factor.",
        "A": "12 is not a prime, but if you consider that you are only allowed to divide by 1 and 5 (the factors of 5), then 12 would be a prime. 12 can be divided by 2, 3, 4, 6, but not by any factor of 5. So 15 and 8 are also relatively prime. You could also say: two numbers are relatively prime if they don t share any factors. For example 15=3*5 and 8=2*2*2",
        "video_name": "jFd-6EPfnec",
        "timestamps": [
            95
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 2:1 l don't understand",
        "A": "What Sal means is that you need a certain amount of cleaning product (bottles in the numerator) to clean a certain fraction of the bathroom. As in the video, you need 1/3 of a bottle of cleaning product to clean 3/5 of the area of a bathroom.",
        "video_name": "2DBBKArGfus",
        "timestamps": [
            121
        ],
        "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": "At 1:48, isn't the vector drawn vector (a+b), not vector (a-b)?",
        "A": "No, it is a - b. To draw a + b, you d put vector b s tail at the head of vector a (i.e. chain them together). Draw some vectors on paper with roughly correct coordinates and give it a go.",
        "video_name": "5AWob_z74Ks",
        "timestamps": [
            108
        ],
        "3min_transcript": "A couple of videos ago we introduced the idea of the length of a vector. That equals the length. And this was a neat idea because we're used to the length of things in two- or three-dimensional space, but it becomes very abstract when we get to n dimensions. If this has a hundred components, at least for me, it's hard to visualize a hundred dimension vector. But we've actually defined it's notion of length. And we saw that this is actually a scalar value. It's just a number. In this video, I want to attempt to define the notion of an angle between vectors. As you can see, we're building up this mathematics of vectors from the ground up, and we can't just say, oh, I know what an angle is because everything we know about angles and even lengths, it just applies to what we associate with two- or three-dimensional space. But the whole study of linear algebra is abstracting these And I haven't even defined what dimension is yet, but I think you understand that idea to some degree already. When people talk about one or two or three dimensions. So let's say that I have some vector-- let's say I have two vectors, vectors a and b. They're nonzero and they're members of Rn. And I don't have a notion of the angle between them yet, but let me just draw them out. Let me just draw them as if I could draw them in two So that would be vector a right there. Maybe that's vector b right there. And then this vector right there would be the vector a minus b. And you can verify that just the way we've learned to add and subtract vectors. Or you know, this is heads to tails. So b plus a minus b is of course, going to be vector a. To help us define this notion of angle, let me construct another triangle that's going to look a lot like this one. But remember, I'm just doing this for our simple minds to imagine it in two dimensions. But these aren't necessarily two-dimensional beasts. These each could have a hundred components. But let me make another triangle. Well, it should look similar. Say it looks like that. And I'm going to define the sides of the triangles to be the lengths of each of these vectors. Remember, the lengths of each of these vectors, I don't care how many components there are, they're just going to be your numbers. So the length of this side right here is just going to be the length of a. The length of this side right here is just going to be the length of vector a minus vector b. And the length of this side right here is going to be the length of vector b."
    },
    {
        "Q": "i kind of get but killo means 1,000 or 1,000,000 on 00:14",
        "A": "Kilo means 1,000.",
        "video_name": "9iulv2QvKwo",
        "timestamps": [
            14
        ],
        "3min_transcript": "What I want to do in this video is convert this amount of kilometers into meters and centimeters. So we'll first start with the 2 kilometers. And I encourage you to now pause this video and try to do this on your own. Well, the one thing that we know is that a kilometer literally means 1,000 meters. So you could literally view this is as 2 times 1,000 meters. Let me write that down. So this is going to be equal to 2 times 1,000 meters, which is equal to 2,000 meters. If we wanted to convert the 11 kilometers into meters, it's the same thing. 11 kilometers-- this right over here-- means 1,000 meters. So you could think of it as 11 times 1,000 meters. So 11 times 1,000 is going to be 11,000 meters. Now let's convert these distances into centimeters. And here we just have to remember that 1 meter is equal to 100 centimeters. Let me write that down. And that's because the prefix centi means one hundredth. Another way you can write it is that one centimeter is equal to one hundredth of a meter. But here we have a certain number of meters, and each of those meters are going to be equivalent to 100 centimeters. So if we wanted to write 2,000 meters as centimeters, we could say, well, we have 2,000 meters. Each of those are going to be equivalent to 100 centimeters. And so this is going to be equal to 2,000 times 100. Well, that's going to be 2, and since where we have the three zeroes from the 2,000. And then every time you multiply by 10, you're going to add another zero. Or you're going to have another zero at the end of it. And we're going to be at multiplying by 100 is equivalent to multiplying by 10 twice. So we're going to have two more zeroes. So this is going to be 200,000 centimeters. with the kilometers, the 11 kilometers, which are 11,000 meters. And once again, I encourage you to pause the video and try to convert it into centimeters. Well, same idea. You have 11,000 meters, and each of them are equivalent to 100 centimeters. So this is going to be 11, and let's see, we have one, two, three, four, five zeroes. So this gets us to 1,100,000 centimeters."
    },
    {
        "Q": "The question is to write an equation, so should we stop at 6:46?",
        "A": "We could have stopped but he wanted to solve for the answer.",
        "video_name": "jQ15tkoXZoA",
        "timestamps": [
            406
        ],
        "3min_transcript": "on the left-hand side just because that might be a little bit more intuitive. So let's subtract 2p squared from both sides. Let's subtract 16p from both sides. We have an 8p and a 12p, and then we're going to subtract a 16p from both sides. And then, actually, let's subtract an 8p as well We have a 16p and an 8p, so that actually works out quite well. So now we've subtracted 8p from both sides, 16p So we've essentially subtracted all of this stuff from both sides. And we are left with-- let's see, I'll do it in degree order. 1.5p squared minus 2p squared is negative 0.5p squared. Now let's see, these cancel out. And then we have plus 64. And then that is going to be equal to 0. And just to simplify this a little bit, or just to make this a little bit cleaner, let's multiply both sides of this equation by negative 2. I want the coefficient over here to be 1. So then we get p squared plus 8p. p squared plus 8p is going to be equal to-- let's see, negative times negative 2. So minus 128-- is going to be equal to 0. So let's see if we can factor this. Can we think of two numbers where if we take their product, we get negative 128? And if we were to add them together, we get positive 8. right over here. So let's see, if we say 12 times-- well, let's see. What numbers could this be? So if we were to think about 128 is the same thing as-- 16, let's see, 16 goes into 128. Let me work through this. 16 goes into 128, does it go 8 times? 8 times 6 is 48. 8 times 10 is 80, plus 40 is 128. Yep, it goes 8 times. So 16 and 8 seem to work. So if you have positive 16 and negative 8, their product would be negative 128. So we can factor this out as p plus 16 times p minus 8 is equal to 0. Now, this is going to be equal to 0 if at least one of these"
    },
    {
        "Q": "2:32 how is the opposite and adjacent line of theta can become sin theta and cos theta? i thought sin is like opposite over adjacent something like that..",
        "A": "In the unit circle, the hypotenuse always equals 1 (it s the radius of the unit circle). Since sin(\u00ce\u00b8)=opposite/hypotenuse, and the hypotenuse equals 1, you can say that sin(\u00ce\u00b8)=opposite/1, or sin(\u00ce\u00b8)=opposite. It s the same idea for cosine. This only works for the unit circle, though. You can watch the videos on the unit circle if you haven t already.",
        "video_name": "Idxeo49szW0",
        "timestamps": [
            152
        ],
        "3min_transcript": "me-- He's just saying the tangent of some angle is equal to x. And I just need to figure out what that angle is. So let's do an example. So let's say I were walk up to you on the street. There's a lot of a walking up on a lot of streets. I would write -- And I were to say you what is the arctangent of minus 1? Or I could have equivalently asked you, what is the inverse tangent of minus 1? These are equivalent questions. And what you should do is you should, in your head-- If you don't have this memorized, you should draw the unit circle. Actually let me just do a refresher of what tangent is even asking us. The tangent of theta-- this is just the straight-up, vanilla, non-inverse function tangent --that's equal to the sine of theta over the cosine of theta. And the sine of theta is the y-value on the unit function-- on the unit circle. 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."
    },
    {
        "Q": "I really don't get how he looks at the length at about 4:40 and says \"So this is square root of 2 over 2\" is it something he memorized or can you calculate it based on the slope?",
        "A": "Sal wasn t looking at the length, really, but rather at the angle Theta (derived from the slope): Theta is -45 degrees. That means that the right triangle he draws is isosceles (with hypotenuse = 1, since this is a unit circle), and a quick calculation (or a good memory) tells us that both of the legs must therefore have a length of sqrt(2)/2.",
        "video_name": "Idxeo49szW0",
        "timestamps": [
            280
        ],
        "3min_transcript": "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 So the tangent of minus 40-- Let me write that down. So if I'm in degrees. And that tends to be how I think. So I could write the tangent of minus 45 degrees it equals this negative value-- minus square root of 2 over 2 over square root of 2 over 2, which is equal to minus 1. Or I could write the arctangent of minus 1 is equal to minus 45 degrees. Now if we're dealing with radians, we just have to convert this to radians. So we multiply that times-- We get pi radians for every 180 degrees. The degrees cancel out. So you have a 45 over 180. This goes four times. So this is equal to-- you have the minus sign-- minus pi over 4 radians."
    },
    {
        "Q": "At 3:53, what does factorial mean?",
        "A": "A FACTORIAL is when a number has the little ! next to it. That means that you multiply that number with every number before it and stop when you get to 1. For example: 5! = 5 x 4 x 3 x 2 x 1 = 120 Hope this helps! :)",
        "video_name": "W7DmsJKLoxc",
        "timestamps": [
            233
        ],
        "3min_transcript": ""
    },
    {
        "Q": "How do you know for sure if it's a conic? I mean, what if you try to simplify it and then you can't multiply it by a number so that it has the form ((x+a)^2)/b)+((y+c)^2)/d)=1? For example, what if, at 3:58, the number on the right wasn't 100? What would you do then?",
        "A": "I m going to call the number on the right z. If z wasn t a square root, Z= (sqrt(z))^2, right? So you work with that. You would go sqrt(z) in the direction needed.",
        "video_name": "cvA4VN1dpuY",
        "timestamps": [
            238
        ],
        "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": "At 0:30 how is 36 the least common multiple of 12 when doesn't it have to not be 1? I don't understand this.",
        "A": "It is not the least common multiple (lcm) of 12 but of 12 and 36. 12=12, 24, 36, 48, 60, 72 36=36, 72 36 is the least multiple 12 and 36 have in common. decompositing method. 12=2*2*3 36=2*2*3*3 So the lcm=2*2*3*3=36",
        "video_name": "znmPfDfsir8",
        "timestamps": [
            30
        ],
        "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": "At 1:39 why did you put 2 times 3 times 3",
        "A": "Because 2 x 3 x 3 = 18. (In more detail, 2x3 = 6. 6x3= 18.) Same reason with 12. 2x2 = 4 4x3=12. So that s why you see: 2x2x3 Why do we need to do this? Because we need to gather some set of numbers that can be multiplied to get both number 12 and 18, in order to find the Least Common Multiple. LCM: 2x2x3x3 = 36 <--- 2x2x3= 12 and 2x3x3 = 18",
        "video_name": "znmPfDfsir8",
        "timestamps": [
            99
        ],
        "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": "at 1:39 why did you did you put 2 times three times three?",
        "A": "2*3*3 is the prime factorization.",
        "video_name": "znmPfDfsir8",
        "timestamps": [
            99
        ],
        "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": "1:24 We're allowed to pick numbers at random? Is that \"legal\"?",
        "A": "The idea is to pick points to plot the graph. Which ones you use isn t important as long as you can draw the plot with enough precision. The more points you pick, the more accurate your graph will be around those points.",
        "video_name": "rgvysb9emcQ",
        "timestamps": [
            84
        ],
        "3min_transcript": "Let's do a couple of problems graphing linear equations. They are a bunch of ways to graph linear equations. What we'll do in this video is the most basic way. Where we will just plot a bunch of values and then connect the dots. I think you'll see what I'm saying. So here I have an equation, a linear equation. I'll rewrite it just in case that was too small. y is equal to 2x plus 7. I want to graph this linear equation. Before I even take out the graph paper, what I could do is set up a table. Where I pick a bunch of x values and then I can figure out what y value would correspond to each of those x values. So for example, if x is equal to-- let me start really low-- if x is equal to minus 2-- or negative 2, I should say-- what is y? Well, you substitute negative 2 up here. It would be 2 times negative 2 plus 7. This is negative 4 plus 7. This is equal to 3. If x is equal to-- I'm just picking x values at random points here. So what happens when x is equal to 0? Then y is going to be equal to 2 times 0 plus 7. Is going to be equal to 7. I just happen to be going up by 2. You could be going up by 1 or you could be picking numbers at random. When x is equal to 2, what is y? It'll be 2 times 2 plus 7. So 4 plus 7 is equal to 11. I could keep plotting points if I like. We should already have enough to graph it. Actually to plot any line, you actually only need two points. So we already have one more than necessary. Actually, let me just do one more just to show you that this really is a line. So what happens when x is equal to 4? Actually, just to not go up by 2, let's do x is equal to 8. Just to pick a random number. Then y is going to be 2 times 8 plus 7, which is-- well this plus 7 is equal to 23. Now let's graph it. Let me do my y-axis right there. That is my y-axis. Let me do my x-axis. I have a lot of positive values here, so a lot of space on the positive y-side. That is my x-axis. And then I use the points x is equal to negative 2. That's negative 1. That's 0, 1, 2, 3, 4, 5, 6, 7, 8. Those are our x values. Then we can go up into the y-axis. I'll do it at a slightly different scale because these numbers get large very quickly. So maybe I'll do it in increments of 2. So this could be 2, 4, 6, 8, 10, 12, 14, 16."
    },
    {
        "Q": "1:16-1:46 he does the translation visually. Is there any way to do it with an equation.",
        "A": "Yes, to translate a figure 2 spots to the right you just add 2 to its x value. So E (3+2,3) -> E (5,3).",
        "video_name": "XiAoUDfrar0",
        "timestamps": [
            76,
            106
        ],
        "3min_transcript": "- [Voiceover] What I hope to introduce you to in this video is the notion of a transformation in mathematics, and you're probably used to the word in everyday language. Transformation means something is changing, it's transforming from one thing to another. What would transformation mean in a mathematical context? Well, it could mean that you're taking something mathematical and you're changing it into something else mathematical, that's exactly what it is. It's talking about taking a set of coordinates or a set of points, and then changing them into a different set of coordinates or a different set of points. For example, this right over here, this is a quadrilateral we've plotted it on the coordinate plane. This is a set of points, not just the four points that represent the vertices of the quadrilateral, but all the points along the sides too. There's a bunch of points along this. You could argue there's an infinite, or there are an infinite number of points along this quadrilateral. This right over here, the point X equals 0I equals negative four, this is a point on the quadrilateral. Now, we can apply a transformation to this, which just means moving all the points in the same direction, and the same amount in that same direction, and I'm using the Khan Academy translation widget to do it. Let's translate, let's translate this, and I can do it by grabbing onto one of the vertices, and notice I've now shifted it to the right by two. Every point here, not just the orange points has shifted to the right by two. This one has shifted to the right by two, this point right over here has shifted to the right by two, every point has shifted in the same direction by the same amount, that's what a translation is. Now, I've shifted, let's see if I put it here every point has shifted to the right one and up one, they've all shifted by the same amount in the same directions. That is a translation, but you could imagine a translation is not the only kind of transformation. In fact, there is an unlimited variation, there's an unlimited number different transformations. I have another set of points here that's represented by quadrilateral, I guess we could call it CD or BCDE, and I could rotate it, and I rotate it I would rotate it around the point. So for example, I could rotate it around the point D, so this is what I started with, if I, let me see if I can do this, I could rotate it like, actually let me see. So if I start like this I could rotate it 90 degrees, I could rotate 90 degrees, so I could rotate it, I could rotate it like, that looks pretty close to a 90-degree rotation. So, every point that was on the original or in the original set of points I've now shifted it relative to that point that I'm rotating around. I've now rotated it 90 degrees, so this point has now mapped to this point over here. This point has now mapped to this point over here, and I'm just picking the vertices because those are a little bit easier to think about."
    },
    {
        "Q": "at 5:09, i dont understand how it is the same distance",
        "A": "Most transformations, translations, rotations, and reflections all end up with an image that is congruent to the pre-image, so same parts of congruent figures are congruent. The video has the rotation slightly off.",
        "video_name": "XiAoUDfrar0",
        "timestamps": [
            309
        ],
        "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": "At 4:30 how does Sal go from B/60*-5 into -b/12?",
        "A": "Multiply: b/60 * -5/1 = (-5b)/60 Cancel out common factor of 5 and you get: -b/12 Hope this helps.",
        "video_name": "Q0tTfe2lKIc",
        "timestamps": [
            270
        ],
        "3min_transcript": "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, This bottom equation, if I multiply both sides by negative five, so both sides by negative five, I'm going to multiply both sides by negative five I'm going to get negative five times W over five is negative W, negative five times B over 60. Let's see, it's gonna be, it's going to be negative five over 60, that's negative 1/12, so this is negative B over 12 and then it is going to be equal to 1.5 times negative five is negative 7.5, negative 7.5. Now we can just add the left and right hand sides of these two equations. Now let me, I can do this a little bit neater, let me actually delete, let me make these line up a little bit better so that we, delete that, make this, so this first equation was, whoops,"
    },
    {
        "Q": "At around 6:11 , how is it possible that X can be less or equal to -3? Wouldn't that make the square root negative?",
        "A": "No, because the absolute value of anything less or equal to -3 is larger than 3, i.e. it is more than 3 removed from zero. Subtracting 3 from something larger than 3 gives a positive outcome, and hence a real solution to the problem. For instance x = -4. Abs value (-4) = 4 4 - 3 = 1 Square root (1) = 1.",
        "video_name": "U-k5N1WPk4g",
        "timestamps": [
            371
        ],
        "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": "Take the answer of problem at 4:51 . And, let x=12. So, f(x)=(sqrt)9 and\n(sqrt) of 9 has 2 values: 3 and -3. Which means, f(12) = -3 or f(12)= 3 which is not possible according to the definition of a function. How can this be explained??",
        "A": "The notation \u00e2\u0088\u009a9 exclusively denotes the POSITIVE value. Its corresponding negative root is denoted as -\u00e2\u0088\u009a9. However, when a positive value does not exist, the radical sign may be used to indicate the negative real value. Eg. ^3\u00e2\u0088\u009a-8 = -2.[The cube root of -8, that can be expressed on a number line, is -2, because (-2)*(-2)*(-2) = -8] That s only when we used symbols; when we say, or write, square root of a certain number, it refers to all applicable values.",
        "video_name": "U-k5N1WPk4g",
        "timestamps": [
            291
        ],
        "3min_transcript": "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 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"
    },
    {
        "Q": "At around 4:25, Sal said the square root couldn't be a negative number, so x would have had to be 3 or more. Wouldn't it have to be 4 or more because 3 - 3 = 0?",
        "A": "As Sal says: there is nothing wrong with the expression sqrt(0). Its just zero.",
        "video_name": "U-k5N1WPk4g",
        "timestamps": [
            265
        ],
        "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": "Mr Sal at 3:43 you said\nF(x)=square root of x-3greater or equal to 0\nHow can it be equal to 0, if I plug 0 in place of x i would get -3 which is wrong??",
        "A": "Notice the subtle distinction. Sal did not say that x was greater than or equal to 0, but that x-3 was. So x must be greater than or equal to 3.",
        "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": "Why Sal did say at 6:28 the domain of function f(x) = sqrt(abs(x)-3) is x\u00e2\u0082\u00acR | x<=-3 and x>=3 ?\nIts a mistake? We can have a positive square root of 0 ?",
        "A": "At this point, Sal says the domain is all real values of x such that x is less than or equal to -3 and greater than or equal to 3. When that value is in the denominator, then the domain is x<=-3 and x>=3 because the denominator cannot be 0. They were two separate equations.",
        "video_name": "U-k5N1WPk4g",
        "timestamps": [
            388
        ],
        "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": "So in 0:38, you said that you can add up all the digits (154 adding up to 10 in the video). Does that method always work? I mean, I understand that 5 would not be able to go into 154 because the result always ends in either 0 or 5 as mentioned, but I'm just a bit confused. Could someone possibly explain if the method really works for all numbers?",
        "A": "Lets tell parts of division 15\u00c3\u00b75=3 15=dividend formula=d \u00c3\u00b7=division sign 5=divisior formula=D 3=quotient formula=q 0=remainder formula=r Euclid was a great mathmatician who made a division statement (Divisior\u00c3\u0097quotient)+remainder =dividend Eg=lets take 67\u00c3\u00b76=11 (6\u00c3\u009711)+1=67",
        "video_name": "5xe-6GPR_qQ",
        "timestamps": [
            38
        ],
        "3min_transcript": "Which of the following numbers is a factor of 154? So when a number is going to be a factor of 154 is if we can divide that number into 154 and not have a remainder. Or another way of thinking about it-- a number is a factor of 154 if 154 is a multiple of that number. So let's look at each of these and see which of these we can rule out or say is a factor. So does 3 divide evenly into 154? Or, another way of thinking about it, is 154 a multiple of 3? Well you'll later learn that you could actually test whether something is divisible by 3 by adding up the digits. And if that's divisible by 3, then it's going to be divisible by 3. And so you see here, 1 plus 5 is 6. 6 plus 4 is 10. 10 is not divisible by 3. But if you didn't want to do that little trick-- and we have other videos where we go into more detail about that trick-- you can actually just divide 3 into 154. 3 doesn't go into 1. It does go into 15 five times. Subtract. Then we have 0. Then you bring down a 4. 3 goes into 4 one time. 1 times 3 is 3. Subtract, and you have a remainder of 1. So 3 is clearly not a factor of 154, so we can rule that out. Now what about 5? Well, any multiple of 5 is either going to have 5 or 0 in the ones place. You see that if we write 5 times 1 is 5, 5 times 2 is 10, 5 times 3 is 15, 5 times 4 is 20, you either have a 5 or a 0 in the ones place. This does not have 5 or a 0 the ones place, so it's not going to be divisible by 5. 5 is not a factor. 154 is not a multiple of 5. Now 6 is interesting. You could do the same thing. You could try to divide 6 into 154. But if something is divisible by 6, it's definitely going to be divisible by 3 as well because 6 is divisible by 3. So we can immediately rule this one out as well. it's also not going to be divisible by 6. And you could try it out if you like. And we could make the same argument for 9. If something is divisible by 9, it's going to be divisible by 3 because 9 is divisible by 3. Well, it's not divisible by 3, so we're going to rule out 9 as well. So we've ruled out everything. It looks like 14 is our only option, but let's actually verify it. Let's actually divide 14 into 154. 14 doesn't go into 1. It goes into 15 exactly one time. 1 times 14 is 14. We subtract. We get 1. Bring down the 4. 14 goes into 14 one time. 1 times 14 is 14. And of course, we have no remainder. So 14 goes into 154 exactly 11 times. Or 11 times 14 is 154. 154 is a multiple of 14. Let's do one more. Which of the following numbers is a multiple of 14?"
    },
    {
        "Q": "I think \"9:50\" does not need a proof as they're just i j k l unit vectors.",
        "A": "9:54 A proof may be simple, but still needed. That is the case here.",
        "video_name": "JUgrBkPteTg",
        "timestamps": [
            590
        ],
        "3min_transcript": "By the same reasoning, no combination of that and that is going to equal this. This is by definition of a pivot entry. When you put it in reduced row echelon form, it's very clear that any pivot column will be the only one to have 1 in that place. So it's very clear that these guys are linearly independent. Now it turns out, and I haven't proven it to you, that the corresponding columns in A-- this is r1, but it's A before we put it in reduced row echelon form-- that these guys right here, so a1, a2, and a4 are also linearly independent. So a1-- let me circle it-- a2, and a4. So if I write it like this, a1, a2, and a4. Let me write it in set notation. These guys are also linearly independant, which I haven't proven. But I think you can kind of get a sense that these row And I'll do a better explanation of this, but I really just wanted you to understand how to develop a basis for the column space. So they're linearly independent. So the next question is do they span our column space? And in order for them to span, obviously all of these 5 vectors, if you have all of them, that's going to span your column space by definition. But if we can show, and I'm not going to show it in this video, but it turns out that you can always represent the non-pivot columns as linear combinations of the pivot columns. And we've kind of touched on that in previous videos where we find a solution for the null space and all that. So these guys can definitely be represented as linear combinations of these guys. I haven't shown you that, but if you take that on faith, then you don't need that column and that column to span. If you did then, or I guess a better way to think it, you are part of the span. Because if you needed this guy, you can just construct him with linear combinations of these guys. So if you wanted to figure out a basis for the column space of A, you literally just take A into reduced row echelon form. You look at the pivot entries in the reduced row echelon form of A, and that's those three. And then you look at the corresponding columns to those pivot columns in your original A. And those form the basis. Because any linear combination of them, or linear combinations of them can be used to construct the non-pivot columns, and they're linearly independant. So I haven't shown you that. But for this case, if you want to know the basis, it's just a1, a2, and a4. And now we can answer another question. So a1, a2, and a4 form a basis for the column space of A,"
    },
    {
        "Q": "at 7:19 sal said it was equal to 112 it is equal to 108 You may do the check if you want 36+108 is 144",
        "A": "Well, yeah, I guess. But I think he was a little rushed.",
        "video_name": "s9t7rNhaBp8",
        "timestamps": [
            439
        ],
        "3min_transcript": "So, we know that C is equal to 5 So, using the Pythagorean Theorem we just figured out that if we know the sides of, one side is 3, the other side is 4 then we can use Pythagorean Theorem for that Hypotenuse of this triangle it has the length of 5 Let's do another example Let's say, once again this is a right angle This side is of length 12, this side is of length 6 and I wanna figure out what this side is So, let's write down the Pythagorean Theorem A squared plus B squared is equal to C squared Where C is the length of the hypotenuse that I just drew Is which side is the hypotenuse? Well this right here, is the right angle So, the hypotenuse is this side right here And we can also eyeball it and say \"oh that's definitely the longest side of this triangle \" So, we know that A squared plus B squared is equal to 12 squared, which is 144 Now, we know that we have one side but we don't have the other side So, I'm gonna ask you as question Does it matter, which side we substitute for A or B? This is because A or B kinda do the same thing in this formula So, we can pick any side to be A other than the hypotenuse And we'll pick the other side to be B So, let's just say that this side is B and let's say this side is A So, we know what A is, so we get 6 squared plus B squared is equal to 144 So, we get 36 plus B squared is equal to 144 Now we got to simplify what the square root of 112 is What we did on those radical modules was probably helpful here So, B is equal to the square root of 112 Let's think about it, how many times does 4 go into a 112? 4 goes into a 125 times, it will go into it 28 times And 4 goes into 28, 7 times I actually think that this is equal to 16 times 7, am I right? 7 times 10 is 70 plus 42 is a 112 So, B equals the square root of 16 times 7 You see, I just factored that as a product of a perfect and a prime number"
    },
    {
        "Q": "During the video at about 7:10 Sal goes a little to fast,I don't understand the\nrest of the formula. I'm struggling with the last few steps.I know 6 squared is\n36 (side A)and the hypotenuse(side c)12 squared is equal to 144.after this point\ni'm lost. I don't know my square roots real well and I struggle with formulas that\nare more than 3 steps. I know w/lots of practice i will get it.any \"positive\" tips\nmuch appreciated.",
        "A": "144-36 = 108, not 112. Other than that, you are correct once you adjust the incorrect values.",
        "video_name": "s9t7rNhaBp8",
        "timestamps": [
            430
        ],
        "3min_transcript": "So, we know that C is equal to 5 So, using the Pythagorean Theorem we just figured out that if we know the sides of, one side is 3, the other side is 4 then we can use Pythagorean Theorem for that Hypotenuse of this triangle it has the length of 5 Let's do another example Let's say, once again this is a right angle This side is of length 12, this side is of length 6 and I wanna figure out what this side is So, let's write down the Pythagorean Theorem A squared plus B squared is equal to C squared Where C is the length of the hypotenuse that I just drew Is which side is the hypotenuse? Well this right here, is the right angle So, the hypotenuse is this side right here And we can also eyeball it and say \"oh that's definitely the longest side of this triangle \" So, we know that A squared plus B squared is equal to 12 squared, which is 144 Now, we know that we have one side but we don't have the other side So, I'm gonna ask you as question Does it matter, which side we substitute for A or B? This is because A or B kinda do the same thing in this formula So, we can pick any side to be A other than the hypotenuse And we'll pick the other side to be B So, let's just say that this side is B and let's say this side is A So, we know what A is, so we get 6 squared plus B squared is equal to 144 So, we get 36 plus B squared is equal to 144 Now we got to simplify what the square root of 112 is What we did on those radical modules was probably helpful here So, B is equal to the square root of 112 Let's think about it, how many times does 4 go into a 112? 4 goes into a 125 times, it will go into it 28 times And 4 goes into 28, 7 times I actually think that this is equal to 16 times 7, am I right? 7 times 10 is 70 plus 42 is a 112 So, B equals the square root of 16 times 7 You see, I just factored that as a product of a perfect and a prime number"
    },
    {
        "Q": "From @7:45 to @10:36 , what are you trying to achieve when you convert the matrix to reduced row-echelon form?\n\nI thought you were trying to turn row 2 and 3 to zeros, like you did with the 2x2 matrix.\n\nWhy do you have to convert it to row-echelon form?\n\nThank for the videos btw, they've been really helpful.",
        "A": "rref is the solution for the system of equations represented by the augmented matrix. You re finding the vectors v such that Av = lambda*v - i.e., the solution for (lambda*I - A)v = 0; i.e., rref ( [ (lambda*I - A) | 0 ] ).",
        "video_name": "3Md5KCCQX-0",
        "timestamps": [
            465,
            636
        ],
        "3min_transcript": "And they're the eigenvectors that correspond to eigenvalue lambda is equal to 3. So if you apply the matrix transformation to any of these vectors, you're just going to scale them up by 3. Let me write this way. The eigenspace for lambda is equal to 3, is equal to the span, all of the potential linear combinations of this guy and that guy. So 1/2, 1, 0. And 1/2, 0, 1. So that's only one of the eigenspaces. That's the one that corresponds to lambda is equal to 3. Let's do the one that corresponds to lambda is equal to minus 3. So if lambda is equal to minus 3-- I'll do it up here, I think I have enough space-- lambda is equal to minus 3. This matrix becomes-- I'll do the diagonals-- minus 3 plus 1 is minus 2. Minus 3 minus 2 is minus 5. 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."
    },
    {
        "Q": "At about 5:30, Sal has 9x36/2, then he divides the 36 by 2, getting 9x18/1. Why doesn't he divide the 9 by 2 also?",
        "A": "36/2 is 36 divided by 2. So then he gets 9 x 18 becuz 36 divided by 2 is 18. Even if you do 9 x 36 and 9 x 2 then you will get 324/18 which is still 18.",
        "video_name": "jFSenp9ueaI",
        "timestamps": [
            330
        ],
        "3min_transcript": "So let's do it that way. If I have 4 and 1/2 of anything, really-- so let me write 4 and 1/2. I'm trying to find a suitable color. So if I have 4 and 1/2, this is the same thing-- 4 is the same thing as 8/2. This is the same thing-- so let me write it this way. 4 and 1/2 is the same thing as 4 plus 1/2, which is the same thing as-- if we want to have the same denominator as this 2 over here or as this 1/2 over here, this is the same thing as 8/2. Or you could say 4/1 is the same thing as 8/2, if we want to have a common denominator, so 8/2 plus 1/2. Actually, let me write it that way, just so you really understand what we're doing. 4 is the same thing as 4/1. So it's 4/1 plus 1/2. If we want to find a common denominator, it's 2. So 4/1 is the same thing as 8/2 plus 1/2, which is equal to 9/2. Now, I did it this way, which takes longer, why it makes-- hopefully conceptually why it just makes intuitive sense, why 4 and 1/2 is the same thing as 9/2. But if you want a simple process for it, you could just say, look, 4 times 2 is 8. 8 plus 1 is 9. And that gives you that 9 right over there, so 9/2. So we have 9/2 yards that we want to convert to inches. Same process-- times 36 inches per yard. Yard in the numerator, yard in the denominator. We are left with 9/2 times 36. We could say times 36/1 if we like, 36 Inches for every 1 yard. 36-- or the number 36 really is the same as 36/1. And then we're left with just inches in our units. We're just left with inches. And over here there's several ways that we can simplify it. Probably the easiest way to simplify it is we So let me write it this way. I don't want to skip steps. So we have 9 times 36 over 2 times 1, or over 2 inches. And we can divide the numerator and the denominator by 2 to simplify it. They're both divisible by 2. 36 divided by 2 is 18. 2 divided by 2 is 1. So we're really just left with 9 times 18 inches. We can just multiply 9 times 18. Let me do it over here. 18 times 9. 8 times 9 is 72. 1 times 9 is 9, plus 7 is 16, so we get 162 inches. So all of this simplifies to 162 inches, and we are done."
    },
    {
        "Q": "At 4:21 , the standard deviation of thesample size is given is given as 0.5s why do we have to determine it.",
        "A": "because the other one we determined is the standard deviation of the sampling distribution of the sample mean.",
        "video_name": "-FtlH4svqx4",
        "timestamps": [
            261
        ],
        "3min_transcript": "So how do we think about this? How do we know whether we should accept the alternative hypothesis or whether we should just default to the null hypothesis because the data isn't convincing? And the way we're going to do it in this video, and this is really the way it's done in pretty much all of science, is you say OK, let's assume that the null hypothesis is true. If the null hypothesis was true, what is the probability that we would have gotten these results with the sample? And if that probability is really, really small, then the null hypothesis probably isn't true. We could probably reject the null hypothesis and we'll say well, we kind of believe in the alternative hypothesis. So let's think about that. Let's assume that the null hypothesis is true. So if we assume the null hypothesis is true, let's try gotten this result, that we would have actually gotten a sample mean of 1.05 seconds with a standard deviation of 0.5 seconds. So I want to see if we assumed the null hypothesis is true, I want to figure out the probability-- and actually what we're going to do is not just figure out the probability of this, the probability of getting something like this or even more extreme than this. So how likely of an event is that? To think about that let's just think about the sampling distribution if we assume the null hypothesis. So the sampling distribution is like this. It'll be a normal distribution. We have a good number of samples, we have 100 samples here. So this is the sampling distribution. It will have a mean. Now if we assume the null hypothesis, that the drug has no effect, the mean of our sampling distribution will be distribution, which would be equal to 1.2 seconds. Now, what is the standard deviation of our sampling distribution? The standard deviation of our sampling distribution should be equal to the standard deviation of the population distribution divided by the square root of our sample size, so divided by the square root of 100. We do not know what the standard deviation of the entire population is. So what we're going to do is estimate it with our sample standard deviation. And it's a reasonable thing to do, especially because we have a nice sample size. The sample size is greater than 100. So this is going to be a pretty good approximator. This is going to be a pretty good approximator for this over here. So we could say that this is going to be approximately equal to our sample standard deviation divided by the square root of 100, which is going to be equal to our sample standard deviation is 0.5, 0.5 seconds, and we want"
    },
    {
        "Q": "At 2:11, how do you know where to subtract the 2 from, like can you subtract it with any of the numbers on the other side of the equation?",
        "A": "It s basic equation balancing. To isolate the variable, you have to get rid of the 2 in the equation. To do so but not change the equation, you have to minus 2 from both sides so that technically it s still the same equation just simplified down so the 2 is gone.",
        "video_name": "qsL_5Y8uWPU",
        "timestamps": [
            131
        ],
        "3min_transcript": "Determine the number of solutions for each of these equations, and they give us three equations right over here. And before I deal with these equations in particular, let's just remind ourselves about when we might have one or infinite or no solutions. You're going to have one solution if you can, by solving the equation, come up with something like x is equal to some number. Let's say x is equal to-- if I want to say the abstract-- x is equal to a. Or if we actually were to solve it, we'd get something like x equals 5 or 10 or negative pi-- whatever it might be. But if you could actually solve for a specific x, then you have one solution. So this is one solution, just like that. Now if you go and you try to manipulate these equations in completely legitimate ways, but you end up with something crazy like 3 equals 5, then you have no solutions. And if you just think about it reasonably, all of these equations are about finding an x that satisfies this. 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."
    },
    {
        "Q": "what is the triangle at 1:00 for?",
        "A": "It is the greek letter delta, which is commonly used today in modern applications of science and mathematics to represent a change or difference. Just like how we use pi in geometry, theta in trigonometry, lambda in calculus, phi to represent the golden ration, and more.",
        "video_name": "f4MYCepzLyQ",
        "timestamps": [
            60
        ],
        "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": "At 3:02, Sal talks about slope-intercept form. Can anyone give me an explanation of what that is and how it can be used to find average rate of change.",
        "A": "y = mx + b is slope-intercept form",
        "video_name": "f4MYCepzLyQ",
        "timestamps": [
            182
        ],
        "3min_transcript": "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. our D intercept, when t is equal to 0, is going to be equal to one. All of this is review and if any of this sounds foreign to you, I encourage you to watch the Khan Academy videos on slope and slope intercept form and things like that. But this is all a background to get us to this curve over here because this is interesting. Because here we're no longer dealing with a line. We actually have a curve and this curve right over here. It's a quadratic, it's a parabola. Let's just say distance as a function of time was defined this way instead of it being 3t plus 1 it is t squared plus 1 and what's interesting here is that the rate of change you can have visualized it is it's always changing for example: If you were to pick an arbitrary point on this curve and if you think about a tangent line to it. A tangent line, a line that has the same slope for just that one moment, it just touches on it, over here"
    },
    {
        "Q": "At 4:00 Sal says 350/360 is = to 35/36 What rules dictate this?",
        "A": "That s how you simplify a fraction. Divide the top and bottom numbers by the same factor (in this case 10) to simplify. Another example: 3/6 = 1/2 because you divide the top and bottom numbers by 3.",
        "video_name": "tVcasOt55Lc",
        "timestamps": [
            240
        ],
        "3min_transcript": "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. subtends this really obtuse angle right over here. Well, same exact logic-- the ratio between our arc length, a, and the circumference of the entire circle, 18 pi, should be the same as the ratio between our central angle that the arc subtends, so 350, over the total number of degrees in a circle, over 360. So multiply both sides by 18 pi. We get a is equal to-- this is 35 times 18 over 36 pi. 350 divided by 360 is 35/36. So this is 35 times 18 times pi over 36. so let's divide them both by 18. And so we are left with 35/2 pi. Let me just write it that way-- 35 pi over 2. Or, if you wanted to write it as a decimal, this would be 17.5 pi. Now does this makes sense? This right over here, this other arc length, when our central angle was 10 degrees, this had an arc length of 0.5 pi. So when you add these two together, this arc length and this arc length, 0.5 plus 17.5, you get to 18 pi, which was the circumference, which makes complete sense because if you add these angles, 10 degrees and 350 degrees, you get 360 degrees in a circle."
    },
    {
        "Q": "Could you cross multiply at 1:32 ?",
        "A": "You mean turning it into a = ====1 ------ ------- 1.8pi 360? If so, then yes",
        "video_name": "tVcasOt55Lc",
        "timestamps": [
            92
        ],
        "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": "4:25 I don't understand why 5(7k(k+3)-(k+3)) becomes 5(k+3)(7k-1)",
        "A": "Try this : If you had for say, 6(k+3) - (k+3), you could clearly say that it is equal to 5(k+3) right? Which is equal to (6-1)(k+3). Same thing is happening in here. You have 7k(k+3)-(k+3) and you are free to transform it into (7k-1)(k+3).",
        "video_name": "R-rhSQzFJL0",
        "timestamps": [
            265
        ],
        "3min_transcript": "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. And so what can we factor out of that group right there? Well, both of these are divisible by 7k, so we can write this as 7k times-- 7k squared divided by 7k, you're just going to have a k left over. And then plus 21k divided by 7k is just going to be a 3. So that factors into that. And then we can look at this group right here. They have a common factor. Well, we can factor out a negative 1 if we like, so this is equal to negative 1 times-- k divided by negative 1 is k. Negative 3 divided by negative 1 is positive 3. And, of course, we have this 5 sitting out there. Now, ignoring that 5 for a second, you see that both of these inside terms have k plus 3 as a factor. So let's ignore this 5 for a second. This inside part right here, the stuff that's inside the parentheses, we can factor k plus 3 out, and it becomes k plus 3, times k plus 3, times 7k minus 1. And if this seems a little bizarre to you, distribute the k plus 3 on to this. K plus 3 times 7k is that term, k plus 3 times negative 1 is that term. And, of course, the whole time you have that 5 sitting outside. You have that 5. We don't even have to put parentheses there. 5 times k plus 3, times 7k minus 1. And we factored it, we're done."
    },
    {
        "Q": "Hi at 1:03 - 1:06, Sal mentions finding a number whose product is 7 * -3. Why isn't he simply referring to the number -3 as his product?\n\nThanks in advance.",
        "A": "The product of two numbers is the number that results from multiplying two numbers together. For example, -3\u00c3\u00977 equals -21. In this instance, -21 is the product. I hope this helps!",
        "video_name": "R-rhSQzFJL0",
        "timestamps": [
            63,
            66
        ],
        "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": "At 1:10, why can we not write {(-1)(-52)}^1/2 = {(-1)^1/2}{(-52)^1/2} ?",
        "A": "I believe it would be mathematically incorrect, because sqrt(-1 x -52) is sqrt(52). However, sqrt(-1) x sqrt(-52) is not the same as sqrt(-1 x -52) because sqrt(-1) x sqrt(-52) is equal to sqrt(-1) x sqrt(52) x sqrt(-1) = -1 x sqrt(52) which is not the same as sqrt(-52). Sorry you got an answer to your question 2 years after you asked it, hopefully this helps! ;-} By the way, I can do the special effects for the iron-man movies, and am coming out with The last Days . if your interested.",
        "video_name": "s03qez-6JMA",
        "timestamps": [
            70
        ],
        "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 3:35, why is it i*\u00e2\u0088\u009a4*13 and not i^2*\u00e2\u0088\u009a4*13? I thought that by definition i^2= -1",
        "A": "i^2 = -1 i = \u00e2\u0088\u009a(-1)",
        "video_name": "s03qez-6JMA",
        "timestamps": [
            215
        ],
        "3min_transcript": "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. And then we're going to multiply that times the square root of 4 times 13. And this is going to be equal to i times the square root of 4. i times the square root of 4, or the principal square root of 4 times the principal square root of 13. The principal square root of 4 is 2. So this all simplifies, and we can switch the order, over here. This is equal to 2 times the square root of 13. 2 times the principal square root of 13, I should say, times i. And I just switched around the order. It makes it a little bit easier to read if I put the i after the numbers over here. But I'm just multiplying i times 2 times the square root of 13. That's the same thing as multiplying 2 times the principal square root of 13 times i. And I think this is about as simplified as we can get here."
    },
    {
        "Q": "At 2:04, Sal says that I can not split sqrt(-1 x -52) into sqrt(-1) x sqrt(-52). Can I go the opposite way? Would it be mathematically correct to simplify: sqrt(-1) x sqrt(-52) into sqrt(-1 x -52)?",
        "A": "I believe it would be mathematically correct, because sqrt(-1 x -52) is sqrt(52). However, sqrt(-1) x sqrt(-52) is not the same as sqrt(-1 x -52) because sqrt(-1) x sqrt(-52) is equal to sqrt(-1) x sqrt(52) x sqrt(-1) = -1 x sqrt(52) which is not the same as sqrt(-52). Sorry you got an answer to your question 2 years after you asked it, hopefully this helps! ;-}",
        "video_name": "s03qez-6JMA",
        "timestamps": [
            124
        ],
        "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 9:01 Sal messed up right? Isn't it supposed to be 365!/362!/(365)to the third??",
        "A": "Yes; he did mess up at 9:01",
        "video_name": "9G0w61pZPig",
        "timestamps": [
            541
        ],
        "3min_transcript": "That'll actually be 30 terms divided by 365 to the 30th power. And you can just type this into your calculator right now. It'll take you a little time to type in 30 numbers, and you'll get the probability that no one shares the same birthday with anyone else. But before we do that let me just show you something that might make it a little bit easier. Is there any way that I can mathematically express this with factorials? Or that I could mathematically express this with factorials? Let's think about it. 365 factorial is what? 365 factorial is equal to 365 times 364 times 363 times-- all the way down to 1. You just keep multiplying. It's a huge number. Now, if I just want the 365 times the 364 in this case, One thing I could do is I could divide this thing by all of these numbers. So 363 times 362-- all the way down to 1. So that's the same thing as dividing by 363 factorial. 365 factorial divided by 363 factorial is essentially this because all of these terms cancel out. So this is equal to 365 factorial over 363 factorial over 365 squared. And of course, for this case, it's almost silly to worry about the factorials, but it becomes useful once we have something larger than two terms up here. So by the same logic, this right here is going to be equal to 365 factorial over 362 factorial over 365 squared. How did we get this 365? Sorry, how did we get this 363 factorial? Well, 365 minus 2 is 363, right? And that makes sense because we only wanted two terms up here. We only wanted two terms right here. So we wanted to divide by a factorial that's two less. And so we'd only get the highest two terms left. This is also equal to-- you could write this as 365 factorial divided by 365 minus 2 factorial 365 minus 2 is 363 factorial and then you just end up with those two terms and that's that there. And then likewise, this right here, this numerator you could rewrite as 365 factorial divided by 365 minus 3-- and we had 3 people-- factorial. And that should hopefully make sense, right? This is the same thing as 365 factorial-- well 365 divided"
    },
    {
        "Q": "on 0:59 he is a horrible stock investor\nVote up if u agree",
        "A": "There isn t enough info to know. -- The problem didn t tell you the time frame. If the portfolio went up in 1 month by 25%, that would be fantastic! -- If the overall market is down for the year by 20% and this person s portfolio went up 25%, that would also be fantastic! -- If it took 20 years for his portfolio to grow 25%, that would not be good, unless his goal was to project his portfolio (low risk). Then, it would be good.",
        "video_name": "X2jVap1YgwI",
        "timestamps": [
            59
        ],
        "3min_transcript": "Let's do some more percentage problems. Let's say that I start this year in my stock portfolio with $95.00. And I say that my portfolio grows by, let's say, 15%. How much do I have now? I think you might be able to figure this out on your own, but of course we'll do some example problems, just in case it's a little confusing. So I'm starting with $95.00, and I'll get rid of the dollar sign. We know we're working with dollars. 95 dollars, right? And I'm going to earn, or I'm going to grow just because I was an excellent stock investor, that 95 dollars So to that 95 dollars, I'm going to add another 15% of 95. So we know we write 15% as a decimal, as 0.15, so 95 plus 0.15 of 95, so this is times 95-- that dot is just a times sign. It's not a decimal, it's a times, it's a little higher than a decimal-- So 95 plus 0.15 times 95 is what we have now, right? Because we started with 95 dollars, and then we made another 15% times what we started with. Hopefully that make sense. Another way to say it, the 95 dollars has grown by 15%. So let's just work this out. This is the same thing as 95 plus-- what's 0.15 times 95? Let's see. So let me do this, hopefully I'll have enough space here. 95 times 0.15-- I don't want to run out of space. 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."
    },
    {
        "Q": "4:48 Hey, does anyone know why Sal puts the = sign like a smiley face? =D",
        "A": "It s not an equal sign, but rather an arrow. He just draws it in a way that it doesn t look quite connected. This is actually a common way to note progression of steps in mathematics.",
        "video_name": "X2jVap1YgwI",
        "timestamps": [
            288
        ],
        "3min_transcript": "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%. 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"
    },
    {
        "Q": "What's the next step if the number doesn't come out to be exact like 13*16 for 208? (2:19 in video)",
        "A": "you go and find the number of ones. example: 114 = 7 * 256 + 16 *13 (D) + 0 * 1 there are 0 1 s. if the number were 115, then it would be 7D1 because there an extra one as supposed to 7D0 = 114. hope this helped",
        "video_name": "NFmDz1dQyPU",
        "timestamps": [
            139
        ],
        "3min_transcript": "-[Voiceover]Let's now try to convert a number from the decimal system to the hexadecimal system. Let's say we want to convert the number 2,000. This is written in decimal form. Let's say we want to write it in hexadecimal form. Like always, I encourage you to pause this video and try to work through it on your own. The key here is to break this down into multiples of powers of 16. Let's just write down our powers of 16 here. 16 to the zero power is one. 16 to the first power is 16. 16 squared is 256. 16 to the third power is 4,096. We've gotten more than large enough. Let's start decomposing. What's the largest power of 16 that is less than or equal to 2,000? It's going to be 256. How many times does 256 go into 2,000? I'll get a calculator out for that. times 256 goes into it, divided by 256. It goes seven times, plus a little bit. What's going to be left over? 2,000 minus seven times 256 is equal to, you're going to have 208 left over. Let me write that. It's going to be seven times 256. Seven, times 256, plus 208 left over. Now let's see if we can decompose 208 into powers of 16. What's the largest power of 16 that is less than or equal to this? Well it's just going to be 16. So how many 16s go into 208? 208, I'll just take that, divided by 16. 13, exactly 13. We lucked out. So this is exactly 13 times 16. This right over here is exactly 13 times 16. Now, we have broken this down into powers of... We have broken this down into multiples of powers of 16. Now we're ready to write it in hexadecimal. We just have to remind ourselves about the place value. This right over here, this is going to be the ones place. Then we're going to have the 16s place. Then we're going to have the 256s place. We know how many 256s we have. We have seven 256s. We have zero ones. And how many 16s do we have? Well we have 13 16s. Well what's the digit for 13? We can just remind ourselves that we obviously have zero through nine and then we have A is equal to 10, B is equal to 11, C is equal to 12,"
    },
    {
        "Q": "At 4:21, Sal says that that 0i is the same thing as 0 + i. Wouldn't be 0 times i? I am confused.",
        "A": "No, he says that i is the same thing as 0 + i , which is true. You are right that 0i is just 0 times i , not 0 + i . So 0i is the same thing as just 0 .",
        "video_name": "A_ESfuN1Pkg",
        "timestamps": [
            261
        ],
        "3min_transcript": "And 9 is a real number. So we could just add those up. So 2 plus negative 7 would be negative 5. Negative 5 plus 9 would be 4. So the real numbers add up to 4. And now we have these imaginary numbers. So 3 times i minus 5 times i. So if you have 3 of something and then I were to subtract 5 of that same something from it, now you're going to have negative 2 of that something. Or another way of thinking about it is the coefficients. 3 minus 5 is negative 2. So three i's minus five i's, that's going to give you negative 2i. Now you might say, well, can we simplify this any further? Well no, you really can't. This right over here is a real number. 4 is a number that we've been dealing with throughout our mathematical careers. And so what we really consider this is this 4 minus 2i, we can now consider this entire expression to really be a number. So this is a number that has a real part and an imaginary part. And numbers like this we call complex numbers. It is a complex number. Why is it complex? Well, it has a real part and an imaginary part. And you might say, well, gee, can't any real number be considered a complex number? For example, if I have the real number 3, can't I just write the real number 3 as 3 plus 0i? And you would be correct. Any real number is a complex number. You could view this right over here as a complex number. a subset of the complex numbers. Likewise, imaginary numbers are a subset of the complex numbers. For example, you could rewrite i as a real part-- 0 is a real number-- 0 plus i. So the imaginaries are a subset of complex numbers. Real numbers are a subset of complex numbers. And then complex numbers also have all of the sums and differences, or all of the numbers that have both real and imaginary parts."
    },
    {
        "Q": "at 0:45 why is i2 + to -1?",
        "A": "The definition of i is : the number that its square is equal to -1. So: i^2 = -1 by the definition of i itself.",
        "video_name": "A_ESfuN1Pkg",
        "timestamps": [
            45
        ],
        "3min_transcript": "Now that we know a little bit about the imaginary unit i, let's see if we can simplify more involved expressions, like this one right over here. 2 plus 3i plus 7i squared plus 5i to the third power plus 9i to the fourth power. And I encourage you to pause the video right now and try to simplify this on your own. So as you can see here, we have various powers of i. You could view this as i to the first power. We have i squared here. And we already know that i squared is defined to be negative 1. Then we have i to the third power. I to the third power would just be i times this, or negative i. And we already reviewed this when we first introduced the imaginary unit, i, but I'll do it again. i to the fourth power is just going to be i times this, which is the same thing as negative 1 times i. That's i to the third power times i again. i times i is negative 1. So that's negative 1 times negative 1, which is equal to 1 So we can rewrite this whole thing as 2 plus 3i. 7i squared is going to be the same thing, so i squared is negative 1. So this is the same thing as 7 times negative 1. So that's just going to be minus 7. And then we have 5i to the third power. i to the third power is negative i. So this could be rewritten as negative i. So this term right over here we could write as minus 5i, or negative 5i, depending on how you want to think about it. And then finally, i to the fourth power is just 1. So this is just equal to 1. So this whole term just simplifies to 9. So how could we simplify this more? Well we have several terms that are not imaginary, that they are real numbers. For example, we have this 2 is a real number. And 9 is a real number. So we could just add those up. So 2 plus negative 7 would be negative 5. Negative 5 plus 9 would be 4. So the real numbers add up to 4. And now we have these imaginary numbers. So 3 times i minus 5 times i. So if you have 3 of something and then I were to subtract 5 of that same something from it, now you're going to have negative 2 of that something. Or another way of thinking about it is the coefficients. 3 minus 5 is negative 2. So three i's minus five i's, that's going to give you negative 2i. Now you might say, well, can we simplify this any further? Well no, you really can't. This right over here is a real number. 4 is a number that we've been dealing with throughout our mathematical careers."
    },
    {
        "Q": "At 5:47 Sal says the x value has to be between 0 and 10. Why isn't it between 0 and 15?",
        "A": "Because the piece of card is only 20 wide, and x can be at most half the width.",
        "video_name": "MC0tq6fNRwU",
        "timestamps": [
            347
        ],
        "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": "At 7:40, why did he set a maximum y value? Isn't that already determined since x cannot be greater than 10?",
        "A": "Right. He knows that the maximum x-value is 10, but he doesn t know what the maximum y-value is. The y-value represents the volume of the box, which can get pretty big depending on the dimensions. (width, depth, height)",
        "video_name": "MC0tq6fNRwU",
        "timestamps": [
            460
        ],
        "3min_transcript": "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 I'm not going to have negative volume, so let me set that equal 0. And my maximum y-value, let's see what would be reasonable here. I'm just going to pick some a random x and see what type of a volume I get. So if x were to be 5, it would be 5 times 20 minus 10, which is 10. So that would be-- did I do that right? Yeah, 20 minus 2 times 5, so that would be 10, and then times 30 minus 2 times 5, which would be 20. So it would be 5 times 10 times 20. So you'd get a volume of 1,000 cubic inches. And I just randomly picked the number 5. So let me give my maximum y-value a little higher than that just in case to that isn't the maximum value. I just randomly picked that. So let's say yMax is 1,500, and if for whatever reason our graph doesn't fit in there, then maybe we can make our yMax even larger. So I think this is going to be a decent range. Now let's actually input the function itself. So our volume is equal to x times 20 And that looks about right. So now I think we can graph it. So 2nd, and I want to select that up there, so I'll do Graph. And it looks like we did get the right range. So this tells us volume is a function of x between x is 0 and x is 10, and it does look like we hit a maximum point right around there. to use the Trace function to figure out roughly what that maximum point is. So let me trace this function. So I can still go higher, higher. So over there my volume is 1,055.5. Then I can get to 1,056. So let's see, this was 1,056.20, this is 1,056.24, then I go back to 1,055. So at least based on the level of zoom that I have my calculator right now,"
    },
    {
        "Q": "At 7:17, is it only a statistical question if you mention \"in 2013\" before you added it in the last example, or can it be statistical either way?",
        "A": "Without the in 2013, it is kind of a complicated case. You have to think about what you want to consider. if you say throughout the history of the schools or since 1800, then this does become a statistical question no matter how you look at it. However, considering other things (like in 2013), you have concrete numbers that you can check the difference of, making it not a statistical question",
        "video_name": "qyYSQDcSNlY",
        "timestamps": [
            437
        ],
        "3min_transcript": "What was the difference in rainfall between Singapore and Seattle in 2013? Well, these two numbers are known. They can be measured. Both the rainfall in Singapore can be measured. The rainfall in Seattle can be measured. And assuming that this has already happened and we can measure them, then we can just find the difference. So you don't need statistics here. You just have to have both of these measurements and subtract the difference. So not a statistical question. In general, will I use less gas driving at 55 miles an hour than 70 miles per hour? This feels statistical, because it probably depends on the circumstance. It might depend on the car. Or even for a given car, when you drive at 55 miles per hour, there's some variation in your gas mileage. It might be how recent an oil change happened, what the wind conditions are like, what the road conditions are like, exactly how you're driving the car. Are you going in a straight line? And same thing for 70 miles an hour. When we're saying in general, there's variation in what the gas mileage is at 55 miles an hour and at 70 miles an hour. What you'd probably want to do is say, well, what's my average mileage when I drive at 55 miles an hour and compare that to the average mileage when I drive at 70. So because we have this variability in each of those cases, this is definitely a statistical question. Do English professors get paid less than math professors? Once again, all English professor don't get paid the same amount, and all math professors don't get paid the same amount. Some English professors might do quite well. Some might make very little. Same thing for math professors. So we'd probably want to find some type of an average to represent the central tendency for each of these. Once again, this is a statistical question. Does the most highly paid English professor at Harvard professor at MIT? Well, now we're talking about two particular individuals. You could go look at their tax forms, see how much each of them get paid, especially if we assume that this is in a particular year. Let's just make it that way, say in 2013, just so that we can remove some variability that they might make from year to year, make it a little bit more concrete. If this was does the most highly paid English professor at Harvard get paid more than the most highly paid math professor at MIT in 2013, then you have an absolute number for each of these people. And then you could just compare them directly. So when we're talking about a particular year, particular people, then it isn't a statistical question."
    },
    {
        "Q": "There is one point at \"2:25\" that I really want it to be clarified. I think that \"integral f(t) d(t) from a to x\" should be \" F(x) - F(a)\" instead of \"F(x)\". Please help me Im so confused!",
        "A": "The integral f(t) d(t) from a to x would equal F(x)-F(a). I believe Sal is simply using F to put the integral as a function of x. Any letter or symbol would do, some calculus books write it as A(x) instead of F(x). They tend to use A because the formula seen in the video is an area accumulation formula (A for area). Whatever x you put into F(x) would give you the amount of area you have accumulated from a to x.",
        "video_name": "C7ducZoLKgw",
        "timestamps": [
            145
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 4:19, Sal explains that every continuous function has an antiderivative, according to the Fundamental Theorem of Calculus. I learned in Calculus A that only continuous functions have derivatives. Is this an inference from the Fundamental Theorem?",
        "A": "No. The fact that every continuous function is a derivative (has an antiderivative) has nothing to do with the fact that only continuous functions have derivatives. The FToC is not necessary to show that continuity is a condition for differentiability.",
        "video_name": "C7ducZoLKgw",
        "timestamps": [
            259
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 0:58, it doesn't really matter what size the camel is, since it's infinite, you'll always end up with the whole page covered.",
        "A": "No, actually it does matter. If you make the next camel too big, then you will end up needing more than the page to draw all the camels. If you make the next camel too small, then you will not use up all the space on the page.",
        "video_name": "DK5Z709J2eo",
        "timestamps": [
            58
        ],
        "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": "If we do not include two set of numbers can we make the brackets face opposite directions? For example at 5:06 we do not include \"-1 and 4\" Sal wrote them like this (-1,4). But instead of writing parentheses can I write them in such form ]-1,4[",
        "A": "no you cannot. If you write it like that, then you re saying that -1 and 4 are not included in the function and the result would be a syntax error. three thumbs up!",
        "video_name": "UJQkqV2zGv0",
        "timestamps": [
            306
        ],
        "3min_transcript": "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. but I'm not going to include them, so I'm going to put the parentheses right over here. Parentheses. So this tells us that we're dealing with an open interval. This right over here, let me make it clear, this is an open interval. Now you're probably wondering, okay, in this case both endpoints were included, it's a closed interval. In this case both endpoints were excluded, it's an open interval. Can you have things that have one endpoint included and one point excluded, and the answer is absolutely. Let's see an example of that. I'll get another number line here. Another number line. And let's say that we want to-- Actually, let me do it the other way around. Let me write it first, and then I'll graph it. So let's say we're thinking about all of the x's that are a member of the real numbers such that let's say negative four is not included, is less than x, is less than or equal to negative one."
    },
    {
        "Q": "I'm sort of curious about why you would write the inclusion of real all real numbers except for one as (-\u00e2\u0088\u009e,1). If you are including all real numbers except for one, would it not look like [-\u00e2\u0088\u009e,1)? @8:42",
        "A": "you can t actually reach infinity, so use ( rather than {",
        "video_name": "UJQkqV2zGv0",
        "timestamps": [
            522
        ],
        "3min_transcript": "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. all the way up to infinity. It needs to be at least open at that endpoint because infinity just keeps going on and on. So you always want to put a parentheses if you're talking about infinity or negative infinity. It's not really an endpoint, it keeps going on and on forever. So you use the notation for open interval, at least at that end, and notice we're not including, we're not including one either, so if x is a member of this interval or that interval, it essentially could be anything other than one. But this would have been the simplest notation to describe that."
    },
    {
        "Q": "At 0:50 can anyone tell me why the yellow line's slope is going to be the negative inverse?",
        "A": "Because they are both perpendicular lines!!",
        "video_name": "0671cRNjeKI",
        "timestamps": [
            50
        ],
        "3min_transcript": "We are asked which of these lines are perpendicular. And it has to be perpendicular to one of the other lines, you can't be just perpendicular by yourself. And perpendicular line, just so you have a visualization for what for perpendicular lines look like, two lines are perpendicular if they intersect at right angles. So if this is one line right there, a perpendicular line will look like this. A perpendicular line will intersect it, but it won't just be any intersection, it will intersect at right angles. So these two lines are perpendicular. Now, if two lines are perpendicular, if the slope of this orange line is m-- so let's say its equation is y is equal to mx plus, let's say it's b 1, so it's some y-intercept-- then the equation of this yellow line, its slope is going to be the negative inverse of this guy. This guy right here is going to be y is equal to negative 1 over mx plus some other y-intercept. perpendicular, the product of their slopes is going to be negative 1. And so you could write that there. m times negative 1 over m, that's going to be-- these two guys are going to cancel out-- that's going to be equal to negative 1. So let's figure out the slopes of each of these lines and figure out if any of them are the negative inverse of any of the other ones. So line A, the slope is pretty easy to figure out, it's already in slope-intercept form, its slope is 3. So line A has a slope of 3. Line B, it's in standard form, not too hard to put it in slope-intercept form, so let's try to do it. So let's do line B over here. Line B, we have x plus 3y is equal to negative 21. Let's subtract x from both sides so that it ends up on the right-hand side. So we end up with 3y is equal to negative x minus 21. and we get y is equal to negative 1/3 x minus 7. So this character's slope is negative 1/3. So here m is equal to negative 1/3. So we already see they are the negative You take the inverse of 3, it's 1/3, and then it's the negative of that. Or you take the inverse of negative 1/3, it's negative 3, and then this is the negative of that. So these two lines are definitely perpendicular. Let's see the third line over here. So line C is 3x plus y is equal to 10. If we subtract 3x from both sides, we get y is equal to"
    },
    {
        "Q": "at 5:08\nisn't this the formula of Explicit geometric sequence ?",
        "A": "In a geometric sequence, the input value is multiplied. Here it is the power.",
        "video_name": "G2WybA4Hf7Y",
        "timestamps": [
            308
        ],
        "3min_transcript": "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. And then times our common ratio to the x power. So once again, initial value, right over there, that's the five. And then our common ratio is the six, right over there. So hopefully that gets you a little bit familiar with some of the parts of an exponential function, why they are called what they are called."
    },
    {
        "Q": "sal at 1:08 instead of moving the 5 cant u move the 4 and then add 8 to the other side",
        "A": "Yes but the way Sal did it is easier.",
        "video_name": "g6nGcnVB8BM",
        "timestamps": [
            68
        ],
        "3min_transcript": "We're asked to solve for x. So we have the square root of the entire quantity 5x squared minus 8 is equal to 2x. Now we already have an expression under a radical isolated, so the easiest first step here is really just to square both sides of this equation. So let's just square both sides of that equation. Now the left-hand side, if you square it, the square root of 5x squared minus 8 squared is going to be 5x squared minus 8. This is 5x squared minus 8. And then the right-hand side, 2x squared is the same thing as 2 squared times x squared or 4x squared. Now we have a quadratic. Now let's see what we can do to maybe simplify this a little bit more. Well, we could subtract 4x squared. Or actually, even better, let's subtract 5x squared from both sides so that we just have all our x terms on the right-hand side. So let's subtract 5x squared from both sides. Subtract 5x squared from both sides of the equation. We're just left with negative 8 is equal to 4x squared minus 5x squared, that's negative 1x squared. Or we could just write negative x squared, just like that. And then we could multiply both sides of this equation by negative 1. That'll make it into positive 8. Or I could divide by negative 1, however you want to view it. Negative 1 times that times negative 1. So we get positive 8 is equal to x squared. And now we could take the square root of both sides of this equation. So let's take the square root of both sides of this equation. The principal square root of both sides of this equation. And what do we get? We get, on the right-hand side, x is equal to the square root of 8. And 8 can be rewritten as 2 times 4. And this can be rewritten as the square root of 2 times the I don't like this green color so much. And what's the square root of 4, the principal root of 4? It's 2. So that right there is 2. So this side becomes 2, this 2, times the square root of 2. And that is equal to x. Now let's verify that this is the solution to our original equation. So let's substitute this in, first to the left-hand side. So on the left-hand side, we have 5 times 2 square roots of 2 squared minus 8. And then we're going to have to take the square root of that whole thing. So this is going to be equal to-- we're just focused on the left-hand side right now. This is equal to the square root of 5 times 2 squared, which is 4, times the square root of 2 squared, which is 2. And then minus 8."
    },
    {
        "Q": "at 3:11 could he use substitution instead of elimination?",
        "A": "Of course. He can do anything.",
        "video_name": "f7cX-Ar2cEM",
        "timestamps": [
            191
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "what does sal mean when he says at 0:21 when he says \"scaling up\"?",
        "A": "By scaling up Sal means multiplying the equation by 2. This scales up all the values in the equation by 2.",
        "video_name": "f7cX-Ar2cEM",
        "timestamps": [
            21
        ],
        "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 3:10 isn't pi 3.14 not 3.5 ?",
        "A": "correct pi is 3.14 rounded to two decimal places. But Sal is showing that 4*pi is less than 14. Since 4*3.5 is 14, and pi is less than 3.5 then 4*pi must be less than 14. Hope that helps",
        "video_name": "EvvxBdNIUeQ",
        "timestamps": [
            190
        ],
        "3min_transcript": "Well, area is equal to pi r squared for a circle, where r is a radius. They gave us the diameter. The radius is half of that. So the radius here is going to be half this distance, or 2r. So the area of our circle is going to be pi times 2r, the whole thing squared. This is the radius, right? So we're squaring the entire radius. So this is going to be equal to pi times 4 times r squared. I'm just squaring each of these terms. Or if we were to change the order, the area of the circle is equal to 4 pi r squared. And we want to find the difference. So to find a difference, It's helpful-- just so we don't end up with a negative number-- to figure out which of these two is larger. So they're telling us that p is greater than 7 r. 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."
    },
    {
        "Q": "i didnt understand from 1:07-1:47 about the circle",
        "A": "it is about pi from 1:07 to 1:47 about the circle",
        "video_name": "EvvxBdNIUeQ",
        "timestamps": [
            67,
            107
        ],
        "3min_transcript": "Write a binomial to express the difference between the area of a rectangle with length p and width 2r and the area of a circle with diameter 4r. And they tell us that p is greater than 7r. So let's first think about the area of a rectangle with length p and width 2r. So this is our rectangle right here. It has a length of p and it has a width of 2r. So what's its area? Well, it's just going to be the length times the width. So the area here is going to be p-- or maybe I should say 2rp. This is the length times the width, or the width times the length. So area is equal to 2rp for the rectangle. Now, we also want to find the difference between this area and the area of a circle with diameter 4r. So what's the area of the circle going to be? So let me draw our circle over here. So our circle looks like that. Its diameter is 4r. Well, area is equal to pi r squared for a circle, where r is a radius. They gave us the diameter. The radius is half of that. So the radius here is going to be half this distance, or 2r. So the area of our circle is going to be pi times 2r, the whole thing squared. This is the radius, right? So we're squaring the entire radius. So this is going to be equal to pi times 4 times r squared. I'm just squaring each of these terms. Or if we were to change the order, the area of the circle is equal to 4 pi r squared. And we want to find the difference. So to find a difference, It's helpful-- just so we don't end up with a negative number-- to figure out which of these two is larger. So they're telling us that p is greater than 7 r. 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."
    },
    {
        "Q": "At 2:42 why is it 14 r squared?",
        "A": "Because when you times a r by a r you get r squared and 7 times 2 you get 14 so 14r^2",
        "video_name": "EvvxBdNIUeQ",
        "timestamps": [
            162
        ],
        "3min_transcript": "Well, area is equal to pi r squared for a circle, where r is a radius. They gave us the diameter. The radius is half of that. So the radius here is going to be half this distance, or 2r. So the area of our circle is going to be pi times 2r, the whole thing squared. This is the radius, right? So we're squaring the entire radius. So this is going to be equal to pi times 4 times r squared. I'm just squaring each of these terms. Or if we were to change the order, the area of the circle is equal to 4 pi r squared. And we want to find the difference. So to find a difference, It's helpful-- just so we don't end up with a negative number-- to figure out which of these two is larger. So they're telling us that p is greater than 7 r. 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."
    },
    {
        "Q": "At 3:55 What does he mean by principle root?",
        "A": "The principal root is just the positive square root. For instance: \u00e2\u0088\u009a9 = \u00c2\u00b13 But the principal square root of 9 is just 3.",
        "video_name": "McINBOFCGH8",
        "timestamps": [
            235
        ],
        "3min_transcript": "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 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."
    },
    {
        "Q": "At 4:28 wouldn't it be 2x=C",
        "A": "You had the equation x^2+x^2 = c^2. Which is 2(x^2)=c^2 which is 2*x*x=c*c When you take the square root of each side, you have to take the square root of both the x*x and the square root of 2. so you get (Square root of 2 * x)=c",
        "video_name": "McINBOFCGH8",
        "timestamps": [
            268
        ],
        "3min_transcript": "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 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."
    },
    {
        "Q": "At 4:59, why is it bn and vn, instead of bk and vk?",
        "A": "There s no error here. b belongs to Rn, otherwise original equation Ax=b, where A is nxk matrix would not make any sense. v is a projection of b to the column space, but it still is a member of Rn, hence it also has n components.",
        "video_name": "MC7l96tW8V8",
        "timestamps": [
            299
        ],
        "3min_transcript": "Or another way to view it, when I say close, I'm talking about length, so I want to minimize the length of-- let me write this down. I want to minimize the length of b minus A times x-star. Now, some of you all might already know where this is going. But when you take the difference between 2 and then take its length, what does that look like? Let me just call Ax. Ax is going to be a member of my column space. Let me just call that v. Ax is equal to v. You multiply any vector in Rk times your matrix A, you're So any Ax is going to be in your column space. And maybe that is the vector v is equal to A times x-star. And we want this vector to get as close as possible to this as long as it stays-- I mean, it has to be in my column space. But we want the distance between this vector and this vector to be minimized. Now, I just want to show you where the terminology for this will come from. I haven't given it its proper title yet. If you were to take this vector-- let just call this vector v for simplicity-- that this is equivalent to the length of the vector. You take the difference between each of the elements. So b1 minus v1, b2 minus v2, all the way to bn minus vn. thing as this. This is going to be equal to the square root. Let me take the length squared, actually. The length squared of this is just going to be b1 minus v1 squared plus b2 minus v2 squared plus all the way to bn minus vn squared. And I want to minimize this. So I want to make this value the least value that it can be possible, or I want to get the least squares estimate here. And that's why, this last minute or two when I was just explaining this, that was just to give you the motivation for why this right here is called the least squares estimate, or the least squares solution, or the least squares approximation for the equation Ax equals b. There is no solution to this, but maybe we can find some"
    },
    {
        "Q": "@ 4:05 he says it goes down 2/3's then he says its 1 and 1/3. How does he go from 2/3's to 1/3rd? I get the whole thing that 1/ 1/3rd makes 4/3rds, but he went down 2/3rds??",
        "A": "When he went down 2/3 it was from 2! The line was 2/3 below 2. So if you look at it in terms of the line being above 1 the line is actually 1/3 above 1. So therefore the line is at 1 1/3.",
        "video_name": "9wOalujeZf4",
        "timestamps": [
            245
        ],
        "3min_transcript": "That's our end point. That's our starting point. So if you simplify this, b minus b is 0. 1 minus 0 is 1. So you get m/1, or you get it's equal to m. So hopefully you're satisfied and hopefully I didn't confuse you by stating it in the abstract with all of these variables here. But this is definitely going to be the slope and this is definitely going to be the y-intercept. Now given that, what I want to do in this exercise is look at these graphs and then use the already drawn graphs to figure out the equation. So we're going to look at these, figure out the slopes, figure out the y-intercepts and then know the equation. So let's do this line A first. So what is A's slope? Let's start at some arbitrary point. Let's start right over there. We want to get even numbers. If we run one, two, three. 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"
    },
    {
        "Q": "At about 10:10, he said that the y-intercept was 0. So why does the graph line still have a slope?",
        "A": "The y-intercept doesn t have to do with the slope; it just shows where the sloped line crosses the y-axis. So, a graph line can have a slope and y-intercept 0, such as y=x, where the slope is 1 and the y-int 0. The line just crosses the y-intercept at the origin.",
        "video_name": "9wOalujeZf4",
        "timestamps": [
            610
        ],
        "3min_transcript": "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. fourth quadrants. Now I'll do one more. Let's do this last one right here. y is equal to 3.75. Now you're saying, gee, we're looking for y is equal to mx plus b. Where is this x term? It's completely gone. Well the reality here is, this could be rewritten as y is equal to 0x plus 3.75. Now it makes sense. The slope is 0. No matter how much we change our x, y does not change. Delta y over delta x is equal to 0. I don't care how much you change your x. Our y-intercept is 3.75. So 1, 2, 3.75 is right around there. You want to get close. 3 3/4. As I change x, y will not change. y is always going to be 3.75. It's just going to be a horizontal line at"
    },
    {
        "Q": "Hold on.. how did he get 4/3 as the y-intercept in 4:05..?",
        "A": "The slope = -2/3. Sal has a point at (-1, 2). He needs to move one uni to the right to get to the y-axis. The slope tells him that as he moves 1 unit to the right, the y-coordinate will decrease by 2/3. Sal finds the y-intercept by doing: 2 - 2/3 = 6/3 - 2/3 = 4/3 Hope this helps.",
        "video_name": "9wOalujeZf4",
        "timestamps": [
            245
        ],
        "3min_transcript": "That's our end point. That's our starting point. So if you simplify this, b minus b is 0. 1 minus 0 is 1. So you get m/1, or you get it's equal to m. So hopefully you're satisfied and hopefully I didn't confuse you by stating it in the abstract with all of these variables here. But this is definitely going to be the slope and this is definitely going to be the y-intercept. Now given that, what I want to do in this exercise is look at these graphs and then use the already drawn graphs to figure out the equation. So we're going to look at these, figure out the slopes, figure out the y-intercepts and then know the equation. So let's do this line A first. So what is A's slope? Let's start at some arbitrary point. Let's start right over there. We want to get even numbers. If we run one, two, three. 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"
    },
    {
        "Q": "At 1:10, is that a different way of expressing it using a sigma?",
        "A": "In this video, he is discussing sequences, which are just lists of numbers. The sigma, or summation notation comes into play when you are calculating series.",
        "video_name": "_cooC3yG_p0",
        "timestamps": [
            70
        ],
        "3min_transcript": "What I want to do in this video is familiarize ourselves with a very common class of sequences. And this is arithmetic sequences. And they are usually pretty easy to spot. They are sequences where each term is a fixed number larger than the term before it. So my goal here is to figure out which of these sequences are arithmetic sequences. And then just so that we have some practice with some of the sequence notation, I want to define them either as explicit functions of the term you're looking for, the index you're looking at, or as recursive definitions. So first, given that an arithmetic sequence is one where each successive term is a fixed amount larger than the previous one, which of these are arithmetic sequences? Well let's look at this first one right over here. To go from negative 5 to negative 3, we had to add 2. Then to go from negative 3 to negative 1, you have to add 2. Then to go from negative 1 to 1, you had to add 2. So this is clearly an arithmetic sequence. And there are several ways that we could define the sequence. We could say it's a sub n. And you don't always have to use k. This time I'll use n to denote our index. From n equals 1 to infinity with-- and there's two ways we could define it. We could either define it explicitly, or we could define it recursively. So if we wanted to define it explicitly, we could write a sub n is equal to whatever the first term is. In this case, our first term is negative 5. It's equal to negative 5 plus-- we're going to add 2 one less times than the term we're at. So for the second term, we add 2 once. For the third term, we add 2 twice. For the fourth term, from our base term, we added 2 three times. So we're going to add 2. We're going to add positive 2 one less than the index that we're looking at-- n minus 1 times. So this is an explicit definition of this arithmetic sequence. 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,"
    },
    {
        "Q": "I dont understand the part at 6:61",
        "A": "Do you mean 7:11? Basically, to prove the quadratic formula Sal is completing the square and then moving everything to one side of the equation.",
        "video_name": "r3SEkdtpobo",
        "timestamps": [
            421
        ],
        "3min_transcript": "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 Let me do this in a new color. So it's orange. Negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. And we are done! By completing the square with just general coefficients in front of our a, b and c, we were able to derive the quadratic formula. Just like that. Hopefully you found that as entertaining as I did."
    },
    {
        "Q": "At 6:00, why did he simply not just take the square root of b squared and leave it as b?",
        "A": "Because you can t distribute square roots like that. You could if the terms in a radical are products or quotients, but when they are sums and differences(i.e. adding and subtracting), you can t distribute a radical.",
        "video_name": "r3SEkdtpobo",
        "timestamps": [
            360
        ],
        "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": "At 5:45, when we take square root of both sides, why does the right side end up with a \"plus or minus\" designation while the left side stays only positive?",
        "A": "| Because the equation on the left has implied positive domain because it is represented with a (term)^2 and that represents that any term within the brackets which were positive or negative will result in a positive output. While on the right, we do not yet know the domain of the right so it is correct to assume that there could be a negative root. I m sorry if i have confused you. :(",
        "video_name": "r3SEkdtpobo",
        "timestamps": [
            345
        ],
        "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": "At 4:00 and 10:00, why is is cos(2a) a minus but sin(2a) is a plus?\n\nAlso which video did I miss? I am so confused here. I don't ever remember learning this in my high school precalc class.",
        "A": "You have to review the formula of cos(a+b) = cos(a)*cos(b) - sin(a)*sin(b) From there, you can write cos(2a) as cos(a+a), then go on. You see why the minus in here. That is trigonometry. Go back to the beginning and learn the formula come from.",
        "video_name": "a70-dYvDJZY",
        "timestamps": [
            240,
            600
        ],
        "3min_transcript": "Because cosine of minus c is the same thing as the cosine of c. Times the cosine of c. And then, minus the sine of c. Instead of writing this, I could write this. Minus the sine of c times the cosine of a. So that we kind of pseudo proved this by knowing this and this ahead of time. Fair enough. And I'm going to use all of these to kind of prove a bunch of more trig identities that I'm going to need. So the other trig identity is that the cosine of a plus b is equal to the cosine of a-- you don't mix up the cosines and the sines in this situation. Cosine of a times the sine of b. And this is minus-- well, sorry. I just said you don't mix it up and then I mixed them up. Times the cosine of b minus sine of a times the sine of b. 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"
    },
    {
        "Q": "at 8:22 how did he get 1-2sin^2(a)?",
        "A": "It s one of your identities. cos(2a) = cos^2(a) - sin^2(a) sin^2(a) + cos^2(a) = 1 => cos^2(a) = 1 - sin^2(a) cos(2a) = 1 - sin^2(a) - sin^2(a) cos(2a) = 1 - 2sin^2(a)",
        "video_name": "a70-dYvDJZY",
        "timestamps": [
            502
        ],
        "3min_transcript": "Now what if I wanted to get an identity that gave me what cosine squared of a is in terms of this? Well we could just solve for that. If we add 1 to both sides of this equation, actually, let me write this. This is one of our other identities. But if we add 1 to both sides of that equation we get 2 times the cosine squared of a is equal to cosine of 2a plus 1. And if we divide both sides of this by 2 we get the cosine squared of a is equal to 1/2-- now we could rearrange these just to do it-- times 1 plus the cosine of 2a. And we're done. And we have another identity. Cosine squared of a, sometimes it's called the power reduction Now what if we wanted something in terms of the sine squared of a? Well then maybe we could go back up here and we know from this identity that the sine squared of a is equal to 1 minus cosine squared of a. Or we could have gone the other way. We could have subtracted sine squared of a from both sides and we could have gotten-- let me go down there. If I subtracted sine squared of a from both sides you could get cosine squared of a is equal to 1 minus sine squared of a. And then we could go back into this formula right up here and we could write down-- and I'll do it in this blue color. We could write down that the cosine of 2a is equal to-- instead of writing a cosine squared of a, I'll write this- is equal to 1 minus sine squared of a minus sine squared of a. So my cosine of 2a is equal to? sine squared of a. So I have 1 minus 2 sine squared of a. So here's another identity. Another way to write my cosine of 2a. We're discovering a lot of ways to write our cosine of 2a. Now if we wanted to solve for sine squared of 2a we could add it to both sides of the equation. So let me do that and I'll just write it here for the sake of saving space. Let me scroll down a little bit. So I'm going to go here. If I just add 2 sine squared of a to both sides of this, I get 2 sine squared of a plus cosine of 2a is equal to 1. Subtract cosine of 2a from both sides. You get 2 sine squared of a is equal to 1 minus cosine of 2a."
    },
    {
        "Q": "At 4:25 isn't cos(2a) = cos(3a - a) = cos(3a)*(cos)a + sin(3a) * sin(a)",
        "A": "You could work it out like that, which would eventually simplify to what he had: cos(2a)=cos^2(a)-sin^2(a); this is one of the double angle formulas",
        "video_name": "a70-dYvDJZY",
        "timestamps": [
            265
        ],
        "3min_transcript": "Because cosine of minus c is the same thing as the cosine of c. Times the cosine of c. And then, minus the sine of c. Instead of writing this, I could write this. Minus the sine of c times the cosine of a. So that we kind of pseudo proved this by knowing this and this ahead of time. Fair enough. And I'm going to use all of these to kind of prove a bunch of more trig identities that I'm going to need. So the other trig identity is that the cosine of a plus b is equal to the cosine of a-- you don't mix up the cosines and the sines in this situation. Cosine of a times the sine of b. And this is minus-- well, sorry. I just said you don't mix it up and then I mixed them up. Times the cosine of b minus sine of a times the sine of b. 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"
    },
    {
        "Q": "1:25 How are two inscribed angles that are subtended by the same arc equal to each other?",
        "A": "For any given arc, there can be any number of inscribed angles that subtend it, but only one central angle will subtend that same arc. Since the inscribed angle theorem tells us that any inscribed angle will be exactly half the measure of the central angle that subtends its arc, it follows that all inscribed angles sharing that arc will be half the measure of the same central angle. Therefore, the inscribed angles must all be congruent. Hope this helps!",
        "video_name": "h-_BDon5oes",
        "timestamps": [
            85
        ],
        "3min_transcript": "- So what I would like you to do is see if you can figure out the measure of angle DEG here. So try to figure out the measure of this angle. I encourage you to pause the video now and try it on your own. All right, now let's work through this together and the key realization here is to think about this angle, it is an inscribed angle, we see it's vertexes sitting on the circle itself. And then think about the arc that it intercepts. And we see, we see that it intercepts, so let me draw these two sides of the angle, we see that it intercepts arc CD. It intercepts arc CD. And so the measure of this angle, since it's an inscribed angle, is going to be half the measure of arc CD. So if we could figure out the measure of arc CD, then we're going to be in good shape. Because if we figure out the measure of arc CD, then we take half of that and we'll figure out what we care about. inscribed angle that also intercepts arc CD. We have this angle right over here. It also intercepts arc CD. So you could call this angle C ... Whoops. You could call this angle CFD. This also intercepts the same arc. So there's two ways you could think about it. Two inscribed angles that intercept the same arc are going to have the same angle measure so just off of that you could say that this is going to be, that these two angles have the same measure, so you could say this is going to be 50 degrees. Or you could go, you could actually solve what the measure of arc CD is. It's going to be twice the measure of the inscribed angle that intercepts it. So the measure of arc CD is going to be 100 degrees. Twice the 50 degrees. And then you use that and you say, Well, if the measure of that arc is 100 degrees, then an inscribed angle that intercepts it it's going to be 50 degrees. So either way we get to 50 degrees."
    },
    {
        "Q": "At 1:47, he does 3*-2, and represents it as -2+-2+-2. Can't he also say 3, -2 times?\nIf you took three twice away from zero, it would still equal -6.",
        "A": "Yes, you are correct, but in this video, he only shows one way but the way you are thinking is definitely correct, they are the same. It is like saying 3*4 is equal to both 3+3+3+3 and 4+4+4. Hope this helps.",
        "video_name": "47wjId9k2Hs",
        "timestamps": [
            107
        ],
        "3min_transcript": "We know that if we were to multiply two times three, that would give us positive six. And so we are going to think about negative numbers in this video. One way to think about it, is that I have a positive number times another positive number, and that gives me a positive number. So if I have a positive times a positive, that would give me a positive number. Now it's mixed up a little bit. Introduce some negative numbers. So what happens if I had negative two times three? Negative two times three. Well, one way to think about it-- Now we are talking about intuition in this video and in the future videos. You could view this as negative two repeatedly added three times. So this could be negative two plus negative two plus negative two-- Not negative six. Plus negative two. which would be equal to-- well, negative two plus negative two is negative four, plus another negative two is negative six. This would be equal to negative six. 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."
    },
    {
        "Q": "At 1:17, I get confused. Can you help?",
        "A": "Ok, so a negative times a negative equals a positive because the negatives cancel out, but if it s a negative times a positive (or vice versa) there s nothing to cancel out that negative so the answer remains negative.",
        "video_name": "47wjId9k2Hs",
        "timestamps": [
            77
        ],
        "3min_transcript": "We know that if we were to multiply two times three, that would give us positive six. And so we are going to think about negative numbers in this video. One way to think about it, is that I have a positive number times another positive number, and that gives me a positive number. So if I have a positive times a positive, that would give me a positive number. Now it's mixed up a little bit. Introduce some negative numbers. So what happens if I had negative two times three? Negative two times three. Well, one way to think about it-- Now we are talking about intuition in this video and in the future videos. You could view this as negative two repeatedly added three times. So this could be negative two plus negative two plus negative two-- Not negative six. Plus negative two. which would be equal to-- well, negative two plus negative two is negative four, plus another negative two is negative six. This would be equal to negative six. 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."
    },
    {
        "Q": "AT 1:39 the regrouping of the problem from -2 x 3 was changed to 3 x -2. Would this be an example of commutative property ?",
        "A": "Yes! The communicative property states that the order does not matter for numbers added or multiplied.",
        "video_name": "47wjId9k2Hs",
        "timestamps": [
            99
        ],
        "3min_transcript": "We know that if we were to multiply two times three, that would give us positive six. And so we are going to think about negative numbers in this video. One way to think about it, is that I have a positive number times another positive number, and that gives me a positive number. So if I have a positive times a positive, that would give me a positive number. Now it's mixed up a little bit. Introduce some negative numbers. So what happens if I had negative two times three? Negative two times three. Well, one way to think about it-- Now we are talking about intuition in this video and in the future videos. You could view this as negative two repeatedly added three times. So this could be negative two plus negative two plus negative two-- Not negative six. Plus negative two. which would be equal to-- well, negative two plus negative two is negative four, plus another negative two is negative six. This would be equal to negative six. 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."
    },
    {
        "Q": "At 2:55 he said that the negative cancels out the negative and makes a positive product. Then according to that, wouldn't a positive cancel out a positive, making a negative product.",
        "A": "No, that isn t how it works. A negative times a negative is a positive, and a positive times a positive is a positive. But a negative times a positive is a negative, and a positive times negatives is a negative. So, basically, if you are multiplying two of the same thing (like two positives, or two negatives), you get a positive. If you are multiplying two different things (one negative and one positive), you get a negative. So no, a positive does not cancel out another positive. I hope this helps.",
        "video_name": "47wjId9k2Hs",
        "timestamps": [
            175
        ],
        "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": "At 3:41 in the graph we can see a vertical asymptote at approximately x=3, why is this and what is its significance?",
        "A": "That is simply due to the fact that 6x^5-100x^2-10 has a zero around 2.57. Because this expression is in the denominator of the rational function, there is a horizontal asymptote (because you can t divide by zero!). However, the vertical asymptotes have nothing to do with the horizontal asymptotes (limits at infinity).",
        "video_name": "gv9ogppphso",
        "timestamps": [
            221
        ],
        "3min_transcript": "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. as x approaches negative infinity, that also looks like it's 2/3. And we can use the exact same logic. When x becomes a very, very, very negative number, as it becomes further and further to the left on the number line, the only terms that are going to matter are going to be the 4x to the fifth and the 6x to the fifth. So this is true for very large x's. It's also true for very negative x's. So we could also say, as x approaches negative infinity, this is also true. And then, the x to the fifth over the x to the fifth is going to cancel out. These are the dominant terms. And we're going to get it equaling 2/3. And once again, you see that in the graph here. We have a horizontal asymptote at y is equal to 2/3. We take the limit of f of x as x approaches infinity, we get 2/3. And the limit of f of x as x approaches negative infinity is 2/3. So in general, whenever you do this, you just have to think about what And focus on those."
    },
    {
        "Q": "why does Sal say at 4:30 \"principal square root\"",
        "A": "Because the \u00e2\u0088\u009a means principal square root, not square root.",
        "video_name": "egNq4tSfi1I",
        "timestamps": [
            270
        ],
        "3min_transcript": "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 so you have four absolute value of x of something, and you have ten of that same something, you could add them up, or another way to think about it is, you could factor out a square root of two either one of those works, so you get four times the absolute value of x, plus ten plus ten times times the principle square root of two, so depending on whether you view this of this more simplified, one of those two will will satisfy you"
    },
    {
        "Q": "So, at 2:50, if we assume the root didn't always indicate the principal square root, could we say that x could be either positive or negative?",
        "A": "Yes it can be either or",
        "video_name": "egNq4tSfi1I",
        "timestamps": [
            170
        ],
        "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": "Im stuck at 3:10, i dont know how sal has p(x)=f(0)+f\"(0)x.\nIs this like linear approximation or something?",
        "A": "First of all, he said p(x)=f(0)+f (0)*x just one prime And yes, this is a linear approximation! Does it remind you of anything? Perhaps point slope form? if y=mx+b, y is p(x), b is f(0), and m is f (0).",
        "video_name": "epgwuzzDHsQ",
        "timestamps": [
            190
        ],
        "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": "Can someone explain the equation for when g(x)?\n5:16",
        "A": "are you confused about why the function is called g(x) and not f(x)? if so they mean the same thing . it is like a name for the function",
        "video_name": "riXcZT2ICjA",
        "timestamps": [
            316
        ],
        "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 8:20, why does G(2) = 1 instead of 4?",
        "A": "Because that s part of the function - it says square x unless x = 2 where g(x) gives you a 1",
        "video_name": "riXcZT2ICjA",
        "timestamps": [
            500
        ],
        "3min_transcript": "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 So my question to you. So there's a couple of things, if I were to just evaluate the function g of 2. Well, you'd look at this definition, OK, when x equals 2, I use this situation right over here. And it tells me, it's going to be equal to 1. Let me ask a more interesting question. Or perhaps a more interesting question. What is the limit as x approaches 2 of g of x. Once again, fancy notation, but it's asking something pretty, pretty, pretty simple. It's saying as x gets closer and closer to 2, as you get closer and closer, and this isn't a rigorous definition, we'll do that in future videos. As x gets closer and closer to 2, what is g of x approaching? So if you get to 1.9, and then 1.999, and then 1.999999, and then 1.9999999, what is g of x approaching. Or if you were to go from the positive direction. If you were to say 2.1, what's g of 2.1,"
    },
    {
        "Q": "At 1:25, why is f(x) = D?",
        "A": "i think you are getting confused by his arrows...",
        "video_name": "riXcZT2ICjA",
        "timestamps": [
            85
        ],
        "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": "Wait a minute, at 9:20 Sal says that when x approaches to 2, the value of g(x) would be getting really close of 4, but shouldn't it be 1? Since when x = 2 <=> g(x) = 1?",
        "A": "g(2) is in fact 1, given by the dot. However, the limit is different. Remeber g(x) = y. To understand limit, try putting your pencil on the graph and trace it. As you trace it toward x=2, you see that y=g(x) is getting toward 4 and not 1.",
        "video_name": "riXcZT2ICjA",
        "timestamps": [
            560
        ],
        "3min_transcript": "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 So my question to you. So there's a couple of things, if I were to just evaluate the function g of 2. Well, you'd look at this definition, OK, when x equals 2, I use this situation right over here. And it tells me, it's going to be equal to 1. Let me ask a more interesting question. Or perhaps a more interesting question. What is the limit as x approaches 2 of g of x. Once again, fancy notation, but it's asking something pretty, pretty, pretty simple. It's saying as x gets closer and closer to 2, as you get closer and closer, and this isn't a rigorous definition, we'll do that in future videos. As x gets closer and closer to 2, what is g of x approaching? So if you get to 1.9, and then 1.999, and then 1.999999, and then 1.9999999, what is g of x approaching. Or if you were to go from the positive direction. If you were to say 2.1, what's g of 2.1, 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."
    },
    {
        "Q": "I don't really get it. At 00:53, why can't you reduce x-1/x-1 to equal to 1?",
        "A": "This is only true if x does not equal 1. If x were to equal 1 then you d have 0/0 which is indeterminate. Therefore the division only works for most cases, but it only takes one counterexample to make a conjecture false. So limits will allow for the case up to but not including the problem point.",
        "video_name": "riXcZT2ICjA",
        "timestamps": [
            53
        ],
        "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": "in the video at 4:24 Sal says that we can get infinitely closer to one. If we can get infinitely closer to one doesn't that mean that we can never approach one?",
        "A": "True. Yes, we can always get closer and closer to one but the function actually never reaches one.",
        "video_name": "riXcZT2ICjA",
        "timestamps": [
            264
        ],
        "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 7:30 Sal has drawn a parabola with a gap at the point (2,4).\n\nA point takes up zero space right? It has no size? So how can something have a gap in it if the gap doesn't cover any actual space? How big is the gap? Surely a gap is between two points.\n\nCan someone please help me understand this?",
        "A": "At this point it has size 0. The gap is just for understanding that there is no defined point at that x coordinate.",
        "video_name": "riXcZT2ICjA",
        "timestamps": [
            450
        ],
        "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": "9:38 he says its 4 we're getting to but how do you get for without a graph",
        "A": "In later lessons you will learn different techniques to do it algebraically. This is just a video introduction of limits, so I recommend you watch other lessons about limits.",
        "video_name": "riXcZT2ICjA",
        "timestamps": [
            578
        ],
        "3min_transcript": "So my question to you. So there's a couple of things, if I were to just evaluate the function g of 2. Well, you'd look at this definition, OK, when x equals 2, I use this situation right over here. And it tells me, it's going to be equal to 1. Let me ask a more interesting question. Or perhaps a more interesting question. What is the limit as x approaches 2 of g of x. Once again, fancy notation, but it's asking something pretty, pretty, pretty simple. It's saying as x gets closer and closer to 2, as you get closer and closer, and this isn't a rigorous definition, we'll do that in future videos. As x gets closer and closer to 2, what is g of x approaching? So if you get to 1.9, and then 1.999, and then 1.999999, and then 1.9999999, what is g of x approaching. Or if you were to go from the positive direction. If you were to say 2.1, what's g of 2.1, 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,"
    },
    {
        "Q": "At 9:20, how is g(x) approaching the value of 4 if it is a hole? Why isn't the limit 1 because it's an actual point?",
        "A": "The limit as x approaches 2 of g(x) does not have to equal g(2). The value of this limit (if any) is determined by the behavior of g(x) near, but not at, x = 2. Since g(x) is near 4 when x is near (but not at) 2, the value of this limit is 4. The fact that g(2) = 1 has no effect on the value of this limit. Note that the fact that the value of this limit does not match the value of g(2) indicates that g(x) is discontinuous at x=2. Have a blessed, wonderful day!",
        "video_name": "riXcZT2ICjA",
        "timestamps": [
            560
        ],
        "3min_transcript": "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 So my question to you. So there's a couple of things, if I were to just evaluate the function g of 2. Well, you'd look at this definition, OK, when x equals 2, I use this situation right over here. And it tells me, it's going to be equal to 1. Let me ask a more interesting question. Or perhaps a more interesting question. What is the limit as x approaches 2 of g of x. Once again, fancy notation, but it's asking something pretty, pretty, pretty simple. It's saying as x gets closer and closer to 2, as you get closer and closer, and this isn't a rigorous definition, we'll do that in future videos. As x gets closer and closer to 2, what is g of x approaching? So if you get to 1.9, and then 1.999, and then 1.999999, and then 1.9999999, what is g of x approaching. Or if you were to go from the positive direction. If you were to say 2.1, what's g of 2.1, 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."
    },
    {
        "Q": "At 1:17; what's the difference between infinity and 1/0 ?",
        "A": "Infinity (symbol: \u00e2\u0088\u009e) is an abstract concept describing something without any bound or larger than any number. In mathematics, infinity is often treated as a number While the expression a/0 has no meaning, as there is no number which, multiplied by 0, gives a (assuming a\u00e2\u0089\u00a00), and so division by zero is undefined. Since any number multiplied by zero is zero, the expression 0/0 also has no defined value; You can also prove it by the concept of limits. Try it out yourself.",
        "video_name": "riXcZT2ICjA",
        "timestamps": [
            77
        ],
        "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 1:33, isn't that the Riemann's zeta function?",
        "A": "Good question! That would be Riemann s zeta function, evaluated at z = 2. In general, this function is sum n=1 to infinity of 1/(n^z), where z is a complex number (that is, a number of the form a+bi where a and b are real numbers and i is the square root of -1). This function is defined when the real part of z (that is, a) is greater than 1. Have a blessed, wonderful day!",
        "video_name": "k9MEOgcc5KY",
        "timestamps": [
            93
        ],
        "3min_transcript": "- [Voiceover] Let's say that you have an infinite series, S, which is equal to the sum from n equals one, let me write that a little bit neater. n equals one to infinity of a sub n. This is all a little bit of review. We would say, well this is the same thing as a sub one, plus a sub two, plus a sub three, and we would just keep going on and on and on forever. Now what I want to introduce to you is the idea of a partial sum. This right over here is an infinite series. But we can define a partial sum, so if we say S sub six, this notation says, okay, if S is an infinite series, S sub six is the partial sum of the first six terms. So in this case, this is going to be we're not going to just keep going on forever, this is going to be a sub one, plus a sub two, plus a sub three, plus a sub four, plus a sub five, And I can make this a little bit more tangible if you like. So let's say that S, the infinite series S, is equal to the sum from n equals one, to infinity of one over n squared. In this case it would be one over one squared, plus one over two squared, plus one over three squared, and we would just keep going on and on and on forever. But what would S sub -- I should do that in that same color. What would S -- I said I would change color, and I didn't. What would S sub three be equal to? The partial sum of the first three terms, and I encourage you to pause the video and try to work through it on your own. Well, it's just going to be the first term one, plus the second term, is going to be the sum of the first three terms, and we can figure that out, that's to see if you have a common denominator here, it's going to be 36. It's going to be 36/36, plus 9/36, plus 4/36, so this is going to be 49/36. 49/36. So the whole point of this video, is just to appreciate this idea of a partial sum. And what we'll see is, that you can actually express what a partial sum might be algebraically. So for example, for example, let's give ourselves a little bit more real estate here. Let's say, let's go back to just saying we have an infinite series, S, that is equal to the sum from n equals one to infinity of a sub n. And let's say we know the partial sum, S sub n, so the sum of the first n terms"
    },
    {
        "Q": "4:06: Wouldn't it be easier to say TRANSPOSE instead of ADJUGATE?! At least would be better to say that T is something that is more used when it comes to studying matrices more? (\u00c3\u008d know its the same thing but might confuce students like me who are having exam about this and never heard of adjugate and instead of TRANSPOSE T)",
        "A": "Adjugate of matrix A is Transpose of Cofactor matrix of A",
        "video_name": "ArcrdMkEmKo",
        "timestamps": [
            246
        ],
        "3min_transcript": "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. so we don't get confused. So we have plus 20 plus 4 minus 6. So what does this get us? 5 plus 20 is 25, plus 4 is 29, minus 6 gets us to 23. So our determinant right over here is equal to 23. So now we are really in the home stretch. The inverse of this matrix is going to be 1 over our determinant times the transpose of this cofactor matrix. And the transpose of the cofactor matrix is called the adjugate. So let's do that. So let's write the adjugate here. This is the drum roll. We're really in the home stretch. C inverse is equal to 1 over the determinant, And so this is going to be equal to 1/23 times the transpose of our cofactor matrix. So we have our cofactor matrix right over here. So each row now becomes a column. So this row now becomes a column. So it becomes 1, negative 7, 5 becomes the first column. The second row becomes the second column-- 18, negative 11, negative 2. And then finally, the third row becomes the third column. You have negative 4, 5, and 3. And now we just have to multiply, or you could say divide, each of these by 23, and we are there."
    },
    {
        "Q": "I rally didn't get what you men't by when the video got to 3:29 to 4:00 that was really confusing to be honest",
        "A": "I think he was just trying to illustrate what he was multiplying. He shows you closer to 4:30 what he means, and how it all works out.",
        "video_name": "j3-XYLnxJDY",
        "timestamps": [
            209,
            240
        ],
        "3min_transcript": "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, just writing it in different ways. These are all different ways to think about it-- 6 plus 2 over 2, and then that times 3. So you could view it as the average of the smaller and larger rectangle. So you multiply each of the bases times the height and then take the average. You could view it as-- well, let's just add up the two base lengths, multiply that times the height, and then divide by 2. Or you could say, hey, let's take the average of the two base lengths and multiply that by 3. And that gives you another interesting way If you take the average of these two lengths, 6 plus 2 over 2 is 4. So that would be a width that looks something like-- let me do this in orange. A width of 4 would look something like this."
    },
    {
        "Q": "At 08:36, why doesn't \"x-1 = +-4\" become \"x = +- 5\"?",
        "A": "@James.p.french: He couldn t simplify to x=+-5 because x is not equal to that. He could ve simplified it to the solution x=5 or x=-3.",
        "video_name": "lGQw-W1PxBE",
        "timestamps": [
            516
        ],
        "3min_transcript": "I know that we're going to be a little bit less than the negative square root, but I'll do it the other way. I'll do it the way I did in the last video. So the other way to think about it is what happens when this term is 0? For this term to be 0, x has to be equal to 1. And does that ever happen? Can x be equal to 1? If x is equal to 1 here this term is 0. And then you have a situation where-- and then you have a minus y squared over 4 would have to equal 1, or this would have to be a negative number. So x could not be equal to 1. So y could be equal to negative 1. Let's try that out. If y is equal to negative 1, this term right here disappears. minus 1 squared over 16 is equal to 1. I just canceled out this term, because I'm saying what happens when y is equal to negative 1. You multiply both sides by 16. Let me do it over here. These get messy. x minus 1 squared is equal to 16. Take the square root of both sides. x minus 1 is equal to positive or negative 4. And so if x is equal to positive 4, if you add 1 to that x would be equal to 5. And then if x minus 1 would be minus 4 and you add 1 to that you will have x is equal to 3. So our 2 points or our 2 points closest to our center are the points 5 comma negative 1 and 3 comma negative 1. Let's plot those 2. So 5, 1 2 3 4 5, negative 1 and 3, negative 1. No, minus 3, because x minus 1 could be minus 4. That's what happens when you skip steps. If you have the minus 4 situation, then x is equal to minus 3. You go 1 2 3 minus 3, minus 1. So those are both points on this hyperbola. And then our intuition was correct, or it was what I said. That-- the positive square root is always going to be slightly below the asymptote, so we get our curve. It's going to look something like this. It's going to get closer and closer, and then here it's going to get closer and closer in that direction. It keeps getting closer and closer to that asymptote. And here, it's going to keep getting closer and closer to"
    },
    {
        "Q": "At 3:45 isn't it possible that b is -2? I mean if you were to ignore the first table and before knowing c and d. Or am I wrong?",
        "A": "Remember b>0 and b=/= 1",
        "video_name": "Iz6IVf8frjw",
        "timestamps": [
            225
        ],
        "3min_transcript": "Now this is an equivalent statement to saying that b to the a power is equal to ... oh sorry, not b to the a power. This is an equivalent statement to saying b to the 0 power is equal to a. This is saying what exponent do I need to raise b to to get a? You raise it to the 0 power. This is saying b to the 0 power is equal to a. Now what is anything to the 0 power, assuming that it's not 0? If we're assuming that b is not 0, if we're assuming that b is not 0, so we're going to assume that, and we can assume, and I think that's a safe assumption because where we're raising b to all of these other powers, we're getting a non-0 value. Since we know that b is not 0, anything with a 0 power is going to be 1. This tells us that a is equal to 1. We got one figured out. Now let's look at this next piece of information right over here. What does that tell us? That tells us that log base b of 2 is equal to 1. This is equivalent to saying the power that I needed to raise b to get to to 2 is 1. Or if I want to write in exponential form, I could write this as saying that b to the first power is equal to 2. I'm raising something to the first power and I'm getting 2? What is this thing? That means that b must be 2. 2 to the first power is 2. So b is equal to 2. b to the first power is equal to 2. You could say b to the first is equal to 2 to the first. That's also equal to 2. So b must be equal to 2. We've been able to figure that out. This is a 2 right over here. It actually makes sense. 2 to the 1.585 power, yeah, that feels right, that that's about 3. Let's see if we can figure out c. Let's look at this column. Let's see what this column is telling us. That column we could read as log base b. Now our y is 2c. Log base b of 2c is equal to 1.585. Or we could read this as b, if we write in exponential form, b to the 1.585 is equal to 2c. Now what's b to the 1.585? They tell us right over here that b to the 1.585 is 3, is 3, so this right over here is equal to 3. We get 2c is equal to 3, or divide both sides by 2, we would get c is equal to 1.5. This is working out pretty well. Now we have this last column,"
    },
    {
        "Q": "1:00 Is doing this method also the same as using the foil method kind of?",
        "A": "Yes, this is basically the intuition behind the foil method. It s showing how the distribution property can be used twice to obtain the quadratic in standard form, which is essentially the foil method :)",
        "video_name": "Xy8NKUoyy98",
        "timestamps": [
            60
        ],
        "3min_transcript": "- [Voiceover] Let's see if we can figure out the product of x minus four and x plus seven. And we want to write that product in standard quadratic form which is just a fancy way of saying a form where you have some coefficient on the second degree term, a x squared plus some coefficient b on the first degree term plus the constant term. So this right over here would be standard quadratic form. So that's the form that we want to express this product in and encourage you to pause the video and try to work through it on your own. Alright, now let's work through this. And the key when we're multiplying two binomials like this, or actually when you're multiplying any polynomials, is just to remember the distributive property that we all by this point know quite well. So what we could view this is as is we could distribute this x minus four, this entire expression over the x and the seven. So we could say that this is the same thing as x minus four times x plus x minus four times seven. So let's write that. So x minus four times x, or we could write this as x times x minus four. that's right there. Plus seven times x minus four. Times x minus four. Notice all we did is distribute the x minus four. We took this whole thing and we multiplied it by each term over here. We multiplied x by x minus four and we multiplied seven by x minus four. Now, we see that we have these, I guess you can call them two seperate terms. And to simplify each of them, or to multiply them out, we just have to distribute. In this first we're going to have to distribute this blue x. And over here we have to distribute this blue seven. So let's do that. So here we can say x times x is going to be x squared. X times, we have a negative here, so we can say negative four is going to be negative four x. And just like, that we get x squared minus four x. And then over here we have seven times x so that's going to be plus seven x. which is negative 28. And we are almost done. We can simplify it a little bit more. We have two first degree terms here. If I have negative four xs and to that I add seven xs, what is that going to be? Well those two terms together, these two terms together are going to be negative four plus seven xs. Negative four plus, plus seven. Negative four plus seven xs. So all I'm doing here, I'm making it very clear that I'm adding these two coefficients, and then we have all of the other terms. We have the x squared. X squared plus this and then we have, and then we have the minus, and then we have the minus 28. And we're at the home stretch! This would simply to x squared. Now negative four plus seven is three, so this is going to be plus three x. That's what these two middle terms simplify to, to three x."
    },
    {
        "Q": "At 0:28 Sal Said that an Odd Function Implies j(a) = - j(-a). Is this equivalent to -j(a) = j(-a) the more well known definition of an odd function? Or did Sal make a mistake?",
        "A": "Multiply both sides by -1. They re the same.",
        "video_name": "zltgXTlUVLw",
        "timestamps": [
            28
        ],
        "3min_transcript": "Which of these functions is odd? And so let's remind ourselves what it means for a function to be odd. So I have a function-- well, they've already used f, g, and h, so I'll use j. So function j is odd. If you evaluate j at some value-- so let's say j of a. And if you evaluate that j at the negative of that value, and if these two things are the negative of each other, then my function is odd. If these two things were the same-- if they didn't have this negative here-- then it would be an even function. So let's see which of these meet the criteria of being odd. So let's look at f of x. So we could pick a particular point. So let's say when x is equal to 2. So we get f of 2 is equal to 2. Now, what is f of negative 2? f of negative 2 looks like it is 6. f of negative 2 is equal to 6. So these aren't the negative of each other. In order for this to be odd, f of negative 2 have had to be equal to negative 2. So f of x is definitely not odd. So all I have to do is find even one case that violated this constraint to be odd. And so I can say it's definitely not odd. Now let's look at g of x. So I could use the same-- let's see, when x is equal to 2, we get g of 2 is equal to negative 7. Now let's look at when g is negative 2. So we get g of negative 2 is also equal to negative 7. So here we have a situation-- and it looks like that's the case for any x we pick-- that g of x is going to be equal to g of negative x. So g of x is equal to g of negative x. It's symmetric around the y-- or I should say the vertical axis-- right over here. So which of these functions is odd? Definitely not g of x. So our last hope is h of x. Let's see if h of x seems to meet the criteria. I'll do it in this green color. So if we take h of 1-- and we can look at it even visually. So h of 1 gets us right over here. h of negative 1 seems to get us an equal amount, an equal distance, negative. So it seems to fit for 1. For 2-- well, 2 is at the x-axis. But that's definitely h of 2 is 0. h of negative 2 is 0. But those are the negatives of each other. 0 is equal to negative 0. If we go to, say, h of 4, h of 4 is this negative number. And h of negative 4 seems to be a positive number of the same magnitude. So once again, this is the negative of this. So it looks like this is indeed an odd function."
    },
    {
        "Q": "At 3:19, why does Sal divide by 7?",
        "A": ".7 x 10 will equal 7. .7 represents 79 percent of the full price.",
        "video_name": "d1oNF88SAgg",
        "timestamps": [
            199
        ],
        "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": "At 1:48 Sal states that m=-4w+11 and proceeds to plug that into the other equation that we derived. But couldn't Sal simply plug that right back into the same equation that we got that from. So couldn't Sal use one equation to solve for two unknowns by plugging back in the m in terms of w?",
        "A": "Try it. 100*(-4w + 11) + 400*w = 1100 -400*w + 1100 + 400*w = 1100 1100 = 1100 That s called a tautology. It s a true statement, but it s not providing any useful information. When you plug the value of m from one equation into another, your w terms don t cancel each other out, leading to a meaningful result of the form w = value",
        "video_name": "2EwPpga_XPw",
        "timestamps": [
            108
        ],
        "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": "What does Mister Khan mean when he says, \"open parentheses\", at, 3:48-3:50? What does he mean specifically? Obviously he means that the parentheses are open. but what does the phenomenon of the open parentheses imply?\n\nThat is my question for the day. I hope it was relevant. Goodbye. Reuben.",
        "A": "Open the parentheses means to open it up and compute the operations inside and operations having to do what s inside the parentheses. The distributive property is an example. Here s to show: -2 ( 9x+4 ) -- Opening the parentheses would be distributing the two to the things inside. => -2 ( 9x+4 ) = -18x-8 I hope I answered your question.",
        "video_name": "rCGHUXSd15s",
        "timestamps": [
            228,
            230
        ],
        "3min_transcript": "Now, if you agree to that, now let's work on this a little bit. So we can start with the absolute value of a minus b. And there's a bunch of ways that we could tackle this. But one way we could approach it, and actually let me write a little lower, so I don't bump into this stuff right over here. So let me start with the absolute value of a minus b. Now a, we could rewrite, this is just a, I don't wanna say it's positive or negative, a could be a negative number. This is just a right over here, but we can rewrite this as being the same thing as negative negative a. Or I guess negative negative a plus, minus b. So let's just think about this a little bit, with what I just did. So this a is the same thing as all of this business. The negative negative a. The negative of a negative is going to be a positive, this is positive a. So I haven't changed this in any dramatic way. But what's this going to be the same thing as? Well we can factor out a negative sign. So let's do that, and actually let me, we could factor out, using a color I haven't used yet, we could factor out the negative signs. So this is going to be equal to the absolute value of, I can factor out the negatives, the negative of, and then I have open parentheses, open parentheses negative a. And if you factor out a negative here, this is going to be positive b. So or plus b. Let me just close the parentheses in the same color, Now why is this interesting? Well we just said the absolute value of a negative is the same thing as, the absolute value of negative x is the same thing as the absolute value of x. So the absolute value of the negative of negative a plus b is going to be equal to, this is going to be equal to the absolute value, let me do that yellow color is going to be equal to the absolute value of this thing without the negative right here. You see that right over here. The absolute value of negative x is the same thing as the absolute value of x. So the absolute value of negative negative a plus b, is the same thing as the absolute value of negative a plus b. And negative a plus b, well that's the same thing as b minus a. I'm just swapping the order right over here. B minus a."
    },
    {
        "Q": "At 2:53, Sal says that the kite has perpendicular lines at an angle of 90 degrees. Do all quadrilaterals with perpendicular lines have to have it at an angle of 90 degrees?",
        "A": "There seem to be a couple of questions in your question. First the definition of perpendicular is any 2 lines that form a 90 degree angle. So anything that refers to the word perpendicular is implying that the angle is 90 degrees. For clarification, the video is talking about the diagonals of the quadrilateral intersecting at 90 degrees. This is a property of a kite. Also squares have this property, so they are also kites. However (non-square) rectangles do no have this property, so are not kites. Hope this helps.",
        "video_name": "inlMrf2d-k4",
        "timestamps": [
            173
        ],
        "3min_transcript": ""
    },
    {
        "Q": "what did sal mean by the congruent side could be opposite to each other on 1:45",
        "A": "This is answered at 2:12 with a color picture of a parallelogram. In a quadrilateral, a shape with four sides, any two sides are either going to be adjacent (share an angle or endpoint) or opposite (do not touch or share an angle or endpoint). We are dealing with 2 pairs of congruent sides. Congruent sides or line segments have the same length. So, if you have two pairs of congruent sides in a quadrilateral, you will either have a parallelogram of a kite.",
        "video_name": "inlMrf2d-k4",
        "timestamps": [
            105
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 1:44, if a pair of congruent sides are adjacent, does it affect the fact that it is congruent?",
        "A": "No. Like in triangles, if it is a scalene triangle, all sides will be adjacent, but sense there are no congruent, or equal, sides, then there will be none. Even in a kite, if you think about it, there will be un-congruent ( I just made that word up ) sides that are, in face, adjacent. Please vote for this answer if it was helpful to you! :D",
        "video_name": "inlMrf2d-k4",
        "timestamps": [
            104
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 1:02,can a kite also be a diamond?",
        "A": "yes, as it is of the same shape",
        "video_name": "inlMrf2d-k4",
        "timestamps": [
            62
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at around 2:00 to 2:11 what is adjacent",
        "A": "Next to or adjoining something else.",
        "video_name": "inlMrf2d-k4",
        "timestamps": [
            120,
            131
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 5:19, how is a rhombus also going to be a kite?",
        "A": "a kite has two pair of congruent (equal) adjacent (next to) sides. A rhombus has 4 congruent sides, which means each pair of adjacent sides is congruent. Therefore every rhombus is a kite.",
        "video_name": "inlMrf2d-k4",
        "timestamps": [
            319
        ],
        "3min_transcript": ""
    },
    {
        "Q": "In 0:39, I don't understand the part where it says that there are a bunch of different Xs. How is that possible?",
        "A": "Multiple variables (Same ones) can be in an inequality. Ex: x + x + 1 =23",
        "video_name": "UTs4uZhu5t8",
        "timestamps": [
            39
        ],
        "3min_transcript": "what i want to do in this video is a handful of fairly simple inequality videos. But the real value of it, I think, will be just to get you warmed up in the notation of inequality. So, let's just start with one. we have x minus 5 is less than 35. So let's see if we can find all of the x's that will satisfy this equation. And that's one of the distinctions of an inequality. In an equation, you typically have one solution, or at least the ones we've solved so far. In the future, we'll see equations where they have more than one solution. But in the ones we've solved so far, you solved for a particular x. In the inequalities, there's a whole set of x's that will satisfy this inequality. So they're saying, what are all the x's, that when you subtract 5 from them, it's going to be less than 35? And we can already think about it. I mean 0 minus 5. That's less than 35. Minus 100 minus 5. That's less than 35. 5 minus 5. That's less than 35. So there's clearly a lot of x's that will satisfy that. essentially encompasses all of the x's. So the way we do that is essentially the same way that we solved any equations. We want to get just the x terms, in this case, on the left-hand side. So I want to get rid of this negative 5, and I can do that by adding 5 to both sides of this equation. So I can add 5 to both sides of this equation. That won't change the inequality. It won't change the less than sign. If something is less than something else, something plus 5 is still going to be less than the other thing plus 5. So on the left-hand side, we just have an x. This negative 5 and this positive 5 cancel out. x is less than 35 plus 5, which is 40. And that's our solution. And to just visualize the set of all numbers that represents, let me draw a number line here. And I'll do it around-- let's say that's 40, 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."
    },
    {
        "Q": "Around 3:00...\nWait, isn't 39.9999(9repeating) equal to 40?\nSince 0.99999(9repeating) is equal to 1,\n39 + 0.99999... =? 39 + 1",
        "A": "No it is not one or 40 because it is a decimal. When it is point somethine it is smaller than the rounded number",
        "video_name": "UTs4uZhu5t8",
        "timestamps": [
            180
        ],
        "3min_transcript": "essentially encompasses all of the x's. So the way we do that is essentially the same way that we solved any equations. We want to get just the x terms, in this case, on the left-hand side. So I want to get rid of this negative 5, and I can do that by adding 5 to both sides of this equation. So I can add 5 to both sides of this equation. That won't change the inequality. It won't change the less than sign. If something is less than something else, something plus 5 is still going to be less than the other thing plus 5. So on the left-hand side, we just have an x. This negative 5 and this positive 5 cancel out. x is less than 35 plus 5, which is 40. And that's our solution. And to just visualize the set of all numbers that represents, let me draw a number line here. And I'll do it around-- let's say that's 40, 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"
    },
    {
        "Q": "4:29 Why did the symbol > did not reversed to < hence you subtracted negative",
        "A": "You re right. The symbol only reverses when multiplying or dividing negative numbers.",
        "video_name": "UTs4uZhu5t8",
        "timestamps": [
            269
        ],
        "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": "At 4:19, why is the square root of 2 x the square root of 2 equal to 2?",
        "A": "Because sqrt(2) * sqrt(2) = sqrt(4) And, sqrt(4) = 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": "At around the 6:15 mark, if the interval of convergence included 1 or -1, would the radius of convergence still be 1?",
        "A": "To determine the radius of convergence, do not worry about whether the endpoints are included or not. The radius of convergence would be one regardless of whether or not the endpoints were included.",
        "video_name": "DlBQcj_zQk0",
        "timestamps": [
            375
        ],
        "3min_transcript": "minus our common ratio. What's our common ratio? Our common ratio in this example is x. Going from one term to the next, we're just multiplying by x. We're just multiplying by x right over there. Now, this is pretty neat, because we're going to be able to use this fact to put more traditionally-defined functions into this form, and then try to expand them out using a geometric series. And this whole idea of using power series, or in this special case, geometric series to represent functions, has all sorts of applications in engineering and finance. Using a finite number of terms of these series, you could kind of approximate the functions in a way that's simpler for the human brain to understand, or maybe a simpler way to manipulate in some way. But what's interesting here is instead of just going from the sum to-- instead of going from this expanded-out version to this kind of finite value, in this form and expand it out into a geometric series. But we have to be careful to make sure that we're only doing it over the interval of convergence. This is only going to be true over the interval of convergence. Now, one other term you might see in your mathematical career is a radius. Radius of convergence. And this is how far-- up to what value, but not including this value. So as long as our x value stays less than a certain amount from our c value, then this thing will converge. Now in this case, our c value is 0. So we could ask ourselves a question. As long as x stays within some value of 0, this thing is going to converge. Well, you see it right over here. As long as x stays within one of 0. as it stays less than 1, or as long as it stays greater than negative 1. It can stray anything less than one away from 0, either in the positive direction or the negative direction. Then this thing will still converge. So we could say that our radius of convergence is equal to 1. Another way to think about it, our interval of convergence-- we're going from negative 1 to 1, not including those two boundaries, so our interval is 2. So our radius of convergence is half of that. As long as x stays within one of 0, and that's the same thing as saying this right over here, this series is going to converge."
    },
    {
        "Q": "In this video Sal uses 6x+2 to get the solution, yet the problem has 6x-2.?? This is at 5:26/10:57 in the video:",
        "A": "The quality of those questions weren t the best, if you look closely you can see that the equation really is 6x+2. I had to tilt my screen a little bit before I could tell if it was a plus or a minus.",
        "video_name": "_HJljJuVHLw",
        "timestamps": [
            326,
            657
        ],
        "3min_transcript": "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? So let's add this up. Minus 6, minus 16, that's minus 22. Plus 2 is minus 20. That's right. And 2x plus 6x that is 8x plus 4 is 12, 12 plus 2 is 14, 14 plus 6 is 20. So we have 20x minus 20 is equal to 540 degrees. 20x minus 20 is equal to 540. Let's divide both sides of this equation by 20. So you get x 1 minus 1 is equal to-- it would be 54 divided by 2, which is equal to 27. And they want to know, what is a measure in degrees of the That's going to be this one. That's the largest one. It's 6 times x plus 2. So 6 times 28, that's 48. 2 times 6 is 12 plus 4 is 168. So it's 168 plus 2. It's 170 degrees. Choice C. Problem 48: What is the measure of angle 1? So this we're going into the angle game. And these are fun, because they are kind of these deductive reasoning problems where you just use a couple of simple rules and just fill in the whole thing. So let's think about it. This is 36 degrees. They tell us that this whole angle right here is a right angle. So this angle right here is going to be the complement to 36 degrees."
    },
    {
        "Q": "8:52, what is equiangular ?",
        "A": "Equiangular means that all the angles of the polygon/shape have equal measure. For example an equiangular triangle would have three angles which all have a degree measure of 60.",
        "video_name": "_HJljJuVHLw",
        "timestamps": [
            532
        ],
        "3min_transcript": "So the answer is A. Problem 49: What is the measure of angle WZX? So they want to know what this angle right here is. Let's do the angle game some more. Let's see, we can immediately figure out what this angle is, because it is the supplement of 132 degrees, so this is going to be 180 minus 132. So this is 48 degrees. This angle plus this angle is going to be equal to this angle. Or this angle plus this angle plus this angle is equal to 180. I don't know what I just said, I think I said something wrong. Write that down. So this angle is going to be equal to 180 minus 52 minus 48. 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."
    },
    {
        "Q": "At 4:00 and again at 5;10, Sal says that 6x+2 is the largest angle, how can he tell?",
        "A": "Among all the angles, ie, 2x, 6x, 4x - 6, 2x - 16 and 6x + 2, In 6x +2 you are multiplying x 6 times and further adding a 2. Try assigning x a value, u will find that 6x + 2 is the largest.",
        "video_name": "_HJljJuVHLw",
        "timestamps": [
            240
        ],
        "3min_transcript": "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? So let's add this up. Minus 6, minus 16, that's minus 22. Plus 2 is minus 20. That's right. And 2x plus 6x that is 8x plus 4 is 12, 12 plus 2 is 14, 14 plus 6 is 20. So we have 20x minus 20 is equal to 540 degrees. 20x minus 20 is equal to 540. Let's divide both sides of this equation by 20. So you get x 1 minus 1 is equal to-- it would be 54 divided by 2, which is equal to 27."
    },
    {
        "Q": "At 4:01, isn't it supposed to be 6x-2 and not 6x+2?",
        "A": "I thought so too. Probably we can t see it clearly, when I rewatched I thought I saw a hint of the vertical line in the +.",
        "video_name": "_HJljJuVHLw",
        "timestamps": [
            241
        ],
        "3min_transcript": "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? So let's add this up. Minus 6, minus 16, that's minus 22. Plus 2 is minus 20. That's right. And 2x plus 6x that is 8x plus 4 is 12, 12 plus 2 is 14, 14 plus 6 is 20. So we have 20x minus 20 is equal to 540 degrees. 20x minus 20 is equal to 540. Let's divide both sides of this equation by 20. So you get x 1 minus 1 is equal to-- it would be 54 divided by 2, which is equal to 27."
    },
    {
        "Q": "at 4:10 Sal mentions that it is ar^k power. Why do the k power instead of nth power?",
        "A": "k is the index, where k=0,1,2,3,...n So it would mean ar\u00e2\u0081\u00b0+ar\u00c2\u00b9+ar\u00c2\u00b2+ar\u00c2\u00b3+...+ar^n",
        "video_name": "CecgFWTg9pQ",
        "timestamps": [
            250
        ],
        "3min_transcript": "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. And I encourage you to pause the video and try it on your own. Well, we could think about it this way. And I'll give you a little hint. You could view this term right over here as a times r to the 0. And let me write it down. This is a times r to the 0. This is a times r to the first, r squared, r third, and now the pattern might be emerging for you. So we can write this as the sum, so capital sigma right over here. We can start our index at 0. So we could say from k equals 0 all the way to k equals n of a times r to the k-th power. a general way to represent a geometric series where r is some non-zero common ratio. It can even be a negative value."
    },
    {
        "Q": "The example at 8:05 if they asked f(-2) would it be does not exist as the line has an open circle",
        "A": "f(-2) would be undefined or does not exist because of the open circle. However, the limit as x->-2 exists and it s 4 as Sal demonstrated.",
        "video_name": "nOnd3SiYZqM",
        "timestamps": [
            485
        ],
        "3min_transcript": "If x is 7.5, f of 7.5 is here. So it looks like our value of f of x is getting closer and closer and closer to 3. So it looks like the limit of f of x, as x approaches 8 from the negative side, is equal to 3. What about from the positive side? What about the limit of f of x as x approaches 8 from the positive direction or from the right side? Well, here we see as x is 9, this is our f of x. As x is 8.5, this is our f of 8.5. It seems like we're approaching f of x equaling 1. So notice, these two limits are different. So the non-one-sided limit, or the two-sided limit, does not exist at f of x or as we approach 8. So let me write that down. The limit of f of x, as x approaches 8-- this does not exist. Let's do one more example. And here they're actually asking us a question. The function f is graphed below. What appears to be the value of the one-sided limit, the limit of f of x-- this is f of x-- as x approaches negative 2 from the negative direction? So this is the negative 2 from the negative direction. So we care what happens as x approaches negative 2. We see f of x is actually undefined right over there. But let's see what happens as we approach from the negative direction, or as we approach from values less than negative 2, or as we approach from the left. As we approach from the left, f of negative 4 is right over here. So this is f of negative 4. f of negative 3 is right over here. f of negative 2.5 seems to be right over here. We seem to be getting closer and closer So I would say that it looks-- at least, graphically-- the limit of f of x, as x approaches 2 from the negative direction, is equal to 4. Now, if we also asked ourselves the limit of f of x, as x approaches negative 2 from the positive direction, we would get a similar result. Now, we're going to approach from when x is 0, f of x seems to be right over here. When x is 1, f of x is right over here. When x is negative 1, f of x is there. When x is negative 1.9, f of x seems to be right over here. So once again, we seem to be getting closer and closer to 4. Because the left-handed limit and the right-handed limit are the same value. Because both one-sided limits are approaching the same thing, we can say that the limit of f of x, as x approaches"
    },
    {
        "Q": "At 5:25,when sal wrote the general limit of F(X) as x approaches 4,didn't he forget the minus sign in front of 5?its -5 and not 5",
        "A": "Yes he did indeed forget the minus sign although he recognized that by editing in a correction box in the bottom right hand of the screen.",
        "video_name": "nOnd3SiYZqM",
        "timestamps": [
            325
        ],
        "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": "I've been thinking about this for a while and I can't figure out how to find the limit of an asymptote. At 6:20 as x=>3- is it negative infinity or undefined?",
        "A": "Recall that an asymptote is just a line that a given function or curve tends to (gets closer and closer to). At 6:20, the asymptote is x = 3. However, if you actually meant to find the limit of f(x) as x -> 3\u00e2\u0081\u00bb, it is -\u00e2\u0088\u009e.",
        "video_name": "nOnd3SiYZqM",
        "timestamps": [
            380
        ],
        "3min_transcript": "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. If x is 7.5, f of 7.5 is here. So it looks like our value of f of x is getting closer and closer and closer to 3. So it looks like the limit of f of x, as x approaches 8 from the negative side, is equal to 3. What about from the positive side? What about the limit of f of x as x approaches 8 from the positive direction or from the right side? Well, here we see as x is 9, this is our f of x. As x is 8.5, this is our f of 8.5. It seems like we're approaching f of x equaling 1. So notice, these two limits are different. So the non-one-sided limit, or the two-sided limit, does not exist at f of x or as we approach 8. So let me write that down. The limit of f of x, as x approaches 8--"
    },
    {
        "Q": "At 5:29, I thought the limit of f(x) as it approaches 4 would not exist because it is a corner? Aren't corners not differentiable? Does it or does it not exist?",
        "A": "The derivative of the function at 4 and limit of the function as x approaches 4 are not the same thing. The derivative does not exist at 4 because of the sharp corner. But the function is continuous at 4, and so the limit is just f(4).",
        "video_name": "nOnd3SiYZqM",
        "timestamps": [
            329
        ],
        "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 12:22 Sal wrote that N(A) = N(rref(A)). Does that stay true if we exchange rows when we are reducing the matrix to the rref?",
        "A": "If you had 2 simultaneous equations (not matrices) in x and y, and you exchanged rows, how would that affect the solution?",
        "video_name": "qvyboGryeA8",
        "timestamps": [
            742
        ],
        "3min_transcript": "These are just random scalars that are a member of-- We can pick any real number for x3 and we could pick any real number for x4. So our solution set is just a linear combination of those two vectors. What's another way of saying a linear combination of two vectors? Let me write this. The null space of A, which is just a solution set of this equation, it's just all the x's that satisfy this equation, it equals all of the linear combinations of this vector and that vector. What do we call all the linear combinations of two vectors? It's the span of those two vectors. So it equals the span of that vector and that vector. Of the vector 1, minus 2, 1, 0, and the vector 2, minus 3, 0, 1. And this is our null space. Before letting you go, let me just point out one interesting 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"
    },
    {
        "Q": "At 5:43 shouldn't the third row be 0 -1 -2 -3? since u r subtracting 4 from each element.",
        "A": "A logical thing to do is what you are describing which is to replace the third row by the third row minus four times the first row. What Sal is doing is kind of the reverse but it works too. He is replacing the third row by the first row multiplied by four and then subtracting the third row. Both ways work and we see that they just differ by a factor of -1. I would have done it the way you suggest in your question and then multiplied by -1 but either way you get the same result.",
        "video_name": "qvyboGryeA8",
        "timestamps": [
            343
        ],
        "3min_transcript": "And then I augment that with the 0 vector. And the immediate thing you should notice is we took the pain of multiplying this times this to equal that, and we wrote this as a system of equations, but now we want to solve the system of equations, we're going back to the augmented matrix world. What does this augmented matrix look like? Well, this is just our matrix A right there. That's just matrix A right there, that's just the 0 vector right there. And to solve this, and we've done this before, we're just going to put this augmented matrix into row echelon form. What you're going to find is when you put it into row echelon form, this right side's not going to change at all, because no matter what you multiply or subtract by, you're just doing it all times 0, so you just keep ending up with 0. So as we put this into reduced row echelon form, were actually just putting matrix A into reduced echelon form. So let me do that, instead of just talking about it. So let me start off by keeping row 1 the same. And then I want to eliminate this 1 right here, so let me replace row 2 with row 2 minus row 1. So 1 minus 1 is 0. 2 minus 1 is 1. 3 minus 1 is 2. 4 minus 1 is 3. 0 minus 0 is 0. You can see the 0's aren't going to change. And then let me replace this guy with 4 times this guy, minus this guy. So I can only get rid of this. So 4 times 1 minus 4 is 0. 4 times 1 minus 3 is 1. 4 times 1 minus 2 is 2. 4 times 1 minus 1 is 3. 4 times 0 minus 0 is 0. Now I want to get rid of, if I want to put this in reduced row echelon form, I want to get rid of that So let me keep my middle row the same. My middle row is 0, 1, 2, 3. So that's 0 on the augmented side of it, although these 0's are never going to change, it's really just a little bit of an exercise just to keep writing them. And my first row, let me replace it with the first row minus the second row, so I can get rid of this 1. So 1 minus 0 is 1. 1 minus 1 is 0. 1 minus 2 is minus 1. 1 minus 3 is minus 2. And 0 minus 0 is 0. And let me replace this last row with the last row minus the middle row. So 0 minus 0 is 0. 1 minus 1 is 0. 2 minus 2 is 0. I think you see where this is going. 3 minus 3 is 0. And obviously 0 minus 0 is 0. So this system of equations has been reduced, just by"
    },
    {
        "Q": "At 15:00, the equation for projv(x) is\nA(AT A)-1 AT x where AT = A transpose and -1 means inverse\nThis formula seems like it should reduce to x since\n\n(ATA)-1 = (A-1)(AT-1) by the rule for inverse of a product\nso (A A-1) (AT-1 AT) = I I and IIx = x?\nWhat am I missing here?",
        "A": "The answer is that A and AT are rectangular, not square so they have no inverse. (ATA) is invertible if A originally had linearly independent columns as per video 106 \u00e2\u0080\u009cLin Alg: Showing that A-transpose x A is invertible\u00e2\u0080\u009d",
        "video_name": "cTyNpXB92bQ",
        "timestamps": [
            900
        ],
        "3min_transcript": "A transpose A inverse, which'll always exist, times A transpose, times x. Now we said the projection of x onto v is going to be equal to A times y, for some y. Well we just solved for the y using our definition of a projection. We just were able to solve for y. So now, we can define our projection of x onto v as a matrix vector product. So we can write the projection onto v of our vector x is equal to A, times y, and y is just equal to that thing right there. So A times A transpose A inverse-- which always exists transpose, times x. And this thing right here, this long convoluted thing, that's just some matrix, some matrix which always exists for any subspace that has some basis. So we've just been able to express the projection of x onto a subspace as a matrix vector product. So anything that can be any matrix vector product transformation is a linear transformation. And not only did we show that it's a linear transformation, we showed that, look, if you can give me the basis for v, I'm going to make those column vectors equal to the column of some matrix A. And then if I take matrix A, if I take its transpose, if I take A transpose times A, and invert it, and if I multiply them all out in this way, I'm going to get the transformation matrix for the projection. do by hand for many, many projections, but this is super useful if you're going to do some three-dimensional graphical programming. Let's say you have some three-dimensional object, and you want to know what it looks like from the point of view of some observer. So let's say you have some observer. Some observer's point of view is essentially going to be some subspace. You want to see what the projection of this cube onto the subspace, how would it look to the person who's essentially on to this flat screen right there. How would that cube look from this point of view? Well if you know the basis for this subspace, you can just apply this transformation. You can make a matrix whose columns are these basis vectors for this observer's point of view. And then you can apply this to every vector in this cube in"
    },
    {
        "Q": "At 2:10, why is e^(-st) * e^(at) combine to e^(a-s)t instead of e^(a-s) ?? Wouldn't the -t and t combine to cancel??",
        "A": "For multiplying terms which have the same base, (x^a) * (x^b) = x^(a + b). So, e^(-st) * e^(at) = e^(-st + at). Take out the common factor, t from (-st + at) => t(a-s), which gives us e^(a-s)t",
        "video_name": "33TYoybjqPg",
        "timestamps": [
            130
        ],
        "3min_transcript": "Let's keep doing some Laplace transforms. For one, it's good to see where a lot of those Laplace transform tables you'll see later on actually come from, and it just makes you comfortable with the mathematics. Which is really just kind of your second semester calculus mathematics, but it makes you comfortable with the whole notion of what we're doing. So first of all, let me just rewrite the definition of the So it's the L from Laverne & Shirley. So the Laplace transform of some function of t is equal to the improper integral from 0 to infinity of e to the minus st times our function. Times our function of t, and that's with respect to dt. So let's do another Laplace transform. Let's say that we want to take the Laplace transform-- and now our function f of t, let's say it is e to the at. Well we just substituted it into this definition of the Laplace transform. And this is all going to be really good integration Especially integration by parts. Almost every Laplace transform problem turns into an integration by parts problem. Which, as we learned long ago, integration by parts is just the reverse product rule. This is equal to the integral from 0 to infinity. e to the minus st times e to the at, right? That's our f of t. dt. Well this is equal to just adding the exponents because we have the same base. The integral from 0 to infinity of e to the a minus stdt. Well that's equal to what? With respect to C. So it's equal to-- a minus s, that's just going to be a So we can just leave it out on the outside. 1/a minus s times e to the a minus st. And we're going to evaluate that from t is equal to infinity or the limit as t approaches infinity to t is equal to 0. And I could have put this inside the brackets, but it's just a constant term, right? None of them have t's in them, so I can just pull them out. And so this is equal to 1/a minus s times-- now we essentially have to evaluate t at infinity. So what is the limit at infinity?"
    },
    {
        "Q": "At 2:50, why is theta taken as the obtuse angle, and not the acute angle?Isn't cos (-60)= -1/2 also?",
        "A": "cos(x) = cos(-x)",
        "video_name": "eTDaJ4ebK28",
        "timestamps": [
            170
        ],
        "3min_transcript": "If I say, you know, what is the inverse cosine of x, my brain says, what angle can I take the cosine of to get x? So with that said, let's try it out on an example. Let's say that I have the arc, I'm told, no, two c's there, I'm told to evaluate the arccosine of minus 1/2. My brain, you know, let's say that this is going to be equal to, it's going to be equal to some angle. And this is equivalent to saying that the cosine of my mystery angle is equal to minus 1/2. And as soon as you put it in this way, at least for my brain, it becomes a lot easier to process. So let's draw our unit circle and see if we can make some headway here. So that's my, let me see if I can draw a little straighter. Maybe I could actually draw, put rulers here, and if I put a ruler here, maybe I can draw a straight line. 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"
    },
    {
        "Q": "At 12:05, what if there is bigger angle like 100 pi instead of 3pi?\nDo you still keep going around the unit circle to find out where you finally land up and then calculate the x coordinate?",
        "A": "You could go around the circle 50 times to finally figure out where you land. However, if you understand that 2\u00cf\u0080 radians is a full circle, then you can save yourself a lot of trouble. 100\u00cf\u0080 = 50\u00e2\u0080\u00a22\u00cf\u0080 This means 100\u00cf\u0080 is the same as 50 full circles, and you end up on the same point on the unit circle as 0 radians. When this happens we say 100\u00cf\u0080 is coterminal to 0.",
        "video_name": "eTDaJ4ebK28",
        "timestamps": [
            725
        ],
        "3min_transcript": "x that you put in here. This is true for any x, any value between negative 1 and 1 including those two endpoints, this is going to be true. Now what if I were ask you what the arccosine of the cosine of theta is? What is this going to be equal to? My answer is, it depends on the theta. So, if theta is in the, if theta is in the range, if theta is between, if theta is between 0 and pi, so it's in our valid a range for, kind of, our range for the product of the arccosine, then this will be equal to theta. If this is true for theta. But what if we take some theta out of that range? Let's try it out. Let's take, so let me do one with theta in that range. one of them that we know. Let's take the cosine of, let's stick with cosine of 2 pi over 3. Cosine of 2 pi over 3 radians, that's the same thing as the arccosine of minus 1/2. Cosine of 2 pi over 3 is minus 1/2. We just saw that in the earlier part of this video. And then we solved this. We said, oh, this is equal to 1 pi over 3. So for in the range of thetas between 0 and pi it worked. And that's because the arccosine function can only produce values between 0 and pi. But what if I were to ask you, what is the arccosine of the cosine of, I don't know, of 3 pi. So if I were to draw the unit circle here, let me draw the unit circle, a real quick one. What's 3 pi? 2 pi is if I go around once. And then I go around another pi, so I end up right here. So I've gone around 1 1/2 times the unit circle. So this is 3 pi. What's the x-coordinate here? It's minus 1. So cosine of 3 pi is minus 1, right? So what's arccosine of minus 1? Arccosine of minus 1. Well remember, the range, or the set of values, that arccosine can evaluate to is in this upper hemisphere. It's between, this can only be between pi and 0. So arccosine of negative 1 is just going to be pi. So this is going to be pi. Arccosine of negative, this is negative 1, arccosine of negative 1 is pi. And that's a reasonable statement, because the difference between 3 pi and pi is just going around the unit"
    },
    {
        "Q": "When he says that it looks more like an ellipse at 2:20, what is an ellipse, please?",
        "A": "Basically, an ellipse is an oval. More technically, here s the definition from wikipedia: In mathematics, an ellipse is a curve on a plane that surrounds two focal points such that the sum of the distances to the two focal points is constant for every point on the curve. As such, it is a generalization of a circle, which is a special type of an ellipse that has both focal points at the same location.",
        "video_name": "eTDaJ4ebK28",
        "timestamps": [
            140
        ],
        "3min_transcript": "I've already made videos on the arcsine and the arctangent, so I've already made videos on the arcsine and the arctangent, so to kind of complete the trifecta, I might as well make a video on the arccosine. And just like the other inverse trigonometric functions, the arccosine is kind of the same thought process. If I were to tell you the arc, no, I'm doing cosine, if our tell you that arccosine of x is equal to theta. This is an equivalent statement to saying that the inverse cosine of x is equal to theta. These are just two different ways of writing the exact same thing. And as soon as I see either an arc- anything, or an inverse trig function in general, my brain immediately rearranges this. My brain immediately says, this is saying that if I take the cosine of some angle theta, that I'm going to get x. Or that same statement up here. If I say, you know, what is the inverse cosine of x, my brain says, what angle can I take the cosine of to get x? So with that said, let's try it out on an example. Let's say that I have the arc, I'm told, no, two c's there, I'm told to evaluate the arccosine of minus 1/2. My brain, you know, let's say that this is going to be equal to, it's going to be equal to some angle. And this is equivalent to saying that the cosine of my mystery angle is equal to minus 1/2. And as soon as you put it in this way, at least for my brain, it becomes a lot easier to process. So let's draw our unit circle and see if we can make some headway here. So that's my, let me see if I can draw a little straighter. Maybe I could actually draw, put rulers here, and if I put a ruler here, maybe I can draw a straight line. 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."
    },
    {
        "Q": "at 9:00, shouldn't we square it and add then take square root?",
        "A": "Only if you are working with the magnitude (unit vector etc.). However in this example Sal is purely working with the vector itself in full (not the unit vector) and he s simply adding the two vectors together to get the resultant.",
        "video_name": "6Kw2nIwWYL0",
        "timestamps": [
            540
        ],
        "3min_transcript": "Scaled up version of the J unit vector and if you add this orange vector to this green vector, you get vector A. Similarly, vector B is equal to the length of the horizontal component and we got to be very careful. It's length is square root of two but it's going in the leftward direction. We're going to put a negative on it times I. If we just squared it, I is doing something like this, square root of two times I would look like that. Negative square root of two would point it to the left. This is negative square root of two times I and then we're going to have plus square root of two times J. Now that we've broken them up in their components, we're ready to figure out what, at least broken up into its components A plus B is equal to, well it's going to be the sum of all of these things. Let me just write that down. It's going to be that, copy and paste. That plus this, let me just copy and paste it and you're going to get that but of course we can simplify this. We can add the I unit vectors to each other. If I have three times the square root of three over two Is and then I have another negative square root of two I, I can add that together. This is going to be equal to three times the square root of three over two minus square root of two times I and then I can add this to this and I'm going to get plus three halves plus square root of two times J. It looks a little bit complicated and get approximations of each of these two values and we essentially have a at least a broken down into its components representation of A plus B. In the next video, we're now going to take this and figure out the actual magnitude and direction of A plus B."
    },
    {
        "Q": "Why is around 2:30 mins in... (1/3) dividing 3x but on the other side it is multiplying!",
        "A": "He s saying that multiplying both sides of the equation by the fraction (1/3) is the same as dividing both sides of the equation by 3.",
        "video_name": "kbqO0YTUyAY",
        "timestamps": [
            150
        ],
        "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": "I got lost at 3:00 when Sal wrote 9 - 2x",
        "A": "pretend 5=x. 9-2x=-1",
        "video_name": "kbqO0YTUyAY",
        "timestamps": [
            180
        ],
        "3min_transcript": "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. minus 2 something is just 1 of something. So you will just have an x there if you get rid of two of them. But on the right-hand side, you're going to get 9 minus 2 So the x's still didn't help you out. You still have a mystery mass on the right-hand side. So that doesn't help. So instead, what we say is-- and we did this the last time. We said, well, what if we took 1/3 of these things? If we take 1/3 of these things and take 1/3 of these things, we should still get the same mass on both sides because the original things had the same mass. And the equivalent of doing that mathematically is to say, why don't we multiply both sides by 1/3? Or another way to say it is we could divide both sides by 3. Multiplying by 1/3 is the same thing as dividing by 3. So we're going to multiply both sides by 1/3. When you multiply both sides by 1/3-- visually over here, if you had three x's, you multiply it by 1/3, you're only going to have one x left. If you have nine of these one-kilogram boxes, you multiply it by 1/3, you're only going to have three left. And over here, you can even visually-- if you divide by 3, which is the same thing as multiplying"
    },
    {
        "Q": "6:54 anyone happen to know what the name of the theorem is or where to find out? I seem to remember better if I have a name for these things. Thanks for any help.",
        "A": "I think it s called the Base Angle Theorem . Hope that helps a lil... :)",
        "video_name": "nMhJLn5ives",
        "timestamps": [
            414
        ],
        "3min_transcript": "Well, hopefully you know the angles in a triangle add up to 180 degrees. If you don't it's my fault because I haven't taught you that already. So let's figure out what the angles of this triangle add up to. Well, I mean we know they add up to 180, but using that information, we could figure out what this angle is. Because we know that this angle is 90, this angle is 45. So we say 45-- lets call this angle x; I'm trying to make it messy --45 plus 90-- this [INAUDIBLE] is a 90 degree angle --plus is equal to 180 degrees. And that's because the angles in a triangle always add up to 180 degrees. So if we just solve for x, we get 135 plus x is equal to 180. Subtract 135 from both sides. We get x is equal to 45 degrees. x is also 45 degrees. So we have a 90 degree angle and two 45 degree angles. Now I'm going to give you another theorem that's not named after the head of a religion or the founder of religion. I actually don't think this theorem doesn't have a name at. All It's the fact that if I have another triangle --I'm going to draw another triangle out here --where two of the base angles are the same-- and when I say base angle, I just mean if these two angles are the same, let's call it a. They're both a --then the sides that they don't share-- these angles share this side, right? --but if we look at the sides that they don't share, we know that these sides are equal. I forgot what we call this in geometry class. Maybe I'll look it up in another presentation; But I got this far without knowing what the name of the theorem is. If I were to change one of these angles, the length would also change. Or another way to think about it, the only way-- no, I don't confuse you too much. But you can visually see that if these two sides are the same, then these two angles are going to be the same. If you changed one of these sides' lengths, then the angles will also change, or the angles will not be equal anymore. But I'll leave that for you to think about. But just take my word for it right now that if two angles in a triangle are equivalent, then the sides that they don't share are also equal in length. Make sure you remember: not the side that they share-- because that can't be equal to anything --it's the side that they don't share are equal in length. So here we have an example where we have to equal angles. They're both 45 degrees. So that means that the sides that they don't share-- this is the side they share, right? Both angle share this side --so that means that the side that"
    },
    {
        "Q": "what 12:00 in night?",
        "A": "12:00 at night is also known to many as midnight.",
        "video_name": "ftndEjAg6qs",
        "timestamps": [
            720
        ],
        "3min_transcript": ""
    },
    {
        "Q": "Is it correct to say three-oh-five for 3:05",
        "A": "I don t know. Though I think it would be five-past-three.",
        "video_name": "ftndEjAg6qs",
        "timestamps": [
            185
        ],
        "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": "how can i know that xy=5. at 5:53",
        "A": "Look at the video. He says, that he has more of these xs and ys. He only gave us a hint how heavy xy is.",
        "video_name": "h9ZgZimXn2Q",
        "timestamps": [
            353
        ],
        "3min_transcript": "Well think about it this way, we know that X plus Y is equal to 5. So if we were to get rid of an X and a Y on this side,on the left hand side of the equation. What would we have to get rid of on the right hand side of the, or if we know if we get rid of X and Y on the left hand side of the scale What would we get rid of the right hand side of the scale to take away the same mass? Well if we take away the X and Y on the left hand side, we know that an X plus Y is 5kg. So, we'll just have to take 5kg from the right hand side. So, lets think about what that would do. Well then I'll just have an X over here, I'll just have some of these masses left over here.Then I would what X is. Now lets think about how we can represent that algebraically Essentially for taking an X and Y from the left hand side. If I'm taking an X and Y from the left hand side. I'm subtracting an X, and I'm subtracting an X. Actually let me think of it this way. I'm subtracting an X plus Y. I'm subtracting an X and Y on the left hand side. Well an X and a Y we know has a mass of 5. So we can subtract 5 from the right hand side. And the only way I'm gonna be able to do this is because of the information that we got from the second scale. So I can take away 5. So this is going to be equal to taking away 5. Taking away X and a Y is equal to taking away 5. And we know that because an X and a Y is equal to 5kg. And if we take away an X and a Y on the left hand side, what do we left with? Well this is gonna be the same thing. Let me rewrite this part. This, taking away an X and a Y is the same thing if you distribute the negative sign as taking away an X and taking away a Y. And so on the left hand side, we're left with just 2X and we have taken away one of the X's, we're left just an X. We see that visually, we're left with just an X here. And what do we have on the right hand side? We had 8 and we know X and Y is equal to 5, so we took away 5. So to keep the scale balance. And so 8 minus 5 is going to be 3. 8 minus 5 is equal to 3 and just like that using this extra information we're able to figure out that the mass of X is equal to 3. Now, one final question. We're able to figure out the mass of X, can you figure out what the mass of Y is. Well we can go back to either one of these scales. Probably be simpler to go back to this one. We know that the mass of X plus the mass of Y is equal to 5. So we could say, one thing we know that X is now is equal to 3."
    },
    {
        "Q": "This makes no sense to me once he jumps to the other scale. Essentially what he did was subtract \"x\" from the left side, and ended up with 5 on the right (in which case, the right should be 8 -x, NOT 5.) We don't have a scale in real life to keep \"adding blocks until it equals out\", this is useless, plus this automatically means x = 3 (and thus y = 2.)\n\nCan someone explain the leap in logic starting at 2:20 in the video?",
        "A": "The way he put the x and y on the left and the 5kg on the right was NOT derived from previous knowledge. In other words, someone came up to you and told you x+y=5 .",
        "video_name": "h9ZgZimXn2Q",
        "timestamps": [
            140
        ],
        "3min_transcript": "so now we have a very very very interesting problem on the left hand side of the scale I have two different types of unknown masses one of these X masses and we know that they have the same identical mass, we call that identical each of them having a mass of X But then we have this other blue thing and that has a mass of Y, which isn't necessarily going to be the same as the mass of X. We have two of these X's and a Y.It seems like the total mass or it definitely is the case, their total mass balance it out to these 8 kg right over here. Each of these is 1kg block and balances them out. So the first question I'm going to ask you is, can you express this Mathematically? Can you express what we're seeing here, the fact that this total mass balances out with this total mass. Can you express that mathematically? Well let's just think about our total mass on this side. We have two masses of mass X so those two are gonna total at 2X, and then you have a mass of Y. 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"
    },
    {
        "Q": "at 1:58 , why did sal divide the numerator by ( x-1 ) , shouldn't it be divided by (x^2 - 1) .\n\nWas this done to actually find factors for which x-1 could be eliminated .",
        "A": "Sal divided the numerator by the factor (x-1) to check whether (x-1) was a factor of the polynomial (x^3 - 1), which it is. He was trying to split the numerator into factors so that he could cancel out the factor (x-1) in the main expression. The reason for doing this is because he does not want the denominator to be 0 when he plugs in x=1 to find the limit of the expression algebraically.",
        "video_name": "rU222pVq520",
        "timestamps": [
            118
        ],
        "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": "And what if at 3:15 x=(-1) ? Is the function undefined there or we can say that as \"x\" does not equal \"1\" it neither equals \"-1\" ?\nThank you.",
        "A": "The function is undefined at x = 1 and x = -1. For purposes of this problem we don t care that the function is undefined at x = -1 because we re taking the limit at x = 1. As Sal points out, we also don t have to be concerned that it isn t defined at x = 1 because the operation involved in taking a limit doesn t require a function to be defined at the point where you re taking the limit.",
        "video_name": "rU222pVq520",
        "timestamps": [
            195
        ],
        "3min_transcript": "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. So that is equal to x squared plus x plus 1 over x plus 1, for x does not equal 1. And that's completely fine, because we're not evaluating x equals 1. We're evaluating as x approaches 1. So this is going to be the same thing as the limit as x approaches 1 of x squared plus x plus 1 over x plus 1. And now this is much easier to find. You could literally just say, well, what happens as we get right to x equals 1? Then you have 1 squared, which is 1 plus 1 plus 1, which is 3, over 1 plus 1, which is 2. So we get that equaling 3/2."
    },
    {
        "Q": "I lost you at 1:53 when you said x goes in x^2 times, then backtracked and said \"actually lemme do..\" I am so confused -_-. How do you factor this problem.",
        "A": "As far as I understood it, he used polynomial division after he had factored the denominator.",
        "video_name": "rU222pVq520",
        "timestamps": [
            113
        ],
        "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": "In 5:02 he says AB is congruent to segment AC but he wrote AC before AB",
        "A": "In Geometry the order only matters when you say AC and AB individually. a is congruent to a, while C is congruent to B. saying CA is congruent to AB would be incorrect. Tell me if it is still unclear, I m not the best explainer",
        "video_name": "7UISwx2Mr4c",
        "timestamps": [
            302
        ],
        "3min_transcript": "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. which we would denote that way, is equal to the length of segment AB. These are essentially equivalent statements. Once again in our toolkit, we have our congruency theorems. But in order to apply them, you really do need to have two triangles. So let's construct two triangles here. And this time, instead of defining another point as the midpoint, I'm going to define D this time as the point that if I were to go straight up, the point that is essentially-- if you view BC as straight horizontal, the point that goes straight down from A. And the reason why I say that is there's some point-- you could call it an altitude-- that intersects BC at a right angle."
    },
    {
        "Q": "Sal uses a calculator throughout the video however, calculators are not allowed in my school so can someone explain how he got 10.7 around the 2:40 mark?",
        "A": "Honestly, if you re not allowed calculators, you should probably just leave the answer in terms of tangent, sine, or cosine unless it s an easy value to find. 65\u00c2\u00b0 isn t an easy value to find, so this should be an acceptable answer: a = 5*tan(65\u00c2\u00b0) (this is actually a more exact answer than 10.7)",
        "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": "At 4:50 why doesn't he multiply both sides by 5 instead of multiplying them by \"b\"?",
        "A": "If you multiply both sides by 5 you would get this: 5cos65 = 25/b Our goal is to get b on its own so multiplying by 5 doesn t help.",
        "video_name": "l5VbdqRjTXc",
        "timestamps": [
            290
        ],
        "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": "At 5:54 he says \" you could of solved this using the Pythagorean theorem... But there is an issue if im not mistaken:\n\n10.7*10.7 + 5*5 does not equal to 11.8?",
        "A": "I don t think there is an issue he just wanted to solve using the trigonometric functions because that is what we had just been working on. Just to show, 10.7*10.7 = 114.49 5*5 = 25 114.49+25=139.49 And the square root of 139.49 = 11.8 a^2+b^2=c^2 so don t forget to square root everything.",
        "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": "At 2:13 don't you mean to say 7 7 and 9",
        "A": "It s just a minor mistake. You can see that it s 7, 7, and 9. Besides, the videos are still awesome at teaching the topic!",
        "video_name": "LEFE1km5ROY",
        "timestamps": [
            133
        ],
        "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": "At 0:00, under the Stem column, why is there a 0?",
        "A": "Hi Nikki. Good question. When you have a stem and leaf plot, you always have to have something on the stem side in order to have something on the leave side. So, when we want to put numbers such as 3, 7, or 9 (one digit numbers) we put a 0 on the stem side so that the value of the number doesn t change, but there is a stem for the leaves be on. :) Hope that makes sense. Syliva",
        "video_name": "LEFE1km5ROY",
        "timestamps": [
            0
        ],
        "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": "At 5:03, Sal says we can find the points (the numbers which will make both sides of the equation equal to 0). Why is it so?",
        "A": "It s the same point he used to get the formula, P = (4, 9) . He used the 2 points, the variable point Z = (x, y) and a given point P = (4, 9) to get (y-9)/(x-4) = -4; so if he lets Z = (4, 9), then Z and P are the same point, and he gets (9-9)/(4-4) = 0/0, or 9-9 = -4(4-4), or 0 = 0. (Why he s saying this, I don t know).",
        "video_name": "LtpXvUCrgrM",
        "timestamps": [
            303
        ],
        "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": "@1:18 why when Sal convert 6/3 + 1/3 to 2 1/3, he emitted the (+) sign ?",
        "A": "because 6/3+1/3 is equal to 2 1/3. 2 1/3 is a mixed number, so even though you say two and one third, the number you write down secludes the and.",
        "video_name": "xiIQQNufFuU",
        "timestamps": [
            78
        ],
        "3min_transcript": "The graph of the line 2y plus 3x equals 7 is given right over here. Determine its x-intercept. The x-intercept is the x value when y is equal to 0, or it's the x value where our graph actually intersects the x-axis. Notice right over here our y value is exactly 0. We're sitting on the x-axis. So let's think about what this x value must be. Well, just trying to eyeball a little bit, it's a little over 2. It's between 2 and 3. It looks like it's less than 2 and 1/2. But we don't know the exact value. So let's go turn to the equation to figure out the exact value. We essentially have to figure out what x value, when y is equal to 0, will have this equation be true. So we could just say 2 times 0 plus 3x is equal to 7. Well, 2 times 0 is just going to be 0, so we have 3x is equal to 7. Then we can divide both sides by 3 to solve for x, Does that look like 7/3? Well, we just have to remind ourselves that 7/3 is the same thing as 6/3 plus 1/3. And 6/3 is 2. So this is the same thing as 2 and 1/3. Another way you could think about it is 3 goes into 7 two times, and then you have a remainder of 1. So you've still got to divide that 1 by 3. It's 2 full times and then a 1/3, so this looks like 2 and 1/3. And so that's its x-intercept, 7/3. If I was doing this on the exercise on Khan Academy, it's always a little easier to type in the improper fraction, so I would put in 7/3."
    },
    {
        "Q": "At 1:00, Sal said \"if y is zero..\" he then plugged in zero to the problem to find 2 1/3. How did he know y was zero. I'm lost.\n\nThanks.",
        "A": "We re looking for the x-intercept. This is the point where the line crosses the x-axis. What do the coordinates of a point on the x-axis look like? They look like (x , 0). Just like for the y-intercept, the coordinates are (0, y).",
        "video_name": "xiIQQNufFuU",
        "timestamps": [
            60
        ],
        "3min_transcript": "The graph of the line 2y plus 3x equals 7 is given right over here. Determine its x-intercept. The x-intercept is the x value when y is equal to 0, or it's the x value where our graph actually intersects the x-axis. Notice right over here our y value is exactly 0. We're sitting on the x-axis. So let's think about what this x value must be. Well, just trying to eyeball a little bit, it's a little over 2. It's between 2 and 3. It looks like it's less than 2 and 1/2. But we don't know the exact value. So let's go turn to the equation to figure out the exact value. We essentially have to figure out what x value, when y is equal to 0, will have this equation be true. So we could just say 2 times 0 plus 3x is equal to 7. Well, 2 times 0 is just going to be 0, so we have 3x is equal to 7. Then we can divide both sides by 3 to solve for x, Does that look like 7/3? Well, we just have to remind ourselves that 7/3 is the same thing as 6/3 plus 1/3. And 6/3 is 2. So this is the same thing as 2 and 1/3. Another way you could think about it is 3 goes into 7 two times, and then you have a remainder of 1. So you've still got to divide that 1 by 3. It's 2 full times and then a 1/3, so this looks like 2 and 1/3. And so that's its x-intercept, 7/3. If I was doing this on the exercise on Khan Academy, it's always a little easier to type in the improper fraction, so I would put in 7/3."
    },
    {
        "Q": "at 3:21, do you always have to flip the numbers over?",
        "A": "nope, but it makes it alot easier",
        "video_name": "Zm0KaIw-35k",
        "timestamps": [
            201
        ],
        "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": "At 2:28, how come he used multiplications to solve pentagon when there was more than 1 of the same numbers, when he didn't use multiplication when he was working out the rectangle",
        "A": "For the rectangle, he could ve done 2*5 + 2*3 to get the perimeter of the rectangle. I think he used multiplication for the pentagon because he would have to write 2 five times, which would take too much space. If we understand that we re adding 2 five times, that just 2 multiplied by five.",
        "video_name": "9uwLgf84p5w",
        "timestamps": [
            148
        ],
        "3min_transcript": "When people use the word \"perimeter\" in everyday language, they're talking about the boundary of some area. And when we talk about perimeter in math, we're talking about a related idea. But now we're not just talking about the boundary. We're actually talking about the length of the boundary. How far do you have to go around the boundary to essentially go completely around the figure, completely go around the area? So let's look at this first triangle right over here. It has three sides. That's why it's a triangle. So what's its perimeter? Well, here, all the sides are the same, so the perimeter for this triangle is going to be 4 plus 4 plus 4, and whatever units this is. If this is 4 feet, 4 feet and 4 feet, then it would be 4 feet plus 4 feet plus 4 feet is equal to 12 feet. Now, I encourage you to now pause the video and figure out the parameters of these three figures. Well, it's the exact same idea. We would just add the lengths of the sides. So let's say that these distances, let's So let's say this is 3 meters, and this is also 3 meters. This is a rectangle here, so this is 5 meters. This is also 5 meters. So what is the perimeter of this rectangle going to be? What is the distance around the rectangle that bounds this area? Well, it's going to be 3 plus 5 plus 3 plus 5, which is equal to-- let's see, that's 3 plus 3 is 6, plus 5 plus 5 is 10. So that is equal to 16. And if we're saying these are all in meters, these are all in meters, then it's going to be 16 meters. Now, what about this pentagon? Let's say that each of these sides are 2-- and I'll make up a unit here. Let's say they're 2 gnus. That's a new unit of distance that I've just invented-- 2 gnus. Well, it's 2 plus 2 plus 2 plus 2 plus 2 gnus. Or we're essentially taking 1, 2, 3, 4, 5 sides. Each have a length of 2 gnus. So the perimeter here, we could add 2 repeatedly five times. Or you could just say this is 5 times 2 gnus, which is equal to 10 gnus, where gnu is a completely made-up unit of length that I just made up. Here we have a more irregular polygon, but same exact idea. How would you figure out its perimeter? Well, you just add up the lengths of its sides. And here I'll just do it unitless. I'll just assume that this is some generic units. And here the perimeter will be 1 plus 4 plus 2 plus 2-- let me scroll over to the right a little bit-- plus 4 plus 6."
    },
    {
        "Q": "Why doesnt he write 5c in 5:31?",
        "A": "It s simpler/cleaner to just call it c. It is a constant of unknown quantity. Sure, it s 5 times something, but it s also 1 times something too.",
        "video_name": "QxbJsg-Vdms",
        "timestamps": [
            331
        ],
        "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": "At 0:24, what does he mean by derivative?",
        "A": "At 0:24, Sal refers to the word derivative. A derivative of a function is the slope of the tangent line to a curve at a particular point on the curve. In this video, Sal shows how to find the antiderivative of a function using the power rule. Keep watching the video to find out what the power rule does.",
        "video_name": "QxbJsg-Vdms",
        "timestamps": [
            24
        ],
        "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": "At 5:09 how do you transition from x^-1/-1 to -x^-1 ?",
        "A": "Dividing by -1 is the same as multiplying by -1, because it only changes the signal of the expression.One easy proof : (-1)^2= 1 =a/a -> {(-1)^2}f(x)=1f(x) -> -f(x).-1=-1f(x)/-1, now divide both sides by -1 and its done -f(x)=f(x)/-1 .",
        "video_name": "QxbJsg-Vdms",
        "timestamps": [
            309
        ],
        "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": "At 2:31, Sal writes x^(n+1). Does the denominator \"n+1\" apply to the whole thing or just x? (i.e. does it read ((x^(n+1))/(n+1) or (x/(n+1))^(n+1)?) Does it matter either way? Thanks!",
        "A": "The whole thing. It is [(x^(n+1)]/(n+1). And yes it matters!",
        "video_name": "QxbJsg-Vdms",
        "timestamps": [
            151
        ],
        "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 0:17, Sal said that if the denominator is the same, you just add the numerator and then the denominator stays the same? I just dont get it. Is it the same on the problems without the same denominators?",
        "A": "no you do not add the denominator you add the numerator",
        "video_name": "EJjnEau6aeI",
        "timestamps": [
            17
        ],
        "3min_transcript": "So we're asked to add 3/15 plus 7/15, and then simplify the answer. So just the process when you add fractions is if they already-- well, first of all, if they're not mixed numbers, and neither of these are, and if they have the same denominator. In this example, the denominators are already the same. The denominator is 15. So if you add these two fractions, your sum is going to have the same denominator, 15, and your numerator is just going to be the sum of the numerator, so it's going to be 3 plus 7, or it's going to be equal to 10/15. Now, if we wanted to simplify this, we'd look for the greatest common factor in both the 10 and the 15, and as far as I can tell, 5 is the largest number that goes into both of them. So divide the 10 by 5 and you divide the 15 by 5, and you get-- 10 divided by 5 is 2 and 15 divided by 5 is 3. You get 2/3. Now, to understand why this works, let's draw it out. So let me split it up into 15 sections. Let me see how well I can do this. Well, actually, even a better way, an easier way might be to draw circles. So let me do the 15 sections. So let me draw. So that is one section right over there. That is one section and then if I copy and paste it, that is a second section, and then a third section, fourth section, and then we have a fifth section. Let me copy and paste this whole thing. So that's five sections right there. Let me copy and then paste that. So that is 10 sections, and then let me do it one more time. So that is 15 sections. So you can imagine this whole thing is like a candy bar or Now, what is 3/15? Well, it's going to be 3 of the 15 sections. So 3/15 is going to be one, two, three: 3/15. Now, to that, were adding 7 of the 1/15 sections, or 7 of the sections. So we're adding 7 of those to it. So that's one, two, three, four, five, six, seven. And you see now, if you take the orange and the blue, you get one, two, three, four, five, six, seven, eight, nine, ten of the sections, or 10 of the 15 sections. And then to see why this is the same thing as 2/3, you can just split this candy bar into thirds, so each third would have five sections in it. So let's do that. One, two, three, four, five, so that is 1/3 right there. One, two, three, four, five, that is"
    },
    {
        "Q": "Does the endpoint always have to be the first letter when naming a ray?\nFor example at 1:27 the ray is labelled ray JH and J is the endpoint.",
        "A": "Yes. Because the arrow always points to the right when naming a ray, the endpoint of the ray must always be the first letter.",
        "video_name": "w9jEq6dmqPg",
        "timestamps": [
            87
        ],
        "3min_transcript": "Identify all the rays shown in the image below. and this is a reminder what a ray is. A ray start at some point and then goes on forever in some direction. In order to find a ray you need that point that you're starting off on so let's that point over there is called X and then you need another point that sits on the ray and the ray is just keep going beyond, we will name that point as \"Y\" and we will call the ray \"XY\" It starts in \"X\" and keeps going in the direction of \"y\" for ever, it crosses \"y\" and keeps going further we need the second point to specify the direction in which the ray is going. So, lets identify all the rays shown in the image below, we can start anywhere, we will start at 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,"
    },
    {
        "Q": "um on 5:05 when you mentioned about GE as a ray, can GC also be a ray?",
        "A": "yeah...... Im pretty sure your right that GC could be a ray too",
        "video_name": "w9jEq6dmqPg",
        "timestamps": [
            305
        ],
        "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": "I don't understand why you need to have two points for it to be a ray. Anyways there are points in between the endpoint A and the arrow (4:41 in the video)",
        "A": "The first point serves to show where the ray is starting from. The second point is needed to show in which direction the ray is going. For example, if we just said ray E, it would be confusing. Is it going towards A, towards, F, or towards some other point not even drawn on the graph? Two points are needed in order to show origin and direction.",
        "video_name": "w9jEq6dmqPg",
        "timestamps": [
            281
        ],
        "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 2:40 he mentions CE and CF. Which one would be better to use? or does it matter at all?",
        "A": "It depends on the direction that you want to go in.",
        "video_name": "w9jEq6dmqPg",
        "timestamps": [
            160
        ],
        "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": "0:25\nIs the line of symmetry compulsory to be diagonal or horizontal to the object?",
        "A": "Any line where you can fold an object and the two sides match one another is a line of symmetry. Go to Google search images and search for line symmetry .",
        "video_name": "LrTn4cvsewk",
        "timestamps": [
            25
        ],
        "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": "at 0:35 , how does he split 60 into 6 * 10 ?",
        "A": "because 6 * 10 is 60, and the order of multiplication doesn t matter.",
        "video_name": "jb8mFpA1YI8",
        "timestamps": [
            35
        ],
        "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": "At 10:04, how do you remember which way to move the decimal if it's a negative exponent?",
        "A": "okay heres an example -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 Negative is backwards, so just think about it like you have 2^-3 that would be 0.125",
        "video_name": "i6lfVUp5RW8",
        "timestamps": [
            604
        ],
        "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": "at 2:11 why is the line dotted instead of solid?",
        "A": "A dotted line symbolizes < or > A solid line symbolizes \u00e2\u0089\u00a4 or \u00e2\u0089\u00a5",
        "video_name": "CA4S7S-3Lg4",
        "timestamps": [
            131
        ],
        "3min_transcript": "Graph the solution set for this system. It's a system of inequalities. We have y is greater than x minus 8, and y is less than 5 minus x. Let's graph the solution set for each of these inequalities, and then essentially where they overlap is the solution set for the system, the set of coordinates that satisfy both. So let me draw a coordinate axes here. So that is my x-axis, and then I have my y-axis. And that is my y-axis. And now let me draw the boundary line, the boundary for this first inequality. So the boundary line is going to look like y is equal to x minus 8. But it's not going to include it, because it's only greater than x minus 8. But let's just graph x minus 8. So the y-intercept here is negative 8. When x is 0, y is going to be negative 8. So just go negative 1, negative 2, 3, 4, 5, 6, 7, 8. 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."
    },
    {
        "Q": "at 1:20 sal says the negative cancel out. What does it mean to cancel out",
        "A": "(+)(+)=(+) (-)(-)=(+) (+)(-)=(-) there is two negative, so he cancel out, because negative times or divid negative is equal to positive",
        "video_name": "bQ-KR3clFgs",
        "timestamps": [
            80
        ],
        "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": "when he takes the two out front of the integral at 2:55, do you have to do that to get the correct answer or does hat just make things easier?",
        "A": "It makes things easier since you would not have to factor it out in the end after taking the integration.",
        "video_name": "n-iEqLhGfd4",
        "timestamps": [
            175
        ],
        "3min_transcript": "and the antiderivative of e to the x is still just e to the x. So let me write this down. So we are saying that f of x-- I'll do it right over here-- f of x is equal to x squared, in which case f prime of x is going to be equal to 2x. And I'm not worrying about the constants right now. We'll just add a constant at the end to make sure that our antiderivative is in the most general form. And then g prime of x is equal to e to the x, which means its antiderivative, g of x, is still equal to e to the x. And now we're ready to apply the right-hand side right over here. So all this thing right over here is going to be equal to f of x, which is x squared-- and let me put it right underneath-- x squared times g of x, which is e to the x, minus. I want to make the colors match up-- minus the antiderivative of f prime of x. Well, f prime of x is 2x, times g of x. g of x is e to the x dx. So you might say, hey, Sal, we're left with another antiderivative, another indefinite integral right over here. How do we solve this one? And as you might guess, the key might be integration by parts again. And we're making progress. This right over here is a simpler expression than this. Notice we were able to reduce the degree of this x squared. It now is just a 2x. And what we can do to simplify this a little bit. Since 2 is just a scalar-- it's a constant, it's multiplying the function-- we can take that out of the integral sign. So let's take it this way. So let me rewrite it this way. that's multiplying the function. So let me put the 2 right out here. And so now what we're concerned about is finding the integral-- let me write it right over here-- the integral of xe to the x dx. And this is another integration by parts problem. And so let's again apply the same principles of integration by parts. What, when I take its derivative, is going to get simpler? Well, x is going to get simpler when I take its derivative. So now, for the purposes of integration by parts, let's redefine f of x to be equal to just x. And then we can still have g prime of x equaling e to the x. And so in this case, let me write it all down. f of x is equal to x. f prime of x is equal to 1. g prime of x is equal to e to the x. g of x, just the antiderivative of this,"
    },
    {
        "Q": "at 6:47\nDose Sal notice this too?: b, h, c, and i are all right angles.",
        "A": "If they were, yes. However, it is not given that those angles are right angles. For all we know, they could be 89 or 91 degrees or anything else close to right. Great observation though.",
        "video_name": "95logvV8nXY",
        "timestamps": [
            407
        ],
        "3min_transcript": "So let's say j is equal to m plus n. And then finally, we can split up h. Remember, it's this whole thing. Let's say that h is the same thing as o plus p plus q. This is o, this is p, this is q. And once again, I wanted to split up these interior angles if they're not already an angle of a triangle. I want to split them up into angles that are parts of these triangles. So we have h is equal to o, plus p, plus q. And the reason why that's interesting is now we can write the sum of these interior angles as the sum of a bunch of angles that are part of these triangles. And then we can use the fact that, for any one triangle, they add up to 180 degrees. So this expression right over her is going to be g. g is that angle right over here. We didn't make any substitutions. let me write the whole thing. So we have 900 minus, and instead of a g, well, actually I'm not making a substitution, so I can write g plus, and instead of an h I can write that h is o plus p plus q. And then plus i. i is sitting right over there. Plus i. And then plus j. And I kind of messed up the colors. The magenta will go with i. And then j is this expression right over here. So j is equal to m plus n instead of writing a j right there. And then finally, we have our f. And f, we've already seen, is equal to k plus l. So plus k plus l. So once again, I just rewrote this part right over here, in terms of these component angles. And now something very interesting is going to happen, because we know what these sums are going to be. They are the measures of the angle for this first triangle over here, for this triangle right over here. So g plus o plus k is 180 degrees. So g-- let me do this in a new color. So for this triangle right over here, we know that g plus o plus k are going to be equal to 180 degrees. So if we cross those out, we can write 180 instead. And then we also know-- let me see, I'm definitely running out of colors-- we know that p, for this middle triangle right over here, we know that p plus l plus m is 180 degrees. So you take those out and you know that sum is going to be equal to 180 degrees. And then finally, this is the home stretch here. We know that q plus n plus i is 180 degrees in this last triangle."
    },
    {
        "Q": "5:44 into the video, why do you put the 2 on the outside of the square root sign?",
        "A": "This problem deals with simplifying a 6th root. So, we have to find factors with an exponent =6. If we find any, then we can find the 6th root for that item. Sal has 6th root( 2^6). This can be simplified to 2. Another way to look at it is to use rational exponents: 6th root( 2^6) = 2^(6/6) = 2 The exponent of 6/6 reduces to 1. 2^1 = 2. Hope this helps.",
        "video_name": "iX7ivCww2ws",
        "timestamps": [
            344
        ],
        "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": "At 0:40, why did Sal say to turn the fraction on it's head?",
        "A": "First off, notice that multiplying a number by 1/5 gives you the same answer as dividing it by 5. 5 and 1/5 are reciprocals, which means that when you multiply them you get 1 (5/1 is 1/5 flipped over). We use this property to help us divide fractions. We just multiply by the reciprocal instead. The easiest way to do that is to just flip over the fraction.",
        "video_name": "tnkPY4UqJ44",
        "timestamps": [
            40
        ],
        "3min_transcript": "Divide and write the answer as a mixed number. And we have 3/5 divided by 1/2. Now, whenever you're dividing any fractions, you just have to remember that dividing by a fraction is the same thing as multiplying by its reciprocal. So this thing right here is the same thing as 3/5 times-- so this is our 3/5 right here, and instead of a division sign, you want a multiplication sign, and instead of a 1/2, you want to take the reciprocal of 1/2, which would be 2/1-- so times 2/1. So dividing by 1/2 is the exact same thing as multiplying by 2/1. And we just do this as a straightforward multiplication problem now. 3 times 2 is 6, so our new numerator is 6. 5 times 1 is 5. So 3/5 divided by 1/2 as an improper fraction is 6/5. Now, they want us to write it as at mixed number. 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."
    },
    {
        "Q": "At 0:16, what does it mean by multiplying its reciprocal?",
        "A": "The reciprocal of a fraction is the inverse of it; you basically take a fraction, flip it over, and that s its reciprocal. 3/4 is the reciprocal of 4/3, and 4/3 is the reciprocal of 3/4. Do you get that? When dividing by fractions, we turn the division problem into a multiplication problem: we take the divisor and find it s reciprocal, and then we just multiply. For example 1/2 \u00c3\u00b7 3/4 = 1/2 x 4/3 = 4/6 or 2/3",
        "video_name": "tnkPY4UqJ44",
        "timestamps": [
            16
        ],
        "3min_transcript": "Divide and write the answer as a mixed number. And we have 3/5 divided by 1/2. Now, whenever you're dividing any fractions, you just have to remember that dividing by a fraction is the same thing as multiplying by its reciprocal. So this thing right here is the same thing as 3/5 times-- so this is our 3/5 right here, and instead of a division sign, you want a multiplication sign, and instead of a 1/2, you want to take the reciprocal of 1/2, which would be 2/1-- so times 2/1. So dividing by 1/2 is the exact same thing as multiplying by 2/1. And we just do this as a straightforward multiplication problem now. 3 times 2 is 6, so our new numerator is 6. 5 times 1 is 5. So 3/5 divided by 1/2 as an improper fraction is 6/5. Now, they want us to write it as at mixed number. 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."
    },
    {
        "Q": "i am confused at 0:35 i think you switch it to multiplication",
        "A": "When you divide by a fraction, it is the same as multiplying by the reciprocal. So if you have 3/8 divided by 1/2, lets say, it would be the same as multiplying 3/8 by 2/1.",
        "video_name": "tnkPY4UqJ44",
        "timestamps": [
            35
        ],
        "3min_transcript": "Divide and write the answer as a mixed number. And we have 3/5 divided by 1/2. Now, whenever you're dividing any fractions, you just have to remember that dividing by a fraction is the same thing as multiplying by its reciprocal. So this thing right here is the same thing as 3/5 times-- so this is our 3/5 right here, and instead of a division sign, you want a multiplication sign, and instead of a 1/2, you want to take the reciprocal of 1/2, which would be 2/1-- so times 2/1. So dividing by 1/2 is the exact same thing as multiplying by 2/1. And we just do this as a straightforward multiplication problem now. 3 times 2 is 6, so our new numerator is 6. 5 times 1 is 5. So 3/5 divided by 1/2 as an improper fraction is 6/5. Now, they want us to write it as at mixed number. 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."
    },
    {
        "Q": "At 0:43, is it necessary to multiply each side of the inequality by the reciprocal of -5 or can you just divide each side by -5 and still get the same answer?",
        "A": "You can divide both sides by -5 and get the same answer.",
        "video_name": "D1cKk48kz-E",
        "timestamps": [
            43
        ],
        "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": "The example Sal gives at 5:51,\n(2(x+3)(x-1))/ (x+3)^2(x+1) I don't understand why x at -3 isn't a removable discontinuity. If you put -3 in the original equation without simplying, you get 0/0.\n\nThank you.",
        "A": "A removable discontinuity would divide COMPLETELY out of the denominator. In this case, after the division there is still a factor of ( x + 3 ) left in the denominator, so that takes precedence over the fact that a single ( x + 3 ) was able to be divided out and indicates an asymptote at x = -3 instead of a removable discontinuity.",
        "video_name": "TX_mx3qULpw",
        "timestamps": [
            351
        ],
        "3min_transcript": "So this would be the two zeros. Two zeros. And so let's look at another similar situation. Let's look at a situation where f of x is equal to, you know, the numerator, x plus three times x minus three. And let's say that we do have, let's say that one of those x values, positive or negative three, do make the denominator equal to zero. So let's say, x plus three and then, say times x plus one. Well, you see here, now since x plus three is both the numerator and the denominator, you could divide x plus three divided by x plus three. They cancel out. And here x equals negative three would be a removable discontinuity. So this would have zero at x equals three and a removable discontinuity And so those values that make the numerator equal to zero we now see it could be a zero or it could represent a removable discontinuity. And here I just pick a removable discontinuity be at negative three or it could be the other way around or it could be at both values. If this was x plus three times x minus three over x plus three times x minus three, then you would have a removable discontinuity at both x is positive three and negative three. And then you could go even further. F of x could look like this. It could be two times x plus three times x minus three over x plus three squared and I'll just make up some other, another expression here, x plus one. So, what's going to happen here? Even if you divide the numerator and the denominator by x plus three, you're still going to have one x plus three left over the denominator. That could cancel with one of the x plus threes but you're still going to have an x plus three. And so in this case, you would have a vertical asymptote. at x is equal to three. And you would have a vertical asymptote. I'll just shorten it shorthand. You would have a vertical, I'll just write that out, vertical asymptote at x is equal to negative three. So these particular examples that I just showed you showed you that any value that makes the numerator equals zero aren't necessarily zeros for the function. They could be zeros. They could be removable discontinuities or they could be vertical asymptotes. But they would all occur at x is equal to positive or negative three. So if that lands, now let's look at the choices again. So choice A has a zero at x equals positive three and it has a vertical asymptote at x equals negative three. So that's actually very consistent with this situation that I just described. So choice A actually is looking pretty good. Choice B has a zero at x equals three but its vertical asymptote looks like it's an x equals two"
    },
    {
        "Q": "At 02:35, Sal says that pi/2 there is equal to 3.5 pi over seven, how does that work out? Really trying to wrap my head around this. Why did he choose the number 3.5?",
        "A": "Sal noted that quadrant I contains angles from 0 (X-axis) to pi/2 radians (Y-axis). Because he was discussing the angle of 2pi / 7 radians, he converted pi / 2 to sevenths. Half of 7 is 3.5. So he checked to see if the given angle was between zero radians and 3.5pi / 7. Because the denominators of the angles were the same, it was then easy to compare the numerators and see that 2 pi / 7 radians is less than 3.5pi / 7 radians . Ttherefore the angle of 2 pi / 7 lies in quadrant 1.",
        "video_name": "fYQ3GRSu4JU",
        "timestamps": [
            155
        ],
        "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": "Infinagons?? 3:05",
        "A": "infinigons are polygons that have an infinite number of sides.",
        "video_name": "D2xYjiL8yyE",
        "timestamps": [
            185
        ],
        "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": "At 4:28 Sal finishes the Pascal Expression for (a+b)^4. He starts at a^0. Why not a^1?",
        "A": "Because that s the level at which Pascal s triangle starts (2:30)",
        "video_name": "v9Evg2tBdRk",
        "timestamps": [
            268
        ],
        "3min_transcript": "or you could go like that. Same exact logic: there's three ways to get to this point. And then there's only one way to get to that point right over there. And so let's add a fifth level because this was actually what we care about when we think about something to the fourth power. This is essentially zeroth power-- binomial to zeroth power, first power, second power, third power. So let's go to the fourth power. So how many ways are there to get here? Well I just have to go all the way straight down along this left side to get here, so there's only one way. There's four ways to get here. I could go like that, I could go like that, I could go like that, and I can go like that. There's six ways to go here. Three ways to get to this place, three ways to get to this place. So six ways to get to that and, if you have the time, you could figure that out. There's three plus one-- four ways to get here. And then there's one way to get there. when I'm taking something to the-- if I'm taking something to the zeroth power. This is if I'm taking a binomial to the first power, to the second power. Obviously a binomial to the first power, the coefficients on a and b are just one and one. But when you square it, it would be a squared plus two ab plus b squared. If you take the third power, these are the coefficients-- third power. And to the fourth power, these are the coefficients. So let's write them down. The coefficients, I'm claiming, are going to be one, four, six, four, and one. And how do I know what the powers of a and b are going to be? Well I start a, I start this first term, at the highest power: a to the fourth. And then I go down from there. a to the fourth, a to the third, a squared, a to the first, and I guess I could write a to the zero which of course is just one. And then b to first, b squared, b to the third power, and then b to the fourth, and then I just add those terms together. And there you have it. I have just figured out the expansion of a plus b to the fourth power. It's exactly what I just wrote down. This term right over here, a to the fourth, that's what this term is. One a to the fourth b to the zero: that's just a to the fourth. This term right over here is equivalent to this term right over there. And so I guess you see that this gave me an equivalent result. Now an interesting question is 'why did this work?' And I encourage you to pause this video and think about it on your own. Well, to realize why it works let's just go to these first levels right over here."
    },
    {
        "Q": "I see many people have questions about this: at approx 4:03, he flipped both sides of the fraction. I understand the concept of doing the same thing to each side of an equation. But what prompted him to do this for the purpose of this problem? I want to understand so I can apply this to other problems.",
        "A": "When your variable is on the bottom(of a fraction), you generally want to get it to the top(of the fraction). When he was done simplifying the left side of the equation, he saw that the variable that he was trying to solve for was on the bottom(of the fraction). This probably prompted him to flip both sides of the equation, or to take the reciprocal of both sides of the equation.",
        "video_name": "gD7A1LA4jO8",
        "timestamps": [
            243
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 4:08 Sal defines the range of f(x) as all real number such that f(x) is greater than or equal to 0. Why is it not all real numbers such that x is greater than or equal to 0? I don't really understand in this case the significance of f(x) as a function vs. a variable.",
        "A": "X has to do with the domain, not the range, so if you were looking for domain, you would be fine. On the other hand, f(x) has to do with the range (this is functional notation f(x)= mx + b rather than perhaps what you are more used to in the slope intercept form y = mx+b), so that is why he says what he did.",
        "video_name": "96uHMcHWD2E",
        "timestamps": [
            248
        ],
        "3min_transcript": "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\" \"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."
    },
    {
        "Q": "At 4:03 Sal places a line after the Real number notation. what does this bar mean?",
        "A": "The bar means such that, so it would read all real numbers such that f(x) is greater than or equal to zero. That is also the language he used in the video.",
        "video_name": "96uHMcHWD2E",
        "timestamps": [
            243
        ],
        "3min_transcript": "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\" \"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."
    },
    {
        "Q": "At 0:30, Mr.Khan asserts that a domain is \"The set of all inputs over which the function is defined.\"\n\nIf that is the case, then in the last video, at the last example, wouldn't the domain be x = 1 and x = 0?\n\nAm I missing something?",
        "A": "No that would have been the range, you put in \u00cf\u0080 and 3, and get out 1 and 0.",
        "video_name": "96uHMcHWD2E",
        "timestamps": [
            30
        ],
        "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": "At 3:17, does the term \"parabola\" refer to the shape of the graph, or something else? I have NEVER heard that term before!",
        "A": "The word parabola refers to the U-shape of the graph. That is the name of that shape. The equations / functions that create parabolas are quadratics. Hope that helps.",
        "video_name": "96uHMcHWD2E",
        "timestamps": [
            197
        ],
        "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 is a \"porabola\"? 3:20",
        "A": "A parabola is a two-dimensional, mirror-symmetrical curve, which is approximately U-shaped.",
        "video_name": "96uHMcHWD2E",
        "timestamps": [
            200
        ],
        "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": "At 3:48, why can't x be negative? .",
        "A": "The domain (values of x) is any real number. It s the range (values of y) that cannot be negative. That s because y = x^2 , and we know that squaring anything (whether x is positive or negative or zero to begin with) cannot produce a negative result.",
        "video_name": "96uHMcHWD2E",
        "timestamps": [
            228
        ],
        "3min_transcript": "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\" \"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."
    },
    {
        "Q": "at 7:20 why do you ignore the denominators when solving the equation?",
        "A": "Both sides of the equation have the same denominator, so multiplying the numerators on both sides by the denominator cancels them out, which has the same effect as ignoring them. Example: x/3 = y/3 Multiply both sides by the denominator 3: 3 * (x/3) = (y/3) * 3 3x/3 = 3y/3 Simplify: x = y It s the same result as ignoring the denominators.",
        "video_name": "S-XKGBesRzk",
        "timestamps": [
            440
        ],
        "3min_transcript": "it actually does add up to this, then I'm done and I will have fully decomposed this fraction. I guess is the way-- I don't know if that's the correct terminology. So let's try to do that. So if I were to add these two terms, what do I get? When you add anything, you find the common denominator, and the common denominator, the easiest common denominator, is to multiply the two denominators, so let me write this here. So a over x plus 5 plus b over x minus 8 is equal to-- well, let's get the common denominator-- it's equal to x plus 5 times x minus 8. And then the a term, we would-- a over x plus 5 is the same thing as a times x minus 8 over this whole thing. I mean, if I just wrote this right here, you would just And then you could add that to the common denominator, x plus 5 times x minus 8, and it would be b times x plus 5. Important to realize, that, look. This term is the exact same thing as this term if you just cancel the x minus 8 out, and this term is the exact same thing as this term if you just cancel the x plus 5 out. But now that we have an actual common denominator, we can add them together, so we get-- let me just write the left side here over-- a over x plus 5-- I'm sorry. I want to write this over here. I want to write x plus 3 over plus 5 times x minus 8 is equal to is equal to the sum of these two things on top. a times x minus 8 plus b times x plus 5, all of that over So the denominators are the same, so we know that this, when you add this together, you have to get this. So if we want to solve for a and b, let's just set that equality. We can ignore the denominators. So we can say that x plus 3 is equal to a times x minus 8 plus b times x plus 5. Now, there's two ways to solve for a and b from this point going forward. One is the way that I was actually taught in the seventh or eighth grade, which tends to take a little longer, then there's a fast way to do it and it never hurts to do the fast way first. If you want to solve for a, let's pick an x that'll make this term disappear. So what x would make this term disappear? Well, if I say x is minus 5, then this becomes 0, and"
    },
    {
        "Q": "4:13 to 4:30 is confusing",
        "A": "double lines mean that that angle is the same as the other angle with the same two lines.",
        "video_name": "wRBMmiNHQaE",
        "timestamps": [
            253,
            270
        ],
        "3min_transcript": "we also know that angle DBA --we know that this is DBA right over here-- we also know that angle DBA and angle DBC are supplementary this angle and this angle are supplementary, their outer sides form a straight angle, they are adjacent so they are supplementary which tells us that angle DBA, this angle right over here, plus angle DBC, this angle over here, is going to be equal to 180 degrees. Now, from this top one, this top statement over here, we can subtract angle DBC from both sides and we get angle CBE is equal to 180 degrees minus angle DBC that's this information right over here, I just put or subtracted it from both sides of the equation and this right over here, if I do the exact same thing, subtract angle DBC from both sides of the equation, I get angle DBA is equal to 180 degrees --let me scroll over to the right a little bit-- is equal to 180 degrees minus angle DBC. So clearly, angle CBE is equal to 180 degrees minus angle DBC angle DBA is equal to 180 degrees minus angle DBC so they are equal to each other! They are both equal to the same thing so we get, which is what we wanted to get, angle CBE is equal to angle DBA. Angle CBE, which is this angle right over here, is equal to angle DBA and sometimes you might see that shown like this; so angle CBE, that's its measure, and you would say that And we have other vertical angles whatever this measure is, and sometimes you will see it with a double line like that, that you can say that THAT is going to be the same as whatever this angle right over here is. You will see it written like that sometimes, I like to use colors but not all books have the luxury of colors, or sometimes you will even see it written like this to show that they are the same angle; this angle and this angle --to show that these are different-- sometimes they will say that they are the same in this way. This angle is equal to this vertical angle, is equal to its vertical angle right over here and that this angle is equal to this angle that is opposite the intersection right over here. What we have proved is the general case because all I did here is I just did two general intersecting lines I picked a random angle, and then I proved that it is equal to the angle that is vertical to it."
    },
    {
        "Q": "why does he put hash marks on the angle markers ? like at 4:40",
        "A": "He puts hash marks on the angle markers to show that the marked angles are congruent. If he didn t put hash marks on the angle markers, the (previously marked) angles would be considered congruent to the other two angles.",
        "video_name": "wRBMmiNHQaE",
        "timestamps": [
            280
        ],
        "3min_transcript": "we also know that angle DBA --we know that this is DBA right over here-- we also know that angle DBA and angle DBC are supplementary this angle and this angle are supplementary, their outer sides form a straight angle, they are adjacent so they are supplementary which tells us that angle DBA, this angle right over here, plus angle DBC, this angle over here, is going to be equal to 180 degrees. Now, from this top one, this top statement over here, we can subtract angle DBC from both sides and we get angle CBE is equal to 180 degrees minus angle DBC that's this information right over here, I just put or subtracted it from both sides of the equation and this right over here, if I do the exact same thing, subtract angle DBC from both sides of the equation, I get angle DBA is equal to 180 degrees --let me scroll over to the right a little bit-- is equal to 180 degrees minus angle DBC. So clearly, angle CBE is equal to 180 degrees minus angle DBC angle DBA is equal to 180 degrees minus angle DBC so they are equal to each other! They are both equal to the same thing so we get, which is what we wanted to get, angle CBE is equal to angle DBA. Angle CBE, which is this angle right over here, is equal to angle DBA and sometimes you might see that shown like this; so angle CBE, that's its measure, and you would say that And we have other vertical angles whatever this measure is, and sometimes you will see it with a double line like that, that you can say that THAT is going to be the same as whatever this angle right over here is. You will see it written like that sometimes, I like to use colors but not all books have the luxury of colors, or sometimes you will even see it written like this to show that they are the same angle; this angle and this angle --to show that these are different-- sometimes they will say that they are the same in this way. This angle is equal to this vertical angle, is equal to its vertical angle right over here and that this angle is equal to this angle that is opposite the intersection right over here. What we have proved is the general case because all I did here is I just did two general intersecting lines I picked a random angle, and then I proved that it is equal to the angle that is vertical to it."
    },
    {
        "Q": "at 1:16 Sal put a decimal point and some zeros on the 63 would he be able to put a decimal and zeros on the 35 without changing the question",
        "A": "yes, the decimals don t change anything because they have no value. 0.00 = 0 35 + 0.00 = 35 the decimals just help with solving the problem",
        "video_name": "xUDlKV8lJbM",
        "timestamps": [
            76
        ],
        "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": "Why e^-u at 2:33.I dont get it!",
        "A": "Basic property of powers in a fraction: those on the bottom are negatives of those on the top. 1/x = x^(-1) Therefore, 1/e^u = e^(-u) to get rid of the fraction.",
        "video_name": "ShpI3gPgLBA",
        "timestamps": [
            153
        ],
        "3min_transcript": "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 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."
    },
    {
        "Q": "Is RSH at 2:39 a real theorem? Or just another name for the HL postulate?",
        "A": "RSH is actually the HL congruence Theorem",
        "video_name": "q7eF5Ci944U",
        "timestamps": [
            159
        ],
        "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": "At 2:25, why did Sal used inverse of Sine to calculate the Radian angle (am I using that terminology right?) instead of just normal Sine? how do you know when to use the inverse calculation?\n\nALSO, we us 2pi because that'll give us a full circle revolution back to the starting, correct? But why is there a need to multiply it with a n integer? Can the answer just be 0.34 + 2pi?",
        "A": "The sine function takes an angle and gives you the ratio of the side lengths. If you have the ratio of the side lengths and you are looking for the angle instead, this is when you would use the inverse of the sine function. And, since sine is a periodic function, the value will repeat infinitely. The 2pi*n is necessary to give ALL of the answers, as opposed to only the answers in the first rotation.",
        "video_name": "NC7iWEQ9Kug",
        "timestamps": [
            145
        ],
        "3min_transcript": "Voiceover:Which of these are contained in the solution set to sine of X is equal to 1/3? Answers should be rounded to the nearest hundredths. Select all that apply. I encourage you to pause the video right now and work on it on your own. I'm assuming you've given a go at it. Let's think about what this is asking. They're asking what are the X values? What is the solution set? What are the possible X values where sine of X is equal to 1/3? To help us visualize this, let's draw a unit circle. That's my Y axis. This right over here is my X axis. This is X set one. This is Y positive one. Negative one along the X axis, and negative one on the Y axis. The unit circle, I'm going to center it at zero. It's going to have a radius of one, a radius of one and we just have to remind ourselves what the unit circle definition of the sine function is. If we have some angle, one side of the angle 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."
    },
    {
        "Q": "At 2:11 in this video, why does Sal use the dot instead of the regular multiplication sign, the regular times symbol?",
        "A": "In *Algebra and above, there are things called variables, which are symbols to represent numbers. x is used as one of these variables, so Algebra people use a dot as the multiplication symbol.",
        "video_name": "xkg7370cpjs",
        "timestamps": [
            131
        ],
        "3min_transcript": "Write 5 and 1/4 as an improper fraction. An improper fraction is just a pure fraction where the numerator is greater than the denominator. This right here, it's not a pure fraction. We have a whole number mixed with a fraction, so we call this a mixed number. So let's think about what 5 and 1/4 represents, and let me rewrite it. So if we're talking about 5 and 1/4, and you can literally think of this as 5 and 1/4 or 5 plus 1/4, that's what 5 and 1/4 represents. So let's think about 5. Five is 5 wholes, or if you're thinking of pie, we could draw literally five pies. Let me just cut up the pies from the get go into four pieces since we're dealing with fourths. So let me just cut up the pies right over here. So that's one pie right over there. Let me copy and paste this. Copy and paste. So I have two pies, and then I have three pies, and then I So this is what the 5 represents. 5 literally represents-- so let me circle all of this together. That is the 5 part right there. That is what 5 literally represents. It represents five whole pies. Now, I have cut up the pies into four pieces, so you can imagine each piece represents a fourth. Now, how many pieces do I have in these five pies? Well, I have four pieces per pie. Let me just right it here. 4 pieces per pie times 5 pies is equal to 20 pieces. this is also equal to 20 times 1/4, or you could just write this as being equal to 20/4. So we have 5 whole pies is equal to 20 fourths. Let me write it like that. 20 fourths. Or we could write it as 20/4. I've kind of done the same thing twice. So that's what the five pies represent. 20/4 or 20 pieces, where each piece is 1/4. Now, the 1/4 right here represents literally one more fourth of a pie or one more piece of a pie, so let me draw another pie here. So that is another pie. Cut it into four pieces. But this 1/4 only represents one of these pieces, right?"
    },
    {
        "Q": "At around 7:05, he said that 2(pi radius) = 360degree, but afterwards he said that 2(pi RADIANS) = 360degree. Why is this so? I know that radian is an angle, and radius is a length, but how can they be used as the same variable in such an equation?",
        "A": "Listen again even more carefully from about 5:40 to 7:15. Sal said that 2pi radiusseses SUBTEND an angle of 360 degrees and that 360 degrees is the SAME as an angle of 2pi radians. Hope this is of use to you!",
        "video_name": "EnwWxMZVBeg",
        "timestamps": [
            425
        ],
        "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 0:07 Sal mentions that Stewart has a negative amount of money in his account; how is that possible?\n\n\nIs it because he's in debt?",
        "A": "It is impossible to achieve a negative amount of money in your account unless you take something that isn t yours. So to answer your question, yes, it means he is in debt",
        "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": "so at (0:19) twice as much means multiply the age or halve it",
        "A": "If the problem says something it s twice as much, it means the amount is two times the value. In this case, Jared is twice as old as Peter. This means he is two times older than peter. If peter is 5, Jared will be 10 since 10 is twice as much as 5.",
        "video_name": "bS6EmYzpou4",
        "timestamps": [
            19
        ],
        "3min_transcript": "We're told that Jared is twice the age of his brother, Peter. So Jared is twice the age of his brother, Peter. Peter is 4 years old. And Jared is twice the age of his brother, Peter. So Jared is twice Peter's age of 4. And then they tell us, Talia's age is 3 times Jared's age. How old are Talia and Jared? So Talia's age is 3 times Jared's age. So let's write this down. We have Peter is 4 years old. So Peter's age, he's 4 years old. Jared is twice the age of his brother, Peter. Let me do that in yellow. That's not yellow. Jared is twice the age of his brother. You could say Jared is twice the age of his brother or Jared is 2 times the age of his brother. Well, his brother is 4. They told us that right over there. So Jared is going to be 2 times 4 years old. And if you know your multiplication tables, you know that's the same thing as 8. 2 times 4 is 8. 4 times 2 is 8. Or you know this is the same thing as Peter's age twice, which is the same thing as 4 plus 4. That would also get you 8. So Jared's age is equal to 8. And then, finally, they say Talia's age-- so let me write Talia right over here. Talia's age and they tell it to us right over here. It's 3 times Jared's age. So Talia's age is 3 times Jared's age. Well, we were able to figure out Jared's age. It is 3 times Jared's age. If you know your multiplication tables, this should maybe jump out at you. And your multiplication tables really are one of those things in mathematics that you should really just know back and forth just because it will make the rest of your life very, very, very simple. Or you could view 3 times 8 as literally 8 three times. So it's 8 plus 8 plus 8. 8 plus 8 is 16, 16 plus 8 is 24. So 3 times 8 is 24. So Peter is 4, Jared is 8, and Talia is 24."
    },
    {
        "Q": "You lost me from 0:36",
        "A": "It is like that because, let s say if x = 0 and we know that y = x+2 so: y = x+2 y=0+2 y=2 So, there the coordinates will be (0,2) Hope it made sense.",
        "video_name": "RLyXTj2j_c4",
        "timestamps": [
            36
        ],
        "3min_transcript": "- We're asked to use the reflect tool to find the image of quadrilateral PQRS, that's this quadrilateral right over here, for a reflection over the line y is equal to x plus two. All right, so let's use the reflect tool. So let me scroll down. So let me click on Reflect, it brings up this tool, and I want to reflect across the line y is equal to x plus two. So let me move this so it is the line y equals x plus two and to think about it, let's see, it's going to have a slope of one, the coefficient on the x term is one, so it's going to have a slope of one, so let me see if I can give this a slope of one. Is this a slope of one? Let me put it a little bit -- yep, it looks like a slope of one as the line moves one to the right. We go from one point of the line to the other, you have to go one to the right, and one up or two to the right, and two up, however much you move to the right in the x direction, you have to move that same amount in the vertical direction. So now it has a slope of one, and the y intercept is going to be the point x equals zero, y equals two, When x is equal to zero, y is going to be equal to two, so let me move this. So we see that we now go through that point. When x is equal to zero, y is equal to two. And now, we just need to reflect PQRS, this quadrilateral, over this line, so let's do that. There you go, we did it. The things, the point, like point S right over here that was to the top and left of the line, its reflection, the corresponding point in the image is now to the right and the bottom of the line. The things that were to the right and the bottom of our line, like point P, it's the corresponding point in the reflection is now on the other side of the line. So there you go, I think we're done, and we got it right."
    },
    {
        "Q": "At 5:02 He says that a 2 - tuple is not a member of 3D real co-ordinate space. Why is that? Can't we just say that a 2 - dimensional vector has zero value for the z-axis? Can't we imagine 3-d space consisting of a lot of planes stacked up?",
        "A": "the keyword is 2 tuple . you can t have a 2 tuple that has x y and z components, that s all. but you can represent a 2 dimensional vector as a 3 tuple, it s just that one component will always be zero.",
        "video_name": "lCsjJbZHhHU",
        "timestamps": [
            302
        ],
        "3min_transcript": "I should make the scale a little bit bigger, so it looks the same-- 1, 2, 3, 4. So it might look something like that. So that would be the vector, negative 3-- let me write a little bit better than that-- negative 3, and negative 4. So if you were to take all of the possible 2-tuples, including the vector 0, 0-- so it has no magnitude, and you could debate what its direction is right over there-- you take all of those combined, and then you have created your two-dimensional real coordinate space. And that is referred to as R2. Now, as you can imagine, the fact that we wrote this 2 here-- we had to specify-- it's like, hey, well, could I put a 3 there? And I would say, absolutely, you could put a 3 there. So R3 would be the three-dimensional real coordinate space. So 3D real coordinate space. the possible real-valued 3-tuples. So, for example, that would be a member of R3. And let me actually label these vectors just so we get in the habit of it. So let's say we call this vector x. Let's say we have a vector b, that looks like this. Negative 1, 5, 3. Both of these would be members of R3. And if you want to see some fancy notation, a member of a set-- so this is a member of R3-- it is a real-valued 3-tuple. Now you say, well, what would not be a member of R3? Well, this right over here isn't a 3-tuple. This right over here is a member of R2. Now, you might be able to extend it in some way, add a zero or something, but formally, Another thing that would not be a member of R3-- let's say someone wanted to make some type of vector that had some imaginary parts in it. So let's say it had i, 0, 1. This is no longer real-valued. We have put an imaginary-- this number up here has an imaginary part. So this is no longer a real-valued 3-tuple. And what's neat about linear algebra is, we don't have to stop there. R3 we can visualize, we can plot these things. In your previous mathematical career, especially if you have some type of a hologram or something, it's not hard to visualize things in three dimensions. But what's neat is that we can keep extending this. We can go into 4, 5, 6, 7, 20, 100 dimensions. And obviously there it becomes much harder, if not impossible, to visualize it. But then we can at least represent it mathematically with an n-tuple of vectors. And so if we were to talk about a real coordinate space generally, you'll often see the notation Rn,"
    },
    {
        "Q": "At 1:55, he says 'neither of these have any imaginary parts'. Does this mean it cannot have x or y or n or any unknown number in it?",
        "A": "No, it means that only Real numbers are components of that vector, rather than the option of complex numbers as well, such as the square root of negative 1.",
        "video_name": "lCsjJbZHhHU",
        "timestamps": [
            115
        ],
        "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": "8:35 does that also mean that g(x)-h(x) and h(x)-g(x) are also solutions?",
        "A": "Yes, it does. 0 - 0 does equal 0. In fact any linear combination of g and h will be solutions. You can do a*g(x) + b*h(x), where a and b are any constants, and that will be a solution.",
        "video_name": "UFWAu8Ptth0",
        "timestamps": [
            515
        ],
        "3min_transcript": "That's just g prime prime, plus h prime prime, plus B times-- the first derivative of this thing-- g prime plus h prime, plus C times-- this function-- g plus h. And now what can we do? Let's distribute all of these constants. We get A times g prime prime, plus A times h prime prime, plus B times the first derivative of g, plus B times the first derivative of h, plus C times g, plus C times h. And now we can rearrange them. And we get A-- let's take this one; let's take all the g terms-- A times the second derivative of g, plus B times the first derivative, plus C times g-- that's these three terms-- plus A times the second derivative of h, plus B And now we know that both g and h are solutions of the original differential equation. So by definition, if g is a solution of the original differential equation, and this was the left-hand side of that differential equation, this is going to be equal to 0, and so is this going to be equal to 0. So we've shown that this whole expression is equal to 0. So if g is a solution of the differential equation-- of this second order linear homogeneous differential equation-- and h is also a solution, then if you were to add them together, the sum of them is also a solution. So in general, if we show that g is a solution and h is a solution, you can add them. And we showed before that any constant times them is also a solution. So you could also say that some constant times g of x And maybe the constant in one of the cases is 0 or something. I don't know. But anyway, these are useful properties to maybe internalize for second order homogeneous linear differential equations. And in the next video, we're actually going to apply these properties to figure out the solutions for these. And you'll see that they're actually straightforward. I would say a lot easier than what we did in the previous first order homogeneous difference equations, or the exact equations. This is much, much easier. I'll see you in the next video."
    },
    {
        "Q": "at 5:58, he says that if g(x) is a solution then c1*g(x) is also one, does this mean that 0 is a solution to all of these?",
        "A": "Homogeneous systems have a few properties, and one of those properties is that they always have at least one solution, the trivial solution, 0. The goal is to find the non-trivial solutions in most cases.",
        "video_name": "UFWAu8Ptth0",
        "timestamps": [
            358
        ],
        "3min_transcript": "prime, plus C times g is equal to 0. Right? These mean the same thing. Now, my question to you is, what if I have some constant times g? Is that still a solution? So my question is, let's say some constant c1 gx-- c1 times g-- is this a solution? Well, let's try it out. Let's substitute this into our original equation. So A times the second derivative of this would just be-- and I'll switch colors here; let me switch to brown-- so A times the second derivative of this would be-- the constant, every time you take a derivative, the constant just carries over-- so that'll just be A times c1 g prime prime, plus-- the same thing for the first different than the c1 c-- times g. And let's see whether this is equal to 0. So we could factor out that c1 constant, and we get c1 times Ag prime prime, plus Bg prime, plus Cg. And lo and behold, we already know. Because we know that g of x is a solution, we know that this is true. So this is going to be equal to 0. Because g is a solution. So if this is 0, c1 times 0 is going to be equal to 0. So this expression up here is also equal to 0. Or another way to view it is that if g is a solution to this second order linear homogeneous differential So this is also a solution to the differential equation. And then the next property I want to show you-- and this is all going someplace, don't worry. The next question I want to ask you is, OK, we know that g of x is a solution to the differential equation. What if I were to also tell you that h of x is also a solution? So my question to you is, is g of x plus h of x a solution? If you add these two functions that are both solutions, if you add them together, is that still a solution of our original differential equation? Well, let's substitute this whole thing into our original differential equation, right? So we'll have A times the second derivative of this thing."
    },
    {
        "Q": "At 0:15, how does the second prize relate to the first prize? Doesn't the first prize have a predetermined ticket, thus making it an independent event?",
        "A": "The first prize is a independent event, yes. But the second prize is dependent on the first prize because the ticket drawn for the first prize is not but back in.",
        "video_name": "Za7G_eWKiF4",
        "timestamps": [
            15
        ],
        "3min_transcript": "The marching band is holding a raffle at a football game with two prizes. After the first ticket is pulled out and the winner determined, the ticket is taped to the prize. The next ticket is pulled out to determine the winner of the second prize. Are the two events independent? Explain. Now before we even think about this exact case, let's think about what it means for events to be independent. It means that the outcome of one event doesn't affect the outcome of the other event. Now in this situation, the first event-- after the first ticket is pulled out and the winner determined-- the ticket is taped to the prize. Then the next ticket is pulled out to determine the winner of the second prize. Now, the winner of the second prize-- the possible winners, the possible outcomes for the second prize, is dependent on who was pulled out for the first prize. You can imagine if there's three tickets, let's say there's tickets A, B, and C in the bag. That's for the first prize. Now, when we think about who could be pulled out for the second prize, it's only going to be tickets B or C. Now the first prize could have gone the other way. It could have been A, B, and C. The first prize could have gone to ticket B. And then the possible outcomes for the second prize would be A or C. So the possible outcomes for the second event, for the second prize, are completely dependent on what happened or what ticket was pulled out for the first prize. So these are not independent events. The second event-- the outcomes for it, are dependent on what happened in the first event. So they are not independent. after the first ticket was pulled out, if they just wrote down the name or something, and then put that ticket back in. Instead they taped it to the prize. But if they put that ticket back in, then the second prize, it would have still had all the tickets there. It wouldn't have mattered who was picked out in the first time because their name was just written down, but their ticket was put back in. And then you would have been independent. So if you had replaced the ticket, you would have been independent. But since they didn't replace the ticket, they taped it to the prize, these are not independent events."
    },
    {
        "Q": "I know this is simple, but I just wanted to reassure myself. At around 8:27 when Sal multiplies the triangles, the formula is b times hieght divided by 2. Now does that only count for that one half of the triangle? Is that why he had to multiply it once more? Also, hypothetically, if I were to encounter a triangle such as this again, would I do the method shown here to find its area?",
        "A": "(base x height)/2 is the formula for the whole triangle, and every other triangle you ll ever see.",
        "video_name": "vaOXkt7uuac",
        "timestamps": [
            507
        ],
        "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": "2:02 i was taught like 29+59+?=180\n? = 180 - (29+59)\n? = 180 - 88\n? = 92 degrees.\n\ncan i do it in this way also?",
        "A": "Yes, although this method might not work with more complicated multi-step equations. This method does work however for interior triangle angles.",
        "video_name": "eTwnt4G5xE4",
        "timestamps": [
            122
        ],
        "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": "In 1:37 Sal wrote 59+29+x=180, and he wrote 180-59-29=x, why he is subtracting if the original operation is addition? Can you add those two numbers, and subtract the total by 180 and you will get the missing angle?",
        "A": "Both ways end up giving you the same answer for x. You can just use the way that is the easiest for you to use.",
        "video_name": "eTwnt4G5xE4",
        "timestamps": [
            97
        ],
        "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": "Is it possible to have a matrix the entries of which are other matrices? Like, in 3:15, when Sal says a matrix can be used to represent the intensity of pixels, maybe you could have a matrix with a cell for each pixel, but the entry is an actual 3x1 matrix that represents the RGB components of this pixel. Does this actually exist?",
        "A": "I m also asking this because when I attended some lectures about the Standard Model of Particle Physics, the lecturer kept talking about polymatrices . I cannot say I fully understood the lectures, not even understood, I attended them just for a challenge. Also, my Linear Algebra concepts where almost none (though they are improving now). But despite all this, I deduced he was talking about matrices inside matrices, although might be I deduced wrong. When I try and look up polymatrix , I find no results.",
        "video_name": "0oGJTQCy4cQ",
        "timestamps": [
            195
        ],
        "3min_transcript": "This is a matrix where 1, 0, negative 7, pi-- each of those are an entry in the matrix. This matrix right over here has two rows. And it has three columns. And because it has two rows and three columns, people will often say that this is a 2 by 3 matrix. Whenever they say it's something by something matrix, they're telling you that it has two rows-- so you see the two rows right over there. And they are telling you that it has three columns. You see the three columns right over there. I could give you other examples of a matrix. So I could have a 1 by 1 matrix. So I could have the matrix 1. This right over here is a 1 by 1 matrix. It has one row, one column. I could have a matrix like this-- 3, 7, and 17. What is this? Well, this has one row. 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."
    },
    {
        "Q": "At 1:08 - 1:13, Sal mentions radians. What are they?",
        "A": "Radians are basically units used for measuring angles. In a circle, if finding the radians of a circle, it would be that particular sectors arc length.",
        "video_name": "D-EIh7NJvtQ",
        "timestamps": [
            68,
            73
        ],
        "3min_transcript": "We already know that an angle is formed when two rays share a common endpoint. So, for example, let's say that this is one ray right over here, and then this is one another ray right over here, and then they would form an angle. And at this point right over here, their common endpoint is called the vertex of that angle. Now, we also know that not all angles seem the same. For example, this is one angle here, and then we could have another angle that looks something like this. And viewed this way, it looks like this one is much more open. So I'll say more open. And this one right over here seems less open. So to avoid having to just say, oh, more open and less open and actually becoming a little bit more exact about it, we'd actually want to measure how open an angle is, or we'd want to have a measure of the angle. Now, the most typical way that angles are measured, The most typical unit is in degrees, but later on in high school, you'll also see the unit of radians being used, especially when you learn trigonometry. But the degrees convention really comes from a circle. So let's draw ourselves a circle right over here, so that's a circle. And the convention is that-- when I say convention, it's just kind of what everyone has been doing. The convention is that you have 360 degrees in a circle. So let me explain that. So if that's the center of the circle, and if we make this ray our starting point or one side of our angle, if you go all the way around the circle, that represents 360 degrees. And the notation is 360, and then this little superscript circle represents degrees. This could be read as 360 degrees. And no one knows for sure, but there's hints in history, and there's hints in just the way that the universe works, or at least the Earth's rotation around the sun. You might recognize or you might already realize that there are 365 days in a non-leap year, 366 in a leap year. And so you can imagine ancient astronomers might have said, well, you know, that's pretty close to 360. And in fact, several ancient calendars, including the Persians and the Mayans, had 360 days in their year. And 360 is also a much neater number than 365. It has many, many more factors. It's another way of saying it's divisible by a bunch of things. But anyway, this has just been the convention, once again, what history has handed us, that a circle is viewed to have 360 degrees."
    },
    {
        "Q": "At 0:47, when they are talking about more open and less open are they ever going to give names to those angles",
        "A": "less open means acute and more open is obtuse, and if the angle forms a right angle it is 90 degrees",
        "video_name": "D-EIh7NJvtQ",
        "timestamps": [
            47
        ],
        "3min_transcript": "We already know that an angle is formed when two rays share a common endpoint. So, for example, let's say that this is one ray right over here, and then this is one another ray right over here, and then they would form an angle. And at this point right over here, their common endpoint is called the vertex of that angle. Now, we also know that not all angles seem the same. For example, this is one angle here, and then we could have another angle that looks something like this. And viewed this way, it looks like this one is much more open. So I'll say more open. And this one right over here seems less open. So to avoid having to just say, oh, more open and less open and actually becoming a little bit more exact about it, we'd actually want to measure how open an angle is, or we'd want to have a measure of the angle. Now, the most typical way that angles are measured, The most typical unit is in degrees, but later on in high school, you'll also see the unit of radians being used, especially when you learn trigonometry. But the degrees convention really comes from a circle. So let's draw ourselves a circle right over here, so that's a circle. And the convention is that-- when I say convention, it's just kind of what everyone has been doing. The convention is that you have 360 degrees in a circle. So let me explain that. So if that's the center of the circle, and if we make this ray our starting point or one side of our angle, if you go all the way around the circle, that represents 360 degrees. And the notation is 360, and then this little superscript circle represents degrees. This could be read as 360 degrees. And no one knows for sure, but there's hints in history, and there's hints in just the way that the universe works, or at least the Earth's rotation around the sun. You might recognize or you might already realize that there are 365 days in a non-leap year, 366 in a leap year. And so you can imagine ancient astronomers might have said, well, you know, that's pretty close to 360. And in fact, several ancient calendars, including the Persians and the Mayans, had 360 days in their year. And 360 is also a much neater number than 365. It has many, many more factors. It's another way of saying it's divisible by a bunch of things. But anyway, this has just been the convention, once again, what history has handed us, that a circle is viewed to have 360 degrees."
    },
    {
        "Q": "AT 3:05 and 3:56 what is the difference between the number lines?",
        "A": "When the inequality is equal or less than/greater than we use a closed circle (solid dot) as in the first number line. When it is less than/greater than we use an open circle as in the second line.",
        "video_name": "ilWDSYnTEFs",
        "timestamps": [
            185,
            236
        ],
        "3min_transcript": "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, is legitimate. And you might say, hey, but what about the situation where it wasn't less than or equal? What about the situation if it was just less than? So let me draw that, too. So going back to where we started, if I were to say C is less than 1,500, the way we would depict that on a number line is-- let's say this is 0, this is 1,500, we want to make it very clear that we're not including the value 1,500. So we would put an open circle around it. Notice, if we're including 1,500, we fill in the circle. If we're not including 1,500, so we're only less than, we were very explicit that we don't color in the circle. But then we show that, look, we can do everything below that. Now, you're probably saying, OK, Sal, you did less than, you did less than or equal, what if you wanted to do it the other way around?"
    },
    {
        "Q": "at the 9:23 shouldn't it be -1 = f\"(y)",
        "A": "He corrects himself at 9:41",
        "video_name": "Pb04ntcDJcQ",
        "timestamps": [
            563
        ],
        "3min_transcript": "respect to y, is just sine of x. Plus the derivative of e to the y is e to the y. x squared So it's just x squared e to the y, plus-- what's the partial of f of y, with respect to y? It's going to be f prime of y. Well, what did we do? We took M, we integrated with respect to x, and we said, well, we might have lost some function of y, so we added that to it. And then we took the partial of that side that we've almost constructed, and we took the partial of that, with respect to y. Now, we know, since this is exact, that that is going to equal our N. So our N is up there. Cosine of x plus-- So that's going to be equal to-- I want to make sure I can read it up there-- to our N, right? Oh no, sorry. N is up here. Our N is up here. Sine of x-- let me write that-- sine of x plus x So sine of x plus x squared, e to the y, minus 1. That was just our N, from our original differential equation. And now we can solve for f prime of y. So let's see, we get sine of x plus x squared, e to the y, plus f prime of y, is equal to sine of x plus x squared, e to the y, minus 1. So let's see, we can delete sine of x from both sides. We can delete x squared e to the y from both sides. And then what are we left with? We're left with f prime of y is equal to 1. And then we're left with f of y is equal to-- well, it So what is our psi now? We wrote our psi up here, and we had this f of y here, so we So psi is a function of x and y-- we're actually pretty much almost done solving it-- psi is a function of x and y is equal to y sine of x, plus x squared, e to the y, plus y-- oh, sorry, this is f prime of y, minus 1. So this is a minus 1. So this is a minus y plus c. So this is going to be a minus y plus c. So we've solved for psi. And so what does that tell us? Well, we said that original differential equation, up here, using the partial derivative chain rule, that original differential equation, can be rewritten now"
    },
    {
        "Q": "1:32 sal said 20 when it is actually -20, isn't it?",
        "A": "It was just a mistake, and they put an infobox at the bottom-right when the mistake comes along to fix what he says.",
        "video_name": "H0q9Fqb8YT4",
        "timestamps": [
            92
        ],
        "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": "According to 3:58, is it safe to say that a ray (or 2 rays closed together) has 0 degrees angle and a line has 180 degrees angle?",
        "A": "You can do that if you show where the vertex is.",
        "video_name": "92aLiyeQj0w",
        "timestamps": [
            238
        ],
        "3min_transcript": "we could say the measure of angle XYZ-- sometimes they'll just say angle XYZ is equal to, but this is a little bit more formal-- the measure of angle XYZ is equal to 77. Each of these little sections, we call them \"degrees.\" So it's equal to 77-- sometimes it's written like that, the same way you would write \"degrees\" for the temperature outside. So you could write \"77 degrees\" like that or you could actually write out the word right over there. So each of these sections are degrees, so we're measuring in degrees. And I want to be clear, degrees aren't the only way to measure angles. Really, anything that measures the openness. So when you go into trigonometry, you'll learn that you can measure angles, not only in degrees, but also using something called \"radians.\" But I'll leave that to another day. So let's measure this other angle, angle BAC. So once again, I'll put A at the center, and then AC I'll put along the 0 degree edge of this half-circle 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.\""
    },
    {
        "Q": "At 3:50 Sal mentioned that we knew b was negative so he changed the inequality. What do we do if we don't know for sure if b is negative or if it could be both?",
        "A": "Can I use a condition |A| =/> |B| ? *absolute value of A has to be same or higher than absolute value of B",
        "video_name": "0_VaUYoNV7Y",
        "timestamps": [
            230
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 1:01 in the video, Sal said you need to find the common denominator, which he made 18. It makes sense, but would it also work to make the denominator 6 instead of reducing at it at the end? So instead of having 3/18 and 12/18, you had 1/6 and 4/6?",
        "A": "Yes... if you reduce 3/18 before you even start, you get 1/6 Then, you can use a common denominator of 6. Great work in seeing that option!",
        "video_name": "8Eb5MWwcMMY",
        "timestamps": [
            61
        ],
        "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 3:10 he breaks 10 down into it's prime numbers. if there is a number all alone do you always have to break it down?",
        "A": "I quess you mean break down 10 in its primenumbers like 10=2times5. this is not necessary. remember that the goal was to make the fraction as simple as possible. In that case you make the denomenator as simple as possible. not the numerator. for instance: 2/4 you make it 1/2 3/9 you make it 1/3 6/3 you make it 3/1 so remember 10 is the same as 10/1 which you would simplify to 10. and besides 10 is easier to write than 2times5.",
        "video_name": "gcnk8TnzsLc",
        "timestamps": [
            190
        ],
        "3min_transcript": ""
    },
    {
        "Q": "where did the 1/3 come from @4:27 ? is it possible to just reduce the fraction\n27/1 * 2/3--> 9/1 * 2/1 = 18 and 8/1 * 2/3--> becomes 16/3=8 so 18/8 is 9/4 ?? would that work for all other problems?",
        "A": "An exponent of 2/3 is not the same as multiplying by 2/3. If you have 9^2, you don t do 9*2. You must do 9*9. An exponent of 2/3 tells you that you need to find the cube root of the base, then square the result. 27^(2/3) = cuberoot(27)^2 = 3^2 = 9 (not the same as your value of 18) 8^(2/3) = cuberoot(8)^2 = 2^2 = 4 (not 16/3, which is your value) Hope this helps.",
        "video_name": "S34NM0Po0eA",
        "timestamps": [
            267
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "Could what he showed at 1:38 be applied to logarithms in the same way it is showed here?",
        "A": "Yes. for example, log(125)25=2/3 because the cube root of the square of 125 (15625) is 25. The cube root of 125 is 5, and 5^2 is 25.",
        "video_name": "S34NM0Po0eA",
        "timestamps": [
            98
        ],
        "3min_transcript": "We've already seen how to think about something like 64 to the 1/3 power. We saw that this is the exact same thing as taking the cube root of 64. And because we know that 4 times 4 times 4, or 4 to the third power, is equal to 64, if we're looking for the cube root of 64, we're looking for a number that that number times that number times that same number is going to be equal to 64. Well, we know that number is 4, so this thing right over here is going to be 4. Now we're going to think of slightly more complex fractional exponents. The one we see here has a 1 in the numerator. Now we're going to see something different. So what I want to do is think about what 64 to the 2/3 power is. And here I'm going to use a property of exponents that we'll study more later on. But this property of exponents is the idea that-- let's say with a simple number-- if I raise something to the third power and then I were to raise that to, say, the fourth power, 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."
    },
    {
        "Q": "at 4:48 does 3/2=9/2 instead of 9/4?",
        "A": "No, 3/2=9/4 because you are squaring both the numerator and the denominator. Therefore: 3^2=9 2^2=4 So, (3/4)^2=9/4",
        "video_name": "S34NM0Po0eA",
        "timestamps": [
            288
        ],
        "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": "at 0:17 if i flip the sides i mean write 11+a would that still be the same as a+11 is int that the commutative law of addition or it doesn't count when we use a variable?",
        "A": "You got it! a + 11 is equivalent to 11 + a via the commutative law.",
        "video_name": "640-86yn2wM",
        "timestamps": [
            17
        ],
        "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": "In 2:40, Sal said if it goes 6 down but moved his cursor 6 sideways. Why?",
        "A": "He said if x goes down by 6. It means 6 in the negative direction of the x axis (left). He should had said decrease instead of down to make it less confusing.",
        "video_name": "uk7gS3cZVp4",
        "timestamps": [
            160
        ],
        "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 0:50 seconds why did you made the slope zero and cancel it out?",
        "A": "He wants to find the y-intercept, which is the point where the line crosses the y-axis. This point obviously has to be exactly on the y-axis. For a point to be exactly on the y-axis its x value has to be exactly zero. So he substitues the 0 for x in the line`s equation to find the y value of this point, which allows him to find the coordinates of the y-intercept.",
        "video_name": "uk7gS3cZVp4",
        "timestamps": [
            50
        ],
        "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 1:11 how do you know that b is on the y intercept and how do you know y will be on zero?",
        "A": "1) What is x where the y axis crosses the x axis? What is the x coordinate of the y axis? 2) What is x where any line intersects the y axis? Now, all the points (x, y) on the line satisfy y = mx + b . What is the x coordinate of the point on the line where the line intersects the y axis? What is the equation for the point on the line at x=0? (y = m*0 + b; or y = b).",
        "video_name": "uk7gS3cZVp4",
        "timestamps": [
            71
        ],
        "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 8:00 Sal shows that lim u->0 of ln (Z)= ln (lim u->0 of Z). Is this a property of limits that I don't remember?",
        "A": "It s the property of limits having to do with continuous functions. If f is continuous, then lim_{x->c}f(g(x)) = f(lim_{x->c}g(x)). basically you can move the limit inside a continuous function.",
        "video_name": "yUpDRpkUhf4",
        "timestamps": [
            480
        ],
        "3min_transcript": "And we know this is an exponent property, which I'll now do in a different color. We know that a to the bc is equal to a to the b to the c power. So that tells us that this me is equal to the limit as u approaches 0 of the natural log of 1 plus u to the 1/u, because this is one over xu, right? 1/u, and then all of that to the 1/x. And how did I do that? Just from this exponent property, right? If I were to simplify this, I would have 1/x times 1/u, and that's where I get this 1 over xu. If I have b to the a I can put that a out front. So I could take this 1/x and put it in front of the natural log. So now what do I have? We're almost there. We have the limit as u approaches 0. Take that 1/x, put it in front of the natural log sign. 1/x times the natural log of 1 plus u to the 1/u. Fair enough. When we're taking the limit as u approaches 0, x, this term doesn't involve it at all. So we could take this out in front, because the limit doesn't affect this term. And then we're essentially saying what happens to this expression as the limit as u approaches 0. So this thing is equivalent to 1/x times the natural log of And by now hopefully you would recognize that this is the definition. This limit comes to e, if you remember anything from compound interest. You might remember it as the limit-- as n approaches infinity of 1 plus 1 over n to the n. But these things are equivalent. If you just took the substitution u is equal to 1/n, you would get this. You would just get this. So this expression right here is e That expression is e. So we're getting close. So this whole thing is equivalent to 1/x times"
    },
    {
        "Q": "How did the substitution that u is equal to 1/n get thrown in? At 9:16, Sal gives the example that the limit as n goes to infinity of (1+1/n)n is e, yet the limit being discussed in the problem is as u goes to 0, not infinity.",
        "A": "As n goes to infinity, 1/n will go to zero (limit as n->infinity of 1/n = 0) because one over a really big number is very close to zero. Therefore, when we substitute u for n, we change the limit from n->infinity to u->0.",
        "video_name": "yUpDRpkUhf4",
        "timestamps": [
            556
        ],
        "3min_transcript": "If I have b to the a I can put that a out front. So I could take this 1/x and put it in front of the natural log. So now what do I have? We're almost there. We have the limit as u approaches 0. Take that 1/x, put it in front of the natural log sign. 1/x times the natural log of 1 plus u to the 1/u. Fair enough. When we're taking the limit as u approaches 0, x, this term doesn't involve it at all. So we could take this out in front, because the limit doesn't affect this term. And then we're essentially saying what happens to this expression as the limit as u approaches 0. So this thing is equivalent to 1/x times the natural log of And by now hopefully you would recognize that this is the definition. This limit comes to e, if you remember anything from compound interest. You might remember it as the limit-- as n approaches infinity of 1 plus 1 over n to the n. But these things are equivalent. If you just took the substitution u is equal to 1/n, you would get this. You would just get this. So this expression right here is e That expression is e. So we're getting close. So this whole thing is equivalent to 1/x times the ways to get to e. So the limit as u approaches 0 of 1 plus u to the 1/u. That is e. And what is the natural log? Well it's the log base e. So you know this is equal to 1/x times the log base e of e. So that's saying e to what power is e. Well e to the first power is e, right? This is equal to 1. So 1 times 1/x is equal to 1/x. There we have it. The derivative of the natural log of x is equal to 1/x, which I find kind of neat, because all of the other exponents lead to another exponent. But all of a sudden in the mix here you have the natural log and the derivative of that is equal to x to the negative 1 or 1/x. Fascinating."
    },
    {
        "Q": "at 3:10 how did we get exactly 1/delta x * ln(1+ delta x/x)??\ni'd like to see more detailed way of getting 1/delta x * ln(1+ delta x/x)",
        "A": "a/b is the same thing as 1/b * a. In this case, the a is a complicated looking thing, but the rule still works. the b is \u00ce\u0094x.",
        "video_name": "yUpDRpkUhf4",
        "timestamps": [
            190
        ],
        "3min_transcript": "So I'm going to take the limit of this whole thing. The natural log ln of x plus delta x-- right, that's like one point that I'm going to take evaluate the function-- minus the ln of x. All of that over delta x. And if you remember from the derivative videos, this is just the slope, and I'm just taking the limit as I find the slope between a smaller and a smaller distance. Hopefully you remember that. So let's see if we can do some logarithm properties to simplify this a little bit. Hopefully you remember-- and if you don't, review the logarithm properties-- but remember that log of a minus log of b is equal to log of a over b, and that comes out of the fact that follow the exponent rules. And if that doesn't make sense to you, you should review those as well. But let's apply this logarithm property to this equation. So let me rewrite the whole thing, and I'm going to keep switching colors to keep it from getting monotonous. So we have the limit as delta x approaches 0 of this big thing. Let's see. So log of a minus b equals log a over b, so this top, the numerator, will equal the natural log of x plus delta x over x. Right? a b a/b, all of that over delta x. And so that equals the limit as delta x approaches 0-- I think it's time to switch colors again-- delta x approaches 0 x out in front. So this is 1 over delta x, and we're going to take the limit of everything. ln x divided by x is 1 plus delta x over x. Fair enough. Now I'm going to throw out another logarithm property, and hopefully you remember that-- and let me put the properties separate so you know it's not part of the proof-- that a log b is equal to log of b to the a. And that comes from when you take something to an exponent, and then to another exponent you just have to multiply those two exponents. I don't want to confuse you, but hopefully you should remember this. So how does apply here? Well this would be a log b. So this expression is the same thing as the limit."
    },
    {
        "Q": "At 5:27, Sal says \u00ce\u0094x=\u00ce\u00bc, but I thought \u00ce\u0094x=x\u00ce\u00bc.",
        "A": "He never said (delta)x = u He said if (delta)x approaches 0, then u also approaches 0, as they are directly proportional.",
        "video_name": "yUpDRpkUhf4",
        "timestamps": [
            327
        ],
        "3min_transcript": "x out in front. So this is 1 over delta x, and we're going to take the limit of everything. ln x divided by x is 1 plus delta x over x. Fair enough. Now I'm going to throw out another logarithm property, and hopefully you remember that-- and let me put the properties separate so you know it's not part of the proof-- that a log b is equal to log of b to the a. And that comes from when you take something to an exponent, and then to another exponent you just have to multiply those two exponents. I don't want to confuse you, but hopefully you should remember this. So how does apply here? Well this would be a log b. So this expression is the same thing as the limit. delta x over x to the 1 over delta x power. And remember all this is the natural log of this entire thing. And then we're going to take the limit as delta x approaches 0. If you've watched the compound interest problems and you know the definition of e, I think this will start to look familiar. But let me make a substitution that might clean things up a little bit. Let me make the substitution, let me call it n-- no, no, no, let me call u-- is equal to delta x over x. sides by x and we get xu is equal to delta x. Or we would also know that 1 over delta x is equal to 1 over xu. These are all equivalent. So let's make the substitution. So if we're taking the limit is delta x approaches 0, in this expression if delta x approaches 0, what does u approach? u approaches 0. So delta x approaching 0 is the same exact thing as taking the limit as u approaches 0. So we can write this as the limit as u approaches 0 of the natural log of 1 plus-- well we did the substitution, delta x over x is now u-- to the 1 over delta x, and that same substitution told us that's the same thing as one over xu."
    },
    {
        "Q": "In 0:45. How could we say that we are multiplying dx and the function if dx is just a notation that we are integrating with respect to x?",
        "A": "Let me explain dx to you. dx is not a notation. It is used like a notation in evaluating intergals, but it is basically an infinetly small value of x (think of it as limit approaching zero). So, f(x)*dx means a rectangle with height of f(x) and base of an infinetly small number, so that it could be used for intergrating area under curve more accurately than using rectangle with base 1 or 2. Hope this helped you a bit. If not, please check out Sal s lesson about intergration.",
        "video_name": "btGaOTXxXs8",
        "timestamps": [
            45
        ],
        "3min_transcript": "Over here I've drawn part of the graph of y is equal to x squared. And what we're going to do is use our powers of definite integrals to find volumes instead of just areas. So let's review what we're doing when we take just a regular definite integral. So if we take the definite integral between, say, 0 and 2 of x squared dx, what does that represent? Well, let's look at our endpoints. So this is x is equal to 0. Let's say that this right over here is x is equal to 2. What we're doing is for each x, we're finding a little dx around it-- so this right over here is a little dx. And we're multiplying that dx times our function, times x squared. So what we're doing is we're multiplying this width times this height right over here. The height right over here is x squared. And we're getting the area of this little rectangle. And the integral sign is literally the sum of all of these rectangles for all of the x's But the limit of that as these dx's get smaller and smaller and smaller, get infinitely small, but not being equal to 0. And we have an infinite number of them. That's the whole power of the definite integral. And so you can imagine, as these dx's get smaller and smaller and smaller, these rectangles get narrower and narrower and narrower, and we have more of them, we are getting a better and better approximation of the area under the curve until, at the limit, we are getting the area under the curve. Now we're going to apply that same idea, not to find the area under this curve, but to find the volume if we were to rotate this curve around the x-axis. So this is going to stretch our powers of visualization here. So let's think about what happens when we rotate this thing around the x-axis. So if we were to rotate it, and I'll look at it and say that we're looking it a little bit from the right. So we get kind of a base that looks something like this. So you have a base that looks something like that. And then the rest of the function, if we just think about between 0 and 2, it looks like one of those pieces from-- I don't know if you ever played the game Sorry-- or it looks like a little bit of a weird hat. So it looks like this, and let me shade it in a little bit so it looks something like that. And just so that we're making sure we can visualize this thing that's being rotated. We care about the entire volume of the thing. Let me draw it from a few different angles. So if I drew from the top, it would look something like this. It'll become a little more obvious that it looks something like a hat. It would point up like this, and it goes down like that. It would look something like that. So in this angle, we're not seeing the bottom of it. And if you were to just orient yourself,"
    },
    {
        "Q": "at 16:14, how do we know that m,n,p must be 527, 11, 40. In other word, how do we know for sure that cannot be any other m,n,p that result the same value of area?",
        "A": "The question indicates that m and p are relatively prime , which means the fraction m/p is not reducible to any other integers. Also, it says that n is not divisible by the square of any prime, which means the number under the radical cannot have any perfect squares factored out. These instructions rule out any other m, n, or p that can give an equivalent correct answer.",
        "video_name": "smtrrefmC40",
        "timestamps": [
            974
        ],
        "3min_transcript": "We also know that CM right over here is 25/2. We also know that sine of theta here is equal to 527/625. That's what we just figured out. And we can use that to figure out the height of this triangle because we know that sine is the opposite over the hypotenuse, if we draw a right triangle right here. So sine of theta, which is 527/625, is equal to the opposite, is equal to the height of this triangle, over the hypotenuse, over 5/2 square roots of 11. So we can multiply both sides of this by 5/2 square roots 11. And we get the height of the triangle is equal to 527/625 times 5/2 square roots of 11. So you get a 1 here. And you have a 125 over here. So this is equal to 527 square roots 11 over 125 times 2, which is 250. That's the height. Now, what's the area of the triangle? It's 1/2 base times height. Area is equal to 1/2 base, which is 25/2, times the height, which we just figured out is 527 square roots of 11 over 250. Let's see. Divide the numerator by 25. Divide the denominator by 25. You get a 10 there. So this is equal to 527 square roots of 11 over 2 times So that's our area. And the whole problem-- they didn't want us to find the area. They wanted us to find m plus n plus p, so they want us to find essentially 527 plus 11 plus 40. So 527 plus 11 is-- let me make sure I do this-- 538, and then plus 40 is 578. And we're done."
    },
    {
        "Q": "At 12:06, how did you get 2 (25/2) ^2 (1 + sine theta) from 2 (25/2) ^2 + 2 (25/2)^2 sine theta?",
        "A": "The expression 2(25/2)^2 + 2(25/2)^2sin theta has two terms. Both of these terms has a common factor 2(25/2)^2. Sal pulled this factor out of both terms. Let s make a substitution so it might be easier to see: let x = 2(25/2)^2. Then the expression becomes x + xsin theta. We pull the x out to become x(1 + sin theta). This is simple factoring out, or Sal sometimes calls it undistributing. We can redistribute the x to both terms by multiplying through to get x + xsin theta.",
        "video_name": "smtrrefmC40",
        "timestamps": [
            726
        ],
        "3min_transcript": "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. 1 plus sine of theta. Or we can just divide both sides of the equation by this here. Actually, let me just simplify it. This thing over here is 625/4, but then we're going to multiply that by 2. So this thing over here is 625/2. So let's divide both sides of this by 625/2. And we get 576 times 2 over 625-- just multiplying both sides by the inverse-- is equal to. When you multiply both sides by the inverse [? like this, ?] it actually cancels out-- is equal to 1 plus sine of theta. Or we just subtract 1 from both sides. We get sine of theta is equal to 576 times 2. Let's see, 76 times 2 is 152, plus 1,000. So it's 1,152/625."
    },
    {
        "Q": "At 5:52 pm what if it ask what is 39 percent of 700 how would you solve that?",
        "A": "Best thing for this problem would be to divide 700 by 100 to get the amount of 1% (7) then multiply 7 by 39 to get 39% of 700 which is 273.",
        "video_name": "FaDtge_vkbg",
        "timestamps": [
            352
        ],
        "3min_transcript": "5 times 16 is 80. You subtract, you have no remainder, and you're done. 4/16 is the same thing as 0.25. Now, 0.25 is the same thing as twenty-five hundredths. Or, this is the same thing as 25/100, which is the same thing as 25%."
    },
    {
        "Q": "at 7:45, why does the Eigenvector equal span (1/2, 1), not span (1, - 1/2)?",
        "A": "these are equivalent, since (1,-1/2) is in the span of (1/2,1) and vice versa. So it doesn t matter which one you choose, both statements are correct.",
        "video_name": "3-xfmbdzkqc",
        "timestamps": [
            465
        ],
        "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 1:24, Where on earth did Sal swipe that 2 from?",
        "A": "I think Sal recognized 30 = 6*5 = 2*3*5. That s why he tried multiplying it by 2. Don t wrap your mind around it too much. I tend to like this approach better: Let a = 5x\u00c2\u00b2 The expression would then be a\u00c2\u00b2 - 6a + 9 You now know how to solve it.",
        "video_name": "o-ZbdYVGehI",
        "timestamps": [
            84
        ],
        "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 1:27 where did the 2 come from?",
        "A": "Sal is using the pattern created by squaring a binomial. Here s the pattern: (a+b)^2 = a^2 + 2ab + b^2 Here s where the 2 comes from... use FOIL and multiply (a+b)(a+b) ... (a+b)(a+b) = a^2 + ab + ab + b^2 Notice... the 2 middle terms match. When you add them you get 2ab. That s where the 2 comes from. Hope this helps.",
        "video_name": "o-ZbdYVGehI",
        "timestamps": [
            87
        ],
        "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 1:00 isn't profit a second variable?",
        "A": "Because the problem gives you the value for Profit, a second variable is not needed. You could use a variable, but you would immediately swap it out because you know the value of the variable. Hope this helps.",
        "video_name": "roHvNNFXr4k",
        "timestamps": [
            60
        ],
        "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": "At 3:21,how is S the same thing as 9/9?",
        "A": "s is not 9/9. if it was, Sal would have gotten rid of the s. any number times 1 is itself, and 9/9 = 1. so 9/9 * s = s * 1",
        "video_name": "vBlR2xNAGmo",
        "timestamps": [
            201
        ],
        "3min_transcript": "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 and on the right hand side you're only going to be left with this four ninths over here Then this four ninths, we can (S is the same thing as nine over nine) write this as nine over nine S minus four ninths S is equal to four ninths So nine over nine minus four over nine of something gives us five over nine So this becomes five ninths S is equal to four ninths Then to solve for S (and this is kind of magical but it's actually quite logical) Multiply both sides times the inverse of this, so times nine fifths on both sides These guys cancel out, and we get S is equal to four fifths That's really neat! We've just shown that this whole thing over here is equal to four fifths"
    },
    {
        "Q": "at 2:10 sal says between that and that why do we have round down or up for example can 1,251 be rounded to 1,250 rather then 1,300",
        "A": "Yes it can! But not if you are rounding to the nearest hundred. If you were rounding to the nearest 50, you WOULD round to 1,250. But 1,250 does not have only zeros after the hundreds place, so in rounding to the nearest hundred that would not be right.",
        "video_name": "fh8gkPW_6g4",
        "timestamps": [
            130
        ],
        "3min_transcript": "Round 423,275 to the nearest thousand. So let me rewrite it: 423,275. And so the thousands place is the 3 right here, and so if we were round it up to the nearest thousand, we would go to 420-- let me write it so we just focus on the 3-- we would go up to 424,000 if we wanted to round up, 424,000, and if we wanted to round down, we would go to 423,000. We would get rid of the 275. 423,000. So this is our choice. Round up to 424,000 or round down to 423,000. And to figure it out, we just look at the digit one place to If that digit is 5 or greater, you round up. So this is 5. So if this is greater than or equal to 5, 5 or greater, you round up. If it's less than 5, you round down. 2 is definitely less than 5, so we just round down, so it is 423,000. Now just to visualize what this means to the nearest thousand, if I were to do a number line-- and you don't We've gotten the answer, but just to have a little bit better visualization of it, if I were to increment by thousands, you might have 422,000, 423,000. You have 424,000, and then maybe over here, you have 425,000, and you could keep going. And so when we round to the nearest thousand, we have to pick between that and that. We see that it much closer to 423,000 than to 424,000, so we round it right there. But you just use the rules we just came up with, and we rounded down to 423,000."
    },
    {
        "Q": "Why did Sal draw two lines over the angles at 1:18?",
        "A": "To show that both angles became one angle",
        "video_name": "jRrRqMJbHKc",
        "timestamps": [
            78
        ],
        "3min_transcript": "All right. We're on problem 26. For the quadrilateral shown below, a quadrilateral has four sides, measure of angle A plus the measure of angle C is equal to what? And here, you should know that the sum of all the angles in a quadrilateral are equal to 360 degrees. And you might say, OK, I'll add that to my memory bank of things to memorize. Like the angles in a triangle are equal to 180. And I'll show you no, you don't have to memorize that. Because if you imagine any quadrilateral, let me draw a quadrilateral for you. And this is true of any polygon. So let's say this is some quadrilateral. You don't have to memorize that the sum of the angles is equal to 360. Although it might be useful for a quadrilateral. But I'll show you how to always prove it for any polygon. You just break it up into triangles. Then you only have to memorize one thing. If you break it up into triangles, this angle plus that angle plus that angle has to be equal to 180. equal to 180. So the angles in the quadrilateral itself are this angle and this angle. And then this angle and this angle. Well this one is just the sum of those two, and this one's just the sum of those two. So if these three added up to 180. And these three added up to 180. This plus this plus this, plus this will add up to 360. And you can do that with an arbitrarily shaped polygon. Let's do five sides, let's do a pentagon. So one, two, three, four, five sides. Wow, how many angles are there in a pentagon. Just break it up into triangles. How many triangles can you fit in it? Let's see. One, two. Each of these triangles, their angles, they add up to 180. So if you want to know that, that, that, plus that, that, that, plus that, that, and that. And that also would be the angle measures of the polygon. Because these three angles add up to that angle. That's that. Those angles add up to that one. Those angles add up to that one, and those angles add up So now hopefully, if I gave you a 20 sided polygon, you can figure out how many times can I fit triangles into it. And you'll know how many angles there are. And the sum of all of them. But anyway, back to the quadrilateral. A quadrilateral, the sum of the angles are going to be 360 degrees. So, if we say, measure of angle A, plus measure of angle C, plus these two angles. Let me write it down. Plus 95 plus 32 is going to be equal to 360. So I'll just write A plus C, just a quick notation. Let's see, 95 plus 32 is 127. Plus 127 is equal to 360. A plus C is equal to 360 minus 127."
    },
    {
        "Q": "At 3:52, why did Sal say 10 to the 3rd power instead of 10 cubed?",
        "A": "10 to the 3rd power is the same thing as 10 cubed. They are just two different ways of saying the same thing. Similarly, 10 to the 2nd power is the same thing as 10 squared.",
        "video_name": "YJdCw2fK-Og",
        "timestamps": [
            232
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 2:16 the sign is little bit confusing. More explanation. Thanks",
        "A": "Sal wrote b < x < c. Between the points b and c on the x-axis, but not including those points, the function is negative. Notice, as Sal mentions, that this portion of the graph is below the x-axis. That is your first clue that the function is negative at that spot. Hope this helps.",
        "video_name": "KxOp3s9ottg",
        "timestamps": [
            136
        ],
        "3min_transcript": "- [Voiceover] What I hope to do in this video is look at this graph y is equal to f of x and think about the intervals where this graph is positive or negative and then think about the intervals when this graph is increasing or decreasing. So first let's just think about when is this function, when is this function positive? Well positive means that the value of the function is greater than zero. It means that the value of the function this means that the function is sitting above the x-axis. So it's sitting above the x-axis in this place right over here that I am highlighting in yellow and it is also sitting above the x-axis over here. And if we wanted to, if we wanted to write those intervals mathematically. Well let's see, let's say that this point, let's say that this point right over here is x equals a. Let's say that this right over here is x equals b and this right over here is x equals c. 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."
    },
    {
        "Q": "why is it not factorial at 2:14?",
        "A": "Because when you are allowed to use a letter more than once, you do not need to take it out of the possibilities for the next letter. You can use HHH, as Sal said. So you don t need to make it 26x25x24 because the numbers reset every time you add a new slot.",
        "video_name": "VYbqG2NuOo8",
        "timestamps": [
            134
        ],
        "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 6:00 he swaps a row with another and add a -ve sign. If further down the computation process, if we have to swap another row, do we add another \"-\" sign thus making the Det. positive again..or does the minus stay no matter the no. of row-swaps?\n\nThanks",
        "A": "Yes, each row or column swap results in the determinant being multiplied by -1. If you do an odd number of swaps, the determinant is therefore negative, and a positive number of swaps keeps it positive.",
        "video_name": "QV0jsTiobU4",
        "timestamps": [
            360
        ],
        "3min_transcript": "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. So there we have it like that. And this guy has two 0's here, so maybe I want to swap some rows. So let me swap some rows. So if we swap rows, what happens? I'm going to swap the middle two rows just for fun. Well, not just for fun. Because I want a pivot entry right here. I shouldn't say a pivot entry. I want to do it in upper triangular form. So I want a non-zero entry here. This is a 0, so I'm going to move this guy down. So I'm going to keep the top row the same. 1, 2, 2, 1. I'm going to keep the bottom row the same. 0, 0, 6, minus 4, 4. And I'm going to swap these guys right here. So this is going to be 0, 3, 1, 0. And then 0, 0, 2, 1. Now, can I just swap entries like that? Well, I can, but you have to remember that when you swap entries, your resulting determinant is going to be the So if we swap these two guys, the determinant of this is going to be the negative of this determinant. When you swap two rows, you just flip the sign of the determinant. That was one of the first videos we did on these, kind of messing with the determinants. Now, what do we want to do here? To get this guy into upper triangular form, it'd be nice to get this to be a 0. So to get that to be a 0, let me keep everything else the same. So I have a 1, 2, 2, 1. I have a 0, 3, 1, 0. The third row is 0, 0, 2, 1. And now this last row, let me replace it with the last row minus 3 times this row. So let me write it like this. I have to carry that negative sign as well. So I'm going to replace this last row with the last row minus 2 times the second row."
    },
    {
        "Q": "At 5:54, why exactly did you combine C1 and C2 to have the constant C? Aren't you supposed to solve for C2 and C1 separately?",
        "A": "C\u00e2\u0082\u0081 and C\u00e2\u0082\u0082 are just constants, numbers, and they are subtracting each other, so not only it makes sense to combine them into a single constant, but it would be impossible to find their values independently. You would end up with and equation in the form C\u00e2\u0082\u0081 - C\u00e2\u0082\u0082 = 0, and there are an infinite number of possibilities here.",
        "video_name": "DL-ozRGDlkY",
        "timestamps": [
            354
        ],
        "3min_transcript": "So if that was a two there and if you don't want to change the value of the integral you put the 1/2 right over there. And so now you could either do U substitution explicitly or you could do it in your head where you said U is equal to negative X squared and then DU will be negative to X, DX or you can kind of do this in your head at this point. So I have something and it's derivative so I really could just integrate with respect to that something too with respect to that U. So this is going to be 1/2. This 1/2 right over here. The anti-derivative. This is E to the negative X squared and then of course, I might have some other constant. I'll just call that C two. And once again, if this part over here what I just did seemed strange, the U substitution, you might want to review that piece. Now, what can I do here? We'll have a constant on the left hand side. It's an arbitrary constant. We don't know what it is. we could call it. So, let me just subtract C one from both sides. So if I just subtract C one from both sides I have an arbitrary so this is gonna cancel, and I have C two, sorry. Let me. So, this is C one. So these are going to cancel and C two minus C one. These are both constants, arbitrary constants and we don't know what they are yet. And so, we could just rewrite this as on the left hand side we have Y squared over two is equal to on the right hand side. I'll write 1/2 E. Let me write that in blue just because I wrote it in blue before. 1/2 E to the negative X squared and I'll just say C two minus C one. Let's just call that C. So if you take the sum of those two things let's just call that C. And so now, this is kind of a general solution. We don't know what this constant is but even in this form we can now find a particular solution using this initial condition. Let me separate it out. This was a part of this original expression right over here but using this initial condition. So, it tells us when X is zero, Y needs to be equal to one. So we would have one squared which is just one over two is equal to 1/2. E to the negative zero squared. Well, that's just going to be either the zero is just one. This is gonna be 1/2 plus C and just like that we're able to figure out if you subtract 1/2 from both sides C is equal to zero. So the relationship between Y and X that goes through this point, we could just set C is equal to zero. So that's equal to zero. That's zero right over there. And so we are left with Y squared over two"
    },
    {
        "Q": "At 4:36 shouldn't the integral be equal to e^ (1 - x^2)/ 2*(1 - x^2) ?",
        "A": "No, I m not sure where you re getting (1 - x\u00c2\u00b2). This is an integral best done with u-substition: u = - x\u00c2\u00b2 du = - 2x dx so 1/2\u00e2\u0088\u00ab -2x e^-x\u00c2\u00b2 dx = 1/2 \u00e2\u0088\u00ab e^u du = 1/2 e^u + C = 1/2 e^-x\u00c2\u00b2 + C",
        "video_name": "DL-ozRGDlkY",
        "timestamps": [
            276
        ],
        "3min_transcript": "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 So if that was a two there and if you don't want to change the value of the integral you put the 1/2 right over there. And so now you could either do U substitution explicitly or you could do it in your head where you said U is equal to negative X squared and then DU will be negative to X, DX or you can kind of do this in your head at this point. So I have something and it's derivative so I really could just integrate with respect to that something too with respect to that U. So this is going to be 1/2. This 1/2 right over here. The anti-derivative. This is E to the negative X squared and then of course, I might have some other constant. I'll just call that C two. And once again, if this part over here what I just did seemed strange, the U substitution, you might want to review that piece. Now, what can I do here? We'll have a constant on the left hand side. It's an arbitrary constant. We don't know what it is. we could call it. So, let me just subtract C one from both sides. So if I just subtract C one from both sides I have an arbitrary so this is gonna cancel, and I have C two, sorry. Let me. So, this is C one. So these are going to cancel and C two minus C one. These are both constants, arbitrary constants and we don't know what they are yet. And so, we could just rewrite this as on the left hand side we have Y squared over two is equal to on the right hand side. I'll write 1/2 E. Let me write that in blue just because I wrote it in blue before. 1/2 E to the negative X squared and I'll just say C two minus C one. Let's just call that C. So if you take the sum of those two things let's just call that C. And so now, this is kind of a general solution. We don't know what this constant is"
    },
    {
        "Q": "Wait. So are all of the foreheads blue, or not? because if they're all blue it should only take 2 times. (as in open lights shut lights open lights shut and everyone is gone.) At 0:09 that's what it said, but... ok this is getting confusing",
        "A": "every person would see a number of blue foreheads. if I HAVE a blue forehead I see one less than the total number of blueforeheads. If I have other than blue forhead I see the total number of blue foreheads. everone would leave on the number of blue foreheads seen plus one. So all blue foreheads would leave on the total number of blue foreheads time the light goes out, while the other foreheads are wating for the total number of blue foreheads plus one.",
        "video_name": "-xYkTJFbuM0",
        "timestamps": [
            9
        ],
        "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": "At 3:32, is anything better than Sal using the word \"DUDE\" in an educational context?",
        "A": "No, no there isn t.",
        "video_name": "-xYkTJFbuM0",
        "timestamps": [
            212
        ],
        "3min_transcript": "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. look around the room. And I look around the room, and I see nobody else, right? And remember, even in the case of one, we've painted everyone's forehead blue. So in this case, this one dude, or me or whoever you want to call him. His forehead is painted blue. So he looks around and he sees no one in the room. But he remembers the statement, and maybe it's even written down on a card for him in case he forgets. That at least one of you has your forehead painted blue. So if he looks around the room and he says, well I'm the only dude in the room. And they told me that at least one of the dudes in the room is going to have their foreheads blue. Well, I'm the only dude in the room. So I must have a blue forehead. So as soon as they turn the lights off, he's going to leave. Fair enough. That's almost trivially simple. And you might say, so how does this apply to 100? Well what happens when there are two people. And once again, both of them have their foreheads painted blue. So let me draw another. I don't want to keep drawing the blue forehead room. So let's put ourselves in the head of this guy. Right behind the blue forehead. That's where we're sitting. So when he enters the room. He says, I either have a blue forehead. I either have a blue forehead, or I don't have a blue forehead. No blue. Right? This is what this guy's thinking. Let me draw him. And he has a blue forehead. But he doesn't know it. He can't see it. That's the whole point about painting the forehead blue, as opposed to another part of the body. So he says, I either have a blue forehead or I don't have a blue forehead. He walks in. Let's say this is this guy. He walks in. The first time the lights get turned on, he sees this other dude there who has a blue forehead."
    },
    {
        "Q": "so does that mean that 1.999....=2? I hope it does, i lost track after 1:26",
        "A": "I think this was already answered for you, but I m not sure so I will. Yes, 1.999.... is equal to 2. Think about it as (1 + .999....) = 2.",
        "video_name": "TINfzxSnnIE",
        "timestamps": [
            86
        ],
        "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 07:42: You will never reach _____ steps or 1/3: What is the blank?",
        "A": "The blank is infinity",
        "video_name": "TINfzxSnnIE",
        "timestamps": [
            462
        ],
        "3min_transcript": "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. 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!"
    },
    {
        "Q": "At 3:24 how does infinity minus 1 still equal infinity?",
        "A": "because it was declared as a rule in math. I know that isn t a great answer but it works. The best answer is because its not a real number. When we add something to a real number we get that number plus 1. Since infinity is not bound by the laws of real numbers it doesn t have to behave that way. I suppose you could say it operates above the laws of real numbers in the hyper-real number space.",
        "video_name": "TINfzxSnnIE",
        "timestamps": [
            204
        ],
        "3min_transcript": "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. Reason number 5, there's infinite 9s. If anyone ever thinks they have the biggest number, well, they don't, because just add 1, or multiply by 2, or whatever, and it's even bigger. Infinity, though, is not a number you can add 1 to to get a bigger number. Adding 1 is an algebra thing that you do with real numbers. Subtracting doesn't work either. Infinity bottles of beer on the wall minus 1 is still infinity bottles of beer. When we did this decimal shift to multiply by 10, unlike un-infinitely many 9s, there's no last 9 that got shifted over to create a 0. Infinite 9s plus another 9 is still infinite 9s, the kind of infuriating property that makes infinity not a real number and makes that proof work. If you're the type of person who is discon-- 9.9 repeating --t with the idea that 9.9 repeating equals 10, you might also feel that 1 divided by 0 should be infinity. And, as it turns out, there is other systems or calculation besides elementary algebra where it does."
    },
    {
        "Q": "At 2:35 she did a thing where she said anything equals anything, when all she said was that:\n\nX = 0\n42 * 0 = 0\n42x = x\n0=0\n\nYou should get 0 = 0, which is true.\nCorrect me if I'm wrong please",
        "A": "Her last step from 42x=x was to divide both sides by x, getting 42=1. Then you can subtract 1 from both sides to get 41=0. Divide by 41 to get 1=0. And multiply by any number c to get c=0. So any number equals 0. But her work is invalid, because she defined x=0. So when she divided by x, she was dividing by 0, which is undefined precisely because of this issue it raises.",
        "video_name": "TINfzxSnnIE",
        "timestamps": [
            155
        ],
        "3min_transcript": "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. Reason number 5, there's infinite 9s. If anyone ever thinks they have the biggest number, well, they don't, because just add 1, or multiply by 2, or whatever, and it's even bigger. Infinity, though, is not a number you can add 1 to to get a bigger number. Adding 1 is an algebra thing that you do with real numbers. Subtracting doesn't work either. Infinity bottles of beer on the wall minus 1 is still infinity bottles of beer. When we did this decimal shift to multiply by 10, unlike un-infinitely many 9s, there's no last 9 that got shifted over to create a 0. Infinite 9s plus another 9 is still infinite 9s, the kind of infuriating property that makes infinity not a real number and makes that proof work. If you're the type of person who is discon-- 9.9 repeating --t with the idea that 9.9 repeating equals 10, you might also feel that 1 divided by 0 should be infinity. And, as it turns out, there is other systems or calculation besides elementary algebra where it does."
    },
    {
        "Q": "at 3:55, how did he come up with 10-9/12? help?",
        "A": "The 10-9/12 came from the right side of the equation: 5/6 - 3/4 and in order to subtract 3/4 from 5/6 we need to have like denominators. The LCM for 5/6 and 3/4 is 12, therefore 5/6 = 10/12 and 3/4 = 9/12 Which gives us 10/12 - 9/12 or 10-9/12 = 1/12",
        "video_name": "DopnmxeMt-s",
        "timestamps": [
            235
        ],
        "3min_transcript": "And actually, well, if we wanted to check it, we could say, well, the original problem was 2x plus 3 equals minus 15. So we could say 2 times minus 9 plus 3. 2 times minus 9 is minus 18 plus 3. Well, that's equal to minus 15, which is equal to what the original equation said, so we know that's right. That's the neat thing about algebra. You can always check your work. Let's do another problem. I'm going to put some fractions in this time, just to show you that it can get a little bit hairy. So let's say I had minus 1/2x plus 3/4 is equal to 5/6. So we'll do the same thing. First, we just want to get this 3/4 out of the left hand 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,"
    },
    {
        "Q": "at 2:19 i didnt hear what sal said what did he say",
        "A": "he said ...times minus nine...",
        "video_name": "DopnmxeMt-s",
        "timestamps": [
            139
        ],
        "3min_transcript": "Welcome to level two linear equations. Let's do a problem. 2x plus 3 is equal to minus 15. Throw the minus in there to make it a little bit tougher. So the first thing we want to do whenever we do any linear equation, is we want to get all of the variable terms on one hand side of the equation and all the constant terms And it doesn't really matter, although I tend to get my variables on the left hand side of the equation. Well, my variables are already on the left hand side of the equation but I have this plus 3 that I somehow want to move to the right hand side of the equation. And the way I can-- you can put it in quotes, move the 3 is I can subtract 3 from both sides of this equation. And look at that carefully as to why you think that works. Because if I subtract 3 from the left hand side, clearly this negative 3 that I'm subtracting and the original 3 will cancel out and become 0. and as long as I do whatever I do on the left hand side, as well, because whatever you do on one side of the equal side, you have to do to the other side, then I'm making a valid operation. So this will simplify to 2x, because the 3's cancel out. They become just 0. Equals minus 15 minus 3. Well, that's minus 18. And now, we're just at a level one problem, and you can just multiply both sides of this equation times the reciprocal on the coefficient of 2x. I mean, some people would just say that we're dividing by 2, which is essentially what we're doing. I like to always go with the reciprocal, because if this 2 was a fraction, it's easier to think about it that way. But either way, you either multiply by the reciprocal, or divide by the number. It's the same thing. So 1/2 times 2x. Well, that's just 1x. So you get x equals, and then minus 18/2. And minus 18/2, well, that just equals minus 9. And actually, well, if we wanted to check it, we could say, well, the original problem was 2x plus 3 equals minus 15. So we could say 2 times minus 9 plus 3. 2 times minus 9 is minus 18 plus 3. Well, that's equal to minus 15, which is equal to what the original equation said, so we know that's right. That's the neat thing about algebra. You can always check your work. Let's do another problem. I'm going to put some fractions in this time, just to show you that it can get a little bit hairy. So let's say I had minus 1/2x plus 3/4 is equal to 5/6. So we'll do the same thing. First, we just want to get this 3/4 out of the left hand"
    },
    {
        "Q": "At 2:34 couldn't you factor out both the 3 and the 5 to get 15? Wouldn't 15(a+b)=2 be correct?",
        "A": "3a+5b=2 To factor, both terms must have a same common factor. Meaning to factor out a 3, the b-term must have a factor of 3 as well. Likewise, to factor out a 5, the a-term must have a factor of 5. Since neither have any common factor, you can t factor it. And certainly, neither of them has a factor of 15.",
        "video_name": "CLQRZ2UbQ4Q",
        "timestamps": [
            154
        ],
        "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": "WHat does recipocral mean? 3:01",
        "A": "the reciprocal means the opposite of a number",
        "video_name": "bAerID24QJ0",
        "timestamps": [
            181
        ],
        "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": "At 4:20 how did you multiply the fraction -4/3 to the fraction 10/13.\nI was quite confused there.....",
        "A": "To multiply fractions: 1) Multiply the numerators: -4 (10) = -40 2) Multiply the denominators: 3(13) = 39 3) This creates the fraction -40/39. 4) To change to a mixed number, divide -40 by 39. The remainder becomes the new numerator: -40/39 = -1 1/39",
        "video_name": "bAerID24QJ0",
        "timestamps": [
            260
        ],
        "3min_transcript": "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. And I like to leave my fractions improper because it's easier to deal with them. But you could also view that-- that's minus-- if you wanted to write it as a mixed number, that's minus 1 and 1/39. I tend to keep it like this. Let's check to make sure that's right. The cool thing about algebra is you can always get your answer and put it back into the original equation to make sure you are right. So the original equation was minus 3/4 times x, and here we'll substitute the x back into the equation. Wherever we saw x, we'll now put our answer. So it's minus 40/39, and our original equation said that equals 10/13. Well, and once again, when I just write the 3/4 right next to the parentheses like that, that's just another way of writing times. So minus 3 times minus 40, it is minus 100--"
    },
    {
        "Q": "At 2:05, how did you get pi. When he said m<ABD + pi/4 + pi/2= pi !",
        "A": "The angle sum of a triangle is 180 degrees, tho convert to radians you multiply by pi/180 which makes it 180 x pi/180 or just pi. Similarly you can find 90 degrees is pi/2",
        "video_name": "KoYZErFpZ5Q",
        "timestamps": [
            125
        ],
        "3min_transcript": "Voiceover:So we have depicted here is a unit circle centered at point A and the point B lies on the circle. And then we dropped a perpendicular from point B to point D, point D lies on the positive X-axis. So they form this triangle ABD, and they tell us at angle BAD, angle bad, it has a measure of pi over four radians. What I want to do, in this video, is use our knowledge of trigonometry and use our knowledge of triangles in order to figure out several things. So the first thing we want to figure out is what's the radian measure of angle ABD? Actually let's just do that first and then I'll talk about the other things. that we need to think about. So I assumed you've paused the video So let's think about what ABD would be. We know that two of the angles of this triangle, so if you know two of the angles of this triangle, you should be able to figure out the third. Now what might be a little bit unfamiliar, we're use to saying that the sum of the interior angles of a triangle add up to 180 degrees, but now we're thinking in terms of radians. So we could say that the sum of the angles of a triangle add up to, instead of saying 180 degrees, 180 degrees is the same thing as pi radians. So this angle plus that angle are going to add up to pi. So lets just say that this right over here, lets just say measure of angle ABD in radians, plus pi over four, plus, this is a right angle. What would that be in radians? Well a right angle in radians, a 90 degree angle in radians, is pi over two radians. So plus pi over two. When you take the sum of them, the interior angles of this triangle, they're going to add up to pi radians. And now we can solve for the measure of angle ABD. Measure of angle ABD is equal to pi minus pi over two, minus pi over four. I just subtracted these two from both the sides. So this is going to be equal to, you can put a common denominator of four, then this is four pi over four, this is minus two pi over four, and this, of course, is minus pi over four. So four minus two minus one is going to get us to one. So this is going to get us to pi over four. So the measure of angle ABD is actually the same as the measure of angle BAD, it is pi over four. So that angle right over there is pi over four. Now what does that help us with? So if we know that this is pi over four and that is pi over four radians, and once again"
    },
    {
        "Q": "At 5:15, Sal finds the side lengths of the isosceles triangle, and that is easy to understand. If the two side lengths are different, how would you find the lengths?",
        "A": "It s still SOH CAH TOA, but you might have trouble getting your calculator to spit out the right answer unless it s set in radians. Eg if your calculator is set in degrees then Cos(pi/3)=0.86 but if set to radians it ll be 0.5 which actually works.",
        "video_name": "KoYZErFpZ5Q",
        "timestamps": [
            315
        ],
        "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": "at 7:30, what happened to the y? the equation was: -2y/x ... Thank you.",
        "A": "Earlier in the video, Sal solved the equation for y, showing that y = 1/x^2. Now, at 7:30, he wants makes a substitution using that equality, plugging in 1/x^2 in place of y. So y goes away and we get 1/x^2 in its place.",
        "video_name": "2CsQ_l1S2_Y",
        "timestamps": [
            450
        ],
        "3min_transcript": "as I go-- we get this thing, which I can rewrite as x times-- well, it's just going to be x in the numerator divided by 2 times the square root of y. y to the negative 1/2 is just the square root of y in the denominator. And 1/2, I just put the 2 in the denominator there-- times dy dx times the derivative of y with respect to x is going to be equal to the negative square root of y. I just subtracted the square root of y from both sides. And actually, this is something that I might actually want to copy and paste up here. So copy and then paste. So let's go back up here, just to continue our simplification solving for dy dx. Well, to solve for dy dx, we just have to divide both sides by x over 2 times the square root So we're left with dy dx-- or dividing both sides by this is the same thing as multiplying by the reciprocal of this-- is equal to 2 times the square root of y over x-- over my yellow x-- times the negative square root of y. Well, what's this going to simplify to? This is going to be equal to y times-- the square root of y times the square root of y is just y. The negative times the 2, you get negative 2. So you get negative 2y over x is equal to the derivative of y with respect to x. Now you might be saying, look, we just figured out the derivative implicitly, and it looks very different than what we did right over here. When we just used the power rule, we got negative 2 x The key here is to realize that this thing right over here we could solve explicitly in terms of-- we could solve for y. So we could just make this substitution back here to see that these are the exact same thing. So if we make the substitution y is equal to 1 over x squared, you would get dy dx, the derivative of y with respect to x, is equal to our negative 2 times 1 over x squared, and then all of that over x, which is equal to negative 2 over x to the third, which is exactly what we have over here, negative 2 x to the negative third power."
    },
    {
        "Q": "At 5:20, how x is divided by 2 square root of y? where does the 2 come from?",
        "A": "It came from d/dy (\u00e2\u0088\u009ay) = 1/2 * y^-0.5 = 1/(2*\u00e2\u0088\u009ay)",
        "video_name": "2CsQ_l1S2_Y",
        "timestamps": [
            320
        ],
        "3min_transcript": "with respect to that something, is going to be 1/2 times that something to the negative 1/2 power. Once again, this right over here is the derivative of the square root of y with respect to y. We've seen this multiple times. If I were to say the derivative of the square root of x with respect to x, you would get 1/2 x to the negative 1/2. Now I'm just doing it with y's. But we're not done yet. Remember, our derivative operator wasn't to say with respect to y. It's with respect to x. So this only gets us with respect to y. We need to apply the entire chain rule. We have to multiply that times the derivative of y with respect to x in order to get the real derivative of this expression with respect to x. So let's multiply times the derivative We don't know what that is. That's actually what we're trying to solve for. But to use the chain rule, we just have say it's the derivative of the square root of y with respect to y times the derivative of y with respect to x. This is the derivative of this thing with respect to x. So we get this on the left-hand side. On the right-hand side, we just have a 0. And now, once again, we can attempt to solve for the derivative of y with respect to x. And maybe the easiest first step is to subtract the square root of y from both sides of this equation. And actually, let me move all of this stuff over, so I have, once again, more room to work with. So let me cut it, actually. And then let me paste it. Let me move it over, right over here. So we went from there to there. I didn't gain a lot of real estate, but hopefully this helps a little bit. And actually, I don't even like that. Let me leave it where it was before. So then, if we subtract the square root of y as I go-- we get this thing, which I can rewrite as x times-- well, it's just going to be x in the numerator divided by 2 times the square root of y. y to the negative 1/2 is just the square root of y in the denominator. And 1/2, I just put the 2 in the denominator there-- times dy dx times the derivative of y with respect to x is going to be equal to the negative square root of y. I just subtracted the square root of y from both sides. And actually, this is something that I might actually want to copy and paste up here. So copy and then paste. So let's go back up here, just to continue our simplification solving for dy dx. Well, to solve for dy dx, we just have to divide both sides by x over 2 times the square root"
    },
    {
        "Q": "At 3:37,why does he subtract 2 times 736 times 915 times the cosine of 3 degrees?",
        "A": "Sal was using the Law of Cosines c^2 = a^2 + b^2 - 2ab*cos(theta) where a and b are 2 sides of the triangle and theta is their included angle.",
        "video_name": "6kGi1dvGZNY",
        "timestamps": [
            217
        ],
        "3min_transcript": "And what we wanna do is figure out the width of Orion's belt, which is the distance between Alnitak and Mintaka. So we need to figure out this distance right over here. And the one thing that they did give us is this angle. They did give us that angle right over there. They said that the angle between these stars in the sky is three degrees. So this is three degrees right over there. So how can we figure out the distance between Alnitak and Mintaka? Let's just say that this is equal to X. This is equal to X. How do we do that? Well if we have two sides and an angle between them, we could use the law of cosines to figure out the third side. So the law of cosines, so let's just apply it. So the law of cosines tells us to the sum of the squares of the other two sides. So it's going to be equal to 736 squared, plus 915 squared, minus two times 736, times 915, times the cosine of this angle. Times the cosine of three degrees. So once again, we're trying to find the length of the side opposite the three degrees. We know the other two sides, so the law of cosines, it essentially... Sorry, I just had to cough off camera because I had some peanuts and my throat was dry. Oh, I was saying, if we know the angle and we know the two sides on either side of the angle, by the law of cosines. Where it essentially starts off not too different than the Pythagorean theorem, but then we give an adjustment because this is not an actual right triangle. And the adjustment... So we have the 736 squared, plus 915 squared, minus two times the product of these sides, times the cosine of this angle. Or another way we could say, think about it is, X, let me write that, X is to equal to the square root of all of this stuff. So, I can just copy and paste that. Copy and paste. X is going to be equal to the square root of that. And so let's get our calculator to calculate it. And let me verify that I'm in degree mode. Yes, I am indeed in degree mode. And so let's exit that. And so I wanna calculate the square root of 736 squared, plus 915 squared,"
    },
    {
        "Q": "At 3:33 What does Sal mean by swapping the inequalities?",
        "A": "When you multiply an inequality by a negative number, you re required to switch the sign in the middle in the other direction. So, if Khan had been multiplying by -4 instead, he d get 8r - 13 > 10 instead of 8r - 13 < 10. Hope this helped.",
        "video_name": "x5EJG_rAtkY",
        "timestamps": [
            213
        ],
        "3min_transcript": "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 And once again, we didn't have to worry about the inequality because we're dividing by a positive number. And we get r is less than 23 over 8. Or, if you want to write that as a mixed number, r is less than-- what is that-- 2 and 7/8. So that's one condition, but we still have to worry about this other condition. There was an and right here. Let's worry about it. So our other condition tells us 2r minus 13 over 4 has to be greater than negative 5/2. Let's multiply both sides of this equation by 4. So 4 times 2r is 8r. 4 times negative 13 over 4 is negative 13, is greater than negative 5/2 times 4 is negative 10. Now we add 13 to both sides of this equation. The left-hand side-- these guys cancel out, you're just"
    },
    {
        "Q": "I'm curious: In the first problem, why didn't we rewrite y=x^x as \"log base x of y = x\" and take the derivative of that? I tried it, but I get a different answer (I get y*ln(x) ).\n\nThen, another question: at 5:09 Sal says \"If we haven't solved this, you can just keep taking the natural log of this...\" - I tried that ( ln(ln(y))=xln(x)+ln(ln(x)) ), and I get a different answer (I get y*ln(y)*(ln(x)+1+1/(x*ln(x)) ).\n\nWhat am I doing wrong?",
        "A": "y = x^x ln(y) = ln(x^x) ln(y) = x\u00e2\u0080\u00a2ln(x) d/dx(ln(y)) = d/dx(x\u00e2\u0080\u00a2ln(x)) d/dx(ln(y)) = d/dx(x)\u00e2\u0080\u00a2ln(x) + x\u00e2\u0080\u00a2d/dx(ln(x)) d/dx(ln(y)) = d/dx(x)\u00e2\u0080\u00a2ln(x) + x\u00e2\u0080\u00a21/x\u00e2\u0080\u00a2d/dx(x) d/dx(ln(y)) = d/dx(x)\u00e2\u0080\u00a2ln(x) + x\u00e2\u0080\u00a21/x\u00e2\u0080\u00a2dx/dx d/dx(ln(y)) = d/dx(x)\u00e2\u0080\u00a2ln(x) + x\u00e2\u0080\u00a21/x d/dx(ln(y)) = d/dx(x)\u00e2\u0080\u00a2ln(x) + 1 d/dx(ln(y)) = dx/dx\u00e2\u0080\u00a2ln(x) + 1 d/dx(ln(y)) = ln(x) + 1 1/y\u00e2\u0080\u00a2d/dx(y) = ln(x) + 1 1/y\u00e2\u0080\u00a2dy/dx = ln(x) + 1 dy/dx = y\u00e2\u0080\u00a2(ln(x) + 1) y = x^x dy/dx = [x^x\u00e2\u0080\u00a2(ln(x) + 1)]",
        "video_name": "N5kkwVoAtkc",
        "timestamps": [
            309
        ],
        "3min_transcript": "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. So we can rewrite this as x to the x times the natural log times the natural log of x. So now our expression our equation is simplified to the natural log of y is equal to x to the x times the natural log of x. But we still have this nasty x to the x here. We know no easy way to take the derivative there, although I've actually just shown you what the derivative of this is, so we could actually just apply it right now. I was going to take the natural log again and it would turn into this big, messy, confusing thing but I realized that earlier in this video I just solved for what the derivative of x to the x is. It's this thing right here. It's this crazy expression right here. So we just have to remember that and then apply and then do our problem. So let's do our problem. an unexpected benefit of doing the simpler version of the problem, you could just keep taking the natural log of this, but it'll just get a little bit messier. But since we already know what the derivative of x to the x is, let's just apply it. So we're going to take the derivative of both sides of the equation. Derivative of this is equal to the derivative of this. We'll ignore this for now. Derivative of this with respect to x is the derivative of the natural log of y with respect to y. So that's 1/y times the derivative of y That's just the chain rule. We learned that in implicit differentiation. And so this is equal to the derivative of the first term times the second term, and I'm going to write it out here just because I don't want to skip steps and confuse people. So this is equal to the derivative with respect to x of x to the x times the natural log of x plus the derivative"
    },
    {
        "Q": "4:24 how does sal equate D/V from V=D/T?",
        "A": "V = D/T Multiply both sides by T: VT = D Divide both sides by V: T = D/V",
        "video_name": "Uc2Tm4Lr7uI",
        "timestamps": [
            264
        ],
        "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": "At 2:58 he says 3/4, Can anyone tell me where he got that from? And also where'd he get 2(distance to gift store) from?\n\nThank you :)",
        "A": "45 minutes * 1 hour/60 minutes = 45/60 hours (since minutes cancel out). Both 45 and 60 are divisible by 15 so you can simplify it to 3/4 (hours). The distance to the gift store is the same as the distance from the gift store since he is taking the same path in both directions, so you can view it as 2 * distance to the gift store.",
        "video_name": "Uc2Tm4Lr7uI",
        "timestamps": [
            178
        ],
        "3min_transcript": "Well, it's time to gift store plus the time coming back from the gift store. Now, we know that the distance to the gift store and the distance back from the gift store is the same. So that's why I just said that the total distance is just going to be two times the distance to the gift store. We don't know-- in fact we know we're going to have different times in terms of times to the gift store and times coming back. How do I know that? Well, she went at different speeds. So it's going to take her-- actually, she went there much slower than she came back. So it would take her longer to get there than it took her to get back. So let's see which of these we can actually-- we already know. So how do we figure out the distance to the gift store? At not point here do they say, hey, the gift store is this far away. But they do tell us, this first sentence right over here, 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."
    },
    {
        "Q": "At 3:05, how do you convert the minutes into hours?",
        "A": "multiply by 60 since there are 60 minutes in an hour. Hope this helps!",
        "video_name": "Uc2Tm4Lr7uI",
        "timestamps": [
            185
        ],
        "3min_transcript": "Well, it's time to gift store plus the time coming back from the gift store. Now, we know that the distance to the gift store and the distance back from the gift store is the same. So that's why I just said that the total distance is just going to be two times the distance to the gift store. We don't know-- in fact we know we're going to have different times in terms of times to the gift store and times coming back. How do I know that? Well, she went at different speeds. So it's going to take her-- actually, she went there much slower than she came back. So it would take her longer to get there than it took her to get back. So let's see which of these we can actually-- we already know. So how do we figure out the distance to the gift store? At not point here do they say, hey, the gift store is this far away. But they do tell us, this first sentence right over here, 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."
    },
    {
        "Q": "in the first example, at 2:11 you have simplified it to -sqrt5/sqrt35, the proceed to rewrite this as -sqrt(5/35) and end with the result of -sqrt(1/7).\n\nWhen solving this on my own, after i reached the step you were at at 2:11, I simplified it to:\n-sqrt5 / (sqrt7)(sqrt5) and cancelled the sqrt5 to end with -1/sqrt(7).\n\nI believe these are both correct, as your example of the whole fraction under the radical could simplify to my form, however which would be considered the most simple answer?",
        "A": "They are both equivalent. It depends on how your instructor wants the answer to be really. Both are simple enough. Later on you ll learn to rationalize the denominator, so you ll most likely require to do that. You simply multiply - 1/\u00e2\u0088\u009a(7) by \u00e2\u0088\u009a(7) / \u00e2\u0088\u009a(7) which will give -\u00e2\u0088\u009a(7)/7. P.S. He mas an error, forgetting to include his negative sign.",
        "video_name": "suwJmCrSDI8",
        "timestamps": [
            131,
            131
        ],
        "3min_transcript": "- [Voiceover] We're asked to simplify the expression by removing all factors that are perfect squares from inside the radicals and combining the terms. So, let's see if we can do it and pause the video and give a-go at it before we do it together. Alright, so let's see how we can re-write these radicals. So, four times the square root of 20. Well, thats the same thing as four times the square root of four, times the square root of five, because 20 is the same thing as four times five. And 45, that's the same thing as nine times five. And the reason why I'm thinking about the four and I'm thinking about the nine is because those are perfect squares, so I could write this as, four times the square root of four times the square root of five. And then I could say, this part right over here is minus three, times the square root of nine, times the square root of five. The square root of 45 is the same thing as the square root of nine times five, which is the same thing as square root of nine times the square root of five, and then all of that is going to be over, 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,"
    },
    {
        "Q": "At 5:30 sal says that the limit of f(c) as x approaches from the negative direction does not exist.\nHow is it?\nWhat does the point over the hole mean?",
        "A": "Think of a simpler formula, where if x < c, then f(x) = 1; and where x >= c then f(x) = 2. If you are determining the limit from the left (ie the lower direction) your limit would equal 1, but the value for c derived from f(x) is actually 2 - thus the limit does not exist, because: lim f(x) (x->c-) = 1 f(c) = 2 -John",
        "video_name": "kdEQGfeC0SE",
        "timestamps": [
            330
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 3:48 I don't understand how the 2 gets under c^2 and disappears from the other side. Is it being added, subtracted, multiplied, or divided?",
        "A": "The 2 dissapears on the left side because the expression is divided by 2 on both sides.",
        "video_name": "tSHitjFIjd8",
        "timestamps": [
            228
        ],
        "3min_transcript": "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. 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"
    },
    {
        "Q": "at 8:21 I don't understand wouldn't you divide instead of multiply?",
        "A": "b/c you re multiplying it by the inverse",
        "video_name": "tSHitjFIjd8",
        "timestamps": [
            501
        ],
        "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": "So at 1:16 I understand that the slope is rise over run and that the line has a negative slope because its going down, but why is going right positive for x and left is negative?",
        "A": "on a graph onthe x axis the right side is positive while the left is negative",
        "video_name": "AqFwKecNaTk",
        "timestamps": [
            76
        ],
        "3min_transcript": "A line has a slope of negative 3/4 and goes through the point 0 comma 8. What is the equation of this line in slope-intercept form? So any line can be represented in slope-intercept form, is y is equal to mx plus b, where this m right over here, that is of the slope of the line. And this b over here, this is the y-intercept of the line. Let me draw a quick line here just so that we can visualize that a little bit. So that is my y-axis. And then that is my x-axis. And let me draw a line. And since our line here has a negative slope, I'll draw a downward sloping line. So let's say our line looks something like that. So hopefully, we're a little familiar with the slope already. The slope essentially tells us, look, start at some point on the line, and go how much you had to move in the x direction, that is your run, and then measure how much you had to move in the y direction, that is your rise. And our slope is equal to rise over run. And you can see over here, we'd be downward sloping. Because if you move in the positive x direction, we have to go down. If our run is positive, our rise here is negative. So this would be a negative over a positive, it would give you a negative number. That makes sense, because we're downward sloping. The more we go down in this situation, for every step we move to the right, the more downward sloping will be, the more of a negative slope we'll have. So that's slope right over here. The y-intercept just tells us where we intercept the y-axis. So the y-intercept, this point right over here, this is where the line intersects with the y-axis. This will be the point 0 comma b. And this actually just falls straight out of this equation. this equation, when x is equal to 0. y will be equal to m times 0 plus b. Well, anything times 0 is 0. So y is equal to 0 plus b, or y will be equal to b, when x is equal to 0. So this is the point 0 comma b. Now, they tell us what the slope of this line is. They tell us a line has a slope of negative 3/4. So we know that our slope is negative 3/4, and they tell us that the line goes through the point 0 comma 8. They tell us we go through the-- Let me just, in a new color. I've already used orange, let me use this green color. They tell us what we go through the point 0 comma 8. Notice, x is 0. So we're on the y-axis. When x is 0, we're on the y-axis."
    },
    {
        "Q": "5:01 what are assymptotes? (I don't think I spelled that right)\nSal keeps mentioning them but I think if I don't know what they are I'm not going anywhere with hyperbolas....",
        "A": "an asymptote is a line that the function gets infinitely close to but never touches. You draw them on your graph before you draw your hyperbola (using dotted lines).",
        "video_name": "pzSyOTkAsY4",
        "timestamps": [
            301
        ],
        "3min_transcript": "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. minus and the b squared go away, and you're just left with y squared is equal to minus b squared. And then minus b squared times a plus, it becomes a plus b squared over a squared x squared. We're almost there. And then you get y is equal to-- and I'm doing this on purpose-- the plus or minus square root, because it can be the plus or minus square root. Of-- and let's switch these around, just so I have the positive term first. b squared over a squared x squared minus b squared. Now you said, Sal, you said this was simple. This looks like a really complicated thing. But remember, we're doing this to figure out asymptotes of the hyperbola, just to kind of give you a sense of where we're going. that other hyperbola. So a hyperbola, if that's the x, that's the y-axis, it has two asymptotes. And the asymptotes, they're these lines that the hyperbola will approach. So if those are the two asymptotes-- and they're always the negative slope of each other-- we know that this hyperbola's is either, and we'll show in a second which one it is, it's either going to look something like this, where as we approach infinity we get closer and closer this line and closer and closer to that line. And here it's either going to look like that-- I didn't draw it perfectly; it never touches the asymptote. It just gets closer and closer and closer, arbitrarily It's either going to look like that, where it opens up to the right and left. Or our hyperbola's going to open up and down. And once again, as you go further and further, and asymptote means it's just going to get closer and closer to one of these lines without ever touching it. It will get infinitely close as you get infinitely far away,"
    },
    {
        "Q": "at 5:15 sal says that the asymptotes are the negative slope of each other. can i say that they are perpendicular lines with the same y-intercept?",
        "A": "yes for the perpendicular but no for same y intercept, because in more complex examples there will be horizontal and vertical shifts which will make the y intercepts of each asymptote different. Try searching for examples of hyperbolas with shifts and see their graphs to understand more.",
        "video_name": "pzSyOTkAsY4",
        "timestamps": [
            315
        ],
        "3min_transcript": "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. minus and the b squared go away, and you're just left with y squared is equal to minus b squared. And then minus b squared times a plus, it becomes a plus b squared over a squared x squared. We're almost there. And then you get y is equal to-- and I'm doing this on purpose-- the plus or minus square root, because it can be the plus or minus square root. Of-- and let's switch these around, just so I have the positive term first. b squared over a squared x squared minus b squared. Now you said, Sal, you said this was simple. This looks like a really complicated thing. But remember, we're doing this to figure out asymptotes of the hyperbola, just to kind of give you a sense of where we're going. that other hyperbola. So a hyperbola, if that's the x, that's the y-axis, it has two asymptotes. And the asymptotes, they're these lines that the hyperbola will approach. So if those are the two asymptotes-- and they're always the negative slope of each other-- we know that this hyperbola's is either, and we'll show in a second which one it is, it's either going to look something like this, where as we approach infinity we get closer and closer this line and closer and closer to that line. And here it's either going to look like that-- I didn't draw it perfectly; it never touches the asymptote. It just gets closer and closer and closer, arbitrarily It's either going to look like that, where it opens up to the right and left. Or our hyperbola's going to open up and down. And once again, as you go further and further, and asymptote means it's just going to get closer and closer to one of these lines without ever touching it. It will get infinitely close as you get infinitely far away,"
    },
    {
        "Q": "At 7:40, Sal got rid of the -b^2, how is that possible?",
        "A": "As x approaches infinity, the b\u00c2\u00b2 term becomes less and less significant. For example, consider x = 1,000,000 and b = 5 .",
        "video_name": "pzSyOTkAsY4",
        "timestamps": [
            460
        ],
        "3min_transcript": "So in order to figure out which one of these this is, let's just think about what happens as x becomes infinitely large. So as x approaches infinity. So as x approaches infinity, or x approaches negative infinity. So I'll say plus or minus infinity, right? It doesn't matter, because when you take a negative, this gets squared. So this number becomes really huge as you approach positive or negative infinity. And you'll learn more about this when we actually do limits, but I think that's intuitive. That this number becomes huge. This number's just a constant. It just stays the same. So as x approaches positive or negative infinity, as it gets really, really large, y is going to be approximately equal to-- actually, I think that's congruent. I always forget notation. Approximately. This just means not exactly but approximately equal to. When x approaches infinity, it's going to be approximately equal to the plus or minus square root of b squared And that is equal to-- now you can take the square root. You couldn't take the square root of this algebraically, but this you can. This is equal to plus or minus b over a x. So that tells us, essentially, what the two asymptotes are. Where the slope of one asymptote will be b over a x. This could give you positive b over a x, and the other one would be minus b over a x. And I'll do this with some example so it makes it a little clearer. But we still know what the asymptotes look like. It's these two lines. Because it's plus b a x is one line, y equals plus b a x. Let's say it's this one. This asymptote right here is y is equal to plus b over a x. I know you can't read that. And then the downward sloping asymptote we could say is y is equal to minus b over a x. So those are two asymptotes. But we still have to figure out whether the hyperbola opens up And there, there's two ways to do this. One, you say, well this is an approximation. This is what you approach as x approaches infinity. But we see here that even when x approaches infinity, we're always going to be a little bit smaller than that number. Because we're subtracting a positive number from this. We're subtracting a positive number, and then we're taking the square root of the whole thing. So we're always going to be a little bit lower than the asymptote, especially when we're in the positive quadrant. So to me, that's how I like to do it. I think, we're always-- at least in the positive quadrant; it gets a little more confusing when you go to the other quadrants-- we're always going to be a little bit lower than the asymptote. So we're going to approach from the bottom there. And since you know you're there, you know it's going to be like this and approach this asymptote. And then since it's opening to the right here, it's also going to open to the left. The other way to test it, and maybe this is more intuitive for you, is to figure out, in the original equation could x or y equal to 0?"
    },
    {
        "Q": "At 4:25 when multiplying b^2, why does the x^2 get moved from the numerator?",
        "A": "x^2 is still part of the numerator - just think of it as x^2/1, multiplied by b^2/a^2. you could also write it as a^2*x^2/b^2, all as one fraction... it means the same thing (multiply x^2 and a^2 and divide by b^2 ->> since multiplication and division occur at the same level of the order of operations, both ways of writing it out are totally equivalent!). hope that helps",
        "video_name": "pzSyOTkAsY4",
        "timestamps": [
            265
        ],
        "3min_transcript": "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. minus and the b squared go away, and you're just left with y squared is equal to minus b squared. And then minus b squared times a plus, it becomes a plus b squared over a squared x squared. We're almost there. And then you get y is equal to-- and I'm doing this on purpose-- the plus or minus square root, because it can be the plus or minus square root. Of-- and let's switch these around, just so I have the positive term first. b squared over a squared x squared minus b squared. Now you said, Sal, you said this was simple. This looks like a really complicated thing. But remember, we're doing this to figure out asymptotes of the hyperbola, just to kind of give you a sense of where we're going."
    },
    {
        "Q": "at 2:30, wouldn't those equations result in the same shape? if not, how are they different?",
        "A": "the equation x2/a2-y2/b2 has the x-axis as the major axis while the equation y2/b2-x2/a2 has the y-axis as the major axis.",
        "video_name": "pzSyOTkAsY4",
        "timestamps": [
            150
        ],
        "3min_transcript": "Let's see if we can learn a thing or two about the hyperbola. And out of all the conic sections, this is probably the one that confuses people the most, because it's not quite as easy to draw as the circle and the ellipse. You have to do a little bit more algebra. But hopefully over the course of this video you'll get pretty comfortable with that, and you'll see that hyperbolas in some way are more fun than any of the other conic sections. So just as a review, I want to do this just so you see the similarity in the formulas or the standard form of the different conic sections. If you have a circle centered at 0, its equation is x squared plus y squared is equal to r squared. And we saw that this could also be written as-- and I'm doing this because I want to show that this is really just the same thing as the standard equation for an ellipse. If you divide both sides of this by r squared, you get x squared over r squared plus y squared over r squared 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."
    },
    {
        "Q": "But when we are graphing g(x)=2-x at 8:00, should't the line be moved more upwards on a y-axis?\nThank you!",
        "A": "If you notice 2 is the y-intercepts in the equation y = -x + 2 in the form y = mx + q",
        "video_name": "rOftmuhGLjY",
        "timestamps": [
            480
        ],
        "3min_transcript": "that's when x is greater than or equal to zero. When the absolute value is non-negative, if you're taking the absolute value of a non-negative number that is just going to be itself. The absolute value of zero, zero. Absolute value of one is one. The absolute value of a hundred is a hundred. Then you could ignore the absolute value for x is greater than or equal to, not greater than or equal to zero, for x is greater than or equal to one. X is greater than or equal to one, this thing right over here is non-negative. It will just evaluate to x minus one. This is going to be x minus one plus one. Which is the same thing as just x, minus one plus one, they just cancel out. Now when this term right over here is negative and that's going to happen for x is less than one. Well then the absolute value You give me the absolute value of a negative number that's going to be the opposite. Absolute value of negative eight is positive eight. It's going to be that the negative of x minus one is one minus x, plus one. Or we could say two minus x. For x is greater or equal to one, we would look at this expression, now what's the slope of that? Well the slope of that is one. We're going to have a curve that looks like or a line I guess we could say that looks like this. For all x is greater than or equal to one. The important thing, remember, we're going to think about the slope of the tangent line when we think about the derivative of g. Slope is equal to one. For x less than one or our slope now, if we look right over here our slope is negative one. It's going to look like this. For the pointing question, if we're thinking about g prime of nine so nine is some place out here, so what is g prime of nine? G prime of nine, let me make it clear, this graph right over here, this is the graph of g of x or we could say y, this is the graph y equals g of x. Y is equal to g of x. What is g prime of nine? Well that's the slope when x is equal to nine. The slope is going to be equal to one. G prime of nine is one. What does this evaluate to? This is going to be three times three, so this part right over here is nine plus two times one, plus two, which is equal to 11. The slope of the tangent line of h when x is equal to nine is 11."
    },
    {
        "Q": "At 3:21, how Sal got 1.5 the original population, did he assumed that or there is some logic behind it.",
        "A": "He said he wanted the population to increase by 50% over 20 years. So, it ll be 150% of the current population, or 1.5.",
        "video_name": "-fIsaqN-aaQ",
        "timestamps": [
            201
        ],
        "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": "Around 4:15 when he found du, why did he keep the ln(2) if it is a constant? Isnt the derivative of a constant 0? Or did he use the product rule?",
        "A": "yes he did use the product rule.",
        "video_name": "1ct7LUx23io",
        "timestamps": [
            255
        ],
        "3min_transcript": "see if we can figure that out. Then we can apply, then we can take, we can evaluate the definite ones. Let's just think about this, let's think about the indefinite integral of x squared times two to the x to the third power d x. I really want to find the anti-derivative of this. Well this is going to be the exact same thing as the integral of, I'll write my x squared still, but instead of two to the x to the third I'm going to write all of this business. Let me just copy and paste that. We already established that this is the same thing as two to the x to the third power. Copy and paste, just like that. Then let me close it with a d x. I was able to get it in terms of e as a base. That makes me a little bit more comfortable but it still seems pretty complicated. You might be saying, \"Okay, look. \"Maybe u substitution could be at play here.\" Because I have this crazy expression, x to the third times Well that's going to be three x squared times the natural log of two, or three times the natural log of two times x squared. That's just a constant times x squared. We already have a x squared here so maybe we can engineer this a little bit to have the constant there as well. Let's think about that. If we made this, if we defined this as u, if we said u is equal to x to the third times the natural log of two, what is du going to be? du is going to be, it's going to be, well natural log of two is just a constant so it's going to be three x squared times the natural log of two. We could actually just change the order we're multiplying a little bit. We could say that this is the same thing as x squared times three natural log of two, which is the same thing just using logarithm properties, as x squared times the natural log of two to the third power. as the natural log of two to the third power. This is equal to x squared times the natural log of eight. Let's see, if this is u, where is du? Oh, and of course we can't forget the dx. This is a dx right over here, dx, dx, dx. Where is the du? Well we have a dx. Let me circle things. You have a dx here, you have a dx there. You have an x squared here, you have an x squared here. So really all we need is, all we need here is the natural log of eight. Ideally we would have the natural log of eight right over here, and we could put it there as long as we also, we could multiply by the natural log of eight as long as we also divide by a natural log of eight. We can do it like right over here, we could divide by natural log of eight. But we know that the anti-derivative of some constant"
    },
    {
        "Q": "At 5:16, Sal puts a negative sign in front of the inverse of ln8. Is it correct that it should be an equal sign? As far as I can tell, the solution works out only if ln8 is not negative.",
        "A": "It actually is an equal sign, but it is very hard to see it the way he wrote it . :)",
        "video_name": "1ct7LUx23io",
        "timestamps": [
            316
        ],
        "3min_transcript": "Well that's going to be three x squared times the natural log of two, or three times the natural log of two times x squared. That's just a constant times x squared. We already have a x squared here so maybe we can engineer this a little bit to have the constant there as well. Let's think about that. If we made this, if we defined this as u, if we said u is equal to x to the third times the natural log of two, what is du going to be? du is going to be, it's going to be, well natural log of two is just a constant so it's going to be three x squared times the natural log of two. We could actually just change the order we're multiplying a little bit. We could say that this is the same thing as x squared times three natural log of two, which is the same thing just using logarithm properties, as x squared times the natural log of two to the third power. as the natural log of two to the third power. This is equal to x squared times the natural log of eight. Let's see, if this is u, where is du? Oh, and of course we can't forget the dx. This is a dx right over here, dx, dx, dx. Where is the du? Well we have a dx. Let me circle things. You have a dx here, you have a dx there. You have an x squared here, you have an x squared here. So really all we need is, all we need here is the natural log of eight. Ideally we would have the natural log of eight right over here, and we could put it there as long as we also, we could multiply by the natural log of eight as long as we also divide by a natural log of eight. We can do it like right over here, we could divide by natural log of eight. But we know that the anti-derivative of some constant times the anti-derivative of that function. We could just take that on the outside. It's one over the natural log of eight. Let's write this in terms of u and du. This simplifies to one over the natural log of eight times the anti-derivative of e to the u, e to the u, that's the u, du. This times this times that is du, du. And this is straightforward, we know what this is going to be. This is going to be equal to, let me just write the one over natural log of eight out here, one over natural log of eight times e to the u, and of course if we're thinking in terms of just anti-derivative there would be some constant out there. Then we would just reverse the substitution."
    },
    {
        "Q": "At 0:52 how is their a 2/3 in 5/3 x 2/5? You aren't multiplying anything by 2/3. I understand there is a 2 and a 3 in the equation but that isn't 2/3 like the fraction.",
        "A": "With the commutative property for multiplication, you can move the factors around within the numerator and denominator, so you can see the 2/3 fraction that way and factor it out 0:51.",
        "video_name": "yUYDhmQsiXY",
        "timestamps": [
            52
        ],
        "3min_transcript": "We have three expressions here. This is 2/3 times 7/8. The second expression is 8/7 times 2/3. This third expression is 5 times 2 over 3 times 5. And what I want you to do is pause this video right now and think about which of these expressions is the largest, which one is in the middle in terms of value, and which one is the smallest. And I want you to think about it without actually doing the calculation. If you could just look at them and figure out which of these is the largest, which of these is the smallest, and which of these is in the middle. So pause the video now. Now, you might have taken a shot at it. And I'll give you a little bit of a hint in case you had trouble with it. All of these involve multiplying something by 2/3. And you see a 2/3 here. You see a 2/3 here. And it might not be as obvious, but you also see a 2/3 here. And let me rewrite that to make it a little bit clearer. So this first expression could be rewritten as 7/8 times 2/3. 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."
    },
    {
        "Q": "At 1:18, are you sure it's 2 by 3 because I'm pretty sure it's 3 by 2?",
        "A": "This way of identifying matrices is arbitrary: we all agree (or: most of us do, I think! :) to name the rows first and the columns second. (The first matrix in the video is 2X3, the second is 3X2.) We (or they) could have reversed the order, but since that s the way others do it, so will I (if I can remember it!).",
        "video_name": "OMA2Mwo0aZg",
        "timestamps": [
            78
        ],
        "3min_transcript": "We're given two matrices over here, matrix E and matrix D. And they ask us, what is ED, which is another way of saying what is the product of matrix E and matrix D? Just so I remember what I'm doing, let me copy and paste this. And then I'm going to get out my little scratch pad. So let me paste that over here. So we have all the information we needed. And so let's try to work this out. So matrix E times matrix D, which is equal to-- matrix E is all of this business. So it is 0, 3, 5, 5, 5, 2 times matrix D, which is all of this. So we're going to multiply it times 3, 3, 4, 4, negative 2, negative 2. Now the first thing that we have to check is whether this is even a valid operation. Now the matrix multiplication is a human-defined operation that happen to have [? new ?] properties. Now the way that us humans have defined matrix multiplication, it only works when we're multiplying our two matrices. So this right over here has two rows and three columns. So it's a 2 by 3 matrix. And this has three rows and two columns, it's 3 by 2. This only works-- we could only multiply this matrix times this matrix, if the number of columns on this matrix is equal to the number of rows on this matrix. And in this situation it is, so I can actually multiply them. If these two numbers weren't equal, if the number of columns here were not equal to the number of rows here, then this would not be a valid operation, at least the way that we have defined matrix multiplication. The other thing you always have to remember is that E times D is not always the same thing as D times E. Order matters when you're multiplying matrices. It doesn't matter if you're multiplying regular numbers, but it matters for matrices. But let's actually work this out. by 2 matrix. But I'm going to create some space here because we're going to have to do some computation. So this is going to be equal to-- I'm going to make a huge 2 by 2 matrix here. So the way we get the top left entry, the top left entry is essentially going to be this row times this product. If you view them each as vectors, and you have some familiarity with the dot product, we're essentially going to take the dot product of that and that. And if you have no idea what that is, I'm about to show you. This entry is going to be 0 times 3, plus 3 times 3, plus 5 times 4. So that is the top left entry. And I already see that I'm going to run out of space here, so let me move this over to the right some space so I have some breathing room."
    },
    {
        "Q": "At 3:36, Sal mentions that there is a positive derivative and then zero and then a negative derivative after the critical point x=2. But the graph has a negative slope. How is it that it's positive then?",
        "A": "The graph is of the derivative not of the original function. This means that when the derivative is positive the function has a positive slope and when the derivative is negative the original function has a negative slope. And the derivative is equal to zero at the function s critical points of +2 and -2.",
        "video_name": "pInFesXIfg8",
        "timestamps": [
            216
        ],
        "3min_transcript": "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 So a minimum. And I just want to make sure we have the correct intuition. If our function, if some function is increasing going into some point, and at that point we actually have a derivative 0-- the derivative could also be undefined-- but we have a derivative of 0 and then the function begins decreasing, that's why this would be a maximum point. Similarly, if we have a situation where the function is decreasing going into a point, the derivative is negative. Remember this is the graph of the derivative. Let me make this clear. This is the graph of y is equal to not f of x, but f prime of x. So if we have a situation we're going into the point, the function has a negative slope, we see we have a negative slope here-- so the function might look something like this. And then right at this point the function is either undefined or has 0 slope. So in this case it has 0 slope."
    },
    {
        "Q": "Other than visually (3:44) how do you know if the derivative is positive or negative before the critical point.",
        "A": "The simple way to do that is to pick a convenient point just before and just after the critical point and plug those values into the first derivative to see whether it is negative or positive.",
        "video_name": "pInFesXIfg8",
        "timestamps": [
            224
        ],
        "3min_transcript": "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 So a minimum. And I just want to make sure we have the correct intuition. If our function, if some function is increasing going into some point, and at that point we actually have a derivative 0-- the derivative could also be undefined-- but we have a derivative of 0 and then the function begins decreasing, that's why this would be a maximum point. Similarly, if we have a situation where the function is decreasing going into a point, the derivative is negative. Remember this is the graph of the derivative. Let me make this clear. This is the graph of y is equal to not f of x, but f prime of x. So if we have a situation we're going into the point, the function has a negative slope, we see we have a negative slope here-- so the function might look something like this. And then right at this point the function is either undefined or has 0 slope. So in this case it has 0 slope."
    },
    {
        "Q": "At 10:40 Sal said that t does not have to be time, so what other things can t be?",
        "A": "Maybe you re trying to describe the path of a complicated music box dancer that moves as you turn a crank. In that case, the independent variable might not represent time but the number of rotations that you gave the crank from its resting place, but traditionally in parametric equations we d call it t even though it isn t about time in that case.",
        "video_name": "57BiI_iD3-U",
        "timestamps": [
            640
        ],
        "3min_transcript": "Cosine of pi over 2 is 0. 0 times 3 is 0. And what's x equal when t is equal to pi? Cosine of pi is minus 1. Minus 1 times 3 is minus 3. Fair enough. Now let's do the y's. When t is 0 what is y? Sine is 0, 0. So 2 times 0 is 0. When t is pi over 2, sine of pi over 2 is 1. 1 times 2 is 2. And when t is pi, sine of pi-- that's sine of 180 degrees-- that's 0. 2 times 0 is 0. So let's plot these points. When time is 0, we're at the point 3, 0. So 3, 0-- 3, 0 is right there. This is t equals 0. When t increases by pi over 2, or if this was seconds, pi over So at t equals pi over 2, we're at the point 0, 2. We're right over here. So this is at t is equal to pi over 2. And then when t increases a little bit more-- when we're at t is equal to pi-- we're at the point minus 3, 0. We're here. So this is t is equal to pi or, you know, we could write 3.14159 seconds. 3.14 seconds. And actually, you know, I want to make the point, t does not have to be time, and we don't have to be dealing with seconds. But I like to think about it that way. I like to think about, maybe this is describing some object in orbit around, I don't know, something else. So now we know the direction. As t increased from 0 to pi over 2 to pi, we went this way. We went counterclockwise. So the direction of t's parametric equations And you might be saying, Sal, you know, why'd we have to do 3 points? We could have just done 2, and made a line. If we just had that point and that point, you might have immediately said, oh, we went from there to there. But that really wouldn't have been enough. Because maybe we got from here to there by going the other way around. So giving that third point lets us know that the direction is definitely counterclockwise. And so what happens if we just take t from 0 to infinity? What happens if we bound t? t is greater than 0 and less than infinity. Well, we're just going to keep going around this ellipse forever. Multiple times. Keep writing over and over, infinite times. If we went from minus infinity to infinity, then we would have always been doing it, I guess is the way to put it. Or if we just wanted to trace this out once, we could go from t is less than or equal to-- or t is greater than or equal to 0. All the way to t is less than or equal to 2 pi."
    },
    {
        "Q": "in 10:24 when Sal uses pi and pi/2 for his time, is there a reason why he couldn't use 90 or 180?",
        "A": "Khan is using angles in radians probably because it is more intuitive rather than in degrees. cos(90\u00c2\u00ba) = 0 = cos(Pi/2) Both expressions are equivalent (just mind the angle units)",
        "video_name": "57BiI_iD3-U",
        "timestamps": [
            624
        ],
        "3min_transcript": "So to do that, let's make our little table. So let's take some values of t. So we'll make a little table. t, x, and y. It's good to pick values of t. Remember-- let me rewrite the equations again, so we didn't lose it-- x was equal to 3 cosine of t, and y is equal to 2 sine of t. It's good to take values of t where it's easy to figure out what the cosine and sine are, and without using a calculator. We're assuming the t is in radiance, just for simplicity. So let's pick t is equal to 0. t is equal to pi over 2. That's 90 degrees in degrees. And t is equal to pi. And so what is x when t is equal to 0? Well, cosine of 0 is 1 times 3, that's 3. Cosine of pi over 2 is 0. 0 times 3 is 0. And what's x equal when t is equal to pi? Cosine of pi is minus 1. Minus 1 times 3 is minus 3. Fair enough. Now let's do the y's. When t is 0 what is y? Sine is 0, 0. So 2 times 0 is 0. When t is pi over 2, sine of pi over 2 is 1. 1 times 2 is 2. And when t is pi, sine of pi-- that's sine of 180 degrees-- that's 0. 2 times 0 is 0. So let's plot these points. When time is 0, we're at the point 3, 0. So 3, 0-- 3, 0 is right there. This is t equals 0. When t increases by pi over 2, or if this was seconds, pi over So at t equals pi over 2, we're at the point 0, 2. We're right over here. So this is at t is equal to pi over 2. And then when t increases a little bit more-- when we're at t is equal to pi-- we're at the point minus 3, 0. We're here. So this is t is equal to pi or, you know, we could write 3.14159 seconds. 3.14 seconds. And actually, you know, I want to make the point, t does not have to be time, and we don't have to be dealing with seconds. But I like to think about it that way. I like to think about, maybe this is describing some object in orbit around, I don't know, something else. So now we know the direction. As t increased from 0 to pi over 2 to pi, we went this way. We went counterclockwise. So the direction of t's parametric equations"
    },
    {
        "Q": "3:20 Commutative property of addition 704 = 700 + 4 or 4 + 700\n3:30 Associative property of addition 18 + (4 + 700) = (18 + 4) + 700",
        "A": "that is 3th grade math for asossitive property and 6th grade math for comunitive property",
        "video_name": "jAfJcgPGqgI",
        "timestamps": [
            200,
            210
        ],
        "3min_transcript": "to get to 500? And I could have done that in my head. Okay, I need to add 20. 355 minus 20 is 335. But now this problem is much, much simpler to compute. 335 plus 500, well, it's going to be three hundreds plus five hundreds, it's going to be equal to 800 and then we have 35. 835. Now, to make it a little bit clearer what we did, remember, we wanted to take 20 from here and put it over here, so we could break up 355, we could say that it's going to be equal to, it's going to be equal to, let's break it up into 335 and 20, and remember, the whole reason why I picked 20 is because I'm going to want to add that to 480, but I'm just doing it step-by-step here, so plus 480, and now I can just change the order with which I add, 335 plus 20 plus 480, and instead of adding the 335 and 20 first, I could add the 20 and 480 first, so I could add these two characters first, and so then I'm going to be left with, this is going to be equal to 335 plus, what's 20 plus 480? Well, that was the whole point. I picked 20 so that I can get to 500. 20 plus 480 is going to be 500 and then, now, you can add them. This is going to be 835. So this is just a longer way of saying what I did here. I took 20 from 355, so I can make the 480 into a 500. Let's do a couple more examples of this, and remember, the key is just thinking about how could I add or take away from one of the numbers to make them simpler? So there's a couple ways we could tackle this. One way, we could try to get the 704 to be equal to 700. So, we could say that this is the same thing as 18 plus, I could write 700 plus 4, or I could write it as 4 plus 700, like that, and then I could put the parentheses around this first, and then I could just switch the order in which I add, so this is the same thing as 18 plus 4 plus 700, and now I could add the 18 and the 4 first. Now what's 18 plus 4? It's 22, and then I have plus 700, and now this is pretty easy to compute, and all of these are going to be equal to each other, so let me just write it like that. 22 plus 700, I could do that in my head."
    },
    {
        "Q": "At 2:18 talking about the passage from the purple curve to the yellow segment of the function, he said a slope is not defined there because we could draw a lot.\nYet a unique limit for that point exists, so it should also exist a derivative right?\n(The same happens between the blue and the orange segments at the end.)",
        "A": "just because a limit exists does not mean that a function is differentiable, although it is one of the conditions of that. For a function to be differentiable, the derivative from the left side of the point must be equal to the derivative from the right using one sided limits.",
        "video_name": "eVme7kuGyuo",
        "timestamps": [
            138
        ],
        "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": "at 5:00 how slope of that line is constant, that line rising up so it shouldn't be constant :S",
        "A": "The slope of any line as always constant! I can show you algebraically: d/dx (ax+b) = a which is a constant. I can also explain it graphically: The slope of a line is always constant. Therefore, its tangent line is also of constant slope. Even if a line rises, the slope is constant. A rising line simply means a positive slope.",
        "video_name": "eVme7kuGyuo",
        "timestamps": [
            300
        ],
        "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": "At 3:55, the red colored line is increasing, but is in f(x)<0. So, I get that it will have a constant f'(x) slope, but shouldn't it be in f'(x)<0.",
        "A": "Yes, f(x) is negative, but f (x) (or F Prime) will be positive, since it is essentially the slope of the line and the slope at that point is positive.",
        "video_name": "eVme7kuGyuo",
        "timestamps": [
            235
        ],
        "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": "At 2:15 it says the numbers are equivalent ,how is that?",
        "A": "As I understand, it means that both formulas will give you the same number; so it doesn t matter how you go about finding the area, either with formula a or b, both will provide the same result and therefore are equivalent.",
        "video_name": "Q3wfb0CPhIY",
        "timestamps": [
            135
        ],
        "3min_transcript": "I have this rectangle here, and I want to figure out its area. I want to figure out how much space it is taking up on my screen right over here. And I encourage you to pause this video and try to figure out the area of this entire rectangle. And when you do it, think about the two different ways you could do it. You could either multiply the length of the rectangle times the entire width, so just figure out the area of the entire thing. Or you could separately figure out the area of this red or this purple rectangle and then separately figure out this blue rectangle and realize that their combined areas is the exact same thing as the entire rectangle. So I encourage you to pause the video and try both of those strategies out. So let's just try them out ourselves. First, let's look at the overall dimensions of the larger rectangle. The length is 9, and we're going to multiply that times the width. But what's width here? Well, the width is going to be 8 plus 12. This entire distance right over here is 8 plus 12. This is one way that we could figure out the area of this entire thing. This is just the length times the width. 8 plus 12 is obviously going to be equal to 20. But the other way that we could do it-- and this must be equivalent, because we're figuring out the area of the same thing-- is to separate out the area of these two sub-rectangles. So let's do that. And this must be equal to this thing. So what's the area of this purple rectangle? Well, it's going to be the length. It's going to be 9. Let me do it in that same color. It's going to be 9 times the width, which is 8. It's going to be 9 times 8. And then what's the area of this the blue rectangle? Well, that's going to be 9 times-- so the height here is 9 still. The height is 9. And what's the width? Well, the width is 12. And what's the area of the combined if you wanted to combine the area of the purple rectangle Well, you'd just add these two things together. And of course, when you add these two things together, you get the area of the entire thing. So these things must be equivalent. They are calculating the same area. Now, what's neat about this is we just showed ourselves the distributive property when we're dealing with these numbers. You could try these out for any numbers. They'll work for any numbers, because the distributive property works for any numbers. You see 9 times the sum of 8 plus 12 is equal to 9 times 8 plus 9 times 12. We essentially have distributed the 9-- 9 times 8 plus 9 times 12. And let's actually calculate it just to satisfy ourselves about the area. So if you multiply the length times the entire width, so that's 9 times 8 plus 12, that's the same thing as 9 times 20, which is 180."
    },
    {
        "Q": "At 2:03, what is the reciprocal?",
        "A": "Reciprocal means to turn the fraction upside-down. For example, the reciprocal of 3/4 is 4/3. The reciprocal of 3 is 1/3, because 3 is really 3/1. Also, if you multiply any number by its reciprocal, you will get one (which I think is what nguyensongthienphuc was saying). So 4/3 * 3/4 = 1.",
        "video_name": "K2b8iMPY11I",
        "timestamps": [
            123
        ],
        "3min_transcript": "- What I hope to do in this video is emphasize the relation, the connection, between fractions and division and then using that knowledge to help us simplify some hairy looking fractions. So, let's say, let's just do a little bit of a review. So, if I say two divided by, two divided by three, two divided by three, I could write this as, I could write this as two over three. Two divided by, two divided by three. So, this expression here, sometimes we might say, \"Hey, this is 2/3. Think of it as a fraction.\" But you can also view this as two divided by three. So, if we went the other way around. If someone were to write four over, actually, let me just do it in a different color, if someone were to say four, 4/9, four, four over nine, we could interpret this, this could be interpreted as four divided by nine. Four divided by nine. this into a decimal that's exactly what we would do. We would calculate what four divided by nine is. We would divide nine into four. Four divided by nine. So, this is all review. So, let's just use this knowledge and I'll think more complex fractions. So, if someone were to say, actually, let me just take this example right over here, if someone were to say two over, instead of saying two over three. If they said two over, I don't know, let me say two over 2/3. Two over 2/3. Well, what would this simplify to? Well, we could go the other way around. This is the same thing as two divided by 2/3. So, let's write it like that. This is the same thing as two divided by, two divided by 2/3, two divided by 2/3, which is the same thing as, dividing by a fraction's the same thing 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,"
    },
    {
        "Q": "Again 11:53:\nHow do I get from [x]B to [Ax]B ?? that is [x]B = [Ax]B algebraically?",
        "A": "Sal is not saying that [Ax]_B = [x]_B. He wrote that D[x]_B = [Ax]_B. We have the rule that some vector v can be expressed in alternative coordinate systems by: C [v]_B = v, and [v]_B = C^-1 v. Ax is some vector. Therefore, we can apply the rule to it. x is also some vector. Therefore, we can apple the rule to it.",
        "video_name": "PiuhTj0zCf4",
        "timestamps": [
            713
        ],
        "3min_transcript": "Let's see what D times xB is equal to. So let's say if we take D times xB, so this thing right here should be equal to D times the representation or the coordinates of x with respect to the basis B. That's what we're claiming. We're saying that this guy is equal to D times the representation of x with respect to the coordinates with respect to the basis B. Let me write all of this down. I'll do it right here, because I think it's nice to have this graphic up here. So we can say that D times xB is equal to this thing right here. It's the same thing as the transformation of x represented in coordinates with respect to B, or in these nonstandard coordinates. represented in this coordinate system, represented in coordinates with respect to B. We see that right there. But what is the transformation of x? That's the same thing as A times x. That's kind of the standard transformation if x was represented in standard coordinates. So this is equal to x in standard coordinates times the matrix A. Then that will get us to this dot in standard coordinates, but then we want to convert it to these nonstandard coordinates just like that. Now, if we have this, how can we just figure out what the vector Ax should look like? What this vector should look like? Well, we can look at this equation right here. We have this. This is the same thing as this. we want to go the other way. We have this. We have that right there. That's this right there. We want to get just this dot represented in regular standard coordinates. So what do we do? We multiply it by C. Let me write it this way. If we multiply both sides of this equation times C, what do we get? We get this right here. I was looking at the right equation the first time. We have this right here, which is the same-- first intuition is always right. We have this, which is the same thing as this right here. So this can be rewritten. This thing can be rewritten as C inverse-- we don't have an x here. We have an Ax here, so C inverse times Ax. The vector Ax represented in these nonstandard coordinates"
    },
    {
        "Q": "At 4:31, is it really possible that there are problems like-(-(-2))?",
        "A": "Yes it is possible, but you don t see it much.",
        "video_name": "3-aryZYsoxU",
        "timestamps": [
            271
        ],
        "3min_transcript": "but let's say that A is some number that is right over here. Well negative A is going to be the opposite of A. So if A is three tick marks to the right negative A is going to be three tick marks to the left. One, two, three. And so the opposite of A is going to be this value right over here. And we can write that, we can write opposite of A over here. We could literally write opposite, opposite of A is that number right over there. Or as a shorthand, we can just write-- We can just write negative. We can just write this is negative A right over here. Negative A. So with that in mind, if we literally view this negative symbol as meaning the opposite of whatever this is, let's try something interesting. Let me do this in another color. What would be the negative of negative three? And I encourage you to pause the video and think about it. Well we just said, this negative means the opposite. So you can think about this as meaning this means the opposite... Opposite of negative-- The opposite of negative three. So what is the opposite of negative three? Well negative three is three to the left of zero. One, two, three. So it's opposite is going to be three to the right of zero. One, two, three. So it's going to be positive three. So this is equal to positive three. Or we could just write positive three like that. So hopefully this gives you a better appreciation for what opposite means actual negative symbol. We could keep going. We could do something like what is the negative of the negative of negative-- Let me do a different number. Of negative two. Well, this part right over here the negative, the opposite of negative two, which is really the opposite of the opposite of two, well that's just going to be two. Every time you say opposite you flip over the number line. So this flips you over the number line two to the left and this flips you back two to the right so all of this is going to be two but then you're going to take the opposite of two so that's just going to be negative two. If you threw another negative in front of this, it would be the opposite of all of this. So it would be the opposite of negative two and then all of a sudden it would become positive. So every time you put that negative in front there you're flipping on the other side of the number line."
    },
    {
        "Q": "what is the angle is an obtuse angle and the protractor isnt big enough to measure it? D:If you dont understand, 3:00\nwell the angle there is acute, but it LOOKS like its obtuse, and then you cant read the angle! D:",
        "A": "you measure whats missing instead. then subtract it from 360. use the protractor to extend a ray so you have a strait line and measure the angle you created and subtract that from 360 to find it in degrees",
        "video_name": "dw41PMWek6U",
        "timestamps": [
            180
        ],
        "3min_transcript": "Let me just keep rotating it. If I could just keep it pressed. That's better. All right. That looks about right. So one side is at the 0 mark. And then my angle, my other side-- or if this was a ray, it points to, looks like, pretty close to the 20 degree mark. So I will type that in off the screen. You don't see that. And that is the right answer. And then we can get another angle. So let's try to measure this one right over here. So once again, place the center of the protractor at the center, at the vertex, of our angle. We can place the 0 degree, the base of the protractor, at this side of the angle. So let's just rotate it a little bit, maybe one more time. That looks about right. And then the angle is now opening up-- let's see, the other side is pointing to 110 degrees. So this is larger than 90 degrees. It's also an obtuse angle. This is obtuse, 110 degrees. More than 90 degrees. So let me type it in. I got the right answer. Let's do a couple more of these. So once again, put the center of the protractor at the vertex of our angle. And now, I want to rotate it. There we go. And this looks like roughly an 80 degree angle, not quite. If I have to be really precise, it looks like it's maybe 81 or 82 degrees. But I'll just go with 80 as my best guess. I got the right answer. Let's do one more of these. So once again, vertex of my angle at the center of my protractor. And then I want to put one side of the angle at the 0 degree. And I want to show you, there's two ways to do that. You could do this. You could do just this. But this isn't too helpful, because the angle is now outside. The other side sits outside of the protractor. So you want the 0 degrees on the side, So let's keep rotating it. There we go. And then our other side opens up or you could say points to 70 degrees. So this is an acute angle right over here. So it is 70 degrees. So I'll leave you with that. Oh look, I'm ready to move on, the exercise tells me. And now we can start talking about more things about angles now that we know how to measure them."
    },
    {
        "Q": "@ 1:09\nWe get to the expression\n-2.7+ -5\nWhat's one thing that tells us not to solve 7-5 first but multiply the first two values and only then take the difference?\nI'd be very thankful",
        "A": "PEMDAS rules always apply. Multiplication is always before subtraction. FYI, your expression as written contains no multiplication. The dot for multiplication has to be raised up: -2 * 7 + (-5). In your expression, you just have a decimal number -2.7 - 5 = -7.7",
        "video_name": "uaPm3Tpuxbc",
        "timestamps": [
            69
        ],
        "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 2:12 he said that he is going to give it to the 10 hundreds. But are you going to give it to the hundreds or tens?",
        "A": "It said he meant 10 tens",
        "video_name": "QOtam19NQcQ",
        "timestamps": [
            132
        ],
        "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": "At around 20:08, Sal denotes that P_1 is a vector and only draws a half arrow, (though I assume he meant to draw a full arrow).\nIs there a significant notational different between a full arrow and a half arrow above a variable?",
        "A": "There is no difference. Some people, myself included, often find themselves writing half arrows , as you labeled them, instead of full arrows , because they are faster to write.",
        "video_name": "hWhs2cIj7Cw",
        "timestamps": [
            1208
        ],
        "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": "At 14:00 why did Sal do b-a instead of b+a ? I thought you had to add vectors together to get the resultant vector. Can someone clarify this for me ?",
        "A": "I have the same question. I don t have a good intuitive answer. But if you plot the vectors mentioned in the video we can see that a-b or b-a is the only vector that passes through the tip of the 2 vectors.",
        "video_name": "hWhs2cIj7Cw",
        "timestamps": [
            840
        ],
        "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 21:30, he mentions that P2 can sub in for P1, and earlier he mentioned that (P1 - P2) can also be (P2-P1). I understand all of that, but can they both sub in at once without any consequences? Can you have (P1 + t(P1 - P2)) and (P1 + t(P2-P1))? And vice versa with P2 as well.",
        "A": "Are you asking if P1 + t(P1 - P2) = P1 + t(P2-P1)? Because if so, the answer is no, the two t s would be different at each point on the line it creates depending on which point is in each position in the equation. If you separate it into two equations using the same points and different t s, then the lines will overlap at every point, but having two equations for the same line is redundant.",
        "video_name": "hWhs2cIj7Cw",
        "timestamps": [
            1290
        ],
        "3min_transcript": "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. So that thing is that vector. And so, our line can be described as a set of vectors, that if you were to plot it in standard position, it would be this set of position vectors. It would be P1, it would be-- let me do that in green-- it would be minus 1, 2, 7. I could've put P2 there, just as easily-- plus t times minus 1, minus 1, 3, where, or such that, t is a member of the real numbers. Now, this also might not be satisfying for you. You're like, gee, how do I plot this in three dimensions? Where's my x, y's, and z's? And if you want to care about x, y's, and z's, let's say that this is the z-axis."
    },
    {
        "Q": "At 0:42why did Sal write 4000+500=3000",
        "A": "Look at the original problem at the top of the screen. Sal is not saying 4000+500=3000. He is writing the problem out using numbers. The original problem as + ? that Sal has not yet written in. But, by the end of the video, he does.",
        "video_name": "a_mzIWvHx_Y",
        "timestamps": [
            42
        ],
        "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": "@ 7:10. Wouldn't x = 2 also cause the answer to be undefined? (2-2)(2+1)=(0)(3)=0?",
        "A": "Yes, but since we re not canceling the (x -2), it s not necessary to include that as a condition.",
        "video_name": "7Uos1ED3KHI",
        "timestamps": [
            430
        ],
        "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 7:19 how did you come up with negative one? is it opposite from x+1? How do you come up with what is not defined?",
        "A": "The fraction is not defined where its denominator goes to zero. You are correct that you set: x+1=0 to see what value of x would cause a problem. x= -1 would make the denominator = 0, so x cannot equal -1.",
        "video_name": "7Uos1ED3KHI",
        "timestamps": [
            439
        ],
        "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": "could you just divided the 3x^2+3x-18 by 3? I was very confused by Khan's whole \"grouping\" process at 10:48",
        "A": "You could divide the whole expression by 3, and then factor from there. However, Sal is trying to explain another way that can be used to factor. On some trinomials, it is difficult to find the correct factors by guessing. A more organized approach is to use grouping.",
        "video_name": "7Uos1ED3KHI",
        "timestamps": [
            648
        ],
        "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": "Why was there no condition in the first problem (at 2:28) ? Shouldn't it have a condition that states x is not equal to -1/3? If x = -1/3, then the denominator would be equal to zero.",
        "A": "wait until you finish the video until commenting",
        "video_name": "7Uos1ED3KHI",
        "timestamps": [
            148
        ],
        "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 6:30 what does he mean?",
        "A": "He is doing what is called FOIL. in the problem (x+5)(x+1), you first multiply the x in both parentheses, then the x in the first and the 1 in the second, then the 5 and the x and then the 5 and 1. the FOIL means: F-first (the two x s) O-outside (the x and the 1) I-inside (5 and x) L-last (5 and 1) Hope this helps clear things a little!",
        "video_name": "7Uos1ED3KHI",
        "timestamps": [
            390
        ],
        "3min_transcript": "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 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"
    },
    {
        "Q": "At 7:28 how come it is x does not equal 1 instead of x does not equal 1 AND 2?",
        "A": "You are correct that x\u00e2\u0089\u00a01 and x\u00e2\u0089\u00a02 and you can list that if you like. However, it is traditional not to list any that remain obvious. But, it is not incorrect to list both of them. I personally prefer to list all of them, despite the tradition, just because I might do some later calculations that makes it difficult to tell where the forbidden values of x are.",
        "video_name": "7Uos1ED3KHI",
        "timestamps": [
            448
        ],
        "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 7:28, (x+5)/(x-2) is not defined for x = -1, but is it defined for x=2?",
        "A": "I believe that when we are deciding which term will make a problem undefined we usually go with the one we are dividing by. So if your are canceling out an (x-1) from the numerator and denominator, even if there is another term present, that is the one that cannot be 0 because it is the one that the problem is divisible by.",
        "video_name": "7Uos1ED3KHI",
        "timestamps": [
            448
        ],
        "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 1:49 how do u get 3(3x+1)",
        "A": "Sal starts with 9x+3 in the numerator of the fraction. He used the distributive property to factor out the GCM = 3. 9x + 3 = 3 ( 9x/3 + 3/3) = 3 (3x + 1). Hope this helps.",
        "video_name": "7Uos1ED3KHI",
        "timestamps": [
            109
        ],
        "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 7:15, Sal said that x cannot equal -1. However, x also cannot be equal to 2, based off of the answer he got. Why wouldn't you put x cannot equal negative 2 in the final answer, as that would also make the expression undefined?",
        "A": "Now that the expression has been simplified to (x+5)(x-2), it is obvious that x cannot be equal to 2, but the fact that x could also not be equal to -1 (otherwise division by 0 resulted) in the original non simplified expression has been lost, so we add the condition as a reminder. You see, we can set x=-1 into (x+5)(x-2) without a problem, but we could not set x=-1 in the original expression - and this simplified expression is based on that.",
        "video_name": "7Uos1ED3KHI",
        "timestamps": [
            435
        ],
        "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 10:45... could you also write (3x-6)(x+3) as 3(x-2)(x+3), or is my thinking quite erred?",
        "A": "You sure can. No, you re right. Take the 3 out of the 3x-6 to get 3(x-2)",
        "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": "How do we do it from slope-intersect form? 00:59",
        "A": "Starting from 5x + 3y = 7, you subtract 5x and divide by 3, so 3y = - 5x + 7 or y = -5/3x + 7/3. He graphed 7/3 as the y intercept, then go down 5 over 3 to get second point. These are not very neat numbers to work with.",
        "video_name": "MRAIgJmRmag",
        "timestamps": [
            59
        ],
        "3min_transcript": "Solve the system of linear equations by graphing, and they give us two equations here. 5x plus 3y is equal to 7, and 3x minus 2y is equal to 8. When they say, \"Solve the system of linear equations,\" they're really just saying find an x and a y that satisfies both of these equations. And when they say to do it by graphing, we're essentially going to graph this first equation. Remember, the graph is really just depicting all of the x's and y's that satisfy this first equation, and then we graph the second equation that's depicting all of the x's and y's that satisfy that one. So if we were looking for an x and a y that satisfies both, that point needs to be on both equations or it has to be on both graphs. So it'll be the intersection of the two graphs. So let's try to see if we can do that. So let's focus on this first equation, and I want to graph it. So I have 5x plus 3y is equal to 7. There's a couple of ways we could graph this. We could put this in slope-intercept form, You just really need two points to graph a line. So let me just set some points over here. Let's say x and y. When x is equal to 0, what does y need to be equal to? So when x is equal to 0, we have-- let me do it over here-- we have 5 times 0, plus 3 times y, is equal to 7. That's just 0 over there. So you have 3y is equal to 7. Divide both sides by 3, you get y is equal to 7/3, which is the same thing as 2 and 1/3 if we want to write it as a mixed number. Now let's set y equal to 0. So if we set y as equal to 0, we get 5x, plus 3 times 0, is equal to 7. Or in this part right over here, it just becomes 0. So we have 5x is equal to 7. Divide both sides by 5, and we get So let's graph both of these points, and then we should be able to graph this line, or at least a pretty good approximation of that line. So we have the point, 0, 2 and 1/3. So that's that point right over there. So I'll call it 0, 7/3 right over there, and then we have the point, 7/5, 0, or 1 and 2/5, 0. So 1 and 2/5. 2/5 is a little less than a half. So 1 and 2/5, 0. So our line is going to look something like this. I just have to connect the dots. It's always hard to draw the straight line. I'll draw it as a dotted line. So it would look something like this. Normally, when you have to solve a system of equations"
    },
    {
        "Q": "at 1:30 why does 1/2 B become 1B, but H does not become 2H?",
        "A": "The commutative law lets us rearrange the right hand side anyway we want. So, it becomes: (2)(1/2)(bh) 2 * 1/2 => 2/2 => 1 So, we end up with: 1*bh which is bh and now we ve got rid of the 1/2 by moving it to the left hand side.",
        "video_name": "eTSVTTg_QZ4",
        "timestamps": [
            90
        ],
        "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": "At 3:53, Where did you get the y=2(2) from, please explain",
        "A": "Because in the Magenta equation we said X=2 you can bring that over into the other equation as X. This only works in systems though",
        "video_name": "GWZKz4F9hWM",
        "timestamps": [
            233
        ],
        "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": "When sal added the two 7x at 2:09, together aren't you supposed to add also the exponents?",
        "A": "No, he is just combining like terms. If you have 7 apples and add 7 apples, you get 14 apples not 14 squared apples. You add exponents when you multiply variables, so (7x)2 = (7x)(7x) = 49 x^2. or 5x * 3x^2 = 15 x^3.",
        "video_name": "xH_GllPuymc",
        "timestamps": [
            129
        ],
        "3min_transcript": "- [Voiceover] Let's see if we can figure out what x plus seven, let me write that a little bit neater, x plus seven squared is. And I encourage you to pause the video and work through it on your own. Alright, now let's work through this together. So we just have to remember, we're squaring the entire binomial. So this thing is going to be the same thing as: x plus seven times x plus seven. I'm gonna write the second x plus seven in a different color, which is going to be helpful when we actually multiply things out. When we see it like this, then we can multiply these out the way we would multiply any binomials. And I'll first do it the, I guess you can say, the slower way, but the more intuitive way, applying the distributive property twice. And then we'll think about maybe some shortcuts or some patterns we might be able to recognize, especially when we are squaring binomials. So let's start with just applying the distributive property twice. So let's distribute this yellow x plus seven over this magenta x plus seven. So we can multiply it by the x, this magenta x, So it's going to be magenta x times x plus seven plus magenta seven times yellow x plus seven. X plus seven, and now we can apply the distributive property again. We can take this magenta x and distribute it over the x plus seven. So x times x is x-squared. X times seven is seven x. And then we can do it again over here. This seven, let me do it in a different color, so this seven times that x is going to be plus another seven x and then the seven times the seven is going to be 49. And we're in the home stretch. We can then simplify it. This is going to be x-squared and then these two middle terms we can add together. seven x plus seven x is going to be 14 x plus 14 x plus 49. Plus 49. And we're done. Now the key question is do we see some patterns here? Do we see some patterns that we can generalize and that might help us square binomials a little bit faster in the future? Well, when we first looked at just multiplying binomials, we saw a pattern like x plus a times x plus b is going to be equal to x-squared, let me write it this way, is going to be equal to x-squared plus a plus b x plus b-squared. And so, if both a and b are the same thing, we can say that x plus a"
    },
    {
        "Q": "At 14:00 could not SQ RT 39/3 be SQ RT 13 ?",
        "A": "sqrt(39)/3 is different from sqrt(39/3). Sqrt(39)/3 means square root of 39 is then divided by 3, while sqrt(39/3) means square root of the value that is 39/3, or 13.",
        "video_name": "i7idZfS8t8w",
        "timestamps": [
            840
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 9:45 why can we not square the negative root thereby getting (36 +- 84)/36?",
        "A": "That is not the square of (6 \u00c2\u00b1 \u00e2\u0088\u009a\u00e2\u0088\u009284)/6 For reference: (n + \u00e2\u0088\u009ap)\u00c2\u00b2 = n\u00c2\u00b2+ p +2n\u00e2\u0088\u009ap (n \u00e2\u0088\u0092 \u00e2\u0088\u009ap)\u00c2\u00b2 = n\u00c2\u00b2+ p \u00e2\u0088\u0092 2n\u00e2\u0088\u009ap [n + \u00e2\u0088\u009a(\u00e2\u0088\u0092p)]\u00c2\u00b2= n\u00c2\u00b2\u00e2\u0088\u0092 p +2n\u00f0\u009d\u0091\u0096\u00e2\u0088\u009ap [n \u00e2\u0088\u0092 \u00e2\u0088\u009a(\u00e2\u0088\u0092p)]\u00c2\u00b2 = n\u00c2\u00b2\u00e2\u0088\u0092 p\u00e2\u0088\u00922n\u00f0\u009d\u0091\u0096\u00e2\u0088\u009ap",
        "video_name": "i7idZfS8t8w",
        "timestamps": [
            585
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 13:34, Sal simplified the fraction, but I''m not clear how he did it. What happened to the two? Does this simplification leave the square root of 39 alone?\nThanks",
        "A": "Yes, just think of the sqrt 39 as some ugly thing multiplied to the 2. When you have (-12 +or- 2*sqrt39) / -6 , there are 3 distinct terms: the neg 12, the 2*sqrt39, and the neg 6. All 3 of those terms have 2 as a factor. That is, neg 12 = 2*neg 6 2*sqrt39 neg 6 = 2*neg 3 So all three terms have a factor of 2, so the 2 can be factored out by dividing each term by 2. That leaves neg 6 sqrt39 neg 3",
        "video_name": "i7idZfS8t8w",
        "timestamps": [
            814
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 13:23, why doesn't he divide the number inside the radical by 2?",
        "A": "Terms are things we add or subtract. They are held together by multiplication and division. The numerator only has 2 terms : -12 and 2*sqrt39. Both of those terms were divided by 2 to get -6 and sqrt39. But, if the numerator had been -12 + 2 + sqrt39, (in other words, 3 terms) and then we divided by 2, we would get -6 + 1 + (sqrt39)/2 ( Of course, that would simplify to -5 + (sqrt39)/2. )",
        "video_name": "i7idZfS8t8w",
        "timestamps": [
            803
        ],
        "3min_transcript": ""
    },
    {
        "Q": "Isn't there a small mistake at 16:10 ? Sal says \"a little less than one\", where the graph shows a little less than 0 ...",
        "A": "he says maybe close to zero but i little bit less than that",
        "video_name": "i7idZfS8t8w",
        "timestamps": [
            970
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 4:16 why did it change to 10 why didn't you put 100?i,m sorry i am so very new to this",
        "A": "the 100 was under a square root and then he took the square root of 100 to get 10",
        "video_name": "i7idZfS8t8w",
        "timestamps": [
            256
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 0:05 Sal says that the quadratic formula is at least top 5 most useful formulas. I wonder what other formulas would be as useful as the quadratic formula.",
        "A": "off the top of my head: the Pythagorean Theorem <-this is certainly one of the most useful if not the most useful formula, Area of a Triangle (you can usually break other shapes into a series of triangles), Combinations/Permutations, and maybe taking an Average.",
        "video_name": "i7idZfS8t8w",
        "timestamps": [
            5
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 7:11, Sal describes the system as existing in R4, but isn't it also safe to describe this as 4 column vectors in R3?",
        "A": "For this system of equations Ax = b , A , not the system, is 4 column vectors. x is an R^4 variable vector, each row is a plane in R^4 , and Ax is constrained to b , an R^3 column vector of constants.",
        "video_name": "JVDrlTdzxiI",
        "timestamps": [
            431
        ],
        "3min_transcript": "entry is in a lower row than that one. So it's in a column to the right of this one right there. And I just inspected, this looks like a-- this column two looks kind of like a free variable-- there's no pivot entry there, no pivot entry there. But let's see, let's map this back to our system of equations. These are just numbers to me and I just kind of mechanically, almost like a computer, put this in reduced row echelon form. Actually, almost exactly like a computer. But let me put it back to my system of linear equations, to see what our result is. So we get 1 times x1, let me write it in yellow. So I get 1 times x1, plus 2 times x2, plus 0 times x3, plus 3 times x4 is equal to 4. Obviously I could ignore this term right there, I didn't even have to write it. Actually. Then I get 0 times x1, plus 0 times x2, plus 1 times x3, so 1 times x3, minus 2 times x4, is equal to 4. And then this last term, what do I get? I get 0 x1, plus 0 x2 plus 0 x3 plus 0 x4, well, all of that's equal to 0, and I've got to write something on the left-hand side. So let me just write a 0, and that's got to be equal to minus 4. Well this doesn't make any sense whatsoever. 0 equals minus 4. This is this is a nonsensical constraint, this is impossible. 0 can never equal minus 4. This is impossible. Which means that it is essentially impossible to find an intersection of these three systems of equations, or a solution set that satisfies all of them. When we looked at this initially, at the beginning of the of the video, we said there are only three equations, we have four unknowns, maybe there's going But turns out that these three-- I guess you can call them these three surfaces-- don't intersect in r4. These are all four dimensional, we're dealing in r4 right here, because we have-- I guess each vector has four components, or we have four variables, is the way you could think about it. And it's always hard to visualize things in r4. But if we were doing things in r3, we can imagine the situation where, let's say we had two planes in r3. So that's one plane right there, and then I had another completely parallel plane to that one. So I had another completely parallel plane to that first one. Even though these would be two planes in r3, so let me give So let's say that this first plane was represented by the equation 3x plus 6y plus 9z is equal to 5, and the second"
    },
    {
        "Q": "11:30. To have an infinte number of solutions does one needs to have free variables AND a row of all Zeroes? The video was unclear on this point but alluded to it.",
        "A": "row of 0 s is not a necessary condition, e.g. x1 - x2 = 5 x1 - x2 + x3 = 3 reduces to 1 -1 0 | 5 0 0 1 | -2 ` with x2 being free. The solution is (x1, x2, x3) = (5, 0, -2) + x2(1, 1, 0) If you think of a row being a constraint to the solution(s), a row of zero s seems to indicate one that is a linear combination of the remaining one s i.e. it is a superfluous constraint.",
        "video_name": "JVDrlTdzxiI",
        "timestamps": [
            690
        ],
        "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": "At about 2:08, are we just supposed to assume the solid has the same volume as the shell?",
        "A": "Conceptually, we begin by estimating the volume of the solid by adding together the volumes of thin shells that make up the solid. As the shells get thinner and thinner, the estimate of the volume of the solid becomes more accurate. The limit of this sum as the thickness of the shells approaches zero is the integral, which is equal to the actual volume of the solid.",
        "video_name": "6Ozz3J-LRrY",
        "timestamps": [
            128
        ],
        "3min_transcript": "I've got the function y is equal to x minus 3 squared times x minus 1. And what I want to do is think about rotating the part of this function that sits right over here between x is equal to 1 and x equals 3. And x equals 3 and x equals 1 are clearly the zeroes of this function right over here. And I want to take this region and rotate it around the y-axis. And if I did that, I'd get a shape that looks something like that. And I want to figure out the volume of that shape. And what we're going to do is a new method called the shell method. And the reason we're going to use the shell method-- you might say, hey, in the past, we've rotated things around a vertical line before. We used the disk method. We wrote everything as a function of y, et cetera, et cetera. We created all of these disks. We figured out the volume of each of those disks. But the problem here is this is hard to express as a function of y. How do you solve explicitly for y right over here? So instead, we're going to keep things in terms of x we can come up with the volume. What we're going to imagine instead-- instead of constructing disks, we're going to construct shells. And what do I mean by a shell? So for each x at the interval, on this kind of cut of it, we can construct a rectangle. And what happens if we were to rotate this rectangle? So this is the rectangle right over here. What happens if we rotate this rectangle around the y-axis along with everything else? I'll try my best attempt to draw it. It's going to look something like this. This is challenging my art skills, but I think I can handle it. So it's going to look something not too dissimilar to that right over there. So it looks like a hollowed-out cylinder. I guess that's why we call it a shell. And it's going to have some depth. The depth is going to be dx. going to be the value of my function. The height is f of x. In this case, f of x is x minus 3 squared times x minus 1. How do we figure out the volume of a cylinder like this? Well, if we can figure out the circumference of the cylinder, and then multiply that circumference times the height of the cylinder, we'd essentially figure out the area of the outside surface of our cylinder. And then if we multiply the area of the outside surface of our cylinder by that infinitesimally small depth, then that'll give us the volume-- I shouldn't say cylinder-- of our shell. So let's try to do it. What is the circumference of a shell? What is the circumference of one shell going to be? Well, it's going to be 2 pi times the radius of that shell."
    },
    {
        "Q": "at 4:32, why doesnt the pi/sqrt(2) distribute to the c? I dont think that it matters, but I would like to know.",
        "A": "You re on the right track when you say I don t think it matters. The variable c stands for an arbitrary (unspecified, unknown) constant, which could be any real number. If you multiply an arbitrary constant by some other constant such as \u00cf\u0080/\u00e2\u0088\u009a2, you re still left with an arbitrary constant, so we continue to call it c and don t bother to multiply it by another constant.",
        "video_name": "8Yl_u_Otcjg",
        "timestamps": [
            272
        ],
        "3min_transcript": "And the square root of cosine squared theta is going to be cosine theta. Now, you might be saying, hey, wait, if I take the square root of something squared, then wouldn't that just be the absolute value of cosine theta? In order to take away the absolute value, I'd have to assume that cosine theta is positive. But we can make that assumption that cosine theta is positive, because if we look right here at this part of our substitution, if we wanted to solve for theta, you'd divide both sides by 2, and you'd get x of 2 is equal to sine of theta. Or we could say that theta is equal to arcsin of x over 2. Now the arcsin function, as it is traditionally defined, will return theta that is between negative pi over 2 and pi over 2. And in that range, cosine of theta is always going to be positive. So we don't have to write the absolute value. We know cosine theta is positive. Cosine theta cancels out with cosine theta. This 2 cancels out with this 2. We can bring this square root of 2 outside. And so we are left with pi over this square root of 2 times the indefinite integral of just d theta. And this is just going to be equal to pi over the square root of 2 times theta plus c. And we're almost done. We just have to rewrite this in terms of x. And we already know that theta is equal to arcsin of x over 2. So we can say that this indefinite integral, or the antiderivative of this expression, is going to be pi over the square root of 2 times arcsin of x over 2 plus c. And we're done. Some people like a square root of 2 in the denominator. If you want to remove it, you can multiply this by square root of 2 over square root of 2, and that will simplify it. But right now, I'll just leave the denominator And this right over here is our antiderivative."
    },
    {
        "Q": "At 1:15 he changes 25% into a decimal. So does that mean that percent is basically the same thing as decimals?",
        "A": "yes, a percentage is always a decimal. You can arrive at the decimal by dividing the number by 100. So 25% = 25/100 = 0.25",
        "video_name": "JaScdH47PYg",
        "timestamps": [
            75
        ],
        "3min_transcript": "We're asked to identify the percent, amount, and base in this problem. And they ask us 150 is 25% of what number? So another way to think about it is 25% times some number. So I'll do 25% in yellow. And 25% times some number is equal to 150. So the percent is pretty easy to spot. We have a 25% right over here. So this is going to be the percent. And we're multiplying the percent times some base number. So this right over here is the base. So we have the percent times the base. We have the percent times the base is equal to some amount. This is essentially saying 25% of some number, 25% times some number is equal to 150. If it helps, we could rewrite this as 0.25, which is the same thing as 25%. 0.25 times some number is equal to 150. And one interesting thing is just to think about, should that number be larger or smaller than 150? Well, if we only take 25% of that number, if we only take 25 hundredths of that number, if we only take 1/4 of that number, because that's what 25 hundredths is, or that's what 25% is, we get 150. So this number needs to be larger than 150. In fact, it has to be larger than 150 by 4. And to actually figure out what the number is, we can actually multiply. Since this, what's on the left-hand side is equal to the right-hand side, if we want to solve this, we can multiply both sides by 4. If we say, look, we have some value over here, In order for it to still be equal, we have to multiply 150 times 4. 4 times 0.25, or 4 times 25%, or 4 times 1/4, this is just going to be 1. And we're going to get our number is equal to 150 times 4 or it is equal to 600. And that makes sense, 25% of 600 is 150. 1/4 of 600 is 150."
    },
    {
        "Q": "At 4:30, why does Sal say that you can't simplify -21/20? Can't you simplify it to\n-1 1/20? Or is it -21 over +20? :(",
        "A": "First of all you can t simplify one part of the fraction when you want to simplify a fraction you must do it for both numerator and denominator and in this fraction you can t simplify it any further because there isn t any common factor between them (21)=3*7 (20)=4*5 hopefully this helps you",
        "video_name": "pi3WWQ0q6Lc",
        "timestamps": [
            270
        ],
        "3min_transcript": "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. But now we have to think about the fact that we're multiplying by a negative times a negative. Now, we remember when you multiply a negative times a negative, it's a positive. The only way that you get a negative is if one of those two numbers that you're taking the product of is negative, not two. If both are positive, it's positive. If both are negative, it's positive. Let's do one more example. Let's take 5-- I'm using the number 5 a lot. So let's do 3/2, just to show that this would work with improper fractions. 3/2 times negative 7/10. I'm arbitrarily picking colors. And so our numerator is going to be 3 times negative 7. 3 times negative 7. 2 times 10. So this is going to be the numerator. Positive times a negative is a negative. 3 times negative 7 is negative 21. Negative 21. And the denominator, 2 times 10. Well, that is just 20. So this is negative 21/20. And you really can't simplify this any further."
    },
    {
        "Q": "In 1:42 I don't get it.",
        "A": "this is totally new exercise as 2 fractions times each other that means u can write them in one fraction and the value is still same. in 2 fractions numerators r always times each other and denominators r always time each other but numerators and denominators r always divide that is why Sal wrote 5 x 3 (numerators) and 9 x 15 (denominators). hope this will help.",
        "video_name": "pi3WWQ0q6Lc",
        "timestamps": [
            102
        ],
        "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": "why at 1:18 does neg. numerator over pos. denominator (-3/+7) become the whole fraction neg. ?",
        "A": "You may be aware of the mathematical rule which dictates that if you divide a negative by a positive, you get a negative. A fraction basically means to divide the top number, or the numerator, by three bottom number, the denominator. EXAMPLE: -2/4 We know that +2/+4 is 1/2, so we can simplify -2/4 to -1/2, or -0.5 I hope this this helps!",
        "video_name": "pi3WWQ0q6Lc",
        "timestamps": [
            78
        ],
        "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": "at 5:05 is he writing out what he did in the model?",
        "A": "Yep, that is correct. Sal (the speaker) just replaced the equation with absolute values, which are the lines on both sides of the subtraction.",
        "video_name": "DPuK6ZgBGmE",
        "timestamps": [
            305
        ],
        "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 4:58, Sal says X squared equals -1. Shouldn't it be 1?",
        "A": "No. He has x^2 + 1 = 0 You must subtract 1 from both sides to isolate x^2 (remember, we use the opposite operation to move items to the other side of an equation). This creates Sal s version: x^2 = -1",
        "video_name": "uFZvWYPfOmw",
        "timestamps": [
            298
        ],
        "3min_transcript": "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. squared is equal to negative 1. x is going to be an imaginary-- or I guess I'll just say it in more general terms-- it's going to be complex valued. So once again, you see you're going to have a pair of complex roots, and you have one real root at x is equal to negative 3."
    },
    {
        "Q": "At the 1:53 mark he says something like \"you could probably draw a better freehand diagram\". My opinion: I definitely cannot. This guy seems to be good at explaining math in these videos-who is he?",
        "A": "He s Richard Rusczyk (yeah, I probably butchered his name), but if you look him up on google you can find more info about him. He s pretty famous.",
        "video_name": "rcLw4BlxaRs",
        "timestamps": [
            113
        ],
        "3min_transcript": "We've got some 3D geometry here so we're going to have read carefully, visualize what's going on because it's kind of hard to draw in 3D. We got six spheres with a radius 1. Their centers are at the vertices of a regular hexagon that has side length 2. So we're starting with a regular hexagon, and we're going to put spheres centered at each of the vertices. And since the radius of each sphere is 1, side length is 2, that means each of these spheres is going to be tangent to its two neighbors. So we start off with a hexagon, six spheres, each one tangent to each of its two neighbors. And then we're going to have a larger sphere centered at the center of the hexagon such that it's tangent to each of the little spheres. Now, each of the little spheres will touch the inside of this giant sphere. And then we bring out an eighth sphere that's externally tangent to the six little ones. So we got out six little ones down here around the hexagon, and we're going to take this new sphere and just set it right on top of those six. And it's going to touch-- right at the top of it, So we have at least somewhat of a picture of what's going on here, and we want the radius of this last sphere that we dropped in at the top there. And now one thing I like to do with these 3D problems is I like to take 2D cross sections, turn 3D problems into 2D problems. So when I have a problem with a whole bunch of spheres, I like to throw my cross sections through centers of those spheres and through points of tangency whenever I have tangent spheres. Now, a natural place to start here, of course, is the hexagon. We take the cross section with the hexagon, because that's going to go through the centers of seven of these spheres and all kinds of points tangency. So to start off, we'll draw a regular hexagon. And you're going to have to bear with me. On the test, of course, you've got your ruler, you got your protractor, you got your compass so you can draw a perfect diagram. You could probably draw a freehand better diagram better than I can, too. But when we take cross sections of our spheres, we make circles. And we include the points of tangency in this cross section. A cross section of that is a circle that touches each of these little circles. All right, and there we go. This is the cross section through the hexagon. Now we can label some lengths. We know that the radii of the little spheres is 1, and one thing that's really nice about regular hexagons is you can break them up into equilateral triangles. So this is an equilateral triangle. Here's the center of the hexagon center and the big circle, and I can extend this out to the point of tangency of small sphere and the big one. So we know this is 1 because it's a radius of the small sphere. This is 1. This is an equilateral triangle so this side is the same as this side. So it tells us that this is 1, and now we know that the radius of the giant sphere is 3. So we've got the radius of the giant sphere. We got the radii of all these little spheres. All we have left is that eighth sphere we sat on top. And, of course, that sphere's not in this diagram. It's sitting right up here."
    },
    {
        "Q": "At 1:15... I know this is right, I just need an explanation.\n\nHow does x^2 go into x^3 and x^4?",
        "A": "5 goes into 15, 3 times (5*3=15) 8 goes into 16, 2 times (8*2=16) 9 goes into 54, 6 times(9*6=54) x^2 goes into x^3, x times (x^2 * x=x^3) x^2 goes into x^4, x^2 times (x^2 * x^2 = x^4)",
        "video_name": "MZl6Mna0leQ",
        "timestamps": [
            75
        ],
        "3min_transcript": "We're told to factor 4x to the fourth y, minus 8x to the third y, minus 2x squared. So to factor this, we need to figure out what the greatest common factor of each of these terms are. So let me rewrite it. So we have 4x to the fourth y, and we have minus 8x to the third y, and then we have minus 2x squared. So in the other videos, we looked at it in terms of breaking it down to its simplest parts, but I think we have enough practice now to be able to do a little bit more of it in our heads. So what is the largest number that divides into all of these? When I say number I'm talking about the actual, I guess, coefficients. We have a 4, an 8, a 2. We don't have to worry about the negative signs just yet. And we say, well, the largest, of, the largest common factor of 2, 8 and 4 is 2. 2 goes into all of them, and obviously that's the largest number that can go into 2. So that is the largest number that's going to be part of the greatest common factor. So it's going to be 2. And then what's the greatest, I guess, factor, what's the greatest degree of x that's divisible into all three of these? Well, x squared goes into all three of these, and obviously that's the greatest degree of x that can be divided into this last term. So x squared is going to be the greatest common x degree in all of them. 2x squared. And then what's the largest degree of y that's divisible into all of them? Well, these two guys are divisible by y, but this guy isn't, so there is no degree of y that's divisible into all of them. So the greatest common factor of all three of these guys right here is 2x squared. So what we can do now is we can think about each of these terms as the product of the 2x squared and something else. And to figure that something else we can literally undistribute the 2x squared, say this is the same thing, or even before we undistribute the 2x squared, we could say look, 4x to the fourth y is the same thing as 2x squared, Right? If you just multiply this out, you get 4xy. Similarly, you could say that 8x to the third y-- I'll put the negative out front-- is the same thing as 2x squared, our greatest common factor, times 8x to the third y, over 2x squared. And then finally, 2x squared is the same thing as if we factor out 2x squared-- so we have that negative sign out front-- if we factor out 2x squared, it's the same thing as 2x squared, times 2x squared, over 2x squared. This is almost silly what I'm doing here, but I'm just showing you that I'm just multiplying and dividing both of these terms by 2x squared. Multiplying and dividing. Here it's trivially simple. This just simplifies to 2x squared right there, or this 2x squared times 1."
    },
    {
        "Q": "At @5:08 you say that the span of T is V but couldn't you also say that it is T itself since a span, unlike a basis, doesn't necessarily need to have linearly independent vectors?",
        "A": "T was just S with one extra vector. V in the example was Span(S) = Span(T). V in theory should have infinitely more vectors than T.",
        "video_name": "zntNi3-ybfQ",
        "timestamps": [
            308
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 4:55, it says a sub i is equal to a sub i minus i times 9/10. Is this supposed to be i-1?",
        "A": "It s confusing... the dot is actually the multiplication symbol, not a dot on a i . So, Sal does have i-1",
        "video_name": "DY9Q3qNmZnw",
        "timestamps": [
            295
        ],
        "3min_transcript": "but notice we have a change in sign here, and the key thing is to say well to go from term to the next what are we multiplying by? Well to go from the first term to the second term, we multiply by negative 0.99. And then, so we're multiplying by negative 0.99. Now to go to the next term, we're again multiplying by negative 0.99, so the common ratio is not positive 0.99, but negative zero, negative 0.99, so let me write that, negative 0.99, and of course that is going to be to the 80th power, all over one minus negative 0.99. And so we could simplify this a little bit, this is all going to be equal to, and so this is going to be one minus, so negative 0.99 to the 80th power, I should put parenthesis there to make sure we are taking the negative 0.99 to the 80th power. Well, we're taking it to an even power, so it's going to be positive, so that's going to be the same thing as 0.99 to the 80th power, and all of that over, well subtracting a negative that's just gonna be adding the positive, so all of that over 1.99, and we could attempt to simplify it more but, if we had a calculator we could actually find this exact value or close value actually, most calculators don't give you the exact value when you take something to the 80th power, but this is what that sum is going to be. Let's do one more of these. Alright, so here we have a series defined recursively and so it's useful to just think about So the first term is 10, and then the next term, so the second term A sub two is equal to A sub one times 9/10, alright. So the next term is gonna be the previous term times 9/10, so it's gonna be 10 times nine over 10, and then the next term is gonna be that times, is gonna be the second term, the third term is the second term times 9/10, so 10 times nine over 10, nine over 10 squared. And the way it's written right now, we don't have it written as a finite geometric series, so let's say we wanna take the sum, let's say we want the sum of first, first I don't know 30 terms, sum of first 30 terms. So what will this be? Well we're gonna take S sub one, S sub 30, oh I wrote ten, S sub 30, the sum of the first 30 terms, is gonna be equal to the first term, we've done this before,"
    },
    {
        "Q": "At 3:30 the instructor immediately went to subtracting. But I thought the order of operations was PEDMAS so shouldn't the division go first? The way I did it, was first doing all the divisions then I found the last equation to be (d/ac)-(d/bc)",
        "A": "The division bar in the middle of a fraction asks as a grouping symbol. When you have 2 terms in the numerator of a fraction or even 2 terms in the denominator, there are implied sets of parentheses the numbers in the numerator / denominator. So, Sal did the parentheses 1st by subtracting the 2 fractions in the numerator, then doing the division. Sometimes it s easier to see a simpler example: (2+3) / 10 You add 1st = 5 / 10 Then, you divide = 0.5 or 1/2",
        "video_name": "_BFaxpf35sY",
        "timestamps": [
            210
        ],
        "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 8:16, how did he get A^2=3/4 h^2?",
        "A": "Simple algebraic solving techniques. if you take h to be c like it is normally represented, and assume a^2 + b^2 = c^2... and take the 30-60-90 side ratio definition where the side opposite the 30 degree angle is c/2, or hypotenuse/2, then (c/2)^2 + a^2 = c^2. Subtract (c/2)^2 from both sides... you get a^2 = c^2 -(c/2)^2. Fraction stuff... a^2 = (4c/4)^2 - (c/2)^2 so a^2 = (3/4c)^2.",
        "video_name": "Qwet4cIpnCM",
        "timestamps": [
            496
        ],
        "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": "at 0:23, where did you get the radical 2 over 2 from?",
        "A": "That was explained in the earlier video about 45-45-90 triangles.",
        "video_name": "Qwet4cIpnCM",
        "timestamps": [
            23
        ],
        "3min_transcript": "Sorry for starting the presentation with a cough. I think I still have a little bit of a bug going around. But now I want to continue with the 45-45-90 triangles. So in the last presentation we learned that either side of a 45-45-90 triangle that isn't the hypotenuse is equal to the square route of 2 over 2 times the hypotenuse. Let's do a couple of more problems. So if I were to tell you that the hypotenuse of this triangle-- once again, this only works for 45-45-90 triangles. And if I just draw one 45 you know the other angle's got to be 45 as well. If I told you that the hypotenuse here is, let me say, 10. We know this is a hypotenuse because it's opposite the right angle. And then I would ask you what this side is, x. Well we know that x is equal to the square root of 2 over 2 times the hypotenuse. So that's square root of 2 over 2 times 10. 10 divided by 2. So x is equal to 5 square roots of 2. And we know that this side and this side are equal. I guess we know this is an isosceles triangle because these two angles are the same. So we also that this side is 5 over 2. And if you're not sure, try it out. Let's try the Pythagorean theorem. We know from the Pythagorean theorem that 5 root 2 squared, plus 5 root 2 squared is equal to the hypotenuse squared, where the hypotenuse is 10. Is equal to 100. Or this is just 25 times 2. So that's 50. But this is 100 up here. Is equal to 100. And we know, of course, that this is true. So it worked. We proved it using the Pythagorean theorem, and that's actually how we came up with this formula in the first place. Maybe you want to go back to one of those presentations if you forget how we came up with this. type of triangle. And I'm going to do it the same way, by just posing a problem to you and then using the Pythagorean theorem to figure it out. This is another type of triangle called a 30-60-90 triangle. And if I don't have time for this I will do another presentation. Let's say I have a right triangle. That's not a pretty one, but we use what we have. That's a right angle. And if I were to tell you that this is a 30 degree angle. Well we know that the angles in a triangle have to add up to 180. So if this is 30, this is 90, and let's say that this is x. x plus 30 plus 90 is equal to 180, because the angles in"
    },
    {
        "Q": "At 7:46 how to you get 4?",
        "A": "(1/2) squared is (1/4). This multiplied by h is (h/4).",
        "video_name": "Qwet4cIpnCM",
        "timestamps": [
            466
        ],
        "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 confused about what happens to the negative at 2:37",
        "A": "you have (2*(1-3^100))/(-2). then the 2s cancel out: (1-3^100)/(-1). negative from the -1 goes to the top: -(1-3^100)/1. 1 in denominator goes away: -(1-3^100). distribute negative: -1+3^100 rewrite order: 3^100-1",
        "video_name": "AXP5PGSaaYk",
        "timestamps": [
            157
        ],
        "3min_transcript": "- [Voiceover] Let's do some examples where we're finding sums of finite geometric series. Now let's just remind ourselves in a previous video we derived the formula where the sum of the first n terms is equal to our first term times one minus our common ratio to the nth power all over one minus our common ratio. So let's apply that to this finite geometric series right over here. So what is our first term and what is our common ratio? And what is our n? Well, some of you might just be able to pick it out by inspecting this here, but for the sake of this example, let's expand this out a little bit. This is going to be equal to two times three to the zero, which is just two, plus two times three to the first power, plus two times three to the second power, I can write first power there, plus two times three to the third power, and we're gonna go all the way to two times three to the 99th power. What is our a? Well, a is going to be two. And we see that in all of these terms here. So a is going to be two. What is r? Well, each successive term, as k increases by one, we're multiplying by three again. So, three is our common ratio. So that right over there, that is r. Let me make sure that we, that is a. And now what is n going to be? Well, you might be tempted to say, well, we're going up to k equals 99, maybe n is 99, but we have to realize that we're starting at k equals zero. So there is actually 100 terms here. Notice, when k equals zero, that's our first term, when k equals one, that's our second term, when k equals two, that's our third term, when k equals three, that's our fourth term, when k equals 99, this is our 100th term, 100th term. So what we really want to find is S sub 100. So let's write that down, S sub 100, for this geometric series is going to be equal to to the 100th power, all of that, all of that over, all of that over one minus three. And we could simplify this, I mean at this point it is arithmetic that you'd be dealing with, but down here you would have a negative two, and so you'd have two divided by negative two so that is just a negative. And so negative of one minus three to the 100th, that's the same thing, this is equal to three to the 100th, three to the 100th power minus one. And we're done."
    },
    {
        "Q": "At 3:10 when Sal factors out the five from the equation, why does he not put the newly factored equation in brackets with a five on the outside?",
        "A": "In order to get rid of the 5, Sal divided both sides of the equation by 5, giving (5x^2 - 20x + 15) / 5 = 0/5, which simplifies to x^2 - 4x + 3 = 0. You don t need to include the 5 as a factor in the left-hand side because you divided both sides by it.",
        "video_name": "MQtsRYPx3v0",
        "timestamps": [
            190
        ],
        "3min_transcript": "when does this equal 0? So I want to figure out those points. And then I also want to figure out the point exactly in between, which is the vertex. And if I can graph those three points then I should be all set with graphing this parabola. So as I just said, we're going to try to solve the equation 5x squared minus 20x plus 15 is equal to 0. Now the first thing I like to do whenever I see a coefficient out here on the x squared term that's not a 1, is to see if I can divide everything by that term to try to simplify this a little bit. And maybe this will get us into a factor-able form. And it does look like every term here is divisible by 5. So I will divide by 5. So I'll divide both sides of this equation by 5. And so that will give me-- these cancel out and I'm left with x squared minus 20 over 5 is 4x. Plus 15 over 5 is 3 is equal to 0 over 5 is just 0. We say are there two numbers whose product is positive 3? The fact that their product is positive tells you they both must be positive. And whose sum is negative 4, which tells you well they both must be negative. If we're getting a negative sum here. And the one that probably jumps out of your mind-- and you might want to review the videos on factoring quadratics if this is not so fresh-- is a negative 3 and negative 1 Negative 3 times negative 1. Negative 3 times negative 1 is 3. Negative 3 plus negative 1 is negative 4. So this will factor out as x minus 3 times x minus 1. And on the right-hand side, we still have that being equal to 0. And now we can think about what x's will make this expression 0, and if they make this expression 0, well they're going to make this expression 0. Which is going to make this expression equal to 0. And so this will be true if either one of these is 0. Or x minus 1 is equal to 0. This is true, and you can add 3 to both sides of this. This is true when x is equal to 3. This is true when x is equal to 1. So we were able to figure out these two points right over here. This is x is equal to 1. This is x is equal to 3. So this is the point 1 comma 0. This is the point 3 comma 0. And so the last one I want to figure out, is this point right over here, the vertex. Now the vertex always sits exactly smack dab between the roots, when you do have roots. Sometimes you might not intersect the x-axis. So we already know what its x-coordinate is going to be. It's going to be 2. And now we just have to substitute back in to figure out its y-coordinate. When x equals 2, y is going to be equal to 5 times 2 squared minus 20 times 2 plus 15, which is equal to-- let's see, this is equal to 2 squared is 4."
    },
    {
        "Q": "Okay, so at 2:36 when he finds integers that multiplied equal 3 and added equal -4, then moves on using that information? What if you have numbers that don't work? I have the equation x^2 +2x+3=0. Nothing works with this. Is there another way to solve this and graph it as a parabola? Please help.",
        "A": "This is because this parabola does not have x-intercepts. You can check it by completing the square: y=(x+1)^2+2 is always positive, which means that every point of the parabola has a positive y value, which means that it is above the x axis. The fastest way to check it though is to evaluate its determinant b^2-4ac and see that it is negative,and therefore has no real zeroes.",
        "video_name": "MQtsRYPx3v0",
        "timestamps": [
            156
        ],
        "3min_transcript": "when does this equal 0? So I want to figure out those points. And then I also want to figure out the point exactly in between, which is the vertex. And if I can graph those three points then I should be all set with graphing this parabola. So as I just said, we're going to try to solve the equation 5x squared minus 20x plus 15 is equal to 0. Now the first thing I like to do whenever I see a coefficient out here on the x squared term that's not a 1, is to see if I can divide everything by that term to try to simplify this a little bit. And maybe this will get us into a factor-able form. And it does look like every term here is divisible by 5. So I will divide by 5. So I'll divide both sides of this equation by 5. And so that will give me-- these cancel out and I'm left with x squared minus 20 over 5 is 4x. Plus 15 over 5 is 3 is equal to 0 over 5 is just 0. We say are there two numbers whose product is positive 3? The fact that their product is positive tells you they both must be positive. And whose sum is negative 4, which tells you well they both must be negative. If we're getting a negative sum here. And the one that probably jumps out of your mind-- and you might want to review the videos on factoring quadratics if this is not so fresh-- is a negative 3 and negative 1 Negative 3 times negative 1. Negative 3 times negative 1 is 3. Negative 3 plus negative 1 is negative 4. So this will factor out as x minus 3 times x minus 1. And on the right-hand side, we still have that being equal to 0. And now we can think about what x's will make this expression 0, and if they make this expression 0, well they're going to make this expression 0. Which is going to make this expression equal to 0. And so this will be true if either one of these is 0. Or x minus 1 is equal to 0. This is true, and you can add 3 to both sides of this. This is true when x is equal to 3. This is true when x is equal to 1. So we were able to figure out these two points right over here. This is x is equal to 1. This is x is equal to 3. So this is the point 1 comma 0. This is the point 3 comma 0. And so the last one I want to figure out, is this point right over here, the vertex. Now the vertex always sits exactly smack dab between the roots, when you do have roots. Sometimes you might not intersect the x-axis. So we already know what its x-coordinate is going to be. It's going to be 2. And now we just have to substitute back in to figure out its y-coordinate. When x equals 2, y is going to be equal to 5 times 2 squared minus 20 times 2 plus 15, which is equal to-- let's see, this is equal to 2 squared is 4."
    },
    {
        "Q": "7:30 in the video sal has written -24-10_/5 should the answer be -(-24-12_/5)",
        "A": "What Sal wrote is correct. After rationalizing the denominator then simplifying we have: 24 + 12sqrt(5) / -1 Because we are dividing the numerator by negative one, we take the opposite of each of its terms: 24 + 12sqrt(5) / -1 -(24 + 12sqrt(5)) -24 - 12sqrt(5)",
        "video_name": "gY5TvlHg4Vk",
        "timestamps": [
            450
        ],
        "3min_transcript": "b, negative b squared. These cancel out and you're just left with a squared minus b squared. So 2 minus the square root of 5 times 2 plus the square root of 5 is going to be equal to 2 squared, which is 4. Let me write it that way. It's going to be equal to 2 squared minus the square root of 5 squared, which is just 5. So this would just be equal to 4 minus 5 or negative 1. If you take advantage of the difference of squares of binomials, or the factoring difference of squares, however you want to view it, then you can rationalize this denominator. So let's do that. Let me rewrite the problem. 12 over 2 minus the square root of 5. In this situation, I just multiply the numerator and the denominator by 2 plus the square root of 5 over 2 plus Once again, I'm just multiplying the number by 1. So I'm not changing the fundamental number. I'm just changing how we represent it. So the numerator is going to become 12 times 2, which is 24. Plus 12 times the square root of 5. Once again, this is like a factored difference of squares. This is going to be equal to 2 squared, which is going to be exactly equal to that. Which is 4 minus 1, or we could just-- sorry. 4 minus 5. It's 2 squared minus square root of 5 squared. So it's 4 minus 5. Or we could just write that as minus 1, or negative 1. Or we could put a 1 there and put a negative sign out in front. And then, no point in even putting a 1 in the denominator. We could just say that this is equal to negative 24 minus 12 square roots of 5. So this case, it kind of did simplify it as well. 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"
    },
    {
        "Q": "how did he turn 1/4 into a whole number at 3:05",
        "A": "he multiplied 12 to 1/4, and that equals 3",
        "video_name": "GmD7Czmol0k",
        "timestamps": [
            185
        ],
        "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 2:25, Sal uses the chain rule for the derivative of (2+x^3)^-1. Would it not work if he just used the power rule and left it at that?",
        "A": "No, the power rule applies only when you have x to the n, not when you have some function of x raised to the n. You can see this with an example like (x^2)^3. If you just apply the power rule, you get 3(x^2)^2, but we know that s wrong because (x^2)^3 is x^6, so the answer has to be 6x^5 or something equivalent. You get the right result when you apply the chain rule.",
        "video_name": "GH8-URjRQpQ",
        "timestamps": [
            145
        ],
        "3min_transcript": "We have the curve y is equal to e to the x over 2 plus x to the third power. And what we want to do is find the equation of the tangent line to this curve at the point x equals 1. And when x is equal to 1, y is going to be equal to e over 3. It's going to be e over 3. So let's try to figure out the equation of the tangent line to this curve at this point. And I encourage you to pause this video and try this on your own first. Well, the slope of the tangent line at this point is the same thing as the derivative at this point. So let's try to find the derivative of this or evaluate the derivative of this function right over here at this point. So to do that, first I'm going to rewrite it. You could use the quotient rule if you like, but I always forget the quotient rule. The product rule is much easier for me to remember. So I can rewrite this as y is equal to-- and I might as well color code it-- is equal to e to the x times 2 plus x And so the derivative of this, so let me write it here. So y prime is going to be equal to the derivative of this part of it, e to the x. So the derivative of e to the x is just e to the x. Just let me write that. So we're going to take the derivative of it. And that's what's amazing about e to the x, is that the derivative of e to the x is just e to the x times this thing. So times 2 plus x to the third to the negative 1. And then to that we're going to add this thing. So not its derivative anymore. We're just going to add e to the x times the derivative of this thing right over here. So we're going to take the derivative. So we can do the chain rule. It's going to be the derivative of 2 power with respect to 2 plus x to the third times the derivative of 2 plus x to the third with respect to x. So this is going to be equal to negative-- I'll write it this way-- negative 2 plus x to the third to the negative 2 power. And then we're going to multiply that times the derivative of 2 plus x to the third with respect to x. Well, derivative of this with respect to x is just 3x squared. And of course, we could simplify this a little bit if we like. But the whole point of this is to actually find the value of the derivative at this point. So let's evaluate. Let's evaluate y prime when x is equal to 1. Y prime of 1 when x is equal to 1."
    },
    {
        "Q": "At 2:25 are you supposed to add the negatives together? Can't you use the old fashioned way too?",
        "A": "If you are talking about the negatives of 70, 10, 15, and 21, keep in mind that you can only add them together when they possess the same exponent amounts.",
        "video_name": "D6mivA_8L8U",
        "timestamps": [
            145
        ],
        "3min_transcript": "We are multiplying 10a minus 3 by the entire polynomial 5a squared plus 7a minus 1. So to do this, we can just do the distributive property. We can distribute this entire polynomial, this entire trinomial, times each of these terms. We could have 5a squared plus 7a minus 1 times 10a. And then 5a squared plus 7a minus 1 times negative 3. So let's just do that. So if we have-- so let me just write it out. Let me write it this way. 10a times 5a squared plus 7a minus 1. That's that right over here. And then we can have minus 3 times 5a squared plus 7a minus 1. And that is this distribution right over here. And then we can simplify it. 10a times 5a squared-- 10 times 5 is 50. 10 times 7 is 70. a times a is a squared. 10a times negative 1 is negative 10a. Then we distribute this negative 3 times all of this. Negative 3 times 5a squared is negative 15a squared. Negative 3 times 7a is negative 21a. Negative 3 times negative 1 is positive 3. And now we can try to merge like terms. This is the only a to the third term here. So this is 50a to the third. I'll just rewrite it. Now we have two a squared terms. We have 70a squared minus 15, or negative 15a squared. So we can add these two terms. 70 of something minus 15 of that something is going to be 55 of that something. So plus 55a squared. We have this negative 10a, and then we have this negative 21a. So if we go negative 10 minus 21, that is negative 31. That is negative 31a. And then finally, we only have one constant term over here. We have this positive 3. So plus 3. And we are done."
    },
    {
        "Q": "at 1:17, what if there isnt a exact weight given?how would you solve?",
        "A": "If there is not an exact weight given then there is to little info to discover an exact answer.",
        "video_name": "z1hz8-Kri1E",
        "timestamps": [
            77
        ],
        "3min_transcript": "An electronics warehouse ships televisions and DVD players in certain combinations to retailers throughout the country. They tell us that the weight of 3 televisions and 5 DVD players is 62.5 pounds, and the weight of 3 televisions and 2 DVD players-- so they're giving us different combinations-- is 52 pounds. Create a system of equations that represents this situation. Then solve it to find out how much each television and DVD player weighs. Well, the two pieces of information they gave us in each of these statements can be converted into an equation. The first one is is that the weight of 3 televisions and 5 DVD players is 62.5 pounds. Then they told us that the weight of 3 televisions and 2 DVD players is 52 pounds. So we can translate these directly into equations. If we let t to be the weight of a television, and d to be the weight of a DVD player, this first statement up here says that 3 times the weight of a television, or 3 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"
    },
    {
        "Q": "what is a parabola 0:00 to 2:46",
        "A": "It is the line you get when you graph a quadratic.",
        "video_name": "v-pyuaThp-c",
        "timestamps": [
            0,
            166
        ],
        "3min_transcript": "So you're me, and you're in math class. And your teacher's ranting on and on about this article about whether algebra should be taught in school \u2013 as if he doesn't realize that what he's teaching isn't even algebra \u2013 which could have been interesting \u2013 but how to manipulate symbols and some special cases of elementary algebra \u2013 which isn't. And so, instead of learning about self-consistent systems and logical thought, you spent all week memorizing how to graph parabolas. News flash: No one cares about parabolas. Which is why half the class is playing Angry Birds under their desks. But, since you don't have a smart phone yet, you have to resort to a more noble and outdated form of boredom relief \u2013 that is, doodling. And you've invented a game of your own. A doodle game that connects the dots in ways your math curriculum never will \u2013 except instead of connecting to the closest dots to discover the mysterious hidden picture you've got this precise method of skipping over some number of dots and connecting them that way. In the past you've characterized how this works if your dots are arranged in a circle \u2013 say 11 dots \u2013 and connect one to the dot four dots over, you get these awesome stars. And you can either draw the lines in the order of the dots, or you can just keep going around and maybe it will hit all the dots, or maybe it won't, 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"
    },
    {
        "Q": "how did she make that shape?! at 4:30",
        "A": "She drew a circle out of dots, then drew more circles with the points being their centers and alternated colors between the shapes made by the circles overlapping.",
        "video_name": "v-pyuaThp-c",
        "timestamps": [
            270
        ],
        "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": "I want so bad to be able to draw the heart thing at 4:29 can someone explain how or make a program maybe??? because I can't understand lol",
        "A": "you could take two circles and roll one around the other tracing one point on the edge; OR you could make dots in the shape of a circle ,choose one dot to be a point on the edge of all circles and use the others as centers",
        "video_name": "v-pyuaThp-c",
        "timestamps": [
            269
        ],
        "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": "At 1:10. why does he make 4.1 to 41?",
        "A": "Moving the decimal around is like multiplying or dividing by 10, so he notes that 4.1 hundredths is the same as 41 thousandths (4.1 x 10^-2 = 41 x 10^-3).",
        "video_name": "ios3QL9t9LQ",
        "timestamps": [
            70
        ],
        "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": "A, B and C are collinear, and B is between A and C. The ratio of AB, B to AC, C is 1:4\nIf A is at (-7,-8) and B is at (-3,-5), what are the coordinates of point C?\nI got the x and y values by subtracting and got 4 and 3. But, when I went to find the ratios of 4 and 3 i'm told to multiply 4 by 1:4 to get 16. And 1:4 by 3 to get 12. But, in the video i'm told to find the ratios by dividing to get 4 1:4=1 so which am I supposed to be doing?",
        "A": "I think you are thinking correctly, but I am not completely sure. Setting up proportions and cross multiplying, For x coordinate, AB/AC = 1/4 and if AB is 4, then 4/AC = 1/4 and AC = 16 in x direction For y coordinate, AB/AC = 1/4 and if AB is 3, then 3/AC= 1/4 and AC = 12 in y direction If I start at A and move these distances, then C (-7+16, -8+12) or (9,4)",
        "video_name": "lEGS5ECgFxE",
        "timestamps": [
            64,
            64,
            64,
            64
        ],
        "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 0:03, the word collinear is mentioned. What exactly is collinear?",
        "A": "It means you can draw a line through all the points.",
        "video_name": "lEGS5ECgFxE",
        "timestamps": [
            3
        ],
        "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": "I was just wondering, at 1:05, why is point B said to be 2/5 of the way from A? Shouldn't it be that the total distance from A to C is 7 (add both sides of the ratio to find the total number of parts) and then B should be 2 parts out of this total distance? Thus, B is 2/7 of the way? I'm not entirely sure of whether or not I am correct, but I'm assuming I've made some error.",
        "A": "Your error was in saying the total distance from A to C is 7. The problem states that AC is 5 when it says the ratio of AB to AC is 2 : 5. In other words, the 2 : 5 ratio is a ratio of a part to the whole, not a ratio of the lengths of 2 portions of the whole.",
        "video_name": "lEGS5ECgFxE",
        "timestamps": [
            65
        ],
        "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 1:35 I solved it as 8^1/4=2, where as in the video its 8^1/3=2. Im Not sure if im missing something of if Sal made a mistake?",
        "A": "8^(1/3) is the cube root of 8, which is 2. (2^3)^(1/3)=2^(3/3)=2^1=2 8^(1/4) is the fourth root of 8, which is not 2. 16^(1/4) is the fourth root of 16, which is 2.",
        "video_name": "eTWCARmrzJ0",
        "timestamps": [
            95
        ],
        "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 the 0:53 problem how does Sal get to an Answer of 1/3 ? What is the work not shown here ?",
        "A": "Let s start with 8^x =2. When the variable is in the exponent, it is useful (where possible) to express both sides of the equation using the same base. Since on the righthand side there is a 2 to the first power, ask yourself whether 8 can be expressed as a power of 2? So we end up with (2^3)^x which is the same as 2^3x. And we then have 2^3x = 2^1 as the equation. The bases are the same, so the exponents must be equal. Therefore 3x = 1 so x = 1/3 Hope you find this useful!",
        "video_name": "eTWCARmrzJ0",
        "timestamps": [
            53
        ],
        "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": "on 2:37 how does negitive turn into a fraction",
        "A": "That is a basic property of exponents. The rule is: a\u00e2\u0081\u00bb\u00e1\u00b5\u0087 = 1/a\u00e1\u00b5\u0087",
        "video_name": "eTWCARmrzJ0",
        "timestamps": [
            157
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "@1:25, I still dont get it... is something wrong with me? 8^1=8 and 8^0=1, but how do you get.... i dont know; is there another video i can watch to buff up my understanding of this?",
        "A": "any number raised to the power 0 is 1, you can think of it like this: x^3 = x*x*x x^2 = x*x which is x^3 divided by x x^1 = x which is x^2 divided by x x^0 = 1 because any number divided by itself is equal to 1. Also, you can think of 8^1 as being 8*1, and 8^2 as being 8*8*1, since any number multiplied by 1 is just itself you don t need to display the 1, but this does explain why 8^1=8 hope this helps :)",
        "video_name": "eTWCARmrzJ0",
        "timestamps": [
            85
        ],
        "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": "Don't get the second problem at 0:42.\nAre there like any practice problems for this?",
        "A": "2^3 = 8; therefore the cube root of 8 is equal to 2, right? Another way of writing cube root of 8 (remember: that s the square root sign with a little 3 above the check mark) is 8^(1/3). Notice that the exponent is 1/3. That is why the Log (base 8) of 2 is 1/3.",
        "video_name": "eTWCARmrzJ0",
        "timestamps": [
            42
        ],
        "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": "The equation is -4x+7. Shortly after the 4:00 mark, Sal replaces the x with -1 and then says, \"4 times -1 = -4\". Shouldn't it be -4 * -1?",
        "A": "He misspoke and says 4*-1=4, but what he really meant is -4*-1=4 and he completes the equation as if he had said that correctly. It does not change the problem because he just misspoke and didn t write the incorrect statement down",
        "video_name": "nGCW5teACC0",
        "timestamps": [
            240
        ],
        "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 1:12, why is the y coordinate of the point g(-1)?",
        "A": "At this point Sal has drawn a blue line representing g(x), and wants to plot the point on that line where x = -1, so the y-coordinate has to be the value that the function g assigns to the x-value -1, which is g(-1). Possibly you just lost Sal s train of thought here, but if this doesn t make sense after this explanation, it might help to review function terminology before you proceed, as similar references will come up often.",
        "video_name": "nGCW5teACC0",
        "timestamps": [
            72
        ],
        "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": "Why does Sal draw a triangle in front of X at 2:16?\ndoes that mean change?",
        "A": "Yes, it does! The triangle is actually the Greek letter Delta. When used this way, it means change .",
        "video_name": "1F7LAJEVp-U",
        "timestamps": [
            136
        ],
        "3min_transcript": "Graph the line that represents a proportional relationship between y and x with a unit rate 0.4. That is, a change of one unit in x corresponds to a change of 0.4 units in y. And they also ask us to figure out what the equation of this line actually is. So let me get my scratch pad out and we could think about it. So let's just think about some potential x and y values here. So let's think about some potential x and y values. So when we're thinking about proportional relationships, that means that y is going to be equal to some constant times x. So if we have a proportional relationship, if you have zero x's, it doesn't matter what your constant here is, you're going to have zero y's. So the point 0, 0 should be on your line. So if this is the point 0, 0, this should be on my line right over there. Now, let's think about what happens as we increase x. So if x goes from 0 to 1, we already know that a change of 1 unit in x corresponds to a change of 0.4 units in y. So if x increases by 1, then y is going to increase by 0.4. The 0.4 is hard to graph on this little tool right over here. So let's try to get this to be a whole number. So then when x increases another 1, y is going to increase by 0.4 again. It's going to get to 0.8. When x increases again by 1, then y is going to increase by 0.4 again. It's going to get to 1.2. If x increases again, y is going to increase by 0.4 again. Notice, every time x is increasing by 1, y is increasing by 0.4. That's exactly what they told us here. Now, if x increases by 1 again to 5, then y is going to increase 0.4 to 2. And I like this point because this is nice and easy to graph. So we see that the point 0, 0 and the point 5 comma 2 should be on this graph. And I could draw it. And I'm going to do it on the tool in a second as well. So it'll look something like this. Notice the slope of this actual graph. If our change in x is 5. So notice, here our change in x is 5. Our change in x is 5. You see that as well. When you go from 0 to 5, this change in x is 5. Change in x is equal to 5. What was our corresponding change in y? Well, our corresponding change in y when our change in x was 5, our change in y was equal to 2. And you see that here, when x went from 0 to 5, y went from 0 to 2. So our change in y in this circumstance is equal to 2. So our slope, which is change in y over change in x, is the rate of change of your vertical axis with respect to your horizontal axis, is going to be equal to 2 over 5,"
    },
    {
        "Q": "At 1:57 Why can he take the radical sign off the right and stick it on the left of the equation? I feel like I missed a step or something. I can't figure out why he did that.",
        "A": "Oh, right. Thank you. It makes total sense now. I must have been trying too hard and missed it.",
        "video_name": "aeyFb2eVH1c",
        "timestamps": [
            117
        ],
        "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": "Can someone elaborate why at 1:50 Sal chose to take the positive square root.",
        "A": "The values here will be positive. The thing squared is (x+2) and the minimum value possible for x is -2. Therefore, 0 (-2+2) and up is the set of possible values. Since the values are positive (well, technically nonnegative) the principal/positive square root is sufficient.",
        "video_name": "aeyFb2eVH1c",
        "timestamps": [
            110
        ],
        "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": "At 2:13, Sal makes note of domain and range. I am familiar with the concept, but am unsure of why it is important.",
        "A": "A domain is a certain range in which the function remains applicable. In many cases, if the domains are not constrained to a certain range, the function could be meaningless. For example, if S(x) in a function stands for the number of students in a classroom, the domain of x must be greater or equal to 0. Otherwise, this functon is meaningless. I hope you see how it works :)",
        "video_name": "aeyFb2eVH1c",
        "timestamps": [
            133
        ],
        "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": "why do we ignore the decimals at 0:40?",
        "A": "thanks to kwymberry",
        "video_name": "JEHejQphIYc",
        "timestamps": [
            40
        ],
        "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 0:17, Sal says that multiplying decimals is the same as multiplying whole numbers. Why is this true?",
        "A": "He means it s ALMOST the same, but not all the same.",
        "video_name": "JEHejQphIYc",
        "timestamps": [
            17
        ],
        "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 2:13 - 2:17, How come they remove the 0 at the very end? I know that it has no meaning but why?",
        "A": "They remove it because it has no meaning and it is simpler to express the number without it.",
        "video_name": "JEHejQphIYc",
        "timestamps": [
            133,
            137
        ],
        "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": "Wait is this stuff a joke? Especially 3:42 are there really such diseases?",
        "A": "Don t worry, all of them are made up, except maybe the mind-blown syndrom. If you show hexaflexagons to your friends, they could very well be disbelieving at the amazing-ness.",
        "video_name": "AmN0YyaTD60",
        "timestamps": [
            222
        ],
        "3min_transcript": "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, become the roundest of circles. Perfectly healthy snakes may turn into snake loops, or worse, become decapitated. Either state is fatal for the snake, as having no head can lead to starvation. This can be avoided by simply marking where connections will be across neighboring triangles first. Afterwards, the lines can be filled in however you like. Be aware that with the trihexaflexagon, there are two variations to each face. So you can simply draw one side where triangles connect, and flip and draw the other. But in the hexa-hexaflexagon, the main three faces each appear four different ways. If you use hexaflexagons, keep an eye out for signs of dependency. Overuse can lead to addiction and possibly an overdose. Some users of hexaflexagons report confusion, mind-blown syndrome, hexaflexaperplexia, hexaflexadyslexia, hexaflexaperfectionism, and hexaflexa-Mexican-food-cravings. If you find yourself experiencing any of these symptoms, stop flexagon use immediately, and see the head of your math department. With proper precautions, flexagating can be a great part of your life. Follow these simple safety guidelines, experience."
    },
    {
        "Q": "At 1:01, Does Vi make two Mobius Strips tied together? Does that mean a hexaflexagon is a mobius strip?",
        "A": "Right-o! A hexaflexagon s unique ability to make that weird twist and end up with multiple sides is caused indirectly by the mobius strip it s made out of!",
        "video_name": "AmN0YyaTD60",
        "timestamps": [
            61
        ],
        "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": "At 1:50 what did you say?",
        "A": "At 1:50 the speaker said So we can change the order of the numbers without changing the answer . Hope this helped ! :)",
        "video_name": "uHHnwafYivk",
        "timestamps": [
            110
        ],
        "3min_transcript": "- [Voiceover] Let's multiply 40 times 70. So 40 times, we have the number 70. So we could actually list that out, the number 70, 40 different times and add it up. But that's clearly a lot of computations to do. And there's gotta be a faster way. So another way is to stick with multiplication, but see if we can break these numbers up, this 40 and this 70, decompose them, break them up in some way to get numbers that might be a little easier to multiply with. For me, multiplying by 10 is the easiest number, because I know the pattern to add a zero. So, I'm gonna break up 40 and say, instead of 40, four times 10. Four times 10 and 40 are equivalent. They're the same thing, so I can replace the 40 with a four times 10. And then for my 70, same thing. I can break this up and write seven times 10. So these two expressions, 40 times 70 and four times 10 times seven times 10, are equal; they're equivalent. So they'll have the same solution. But for me, this one down here is simpler to work out because of these times 10s. So I'll solve this one, knowing that I'll get the same solution as I would have for this top expression. So what we can do is we can re-order these numbers in a different order to, again, continue making this question easier for us to solve. Because in multiplication, the order doesn't matter. If we have five times two, for example, that would be the same as two times five. They're both 10. Five twos or two fives, either way, it's 10. So we can change the order of the numbers without changing the answer. So again, we're going to change our expression a little bit, but what we're not going to change is the solution. So I'm gonna put my one-digit numbers first. And then, I'll put the two-digit numbers, the 10s, times 10 and the other times 10. So we have all the same factors, all the same numbers, in both of these expressions. They've just been re-ordered. And now, I'll solve going across. Four times seven is 28. And now we have 28 times 10, and times another 10. Well, the pattern for times 10 that we know is when we multiply a whole number like 28 times 10, we will add a zero to the end. One zero for that zero in 10, because 28 times 10 is 28 10s, 28 10s, or 280. And that multiplied 28 times 10, and then if we multiply by this other 10, well we have to add another zero."
    },
    {
        "Q": "At 1:43 what is the difference in tangent line and secant line?",
        "A": "A tangent line touches the curve at one point whereas the secant line intersects at two points. The secant slope can be found by simple slope eq: y2 - y1/x2-x1 The tangent slope is found by f(x) - f(a)/ x - a. Hope this helps",
        "video_name": "BYTfCnR9Sl0",
        "timestamps": [
            103
        ],
        "3min_transcript": "What I want to do in this video is to see whether the power rule is giving us results that at least seem reasonable. This is by no means a proof of the power rule, but at least we'll feel a little bit more comfortable using it. So let's say that f of x is equal to x. The power rule tells us that f prime of x is going to be equal to what? Well, x is the same thing as x to the first power. So n is implicitly 1 right over here. So we bring the 1 out front. It'll be 1 times x to the 1 minus 1 power. So it's going to be 1 times x to the 0 power. x to the 0 So it's just going to be equal to 1. Now, does that makes conceptual sense if we actually try to visualize these functions? So let me actually try to graph these functions. So that's my y-axis. This is my x-axis. And let me graph y equals x. So y is equal to f of x here. So y is equal to x. So it looks something like that. Or this is f of x is equal to x, or y is equal to this f of x right over there. Now, actually, let me just call that f of x just to not confuse you. So this right over here is f of x is equal to x that I graphed right over here. y is equal to f of x, which is equal to x. And now, let me graph the derivative. Let me graph f prime of x. That's saying it's 1. That's saying it's 1 for all x. Regardless of what x is, it's going to be equal to 1. Is this consistent with what we know about derivatives and slopes and all the rest? Well, let's look at our function. What is the slope of the tangent line right at this point? Well, right over here, this has slope 1 continuously. Or it has a constant slope of 1. Slope is equal to 1 no matter what x is. It's a line. And for a line, the slope is constant. So over here, the slope is indeed 1. If you go to this point over here, the slope is indeed 1. If you go over here, the slope is indeed 1. So we've got a pretty valid response there. So let's say I have g of x is equal to x squared. The power rule tells us that g prime of x would be equal to what? Well, n is equal to 2. So it's going to be 2 times x to the 2 minus 1. Or it's going to be equal to 2 x to the first power. It's going to be equal to 2x. So let's see if this makes a reasonable sense. And I'm going to try to graph this one a little bit more precisely. Let's see how precisely I can graph it. So this is the x-axis, y-axis. Let me mark some stuff off here. So this is 1, 2, 3, 4, 5. This is 1, 2, 3, 4. 1, 2, 3, 4. So g of x."
    },
    {
        "Q": "8:20 but 1/2 squared is 1/4 :'<",
        "A": "The ambiguity of not having parenthesis. If you are referring to the blue writing I think it is (1^2)/ 2 and not (1/2)^2 since he substituted in 1 for x in x^2/2.",
        "video_name": "vhMl755vR5Q",
        "timestamps": [
            500
        ],
        "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 1:28 do you have to do it height times width times depth in that order?",
        "A": "Yes, the formula for volume in rectangular prisms is: V=B*L*H V=volume B=base L=length H=height.",
        "video_name": "I9efKVtLCf4",
        "timestamps": [
            88
        ],
        "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": "0:39 i dont understand the math from here on",
        "A": "the video shows how to find the volume from there on for example lxwxh= 2x4x3",
        "video_name": "I9efKVtLCf4",
        "timestamps": [
            39
        ],
        "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 1:05, Can we use any symbol (\u00c3\u00ab, \u00c3\u00a6, \u00c2\u00b4\u00c2\u00ac or \u00c2\u00a4) represent an unknown number?",
        "A": "I think that s a yes, since Mr. Khan says you could use a smiley face.",
        "video_name": "Tm98lnrlbMA",
        "timestamps": [
            65
        ],
        "3min_transcript": "I'm here with Jesse Ro, whose a math teacher at Summit San Jose and a Khan Academy teaching fellow and you had some interesting ideas or questions. Yeah, one question that students ask a lot when they start Algebra is why do we need letters, why can't we just use numbers for everything? Why letters? So why do we have all these Xs and Ys and Zs and ABCs when we start dealing with Algebra? Yeah, exactly. That's interesting, well why don't we let people think about that for a second. So Sal, how would you answer this question? Why do we need letters in Algebra? So why letters. So there are a couple of ways I'd think about it. One is if you have an unknown. So if I were to write X plus three is equal to ten the reason why we're doing this is that we don't know what X is It's literally an unknown. And so we're going to solve for it in some way. But it did not have to be the letter X. We could have literally written blank plus three is equal to ten. Or we could have written Question Mark plus three is equal to ten. So it didn't have to be letters, but we needed some type of symbol. But until you know it, you need some type of a symbol to represent whatever that number is. Now we can go and solve this equation and then know what that symbol represents. But if we knew it ahead of time, it wouldn't be an unknown. It wouldn't be something that we didn't know. So that's one reason why I would use letters and where just numbers by itself wouldn't be helpful. The other is when you're describing relationships between numbers. So I could do something like - I could say - that whenever you give me a three, I'm going to give you a four. And I could say, if you give me a five, I'm going to give you a six. And i could keep going on and on forever. If you give me a 7.1, I'm going to give you an 8.1. And I could keep listing this on and on forever. Maybe you could give me any number, and I could tell you what I'm going to give you. But I would obviously run out of space and time if I were to list all of them. And we could do that much more elegantly if we used letters to describe the relationship. And so I say, look, whatever you give me, I'm going to add one to it. And that's what I'm going to give back to you. And so now, this very simple equation here can describe an infinite number of relationships between X or an infinite number of corresponding Ys and Xs. So now someone knows whatever X you give me you give me three, I add one to it, and I'm going to give you four. You give me 7.1, I'm going to add one to it and give you 8.1. So there is no more elegant way that you could've done it than by using symbols. With that said, I didn't have to use Xs and Ys. This is just a convention that kind of comes to use from history. I could've defined what you give me as Star and what I give you as Smiley Face and this also would've been a valid way to express this. So the letters are really just symbols. Nothing more."
    },
    {
        "Q": "At 2:16, it could have been (Star) + 1 = ... I don't know... Smiley Face). We didn't have to say \"y = x + 1\".",
        "A": "It could have been. But people thought letters were more simpler.",
        "video_name": "Tm98lnrlbMA",
        "timestamps": [
            136
        ],
        "3min_transcript": "I'm here with Jesse Ro, whose a math teacher at Summit San Jose and a Khan Academy teaching fellow and you had some interesting ideas or questions. Yeah, one question that students ask a lot when they start Algebra is why do we need letters, why can't we just use numbers for everything? Why letters? So why do we have all these Xs and Ys and Zs and ABCs when we start dealing with Algebra? Yeah, exactly. That's interesting, well why don't we let people think about that for a second. So Sal, how would you answer this question? Why do we need letters in Algebra? So why letters. So there are a couple of ways I'd think about it. One is if you have an unknown. So if I were to write X plus three is equal to ten the reason why we're doing this is that we don't know what X is It's literally an unknown. And so we're going to solve for it in some way. But it did not have to be the letter X. We could have literally written blank plus three is equal to ten. Or we could have written Question Mark plus three is equal to ten. So it didn't have to be letters, but we needed some type of symbol. But until you know it, you need some type of a symbol to represent whatever that number is. Now we can go and solve this equation and then know what that symbol represents. But if we knew it ahead of time, it wouldn't be an unknown. It wouldn't be something that we didn't know. So that's one reason why I would use letters and where just numbers by itself wouldn't be helpful. The other is when you're describing relationships between numbers. So I could do something like - I could say - that whenever you give me a three, I'm going to give you a four. And I could say, if you give me a five, I'm going to give you a six. And i could keep going on and on forever. If you give me a 7.1, I'm going to give you an 8.1. And I could keep listing this on and on forever. Maybe you could give me any number, and I could tell you what I'm going to give you. But I would obviously run out of space and time if I were to list all of them. And we could do that much more elegantly if we used letters to describe the relationship. And so I say, look, whatever you give me, I'm going to add one to it. And that's what I'm going to give back to you. And so now, this very simple equation here can describe an infinite number of relationships between X or an infinite number of corresponding Ys and Xs. So now someone knows whatever X you give me you give me three, I add one to it, and I'm going to give you four. You give me 7.1, I'm going to add one to it and give you 8.1. So there is no more elegant way that you could've done it than by using symbols. With that said, I didn't have to use Xs and Ys. This is just a convention that kind of comes to use from history. I could've defined what you give me as Star and what I give you as Smiley Face and this also would've been a valid way to express this. So the letters are really just symbols. Nothing more."
    },
    {
        "Q": "For the graph that Sal draws beginning at 1:35, what does the Y axis represent? I understand how the probability of the event is represented by the area under the curve, but doesn't that mean the Y axis doesn't chart probability, but something different? Is it the probability of the probability? ;)",
        "A": "Yes, the y axis charts something different. The y value at each value of x (or possible outcome) is the rate of change of the area under the curve as the interval (range of possible outcomes) increases or decreases. This ensures that the area under the curve in any interval (range of outcomes) is always equal to the probability for that interval.",
        "video_name": "Fvi9A_tEmXQ",
        "timestamps": [
            95
        ],
        "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": "At 5:07 Sal says that the two statements P(|Y-2|<.1) and P(1.9<Y<2.1) are the same. Why?",
        "A": "|Y-2|<.1 is the same as 1.9<Y<2.1 because : solving |Y-2|<.1 for (Y-2) >= 0 gives Y-2<.1 wich gives Y < 2.1 solving |Y-2|<.1 for (Y-2) < 0 gives -Y+2<.1 wich gives -Y<-1.9 wich gives Y > 1.9 Search for lecture about absolute value for more explanation.",
        "video_name": "Fvi9A_tEmXQ",
        "timestamps": [
            307
        ],
        "3min_transcript": "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? I'll let you think about it a little bit. But now this starts to make a little bit of sense. Now we have an interval here. So we want all Y's between 1.9 and 2.1. So we are now talking about this whole area. And area is key. So if you want to know the probability of this occurring, you actually want the area under this curve from this point to this point. And for those of you who have studied your calculus, that would essentially be the definite integral of this probability density function from this point to this point. So from-- let me see, I've run out of space down here. So let's say if this graph-- let me draw it in a different color. If this line was defined by, I'll call it f of x. I could call it p of x or something."
    },
    {
        "Q": "At 1:23, why is it plus or minus the square root? Why not just square root?",
        "A": "A square root undoes a power term, but a power term can hide information. Think what happens if you have x^2 and x = 1, you get 1 right? But if x were to equal -1 you would still get one. So if you were to solve with out putting the + - you only get half of the answer. This will become really big when you start solving quadratic equations.",
        "video_name": "RweAgQwLdMs",
        "timestamps": [
            83
        ],
        "3min_transcript": "We're asked to solve the equation 2x squared plus 3 is equal to 75. So in this situation, it looks like we might be able to isolate the x squared pretty simply. Because there's only one term that involves an x here. It's only this x squared term. So let's try to do that. So let me just rewrite it. We have 2x squared plus 3 is equal to 75. And we're going to try to isolate this x squared over And the best way to do that, or at least the first step, would be to subtract 3 from both sides of this equation. So let's subtract 3 from both sides. The left hand side, we're just left with 2x squared. That was the whole point of subtracting 3 from both sides. And on the right hand side, 75 minus 3 is 72. Now, I want to isolate this x squared. I have a 2x squared here. So I could have just an x squared here if I divide this side or really both sides by 2. Anything I do to one side, I have to do to the other side if I want to maintain the equality. So the left side, just becomes x squared. So we're left with x squared is equal to 36. And then to solve for x, we can take the positive, the plus or minus square root of both sides. So we could say the plus or-- let me write it this way-- If we take the square root of both sides, we would get x is equal to the plus or minus square root of 36, which is equal to plus or minus 6. Let me just write that on another line. So x is equal to plus or minus 6. And remember here, if something squared is equal to 36, that something could be the negative version or the positive version. It could be the principal root or it could be the negative root. Both negative 6 squared is 36 and positive 6 squared is 36, so both of these work. And you could put them back into the original equation to verify it. Let's do that. If you say 2 times 6 squared plus 3, that's 2 times 36, which is 72 plus 3 is 75. going to get the exact same result. Because negative 6 squared is also 36. 2 times 36 is 72 plus 3 is 75."
    },
    {
        "Q": "How does Sal know at 6:34, 6:38, and 6:46 that y=x^2, xy=12, and 5/x+y=10 are not linear equations without graphing them first?",
        "A": "Linear equations have specific formats. For example, here are some of their formats: 1) Ax + By = C where A, B and C are integers 2) y = mx + b where m and b are numbers None of Sal s equations look like the examples above. y = x^2: this has an exponent on x which makes it non-linear xy = 12: this has x and y being multiplied which doesn t occur in a linear equation 5/x + y = 10: this has x in a denominator which doesn t occur in a linear equation Hope this helps.",
        "video_name": "AOxMJRtoR2A",
        "timestamps": [
            394,
            398,
            406
        ],
        "3min_transcript": "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, linear equations. One way to think about is it's an equation that if you were to graph all of the x and y pairs that satisfy this equation, you'll get a line. And that's actually literally where the word linear equation comes from. But another way to think about it is it's going to be an equation where every term is either going to be a constant, so for example, twelve is a constant. It's not going to change based on the value of some variable, twelve is twelve. Or negative three is negative three. So every term is either going to be a constant or it's going to be a constant times a variable raised to the first power. So this is the constant two times x to the first power. This is the variable y raised to the first power. You could say that bceause this is just one y. We're not dividing by x or y, we're not multiplying, we don't have a term that has x to the second power, or x to the third power, or y to the fifth power. We just have y to the first power, we have x to the first power. We're not multiplying x and y together like we did over here."
    },
    {
        "Q": "At 4:55, on the sketched graph I noticed that some values of h(x) were actually smaller than some values of f(x)(check out the minimum of h(x) and the maximum of f(x)), if the inequallity is right, shouldn't the values of h(x) always be bigger than f(x)'s?",
        "A": "This means above and below,(Recall [f(x)=y] so its like saying f(x) s y value < g(x) s y value < h(x) s y value Were only comparing the function evaluated at the same x-values (x is the input y [which is equal to f(x)] is the output)",
        "video_name": "WvxKwRcHGHg",
        "timestamps": [
            295
        ],
        "3min_transcript": "and Diya's calories as a function of the day is always going to be in between those. So now let's make this a little bit more mathematical. So let me clear this out so we can have some space to do some math in. So let's say that we have the same analogy. So let's say that we have three functions. Let's say f of x over some interval is always less than or equal to g of x over that same interval, which is always less than or equal to h of x over that same interval. So let me depict this graphically. So that is my y-axis. This is my x-axis. in the x-axis right over here. So let's say h of x looks something like that. Let me make it more interesting. This is the x-axis. So let's say h of x looks something like this. So that's my h of x. Let's say f of x looks something like this. Maybe it does some interesting things, and then it comes in, and then it goes up like this, so f of x looks something like that. And then g of x, for any x-value, g of x is always in between these two. And I think you see where the squeeze is happening and where the sandwich is happening. If h of x and f of x were bendy pieces of bread, g of x would be the meat of the bread. So it would look something like this. Now, let's say that we know-- this is the analogous thing. On a particular day, Sal and Imran ate the same amount. the limit as f and h approach that x-value, they approach is the same limit. So let's take this x-value right over here. Let's say the x-value is c right over there. And let's say that the limit of f of x as x approaches c is equal to L. And let's say that the limit as x approaches c of h of x is also equal to L. So notice, as x approaches c, h of x approaches L. As x approaches c from either side, f of x approaches L. So these limits have to be defined. Actually, the functions don't have to be defined at x approaches c. Just over this interval, they have to be defined as we approach it."
    },
    {
        "Q": "What does Sal mean by \"wacky functions\" at 6:54?",
        "A": "He means functions that are complex or difficult to manipulate algebraically.",
        "video_name": "WvxKwRcHGHg",
        "timestamps": [
            414
        ],
        "3min_transcript": "the limit as f and h approach that x-value, they approach is the same limit. So let's take this x-value right over here. Let's say the x-value is c right over there. And let's say that the limit of f of x as x approaches c is equal to L. And let's say that the limit as x approaches c of h of x is also equal to L. So notice, as x approaches c, h of x approaches L. As x approaches c from either side, f of x approaches L. So these limits have to be defined. Actually, the functions don't have to be defined at x approaches c. Just over this interval, they have to be defined as we approach it. And if these limits right over here are defined and because we know that g of x is always sandwiched in between these two functions, therefore, on that day or for that x-value-- I should get out of that food-eating analogy-- this tells us if all of this is true over this interval, this tells us that the limit as x approaches c of g of x must also be equal to L. And once again, this is common sense. f of x is approaching L, h of x is approaching L, g of x is sandwiched in between it. So it also has to be approaching L. And you might say, well, this is common sense. Why is this useful? Well, as you'll see, this is useful for finding the limits of some wacky functions. If you can find a function that's always greater than it and a function that's always less than it, and you can find the limit as they approach some c, know that that wacky function in between is going to approach that same limit."
    },
    {
        "Q": "Wait, I don't understand something. At 2:50, Sal squared the numbers 15 and 20. Why? What was the need to square them? He then added them, got 625 as a sum, and then found the positive square root of it--25--and deemed that number the length of the rope. Isn't there a more simpler way to solve the problem?",
        "A": "Pythagorean theorem: a^2+b^2=c^2 in this case: 15^2+20^2=c^2 you re trying to figure out c (length of rope) so 625 answer -- from that square root of 625 is 25",
        "video_name": "JVrkLIcA2qw",
        "timestamps": [
            170
        ],
        "3min_transcript": "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. hypotenuse. The hypotenuse will always be the longest side. Let's say the hypotenuse is in green just so we get our color coding nice. That is going to be equal to the rope squared. Or the length of the rope. Let's call this distance right here r. r for rope. So 15 squared plus 20 squared is going to be equal to r squared. And what's 15 squared? It's 225. 20 squared is 400. And that's going to be equal to r squared. Now 225 plus 400 is 625. 625 is equal to r squared. And then we can take the principal root of both sides of this equation. Because we're talking about distances, we want the positive square root."
    },
    {
        "Q": "In 2:05 when Sal said that... when you know two sides of the triangle you calculate the the hypotenuse...My question is that can you calculate the hypotenuse or any one of the sides if you only know ONE side of the triangle?",
        "A": "In general, no, unless you know something else about the triangle. There are techniques using trigonometry that will let you find sides if you know one side and one angle (other than the right angle).",
        "video_name": "JVrkLIcA2qw",
        "timestamps": [
            125
        ],
        "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": "at 1:44 he shows that (7) (5) means the same thing is 7x5\nwhat is the reasoning or history for why a number inside a ( ) means to multiply. I would love to understand the thinking behind that better.",
        "A": "Like most shortcuts... laziness. I suspect that if you look back far enough, some symbol was regularly used for multiplication... but anything you do that much begs for simplification so at some point someone said how about we call it multiply if there is just no operator and put in the operator if it is anything else ... This works pretty well so we kept it.",
        "video_name": "Yw3EoxC_GXU",
        "timestamps": [
            104
        ],
        "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": "I thought that dx meant that the integral was done relative to the variable x. So at around 3:16 he brings the dx to the middle of the equation. Doesn't that affect the e to the power of .... somehow? Also why is dx being treated as a variable?",
        "A": "It doesn t affect the integral at all. dx is essentially meaning a small change in x (think of a minuscule delta-x). It is for all intents and purposes in integration a dummy variable: like any other variable in an expression, you can move it around subject to the laws of multiplicative association. dx is what ultimately creates the +C at the end as well, because of this treatment as a dummy variable. Yet another question in this section that goes to blaming Leibniz.",
        "video_name": "b76wePnIBdU",
        "timestamps": [
            196
        ],
        "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": "At 3:26, where did the dx go? Can I just leave it out? And if yes, why?",
        "A": "The dx is still there at 3:26. It s between the two functions in the integral.",
        "video_name": "b76wePnIBdU",
        "timestamps": [
            206
        ],
        "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": "@2:01 Can you please explain how you used the notation du/dx as a fraction?\nHow can we pretend that it's a fraction?",
        "A": "Remember that slope is \u00ce\u0094y/\u00ce\u0094x, change in y over change in x, and that IS a fraction. Well dy/dx (or du/dx) is the same deal, just that we are dealing with infinitesimals, which we call the differential difference, that is dx is what happens to \u00ce\u0094x in the limit when \u00ce\u0094x approaches zero, thus dy/dx, du/dx are fractions.",
        "video_name": "b76wePnIBdU",
        "timestamps": [
            121
        ],
        "3min_transcript": "Let's say that we have the indefinite integral, and the function is 3x squared plus 2x times e to x to the third plus x squared dx. So how would we go about solving this? So first when you look at it, it seems like a really complicated integral. We have this polynomial right over here being multiplied by this exponential expression, and over here in the exponent, we essentially have another polynomial. It seems kind of crazy. And the key intuition here, the key insight is that you might want to use a technique here called u-substitution. And I'll tell you in a second how I would recognize that we have to use u-substitution. And then over time, you might even be able to do this type of thing in your head. u-substitution is essentially unwinding the chain rule. And the chain rule-- I'll go in more depth in another video, 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."
    },
    {
        "Q": "i do not understand 4:09 it is a little hazy can someone please explain it to me??",
        "A": "it s to easy he seperates the inequality into two peices so it will be more easy to solve",
        "video_name": "A3xPhzs-KBI",
        "timestamps": [
            249
        ],
        "3min_transcript": "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 We're going to circle it in because we have a greater than or equal to. And then x is greater than that, but it has to be less than or equal to 17. So it could be equal to 17 or less than 17. So this right here is a solution set, everything that I've shaded in orange. And if we wanted to write it in interval notation, it would be x is between negative 1 and 17, and it can also equal negative 1, so we put a bracket, and it can also equal 17. So this is the interval notation for this compound inequality right there. Let's do another one. Let me get a good problem here. Let's say that we have negative 12. I'm going to change the problem a little bit from the one that I've found here. Negative 12 is less than 2 minus 5x, which is less than I want to do a problem that has just the less than and a less than or equal to. The problem in the book that I'm looking at has an equal sign here, but I want to remove that intentionally because I want to show you when you have a hybrid situation, when you have a little bit of both. So first we can separate this into two normal inequalities. You have this inequality right there. We know that negative 12 needs to be less than 2 minus 5x. That has to be satisfied, and-- let me do it in another color-- this inequality also needs to be satisfied. 2 minus 5x has to be less than 7 and greater than 12, less than or equal to 7 and greater than negative 12, so and 2 minus 5x has to be less than or equal to 7."
    },
    {
        "Q": "at 8:30 i dont get the dif. between \"and\" and \"or\". i went on the practice questions and it said something about \" the first inequality is included........\". im so confused i kept on getting it wrong help!",
        "A": "OK. What you have to do is to first solve the first inequality and then the second.",
        "video_name": "A3xPhzs-KBI",
        "timestamps": [
            510
        ],
        "3min_transcript": "I just wrote this improper fraction as a mixed number. Now let's do the other constraint over here in magenta. So let's subtract 2 from both sides of this equation, just like we did before. And actually, you can do these simultaneously, but it becomes kind of confusing. So to avoid careless mistakes, I encourage you to separate it out like this. So if you subtract 2 from both sides of the equation, the left-hand side becomes negative 5x. The right-hand side, you have less than or equal to. The right-hand side becomes 7 minus 2, becomes 5. Now, you divide both sides by negative 5. On the left-hand side, you get an x. On the right-hand side, 5 divided by negative 5 is negative 1. And since we divided by a negative number, we swap the inequality. It goes from less than or equal to, to greater than or equal to. So we have our two constraints. x has to be less than 2 and 4/5, and it has to be greater So we could write it like this. x has to be greater than or equal to negative 1, so that would be the lower bound on our interval, and it has to be less than 2 and 4/5. And notice, not less than or equal to. That's why I wanted to show you, you have the parentheses there because it can't be equal to 2 and 4/5. x has to be less than 2 and 4/5. Or we could write this way. x has to be less than 2 and 4/5, that's just this inequality, swapping the sides, and it has to be greater than or equal to negative 1. So these two statements are equivalent. And if I were to draw it on a number line, it would look like this. So you have a negative 1, you have 2 and 4/5 over here. Obviously, you'll have stuff in between. We have to be greater than or equal to negative 1, so we can be equal to negative 1. And we're going to be greater than negative 1, but we also have to be less than 2 and 4/5. So we can't include 2 and 4/5 there. We can't be equal to 2 and 4/5, so we can only be less than, so we put a empty circle around 2 and 4/5 and then we fill in everything below that, all the way down to negative 1, and we include negative 1 because we have this less than or equal sign. So the last two problems I did are kind of \"and\" problems. You have to meet both of these constraints. Now, let's do an \"or\" problem. So let's say I have these inequalities. Let's say I'm given-- let's say that 4x minus 1 needs to be greater than or equal to 7, or 9x over 2 needs to be less than 3. So now when we're saying \"or,\" an x that would satisfy these"
    },
    {
        "Q": "At 1:58, why does he divide by ten?",
        "A": "he divided by ten because if you look after you solve the 9x6 which was what was in the parentheses you then solve the 54/10 because in order of operation division is after parentheses, exponents, and multiplication and since there is none of that you divide which comes after multiplication in order of operations.",
        "video_name": "STyoP3rCmb0",
        "timestamps": [
            118
        ],
        "3min_transcript": "Let's see if we can multiply 9 times 0.6. Or another way to write it, we want to calculate 9 times 0.6. I'll write it like this-- 0.6. We want to figure out what this is equal to. And I encourage you to pause the video and try to figure it out on your own. And I'll give you a little bit of a hint. 0.6 is the same thing as 6 divided by 10. We know that if we start with 6, which we could write as 6.0, and if you were to divide it by 10, dividing by 10 is equivalent to moving the decimal place one place to the left. So 6 divided by 10 is 0.6. We are moving the decimal one place to the left. So I'm assuming you given a go at it. But what I'm going to do is use this that we already know to rewrite what we're trying to multiply. 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."
    },
    {
        "Q": "At 1:15 Sal says \"We could either do the 6 divided by the 10 first ... or we could do the 9*6 first\"\n\nWhat happened to BIDMAS?\n\nCan someone explain?\n\nI know the answers will be the same but then where does the principle apply then and how?\n\nThanks",
        "A": "Im not familiar with the term BIDMAS but we used PEMDAS meaning parentheses are done first followed by exponents then you can do either division OR multiplication with addition or subtraction as the last operation. I was taught that multiplication and division are equal as far as the sequence of their operation is performed. The same goes for addition and subtraction. It just depends on what operation you are most comfortable performing first.",
        "video_name": "STyoP3rCmb0",
        "timestamps": [
            75
        ],
        "3min_transcript": "Let's see if we can multiply 9 times 0.6. Or another way to write it, we want to calculate 9 times 0.6. I'll write it like this-- 0.6. We want to figure out what this is equal to. And I encourage you to pause the video and try to figure it out on your own. And I'll give you a little bit of a hint. 0.6 is the same thing as 6 divided by 10. We know that if we start with 6, which we could write as 6.0, and if you were to divide it by 10, dividing by 10 is equivalent to moving the decimal place one place to the left. So 6 divided by 10 is 0.6. We are moving the decimal one place to the left. So I'm assuming you given a go at it. But what I'm going to do is use this that we already know to rewrite what we're trying to multiply. 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."
    },
    {
        "Q": "At 3:00 how is how is 9 * 0.6=5.4 . isn't the product of a multiplication problem always bigger then the numbers you are multiplying?",
        "A": "not always because when you multiply by a number less than 1 in a multiplication the biggest number gets smaller. 0.6 is 6/10 * 9 is 54/10 and that is 5.4",
        "video_name": "STyoP3rCmb0",
        "timestamps": [
            180
        ],
        "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": "in the video @ 4:54 i do not understand the part where the instructor of the video multiplied 4*84 to get the sum of the previous four tests.\n\n80 + 81 + 87 + 88 = 4 * 84 ? <--- how is this possible, i think its cool but do not understand how that works. thanks!!",
        "A": "80+81+87+88=80+80+1+80+7+80+8=(80+80+80+80)+(0+1+7+8)=4*80+16=4*(80+4)=4*84",
        "video_name": "9VZsMY15xeU",
        "timestamps": [
            294
        ],
        "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": "at 1:27, what in the world is he doing??? I mean, i don't get it. Where did the .25 come from??",
        "A": "Did you study long division yet? 4 s into 29 go 7 times with 1 left over. So we do not throw the 1 away, but instead put a decimal point and carry on. We always put a 0 after whatever digit we are left with, so the 1, becomes 10, so how many 4 s go into 10.....? two 4 s go into 10 with 2 left over. Once again put a 0 after the 2 and we get 20, 4 s into 20 go 5 times with none left over. Our final answer is 7.25. :)",
        "video_name": "9VZsMY15xeU",
        "timestamps": [
            87
        ],
        "3min_transcript": "Welcome to the presentation on averages. Averages is probably a concept that you've already used before, maybe not in a mathematical way. But people will talk in terms of, the average voter wants a politician to do this, or the average student in a class wants to get out early. So you're probably already familiar with the concept of an average. And you probably already intuitively knew that an average is just a number that represents the different values that a group could have. But it can represent that as one number as opposed to giving all the different values. And let's give a couple of examples of how to compute an average, and you might already know how to do this. So let's say I had the numbers 1, 3, 5, and 20. And I asked you, what is the average of these four numbers? Well, what we do is, we literally just add up the numbers. And then divide by the number of numbers we have. So we say 1 plus 3 is 4. 4 plus 5 is 9. 9 plus 20 is 29. And we had 4 numbers; one, two, three, four. So 4 goes into 29. And it goes, 7, 7, 28. And then we have 10, I didn't have to do that decimal there, oh well. 2, 8, 25. So 4 goes into 29 7.25 times. So the average of these four numbers is equal to 7.25. And that might make sense to you because 7.25 is someplace in between these numbers. And we can kind of view this, 7.25, as one way to represent these four numbers without having to list these Like the mode. You'll also the mean, which we'll talk about later, is actually the same thing as the average. But the average is just one number that you can use to represent a set of numbers. So let's do some problems which I think are going to be close to your heart. Let's say on the first four tests of an exam, I got a-- let's see, I got an 80, an 81. An 87, and an 88. What's my average in the class so far? Well, all I have to do is add up these four numbers. So I say, 80 plus 81 plus 87 plus 88. Well, zero plus 1 is 1."
    },
    {
        "Q": "The ratio of chocolate bars to taffy pieces in a candy shop is 7:3, the total amount of candy is 3000. Then the candy store buys 32 more chocolate for there new daily order. How meny chocolate bars and taffy pieces are there?",
        "A": "3000*Candy=3*700*Chocolate+3*300*Taffy=2100*Chocolate+900*Taffy 2100+32=2132 the answer is 2132 chocolate bars and 900 taffy pieces",
        "video_name": "MaMk6-f3T9k",
        "timestamps": [
            423
        ],
        "3min_transcript": "Table 2-- 11 to 4 and then 12 to 5. Here, it's just incrementing by 1, but the ratios are not the same. 11 to 4 is not the same thing as 12 to 5. So we're not going to be able to-- this right over here is not a legitimate table. Table 3-- so 1 to 1. Then when you double the distance, we double the time. When you triple the distance from 1, you didn't triple the time. So table 3 doesn't seem to make sense, either. Table 4-- so 14 to 10. So that's the same thing as-- let's see, that's the same ratio as, if we were to divide by 2, as 7 to 5 ratio. If we divide both of these by 3, this is also a 7 to 5 ratio. And if you divide both of these by 7, this is also a 7 to 5 So table 4 seems like a completely reasonable scenario."
    },
    {
        "Q": "At 14:50, you do what you've done many times before and write matrix multiplication as the sum of the each product of the column in the matrix and the row (1 scalar value) in the vector. However, this multiplication is actually defined in the opposite way (first row * first column = entry1,1) etc. Why are you using a different definition for matrix vector multiplication?",
        "A": "The two definitions of matrix multiplication are actually equivalent. The definition Sal uses is, however, generally more compact and also carries useful geometric intuition (although not necessarily easier for numerical calculations). From your typical definition, you would find that the first element of the product is v1,1*x1 + v2,1*x2 + ... + vn,1*xn. However, by taking the vector sum in the definition Sal uses, you also obtain the same result.",
        "video_name": "ondmopWLiEg",
        "timestamps": [
            890
        ],
        "3min_transcript": "So it just equals-- I'll write a little bit lower. That equals c a1 times this column vector, times v1. Plus c a2 times v2 times this guy, all the way to plus c a n times vn. And if you just factor this c out, once again, scalar multiplication times vectors exhibits the distributive property. I believe I've done a video on that, but it's very easy to prove. So this will be equal to c times-- I'll just stay in one color right now-- a1 v1 plus a2 v2 plus all the way to a n vn. And what is this thing equal to? Well that's just our matrix A times our vector-- or our Maybe I'm overloading the letter A. My matrix uppercase A times my vector lowercase a. Where the lowercase a is just this thing right here, a1, a2 This thing up here was the same thing as that. So I just showed you that if I take my matrix and multiply it times some vector that was multiplied by a scalar first, that's equivalent to first multiplying the matrix times a vector and then multiplying by the scalar. So we've shown you that matrix times vector products or matrix vector products satisfied this condition of linear transformations and this condition. So the big takeaway right here is matrix multiplication. And this is a important takeaway. Matrix multiplication or matrix products with vectors And this is a bit of a side note. In the next video I'm going to show you that any linear transformation-- this is incredibly powerful-- can be represented by a matrix product or by-- any transformation on any vector can be equivalently, I guess, written as a product of that vector with a matrix. Has huge repercussions and you know, just as a side note, kind of tying this back to your everyday life. You have your Xbox, your Sony Playstation and you know, you have these 3D graphic programs where you're running around and shooting at things. And the way that the software renders those programs where you can see things from every different angle, you have a cube then if you kind of move this way a little bit, the cube will look more like this and it gets rotated, and you move up and down, these are all transformations of matrices. And we'll do this in more detail."
    },
    {
        "Q": "Where does the (x-a)^2 come from? [at 1:35]",
        "A": "We put it in ourselves to create an equation that could be used to solve for the question mark.",
        "video_name": "bNQY0z76M5A",
        "timestamps": [
            95
        ],
        "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": "At 9:11 why is +9/4 when moved to the right side of the equation not a -9/4?",
        "A": "Well, as you probably know from the video, this method is call completing the square . In this step, Sal took the coefficient in front of the x, which is -3. To find the constant we must add to the left-hand side, we do ((coefficient/2)^2), which in this case gives ((-3/2)^2) = 9/4. When adding 9/4 to the left-hand side, in order to keep the equality between both sides of the equal sign, we must thus add 9/4 to the right-hand side of the equation.",
        "video_name": "bNQY0z76M5A",
        "timestamps": [
            551
        ],
        "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": "at 12:30 wouldn't the answer be 4.55?",
        "A": "These are the two values we have to get the answer for x: 3/2 and \u00e2\u0088\u009a(61/20) and we know that x can equal 3/2 plus or minus the value of the square root. so, 3/2 +- \u00e2\u0088\u009a(61/20). We can just get the values of both of the terms and add/subtract appropriately. 3/2 is 1.5 and \u00e2\u0088\u009a(61/20) is 1.74642... So we can solve thusly x = 1.5 + 1.74642 x = 3.246 or x = 1.5 - 1.74642 x = -0.246",
        "video_name": "bNQY0z76M5A",
        "timestamps": [
            750
        ],
        "3min_transcript": "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 Let's do the subtraction version. So we can actually put our entry-- if you do second and then entry, that we want that little yellow entry, that's why I pressed the second button. So I press enter, it puts in what we just put, we can just change the positive or the addition to a subtraction and you get negative 0.246. So you get negative 0.246. And you can actually verify that these satisfy our original equation. Our original equation was up here. Let me just verify for one of them. So the second answer on your graphing calculator is the last answer you use. So if you use a variable answer, that's this number right here. So if I have my answer squared-- I'm using answer represents negative 0.24."
    },
    {
        "Q": "At 4:12, why is the square root of 9 equal to +/- 3?",
        "A": "because, nomatter the numbers is positive or nagetive if it be squared, it will become positive. so square +3 =9 square -3 =9 =====\u00e3\u0080\u008b square root of 9 =+3/-3",
        "video_name": "bNQY0z76M5A",
        "timestamps": [
            252
        ],
        "3min_transcript": "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. minus 2 is equal to plus or minus 3. Add 2 to both sides, you get x is equal to 2 plus or minus 3. That tells us that x could be equal to 2 plus 3, which is 5. Or x could be equal to 2 minus 3, which is negative 1. And we are done. Now I want to be very clear. You could have done this without completing the square. We could've started off with x squared minus 4x is equal to 5. We could have subtracted 5 from both sides and gotten x squared minus 4x minus 5 is equal to 0. And you could say, hey, if I have a negative 5 times a positive 1, then their product is negative 5 and their sum is negative 4. So I could say this is x minus 5 times x plus 1 is equal to 0."
    },
    {
        "Q": "at time 10:45, can anyone explain to me how we get the +/- square root of 61/20. specifically, the reason why we have +/- the square root. what rule is that?",
        "A": "I don t know if this will help or not but I ll try to explain. So basically you re wondering why should there be a positive + and negative - square root, right? Think about any squared number really does have two possible square roots, the positive and the negative ones. For instance \u00c2\u00b1\u00e2\u0088\u009a9= -3 or 3 Because if we square (-3)^2=9 (\u00e2\u0086\u0092Notice that this different from -3^2 which is equal to =-9) and 3^3=9",
        "video_name": "bNQY0z76M5A",
        "timestamps": [
            645
        ],
        "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 4:05, why can't you solve the equation as\n(x-2) = 9 (x-2) = 9\nx = 11 x = 11\nWhy do we instead need to take the square root of (x-2)^2 and of 9 to get\n(x-2) = +-3?\n\nIsn't the former how we've been solving these equations up until this point?",
        "A": "The ones that you solved before always had the other side was 0 which would allow you to do this. If you have (x-3)(x+4) = 12, you should not have said x - 3 = 12 or x + 4 = 12 which would give you wrong answers. You would have to multiply to get a trinomial, subtract 12, and refactor if you could or use quadratic formula to see if there are factors. Also, if you put your attempted answer back in as a check (x-2)^2 = 9, you would have (11-2)^2 = 9 or 81 = 9 which is an impossibility.",
        "video_name": "bNQY0z76M5A",
        "timestamps": [
            245
        ],
        "3min_transcript": "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. minus 2 is equal to plus or minus 3. Add 2 to both sides, you get x is equal to 2 plus or minus 3. That tells us that x could be equal to 2 plus 3, which is 5. Or x could be equal to 2 minus 3, which is negative 1. And we are done. Now I want to be very clear. You could have done this without completing the square. We could've started off with x squared minus 4x is equal to 5. We could have subtracted 5 from both sides and gotten x squared minus 4x minus 5 is equal to 0. And you could say, hey, if I have a negative 5 times a positive 1, then their product is negative 5 and their sum is negative 4. So I could say this is x minus 5 times x plus 1 is equal to 0."
    },
    {
        "Q": "At 10:52, there is a fraction underneath a radical, the square root of 61/20. While Sal uses his calculator to solve it, I want to know how you would simplify a fraction under a radical like that without a calculator. My math teacher takes off points on tests for leaving answers as fractions under radicals, but it's not quite the same as simplifying a radical that is just a whole number, so I'm confused on how to do this. Also, is this process similar if you take the cube of a fraction instead of the square?",
        "A": "First of all, 61 is a prime number so in a radical it remains. With 20 you can factor it into 4 and 5, so it can be written as \\ \u00e2\u0088\u009a20 = 2\u00e2\u0088\u009a5 So now we have \u00e2\u0088\u009a61 / 2\u00e2\u0088\u009a5, which is an OK answer as long as radicals in the denominator are permitted. To get rid of the radical in the denominator, multiply \u00e2\u0088\u009a61 / 2\u00e2\u0088\u009a5 by \u00e2\u0088\u009a5/\u00e2\u0088\u009a5 and we have: (\u00e2\u0088\u009a5)(\u00e2\u0088\u009a61)/(2)(\u00e2\u0088\u009a5)(\u00e2\u0088\u009a5), which can be simplified to \u00e2\u0088\u009a(61*5)/10 = \u00e2\u0088\u009a305/10. So now you only have a radical in the numerator.",
        "video_name": "bNQY0z76M5A",
        "timestamps": [
            652
        ],
        "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 5:48 why do they equal zero",
        "A": "That s simply the problem statement; at 5:37 Sal starts a new quadratic equation as another example.",
        "video_name": "bNQY0z76M5A",
        "timestamps": [
            348
        ],
        "3min_transcript": "minus 2 is equal to plus or minus 3. Add 2 to both sides, you get x is equal to 2 plus or minus 3. That tells us that x could be equal to 2 plus 3, which is 5. Or x could be equal to 2 minus 3, which is negative 1. And we are done. Now I want to be very clear. You could have done this without completing the square. We could've started off with x squared minus 4x is equal to 5. We could have subtracted 5 from both sides and gotten x squared minus 4x minus 5 is equal to 0. And you could say, hey, if I have a negative 5 times a positive 1, then their product is negative 5 and their sum is negative 4. So I could say this is x minus 5 times x plus 1 is equal to 0. 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"
    },
    {
        "Q": "At 4:10, why does it become +/- 3, instead of just 3?",
        "A": "the sqare root of 9 = +/-3 because in algebra, whenever you square a positive OR negative number, the answer is always positive.",
        "video_name": "bNQY0z76M5A",
        "timestamps": [
            250
        ],
        "3min_transcript": "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. minus 2 is equal to plus or minus 3. Add 2 to both sides, you get x is equal to 2 plus or minus 3. That tells us that x could be equal to 2 plus 3, which is 5. Or x could be equal to 2 minus 3, which is negative 1. And we are done. Now I want to be very clear. You could have done this without completing the square. We could've started off with x squared minus 4x is equal to 5. We could have subtracted 5 from both sides and gotten x squared minus 4x minus 5 is equal to 0. And you could say, hey, if I have a negative 5 times a positive 1, then their product is negative 5 and their sum is negative 4. So I could say this is x minus 5 times x plus 1 is equal to 0."
    },
    {
        "Q": "At 0:35 why does it have to be greater or equal to? Why not less than or equal to or equal to.",
        "A": "Because you can t take the square root of anything less than 0.",
        "video_name": "4h54s7BBPpA",
        "timestamps": [
            35
        ],
        "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": "i do not understand from 0:22 to 0:56",
        "A": "Sal realised that he ll be unable to find the square root of the quantity within the radical if it s negative. So he sets up the inequality to show that that quantity (2x-8) has to be either zero or positive.",
        "video_name": "4h54s7BBPpA",
        "timestamps": [
            22,
            56
        ],
        "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 1:45 i still do not really understand why after adding 64 you must also subtract 64 from 9. Could someone explain I'm a bit confused.",
        "A": "When we make changes to an expression, we need to make sure that the new version is equivalent to the prior version. If you just add 64, you have changed the value of the original expression. The new version would not longer be equal to the original. The identify property of addition says we can add zero to any expression and it doesn t change the value of the expression. Thus, by using: +64 and -64, Sal is adding zero to the expression (+64 - 64 =0). Hope this helps.",
        "video_name": "sh-MP-dVhD4",
        "timestamps": [
            105
        ],
        "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": "In 0:44 , how did Khan get 2ax from (x+a)^2? Can someone explain how and why it works?",
        "A": "(x+a)\u00c2\u00b2=(x+a)(x+a). When you multiply it out, you get x\u00c2\u00b2+ax+ax+a\u00c2\u00b2. Simplify it you get x\u00c2\u00b2+2ax+a\u00c2\u00b2",
        "video_name": "sh-MP-dVhD4",
        "timestamps": [
            44
        ],
        "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": "At 3:18 of the video, how does Sal get that x= \u00e2\u0080\u00934 or x= 3 from the \"(x+4)(x-3)=0?\"",
        "A": "When we get down to the factored formula (x+4)(x-3)=0, then we use the zero property principle that the others have described to finish the solution. The next step is to set each factor equal to zero by itself and solve. (x+4) = 0 x + 4 -4 = 0 -4 x = -4 First solution for x (x-3) = 0 x-3 +3 = +3 x = 3 Second solution for x",
        "video_name": "swFohliPgmQ",
        "timestamps": [
            198
        ],
        "3min_transcript": "So this is 1, 2, 3, 4. y-intercept is there and has a slope of 1 so it looks something like this. So when we're looking for the solutions, we're looking for the points that satisfy both. The points that satisfy both are the points that sit on both. So it's that point-- let me do it in green-- It's this point and it's this point right over here. So how do we actually figure that out? Well, the easiest way is to-- well, sometimes the easiest way is to substitute one of these constraints into the other constraint. And since they've already solved for y here, we can substitute y in the blue equation with x plus 1 with this constraint right over here. So instead of saying x squared plus y squared equals 25, we can say x squared plus, and instead of writing a y, we're adding the constraint that y must be x plus 1. So x squared plus x plus 1 squared must be equal to 25. So we get x squared plus-- now we square this. We'll get-- let me write in magenta-- we'll get x squared plus 2x plus 1, and that must be equal to 25. We have 2x squared-- now, I'm just combining these two terms-- 2x squared plus 2x plus 1 is equal to 25. Now, we could just use the quadratic formula to find the-- well, we have to be careful. We have to set this equal to 0, and then use the quadratic formula. So let's subtract 25 from both sides, and you could get 2x squared plus 2x minus 24 is equal to 0. And actually, let's-- just to simplify this-- let's divide both sides by 2, and you get x squared plus x minus 12 is equal to 0. And actually, we don't even have to use the quadratic formula, we can factor this right over here. What are two numbers that when we take their product Well, positive 4 and negative 3 would do the trick. So we have x plus 4 times x minus 3 is equal to 0. So x could be equal to-- well, if x plus 4 is 0 then that would make this whole thing true. So x could be equal to negative 4 or x could be equal to positive 3. So this right over here is a situation where x is negative 4. This right over here is a situation where x is 3. So we're almost done, we just have to find the corresponding y's. And for that, we can just resort to the simplest equation right over here, y is x plus 1. So in this situation when x is negative 4, y is going to be that plus 1. So y is going to be negative 3. This is the point negative 4 comma negative 3. Likewise, when x is 3, y is going to be equal to 4."
    },
    {
        "Q": "At 0:55, Sal draws a circle, but from an equation like the one in the video, is there any way to figure out exactly how large the circle is?",
        "A": "The constant term is 25, so the radius of the circle is sqrt 25 = 5.",
        "video_name": "swFohliPgmQ",
        "timestamps": [
            55
        ],
        "3min_transcript": "What are the solutions to the system of equations y is equal to x plus 1, and x squared plus y squared is equal to 25? So let's first just visualize what we're trying to do. So let me try to roughly graph these two equations. So my y-axis, this is my x-axis. This right over here, x squared plus y squared is equal to 25, that's going to be a circle centered at 0 with radius 5. You don't have to know that to solve this problem, but it helps to visualize it. So if this is 5, this is 5, 5, 5. This right over here is negative 5. This right over here is negative 5. This equation would be represented by this set of points, or this is a set of points that satisfy this equation. So let me-- there you go. Trying to draw it as close to a perfect circle as I can. And then y equals x plus 1 is a line So this is 1, 2, 3, 4. y-intercept is there and has a slope of 1 so it looks something like this. So when we're looking for the solutions, we're looking for the points that satisfy both. The points that satisfy both are the points that sit on both. So it's that point-- let me do it in green-- It's this point and it's this point right over here. So how do we actually figure that out? Well, the easiest way is to-- well, sometimes the easiest way is to substitute one of these constraints into the other constraint. And since they've already solved for y here, we can substitute y in the blue equation with x plus 1 with this constraint right over here. So instead of saying x squared plus y squared equals 25, we can say x squared plus, and instead of writing a y, we're adding the constraint that y must be x plus 1. So x squared plus x plus 1 squared must be equal to 25. So we get x squared plus-- now we square this. We'll get-- let me write in magenta-- we'll get x squared plus 2x plus 1, and that must be equal to 25. We have 2x squared-- now, I'm just combining these two terms-- 2x squared plus 2x plus 1 is equal to 25. Now, we could just use the quadratic formula to find the-- well, we have to be careful. We have to set this equal to 0, and then use the quadratic formula. So let's subtract 25 from both sides, and you could get 2x squared plus 2x minus 24 is equal to 0. And actually, let's-- just to simplify this-- let's divide both sides by 2, and you get x squared plus x minus 12 is equal to 0. And actually, we don't even have to use the quadratic formula, we can factor this right over here. What are two numbers that when we take their product"
    },
    {
        "Q": "At 3:27 Sal is telling that angle AEB and angle ECB are congruent because they are corresponding angles but the line l and line m are not parallel. If they are corresponding angles, they have to be parallel. But how he did this?",
        "A": "Corresponding angles are actually any two angles in that position. They are only congruent if the lines are parallel, and vice versa. Also, if two corresponding angles are labeled as congruent, the lines are parallel, end of story",
        "video_name": "6dIMIBO_2mc",
        "timestamps": [
            207
        ],
        "3min_transcript": "let's assume that it is. So let's assume that we do have a quadrilateral where this is indeed the case. Where AEB, this angle right over here, is congruent to angle ECB. We're just going to assume that right from the get-go. Now let's try to visualize this whole thing a little bit differently. DB is a segment, but it's a segment of a larger line. So we could keep extending it off like this. And let's call that larger line, line l. Let me draw it like this. So this right over here is line l. DB is a subset of line l. And CB, which is a segment, is also a subset of a line. And we could call the line, if we were to keep extending it, let's call that line m. So I'll draw it like this. So line m. And then we also have the segment AC. And AC, once again, is a segment but we can view it as a subset of a larger line. And we'll call that larger line, line n. And so let me draw a line n like this. And we see that it intersects both line l and line m. So let me draw it like this, so it looks something like that. That is line n. And we've assumed that angle AEB is congruent to angle ECB. Well point E is just where n and l intersect. So this right over here is point E. And point C is where lines n and m intersect. So this is point C. So this assumption right over here, that we're assuming from the get-go, is saying that this angle AEB-- I'll ECB is congruent to that angle. These angles are these angles right over here. Now what does that tell us about lines m and l? Well the way we've set it up, we have two lines, lines l and m. Line n is a transversal. And now we have two corresponding angles are congruent. We assumed that from the get-go that we could find two quadrilateral, where these two corresponding angles are congruent. But if you have two corresponding angles congruent like this, that means that these two lines must be parallel. So this tells us that line l is parallel to line m. Line l is parallel to line m. Which means that they will never intersect. But we have a contradiction showing up. And I'll let you think about what that is for a second."
    },
    {
        "Q": "At 7:27, shouldn't it be 1 /\u00e2\u0088\u009a2, rather than \u00e2\u0088\u009a2/2?",
        "A": "The second fraction \u00e2\u0088\u009a2/2 is the same as the first fraction 1 /\u00e2\u0088\u009a2, just with a rationalized denominator. The 2 forms are equivalent.",
        "video_name": "mSVrqKZDRF4",
        "timestamps": [
            447
        ],
        "3min_transcript": "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 But we got our derivative in terms of an x and a y. Not just only in terms of an x. But what does this mean? Well, if we wanted to find, let's say we wanted to find the derivative at this point right over here. Which, if you're familiar with the unit circle, so if this was a 45 degree angle, this would be the square root of 2 over 2 comma the square root of 2 over 2. What is the slope of the tangent line there? Well, we figured it out. It's going to be negative x over y. So the slope of the tangent line here is going to be equal to negative x. So negative square root of 2 over 2 over y. Over square root of 2 over 2, which is equal to negative 1. And that looks just about right."
    },
    {
        "Q": "What if we tried to find dy/dx by explicitly defining y in terms of x first, just like Sal did at 0:47. Would we have to find dy/dx separately for both the positive and negative roots? What would we do afterwards?",
        "A": "Yes, since when we define it explicitly we get 2 equations (for a different one, we could get 4, there s an example of this, called a... fleur or something, it s 4 loops, obviously to define it as a correct function you have to limit it, since it loops back on itself). The two functions give you different slopes in terms of which one you re using. When you use y = +sqrt(1 - x^2) you re getting dy/dx for that function, the top half of our circle.",
        "video_name": "mSVrqKZDRF4",
        "timestamps": [
            47
        ],
        "3min_transcript": "So we've got the equation x squared plus y squared is equal to 1. I guess we could call it a relationship. And if we were to graph all of the points x and y that satisfied this relationship, we get a unit circle like this. And what I'm curious about in this video is how we can figure out the slope of the tangent line at any point of this unit circle. And what immediately might be jumping out in your brain is, well a circle defined this way, this isn't a function. It's not y explicitly defined as a function of x. For any x value we actually have two possible y's that satisfy this relationship right over here. So you might be tempted to maybe split this up into two separate functions of x. You could say y is equal to the positive square root of 1 minus x squared. And you could say y is equal to the negative square root of 1 minus x squared. Take the derivatives of each of these separately. or the derivative of the slope of the tangent line at any point. But what I want to do in this video is literally leverage the chain rule to take the derivative implicitly. So that I don't have to explicitly define y is a function of x either way. And the way we do that is literally just apply the derivative operator to both sides of this equation. And then apply what we know about the chain rule. Because we are not explicitly defining y as a function of x, and explicitly getting y is equal to f prime of x, they call this-- which is really just an application of the chain rule-- we call it implicit differentiation. And what I want you to keep in the back of your mind the entire time is that it's just an application of the chain rule. So let's apply the derivative operator to both sides of this. So it's the derivative with respect to x of x squared plus y squared, on the left hand And then that's going to be equal to the derivative with respect to x on the right hand side. I'm just doing the same exact thing to both sides of this equation. Now if I take the derivative of the sum of two terms, that's the same thing as taking the sum of the derivative. So this is going to be the same thing as the derivative with respect to x of x squared, plus the derivative with respect to x of y squared. I'm writing all my orange stuff first. This is going to be x squared, this is going to be y squared. And then this is going to be equal to the derivative with respect to x of a constant. This isn't changing with respect to x. So we just get 0. Now this first term right over here, we have done many, many, many, many, many, many times. The derivative with respect to x of x squared is just the power rule here. It's going to be 2 times x to the first power."
    },
    {
        "Q": "at about 4:30, when Sal is taking the derivative of y^2, why is he using the chain rule and not the power rule?",
        "A": "He s using both. The chain rule is required here because we re differentiating with respect to x and the expression includes y. We re able to do this by treating y as a function of x. This means that y\u00c2\u00b2 is actually a composition of two functions: the squaring function applies to whatever function would turn x into y. We don t necessarily know what that function is, but that s okay, it s why we re doing implicit differentiation instead of normal (explicit) differentiation.",
        "video_name": "mSVrqKZDRF4",
        "timestamps": [
            270
        ],
        "3min_transcript": "And then that's going to be equal to the derivative with respect to x on the right hand side. I'm just doing the same exact thing to both sides of this equation. Now if I take the derivative of the sum of two terms, that's the same thing as taking the sum of the derivative. So this is going to be the same thing as the derivative with respect to x of x squared, plus the derivative with respect to x of y squared. I'm writing all my orange stuff first. This is going to be x squared, this is going to be y squared. And then this is going to be equal to the derivative with respect to x of a constant. This isn't changing with respect to x. So we just get 0. Now this first term right over here, we have done many, many, many, many, many, many times. The derivative with respect to x of x squared is just the power rule here. It's going to be 2 times x to the first power. Now what's interesting is what we're doing right over here. The derivative with respect to x of y squared. And the realization here is to just apply the chain rule. If we're taking the derivative with respect to x of this something, we just have to take the derivative-- let me make it clear-- we're just going to take the derivative of our something. The derivative of y squared-- that's what we're taking, you can kind of view that as a function-- with respect to y and then multiply that times the derivative of y with respect to x. We're assuming that y does change with respect to x. y is not some type of a constant that we're writing just an abstract terms. So we're taking the derivative of this whole thing with respect to y. Once again, just the chain rule. And then we're taking the derivative of y 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."
    },
    {
        "Q": "at 3:14 can someone explain how d/dx (y^2) = dy^2/dy ? how does the denominator of dx turn into dy?",
        "A": "First, notice that d/dx (y^2) has a different variable for the function, and the variable we re taking the derivative with respect to. This brings to mind chain rule, that is, if y is a function of x, like y(x). The chain rule says in this example says that d/dx (y(x)^2) = d/dy (y^2) * d/dx (y(x).",
        "video_name": "mSVrqKZDRF4",
        "timestamps": [
            194
        ],
        "3min_transcript": "or the derivative of the slope of the tangent line at any point. But what I want to do in this video is literally leverage the chain rule to take the derivative implicitly. So that I don't have to explicitly define y is a function of x either way. And the way we do that is literally just apply the derivative operator to both sides of this equation. And then apply what we know about the chain rule. Because we are not explicitly defining y as a function of x, and explicitly getting y is equal to f prime of x, they call this-- which is really just an application of the chain rule-- we call it implicit differentiation. And what I want you to keep in the back of your mind the entire time is that it's just an application of the chain rule. So let's apply the derivative operator to both sides of this. So it's the derivative with respect to x of x squared plus y squared, on the left hand And then that's going to be equal to the derivative with respect to x on the right hand side. I'm just doing the same exact thing to both sides of this equation. Now if I take the derivative of the sum of two terms, that's the same thing as taking the sum of the derivative. So this is going to be the same thing as the derivative with respect to x of x squared, plus the derivative with respect to x of y squared. I'm writing all my orange stuff first. This is going to be x squared, this is going to be y squared. And then this is going to be equal to the derivative with respect to x of a constant. This isn't changing with respect to x. So we just get 0. Now this first term right over here, we have done many, many, many, many, many, many times. The derivative with respect to x of x squared is just the power rule here. It's going to be 2 times x to the first power. Now what's interesting is what we're doing right over here. The derivative with respect to x of y squared. And the realization here is to just apply the chain rule. If we're taking the derivative with respect to x of this something, we just have to take the derivative-- let me make it clear-- we're just going to take the derivative of our something. The derivative of y squared-- that's what we're taking, you can kind of view that as a function-- with respect to y and then multiply that times the derivative of y with respect to x. We're assuming that y does change with respect to x. y is not some type of a constant that we're writing just an abstract terms. So we're taking the derivative of this whole thing with respect to y. Once again, just the chain rule. And then we're taking the derivative of y"
    },
    {
        "Q": "At 2:20, you said \"the length of u1\" but you just showed u1. Does it matter which way you do it?",
        "A": "Here the letter u is meant to stand for unit vector , so you don t need to write ||u1|| .",
        "video_name": "rHonltF77zI",
        "timestamps": [
            140
        ],
        "3min_transcript": "Let's say I have a set of linearly independent vectors, V1, V2, all the way to Vk, that are a basis for V. We've seen this many times before. That's nice. It's a basis, but we've learned over the last few videos it would be even better to have an orthonormal basis for V. What we're going to address in this video is, is there some way that we can construct an orthonormal basis for V, given that we already have this basis, which I'm assuming isn't orthonormal. It'll work whether or not it's orthonormal. It'll just generate another orthonormal basis. But can we somehow, just given any basis, generate an orthonormal basis for V, and then be able to benefit from all of the properties of having an orthonormal basis? Let's see if we can make some headway here. Let's say that I have a one-dimensional subspace. Let's call that V1, for it's a one-dimensional subspace. I'm going to say it's the span of just this vector V1. Now, or you could say the vector V1 is the basis for the subspace V1. If I had this simple case, how can I ensure that this is orthonormal? What I could do is I could define some vector, let's call it U1. Obviously, this is orthogonal to all of the other guys. There aren't any other guys here. [INAUDIBLE] vectors there, but there aren't any other members of this set, so it's orthogonal to everything else because there isn't anything else. And then to make its length equal one, you can take this vector and divide it by its length. the length of V1, then the length of U1 is going to be what? It's going to be the length of V1 over the length of V1 like this. This is just a constant right here. This is going to be 1 over the length of V1. That could be 5, times length of V1, which is just equal to 1. This one will have a length of 1. So just like that, if we have the set of just U1, we could say that just the set of U1 is an orthonormal basis for V1, for the subspace V1. Now, that was trivially easy."
    },
    {
        "Q": "what is the thing that sal did 2:43 called?",
        "A": "I think improper fraction or mixed number/fraction.",
        "video_name": "RPhaidW0dmY",
        "timestamps": [
            163
        ],
        "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 2:18 what if you have a number that can't be divided normally",
        "A": "At 2:18, Sal cancelled out a common factor. If there are no common factors to cancel, then you just multiply. For example: 7/4 * 11/5 = 77/20 = 6 17/20 Hope this helps.",
        "video_name": "RPhaidW0dmY",
        "timestamps": [
            138
        ],
        "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 3:19 he said 2x5. = 12 and I am pretty sure it = 10 so I'm just confused",
        "A": "Yeah, your right, he made a mistake, but if you watch for about two seconds longer, he actually corrects himself. \u00f0\u009f\u0091\u008dgood job on catching his mistake though.",
        "video_name": "RPhaidW0dmY",
        "timestamps": [
            199
        ],
        "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 1:51 you simplify the numerator of one fraction, but then simplify the denominator of the other fraction. How is it that you can do this? I thought you had to simplify the denominator of the same fraction.",
        "A": "You can do this if you are multiplying fractions. If you multiply first, you get (7/4)*(36/5)=(7*36)/(4*5). Now you have all the numbers from the original two fractions in the same fraction. So it really doesn t matter at what stage you simplify. I hope this helped you. And remember that this is allowed only in multiplication, not in any other operation.",
        "video_name": "RPhaidW0dmY",
        "timestamps": [
            111
        ],
        "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 2:37 you only did the denominator for the first fraction and for the second fraction you only did the numerator, why didnt you do all of both fractions?",
        "A": "At this point in the video, Sal is cancelling out (dividing out) a common factor. As long as the fractions are being multiplied, we can cancel out a common factor from any numerator with the same factor in any denominator. The only numerator and denominator values that share a common factor are the 36 and the 4. Five (5), the remaining number in a denominator is not a common factor of 7 or of 9. This is why no other numbers were cancelled. out. hope this helps.",
        "video_name": "RPhaidW0dmY",
        "timestamps": [
            157
        ],
        "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 0:34 what do you mean by geometric and arethmatic mean because I'm confused?",
        "A": "Geometric and arithmetic mean have different meanings in some areas of math. For now just use the mean you ve learned so far.",
        "video_name": "k5EbijWu-Ss",
        "timestamps": [
            34
        ],
        "3min_transcript": "Let's say you're trying to design some type of a product for men. One that is somehow based on their height. And the product is for the United States. So ideally, you would like to know the mean height of men in the United States. Let me write this down. So how would you do that? And when I talk about the mean, I'm talking about the arithmetic mean. If I were to talk about some other types of means-- and there are other types of means, like the geometric mean-- I would specify it. But when people just say mean, they're usually talking about the arithmetic mean. So how would you go about finding the mean height of men in the United States? Well, the obvious one is, is you go and ask every or measure every man in the United States. Take their height, add them all together, and then divide by the number of men there are in the United States. is whether that is practical. Because you have on the order-- let's see, there's about 300 million people in the United States. Roughly half of them will be men, or at least they'll be male, and so you will have 150 million, roughly 150 million men in the United States. So if you wanted the true mean height of all of the men United States, you would have to somehow survey-- or not even survey. You would have to be able to go and measure all 150 million men. And even if you did try to do that, by the time you're done, many of them might have passed away, new men will have been born, and so your data will go stale immediately. So it is seemingly impossible, or almost impossible, to get the exact height of every man in the United States in a snapshot of time. And so, instead, what you do is say, well, look, OK, I can't get every man, but maybe I can take a sample. I could take a sample of the men in the United States. And I'm going to make an effort that it's a random sample. happen to play basketball, or played basketball for their college. I don't want to go sample 100 people who are volleyball players. I want to randomly sample. Maybe the first person who comes out of the mall in a random town, or in several towns, or something like that. Something that should not be based in any way, or skewed in any way, by height. So you take a sample and from that sample you can calculate a mean of at least the sample. And you'll hope that that is indicative of-- especially if this was a reasonably random sample-- you'll hope that was indicative of the mean of the entire population. And what you're going to see in much of statistics it is all about using information, using things that we can calculate about a sample, to infer things about a population. Because we can't directly measure the entire population."
    },
    {
        "Q": "WHY is 5.5 = to 5 feet,6inches? SAL said so at 3:31-3:37",
        "A": "Because 1 foot is 12 inches. Compare this to for example hours and minutes. 1.5 hours is one hour plus half an hour, or 1 hour 30 minutes. Likewise, 5.5 feet is 5 feet and half a foot, or 5 feet and 6 inches.",
        "video_name": "k5EbijWu-Ss",
        "timestamps": [
            211,
            217
        ],
        "3min_transcript": "happen to play basketball, or played basketball for their college. I don't want to go sample 100 people who are volleyball players. I want to randomly sample. Maybe the first person who comes out of the mall in a random town, or in several towns, or something like that. Something that should not be based in any way, or skewed in any way, by height. So you take a sample and from that sample you can calculate a mean of at least the sample. And you'll hope that that is indicative of-- especially if this was a reasonably random sample-- you'll hope that was indicative of the mean of the entire population. And what you're going to see in much of statistics it is all about using information, using things that we can calculate about a sample, to infer things about a population. Because we can't directly measure the entire population. trying to do this, I would recommend doing at least 100 data points, or 1,000, and later on we'll talk about how you can think about whether you've measured enough or how confident you can be. But let's just say you're a little bit lazy, and you just sample five men. And so you get their five heights. Let's say one is 6.2 feet. Let's say one is 5.5 feet-- 5.5 feet would be 5 foot, 6 inches. Let's say one ends up being 5.75 feet. Another one is 6.3 feet. Another is 5.9 feet. Now, if these are the ones that you happen to sample, what would you get for the mean of this sample? Well let's get our calculator out. And we get 6.2 plus 5.5 plus 5.75 plus 6.3 plus 5.9. And then we want to divide by the number of data points we have. So we have five data points. So let's divide 29.65 divided by 5, and we get 5.93 feet. So here, our sample mean-- and I'm going to denote it with an x with a bar over it, is-- and I already forgot the number-- 5.93 feet. This is our sample mean, or, if we want to make it clear, sample arithmetic mean. And when we're taking this calculation based on a sample, and somehow we're trying to estimate it for the entire population, we call this right over here,"
    },
    {
        "Q": "At 4:43, is a capital Sigma only used for sample sums?",
        "A": "Per Wiki, a capital Sigma when used by mathematicians indicates summation. A capital Sigma has other uses in other fields like physics and economics. Here s the Wiki link for a more nuanced answer.",
        "video_name": "k5EbijWu-Ss",
        "timestamps": [
            283
        ],
        "3min_transcript": "trying to do this, I would recommend doing at least 100 data points, or 1,000, and later on we'll talk about how you can think about whether you've measured enough or how confident you can be. But let's just say you're a little bit lazy, and you just sample five men. And so you get their five heights. Let's say one is 6.2 feet. Let's say one is 5.5 feet-- 5.5 feet would be 5 foot, 6 inches. Let's say one ends up being 5.75 feet. Another one is 6.3 feet. Another is 5.9 feet. Now, if these are the ones that you happen to sample, what would you get for the mean of this sample? Well let's get our calculator out. And we get 6.2 plus 5.5 plus 5.75 plus 6.3 plus 5.9. And then we want to divide by the number of data points we have. So we have five data points. So let's divide 29.65 divided by 5, and we get 5.93 feet. So here, our sample mean-- and I'm going to denote it with an x with a bar over it, is-- and I already forgot the number-- 5.93 feet. This is our sample mean, or, if we want to make it clear, sample arithmetic mean. And when we're taking this calculation based on a sample, and somehow we're trying to estimate it for the entire population, we call this right over here, Now, you might be saying, well, what notation do we use if, somehow, we are able to measure it for the population? Let's say we can't even measure it for the population, but we at least want to denote what the population mean is. Well if you want to do that, the population mean is usually denoted by the Greek letter mu. And so in a lot of statistics, it's calculating a sample mean in an attempt to estimate this thing that you might not know, the population mean. And these calculations on the entire population, sometimes you might be able to do it. Oftentimes, you will not be able to do it. These are called parameters. So what you're going to find in much of statistics, it's all about calculating statistics for a sample, finding these sample statistics in order"
    },
    {
        "Q": "At 1:23 , how many ounces is a ton?",
        "A": "There is 32,000 ounces in a ton.",
        "video_name": "Dj1rbIP8PHM",
        "timestamps": [
            83
        ],
        "3min_transcript": "Let's talk a little bit about the US customary units of weight. So the one that's most typically used is the pound, especially for things of kind of a human scale. And to understand what a pound is, most playing balls are roughly about a pound. So, for example, a soccer ball-- my best attempt to draw a soccer ball. So let's say that this is a soccer ball right over here. And then of course it has some type of pattern on it. So you could imagine a soccer ball is about a pound. So it's roughly one pound. And a pound will often be shorthanded with this \"lb.\" right over here. So it's about a pound. A football, an American football, is also a little under a pound. But we could say it is about a pound, just so we get a sense of what a pound actually represents. Now, if you want to go to scales smaller than a pound, you would think about using the ounce. thinking about weight is that one pound-- let me write this-- is equal to 16 ounces. Or another way of thinking about it is that 1 ounce is equal to 1/16 of a pound. And if you want to visualize things that weigh about an ounce, you could imagine a small box of matches weigh about an ounce. So a small box of matches might weigh about an ounce. Maybe a small AA battery would weigh about an ounce. But that gives you a sense of it. So if you were to take 16 of these together, they would be about the weight of a soccer ball. 16 of these things together, they would be about-- they would be about the weight of a soccer ball or a football. that are larger than a pound, then we would go to the ton. And a ton is equal to-- 1 ton is equal to 2,000 pounds. And you have to be a little bit careful with the ton. We're talking about the US customary units, and this is where we're talking about 2,000 pounds. But when we're talking on a more international level, this is sometimes called the short ton. There's also a long ton. There's also the metric ton. But here we're talking about US customary units, which is the short ton. So one ton is 2,000 pounds. And to get a sense of something that weighs 2,000 pounds, or to get a sense of what 2,000 pounds is like, or what might be measured in tons, a car is a good example. Your average midsize sedan would weigh about a little under to a little over 2 tons, so a little"
    },
    {
        "Q": "At 6:46, Sal says the function would be defined or all other real numbers except the ones specified. But shouldn't x not be equal to 0 either? Because if x=0, then the denominator would again become 0, hence resulting in an undefined solution. Is that correct?",
        "A": "We have h(x)=(x+10)/((x+10)(x-9)(x-5)). In that case, h(0) = (0+10)/((0+10)(0-9)(0-5)) = 10/((10)(-9)(-5) = 1/45",
        "video_name": "n17q8CBiMtQ",
        "timestamps": [
            406
        ],
        "3min_transcript": "But remember when x is equal 5, we don't look at this part of the compound definition. We look at this part. So it's true that up here you would be dividing by 0 if x is equaling 5 but x equaling 5, you wouldn't even look at -- look there. For input is 5, you use this part of the definition. So you'd divide by 0. Maybe I should write it this way, divide by 0 on, I guess you could say the top -- the top clause or the top part a definition. Part of the definition. If x equals 9, X equals negative 10 or -- and that's it because x equal 5 doesn't apply to this top part. If this clause wasn't here then yes, you would write x equals 5. Now we're almost done, but some of you might say, wait, wait, wait, but look, can't I simplify this? I have x plus 10 in the numerator and x plus 10 in the denominator. Can I just simplify this and that will disappear? And you could except, if you did that, you are now creating a different function definition. Because if you just simplify this, you just said 1 over x minus 9 -- That one actually would be defined at x equals negative 10. But the one that we have started with, this one is not. This is -- you're gonna end up with 0 over 0. You gonna end up with that indeterminate form. So for this function, exactly the way it's written, it's not going to be defined with x is equal to 9 or x is equal to negative 10. So once again if you want a fan -- write in our fancy domain set notation. The domain is going to be x all the x's that are a member of the real such that x does not equal 9 and x does not equal negative 10. Any other real number x -- it's going to work including 5. If x equals 5, h of -- h of 5 is going to be equal to pi, because you default to this one over here. h of 5 so OK x is equal 5, we do that one right over there. Now if you gave x equals 9 by 0, but that's gonna work for anything else. So that right over there is the domain."
    },
    {
        "Q": "At 1:25, Sal shows us the outer angle. What is the use for these outer angles in the real world?",
        "A": "In some math problems, if you are only given the outer angle, you can still solve for the remaining angles. Angles help in many things such as the job of construction workers, carpenters when deciding table lengths, etc.",
        "video_name": "QmfoIvgIVlE",
        "timestamps": [
            85
        ],
        "3min_transcript": "We're asked to construct a 10 degree angle. So we have this little angle tool here that we can use to construct an angle. So just like that. And then they give us a protractor to actually measure the angle. So let's set it up. Let's put the protractor here. Let's put the vertex of the angle at the center of the protractor, in other words, center the protractor at the center of the vertex. Looks like the protractor's a little bit easier to manipulate. So let's do that there. Now let's put one of the rays here at 0 degrees. And now I'm going to put the other ray at 10 degrees. And it looks like I am done. I have constructed a 10 degree angle. And you want to be careful here when you use this tool because the angle in question-- let me move the protractor show you what I'm talking about-- that this right over here is the angle that we're talking about. If I'd switched these two rays, the way the tool is set up, it might have interpreted the angle as this outer angle right over here. So be careful to look at which angle we're actually measuring. So let me check my answer. And just to be clear what I'm talking about. If you did the 10 degree angle like this, then it definitely would have marked it wrong even though the interior angle right over here or this angle right here might be 10 degrees. The way the tool is set up, you see from this circle, that it thinks that you're looking at this outer angle. So it's important to make sure, at least for the sake of this exercise, that the system, that the computer program knows which angle you're going to talk about. Let's do one more of these. 155 degree angle. And this one's interesting because this is an obtuse angle. So once again, let us put the protractor at the vertex of the angle. So just like that. And now that seems pretty good. Now let's take one ray and put it at 0 degrees, and then let's take the other one and put it at 155. 90-- that gets us to a right angle. Then we'll start getting into obtuse angles, 100, 110, 120, 130, 140, 150. And just to make sure that blue arc is measuring this angle right over here, not the outer one. And let me move the protractor out of the way so we can get a good look at it. And we got it wrong. So let's see what we-- oh, 155 degree angle, not 150 degree angle. Let me fix that. 155 degree angle. Now we got it right."
    },
    {
        "Q": "What does he mean at 0:31 when he says fair coin?",
        "A": "He means a coin with one head and one tail that has an equal chance of flipping one or the other.",
        "video_name": "cqK3uRoPtk0",
        "timestamps": [
            31
        ],
        "3min_transcript": "Voiceover:Let's say we define the random variable capital X as the number of heads we get after three flips of a fair coin. So given that definition of a random variable, what we're going to try and do in this video is think about the probability distributions. So what is the probability of the different possible outcomes or the different possible values for this random variable. We'll plot them to see how that distribution is spread out amongst those possible outcomes. So let's think about all of the different values that you could get when you flip a fair coin three times. So you could get all heads, heads, heads, heads. You could get heads, heads, tails. You could get heads, tails, heads. You could get heads, tails, tails. You could have tails, heads, heads. You could have tails, head, tails. You could have tails, tails, heads. And then you could have all tails. when you do the actual experiment there's eight equally likely outcomes here. But which of them, how would these relate to the value of this random variable? So let's think about, what's the probability, there is a situation where you have zero heads. So what's the probably that our random variable X is equal to zero? Well, that's this situation right over here where you have zero heads. It's one out of the eight equally likely outcomes. So that is going to be 1/8. What's the probability that our random variable capital X is equal to one? Well, let's see. Which of these outcomes gets us exactly one head? We have this one right over here. We have that one right over there. We have this one right over there. So three out of the eight equally likely outcomes provide us, get us to one head, which is the same thing as saying that our random variable equals one. So this has a 3/8 probability. So what's the probability, I think you're getting, maybe getting the hang of it at this point. What's the probability that the random variable X is going to be equal to two? Well, for X to be equal to two, we must, that means we have two heads when we flip the coins three times. So that's this outcome meets this constraint. This outcome would get our random variable to be equal to two. And this outcome would make our random variable equal to two. And this is three out of the eight equally likely outcomes. So this has a 3/8 probability. And then finally we could say what is the probability that our random variable X is equal to three? Well, how does our random variable X equal three? Well we have to get three heads when we flip the coin. So there's only one out of the eight"
    },
    {
        "Q": "At 3:28 why is the probability range between 0 and 1? I understand that beyond 1 we have a certainty of something happening, but why 1?",
        "A": "In statistics, 1=100%. One hundred percent is is absolute certainty. You can t be more certain than that.",
        "video_name": "cqK3uRoPtk0",
        "timestamps": [
            208
        ],
        "3min_transcript": "when you do the actual experiment there's eight equally likely outcomes here. But which of them, how would these relate to the value of this random variable? So let's think about, what's the probability, there is a situation where you have zero heads. So what's the probably that our random variable X is equal to zero? Well, that's this situation right over here where you have zero heads. It's one out of the eight equally likely outcomes. So that is going to be 1/8. What's the probability that our random variable capital X is equal to one? Well, let's see. Which of these outcomes gets us exactly one head? We have this one right over here. We have that one right over there. We have this one right over there. So three out of the eight equally likely outcomes provide us, get us to one head, which is the same thing as saying that our random variable equals one. So this has a 3/8 probability. So what's the probability, I think you're getting, maybe getting the hang of it at this point. What's the probability that the random variable X is going to be equal to two? Well, for X to be equal to two, we must, that means we have two heads when we flip the coins three times. So that's this outcome meets this constraint. This outcome would get our random variable to be equal to two. And this outcome would make our random variable equal to two. And this is three out of the eight equally likely outcomes. So this has a 3/8 probability. And then finally we could say what is the probability that our random variable X is equal to three? Well, how does our random variable X equal three? Well we have to get three heads when we flip the coin. So there's only one out of the eight So it's a 1/8 probability. So now we just have to think about how we plot this, to see how this is distributed. So let me draw... So over here on the vertical axis this will be the probability. Probability. And it's going to be between zero and one. You can't have a probability larger than one. So just like this. So let's see, if this is one right over here, and let's see everything here looks like it's in eighths so let's put everything in terms of eighths. So that's half. This is a fourth. That's a fourth. That's not quite a fourth. This is a fourth right over here. And then we can do it in terms of eighths. So that's a pretty good approximation. And then over here we can have the outcomes."
    },
    {
        "Q": "At 4:36, why didn't Sal distribute the 2 to AC and BD?",
        "A": "The Distributive Law only applies to cases of multiplication over addition. The distributive property would not apply in this case, as it is 1/2AC*BD, not 1/2(AC+BD)",
        "video_name": "3FManXv4mZM",
        "timestamps": [
            276
        ],
        "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": "0:03 how do you now its a rhombus",
        "A": "And it was a given.",
        "video_name": "3FManXv4mZM",
        "timestamps": [
            3
        ],
        "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": "In 1:12, How did Sal get 1 from i^25?",
        "A": "Sal got one from i^100, not i^25. he said that since 4x25=100, then i^100 is the same thing as (i^4)^25. i^4 is equal to 1, so 1^25=1. Finally, Sal deduced that i^100=1",
        "video_name": "QiwfF83NWNA",
        "timestamps": [
            72
        ],
        "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": "At 1:12, isn't 100 also divisible by 5?, and since 5 is larger, wouldn't i^100 be i?",
        "A": "When working with powers of i, we always use 4 to divide the exponent (regardless of what factors the exponent has) because the values of the powers of i repeat in groups of 4. That way, by using the fact that i^4 is 1, it makes the problem simpler.",
        "video_name": "QiwfF83NWNA",
        "timestamps": [
            72
        ],
        "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": "At 5:20 Sal says regarding i^96 that \"This is i^4, and then that to the 16th power\". Shouldn't he have said that \"This is i^4, and then that to the 24th power\" instead?",
        "A": "Yeah... But the result wasn t wrong.",
        "video_name": "QiwfF83NWNA",
        "timestamps": [
            320
        ],
        "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 5:18 he said 8 games and 16 somgs. How does this meets the first constranit?",
        "A": "The first constraint is songs + games must be greater than or equal to 15. 8+16=24 which is greater than 15.",
        "video_name": "BUmLw5m6F9s",
        "timestamps": [
            318
        ],
        "3min_transcript": "So let's see if we can graph these two constraints. Well, this first constraint, s plus g is going to be greater than or equal to 15. The easiest way to think about this-- or the easiest way to graph this is to really think about the intercepts. If g is 0, what is s? Well, s plus 0 has to be greater than or equal to 15. So if g is 0, s is going to be greater than or equal to 15. Let me put it this way. So if I'm going to graph this one right here. If g is 0, s is greater than or equal to 15. So g is 0, s, 15, let's see, this is 12, 14, 15 is right over there. And s is going to be all of the values equivalent to that or greater than for g equal to 0. If s is equal to 0, g is greater than or equal to 15. So if s is equal to 0, g is greater than or equal to 15. So the boundary line, s plus g is equal to 15, we would just have to connect these two dots. Let me try my best to connect these dots. So it would look something like this. This is always the hardest part. Let me see how well I can connect these two dots. Nope. Let me see. I should get a line tool for this. So that's pretty good. So that's the line s plus g is equal to 15. And we talk about the values greater than 15, we're going And you saw that when g is equal to 0, s is greater than or equal to 15. It's all of these values up here. And when s was 0, g was greater than or equal to 15. So this constraint right here is all of this. All of this area satisfies this. All of this area-- if you pick any coordinate here, it coordinates, because we're not going to buy parts of games. But if you think about all of the integer coordinates here, they represent combinations of s and g, where you're buying at least 15 games. For example here, you're buying 8 games and 16 songs. That's 24. So you're definitely meeting the first constraint. Now the second constraint. 0.89s plus 1.99g is less than or equal to 25. This is a starting point. Let's just draw the line 0.89s plus 1.99 is equal to 25. And then we could think about what region the less than would represent. Oh, 1.99g. And the easiest way to do this, once again, we could do slope y-intercept all that type of thing. But the easiest way is to just find the s- and the g-intercepts. So if s is equal to 0 then we have 1.99g is equal to 25 or g"
    },
    {
        "Q": "Hay,\nI was just wondering, at 5:24 Khan began to solve for the variables using the standard form. He started to graph it at 7:22. If someone has to graph an inequality converting it to standard form, do you have to use the same interval as in the standard form or can you adjust it due to the converted slope-intercept form?",
        "A": "Just use intecepts",
        "video_name": "BUmLw5m6F9s",
        "timestamps": [
            324,
            442
        ],
        "3min_transcript": "So let's see if we can graph these two constraints. Well, this first constraint, s plus g is going to be greater than or equal to 15. The easiest way to think about this-- or the easiest way to graph this is to really think about the intercepts. If g is 0, what is s? Well, s plus 0 has to be greater than or equal to 15. So if g is 0, s is going to be greater than or equal to 15. Let me put it this way. So if I'm going to graph this one right here. If g is 0, s is greater than or equal to 15. So g is 0, s, 15, let's see, this is 12, 14, 15 is right over there. And s is going to be all of the values equivalent to that or greater than for g equal to 0. If s is equal to 0, g is greater than or equal to 15. So if s is equal to 0, g is greater than or equal to 15. So the boundary line, s plus g is equal to 15, we would just have to connect these two dots. Let me try my best to connect these dots. So it would look something like this. This is always the hardest part. Let me see how well I can connect these two dots. Nope. Let me see. I should get a line tool for this. So that's pretty good. So that's the line s plus g is equal to 15. And we talk about the values greater than 15, we're going And you saw that when g is equal to 0, s is greater than or equal to 15. It's all of these values up here. And when s was 0, g was greater than or equal to 15. So this constraint right here is all of this. All of this area satisfies this. All of this area-- if you pick any coordinate here, it coordinates, because we're not going to buy parts of games. But if you think about all of the integer coordinates here, they represent combinations of s and g, where you're buying at least 15 games. For example here, you're buying 8 games and 16 songs. That's 24. So you're definitely meeting the first constraint. Now the second constraint. 0.89s plus 1.99g is less than or equal to 25. This is a starting point. Let's just draw the line 0.89s plus 1.99 is equal to 25. And then we could think about what region the less than would represent. Oh, 1.99g. And the easiest way to do this, once again, we could do slope y-intercept all that type of thing. But the easiest way is to just find the s- and the g-intercepts. So if s is equal to 0 then we have 1.99g is equal to 25 or g"
    },
    {
        "Q": "in 9:37 I thought the question said that a total of 15 games should be purchased, i was just wondering if it was relevant?",
        "A": "Sal actually said that at least 15 items were purchased. Like English, mathematics is a language. It is important to read the words very carefully, especially in word problems.",
        "video_name": "BUmLw5m6F9s",
        "timestamps": [
            577
        ],
        "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": "What does Congruent mean? 1:25",
        "A": "Congruent means that 2 or more shapes look exactly the same in shape and size.",
        "video_name": "tFhBAeZVTMw",
        "timestamps": [
            85
        ],
        "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 7:46, I would like to make sure that I have the correct answer for the \"cliffhanger\". Is it 20x^9\nThanks",
        "A": "Yes, you have the correct answer. Good job. Looks like they cut the video off too quickly.",
        "video_name": "iHnzLETGz2I",
        "timestamps": [
            466
        ],
        "3min_transcript": "Negative 9x to the fifth power times negative three, use parentheses there, when you have a negative in front, you always wanna use parentheses. Let's do x to the 107th power. If I would have showed you this before this video, you would have said oh my goodness, there's nothing I can do, I'm boxed, there's no way out. But now you know that it's as simple as follow the rules. We're going to multiply the coefficients, negative nine times negative three is 27. Two negatives is a positive and nine times three is 27. I'm gonna add my powers. Five plus 107 is a hundred, ooh, not two, that was almost a mistake I made there. Let's get rid of that, give me a second chance here. Life's all about second chances, And so, this crazy expression, which is two monomials, here's the first, here's the second, when we multiply and simplify we get another monomial, which is 27x to the 112th. I'm gonna leave you on a cliffhanger here. Which, I'm gonna show you a problem. What variable should we use? You notice I've been trying to vary the variables up to show you that it just doesn't matter. That's an ugly five, let's get rid of that. Give me a second chance with that one too. So let's look at 5x to the third power, times 4x to the sixth power. And I'm gonna show you a wrong answer. I had a student that asked to do this, and here's the wrong answer that they gave me. They told me 9x to the 18th power. What did they do wrong? What did they do wrong? I want you to think to yourself, what have we been talking about? What did they do with the five and the four to get the nine? What should they have done? What did they do with the three and the six to get the 18, and what should they have done? That's multiplying monomials by monomials."
    },
    {
        "Q": "at 2:04 when p^2 = 2p, why wouldn't you solve it as sqrt(p^2) = sqrt(2p)",
        "A": "You could do it that way, you d get p = \u00e2\u0088\u009a(2p). The solution would be the same (0 = \u00e2\u0088\u009a0, 2 = \u00e2\u0088\u009a4). It s just easier the way Sal does it, p(p-2) = 0, where you can clearly see the solutions are 0/2.",
        "video_name": "ZIqW_sXymrM",
        "timestamps": [
            124
        ],
        "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": "At 4:06, why did Sal multiply 3x and 12 by 1/3?\n\nCouldn't you divide each side by 3 to isolate the variable?\n\nI was taught to use inverse operations. The inverse operation of multiplication is division.\n\nSo why did Sal not use inverse operations to solve 3x=12?",
        "A": "Multiplying by 1/3 is the exact same thing as dividing by 3. Remember how to divide fractions? If you have 2/3 divided by 5/6 it is the same as 2/3 times 6/5, right. Remember you can write whole numbers, such as 12 and 3 as 12/1 and 3/1, So 12/3 = 12/1 / 3/1 = 12/1 * 1/3 = 12 * 1/3. So you see, multiplying by 1/3 is the same a dividing by 3. Great Question! Keep Studying! 12/3 = 12/1 * 1/3",
        "video_name": "_y_Q3_B2Vh8",
        "timestamps": [
            246
        ],
        "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 4:05 why didn't he do 3x/3 and 12/3",
        "A": "Multiplying by 1/3 and dividing by 3 are the same operation, so your way is the same as Sal s, it just looks a little different",
        "video_name": "_y_Q3_B2Vh8",
        "timestamps": [
            245
        ],
        "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 \"2:03\" is the answer Sal has not the correct one, so he has to use another formula to finish the problem?",
        "A": "Sal explains earlier (at 0:28) that this problem will have only one solution because there s only one way the absolute value of something can be 0 since the only number with an absolute value of 0 is 0.",
        "video_name": "GwjiR2_7A7Y",
        "timestamps": [
            123
        ],
        "3min_transcript": "We're told, solve the absolute value of 3x minus 9 is equal to 0, and graph the solution on a number line. So let's just rewrite the absolute value equation. They told us that the absolute value of 3x minus 9 is equal to 0. So we're told that the absolute value of the something-- in this case the something is 3x minus 9-- is equal to 0. If I told you that the absolute value of something is equal to 0, I'm telling you that something has to be exactly 0 away from 0, or 0 away from the origin on the number line. So the only thing that that something could be is 0. If I told you that the absolute value of x is equal to 0, you know that x has to be equal to 0. That's the only value whose absolute value is 0. So if I told you that the absolute value of 3x minus 9 is 0, than we know that 3x minus 9 has to be equal to 0, and that's kind of unique about the 0 is that, it's the absolute value of 0. If you had, say, a 1 here, you could say, oh well, then this thing could be a 1 or a negative 1. But here, if you have a 0, this thing can only be 0. So solving this equation is fairly straightforward. If we want to isolate the 3x, get rid of the negative 9 on the left-hand side, we add 9 to both sides of the equation. Add 9 to both sides of the equation, these 9's cancel out. That's the whole point. On the left-hand side, you're just left with 3x, and on the right-hand side, you are just left with 9. Now we want to solve for x, so we have 3 times x. Let's divide it by 3, because 3 times x divided by 3 is just going to be x. But if we divide the left side by 3, we have to divide the right side by 3. So we are left with-- these guys cancel out. x is equal to 9 over 3, which is 3. And that's our solution. Let's substitute it back into our original equation. So we have the absolute value of 3 times x. Instead of x, I'll just put in our actual answer that we got, 3 times 3 minus 9 has got to be equal to 0. So what's this going to be equal to? 3 times 3 is 9. So it's the absolute value of 9 minus 9, which is the absolute value of 0, which is, indeed, 0. So it does, indeed, equal 0, and we are done."
    },
    {
        "Q": "How come at 5:37 you had to take the inverse cosine and could not leave the answer as cos(19/20)?",
        "A": "In this example we are solving for theta. Notice that theta is stuck inside the COS function when we say: cos(theta) = 19/20 To get theta out in the open we take the inverse cos of both sides: arccos(cos(theta)) = arccos(19/20) theta = acrcos(19/20) theta = 18.19 degrees Notice that arccos and cos are inverse function as they undo each other.",
        "video_name": "Ei54NnQ0FKs",
        "timestamps": [
            337
        ],
        "3min_transcript": "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. We get 18.19 degrees. And I already verified that my calculator is in degree mode. So it gets 18.19 degrees. So if we wanted to round, this is approximately equal to 18.2 degrees, if we wanna round to the nearest tenth. So that essentially gives us a sense of how steep this slope actually is."
    },
    {
        "Q": "What is an arbitrary triangle? it mentions that term in the video at 1:23. Thank you!",
        "A": "It basically means a random triangle. He is paying no attention to side lengths or angle measures. It s not a term you need to know, don t worry.",
        "video_name": "Ei54NnQ0FKs",
        "timestamps": [
            83
        ],
        "3min_transcript": "Voiceover:Let's say you're studying some type of a little hill or rock formation right over here. And you're able to figure out the dimensions. You know that from this point to this point along the base, straight along level ground, is 60 meters. You know the steeper side, steeper I guess surface or edge of this cliff or whatever you wanna call it, is 20 meters. And then the longer side here, I guess the less steep side, is 50 meters long. So you're able to measure that. But now what you wanna do is use your knowledge of trigonometry, given this information, to figure out how steep is this side. What is the actual inclination relative to level ground? Or another way of thinking about it, what is this angle theta right over there? And I encourage you to pause the video and think about it on your own. Well it might be ringing a bell. Well you know three sides of a triangle and then we want to figure out an angle. And so the thing that jumps out in my head, well maybe the law of cosines could be useful. Let me just write out the law of cosines, before we try to apply it to this So the law of cosines tells us that C-squared is equal to A-squared, plus B-squared, minus two A B, times the cosine of theta. And just to remind ourselves what the A, B's, and C's are, C is the side that's opposite the angle theta. So if I were to draw an arbitrary triangle right over here. And if this is our angle theta, then this determines that C is that side, and then A and B could be either of these two sides. So A could be that one and B could be that one. Or the other way around. As you can see, A and B essentially have the same role in this formula right over here. This could be B or this could be A. So what we wanna do is somehow relate this angle... We wanna figure out what theta is in our little hill example right over here. So if this is going to be theta, what is C going to be? Well C is going to be this 20 meter side. 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."
    },
    {
        "Q": "At 2:25, can't you also add 9x to both sides first instead of subtracting 12x?",
        "A": "Yes! You can also add 9x to both sides, instead of subtracting 12x. The reason why Sal had subtracted 12x from both sides was that it is often preferred to have x on the left side. He ends up with: -21x - 3 = 18",
        "video_name": "YZBStgZGyDY",
        "timestamps": [
            145
        ],
        "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 1:17. Sal says that 12x + 18 cannot happen because x has a coefficient and 18 is a regular number. If this is correct, how did he get 3 + 4x = 12x?",
        "A": "He did 3*4x, not 3+4x.",
        "video_name": "YZBStgZGyDY",
        "timestamps": [
            77
        ],
        "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 5:10, there should and maybe there is only one x for ( x^2=9 ) as, x = sqrt(9), which is equal to 3, ( by principal squareroot ), not -3!, -3 = -x right, as its the same number 3, & hence, -3 ^ 2 = 9 should always be written as -x ^ 2, for any*, any number even for infinity",
        "A": "That is not correct. -3^2 = -9 and you cannot take the square root of a negative number. (-3)^2 = 9. If you have -3 = -x, that is the same as x = 3, so it still is the primary square root. The reason this is one place where we do not just take the primary square root is because we are talking about x intercepts of a quadratic, so the answer is x = +/- 3.",
        "video_name": "mbc3_e5lWw0",
        "timestamps": [
            310
        ],
        "3min_transcript": "And another way to think about it, it's the positive, this is going to be the positive square root. If someone wants the negative square root of nine, they might say something like this. They might say the negative, let me scroll up a little bit, they might say something like the negative square root of nine. Well, that's going to be equal to negative three. And what's interesting about this is, well, if you square both sides of this, of this equation, if you were to square both sides of this equation, what do you get? Well negative, anything negative squared becomes a positive. And then the square root of nine squared, well, that's just going to be nine. And on the right-hand side, negative three squared, well, negative three times negative three is positive nine. So, it all works out. Nine is equal, nine is equal to nine. And so this is an interesting thing, actually. Let me write this a little bit more algebraically now. the principal root of nine is equal to x. This is, there's only one possible x here that satisfies it, because the standard convention, what most mathematicians have agreed to view this radical symbol as, is that this is a principal square root, this is the positive square root, so there's only one x here. There's only one x that would satisfy this, and that is x is equal to three. Now, if I were to write x squared is equal to nine, now, this is slightly different. X equals three definitely satisfies this. This could be x equals three, but the other thing, the other x that satisfies this is x could also be equal to negative three, 'cause negative three squared is also equal to nine. So, these two things, these two statements, are almost equivalent, although when you're looking at this one, there's two x's that satisfy this one, because this is a positive square root. If people wanted to write something equivalent where you would have two x's that could satisfy it, you might see something like this. Plus or minus square root of nine is equal to x, and now x could take on positive three or negative three."
    },
    {
        "Q": "at 5:11 why did he put a 1/2 sign as the exponent",
        "A": "He used 1/2 as the exponent to explain to us that polynomials can t have fractional or decimal exponents (although other numbers can have them).",
        "video_name": "Vm7H0VTlIco",
        "timestamps": [
            311
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "At 0:16, How would you turn that into a linear equation?",
        "A": "Good Question. first take the change of your y and x points. which is y( 7) and x(4). now we need to find the slope. to find the slope lest divide our y difference by our x difference: 7/4 now we have our slope! now so far we have Y=7/4x+b at 2:30 Sal was confirming about the dotted line. as we know we are now trying to find the y-Intercept. looking at the graph we can see that the y-Intercept is -7 so now we get!( Drum roll!) Y=7/4X+(-7) or: Y=7/4X-7 Hope This Helps! =)",
        "video_name": "wl2iQAuQl7Y",
        "timestamps": [
            16
        ],
        "3min_transcript": "f is a linear function whose table of values is shown below. So they give us different values of x and what the function is for each of those x's. Which graphs show functions which are increasing at the same rate as f? So what is the rate at which f is increasing? When x increases by 4, we have our function increasing by 7. So we could just look for which of these lines are increasing at a rate of 7/4, 7 in the vertical direction every time we move 4 in the horizontal direction. And an easy way to eyeball that would actually be just to plot two points for f, and then see what that rate looks like visually. So if we see here when x is 0, f is negative 1. When x is 0, f is negative 1. So when x is 0, f is negative 1. And when x is 4, f is 6, so 1, 2, 3, 4, 5, 6, And two points specify a line. We know that it is a linear function. You can even verify it here. When we increase by 4 again, we increase our function by 7 again. We know that these two points are on f and so we get a sense of the rate of change of f. Now, when you draw it like that, it immediately becomes pretty clear which of these has the same rate of change of f. A is increasing faster than f. C is increasing slower. A is increasing much faster than f. C is increasing slower than f. B is decreasing, so that's not even close. But D seems to have the exact same inclination, the exact same slope, as f. So D is what we would go with. And we could even verify it, even if we didn't draw it in this way. Our change in f for a given change in x our function changed plus 7. It is equal to 7/4. And we can verify that on D, if we increase in the x-direction by 4, so we go from 4 to 8, then in the vertical direction we should increase by 7, so 1, 2, 3, 4, 5, 6, 7. And it, indeed, does increase at the exact same rate."
    },
    {
        "Q": "At 4:02, we go from d=5m/s * 1h to 5 m*h/s which is cool because we did 5m/s * h/1. But then he just writes 3600/1 s/h without any symbols. Are we just multiplying 5m*h/s * 3600/1 * s/h or what is happening here. Im not really used to adding things into a middle of equations without reasons like simplification etc.",
        "A": "Right, there are 3600 seconds in an hour (because there are 60 seconds in a minute and 60 minutes in an hour so 60*60=3600 seconds in an hour). It s 3600 seconds/1 hour, or 3600 s/1 h.",
        "video_name": "hIAdCTNi1S8",
        "timestamps": [
            242
        ],
        "3min_transcript": "Now you're saying, \"OK, that's cute and everything, \"but this seems like a little bit of too much overhead \"to worry about when I'm just doing \"a simple formula like this.\" But what I want to show you is that even with a simple formula like distance is equal to rate times time, what I just did could actually be quite useful, and this thing that I'm doing is actually called dimensional analysis. It's useful for something as simple as distance equals rate times time, but as you go into physics and chemistry and engineering, you'll see much, much, much more, I would say, hairy formulas. When you do the dimensional analysis, it makes sure that the math is working out right. It makes sure that you're getting the right units. But even with this, let's try a slightly more complicated example. Let's say that our rate is, let's say, let's keep our rate at 5 meters per second, but let's say that someone gave us the time. Instead of giving it in seconds, they give it in hours, so they say the time is equal to 1 hour. We're going to get distance is equal to 5 meters per second, 5 meters per second times time, which is 1 hour, times 1 hour. What's that going to give us? The 5 times the 1, so we multiply the 5 times the 1, that's just going to give us 5. But then remember, we have to treat the units algebraically. We're going to do our dimensional analysis, so it's 5, so we have meters per second times hours, times hours, or you could say 5 meter hours per second. Well, this doesn't look like a ... This isn't a set of units that we know that makes sense to us. This doesn't feel like our traditional units of distance, so we want to cancel this out in some way. It might jump out of you, well, if we can get rid of this hours, if we can express it in terms of seconds, then that would cancel here, and we'd be left with just the meters, which is a unit of distance that we're familiar with. So how do we do that? We'd want to multiply this thing and seconds in the numerator, times essentially seconds per hour. How many seconds are there per hour? Well, there are 3,600 ... Let me do this in a ... I'll do it in this color. There are 3,600 seconds per hour, or you could even say that there are 3,600 seconds for every 1 hour. Now when you multiply, these hours will cancel with these hours, these seconds will cancel with those seconds, and we are left with, we are left with 5 times 3,600. What is that? That's 5 times 3,000 would be 15,000, 5 times 600 is another 3,000, so that is equal to 18,000. The only units that we're left with, we just have the meters there. 18- Oh, it's 18,000, 18,000, 18,000 meters."
    },
    {
        "Q": "At 2:30 and 2:37, didn't Sal mean a1, a2 and a4 rather than a1, a2 and aN?",
        "A": "I believe so. Notice that he switched back. His 4 s do look like n s.",
        "video_name": "CkQOCnLWPUA",
        "timestamps": [
            150,
            157
        ],
        "3min_transcript": "Two videos ago we asked ourselves if we could find the basis for the columns space of A. And I showed you a method of how to do it. You literally put A in reduced row echelon form, so this matrix R is just a reduced row echelon form of A. And you look at its pivot columns, so this is a pivot column. It has a 1 and all 0's, this is a pivot column, 1 and all 0's, and the 1 is the leading non-zero term in its row. And this is a pivot column, let me circle them, these guys are pivot columns, and this guy's a pivot column right there. You look at those in the reduced row echelon form of the matrix, and the corresponding columns in the original matrix will be your basis. So this guy, this guy, so the first, second, and forth columns. So if we call this a1, this is a2, and let's call this a4, this would be a3, and this is a5. So we could say that a1, a2, and a4 are a basis for the And I didn't show you why two videos ago. I just said this is how you do it. You have to take it as a bit of an article of faith. Now in order for these to be a basis, two things have to be true. They have to be linearly independent, and I showed you in the very last video, the second in our series dealing I showed you that by the fact that this guy is r1, this guy is r2, and this guy is r4, it's clear that these guys are linearly independent. They each have a 1 in a unique entry, and the rest of their entries are 0. We're looking at three pivot columns right now, but it's true if we had n pivot columns. That each pivot column would have a 1 in a unique place, and all the other pivot columns would have 0 in that entry. So there's no way that the other pivot columns, any to each of them. So these are definitely linearly independent. And I showed you in the last video that if we know that these are linearly independent, we do know that they are, given that R has the same null space as A, we know that these guys have to be linearly independant, I did that in the very last video. Now the next requirement for a basis, we checked this one off, is to show that a1 a2 and an, that their span equals the column space of A. Now the column space of A is a span of all five of these vectors, so I had to throw a3 in there and a5. So to show that just these three vectors by themselves span our column space, we just have to show that I can represent a3 and a5 as linear combinations"
    },
    {
        "Q": "At 3:30 just when I learned that parenthesis most be solved first. But, I also learned about the Distributed Property. SMH",
        "A": "Whenever you have parenthesis, do the math inside the parenthesis first, then use the distributive property. 4(11-3)= 4(8)=32 If you want something to help you remember the order of operations, try PEMDAS. Parentheses Exponents (if necessary) Multiplication (this includes the distributive property) Division Addition Subtraction",
        "video_name": "tuVd355R-OQ",
        "timestamps": [
            210
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "In 9:09, you write an equation which I don't understand. Could you explain it again, please?",
        "A": "But why does he put the twelve in brackets then?",
        "video_name": "tuVd355R-OQ",
        "timestamps": [
            549
        ],
        "3min_transcript": "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 I just don't like that 12 sitting there, so I'm going to multiply both sides by 12. So these are going to cancel out, and you're going to be left with 5 times q minus 7 is equal to 2/3 times 12. That's the same thing as 24 over 3. So this is, let me write this. 2 over 3 times 12 over 1 is equal to-- if you divide that by 3, you get a 4, divide that by 3, you get a 1-- is equal to 8. So you get 5 times q minus 7 is equal to 8. And then instead of dividing both sides by 5, which would get us pretty close to what we were doing over here, let me distribute this 5, I just want to show you, you can do it multiple legitimate ways. So 5 times q is 5q. 5 times negative 7 is minus, or negative 35, is equal to 8. 5q minus 35 is equal to 8. Now, if I want to get rid of that minus 35, or that negative 35, the best way to do it is to"
    },
    {
        "Q": "8:19 How can 7 be written as 35/5?",
        "A": "35/5 is the same thing as 7.",
        "video_name": "tuVd355R-OQ",
        "timestamps": [
            499
        ],
        "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": "@4:42 I didn't understand how you got 3/2. The way i did it was i divided 3 by 2 and got 1.5 .",
        "A": "they re really the same thing, 1.5=3/2",
        "video_name": "tuVd355R-OQ",
        "timestamps": [
            282
        ],
        "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": "at 8:19 why is seven written as 35? I thought it would be 8/5 plus 7/1",
        "A": "Seven is written as 35/5, not 35. Remember that if I have 35 and divide it by 5 I get 7. It s the same as when you re reducing a fraction: When you reduce a fraction, you divide the top and bottom by the same number. You can also multiply the top and bottom by the same number. In this case, Sal multiplied the top and bottom of 7/1 by 5, which gives (7*5)/(1*5) = 35/5. Hope this helps",
        "video_name": "tuVd355R-OQ",
        "timestamps": [
            499
        ],
        "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 2:11 Khan says that 0.6 goes into 12 two times but doesn't 6.0 not 0.6 go into 12 two times..?",
        "A": "he moved the decimal point one place to the right, so instead of it being .6 into 12 it was 6 into 120. He was doing the first part of the problem , 6 into 12 goes twice, so you put a 2, then six goes into 0 zero times so you put a zero. The 2 and the zero together is 20.",
        "video_name": "tuVd355R-OQ",
        "timestamps": [
            131
        ],
        "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": "At 1:40 shouldn't it be (y^3)/2 (instead of (y^3)/3)?",
        "A": "Okay, this is corrected later at 2:46",
        "video_name": "0pv0QtOi5l8",
        "timestamps": [
            100
        ],
        "3min_transcript": "Welcome back. In the last video we were just figuring out the volume under the surface, and we had set up these integral bounds. So let's see how to evaluate it now. And look at this. I actually realized that I can scroll things, which is quite useful because now I have a lot more board space. So how do we evaluate this integral? Well, the first integral I'm integrating with respect to x. I'm adding up the little x sums. So I'm forming this rectangle right here. Or you could kind of view it I'm holding y constant and integrating along the x-axis. I should switch colors. So what's the antiderivative of x y squared with respect to x? Well it's just x squared over 2. And then I have the y squared-- that's just a constant-- all over 2. And I'm going to evaluate that from x is equal to 1 to x is equal to the square root of y, which you might be daunted by. But you'll see that it's actually not that bad once you evaluate them. This is y is equal to 0 to y is equal 1. dy. Now, if x is equal to 1 this expression becomes y squared over 2. Right? y squared over 2, minus-- now if x is equal to square root of y, what does this expression become? If x is equal to the square root of y, then x squared is just y. And then y times y squared is y to the third. Right? So it's y to the third over 3. And now I take the integral with respect to y. So now I sum up all of these rectangles in the y direction. 0, 1. This is with respect to y. And that's cool, right? Because when you take the first integral with respect to x you end up with a function of y anyway, so you might as well have your bounds as functions of y's. It really doesn't make it any more difficult. But anyway, back to the problem. What is the antiderivative of y squared over 2 minus Well the antiderivative of y squared-- and you have to divide by 3, so it's y cubed over 6. Minus y to the fourth-- you have to divide by 4. Minus y to the fourth over-- did I mess up some place? No, I think this is correct. y to the fourth over 12. How did I get a 3 here? That's where I messed up. This is a 2, right? Let's see. x is square root of y. Yeah, this is a 2. I don't know how I ended up. Square root of y squared is y, times y squared y to the third over 2. Right. And then when I take the integral of this it's 4 times 2."
    },
    {
        "Q": "I'm a little confused about Sal's second table that he draws in green later in the video. Does this signify that the inputs and outputs have changed? At 2:56 you can see the table has written that x is 5, and f \u00e2\u0081\u00bb\u00c2\u00b9(x) is -9. Is 5 now the input, and -9 is the output? Or is x the range and f \u00e2\u0081\u00bb\u00c2\u00b9(x) is the domain?",
        "A": "In the inverse function, the domain and the range are switched. domain->range range->domain",
        "video_name": "KzaPBzFFLRM",
        "timestamps": [
            176
        ],
        "3min_transcript": "With that in mind, let's see if we can evaluate something like f inverse of 8. What is that going to be? I encourage you to pause the video and try to think about it. So f of x, just as a reminder of what functions do, f of x is going to map from this domain, from a value in its domain to a corresponding value in the range. So this is what f does, this is domain... and this right over here is the range. Now f inverse, if you pass it, the value and the range, it'll map it back to the corresponding value in the domain. But how do we think about it like this? so if this was 8, we'd have to say, well, what mapped to 8? We see here f of 9 is 8, so f inverse of 8 is going to be equal to 9. If it makes it easier, we could construct a table, where I could say x and f inverse of x, and what I'd do is swap these two columns. f of x goes from -9 to 5, f inverse of x goes from 5 to -9. All I did was swap these two. Now we're mapping from this to that. So f inverse of x is going to map from 7 to -7. Notice, instead of mapping from this thing to that thing, we're now going to map from that thing to this thing. It's going to map from -7 to 6. It's going to map from 8 to 9, and it's going to map from 12 to 11. Looks like I got all of them, yep. So all I did was swap these columns. The f inverse maps from this column to that column. So I just swapped them out. Now it becomes a little clearer. You see it right here, f inverse of 8, if you input 8 into f inverse, you get 9. Now we can use that to start doing fancier things. We can evaluate something like f of f inverse of 7. f of f inverse of 7."
    },
    {
        "Q": "At 8:34, she sang something about unary. i dont that that makes sense. help me!",
        "A": "Pick a number. 13 for example. In the decimal system (base ten) it is written 13, 1 ten and three 1s. In the binary system (base two) it is written 1101, one 8, one 4, no 2s, one 1. In the unary system (base one) it is written 0000000000000, one 1, another 1, another 1, another 1, another 1, another 1, another 1, another 1, another 1, another 1, another 1, another 1, and another 1.",
        "video_name": "sxnX5_LbBDU",
        "timestamps": [
            514
        ],
        "3min_transcript": "backwards E, half a heart on a plate, and a simple, boring, short and straight line. So yeah. If you want to add up all the numbers between 1 and anything, this trick works. For example, all the numbers between 1 and 12. 1 plus 12 is 13. 2 plus 11 is 13. All the way in to 6 plus 7. That's 6 times 13, which I do my head as 60 plus 18 equals 78. So 78 is the 12th triangular number, while 5,050 is the 100th triangular number. At least it's not \"100 Days of Christmas.\" I'll stick with \"Bottles of Beer.\" [SINGING] On the tenth day of Christmas, my true love gave to me, the base of our Arabic numeral system, the base of a nonary numeral and the base of senary, and a number five, quaternary's base, ternary's base also, and binary two, and the base of unary. Here's my favorite way of visualizing the triangular numbers. Say you've got them in this configuration where they make a nice right equilateral triangle, or half a square. Finding the area of a square is easy, because you just square the length of it. In this case, 12 times 12. And the triangle is half of that. Only not really, because half the square means you only get half of this diagonal, so you've got to add back in the other half. But that's easy because there's 12 things in the diagonal, So to get the n-th triangular number, just take n squared over 2 plus n over 2, or n squared plus n over 2. [SINGING] On the eleventh day of Christmas, my true love gave to me the number my amp goes up to, the number of fingers on my hands, the German word for no, what I did after I eat, the number of heads on a hydra, at least until you start cutting them off, the number of strings on a guitar, the number I like to do high, the amount of horsemen of the Apocalypse, the number of notes in a triad, the number of pears in a pair of pears, and the number of partridges in a pear tree. [SINGING] On the twelfth day of Christmas,"
    },
    {
        "Q": "at 0:40 why is y make x =0",
        "A": "The y-intercept just means where the line crosses the y axis. Any point on the y-axis has a x value of 0, because you aren t moving left or right, you are just going either up or down. So the y-intercept is where the line crosses the y axis, and that point will have a x value of 0.",
        "video_name": "405boztgZig",
        "timestamps": [
            40
        ],
        "3min_transcript": "We're told to find the x- and y-intercepts for the graph of this equation: 2 y plus 1/3x is equal to 12. And just as a bit of a refresher, the x-intercept is the point on the graph that intersects the x-axis. So we're not above or below the x-axis, so our y value must be equal to 0. And by the exact same argument, the y-intercept occurs when we're not to the right or the left of the y-axis, so that's when x is equal to 0. So let's set each of these values to 0 and then solve for what the other one has to be at that point. So for the x-intercept, when y is equal to 0, let's solve this. So we get 2 times 0, plus 1/3x is equal to 12. I just set y is equal to 0 right there, right? I put 0 for y. Well, anything times 0 is just 0, so you're just left with 1/3x is equal to 12. sides by 1/3, or we can multiply both sides by the reciprocal of 1/3. And the reciprocal of 1/3 is 3, or you can even think of it as 3 over 1. So times 3 over 1. And so we're left with 3 times 1/3, that just cancels out, so you're left with x is equal to 12 times 3, or x is equal to 36. So when y is equal to 0, x is 36. So the point 36 comma 0 is on the graph of this equation. And this is also the x-intercept. Now, let's do the same thing for the y-intercept. So let's set x equal 0, so you get 2y plus 1/3, times 0 is equal to 12. Once again, anything times 0 is 0. So that's 0, and you're just left with 2y is equal to 12. with y is equal to 12 over 2, is 6. So the y-intercept is when x is equal to 0 and y is equal to 6. So let's plot these two points. I'll just do a little hand-drawn graph, and make it clear what the x- and the y-intercepts are. So let me draw-- that's my vertical axis, and that is my horizontal axis-- and we have the point 36 comma 0. So this is the origin right here, that's the x-axis, that's the y-axis. The point 36 comma 0 might be all the way over here. So that's the point 36 comma 0. And if that's 36, then the point 0, 6 might be right about there. So that's the point 0, 6. And the line will look something like this."
    },
    {
        "Q": "Why are you multiplying it by 0? 0:22 of the video.",
        "A": "I didn t see any multiplication by zero at 0:22. There is some at 0:49. At this point Sal is looking for the x-intercept (where the line crosses the x-axis) which is when y is zero, so he s plugging in 0 for y in the given equation 2y + \u00e2\u0085\u0093x = 12 and solving for x.",
        "video_name": "405boztgZig",
        "timestamps": [
            22
        ],
        "3min_transcript": "We're told to find the x- and y-intercepts for the graph of this equation: 2 y plus 1/3x is equal to 12. And just as a bit of a refresher, the x-intercept is the point on the graph that intersects the x-axis. So we're not above or below the x-axis, so our y value must be equal to 0. And by the exact same argument, the y-intercept occurs when we're not to the right or the left of the y-axis, so that's when x is equal to 0. So let's set each of these values to 0 and then solve for what the other one has to be at that point. So for the x-intercept, when y is equal to 0, let's solve this. So we get 2 times 0, plus 1/3x is equal to 12. I just set y is equal to 0 right there, right? I put 0 for y. Well, anything times 0 is just 0, so you're just left with 1/3x is equal to 12. sides by 1/3, or we can multiply both sides by the reciprocal of 1/3. And the reciprocal of 1/3 is 3, or you can even think of it as 3 over 1. So times 3 over 1. And so we're left with 3 times 1/3, that just cancels out, so you're left with x is equal to 12 times 3, or x is equal to 36. So when y is equal to 0, x is 36. So the point 36 comma 0 is on the graph of this equation. And this is also the x-intercept. Now, let's do the same thing for the y-intercept. So let's set x equal 0, so you get 2y plus 1/3, times 0 is equal to 12. Once again, anything times 0 is 0. So that's 0, and you're just left with 2y is equal to 12. with y is equal to 12 over 2, is 6. So the y-intercept is when x is equal to 0 and y is equal to 6. So let's plot these two points. I'll just do a little hand-drawn graph, and make it clear what the x- and the y-intercepts are. So let me draw-- that's my vertical axis, and that is my horizontal axis-- and we have the point 36 comma 0. So this is the origin right here, that's the x-axis, that's the y-axis. The point 36 comma 0 might be all the way over here. So that's the point 36 comma 0. And if that's 36, then the point 0, 6 might be right about there. So that's the point 0, 6. And the line will look something like this."
    },
    {
        "Q": "At around (0:17) he says that the points aren't above or below the x axis....how do you figure out whether or not the point is above or below the x axis?",
        "A": "Hi Alice. What Sal means by the x-intercept not being above or below the x-axis is that this point is the point on the line where the line intercepts the x-axis. Finding this means that we have to set y equal to 0, and doing so means that we do not move above or below the x-axis, and that we only move along the x-axis. Hope that helped!",
        "video_name": "405boztgZig",
        "timestamps": [
            17
        ],
        "3min_transcript": "We're told to find the x- and y-intercepts for the graph of this equation: 2 y plus 1/3x is equal to 12. And just as a bit of a refresher, the x-intercept is the point on the graph that intersects the x-axis. So we're not above or below the x-axis, so our y value must be equal to 0. And by the exact same argument, the y-intercept occurs when we're not to the right or the left of the y-axis, so that's when x is equal to 0. So let's set each of these values to 0 and then solve for what the other one has to be at that point. So for the x-intercept, when y is equal to 0, let's solve this. So we get 2 times 0, plus 1/3x is equal to 12. I just set y is equal to 0 right there, right? I put 0 for y. Well, anything times 0 is just 0, so you're just left with 1/3x is equal to 12. sides by 1/3, or we can multiply both sides by the reciprocal of 1/3. And the reciprocal of 1/3 is 3, or you can even think of it as 3 over 1. So times 3 over 1. And so we're left with 3 times 1/3, that just cancels out, so you're left with x is equal to 12 times 3, or x is equal to 36. So when y is equal to 0, x is 36. So the point 36 comma 0 is on the graph of this equation. And this is also the x-intercept. Now, let's do the same thing for the y-intercept. So let's set x equal 0, so you get 2y plus 1/3, times 0 is equal to 12. Once again, anything times 0 is 0. So that's 0, and you're just left with 2y is equal to 12. with y is equal to 12 over 2, is 6. So the y-intercept is when x is equal to 0 and y is equal to 6. So let's plot these two points. I'll just do a little hand-drawn graph, and make it clear what the x- and the y-intercepts are. So let me draw-- that's my vertical axis, and that is my horizontal axis-- and we have the point 36 comma 0. So this is the origin right here, that's the x-axis, that's the y-axis. The point 36 comma 0 might be all the way over here. So that's the point 36 comma 0. And if that's 36, then the point 0, 6 might be right about there. So that's the point 0, 6. And the line will look something like this."
    },
    {
        "Q": "7:43. Isn't 1 -3 -2, not 2, or are you saying the absolute value?",
        "A": "Sal is using absolute value in this video, and that is why your a little confused. \u00f0\u009f\u0098\u0089",
        "video_name": "GdIkEngwGNU",
        "timestamps": [
            463
        ],
        "3min_transcript": "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. We see this six is three to the right of the mean. We don't care whether it's to the right or the left. And then four minus three. Four minus three is one, absolute value is one. And we see that. It is one to the right of three. And so what do we have? We have two plus two is four, plus three is seven, plus one is eight, over four, which is equal to two. So the Mean Absolute Deviation ... It fell off over here. Here, for this data set, the Mean Absolute Deviation is equal to two, while for this data set, the Mean Absolute Deviation is equal to one. And that makes sense. They have the exact same means. They both have a mean of three. But this one is more spread out. The one on the right is more spread out because, on average, each of these points are two away from three, while on average, each of these points are one away from three. The means of the absolute deviations on this one is one."
    },
    {
        "Q": "When Sal say's: du is going to be equal to the derivative of x^4+7 \"with respect to x\" (1:16). What does he mean with the last bit? Would it be possible to have a different variable than x and if so what would happen then?",
        "A": "With respect to x means that the equation is being derivated in the form d(u)/d(x), as the equation is written in terms of x. It is possible to have a variable other than x- If we had an equation that was, for example, y=3z-1, we would derivate y with respect to z because the equation is written in terms of z. hence dy/dz = 3. Hope this helps!",
        "video_name": "Zp5z0wa0kgo",
        "timestamps": [
            76
        ],
        "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": "At 3:10 why does Sal use the \"ln\"? i know ln means natural log but why is this used?",
        "A": "All the formulas in calculus involving log use the log with base e or ln. The formula used here is \u00e2\u0088\u00ab (1/x) = ln|x|",
        "video_name": "Zp5z0wa0kgo",
        "timestamps": [
            190
        ],
        "3min_transcript": "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? and we use the absolute value so it'll be defined even for negative use -- and it actually does work out. I'll do another video where I'll show you it definitely does. The natural log of the absolute value of u and then we might have a constant there that was lost when we took the derivative. So that's essentially our answer in terms of u. But now we need to un-substitute the u. So what happens when we un-substitute the u? Well, then we are left with -- this is going to be equal to -- the natural log of the absolute value of -- well, u is x^4 plus 7 -- not C, plus 7 -- and then we can't forget our plus C out here. And we are done!"
    },
    {
        "Q": "At 2:56, I'm really confused about where the 1 came from. Can someone please explain? Thank you",
        "A": "Example: 5/3 = (1/3)*5, right? You had du/u and did the same thing as above, so du/u = (1/u)*du.",
        "video_name": "Zp5z0wa0kgo",
        "timestamps": [
            176
        ],
        "3min_transcript": "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? and we use the absolute value so it'll be defined even for negative use -- and it actually does work out. I'll do another video where I'll show you it definitely does. The natural log of the absolute value of u and then we might have a constant there that was lost when we took the derivative. So that's essentially our answer in terms of u. But now we need to un-substitute the u. So what happens when we un-substitute the u? Well, then we are left with -- this is going to be equal to -- the natural log of the absolute value of -- well, u is x^4 plus 7 -- not C, plus 7 -- and then we can't forget our plus C out here. And we are done!"
    },
    {
        "Q": "At 3:24 where does the du go? I do understand that 1/u = ln |u| then we put the constant +C . Bu I can not seem to understand what happens to du and where it goes...? Thank you!",
        "A": "It disappears the same way dx does when you do a regular integration without a u-substitution. For example, the integral of x dx is (x^2)/2 + C, and the integral of (1/x)dx is ln|x| + C. We re using the same process when we integrate after a u-substitution, but now we re integrating with respect to u, so the integration needs du to work instead of dx.",
        "video_name": "Zp5z0wa0kgo",
        "timestamps": [
            204
        ],
        "3min_transcript": "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? and we use the absolute value so it'll be defined even for negative use -- and it actually does work out. I'll do another video where I'll show you it definitely does. The natural log of the absolute value of u and then we might have a constant there that was lost when we took the derivative. So that's essentially our answer in terms of u. But now we need to un-substitute the u. So what happens when we un-substitute the u? Well, then we are left with -- this is going to be equal to -- the natural log of the absolute value of -- well, u is x^4 plus 7 -- not C, plus 7 -- and then we can't forget our plus C out here. And we are done!"
    },
    {
        "Q": "At 10:45 he takes the \u00e2\u0088\u009a200 and converts it to 10\u00e2\u0088\u009a2. That lost me completely. What math should I learn so that makes sense? Thank you in advance!",
        "A": "The key to simplifying square roots is to take out any perfect squares and leave any non perfect squares inside. So if we notice that 200 = 100 \u00e2\u0080\u00a2 2, we see that 10^2 can be taken out as a 10 and 2 has to stay in. If we prime factor it, we have 2 \u00e2\u0080\u00a2 2 \u00e2\u0080\u00a2 2 \u00e2\u0080\u00a2 5 \u00e2\u0080\u00a2 5, and again \u00e2\u0088\u009a200 = \u00e2\u0088\u009a4 \u00e2\u0080\u00a2 \u00e2\u0088\u009a2 \u00e2\u0080\u00a2 \u00e2\u0088\u009a25 or 2 \u00e2\u0080\u00a2 5 \u00e2\u0088\u009a2",
        "video_name": "E4HAYd0QnRc",
        "timestamps": [
            645
        ],
        "3min_transcript": "So this is going to be--all right, this is 10/5, which is equal to 2. So the variance here-- let me make sure I got that right. Yes, we have 10/5. So the variance of this less-dispersed data set is a lot smaller. The variance of this data set right here is only 2. So that gave you a sense. That tells you, look, this is definitely a less-dispersed data set then that there. Now, the problem with the variance is you're taking these numbers, you're taking the difference between them and the mean, then you're squaring it. It kind of gives you a bit of an arbitrary number, and if you're dealing with units, let's say if these are distances. So this is negative 10 meters, 0 meters, 10 meters, this is 8 meters, so on and so forth, then when you square it, you get your variance in terms of meters squared. It's kind of an odd set of units. So what people like to do is talk in terms of standard or the square root of sigma squared. And the symbol for the standard deviation is just sigma. So now that we've figured out the variance, it's very easy to figure out the standard deviation of both of these The standard deviation of this first one up here, of this first data set, is going to be the square root of 200. The square root of 200 is what? The square root of 2 times 100. This is equal to 10 square roots of 2. That's that first data set. Now the standard deviation of the second data set is just going to be the square root of its variance, which is just 2. this first data set. This is 10 roots of 2, this is just the root of 2. So this is 10 times the standard deviation. And this, hopefully, will make a little bit more sense. This has 10 times more the standard deviation than this. And let's remember how we calculated it. Variance, we just took each data point, how far it was away from the mean, squared that, took the average of those. Then we took the square root, really just to make the units look nice, but the end result is we said that that first data set has 10 times the standard deviation as the second data set. So let's look at the two data sets. This has 10 times the standard deviation, which makes sense"
    },
    {
        "Q": "i have doubt about these kind of linear equation which at 7:12 have many solution due to rank of matrix is 2 and order of matrix is 3 * 2",
        "A": "No, it s a simple system of two variables. There is one solution for A and another for B.",
        "video_name": "hbJ2o9EUmJ0",
        "timestamps": [
            432
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 5:21 you said that unbiased variance = (n-1)*\u00cf\u0083\u00c2\u00b2/n. we know that (n-1)/n is a increasing function. So how come value decrease as n at 9 compared to at 8",
        "A": "That s because the graph is from randomly picked samples and there is some error in this process.",
        "video_name": "Cn0skMJ2F3c",
        "timestamps": [
            321
        ],
        "3min_transcript": "while the bluer dots are the ones of a larger sample size. You see here these two little, I guess the tails ,so to speak, of this hump, that these ends, are more of a reddish color. that most of the blueish or the purplish dots are focused right in the middle right over here, that they are giving us better estimates. There are some red ones here, and that's why it gives us that purplish color, but out here on these tails, it's almost purely some of these red. Every now and then by happenstance you get a little blue one, but it's disproportionately far more red, which really makes sense when you have a smaller sample size, you are more likely to get a sample mean that is a bad estimate of the population mean, that's far from the population mean, and you're more likely to significantly underestimate the sample variance. Now this next chart really gets to the meat of the issue, that for each of these sample sizes, so this right over here for sample size two, if we keep taking sample size two, and we keep calculating the biased sample variances and dividing that by the population variance, and finding the mean over all of those, you see that over many, many, many trials, and many, many samples of size two, that that biased sample variance over population variance, it's approaching half of the true population variance. When sample size is three, it's approaching 2/3, 66 point six percent, of the true population variance. When sample size is four, it's approaching 3/4 of the true population variance. So we can come up with the general theme that's happening. When we use the biased estimate, we're not approaching the population variance. We're approaching n minus one over n When n was two, this approached 1/2. When n is three, this is 2/3. When n is four, this is 3/4. So this is giving us a biased estimate. So how would we unbias this? Well, if we really want to get our best estimate of the true population variance, not n minus one over n times the population variance, we would want to multiply, I'll do this in a color I haven't used yet, we would want to multiply times n over n minus one. to get an unbiased estimate. Here, these cancel out and you are just left with your population variance. That's what we want to estimate. Over here you are left with our unbiased estimate of population variance,"
    },
    {
        "Q": "Mr. Khan,\nAt 2:52 of the video you mentioned how to write domain. Does {x/x\u00e2\u0082\u00acR} work too to write the domain because I was taught this way?",
        "A": "Yes, you could use that when the domain = (-infinity, infinity). The domain and range are sets, so either interval notation is used to describe the set. Or, you could also use set notation which is what you have.",
        "video_name": "-DTMakGDZAw",
        "timestamps": [
            172
        ],
        "3min_transcript": "over pi. We're able to find the output pretty easily. But I want to do something interesting. Let's attempt to input 0 into the function. If we input 0 then the function tells us what we need to output. Does this definition tell us what we need to output? So if I attempt to put x equal 0, then this definition would say f of 0 be 2 over 0, but 2 over 0 is undefined. Rewrite this -- 2 over 0. This is undefined. This function definition does not tell us what to actually do with 0. It gives us an undefined answer. So this function is not defined here. It gives a question mark. So this gets to the essence of what domain is. Domain is the set of all inputs over which the function is defined. So the domain of this function f would be all real numbers except for x equals 0. So we write down these, these big ideas. This is the domain -- the domain of a function -- A domain of a function is the set of all inputs -- inputs over which the function is defined -- over which the function is defined, or the function has defined outputs over which the function has defined outputs. So the domain for this f in particular -- so the domain for this one -- if I want to say its domain, I could say, look, it's going to be the set of these curly brackets. These are kind of typical mathy set notation. I said OK , it could be the set of -- I gonna put curly brackets like that. Well, x can be a member So this little symbol means a member of the real numbers. But it can't be 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"
    },
    {
        "Q": "I'm a little confused what the (base) is... 0:53",
        "A": "Usually, the bottom side of a parallelogram is thought of as the base, but any side of the parallelogram can be chosen as the base when using the A = bh formula. Once a side is chosen as the base (b), the height (h) must be the perpendicular distance from the side chosen as the base to the side parallel to this base. (Note that the height, h, is usually not the length of a side of the parallelogram.) Have a blessed, wonderful day!",
        "video_name": "hm17lVaor0Q",
        "timestamps": [
            53
        ],
        "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": "At 1:05 What does the triangle on x+3 stand for?",
        "A": "It s not a triangle, it s delta . In this case it means that you have to find the absolute value of x1 (which is -3) minus x2 (which is 0). The result is 3.",
        "video_name": "81SseQCpGws",
        "timestamps": [
            65
        ],
        "3min_transcript": "Find the slope of the line pictured on the graph. So the slope of a line is defined to be rise over run. Or you could also view it as change in y over change in x. And let me show you what that means. So let's start at some arbitrary point on this line, and they highlight some of these points. So let's start at one of these points right over here. So if we wanted to start one of these points-- and let's say we want to change our x in the positive direction. So we want to go to the right. So let's say we want to go from this point to this point over here. How much do we have to move in x? So if we want to move in x, we have to go from this point We're going from negative 3 to 0. So our change in x-- and this triangle, that's delta. That means \"change in.\" Our change in x is equal to 3. Well, when we moved from this point to this point, our x-value changed by 3, but what happened to our y-value? Well, our y-value went down. It went from positive 3 to positive 2. Our y-value went down by 1. So our change in y is equal to negative 1. So we rose negative 1. We actually went down. So our rise is negative 1 when our run-- when our change in x-- is 3. So change in y over change in x is negative 1 over 3, or we could say that our slope is negative 1/3. Let me scroll over a little bit. It is negative 1/3. And I want to show you that we can do this with any two points on the line. We could even go further than 3 in the x-direction. Let's start at this point right over here and then move backwards to this point over here, just to show you that we'll still get the same result. So to go from this point to that point, what is our change in x? So our change in x is this right over here. Our change in x is that distance right over there. We started at 3, and we went to negative 3. We went back 6. Over here, our change in x is equal to negative 6. We're starting at this point now. So over here our change in x is negative 6. And then when our change in x is negative 6, when we start at this point and we move 6 back, what is our change of y to get to that point? Well, our y-value went from 1. That was our y-value at this point. And then when we go back to this point, our y-value is 3. So what did we do? We moved up by 2. Our change in y is equal to 2."
    },
    {
        "Q": "At 5:41 why do we multiply P(A) by P(B)",
        "A": "As the events A and B are independent, meaning the outcome of event A does not affect the outcome of event B, then we can calculate the probability of both A and B taking place by multiplying the P(A) with the probability of B. This holds for independent events. But we must be careful about what we are calculating according to the problem we are trying to solve.",
        "video_name": "RI874OSJp1U",
        "timestamps": [
            341
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 0:30 what is sal trying to say??",
        "A": "At 0:30 Sal is explaining the significance of the constant of integration added in the end. \u00e2\u0088\u00ab2xdx = x^2 + C The +C is added as the derivative of a constant is 0. So the derivative of x^2, x^2 + 1 or x^2 + e or x^2 + \u00cf\u0080 or in general x^2+C is the same each time.",
        "video_name": "MMv-027KEqU",
        "timestamps": [
            30
        ],
        "3min_transcript": "We know how to take derivatives of functions. If I apply the derivative operator to x squared, I get 2x. Now, if I also apply the derivative operator to x squared plus 1, I also get 2x. If I apply the derivative operator to x squared plus pi, I also get 2x. The derivative of x squared is 2x. Derivative, with respect to x of pi of a constant, is just 0. Derivative, with respect to x of 1, is just a constant, is just 0. So once again, this is just going to be equal to 2x. In general, the derivative, with respect to x of x squared plus any constant, is going to be equal to 2x. The derivative of x squared, with respect to x, is 2x. Derivative of a constant, with respect to x, a constant does not change with respect to x, so it's just equal to 0. So you have-- You apply the derivative operator 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."
    },
    {
        "Q": "i don't really get what sal says at 4:05 can some one help",
        "A": "That order doesn t matter in combinations.",
        "video_name": "iKy-d5_erhI",
        "timestamps": [
            245
        ],
        "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": "How do I solve this problem?\n: a ski resort has started to keep track of the number of skiers and snowboarders who bought season passes. The ratio of the number of skiers who bought season passes to the number of snowboarders who bought season passes is 1:2. If 1250 more snowboarders bought season passes than skiers, how many snowboarders and how many skiers bought season passes?",
        "A": "ratio of skiers to snowboarders 1:2 nbr of skiers = 1 * x = 1x nbr of snowborarders = 2 * x = 2x how many more snow boarders than skiers = 2x -1x = x so x should be equal to 1250 so nbr of skiers = 1x = 1 * 1250 =1250 nbr of snow boarders = 2x = 2 * 1250 = 2450",
        "video_name": "tOd2T72eJME",
        "timestamps": [
            62
        ],
        "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": "i don't get 0:32",
        "A": "Think of it this way: You take a ruler and measure 7cm on the map. That whole 7cm is 10km in the real world. :)",
        "video_name": "tOd2T72eJME",
        "timestamps": [
            32
        ],
        "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": "What about a ratio like 1/2:3/4",
        "A": "A ratio of 1/2:3/4 is a complex fraction (fractions within a fraction). It would need to be simplified by dividing the 2 fractions. 1/2 / 3/4 = 1/2 * 4/3 = 2/3",
        "video_name": "tOd2T72eJME",
        "timestamps": [
            123
        ],
        "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": "2:22\nWhy can't we count the Probability of \"not catching a sunfish 3 times\" as well 1/2*1/2*1/2?\nAs each time of not catching a sunfish means catching a trout, which has the same probablity.\nThank's!",
        "A": "There are 8 equally likely possibilities: SSS, SST, STT, STS, TTT, TTS, TSS, and TST. So the probability you said: 1/2*1/2*1/2 could also be said as: P(not get a sunfish)*P(not get a sunfish)*P(not get a sunfish) which would be the fifth possibility: TTT. Not getting a sunfish three times in a row, however, could be any of the last seven possibilities. I hope this helps! :)",
        "video_name": "86nb02Bx_5w",
        "timestamps": [
            142
        ],
        "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": "At 0:24, why did Sal say 10+10+6? That's 26, instead of 16...",
        "A": "Oh, no. Sal said 10+6, not 10+10+6. That s where your trip was.",
        "video_name": "vbGwcvXgDlg",
        "timestamps": [
            24
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 7:35, Sal enters some values into the calculator which are close to -4. What would happen on the calculator if we tried to enter -4 into the function replacing x??",
        "A": "When we substitute -4 for x, the equation becomes (-4)^2/(-4)^2-16. We can simplify this expression to 16/16-16 which then becomes 16/0. It s impossible to divide by 0 because nothing times 0 is equal to 16. There is no way to get 16 (or any other number) by multiplying by 0. The expression x^2/x^2-16 becomes undefined when x=-4 and when x=4 because there are no possible outputs from those inputs.",
        "video_name": "2N62v_63SBo",
        "timestamps": [
            455
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 16:59 Sal explains how to get the horizontal asymptote. Is it possible to ever have a negative horizontal asymptote?",
        "A": "Yes, a horizontal asymptote can be any y value. It s location is determined by the coefficients of the variables, which can be any number (the coefficients that is).",
        "video_name": "2N62v_63SBo",
        "timestamps": [
            1019
        ],
        "3min_transcript": ""
    },
    {
        "Q": "In 3:40, Sal writes |-2-3|, while on the number line, we normally subtract the lesser number from the greater one, i.e. |3-(-2)|, right?",
        "A": "As long as you subtract the 2 numbers and take their absolute value, you will create the same result: |-2-3| = |-5| = 5 |3-(-2)| = |5| = 5 So, pick the version you are more comfortable with.",
        "video_name": "t4xOkpP8FgE",
        "timestamps": [
            220
        ],
        "3min_transcript": "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. This distance right over here, this distance right over here is equal to five. One, two, three, four, five. Or you'd have to go five backwards to go from three to negative two. But let's see that what I just wrote actually applies right over here. So if we took negative two to be our a and three to be our b, then we could write this as the absolute value of negative two minus three, what is this going to be equal to? Well this is going to be equal to negative two minus three is negative five, so it's the absolute value of negative five. So this indeed equals five. So notice I subtracted the larger number from the smaller number. I got a negative value, but then I took the absolute value of it. That gave me the actual distance Now what if I did it the other way around? What if I took three minus negative two? So it's going to be the absolute value of three, let me do it in the blue color, the absolute value of three minus, and in parentheses I'll write the negative two. Negative two. Now if you subtract a smaller number from a larger number, you should get a positive value. So the absolute value sign here is just kind of extra. You don't really need it, unless to verify that that's true. This is going to be three minus negative two. That's the same thing as three plus positive two, or five. So this is just going to be the absolute value of five, which of course, is equal to five. So hopefully this makes you feel good that if you want the distance between two numbers, you subtract one from the other, and it doesn't matter which order you do it. You could subtract three from negative two, or negative two from three, be careful with the negative symbols here, and then take the absolute value,"
    },
    {
        "Q": "At around 7:10, why did Sal do 16-2=14 instead of 2+16=18?",
        "A": "At 3:00 and at 4:12 or so, he explanes that the change in y is 14 because one point is at +2 and the other is at +16. and if you count that on the graph they are 14 spaces apart.",
        "video_name": "WkspBxrzuZo",
        "timestamps": [
            430
        ],
        "3min_transcript": "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. And just to visualize our change in x's and our change in y's that we dealt with here, when we started at 4 and we ended at-- or when we started at 4 comma 2 and ended at negative 3 comma 16, that was analogous to starting here and ending over there. And we said our change in x was negative 7. We had to move back. Our run we had to move in the left direction by 7. That's why it was a negative 7. And then we had to move in the y-direction. We had to move in the y-direction positive 14. So that's why our rise was positive. So it's 14 over negative 7, or negative 2. When we did it the other way, we started at this point. We started at this point, and then ended at this point. Started at negative 3, 16 and ended at that point. So in that situation, our run was positive 7. And now we have to go down in the y-direction since we switched the starting and the endpoint. And now we had to go down negative 14. Either way, we got the same slope."
    },
    {
        "Q": "in 1:16 does m equal slope?",
        "A": "In y=mx + b, m is always the slope, and b is always the y-intercept",
        "video_name": "WkspBxrzuZo",
        "timestamps": [
            76
        ],
        "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 0:32 Sal said \"all the rectangles\" but there was actually one square .",
        "A": "A rectangle just means a shape with four sides & four right angles, so technically all squares are rectangles.",
        "video_name": "nLY2bzRfQyo",
        "timestamps": [
            32
        ],
        "3min_transcript": "So I have this yellow rectangle here, and we know two things about this yellow rectangle. We know that it has a length of 10, that the length of this side right over here is 10. And we also know that this yellow rectangle has an area of 60 square units, whatever units we're measuring this 10 in. So what I want you to do is now pause this video, and based on the information given on these other rectangles-- so in some of them we give you two of their dimensions, in some of them we give you something like the perimeter and one of the dimensions-- I want you to pause the video and think about which of these rectangles, if any of them, have either the same area or the same perimeter as this yellow rectangle. So pause the video right now. Well, the best way to figure out which of these have the same area or perimeter as this original yellow rectangle is to just figure out the area and the perimeter for all of these rectangles and see which ones of them are equivalent. So we already know the area for this one, So how do we figure that out? Well, to figure out perimeter, we would need to know the lengths of all the sides. Well, if the area is 60 square units, that means the length times the width is equal to 60, that 10 times this width right over here is going to be equal to 60. So 10 times what is equal to 60? Well, 10 times 6 is equal to 60. 10 times 6 is equal to 60 square units. 10 units times 6 units is equal to 60 square units. Fair enough. So how do we figure out the perimeter now? Well, this is a rectangle. So we know if this length is 10, then this length must also be 10. And if this width is 6, then this width must be 6 as well. And now we can figure out the perimeter. It's 10 plus 10 plus 6 plus 6, which is 32. So let me write that down. The perimeter of our original yellow rectangle is equal to 32. and figure out what their perimeters and the areas are. We already know the perimeter for this purple or mauve rectangle, but we need to figure out its area. In order to figure out its area, we can't just rely on this one dimension just on its width. We have to figure out its length as well. So how do we figure that out? Well, one way to realize it is that the perimeter is the distance all the way around the rectangle. So what would be the distance halfway around the rectangle? So let me see if I can draw it. What would be the distance of this side, our length, plus this side? Well, it would be half the perimeter. 5 plus something is going to be equal to half the perimeter. Remember, the perimeter is all four sides. If we just took these two sides, which would be half the perimeter. So, these two sides must be equal to, when you take their sum, must be equal to 17, half the perimeter. So 5 plus what is equal to 17? 5 plus this question mark is equal to 17."
    },
    {
        "Q": "At 1:41, it is 4.",
        "A": "Yes, 4 would be the correct opposite for -4.",
        "video_name": "2Zk6u7Uk5ow",
        "timestamps": [
            101
        ],
        "3min_transcript": "- [Voiceover] What I want to do in this video is think about what it means to have an opposite of a number. Let me draw a number line here. Let's draw a number line. And let's put some numbers on this number line. We can start at zero, and if we go to the right we have positive numbers. One, two, three, four, five. As we go to the left we get more and more negative. So negative one, negative two, negative three, negative four, and I can keep going on and on and on. Let's pick one of these numbers. Let's say that we pick the number three. What is going to be the opposite of the number three? Well the opposite of the number is a number that's the same distance from zero but on the other side. So three is three to the right of zero. One, two, three. So its opposite is going to be three to the left of zero. So the opposite of three is negative three. Let me make a little table here. If we have the number, the number. And then we have its opposite. We have its opposite. So we just figured out that if you have the number three, its opposite is going to be negative three. Now what if your number is negative? What if your number, let's say the number negative four. What's the opposite of that? And I encourage you to pause the video and try to think about it on your own. Well, you say, okay, negative four is right over here. That's negative four. It is four to the left of zero. One, two, three, four to the left of zero. So its opposite is going to be four to the right of it. So one, two, three, four. It's going to be positive. So you're probably starting to see a pattern here. The opposite of a number is going to be the opposite sign of that number. If you have a positive three here, its opposite is going to be negative three. If you start with negative four, its opposite is going to be positive four. One way to think about it, it's going to have the same absolute value but have a different sign. Or another way to think about it is, however if this is three to the right of zero, its opposite is going to be three to the left of zero. Or if the number is four to the left of zero, its opposite is going to be four to the right of zero. So we'll do one last one. What is the opposite of... What's going to be the opposite of one? Well one is one to the right of zero, so its opposite is going to be one to the left of zero, or negative one. Or another way to think about it, one is positive"
    },
    {
        "Q": "I am confused at 4:35 the video lost me, i am so confused",
        "A": "He s basically saying you need all of the factors for both of the numbers and if you leave one out or forget one your answer will be wrong. Hope this helps!",
        "video_name": "1Vb8t7Y-pI0",
        "timestamps": [
            275
        ],
        "3min_transcript": "find the least common multiple other than just looking at the multiples like this. You could look at it through prime factorization. 30 is 2 times 15, which is 3 times 5. So we could say that 30 is equal to 2 times 3 times 5. And 24-- that's a different color than that blue-- 24 is equal to 2 times 12. 12 is equal to 2 times 6. 6 is equal to 2 times 3. So 24 is equal to 2 times 2 times 2 times 3. So another way to come up with the least common multiple, if we didn't even do this exercise up here, says, look, the number has to be divisible by both 30 and 24. If it's going to be divisible by 30, it's going to have to have 2 times 3 times 5 in its prime factorization. So this makes it divisible by 30. And say, well in order to be divisible by 24, its prime factorization is going to need 3 twos and a 3. Well we already have 1 three. And we already have 1 two, so we just need 2 more twos. So 2 times 2. So this makes it-- let me scroll up a little bit-- this right over here makes it divisible by 24. And so this is essentially the prime factorization of the least common multiple of 30 and 24. You take any one of these numbers away, you are no longer going to be divisible by one of these two If you take a two away, you're not going to be divisible by 24 If you take a two or a three away. If you take a three or a five away, you're not going to be divisible by 30 anymore. And so if you were to multiply all these out, this is 2 times 2 times 2 is 8 times 3 is 24 times 5 is 120. Umama just bought one package of 21 binders. Let me write that number down. 21 binders. She also bought a package of 30 pencils. She wants to use all of the binders and pencils to create identical sets of office supplies for her classmates. What is the greatest number of identical sets Umama can make using all the supplies? So the fact that we're talking about greatest is clue that it's probably going to be dealing with greatest common divisors. And it's also dealing with dividing these things. We want to divide these both into the greatest number of identical sets. So there's a couple of ways we could think about it. Let's think about what the greatest common divisor of both these numbers are. Or I could even say the greatest common factor. The greatest common divisor of 21 and 30. So what's the largest number that divides into both of them?"
    },
    {
        "Q": "um... in 3:6-11 what do you mean by distribute? that i so confused",
        "A": "Distributing is multiplying the number outside the parentheses by everything inside the parentheses 9(1+99) Aka 9*1 and 9*99",
        "video_name": "XAzFGx3Ruig",
        "timestamps": [
            186
        ],
        "3min_transcript": "or seven, or 11, or 17, why does it work for nine? And actually also works for three, and we'll think about that in the future video. And to realize that, we just have to rewrite 2943. So the two in 2943, it's in the thousands place, so we can literally rewrite it as two times, two times 1000, the nine is in the hundreds place, so we can literally write it as nine times 100, the four is in the tens place, so it's literally the same thing as four times 10, and then finally we have our three in the ones place, we could write it as three times one or just three. This is literally 2000, 900, 40, and three, 2943. But now we can rewrite each of these things, this thousand, this hundred, this ten, as a sum of one plus something that is divisible by nine, so 1000, 1000 I can rewrite as one plus 999, I can rewrite 100 as one plus 99, one plus 99, I can rewrite 10 as one plus 9, and so two times 1000 is the same thing as two times one plus 999, nine times 100 is the same thing as nine times one plus 99, four times 10 is the same thing as four times one plus nine, and then I have this plus three over here. But now I can distribute, I can say, well, this over here is the same thing as two times one, which is just two, plus two times 999, this thing right over here is the same thing as nine times one, just to be clear what I'm doing, I'm distributing the two over the first parenthesis, these first two terms, then the nine, I'm gonna distribute again, so it's gonna be nine, nine times one plus nine times 99, and then over here, I forgot the plus sign right over here, I'm gonna distribute the four, four times one, so plus four, and then four times nine, so plus four times nine, and then finally I have this positive three, or plus three right over here. Now I'm just gonna rearrange this addition, so I'm gonna take all the terms we're multiplying by 999, and I'm going to do that in orange. So I'm gonna take this term, this term, and this term right over here, and so I have two times 999, that's that there, plus nine times 99, plus four times nine, plus four times nine, so that's those three terms, and then I have plus two, plus two, plus nine, plus nine, plus four, plus four, and plus three, plus three. And this is interesting, this is just the sum of our digits,"
    },
    {
        "Q": "At 4:15 mins into the video, Sal says that since 999 is divisible by 9, everything before it is divisible by 9, hence 2.999 is divisible by 9. But 4.9 is not divisible by 9. What am I missing here?",
        "A": "let s look here. i think 4*9 is indeed divisible by 9. 4*9 = 36. is 36 divisible by 9? yes, because 36/9 is 4, a whole number. also to do a multiplication sign, use an asterisk (shift-8) on your keyboard instead of a decimal.",
        "video_name": "XAzFGx3Ruig",
        "timestamps": [
            255
        ],
        "3min_transcript": "I can rewrite 100 as one plus 99, one plus 99, I can rewrite 10 as one plus 9, and so two times 1000 is the same thing as two times one plus 999, nine times 100 is the same thing as nine times one plus 99, four times 10 is the same thing as four times one plus nine, and then I have this plus three over here. But now I can distribute, I can say, well, this over here is the same thing as two times one, which is just two, plus two times 999, this thing right over here is the same thing as nine times one, just to be clear what I'm doing, I'm distributing the two over the first parenthesis, these first two terms, then the nine, I'm gonna distribute again, so it's gonna be nine, nine times one plus nine times 99, and then over here, I forgot the plus sign right over here, I'm gonna distribute the four, four times one, so plus four, and then four times nine, so plus four times nine, and then finally I have this positive three, or plus three right over here. Now I'm just gonna rearrange this addition, so I'm gonna take all the terms we're multiplying by 999, and I'm going to do that in orange. So I'm gonna take this term, this term, and this term right over here, and so I have two times 999, that's that there, plus nine times 99, plus four times nine, plus four times nine, so that's those three terms, and then I have plus two, plus two, plus nine, plus nine, plus four, plus four, and plus three, plus three. And this is interesting, this is just the sum of our digits, and you might see where all of this is going, this orange stuff here, is this divisible by nine? What? Sure, it will definitely, 999, that's divisible by nine, so anything this is multiplying by, it's divisible by nine, so this is divisible by nine, this is definitely divisible by nine, 99, regardless of whether is being multiplied by nine, whatever is multiplying by nine, whatever is multiplying 99 is gonna be divisible by nine, because 99 is divisible by nine, so this works out, and same thing over here, you're always going to be multiplying by a multiple of nine, so all of this business all over here is definitely going to be divisible by nine. And so in order for the whole thing, and all I did is I rewrote 2943 like this right over here, in order for the whole thing to be divisible by nine, this part definitely is divisible by nine, in order for the whole thing does the rest of this sum, it has to be divisible by nine as well. So in order for this whole thing,"
    },
    {
        "Q": "Around the 1:36 mark in your video, I did not understand it when you said that you would multiply the number by 3 jumps. I get the jump part, but i do not get the multiplication part. By that I mean, why would you multiply 8/3 by 3? I would appreciate it very much if anyone could help me figure this out!",
        "A": "You have to flip one of the numbers you are dividing and then multiply the flipped number with the other number(in this case 3) to get your answer",
        "video_name": "f3ySpxX9oeM",
        "timestamps": [
            96
        ],
        "3min_transcript": "Let's think about what it means to take 8/3 and divide it by 1/3. So let me draw a number line here. So there is my number line. This is 0. This is 1. And this is 2. Maybe this is 3 right over here. And let me plot 8/3. So to do that, I just need to break up each whole into thirds. So let's see. That's 1/3, 2/3, 3/3, 4/3, 5/3, 6/3, 7/3, 8/3. So right over here. And then of course, 9/3 would get us to 3. So this right over here is 8/3. Now, one way to think about 8/3 divided by 3 is what if we take this length. And we say, how many jumps would it take to get there, if we're doing it in jumps of 1/3? Or essentially, we're breaking this up. If we were to break up 8/3 into sections of 1/3, how many sections would I have, or how many jumps would I have? If we're trying to take jumps of 1/3, we're going to have to go 1, 2, 3, 4, 5, 6, 7, 8 jumps. So we could view this as-- let me do this in a different color. I'll do it in this orange. So we took these 8 jumps right over here. So we could view 8/3 divided by 1/3 as being equal to 8. Now, why does this actually make sense? Well, when you're dividing things into thirds, for every whole, you're now going to have 3 jumps. So whatever value you're trying to get to, you're going to have that number times 3 jumps. So another way of thinking about it is that 8/3 divided by 1/3 is the same thing as 8/3 times 3. 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."
    },
    {
        "Q": "at 2:20 , how are we telling that they are parallel",
        "A": "Because of the right angles. It requires a 180 degree angle to make a line fold back over itself, or to be parallel. At 2:20, the top of the square forms a 90 degree angle with the side, and the side forms another 90 degree angle with the bottom. Adding these up, we get 180 degrees, meaning the top and bottom lines are parallel. The same thing is true with the two sides. I hope this answers your question! :)",
        "video_name": "wPZIa3SjPF0",
        "timestamps": [
            140
        ],
        "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 0:13, how are you able to tell that the opposite sides are parallel?",
        "A": "107+73=180 degrees, so the opp sides are parallel by same-side interior/consecutive interior angles",
        "video_name": "wPZIa3SjPF0",
        "timestamps": [
            13
        ],
        "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 1:45, how can we say that two pairs of opposite sides are parallel ?\nIs it just by observation or anything else ?",
        "A": "If they ever touch, then it won t be parallel, but they have to continuously have the same space between them.",
        "video_name": "wPZIa3SjPF0",
        "timestamps": [
            105
        ],
        "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 1:24, Sal mentions that 2 is a factor of (x^3 - 8). How did he come up with that?",
        "A": "If 2 is a root of a polynomial, then (x-2) is a factor. The polynomial is x^3-8. Set that equal to 0 and solve: x^3-8=0 x^3=8 (x^3)^(1/3)=8^(1/3) x=2. So 2 must be a root, and therefore (x-2) is a factor. You can divide x^3-8 by (x-2) and get the quotient of x^2+2x+4. This is how you derive the difference of cubes formula.",
        "video_name": "6FrPLJY0rqM",
        "timestamps": [
            84
        ],
        "3min_transcript": "Let's see if we can tackle a more complicated partial fraction decomposition problem. I have 10x squared plus 12x plus 20, all of that over x to the third minus 8. The first thing to do with any of these rational expressions that you want to decompose is to just make sure that the numerator is of a lower degree than the denominator, and if it's not, then you just do the algebraic long division like we did in the first video. But here, you can do from [UNINTELLIGIBLE] the highest degree term here is a second-degree term, here it's a third-degree term, so we're cool. This is a lower degree than that one. If it was the same or higher, we would do a little long division. The next thing to do, if we're going to decompose this into its components, we have to figure out the factors of the denominator right here, so that we can use those factors as the denominators in each of the components, and a third-degree polynomial is much, much, much harder to factor than a second-degree polynomial, normally. pop out at you-- if it doesn't immediately, hopefully what I'm about to say will make it pop out at you in the future-- is you should always think about what number, when you substitute into a polynomial, will make it equal to 0. And in this case, what to the third power minus 8 equals 0? And hopefully 2 pops out at you. And this is something you can only do really through inspection or through experience. And you'll immediately see 2 to the third minus 8 is 0, so 2 is a 0 of this, or 2 makes this expression 0, and that tells us that x minus 2 is a factor. So we can rewrite this right here as 10x squared plus 12x plus 20 over x minus 2 times something something We don't know what that something is yet. And I just want to hit the point home of why this is true, or what the intuition vibe has true. expression 0. And we know that 2 would make this factored expression 0, because when you put a 2 right here, this factor become 0, so it'll make the whole thing 0. And so that's why, that's the intuition where, if you substitute a number here and it makes this 0, you do x minus that number here, and we know that that will be a factor of the thing. Well, anyway, the next step if we really want to decompose this rational expression is to figure out what this part of it is, and the way to do that is with algebraic long division. We essentially just divide x minus 2 into x to the third minus 8 to get this, so let's do that. So you get x minus 2 goes into x to the third-- and actually, what I'm going to do is, I'm going to write-- I leave space for the second-degree term, which is 0, the first-degree term, and then minus 8 is the constant term, so minus 8-- I"
    },
    {
        "Q": "In 2:45, who is blaise pascal?",
        "A": "Blaise Pascal (1623-1662) was a French mathematician, physicist, writer, philosopher, and inventor. Pascal s earliest work was in the field of fluids, where the Pascal (Pa), a unit of pressure, was named after him. He then worked to invent the calculator. He became the second inventor of a calculator. Additionally, he basically created the field of projective geometry and worked on probabilities.",
        "video_name": "Yhlv5Aeuo_k",
        "timestamps": [
            165
        ],
        "3min_transcript": "\"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 don't have to do all the adding. I don't know if this was discovered through doodling but it was discovered independantly in: France, Italy, Persia, China and probably other places too so it's possible someone did. Right... so I don't actually care about the individual numbers right now. So, if you still Ulam, you pick a property and highlight it(e.g. if it's even or odd) If you circle all the odd numbers you'll get a form which might be starting to look familiar. And it makes sense you'd get Sierpinski's Triangle because when you add an odd number and an even number, you get an odd number. (odd + odd) = even and (even + even) = even... So it's just like the crash and burn binary tree game. The best part about it is that, if you know these properties, you can forget about the details of the numbers You don't have to know that a space contains a 9 to know that it's going to be odd. Now, instead of two colours, let's try three. we'll colour them depending on what the remainder is when you divide them by three(instead of by two). Here's a chart! :) So, all the multiples of three are coloured red, remainder of one will be coloured black and"
    },
    {
        "Q": "At 1:34 how come that's the highest prime number known to man isn't there a higher prime number?",
        "A": "There are an infinite number of primes, but it is very difficult to prove that a really large number is prime. So, we know that there is a higher prime number, but we don t know what it is. This video is slightly out of date, though, because the highest prime known to man is now 2^57,885,161 \u00e2\u0088\u0092 1.",
        "video_name": "Yhlv5Aeuo_k",
        "timestamps": [
            94
        ],
        "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": "At 0:14, did she choose Ulam for any specific reason? Is he just a random person that she chose or does he have something to do with sick number games?",
        "A": "Ulam is a famous mathematician",
        "video_name": "Yhlv5Aeuo_k",
        "timestamps": [
            14
        ],
        "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": "For 1:40, do you have to represent if it's a natural number, integer, etc..",
        "A": "When using sigma notation to express sums, it is implied that we re working with integers (which are a superset of natural numbers). Usually we re only working with the natural numbers, but there s nothing preventing starting at a negative index.",
        "video_name": "5jwXThH6fg4",
        "timestamps": [
            100
        ],
        "3min_transcript": "What I want to do in this video is introduce you to the idea of Sigma notation, which will be used extensively through your mathematical career. So let's just say you wanted to find a sum of some terms, and these terms have a pattern. So let's say you want to find the sum of the first 10 numbers. So you could say 1 plus 2 plus 3 plus, and you go all the way to plus 9 plus 10. And I clearly could have even written this whole thing out, but you can imagine it becomes a lot harder if you wanted to find the sum of the first 100 numbers. So that would be 1 plus 2 plus 3 plus, and you would go all the way to 99 plus 100. So mathematicians said, well, let's find some notation, instead of having to do this dot dot dot thing-- which you will see sometimes done-- so that we can more cleanly express these types of sums. And that's where Sigma notation comes from. So this sum up here, right over here, this first one, it could be represented as Sigma. And what you do is you define an index. And you could start your index at some value. So let's say your index starts at 1. I'll just use i for index. So let's say that i starts at 1, and I'm going to go to 10. So i starts at 1, and it goes to 10. And I'm going to sum up the i's. So how does this translate into this right over here? Well, what you do is you start wherever the index is. If the index is at 1, set i equal to 1. Write the 1 down, and then you increment the index. And so i will then be equal to 2. i is 2. Put the 2 down. And you're summing each of these terms as you go. And you go all the way until i is equal to 10. So given what I just told you, I encourage you to pause this video and write the Sigma notation for this sum right over here. well, this would be the sum. The first term, well, it might be easy to just say we'll start at i equals 1 again. But now we're not going to stop until i equals 100, and we're going to sum up all of the i's. Let's do another example. Let's imagine the sum from i equals 0 to 50 of-- I don't know, let me say-- pi i squared. What would this sum look like? And once again, I encourage you to pause the video and write it out, expand out this sum. Well, let's just go step by step. When i equals 0, this will be pi times 0 squared. And that's clearly 0, but I'll write it out. pi times 0 squared."
    },
    {
        "Q": "I don't get the part on 4:34.\n\nOk if you are taking the -sqrt of (x-1)^2 doesn't that equal -(x-1), thus the final value equal (-x+1).\n\nFor example the negative sqrt of 9:\n\n=-sqrt(9)\n=-(3)\n=-3\n\nSomeone tell me if I'm wrong cause I'm confused on that part on how even with negative square root he still gets (x-1).",
        "A": "So I read the answer above and yh... I still don t exactly get it. Does it have to do with the fact that x<= 1 ensuring a negative number so when you get a -sqrt of (x-1) you keep it like that so the answer doesn t go positive or is that completely unrelated.",
        "video_name": "Bq9cq9FZuNM",
        "timestamps": [
            274
        ],
        "3min_transcript": "sides of this equation. And our goal is to get back to negative 3. If we take the positive square root. If we just take the principle root of both sides of that, we would get 3 is equal to 3. But that's not our goal. We want to get back to negative 3. So we want to take the negative square root of our square. So because this expression is negative and we want to get back to this expression, we want to get back to this x minus 1, we need to take the negative square root of both sides. You can always -- every perfect square has a positive or a negative root. The principle root is a positive root. But here we want to take the negative root because this expression right here is going to be negative. And that's what we want to solve for. So let's take the negative root of both sides. So you get the negative square root of y plus 2 is equal to -- and I'll just write this extra step here, just so you Is equal to the negative square root, the negative square root of x minus 1 squared. For y is greater than or equal to negative 2. And x is less than or equal to 1. That's why the whole reason we're going to take the negative square there. And then this expression right here -- so let me just write the left again. Negative square root of y plus 2 is equal to the negative square root of x minus 1 squared is just going to be x minus 1. It's just going to be x minus one. x minus 1 squared is some positive quantity. The negative root is the negative number that you have to square to get it. To get x minus 1 squared. So that just becomes x minus 1. Hopefully that doesn't confuse you too much. We just want to get rid of this squared sign. We want to make sure we get the negative version. We don't want the positive version, which would have been 1 minus x. Don't want to confuse you. So here, we now just have to solve for x. And let me write the 4. y is greater than or equal to negative 2. 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"
    },
    {
        "Q": "At 4:57 how do you know when to take the negative square root or positive one? I didn't get how in the video. Also if you could tell me, is this level of math algebra 1 or 2? They should do a better job of subdividing the algebra category between the 1 and 2.",
        "A": "Whether you take the positive or negative root depends on which side of the graph is present. That s why Sal kept the for y >= -2 and x <= 1 pulled along to every step, to remember which side to use.",
        "video_name": "Bq9cq9FZuNM",
        "timestamps": [
            297
        ],
        "3min_transcript": "sides of this equation. And our goal is to get back to negative 3. If we take the positive square root. If we just take the principle root of both sides of that, we would get 3 is equal to 3. But that's not our goal. We want to get back to negative 3. So we want to take the negative square root of our square. So because this expression is negative and we want to get back to this expression, we want to get back to this x minus 1, we need to take the negative square root of both sides. You can always -- every perfect square has a positive or a negative root. The principle root is a positive root. But here we want to take the negative root because this expression right here is going to be negative. And that's what we want to solve for. So let's take the negative root of both sides. So you get the negative square root of y plus 2 is equal to -- and I'll just write this extra step here, just so you Is equal to the negative square root, the negative square root of x minus 1 squared. For y is greater than or equal to negative 2. And x is less than or equal to 1. That's why the whole reason we're going to take the negative square there. And then this expression right here -- so let me just write the left again. Negative square root of y plus 2 is equal to the negative square root of x minus 1 squared is just going to be x minus 1. It's just going to be x minus one. x minus 1 squared is some positive quantity. The negative root is the negative number that you have to square to get it. To get x minus 1 squared. So that just becomes x minus 1. Hopefully that doesn't confuse you too much. We just want to get rid of this squared sign. We want to make sure we get the negative version. We don't want the positive version, which would have been 1 minus x. Don't want to confuse you. So here, we now just have to solve for x. And let me write the 4. y is greater than or equal to negative 2. 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"
    },
    {
        "Q": "I'm confused. Khan says there is only one x term at 1:55. Can someone help me with this?",
        "A": "Look at the second line. There is a 3x, a -3x, and there is a x by itself near the middle. Since the -3x cancels out the 3x, the x by itself is left. hope this helps :)",
        "video_name": "DMyhUb1pZT0",
        "timestamps": [
            115
        ],
        "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": "I understood everything until 1:24. Why does he insert a \"1\" in front of the parenthesis? I mean, where does that come from? He then multiplies \"-1 * -4\", to get a value of 4.... I'm confused why is the 3x ignored? Thanks for any clarification you can provide.",
        "A": "He inserted a 1 to show that the ... - ( ... is also equal to ... -1 ( ... The 3x was not ignored because it was already distributed with the - (or -1), making -3x, as shown in 1:20.",
        "video_name": "DMyhUb1pZT0",
        "timestamps": [
            84
        ],
        "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": "At 5:30 you gave the formula of how to find the area of a quadrilateral but it was quite confusing. Could you please explain it again?",
        "A": "Pretty much, with the edges, you form right triangles. Then, you find their area and then add that to the area of the regular quadrilateral. :0)",
        "video_name": "gkifo46--JA",
        "timestamps": [
            330
        ],
        "3min_transcript": "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": "At 5:07 Sal says S parameter describes the angle between radius and x-z plane. Has he mistaken it for x-y plane, considering the way he uses the parameter afterwards?",
        "A": "Yes, he meant to say x-y plane. A little box pops up in the lower right corner of the screen with the correction.",
        "video_name": "owKAHXf1y1A",
        "timestamps": [
            307
        ],
        "3min_transcript": "of the torus, or from the z-axis it's a distance of b. It's always going to be a distance of b. It's always, if you imagine the top of the doughnut, let me draw the top of the doughnut. If you're looking down on a doughnut, let me draw a doughnut right here, if you're looking down on a doughnut, it just looks something like that. The z-axis is just going to be popping straight out. The x-axis would come down like this, and then the y-axis would go to the right, like that. So you can imagine, I'm just flying above this. I'm sitting on the z-axis looking down at the doughnut. It will look just like this. And if you imagine the cross section, this circle right here, the top part of the circle if you're looking down, would look just like that. And this distance b is a distance from the z-axis So this distance, let me draw it in the same color, from the center to the center of these circles, that is going to be b. It's just going to keep going to the center of the circles, b. That's going to be b, that's going to be b. That's going to be b. From the center of our torus to the center of our circle that defines the torus, it's a distance of b. So this distance right here, that distance right there is b. And from b, we can imagine we have a radius. A radius of length a. So these circles have radius of length a. So this distance right here is a, this distance right here is a, this distance right there is a, that distance right there is a. If I were look at these circles, these circles have radius a. And what we're going to do is have two parameters. One is the angle that this radius makes with the x-z Let me do that in the same color. You can imagine the x-axis coming out here. So this is the x-z plane. So one parameter is going to be the angle between our radius and the x-z plane. We're going to call that angle, or that parameter, we're going to call that s. And so as s goes between 0 and 2 pi, as s goes between 0 in 2 pi, when the 0 is just going to be at this point right here, and then as it goes to 2 pi, you're going to trace out a circle that looks just like that. Now, we only have one parameter. What we want to do is then spin this circle around. What I just drew is that circle right there. What we want to do is spin the entire circle around. So let's define another parameter. We'll call this one t, and I'll take the top view again."
    },
    {
        "Q": "at 7:36 how do you get your limits of integration? I understand that you did by inspection, but how would you do it by making the two equations equal to each other ?",
        "A": "In that example, Sal is integrating with respect to y. To solve for zero and one analytically, we could make the equations equivalent to each other and solve. Here s the example: sqrt(x) = x^2 x = x^4 This last equation is true only when x = 0 or x = 1.",
        "video_name": "WAPZihVUmzE",
        "timestamps": [
            456
        ],
        "3min_transcript": "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. So we're going to sum up all of the rings over the interval. And when you take the integral sign, it's a sum where you're taking the limit as you have an infinite number of rings that become infinitesimally small in height or depth, depending on how you view it. And what's our interval? So we've looked at this multiple times. These two graphs-- you could do it by inspection. You could try to solve it in some way, but it's pretty obvious that they intersect at-- remember we care about our y interval. They intersect at y is equal to 0 and y is equal to 1. And there you have it. We've set up our integral for the volume of this shape right over here. I'll leave you there, and in this next video, we will just evaluate this integral."
    },
    {
        "Q": "At 1:46. I do not understand the purpose of the 1/7 * 7, nor where they even came from, nor what allows him to write them there to begin with.",
        "A": "To use u-substitution you have to identify a function u such that a constant multiple of the derivative of u appears in the integrand. In this case the function u is 7x+9. The derivative of 7x+9 is 7. Therefore a constant multiple of 7 must appear in the integrand. Because it does not, Sal multiples the expression by 1 in the form of (1/7 * 7 = 1). In this way he makes sure 7 appears in the integrand therefore facilitating u substitution.",
        "video_name": "oqCfqIcbE10",
        "timestamps": [
            106
        ],
        "3min_transcript": "Let's take the indefinite integral of the square root of 7x plus 9 dx. So my first question to you is, is this going to be a good case for u-substitution? Well, when you look here, maybe the natural thing to set to be equal to u is 7x plus 9. But do I see its derivative anywhere over here? If we set u to be equal to 7x plus 9, what is the derivative of u with respect to x going to be? Derivative of u with respect to x is just going to be equal to 7. Derivative of 7x is 7. Derivative of 9 is 0. So do we see a 7 lying around anywhere over here? Well, we don't. But what could we do in order to have a 7 lying around, but not change the value of the integral? Well, the neat thing-- and we've seen this multiple times-- is when you're evaluating integrals, scalars can go in and outside of the integral very easily. some scalar a times f of x dx, this is the same thing as a times the integral of f of x dx. The integral of the scalar times a function is equal to the scalar times the integral of the functions. So let me put this aside right over here. So with that in mind, can we multiply and divide by something that will have a 7 showing up? Well, we can multiply and divide by 7. So imagine doing this. Let's rewrite our original integral. So let me draw a little arrow here just to go around that aside. We could rewrite our original integral as being 9 to the integral of times 1/7 times 7 times the square root of 7x plus 9 dx. And if we want to, we could take the 1/7 outside We don't have to, but we can rewrite times the square root of 7x plus 9 dx. So now if we set u equal to 7x plus 9, do we have its derivative laying around? Well, sure. The 7 is right over here. We know that du-- if we want to write it in differential form-- du is equal to 7 times dx. So du is equal to 7 times dx. That part right over there is equal to du. And if we want to care about u, well, that's just going to be the 7x plus 9. That is are u. So let's rewrite this indefinite integral in terms of u. It's going to be equal to 1/7 times the integral of-- and I'll just take the 7 and put it in the back. So we could just write the square root of u du, 7 times dx is du."
    },
    {
        "Q": "Are du and dx able to just be treated like normal variables?\n@5:30 he treats the dx as another variable, to be multiplied with the 7",
        "A": "Yes, when using Leibnitz notation, you can think of them as quantifiable numbers. What they really mean is this - an infinitely small change in the given variable. For example, du is an infinitely small change in u. dx is an infinitely small change in x. Then, if we think of it this way, du/dx as giving the slope makes sense! The change in u over the change in x gives slope! Long story short, you can treat them as actual numbers - infinitely small changes in a variable.",
        "video_name": "oqCfqIcbE10",
        "timestamps": [
            330
        ],
        "3min_transcript": "It makes it a little bit easier for us to kind of do the reverse power rule here. So we can rewrite this as equal to 1/7 times the integral of u to the 1/2 power du. And let me just make it clear. This u I could have written in white if I want it the same color. And this du is the same du right over here. So what is the antiderivative of u to the 1/2 power? Well, we increment u's power by 1. So this is going to be equal to-- let me not forget this 1/7 out front. So it's going to be 1/7 times-- if we increment the power here, it's going to be u to the 3/2, 1/2 plus 1 is 1 and 1/2 or 3/2. So it's going to be u to the 3/2. times the reciprocal of 3/2, which is 2/3. And I encourage you to verify the derivative of 2/3 u to the 3/2 is indeed u to the 1/2. And so we have that. And since we're multiplying 1/7 times this entire indefinite integral, we could also throw in a plus c right over here. There might have been a constant. And if we want, we can distribute the 1/7. So it would get 1/7 times 2/3 is 2/21 u to the 3/2. And 1/7 times some constant, well, that's just going to be some constant. And so I could write a constant like that. I could call that c1 and then I could call this c2, but it's really just some arbitrary constant. Oh, actually, no we aren't done. We still just have our entire thing in terms of u. So now let's unsubstitute it. So this is going to be equal to 2/21 times u to the 3/2. u is equal to 7x plus 9. Let me put a new color here just to ease the monotony. So it's going to be 2/21 times 7x plus 9 to the 3/2 power plus c. And we are done. We were able to take a kind of hairy looking integral and realize that even though it wasn't completely obvious at first, that u-substitution is applicable."
    },
    {
        "Q": "You took the average change from 2000-2003 and then 2003-2004 and averaged it out, but why couldn't you just use 2002-2004? 2:27",
        "A": "It s mathematically equivalent (and easier) to take the average for 2002 to 2004 (which would be (S(2004) - S(2002))/2) as you suggest. However, the problem specifies that we are to solve by finding the slope of the two secant lines and averaging those slopes, so that is the approach used in the video.",
        "video_name": "fI6w2kL295Y",
        "timestamps": [
            147
        ],
        "3min_transcript": "The table shows the number of stores of a popular US coffee chain from 2000 to 2006. The number of stores recorded is the number at the start of each year on January 1. So in 2000, there was 1,996 stores, in 2005, 6,177, so on and so forth. Determine a reasonable approximation for the instantaneous rate of change, in coffee stores per year at the beginning of 2003-- so we care about 2003-- by taking the average of two nearby secant slopes. So let's visualize this. So this right over here, I've plotted all of the points. Now let me make sure that the axes are clear. This horizontal axis, this is my t-axis that tells us the year. And then the vertical axis is the number of stores. And we could even say that it is a function of time. So you see in the year 2000, there was 1,996 stores-- 2003, 4,272. 2003, 4,272 stores. Now if you could imagine that they're constantly adding stores, you could even imagine minute by minute they're adding stores. So this is just sampling the number of stores they had on January 1. But if you were to really plot it as a more continuous function, it might look something like this. I'll do my best to approximate it. It might be more of some type of curve that looks something like this. And once again, I'm just approximating what it might actually look like. So when they're saying the instantaneous rate of change in coffee stores per year, so this is the change, the instantaneous rate of change of stores per time. They're really saying, we need to approximate the slope of the tangent line in 2003, when time is 2003. So the tangent line might look something like that. is tangent. Now, they say approximate. We don't have the information to figure it out exactly. But we have some data around it, and we can figure out the slopes of the secant lines between this point and those points. And then we can take the average of the slopes of the secant lines to approximate the slope of this tangent line. So for example, we could find the slope of this secant line right over here as we go from 2002 to 2003. And then we can find the slope of this secant line as we go from 2003 to 2004. And if we average those, that should be a pretty good approximation for the instantaneous rate of change in 2003. So let's do that. So the slope of this pink secant line, as we go from 2002 to 2003, that's going to be the number of stores in 2003 minus the number of stores in 2002. So that's the change in our number of stores over the change in years, or the change in time."
    },
    {
        "Q": "\"it is a function of time\" can someone rephrase that please, in a very simple way @1:00",
        "A": "A function of time meaning that as time passes the number of things change. In this example as the year goes by the number of stores built is increased which can be represented by a function with respect to time. A function being something along the lines of y = x + 1.",
        "video_name": "fI6w2kL295Y",
        "timestamps": [
            60
        ],
        "3min_transcript": "The table shows the number of stores of a popular US coffee chain from 2000 to 2006. The number of stores recorded is the number at the start of each year on January 1. So in 2000, there was 1,996 stores, in 2005, 6,177, so on and so forth. Determine a reasonable approximation for the instantaneous rate of change, in coffee stores per year at the beginning of 2003-- so we care about 2003-- by taking the average of two nearby secant slopes. So let's visualize this. So this right over here, I've plotted all of the points. Now let me make sure that the axes are clear. This horizontal axis, this is my t-axis that tells us the year. And then the vertical axis is the number of stores. And we could even say that it is a function of time. So you see in the year 2000, there was 1,996 stores-- 2003, 4,272. 2003, 4,272 stores. Now if you could imagine that they're constantly adding stores, you could even imagine minute by minute they're adding stores. So this is just sampling the number of stores they had on January 1. But if you were to really plot it as a more continuous function, it might look something like this. I'll do my best to approximate it. It might be more of some type of curve that looks something like this. And once again, I'm just approximating what it might actually look like. So when they're saying the instantaneous rate of change in coffee stores per year, so this is the change, the instantaneous rate of change of stores per time. They're really saying, we need to approximate the slope of the tangent line in 2003, when time is 2003. So the tangent line might look something like that. is tangent. Now, they say approximate. We don't have the information to figure it out exactly. But we have some data around it, and we can figure out the slopes of the secant lines between this point and those points. And then we can take the average of the slopes of the secant lines to approximate the slope of this tangent line. So for example, we could find the slope of this secant line right over here as we go from 2002 to 2003. And then we can find the slope of this secant line as we go from 2003 to 2004. And if we average those, that should be a pretty good approximation for the instantaneous rate of change in 2003. So let's do that. So the slope of this pink secant line, as we go from 2002 to 2003, that's going to be the number of stores in 2003 minus the number of stores in 2002. So that's the change in our number of stores over the change in years, or the change in time."
    },
    {
        "Q": "At 0:46 Sal Khan said 30.42 but meant 30.24",
        "A": "This is a known error in the video. There is a box that pops up and tells you Sal made an error and provides the correct info.",
        "video_name": "Hrjr5f5pZ84",
        "timestamps": [
            46
        ],
        "3min_transcript": "Let's see if we can divide 30.24 divided by 0.42. And try pausing the video and solving it on your own before I work through it. So there is a couple of ways you can think about it. We could just write it as 30.24 divided by 0.42. But what do you do now? Well the important realization is, is when you're doing a division problem like this, you will get the same answer as long as you multiply or divide both numbers by the same thing. And to understand that, rewrite this division as 30.42 over 0.42. We could write it really as a fraction. And we know that when we have a fraction like this we're not changing the value of the fraction if we multiple the numerator and the denominator by the same quantity. to make it a whole number? Well we can multiply it by 10 and then another 10. So we can multiply it by a 100. If we multiply the denominator by 100 in order to not change the value of this, we also need to multiply the numerator by 100. We are essentially multiplying by 100 over 100, which is just 1. So we're not changing the value of this fraction. Or, you could view this, this division problem. So this is going to be 30.42 times 100. Move the decimal two places to the right, gets you 3,042. The decimal is now there if you care about it. And, 0.42 times 100. Once again move the decimal one, two places to the right, it is now 42. So this is going to be the exact same thing as 3,042 divided by 42. So once again we can move the decimal here, two to the right. then we can move this two to the right. Or we need to move this two to the right. And so this is where, now the decimal place is. You could view this as 3,024. Let me clear that 3024 divided by 42. Let me clear that. And we know how to tackle that already, but lets do it step by step. How many times does 42 go into 3? Well it does not go at all, so we can move on to 30. How many times does 42 go into 30? Well it does not go into 30 so we can move on to 302. How many times does 42 go into 302? And like always this is a bit of an art when your dividing by a two-digit or a multi-digit number, I should say. So lets think about it a little bit. So this is roughly 40. This is roughly 300. So how many times does 40 go into 300? Well how many times does 4 go into 30? Well, it looks like it's about seven times, so I'm going to try out a 7, see if it works out. 7 times 2 is 14."
    },
    {
        "Q": "At 1:10, Sal says that the perimeter is length+width+length+width, can't you do length*2 and width*2 ?",
        "A": "Yup... depends on weather you are faster at addition or multiplication:)",
        "video_name": "CDvPPsB3nEM",
        "timestamps": [
            70
        ],
        "3min_transcript": "The perimeter of a rectangular fence measures 0.72 kilometers. The length of the fenced area is 160 meters. What is its width? So we have a rectangular fence. Let me draw it. So that is a rectangle. You can imagine we're looking from above. This line is the top of the fence. And if you take its perimeter, the perimeter is the distance around the fence. If you take this distance plus this distance plus that distance plus that distance, it's going to be 0.72 kilometers. That's the total distance of all of the sides. Now, the length of the fenced area is 160 meters. Let's call this the length. So this distance right here is 160 meters. Since it's a rectangle, that distance over there is also going to be 160 meters, and we need to figure out its width. They want us to figure out the width. The width is this distance right over here, which is also Now, what is the perimeter of the fence? The perimeter is the sum of this length plus this length plus that length plus that length. So the perimeter-- let me write it in orange-- is going to be equal to w, the width, plus 160 meters, plus the width, plus 160 meters. And let's assume w is in meters. So we could add it all up. So if you were to add all of these up, you'd get a certain If you were to add all of these up, you have a w plus a w, so you'd be 2w, plus-- 160 plus 160 is 320-- so plus 320. We're assuming that w is in meters. Now, they also told us that the perimeter of the fence is 0.72 kilometers. So the perimeter-- let me do it in that same orange color. The perimeter is also equal to 0.72 kilometers, and we can abbreviate that with km. Now, if we wanted to solve an equation, if we wanted to set this thing equal to the perimeter they gave us, we have to make sure that the units are the same. Here it's in meters. Here it's in kilometers. So let's convert this 0.72 kilometers to meters. And the way to do that, we want the kilometers in the denominator so it cancels out with the kilometers, and you want meters in the numerator. Now, how many meters equal a kilometer? Well, it's 1,000 meters for every 1 kilometer."
    },
    {
        "Q": "At 8:24 Sal combines X(-b-2a) and X(-a-2b) and is left with only a single X coefficient, I was sure this would be 2X. Can someone please explain why it remains a single X coefficient? Thanks.",
        "A": "Because the both 2s in -b-2a and -a-2b would be simplified and that would make it a single X coefficient.",
        "video_name": "Vc09LURoMQ8",
        "timestamps": [
            504
        ],
        "3min_transcript": "as being equal to that, so let's do that. We get x plus b times c, over a plus 2b is equal to x minus a times c over negative b minus 2a. Well, both sides are divisible by c, we can assume that c is non-zero, so this triangle actually exists, it actually exists in two dimensions, so we can divide both sides by c and we get that. Let's see, now we can cross-multiply. We can multiply x plus b times this quantity right here, and that's going to be equal to a plus 2b times this quantity over here. We can just distribute it, so it's going to be x times all of this business, it's going to be x times negative b minus 2a, plus b times all of this, so I just multiply that times that, is going to equal to this times that, so it's going to be x times all of this business. X times a plus 2b, minus a times that, so distribute the a, minus a squared minus 2ab. Let's see if we can simplify this. We have a negative 2ab on both sides, so let's just subtract that out, so we cancelled things out, let's subtract this from both sides of the equation, so we'll have a minus x times a plus 2b, and I'll subtract that from here, but I'll write it a little different, it'll simplify things, so this will become x times, I'll distribute the negative inside the a plus 2b, so negative a minus 2b, and let's add a b squared to both sides, so plus b squared, plus b squared. and all of the constants on the other. This becomes on the left-hand side, the coefficient on x, I have negative b minus 2b is negative 3b negative 2a minus a is negative 3a, times x is equal to, these guys cancel out, and these guys cancel out, is equal to b squared minus a squared. Let's see if we can factor, this already looks a little suspicious in a good way, something that we should be able to solve. This can be factored. We can factor out a negative three. We can actually factor out a three. We'll get three times b minus a, actually, let's factor out a negative three, so negative three times b plus a times x is equal to, now this is the same thing as b plus a"
    },
    {
        "Q": "Why does Sal use (1 2, -1 0) matrix at 6:21 ?",
        "A": "He s just giving an example of a transformation. Since any matrix can be seen as a transformation, he just choose one at random.",
        "video_name": "MIAmN5kgp3k",
        "timestamps": [
            381
        ],
        "3min_transcript": "line segment. So that's our L0. It's just a set of vectors. Now we can do the same exercise if we wanted to find out the line, the equation of the line, that goes between x1 and x2. If we wanted to find the equation of that line, we could call this L1. And L1 would be equal to x1 plus t times x2 minus x1 for 0 is less than or equal to t is less than or equal to 1. That's L1. And then finally, if we want to make a triangle out of this, let's define this line right here. Let's define that as L2. L2 would be equal to the set of all of vectors where you start off at x2. Set of all of vectors that are x2 plus some scaled up sum of x0 minus x2 is this vector right here. So x0 minus x2 such that 0 is less than or equal to t is less than or equal to 1. And so if you take the combination, if you were to define kind of a super set-- I could have defined my shape as-- let's say it's the union of all of those guys. Well, let me just write it. L0, L1, and L2. Then you'd have a nice triangle here. If you take the union of all of these three sets, you get that nice triangle there. Now, what I want to do in this video, I think this is all a bit of review for you. But it's maybe a different way of looking at things than we've done in the past. Is I want to understand what happens to this set right here when I take a transformation, a linear transformation, of it? I'll make it a fairly straightforward transformation. Let me define my transformation of x, of any x, to be equal to the matrix 1 minus 1, 2, 0 times whatever vector x. So times x1, x2. And we know that any linear transformation can actually be written as a matrix and vice versa. So you might have said, hey, you know, you're giving an example with the matrix, what about all those other ways to write in your transformation? You can write all of those as a matrix. So let's translate-- let's try to figure out what this is going to look like. What our triangle is going to look like when we transform every point in it. Let me take the transformation first. The transformation of L0 is equal to the transformation of this thing. This is just one of the particular members."
    },
    {
        "Q": "Would it work if at 1:50, instead of factoring as he did, could we just FOIL? I would only think to do it that way if I knew beforehand that the result should look a certain way.",
        "A": "I don t think I understand your question too well... Sal didn t factor at that point in the problem, he just distributed the entire sin expression with the (5 - 3y )",
        "video_name": "-EG10aI0rt0",
        "timestamps": [
            110
        ],
        "3min_transcript": "Let's say we have the relationship y is equal to cosine of 5x minus 3y. And what I want to find is the rate at which y is changing with respect to x. And we'll assume that y is a function of x. So let's do what we've always been doing. Let's apply the derivative operator to both sides of this equation. On the left-hand side, right over here, we get dy/dx is equal to-- now here on the right hand side, we're going to apply the chain rule. The derivative of the cosine of something, with respect to that something, is going to be equal to negative sine of that something. So negative sine of 5x minus 3y. And then we have to multiply that by the derivative of that something with respect to x. So what's the derivative of the something with respect to x? Well the derivative of 5x with respect to x is just equal to 5. And the derivative of negative 3y with respect to x 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--"
    },
    {
        "Q": "At 2:35, when Sal subtracts the 3Sin(5x-3y), what happened to the dy/dx that was being multiplied to it? Could the 3Sin(5x-3y) even be subtracted without the dy/dx?",
        "A": "Yes, you are correct. We are subtracting 3sin(5x-3y)dy/dx. This is exactly what Sal did. He just didn t mention dy/dx out loud.",
        "video_name": "-EG10aI0rt0",
        "timestamps": [
            155
        ],
        "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": "at 1:24, how can there be 4 possibilities, one person only have 3 options and not his own self. right?",
        "A": "The 4 means that the first person involved in shaking hands can be any of the 4 people. The 3 means that the second person involved in shaking hands can be any of the remaining 3 people not counting the person identified as the first person.",
        "video_name": "boH4l1SgJbM",
        "timestamps": [
            84
        ],
        "3min_transcript": "- Let's say that there are four people in a room. And you're probably tired of me naming the people with letters, but I'm going to continue doing that. So the four people in the room are people A, B, C, and D. And they are all told, \"You don't know each other. \"So I want you to all meet each other. \"You need to shake the hand, exactly once, \"of every other person in the room so that you all meet.\" So my question to you is, if each of these people need to shake the hand of every other person exactly once, how many handshakes are going to occur? The number of handshakes that are going to occur. So, like always, pause the video and see if you can make sense of this. Alright, I'm assuming you've had a go at it. So one way to think about it is, if you say there's a handshake, two people are party to a handshake. We're not talking about some new three-person handshake or four-person handshake, we're just talking about the traditional, two people shake their right hands. another person in this party. There's four possibilities of one party. And if we assume people aren't shaking their own hands, which we are assuming, they're always going to shake someone else's hand. For each of these four possibilities who's this party, there's three possibilities who's the other party. And so you might say that there's four times three handshakes. Since there's four times three possible handshakes. And what I'd like you to do is think a little bit about whether this is right, whether there would actually be 12 handshakes. You might have thought about it, and you might say, there's four times three, this is actually counting the permutations. This is counting how many ways can you permute four people into two buckets, the two buckets of handshakers, where you care about which bucket they are in. Whether they're handshaker number one, or handshaker number two. and B being the number two handshaker as being different than B being the number one handshaker and A being the number two handshaker. But we don't want both of these things to occur. We don't want A to shake B's hand, where A is facing north and B is facing south. And then another time, they shake hands again where now B is facing north and A is facing south. We only have to do it once. These are actually the same thing, so there's no reason for both of these to occur. So we are going to be double counting. So what we really want to do is think about combinations. One way to think about it is, you have four people. In a world of four people or a pool of four people, how many ways can you choose two?"
    },
    {
        "Q": "1:43\n\nWe were taught about radial symmetry. Is rotational and radial symmetry the same?",
        "A": "rotational and radial symmetry is generally the same, but radial symmetry is usually used in biological contexts",
        "video_name": "toKu2-qzJeM",
        "timestamps": [
            103
        ],
        "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": "1:23 how did he get 6 as his exponent ?",
        "A": "at 1:23 he was subtracting 6 from 2 because there was 2 exponents",
        "video_name": "AR1uqNbjM5s",
        "timestamps": [
            83
        ],
        "3min_transcript": "Let's do some exponent examples that involve division. Let's say I were to ask you what 5 to the sixth power divided by 5 to the second power is? Well, we can just go to the basic definition of what an exponent represents and say 5 to the sixth power, that's going to be 5 times 5 times 5 times 5 times 5-- one more 5-- times 5. 5 times itself six times. And 5 squared, that's just 5 times itself two times, so it's just going to be 5 times 5. Well, we know how to simplify a fraction or a rational expression like this. We can divide the numerator and the denominator by one 5, and then these will cancel out, and then we can do it by another 5, or this 5 and this 5 will cancel out. And what are we going to be left with? 5 times 5 times 5 times 5 over 1, or you could say that this is just 5 to the fourth power. Essentially we started with six in the numerator, six 5's multiplied by themselves in the numerator, and then we subtracted out. We were able to cancel out the 2 in the denominator. So this really was equal to 5 to the sixth power minus 2. So we were able to subtract the exponent in the denominator from the exponent in the numerator. Let's remember how this relates to multiplication. If I had 5 to the-- let me do this in a different color. 5 to the sixth times 5 to the second power, we saw in the last video that this is equal to 5 to the 6 plus-- I'm trying to make it color coded for you-- 6 plus 2 power. Now, we see a new property. And in the next video, we're going see that these aren't really different properties. They're really kind of same sides of the same coin when we learn about negative exponents. divided by 5 to the second power-- let me do it in a different color-- is going to be equal to 5 to the-- it's time consuming to make it color coded for you-- 6 minus 2 power or 5 to the fourth power. Here it's going to be 5 to the eighth. So when you multiply exponents with the same base, you add When you divide with the same base, you subtract the denominator exponent from the numerator exponent. Let's do a bunch more of these examples right here. What is 6 to the seventh power divided by 6 to the third power?"
    },
    {
        "Q": "AT 9:20 would 5/4 x^-1 y also be correct",
        "A": "It generally would be considered incomplete. Final answers should have positive exponents. And, we would multiply the items together to show them as one fraction. This is why Sal is showing the answer as 5y / (4x)",
        "video_name": "AR1uqNbjM5s",
        "timestamps": [
            560
        ],
        "3min_transcript": "And we would have simplified this about as far as you can go. Let's do one more of these. I think they're good practice and super-valuable experience later on. Let's say I have 25xy to the sixth over 20y to the fifth x squared. So once again, we can rearrange the numerators and the denominators. So this you could rewrite as 25 over 20 times x over x squared, right? We could have made this bottom 20x squared y to the fifth-- it doesn't matter the order we do it in-- times y to the sixth over y to the fifth. actually just simplify fractions. 25 over 20, if you divide them both by 5, this is equal to 5 over 4. x divided by x squared-- well, there's two ways you could think about it. That you could view as x to the negative 1. You have a first power here. 1 minus 2 is negative 1. So this right here is equal to x to the negative 1 power. Or it could also be equal to 1 over x. These are equivalent. So let's say that this is equal into 1 over x, just like that. And it would be. x over x times x. One of those sets of x's would cancel out and you're just left with 1 over x. And then finally, y to the sixth over y to the fifth, that's y to the 6 minus 5 power, which is just y to the first power, or just y, so times y. rational expression, you have 5 times 1 times y, which would be 5y, all of that over 4 times x, right? This is y over 1, so 4 times x times 1, all of that over 4x, and we have successfully simplified it."
    },
    {
        "Q": "if at 1:05 L = 1 and R = 0 then that is 01001100110 - a binary number, will someone find what it represents?",
        "A": "Bin 01001100110 = Dec 614 If you were to switch them, with L=0, R=1, Bin 10110011001 = Dec 1433",
        "video_name": "Gx5D09s5X6U",
        "timestamps": [
            65
        ],
        "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 2:06, could we take the y and pull it to the front similarly to how we pulled x to the front earlier in the video since it's an exponent?",
        "A": "The Y is not an exponent. We re not raising 5 to the Y power. Rather, it is the result of raising the 5 to some power, i.e., 5 to the what power equals y? So the answer is no.",
        "video_name": "RhzXX5PbsuQ",
        "timestamps": [
            126
        ],
        "3min_transcript": "We're asked to simplify log base 5 of 25 to the x power over y. So we can use some logarithm properties. And I do agree that this does require some simplification over here, that having this right over here inside of the logarithm is not a pleasant thing to look at. So the first thing that we realize-- and this is one of our logarithm properties-- is logarithm for a given base-- so let's say that the base is x-- of a/b, that is equal to log base x of a minus log base x of b. And here we have 25 to the x over y. So we can simplify. So let me write this down. I'll do this in blue. Log base 5 of 25 to the x over y using this property means that it's the same thing as log base 5 of 25 to the x power minus log base 5 of y. Now, this looks like we can do a little bit of simplifying. is if I have log base x of a to the b power, that's the same thing as b times log base x of a, that this exponent over here can be moved out front, which is what we did it right over there. So this part right over here can be rewritten as x times the logarithm base 5 of 25. And then, of course, we have minus log base 5 of y. And this is useful because log base 5 of 25 is actually fairly easy to think about. This part right here is asking us, what power do I have to raise 5 to to get to 25? So we have to raise 5 to the second power to get to 25. So this simplifies to 2. So then we are left with, this is equal to-- and I'll write it in front of the x now-- 2 times x minus log base 5 of y."
    },
    {
        "Q": "At 0:20 can we move the x out in front first, instead of applying the quotient log property first? And then apply the quotient log property?\n\nIf not...is there a certain order in which log properties should be applied (like BEDMAS)??? Thanks in advance :)",
        "A": "You can t move the x in front first, because not the whole factor is to the xth power. if we had log(25/y)^x we could. otherwise the whole term, so both log25 and log5 would be times 5, which is wrong. I hope this helps you, English is not my first language, so I don t know if I explained it well enough!",
        "video_name": "RhzXX5PbsuQ",
        "timestamps": [
            20
        ],
        "3min_transcript": "We're asked to simplify log base 5 of 25 to the x power over y. So we can use some logarithm properties. And I do agree that this does require some simplification over here, that having this right over here inside of the logarithm is not a pleasant thing to look at. So the first thing that we realize-- and this is one of our logarithm properties-- is logarithm for a given base-- so let's say that the base is x-- of a/b, that is equal to log base x of a minus log base x of b. And here we have 25 to the x over y. So we can simplify. So let me write this down. I'll do this in blue. Log base 5 of 25 to the x over y using this property means that it's the same thing as log base 5 of 25 to the x power minus log base 5 of y. Now, this looks like we can do a little bit of simplifying. is if I have log base x of a to the b power, that's the same thing as b times log base x of a, that this exponent over here can be moved out front, which is what we did it right over there. So this part right over here can be rewritten as x times the logarithm base 5 of 25. And then, of course, we have minus log base 5 of y. And this is useful because log base 5 of 25 is actually fairly easy to think about. This part right here is asking us, what power do I have to raise 5 to to get to 25? So we have to raise 5 to the second power to get to 25. So this simplifies to 2. So then we are left with, this is equal to-- and I'll write it in front of the x now-- 2 times x minus log base 5 of y."
    },
    {
        "Q": "at 2:00, why x(2 - log_5(y)) = 2x - log_5(y) and not 2x - xlog_5(y)?",
        "A": "The 2 resulted from the expression log_5(25) , which was being multiplied by x. There were no parentheses to distribute the x to the second term ( -log_5(y) ), so it was unaffected by it.",
        "video_name": "RhzXX5PbsuQ",
        "timestamps": [
            120
        ],
        "3min_transcript": "We're asked to simplify log base 5 of 25 to the x power over y. So we can use some logarithm properties. And I do agree that this does require some simplification over here, that having this right over here inside of the logarithm is not a pleasant thing to look at. So the first thing that we realize-- and this is one of our logarithm properties-- is logarithm for a given base-- so let's say that the base is x-- of a/b, that is equal to log base x of a minus log base x of b. And here we have 25 to the x over y. So we can simplify. So let me write this down. I'll do this in blue. Log base 5 of 25 to the x over y using this property means that it's the same thing as log base 5 of 25 to the x power minus log base 5 of y. Now, this looks like we can do a little bit of simplifying. is if I have log base x of a to the b power, that's the same thing as b times log base x of a, that this exponent over here can be moved out front, which is what we did it right over there. So this part right over here can be rewritten as x times the logarithm base 5 of 25. And then, of course, we have minus log base 5 of y. And this is useful because log base 5 of 25 is actually fairly easy to think about. This part right here is asking us, what power do I have to raise 5 to to get to 25? So we have to raise 5 to the second power to get to 25. So this simplifies to 2. So then we are left with, this is equal to-- and I'll write it in front of the x now-- 2 times x minus log base 5 of y."
    },
    {
        "Q": "At 1:46, I got confused. If something is \"over\" something else, does that mean divide that number by the bottom number?",
        "A": "Yes but do you see that x was being divided by 4 then multiplied by 4 so there is no reason to go through all that math when there opposites it would be like adding 4 to five and getting nine then subtracting $ and getting five again",
        "video_name": "p5e5mf_G3FI",
        "timestamps": [
            106
        ],
        "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 2:05 it says add one shouldn't it say add 3?????",
        "A": "Hey, you re right! That s funny.... well, he got the answer right anyhow.",
        "video_name": "p5e5mf_G3FI",
        "timestamps": [
            125
        ],
        "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 3:30, how does he know that the question is linear?",
        "A": "He knows at 3:30 because the equation can be plotted as a line. This means it will be straight no matter how far you extend it, and will not curve or bend anywhere. It is constant and will always remain so. I hope that helps :)",
        "video_name": "OWPVZoxNe-U",
        "timestamps": [
            210
        ],
        "3min_transcript": "So if we have the minutes they play time playing Bologna and the time playing Cut Your Wire. If I were to add those two together, so if I say x plus y. I'll write that plus in a neutral color, x plus y. What does this need to be equal to? The time I play Bologna Man plus the time I play Have to Cut the Wire. Well if I add them together, they want to spend exactly 45 minutes playing both games. So this is going to be equal to 45 minutes. Now we have set up an equation that relates the time playing Bologna Man and the time playing You Have to Cut the Wire. But now we have to think about is this a linear relationship? for a linear relationship is if you can write it in the traditional form of a line. So if you can write it in the y is equal to mx plus b form, where m is the slope of the line and b is the y-intercept. So let's see if we can do that. Well if we want to do that here, we could just subtract x from both sides. You subtract x from both sides, you get-- so let's subtract an x over here, let's subtract an x over here. So negative x plus this and then subtract an x there. Well, that's going to cancel, and you're going to be left with-- and I'm going try to write it in this form right over here-- y is equal to 45. Let me do that in the same color, just to make it not be confusing. y is equal to, and I'll write the negative x first because we have the x term right over here first. So y is equal to negative x plus 45. But you see here, it has that form. And you might say wait what is m here? I see that b is 45. Well if I write negative x, that's the same thing as writing negative 1x. So this is definitely a line. I was able to write it in this form right over here. So can this relationship be represented using a linear equation? Absolutely, absolutely yes."
    },
    {
        "Q": "At 4:46, Sal says that there is only one way to get to the cube below the starting point. Isn't that logic flawed because it is a three dimensional object? Couldn't he go to the right an infinite number of times and then go down? Or go to the right one, go down one, and wrap around the cube?",
        "A": "You are not travelling along the surface of the cube, you are traveling through the cube itself. Therefore, there is no wrapping around the cube. There is no going right an infinite number of times, because you can only go right twice. You are able to move right twice, down twice, and forward twice. All that matters is the order in which you do those moves.",
        "video_name": "wRxzDOloS3o",
        "timestamps": [
            286
        ],
        "3min_transcript": "Let me draw some squares in here. It's like that, and like that, and like that, and like that. And then, the middle one was the mauve layer. We'll draw that. The mauve layer looks something like that. And you can imagine I'm slicing it, and just looking at it from above. That's the idea here. And it's going to help us visualize this problem. So the mauve layer looks something like that. And then finally the orange layer. Looks like this. And we're almost ready to actually start doing the problem. Good enough. So just to make sure we understand our visualization, this layer up here-- we call that layer one. We could but this as box one. So I'll put a little two here. And I don't want to get these confused with the paths and all that, so I'm writing it really small. And this is layer three, or level three. And that's right there. And just to make sure you understand, this corner right there, this is our start point. And that's right there. Because this is a whole top. So this is the back left of the top. And are finish point, the bottom right, is right here. So, essentially, our problem goes from, how many ways to get from there to there, to how many ways to get from there to there? So let's just stay within a layer. So how many ways can I get to this point right here? Well I can only go from this point, and go straight in the layer like that. So there's only one way to get there. That movement is the exact same movement as this right here. Going from this box to this box. So there's one way to get there. That's the same thing as there. And similarly, I could go there. And I can just go one more step. So there's only one way to get there. And that's like going there and then there. 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."
    },
    {
        "Q": "At 8:43, how does Sal automatically determine that x is 8 or 2. In the equation above, shouldn't x be -8 or -2 since x is getting subtracted by those two numbers?",
        "A": "No. Since it is -5, he would have to add 5 to both sides to isolate x. Therefore, it would be 5 plus or minus 3, giving him 8 and 2. If he subtracted 5, the left side would have -10 + x and the right side would have -5 plus or minus 3. This would not be possible. He would have to add 10 to both sides to isolate x, giving him x=5 plus or minus 3.",
        "video_name": "55G8037gsKY",
        "timestamps": [
            523
        ],
        "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": "why do you find the square root of (4x+1 )squared and 8 around 1:30?",
        "A": "Since we are trying to solve for x, we want to isolate x on one side of the equation. in order to do that, we have to cancel out as much stuff on the left hand side of the equation as possible. In this example, if we take (4x+1)^2 = 8 and we take the square root of both sides, we are left with 4x+1=sqrt(8), and we are one step closer to isolating x.",
        "video_name": "55G8037gsKY",
        "timestamps": [
            90
        ],
        "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 1:40, what does Sal mean by positive or negative square root. If the square root on one side is positive, shouldn't the square root on the other side be positive, too?",
        "A": "Technically, we would get two equations: 4\u00f0\u009d\u0091\u00a5 + 1 = \u00c2\u00b1\u00e2\u0088\u009a8 -(4\u00f0\u009d\u0091\u00a5 + 1) = \u00c2\u00b1\u00e2\u0088\u009a8 Since the two equations are equivalent, we can discard one of them, leaving us with: 4x + 1 = \u00c2\u00b1\u00e2\u0088\u009a8",
        "video_name": "55G8037gsKY",
        "timestamps": [
            100
        ],
        "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 10:09 why is it plus or minus 1, when we square?",
        "A": "When you square, you multiply a number by itself. And -3*-3 = 9 as well as 3*3 = 9. So there can be two solutions, when you take the square root of something (9 in my example). It s different when you cube a number. -3*-3*-3 = -27 which is not the same as 3*3*3 = 27.",
        "video_name": "55G8037gsKY",
        "timestamps": [
            609
        ],
        "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": "At 1:50, why does Sal write \u00e2\u0088\u009a8 as \u00e2\u0088\u009a4*2, and how would I know when I should?",
        "A": "You can simplify any square roots if they are the product of a perfect square and another number. For example, the number 24. The \u00e2\u0088\u009a24 is the same thing as the (\u00e2\u0088\u009a4*6). We know that \u00e2\u0088\u009a4=2 so, we bring it outside the \u00e2\u0088\u009a sign. We result with 2\u00e2\u0088\u009a6. Same thing Sal did, just with different numbers.",
        "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 5:30 can you simply the fraction even more than that? Can you separate the fraction into 2 parts?",
        "A": "It appears you are talking more @4:10 where he ends with the fraction (-1 +/- 2 \u00e2\u0088\u009a2)/4. You cannot simplify the fraction anymore than that, but you could separate it into parts, and you could separate it into two different solutions. Separating into parts to get -1/4 +/- \u00e2\u0088\u009a2/2 does not look as neat and compact as what he has, but it is equivalent.",
        "video_name": "55G8037gsKY",
        "timestamps": [
            330
        ],
        "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": "What does Sal mean when he says, \"vanilla like\"? I am guessing he means not-so-complex-looking, like at 4:48.",
        "A": "In the U.S, an object can be described to be as plain as vanilla , which means that something is plain and simple like vanilla ice cream. You are correct in that he means not-so-complex-looking.",
        "video_name": "55G8037gsKY",
        "timestamps": [
            288
        ],
        "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 5:29, how did he get 4 times 2?",
        "A": "It s because 2 and sqrt(2) were squared separately, therefore you have 4 times 2. You can also think of 2sqrt(2) as sqrt(8) and when you square it, you ll get 8 which is 4 times 2.",
        "video_name": "55G8037gsKY",
        "timestamps": [
            329
        ],
        "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 2:35, how does drawing a perpendicular line to the base prove that it is an equilateral triangle?",
        "A": "I think he started the video already saying that the triangle was equilateral. He is not trying to prove that part.",
        "video_name": "SFL4stapeUs",
        "timestamps": [
            155
        ],
        "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 0:58 how did he get 90, and if its 90/100 why couldn't the other fraction be 80/100?",
        "A": "1st question: he got 90/100 because that = 9/10. he has to + them and you have to have a common denominator. 2nd question: because 8 is in the 100ths place. so you could have it 80/1000. so 9/10 + 8/100 = 90/100 + 8/100 = 98/100 sal makes it easy to understand.",
        "video_name": "qbMe4f2yvKs",
        "timestamps": [
            58
        ],
        "3min_transcript": "Let's see if we can write 12.98 as a mixed number. So the first thing you might want to realize here is that this is the exact same thing as 12 plus 0.98. And this simplifies it, because then we just have to write it as 12, and to some fraction, that's the same thing as 0.98. So if we could write 0.98 as a fraction, then we're almost done. So let's see if we can do that. So this 9 right over here is in the tenths place. I'll just write it like that. Tenths place. And this right over here is in the hundredths place. So you could view 0.98 two different ways. You could view it as 9/10. That's this part right over here. 9/10 plus 8/100. And if you want to find a common denominator that would be the same thing as 90 over 100 plus 8 over 100, And so 0.98 is 98/100. And another way you could have said that is look, the space right over here is in the hundredths place, and so this is 98/100 or 98/100. So you could have skipped this right over here. So if we just wanted to write it as a mixed number, we could just write it as 12 and, instead of 0.98, 12 and 98/100. Now we haven't reduced this to lowest terms yet. So let's see if we can simplify this. 98 is divisible by 2, and so is 100. So let's divide both of them by 2. They have that common factor. So we're going to divide both of them by 2. And so this is the same thing as 12, and 98 divided by 2 is 49. 100 divided by 2 is 50. And I think that's about as far as we can do. 49 factors, it's divisible by 7, but 50 isn't, so we've So 12.98 can be written as a mixed number, 12 and 49/50."
    },
    {
        "Q": "At 3:06, he switched but i didn't get why?",
        "A": "I believe Sal is trying to show you that you can multiply the 2 numbers ignoring the decimal points. Then, you can set the decimal point in your answer after completing the multiplication.",
        "video_name": "4IWfJ7-CYfE",
        "timestamps": [
            186
        ],
        "3min_transcript": "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. 2 times 9 is 18. Carry the 1. 2 times 2 is 4, plus 1 is 5. And now we can think about the 3. 3 times 1-- oh, let me throw a 0 here. Because this isn't a 3. This is now a 30. So this is in the tens place. So that's why I put a 0 there. So 30 times 1 is 30. That's why we say 3 times 1 is 3, but notice, it's in the tens place right now. And then 3 times 9 is 27. Carry the 2. 3 times 2 is 6, plus 2 is 8. And now we can add. And we would get 2. 8 plus 3 is 11. 6 plus 3 is 13. And then you get 9. So you get 9,312. So this is going to be equal to 9,312 divided by 1,000."
    },
    {
        "Q": "at 0:04 why are the decimals not lined up? aren't they supposed to be lined up when you are multiplying?",
        "A": "No, when you multiply, you line it up by how many numbers there are, so in 3.14 and 53.4, you line up the 4 with 4, 1 with 3, and 3 with 5, it doesn t matter where the decimal is. You only line your digits up when you are adding or subtracting.",
        "video_name": "4IWfJ7-CYfE",
        "timestamps": [
            4
        ],
        "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": "I couldn't hear you at 0:25",
        "A": "0:20 - 0:30 is What i want to do in this video is test whether any of these four numbers satisfy either of these inequalities. i encourage you to pause the video and try these numbers out, does zero satisfy this inequality? to see what he is saying pause the video and press c on the keyboard for captions or you can scroll down a tiny bit and next to About there is Transcript. Click Transcript to see what he is saying.",
        "video_name": "Yh4TXMVq9eg",
        "timestamps": [
            25
        ],
        "3min_transcript": "- [Voiceover] We have two inequalities here, the first one says that x plus two is less than or equal to two x. This one over here in I guess this light-purple-mauve color, is three x plus four is greater than five x. Over here we have four numbers and what I want to do in this video is test whether any of these four numbers satisfy either of these inequalities. I encourage you to pause this video and try these numbers out, does zero satisfy this inequality? Does it satisfy this one? Does one satisfy this one? Does it satisfy that one? I encourage you to try these four numbers out on these two inequalities. Assuming you have tried that, let's work through this together. Let's say, if we try out zero on this inequality right over here, let's substitute x with zero. So, we'll have zero plus two needs to be less than or equal to two times zero. Well, on the left hand side, this is two needs to be less than or equal to zero. Is that true, is two less than or equal to zero? No, two is larger than zero. So this is not going to be true, this does not satisfy the left hand side inequality, let's see if it satisfies this inequality over here. In order to satisfy it, three times zero plus four needs to be greater than five times zero. Well three times zero is just zero, five times zero is zero. So four needs to be greater than zero, which is true. So it does satisfy this inequality right over here so zero does satisfy this inequality. Let's try out one. To satisfy this one, one plus two needs to be less than One plus two is three, is three less than or equal than two? No, three is larger than two. This does not satisfy the left hand inequality. What about the right hand inequality right over here? Three times one plus four needs to be greater than five times one. So three times one is three, plus four. So seven needs to be greater than five, well that's true. Both zero and one satisfy three x plus four is greater than five x, neither of them satisfy x plus two is less than or equal to two x. Now let's go to the two. I know it's getting a little bit unaligned, but I'll just do it all in the same color so you can tell. Let's try out two here, two plus two needs to be less than or equal to two times two. Four needs to be less than or equal to four. Well four is equal to four and it just has to be"
    },
    {
        "Q": "Around 3:10 why not take a proportion like 37% instead of 30%?",
        "A": "Look at the null hypothesis. A value of 37% does not satisfy the null hypothesis of p<=0.30. When choosing a value of p, we need to satisfy the null hypothesis, and to choose a conservative number (one which will be less likely to reject the null hypothesis. In this case the largest value allowed by the null, which is 0.30).",
        "video_name": "1JT9oODsClE",
        "timestamps": [
            190
        ],
        "3min_transcript": "hypothesis for the population. And the given that assumption, what is the probability that 57 out of 150 of our samples actually have internet access. And if that probability is less than 5%, if it's less than our significance level, then we're going to reject the null hypothesis in favor of the alternative one. So let's think about this a little bit. So we're going to start off assuming-- we're going to assume the null hypothesis is true. And in that assumption we're going to have to pick a population proportion or a population mean-- we know that for Bernoulli distributions do the same thing. And what I'm going to do is I'm going to pick a proportion so high so that it maximizes the probability of getting this over here. And we actually don't even know what that number is. And actually so that we can think about a little more proportion even is. We had 57 people out of 150 having internet access. So 57 households out of 150. So our sample proportion is 0.38, so let me write that over here. Our sample proportion is equal to 0.38. So when we assume our null hypothesis to be true, we're going to assume a population proportion that maximizes the probability that we get this over here. So the highest population proportion that's within our null hypothesis that will maximize the probability of getting this is actually if we are right at 30%. So if we say our population proportion, we're going to assume this is true. This is our null hypothesis. We're going to assume that it is 0.3 or 30%. And I want you understand that-- 29% would have been a 28% that would have been a null hypothesis. But for 29% or 28%, the probability of getting this would have been even lower. So it wouldn't have been as strong of a test. If we take the maximum proportion that still satisfies our null hypothesis, we're maximizing the probability that we get this. So if that number is still low, if it's still less than 5%, we can feel pretty good about the alternative hypothesis. So just to refresh ourselves we're going to assume a population proportion of 0.3, and if we just think about the distribution-- sometimes it's helpful to draw these things, so I will draw it. So this is what the population distribution looks like based on our assumption, based on this assumption right over here. Our population distribution has-- or maybe I should write 30% have internet access. And I'll signify that with a 1. And then the rest don't have internet access. 70% do not have internet access."
    },
    {
        "Q": "at 1:50 how do you divide?",
        "A": "You don t, you multiply the improper fraction by the recipricole of 1/5, which is 5/1",
        "video_name": "xoXYirs2Mzw",
        "timestamps": [
            110
        ],
        "3min_transcript": "Tracy is putting out decorative bowls of potpourri in each room of the hotel where she works. She wants to fill each bowl with 1/5 of a can of potpourri. If Tracy has 4 cans of potpourri, in how many rooms can she place a bowl of potpourri? So she has 4 cans, and she wants to divide this 4 cans into groups of 1/5 of a can. So if you have 4 of something and you're trying to divide it into groups of a certain amount, you would divide by that amount per group. So you want to divide 4 by 1/5. You want to divide 4 cans of potpourri into groups of 1/5. So let's visualize this. Let me draw one can of potpourri right over here. So one can of potpourri can clearly be cut up into 5/5. We have it right over here. 1, 2, 3, 4, 5. So 1 can of potpourri can fill 5 bowls Now, we have 4 cans. So let me paste these. So 2, 3, and 4. So how many total bowls of potpourri can Tracy fill? Well, she's got 4 cans. So this is going to be equal to-- let me do this is the right color-- this is going to be equal to, once again, she has 4 cans. And then for each of those cans, she can fill 5 bowls of potpourri because each bowl only requires a 1/5 of those cans. So this is going to be the same thing as 4 times 5. Or we can even write this as 4 times 5 over 1. 5 is the same thing as 5/1, which is the same thing as 4 times 5, which, of course, is equal to 20. she can fill 20 bowls of potpourri. Now, just as a review here, we've already seen that dividing by a number is equal to multiplying by its reciprocal. And we see that right over here. Dividing by 1/5 is the same thing as multiplying by the reciprocal of 1/5, which is 5/1. So she could fill up 20 bowls of potpourri."
    },
    {
        "Q": "cane someone pls explain wat sal is trying to say at 3:45 - 4:52?",
        "A": "Hi, We need to divide 9.2 by 11.5. As division with decimals get a bit confusing, Sal does the below-mentioned step. 9.2/11.5 = 9.2 *10 / 11.5 *10 ==>92/115 (Multiplying both numerator and denominator by 10, thereby we are not changing the value of the fraction). If you are still confused, take out a calculator and perform both the functions [9.2/11.5 and 92/115]. You will get the same answer. This is the trick in fractions. Hope this helped you.",
        "video_name": "EbmgLiSVACU",
        "timestamps": [
            225,
            292
        ],
        "3min_transcript": "So here, we have 4.6 times 2. So 4.6 times 2 is 9.2. So that's 9.2. And then 10 to the sixth times 10 to the negative 1-- we have the same base. We're taking the product. We can add the exponents-- is going to be 10 to the 6 minus 1 or 10 to the fifth power. So we've simplified our numerator. And now in our denominator, let's see. 5 times 2.3-- 5 times 2 is 10. 5 times 0.3 is 1.5, so it's going to be 11.5. So this is going to be 11.5. And then if I multiply 10 to the fourth times 10 to the negative 2, that's going to be 10 to the 4 minus 2 or 10 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."
    },
    {
        "Q": "at 5:07 i got confused.",
        "A": "There, 15 goes into 75 ----> 5 times r 1. To review it, when he was writing 5*5 = 25 in the division table, after writing 5, he takes 2 to add with (1*5). But, instead of writing 2 there, he faultily wrote it as 7. Then, after finding the error, he rectified it.",
        "video_name": "gHTH6PKfpMc",
        "timestamps": [
            307
        ],
        "3min_transcript": "multiply 5 times 5 is 25. 5 times 2 is 10 plus 2, 125. So it goes in exact. So 125 minus 125 is clearly 0. Then we bring down this 0. And 25 goes into 0 zero times. 0 times 25 is 0. Remainder is 0. So we see that 25 goes into 6,250 exactly 250 times. Let's do another problem. Let's say I had-- let me pick an interesting number. Let's say I had 15 and I want to know how many times it goes into 2,265. We say OK, does 15 go into 2? No. So does 15 go into 22? Sure. 15 goes into 22 one time. Notice we wrote the 1 above the 22. If it go had gone into 2 we would've written the 1 here. But 15 goes into 22 one time. 1 times 15 is 15. 22 minus 15-- we could do the whole carrying thing-- 1, 12. 12 minus 5 is 7. 1 minus 1 is 0. 22 minus 15 is 7. Bring down the 6. OK, now how many times does 15 go into 76? Once again, there isn't a real easy mechanical way to do it. You can kind of eyeball it and estimate. Well, 15 times 2 is 30. 15 times 4 is 60. 15 times 5 is 75. So 5 times 5 once again, I already figured it out in my head, but I'll just do it again. 5 times 1 is 5. Plus 7. Oh, sorry. 5 times 5 is 25. 5 times 1 is 5. Plus 2 is 7. Now we just subtract. 76 minus 75 is clearly 1. Bring down that 5. Well, 15 goes into 15 exactly one time. 1 times 15 is 15. Subtract it and we get a remainder of 0. So 15 goes into 2,265 exactly 151 times. So just think about what we're doing here and why it's a"
    },
    {
        "Q": "at 0:24, he asks what 6250 divided by 25 is. What's the answer?",
        "A": "250. @Victoria Fine, you mean pie as in I LIKE PIE or pi as in 3.141592465?",
        "video_name": "gHTH6PKfpMc",
        "timestamps": [
            24
        ],
        "3min_transcript": "Welcome to the presentation on level 4 division. So what makes level 4 division harder than level 3 division is instead of having a one-digit number being divided into a multi-digit number, we're now going to have a two-addition number divided into a multi-digit number. So let's get started with some practice problems. So let's start with what I would say is a relatively straightforward example. The level 4 problems you'll see are actually a little harder than this. But let's say I had 25 goes into 6,250. So the best way to think about this is you say, OK, I have 25. Does 25 go into 6? Well, no. Clearly 6 is smaller than 25, so 25 does not go into 6. So then ask yourself, well, then if 25 doesn't go into 6, does 25 go into 62? Well, sure. 62 is larger than 25, so 25 will go into 62? 25 times 1 is 25. 25 times 2 is 50. So it goes into 62 at least two times. And 25 times 3 is 75. So that's too much. So 25 goes into 62 two times. And there's really no mechanical way to go about figuring this out. You have to kind of think about, OK, how many times do I think 25 will go into 62? And sometimes you get it wrong. Sometimes you'll put a number here. Say if I didn't know, I would've put a 3 up here and then I would've said 3 times 25 and I would've gotten a 75 here. And then that would have been too large of a number, so I would have gone back and changed it to a 2. Likewise, if I had done a 1 and I had done 1 tmes 25, when I subtracted it out, the difference I would've gotten would be larger than 25. And then I would know that, OK, 1 is too small. I have to increase it to 2. I hope I didn't confuse you too much. I just want you to know that you shouldn't get nervous if you're like, boy, every time I go through the step it's kind of like- I kind of have to guess what the numbers is as And that's true; everyone has to do that. So anyway, so 25 goes into 62 two times. Now let's multiply 2 times 25. Well, 2 times 5 is 10. And then 2 times 2 plus 1 is 5. And we know that 25 times 2 is 50 anyway. Then we subtract. 2 minus 0 is 2. 6 minus 5 is 1. And now we bring down the 5. So the rest of the mechanics are pretty much just like a level 3 division problem. Now we ask ourselves, how many times does 25 go into 125? Well, the way I think about it is 25-- it goes into 100 about four times, so it will go into 125 one more time. It goes into it five times. If you weren't sure you could try 4 and then you would see that you would have too much left over. Or if you tried 6 you would see that you would actually get 6 times 25 is a number larger than 125. So you can't use 6."
    },
    {
        "Q": "At 2:50 couldnt you multiply both sides by 5, and then subtract by 0.6?",
        "A": "Yes, you could. You are right. Good observation. There are multiple ways you can solve this problem.",
        "video_name": "BOIA9wsM4ok",
        "timestamps": [
            170
        ],
        "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": "Can someone help me here?\nAt 5:15 rather than distributing the 0.5 to both r and 2.75, he just divided r and 3 by 0.5. If 0.5 can be only divided by r and 3 then why is 2.75 inside the parentheses with r? Wouldn't dividing 0.5 only by r and not by 2.75 create an imbalance?",
        "A": "Sal just liked to show you another way to solve that problem, and it is more simple way too.",
        "video_name": "BOIA9wsM4ok",
        "timestamps": [
            315
        ],
        "3min_transcript": "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, is going to be just n. But I can't just multiply the left-hand side I have to multiply the right-hand side by five as well. And so what is that going to get us? We are going to get n = 1.4 x 5. 1.4 x 5. Now you might be able to do this in your head, 'cause this is one and 2/5. So this thing should all be equal to seven, but I'll just do it this way as well. Five times four is 20. Re-group the two. One times five is five plus two is seven. And when I look at all the numbers that I'm multiplying, I have one digit to the right of the decimal point. So my answer will have one digit to the right of the decimal point. So it's 7.0, or just 7. n = 7. And you can verify this works, 'cause seven divided by five is going to be equal to 1.4, plus 0.6 is equal to two. Let's do one more example. Alright. 0.5 times the whole quantity (r + 2.75) = 3. Now there's a bunch of ways that you could tackle this. A lot of times when you see something like this, your temptation might be, \"Let's distribute the 0.5.\" But that makes it a little bit hairy, 'cause 0.5 times 2.75. You could calculate that, and you will get the right answer if you do it correctly. But a simpler thing might be, well, let's just divide both sides by 0.5. That way I'm going to get more whole numbers involved. So if I divide. Remember, whatever I do to the left-hand side I have to do to the right-hand side. And the way my brain thought about it was, well, if I divide by 0.5 on the left-hand side I can get rid of this. And if I divide by 0.5 on the right-hand side I'm still going to get an integer. Three divided by 0.5 is six. It's the same thing as three divided by a half. How many halves fit into three? Six halves fit into three. So this going to be six right over here. And then this is going to be equal to six. So the whole thing is simplified now"
    },
    {
        "Q": "At 0:47, what does it exactly mean to have a slope of 1?",
        "A": "A slope of 1 means that for every unit the line goes along the x axis, it goes up the y axis 1 unit.",
        "video_name": "5a6zpfl50go",
        "timestamps": [
            47
        ],
        "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 0:06 y=x+3 how do i sketch the graph if x is x square",
        "A": "You can t. It s not a linear problem then. The way to graph it is like the quadratics, or other x^2. The y intercept is now the vertex. the line will look like a U.",
        "video_name": "5a6zpfl50go",
        "timestamps": [
            6
        ],
        "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": "i don't get the whole rate system. if you have the ratio how do you get the rate? for ex: 8 boxes can hold 48 books the ratio is 48:8 but how do you figure out the rate?",
        "A": "Your rate will be 6 because how much does 1 box hold it holds 6 if you divide,if you need any help comment! :D",
        "video_name": "qGTYSAeLTOE",
        "timestamps": [
            2888
        ],
        "3min_transcript": ""
    },
    {
        "Q": "By saying that one is going 35 miles per hour, in ratio form would be 35:1, but could it also be expressed the other way around? for example, 1:35, saying that in every hour one is going 35 miles. Is is the same thing?",
        "A": "yes it would be the same thing except it has to due with how it is worded. say if the problem said miles per hour it is 29:1 58:2 and so on but if it is hours per amount of miles 1:29 2:29",
        "video_name": "qGTYSAeLTOE",
        "timestamps": [
            2101,
            95
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 1:13, why is theta written in radians?",
        "A": "Because it is easier to measure sin cos and tan that way.",
        "video_name": "sjUhr0HkLUg",
        "timestamps": [
            73
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 8:15, Isn't PI transcendental number, so it is not a real number.\nSo is the graph wrong ?",
        "A": "\u00cf\u0080 is both transcendental and real. Transcendental just means that the number isn t the root of a polynomial with rational coefficients. \u00e2\u0088\u009a2 is irrational, but not transcendental, since it s the root of x\u00c2\u00b2-2=0.",
        "video_name": "sjUhr0HkLUg",
        "timestamps": [
            495
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 3:18 , Sal goes 3pi/2. Isnt it 3pi/4?\nIf im wrong can someone tell me why?",
        "A": "Sal is right, it s 3pi/2. Maybe you got confused because Sal mentioned that we got 3/4 of the way , but you have to remember that the whole way is 2pi. So each quarter is pi/2 and 3*pi/2 is 3pi/2",
        "video_name": "sjUhr0HkLUg",
        "timestamps": [
            198
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 5:50, on the mounting of the graph, why the connection between the points [\u00ce\u00b8,sin(\u00ce\u00b8)] of the table, was made for curved lines instead of straight lines?\nIn other words, why the graph of sin(\u00ce\u00b8) is not like this: /\\/\\/\\/\\/\\/\\/\\/\\/\\/\\ ?",
        "A": "You can check this in different ways. If you want to be really confident, just take many points between 0 and pi and take the sine, you will get this smooth line. It s easier to follow the unit circle, as seen from 2:20 - to get from 0 to pi/2 Sal goes along a curved path - this has to be on the graph of the sine als well.",
        "video_name": "sjUhr0HkLUg",
        "timestamps": [
            350
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 6:48 sal says with arcsin we only have to draw the first and fourth quadrants. why is this? I saw somewhere else that sine is between 1 and -1, but I still don't understand this.",
        "A": "arc sin is a function, and as such it cannot have different outputs for the same input. If we included the second and third quadrant, we would get two solutions for most input values, so we wouldn t have a function.",
        "video_name": "JGU74wbZMLg",
        "timestamps": [
            408
        ],
        "3min_transcript": "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? that means the y-coordinate on the unit circle is minus square root of 3 over 2. So it means we're right about there. So this is minus the square root of 3 over 2. This is where we are. Now what angle gives me that? Let's think about it a little bit. My y-coordinate is minus square root of 3 over 2. This is the angle. It's going to be a negative angle because we're going below the x-axis in the clockwise direction. And to figure out-- Let me just draw a little triangle here. Let me pick a better color than that. That's a triangle. Let me do it in this blue color. So let me zoom up that triangle. Like that. This is theta. That's theta. And what's this length right here? Well that's the same as the y-height, I guess"
    },
    {
        "Q": "At 7:14, how does Sal know to put the y-axis point in that spot? Is it just an estimate? It seems that the result of the problem would depend on the accuracy of the placement of that point and therefore an estimate would not work.",
        "A": "In this example, it is a 30-60-90 triangle. He draws the triangle as a visual aid, not as a means of actually calculating the angle. More generally, you might use a sketch like that to estimate the quadrant but use a calculator if you needed the actual result.",
        "video_name": "JGU74wbZMLg",
        "timestamps": [
            434
        ],
        "3min_transcript": "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? that means the y-coordinate on the unit circle is minus square root of 3 over 2. So it means we're right about there. So this is minus the square root of 3 over 2. This is where we are. Now what angle gives me that? Let's think about it a little bit. My y-coordinate is minus square root of 3 over 2. This is the angle. It's going to be a negative angle because we're going below the x-axis in the clockwise direction. And to figure out-- Let me just draw a little triangle here. Let me pick a better color than that. That's a triangle. Let me do it in this blue color. So let me zoom up that triangle. Like that. This is theta. That's theta. And what's this length right here? Well that's the same as the y-height, I guess"
    },
    {
        "Q": "Hi. In the video you were trying to find the arcsin of rad3/2. Since sin = y/r would not the hypotenuse(r) of the triangle at 8:13 be 2 and not 1? That would make the other sides = rad3 and 1 respectively. 1^2 + rad3^2 = 2^2. I don't understand why one side of the triangle you drew would be rad3/2.",
        "A": "the hypotenuse is the radius of the unit circle, which is always 1",
        "video_name": "JGU74wbZMLg",
        "timestamps": [
            493
        ],
        "3min_transcript": "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? that means the y-coordinate on the unit circle is minus square root of 3 over 2. So it means we're right about there. So this is minus the square root of 3 over 2. This is where we are. Now what angle gives me that? Let's think about it a little bit. My y-coordinate is minus square root of 3 over 2. This is the angle. It's going to be a negative angle because we're going below the x-axis in the clockwise direction. And to figure out-- Let me just draw a little triangle here. Let me pick a better color than that. That's a triangle. Let me do it in this blue color. So let me zoom up that triangle. Like that. This is theta. That's theta. And what's this length right here? Well that's the same as the y-height, I guess It's a minus because we're going down. But let's just figure out this angle. And we know it's a negative angle. So when you see a square root of 3 over 2, hopefully you recognize this is a 30 60 90 triangle. The square root of 3 over 2. This side is 1/2. And then, of course, this side is 1. Because this is a unit circle. So its radius is 1. So in a 30 60 90 triangle, the side opposite to the square root of 3 over 2 is 60 degrees. This side over here is 30 degrees. So we know that our theta is-- This is 60 degrees. That's its magnitude. But it's going downwards. So it's minus 60 degrees. So theta is equal to minus 60 degrees. But if we're dealing in radians, that's not good enough. So we can multiply that times 100-- sorry --pi radians for every 180 degrees. Degrees cancel out. And we're left with theta is equal to minus pi over 3 radians."
    },
    {
        "Q": "At 6:02 he writes the range for a general arcsin function, but just to confirm, that's the range for all arcsin functions?? Or am I missing something. Because for the example problem he did at the end, he didn't state the domain and range, but I thought restricting the range was necessary to make the sine function valid. Could someone please clarify?",
        "A": "i think you mean all inverse trig functions rather than all arcsin functions. and to answer your question, the range of all inverse trig functions are not the same due to the shape of the graph.",
        "video_name": "JGU74wbZMLg",
        "timestamps": [
            362
        ],
        "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": "What are Radians? 00:14",
        "A": "Radians are a different form of measurement of an angle. They act no differently than degrees. They are much like different forms of currency. The Canadian Dollar and the U.S. Dollar serve the same purpose, but have different values. In the same way that a U.S. Dollar has more value than a Canadian Dollar, Radians have more value than degrees. 180 degrees is equal to 1 pi radians.",
        "video_name": "JGU74wbZMLg",
        "timestamps": [
            14
        ],
        "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": "Why didn't Sal round up to 1.4 from 1.47? Wouldn't that be more accurate? 10:01",
        "A": "Sal DID round UP. The calculator showed a NEGATIVE 1.047..... which is between - 1.05 and -1.04 The larger number (the one furthest to the right on the number line) is -1.04..",
        "video_name": "JGU74wbZMLg",
        "timestamps": [
            601
        ],
        "3min_transcript": "It's a minus because we're going down. But let's just figure out this angle. And we know it's a negative angle. So when you see a square root of 3 over 2, hopefully you recognize this is a 30 60 90 triangle. The square root of 3 over 2. This side is 1/2. And then, of course, this side is 1. Because this is a unit circle. So its radius is 1. So in a 30 60 90 triangle, the side opposite to the square root of 3 over 2 is 60 degrees. This side over here is 30 degrees. So we know that our theta is-- This is 60 degrees. That's its magnitude. But it's going downwards. So it's minus 60 degrees. So theta is equal to minus 60 degrees. But if we're dealing in radians, that's not good enough. So we can multiply that times 100-- sorry --pi radians for every 180 degrees. Degrees cancel out. And we're left with theta is equal to minus pi over 3 radians. arcsine of minus square root of 3 over 2 is equal to minus pi over 3 radians. Or we could say the inverse sign of minus square root of 3 over 2 is equal to minus pi over 3 radians. And to confirm this, let's just-- Let me get a little calculator out. I put this in radian mode already. You can just check that. Per second mode. I'm in radian mode. So I know I'm going to get, hopefully, the right answer. And I want to figure out the inverse sign. So the inverse sine-- the second and the sine button --of the minus square root of 3 over 2. It equals minus 1.04. So pi over 3 must be equal to 1.04. Let's see if I can confirm that. So if I were to write minus pi divided by 3, what do I get? I get the exact same value. So my calculator gave me the exact same value, but it might have not been that helpful because my calculator doesn't tell me that this is minus pi over 3."
    },
    {
        "Q": "At 5:58 it says that the range of theta/arcsin is being less than or equal to pi over 2 and greater than or equal to minus pi over 2. Can it be equal to these numbers? Can the range equal positive pi over 2 or negative pi over 2, or was this a mistake?",
        "A": "the range can be pi/2 or -pi/2 because it s still a one to one function",
        "video_name": "JGU74wbZMLg",
        "timestamps": [
            358
        ],
        "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 1:50, Sal says that the sin is the height, but I'm pretty sure that is not usually sin. What special circumstances makes sin the height?",
        "A": "Hello Kevin, Height, including negative heights, is an appropriate description for SIN(X). You see, SIN(X) gives the vertical component and COS(X) gives the horizontal component. If this is a fuzzy answer it will make more sense when you get to the unit circle. Regards, APD",
        "video_name": "JGU74wbZMLg",
        "timestamps": [
            110
        ],
        "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": "Don't get 2:55 to 3:55",
        "A": "That s an example of using Heron s Formula.",
        "video_name": "-YI6UC4qVEY",
        "timestamps": [
            175,
            235
        ],
        "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 1:36 what does he mean by \"third variable S\"",
        "A": "I think he meant to say the fourth variable (three sides of the triangle being the first three), but the ordinal number used to refer to it is irrelevant. s = the semi-perimeter of the triangle. That would be (a + b + c) / 2.",
        "video_name": "-YI6UC4qVEY",
        "timestamps": [
            96
        ],
        "3min_transcript": "I think it's pretty common knowledge how to find the area of the triangle if we know the length of its base and its height. So, for example, if that's my triangle, and this length right here-- this base-- is of length b and the height right here is of length h, it's pretty common knowledge that the area of this triangle is going to be equal to 1/2 times the base times the height. So, for example, if the base was equal to 5 and the height was equal to 6, then our area would be 1/2 times 5 times 6, which is 1/2 times 30-- which is equal to 15. Now what is less well-known is how to figure out the area of a triangle when you're only given the sides of the triangle. When you aren't given the height. So, for example, how do you figure out a triangle where I just give you the lengths of the sides. Let's say that's side a, side b, and side c. a, b, and c are the lengths of these sides. 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."
    },
    {
        "Q": "I was wondering. How come at 4 29 after he has worked it down to the square roots of all of them. How come it is 7? What happens if each number was to not have a perfact sqaure in it? Like a triangle with the sides: 8:10:12? What would happen? What one would you pick?",
        "A": "Hi Fuller! If the triangle had sides 8, 10, and 12, then the semiperimeter would equal 15. So, the answer is the square root of 8x10x12 which is the square root of 2x2x2x2x5x2x3x2 which is 2x2x2x the square root of 3x5 which is 8 times the square root of 15.",
        "video_name": "-YI6UC4qVEY",
        "timestamps": [
            490
        ],
        "3min_transcript": ""
    },
    {
        "Q": "4:12-4:20 i got confused who can help me ???",
        "A": "oh ok thanks for the help ..i really appreciated it",
        "video_name": "-YI6UC4qVEY",
        "timestamps": [
            252,
            260
        ],
        "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": "at 1:29 what does Sal mean by \"nice trick\"?",
        "A": "He s referring to the formula (Heron s formula) itself.",
        "video_name": "-YI6UC4qVEY",
        "timestamps": [
            89
        ],
        "3min_transcript": "I think it's pretty common knowledge how to find the area of the triangle if we know the length of its base and its height. So, for example, if that's my triangle, and this length right here-- this base-- is of length b and the height right here is of length h, it's pretty common knowledge that the area of this triangle is going to be equal to 1/2 times the base times the height. So, for example, if the base was equal to 5 and the height was equal to 6, then our area would be 1/2 times 5 times 6, which is 1/2 times 30-- which is equal to 15. Now what is less well-known is how to figure out the area of a triangle when you're only given the sides of the triangle. When you aren't given the height. So, for example, how do you figure out a triangle where I just give you the lengths of the sides. Let's say that's side a, side b, and side c. a, b, and c are the lengths of these sides. 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."
    },
    {
        "Q": "Principal root of a aquare root fnction? What does he mean by this? It is approximately 5:45...",
        "A": "Principal root is the positive one of the two roots",
        "video_name": "rYG1D5lUE4I",
        "timestamps": [
            345
        ],
        "3min_transcript": "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. then negative x is clearly a negative number, or I guess it could also be 0. It's a negative number. And then we can apply this right over here. If x was less than 0, then we would be doing all of this nonsense up here. And we start to get nonsensical answers. And if you look at it this way and you say hey, look, i can be the square root of negative one, if it's the principal branch of the complex square root function. Then you could rewrite this right over here as the square root of negative 1 times the square root of x. And so really the real fault in this logic when people say, hey, negative 1 can't be equal to 1, the real fault is using this property when both a and b, where both of these are negative numbers. That will come up with something that is unambiguously false. If you expand your definition of the complex to include negative numbers in the domain and to include imaginary numbers, then you can do this. You can say the square root of negative x is the square root of negative 1 times-- Or you should say the principal square root of negative x-- I should be particular my words-- is the same thing as the principal square root of negative 1 times the principal square root of x when x is greater than or equal to 0. And I don't want to confuse you, if x is greater than or equal to 0, this negative x, that is clearly a negative, or I guess you should say a non positive number."
    },
    {
        "Q": "I think Sal make a mistake on (vid @ 5:11) when he write the greater than sign! it should be Less than",
        "A": "No, Sal is correct. If he had: i sqrt(x) where X<0, then X is negative. Backup thru Sal steps. If X is negative * (-1) = +X. And he would have started with sqrt(x), not sqrt(-x). He is also trying to highlight that if you had something like: sqrt(12), you would not make this into i sqrt(-12). The imaginary number is not needed if the radical contains a positive number to start with.",
        "video_name": "rYG1D5lUE4I",
        "timestamps": [
            311
        ],
        "3min_transcript": "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. then negative x is clearly a negative number, or I guess it could also be 0. It's a negative number. And then we can apply this right over here. If x was less than 0, then we would be doing all of this nonsense up here. And we start to get nonsensical answers. And if you look at it this way and you say hey, look, i can be the square root of negative one, if it's the principal branch of the complex square root function. Then you could rewrite this right over here as the square root of negative 1 times the square root of x. And so really the real fault in this logic when people say, hey, negative 1 can't be equal to 1, the real fault is using this property when both a and b, where both of these are negative numbers. That will come up with something that is unambiguously false. If you expand your definition of the complex"
    },
    {
        "Q": "Does the rule at 2:32 apply if, for example, a is positive and b is negative?",
        "A": "Bobo, \u00e2\u0088\u009aa*\u00e2\u0088\u009ab = \u00e2\u0088\u009a(a*b) applies if both a and b are positive or it either a or b are negative. It does not apply if both a and b are negative.",
        "video_name": "rYG1D5lUE4I",
        "timestamps": [
            152
        ],
        "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": "At , 3:03 Sal says that the people who say that\ni\nisn't the principal square root of -1 are wrong. Is this just Sal's personal thinking, or is it a thing that mathematicians have decided upon? Basically, is it official that\ni\nisn't the principal square root of -1?",
        "A": "It honestly depends on who you ask. However, i is, in fact, widely believed to be rad(-1). However, NOT considering i to be such a thing prevents you from solving ( or even understanding ) many problems where using complex numbers is necessary, a lot of which are applicable to real life. But really, like with everything else, it s up to each individual person to decide what they think.",
        "video_name": "rYG1D5lUE4I",
        "timestamps": [
            183
        ],
        "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": "At 6:41 how can it only apply when x is greater or equal to zero, given that it is a negative number?",
        "A": "When the number is positive, it applies as a normal (square) root. For example, the square root of 4 is 2, but the square root of -4 could be seen as 2i, because i=-1. Therefore, x must be greater than or equal to zero in order to have that negative number, and in turn, contain i. Like he says at 5:58, the two negative numbers are where it goes wrong. x cannot be negative.",
        "video_name": "rYG1D5lUE4I",
        "timestamps": [
            401
        ],
        "3min_transcript": "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. then negative x is clearly a negative number, or I guess it could also be 0. It's a negative number. And then we can apply this right over here. If x was less than 0, then we would be doing all of this nonsense up here. And we start to get nonsensical answers. And if you look at it this way and you say hey, look, i can be the square root of negative one, if it's the principal branch of the complex square root function. Then you could rewrite this right over here as the square root of negative 1 times the square root of x. And so really the real fault in this logic when people say, hey, negative 1 can't be equal to 1, the real fault is using this property when both a and b, where both of these are negative numbers. That will come up with something that is unambiguously false. If you expand your definition of the complex to include negative numbers in the domain and to include imaginary numbers, then you can do this. You can say the square root of negative x is the square root of negative 1 times-- Or you should say the principal square root of negative x-- I should be particular my words-- is the same thing as the principal square root of negative 1 times the principal square root of x when x is greater than or equal to 0. And I don't want to confuse you, if x is greater than or equal to 0, this negative x, that is clearly a negative, or I guess you should say a non positive number."
    },
    {
        "Q": "At 14:21 Sal points out that S = {[x1, x2] \u00e2\u0088\u0088 R2 | x1 \u00e2\u0089\u00a5 0} is not a subspace because it is not closed under multiplication. Then, at 18:07 he says that the span of any set of vectors is a valid subspace. However, I thought that since the definition of Span is the set of all linear combinations of a set of vectors, there must be a Span ([x1, x2] \u00e2\u0088\u0088 R2 | x1 \u00e2\u0089\u00a5 0). So, there seems to be a contradiction.",
        "A": "From what I understand, it must be able to span all R2 vectors in order to be consider a valid subspace of R2 but with the given requirement which is x1\u00e2\u0089\u00a5 0, that is not possible with negative scalar. Thus it is not a span of R2. Am I correct?",
        "video_name": "pMFv6liWK4M",
        "timestamps": [
            861,
            1087
        ],
        "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 about 5:30 he was talking about the unit circle. How did he come up with 2pi/3 instead of 2pi?",
        "A": "Instead of having cos(x), the problem involved cos(3x). In other words, there were 3 repetitions within the usual period of 2\u00cf\u0080. Therefore, each of those periods was of length 2\u00cf\u0080/3.",
        "video_name": "SBqnRja4CW4",
        "timestamps": [
            330
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 5:36 , how did you plot g inverse (x) on the graph, and how did slope come in there ?",
        "A": "Well, because y = -x/2 - 1/2 is a linear function, you only need two points to draw it. You can use the y-intercept (x=0): y = - 0/2 - 1/2 = - 1/2. That s the first point. Secondly, you can write -x/2 = -(1*x)/2 = -1/2 x, so -1/2 is the slope of the function. with any x your getting your second point (e.g. x=1 -> y = -1/2 - 1/2 = -1). Draw a line and you re finished.",
        "video_name": "wSiamij_i_k",
        "timestamps": [
            336
        ],
        "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 1:30 Sal says that f^-1(x)= -x+4, but in the start f(x) is also equal to -x+4. How can it be the same?\n\nThis may seem silly but i am confused about this question!",
        "A": "okay, if we have y=-x + 4, then x = -y +4, that s why the inverse function is same. Another example can be if y=x, then x=y, so the inverse function is same as the function.",
        "video_name": "wSiamij_i_k",
        "timestamps": [
            90
        ],
        "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": "At 2:41, what does Sal mean by \" And notice, both of these numbers are exactly 10 away from the number 5?\" Why are the numbers 10 units away from 5? Why isn't it 10 away from 0?",
        "A": "To be 10 away from zero, the problem would need to be: |x| = 10. Notice the original equation: |x - 5| = 10. You need to take into account the -5 inside the absolute value. That s where the 5 comes from. Hope this helps.",
        "video_name": "u6zDpUL5RkU",
        "timestamps": [
            161
        ],
        "3min_transcript": "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 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?"
    },
    {
        "Q": "At about 5:30, Could you write that as -4 1/2? Or is that not allowed?",
        "A": "That is allowed, but it just makes it easier to write it like -9/2.",
        "video_name": "u6zDpUL5RkU",
        "timestamps": [
            330
        ],
        "3min_transcript": "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 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"
    },
    {
        "Q": "At 1:39, Sal said that 0 to the zeroth power is undefined. Why?",
        "A": "Well, any number to the zeroth power is 1 right? So 0 to the zeroth power would be 1. But zero to any power is 0. So 0 to the zeroth power would be 0. Which is it? It can t be both, so it is undefined.",
        "video_name": "NEaLgGi4Vh4",
        "timestamps": [
            99
        ],
        "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 1:58 why is 0 to the 0 power undefined?",
        "A": "it is 1 any number to the power of 0 has to be one",
        "video_name": "NEaLgGi4Vh4",
        "timestamps": [
            118
        ],
        "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 1:17 Sal said \"anything to the zeroth power is equal to 1. What if the number is a negative?\nWouldn't that be -1?",
        "A": "It depends on how it s written. If it is expressed as -x^0, or negative x to the power of 0, then it is still equal to 1. If there is a negative sign outside of the term, such as 5 - x^0, you do the exponent first and get 5 - 1.",
        "video_name": "NEaLgGi4Vh4",
        "timestamps": [
            77
        ],
        "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": "I don't understand. At 0:51, Sal says 1/0 - 1/0 is undefined. And then, at 3:11, he says it's indeterminate. Are undefined and indeterminate supposed to be equal to each other? Because the function's the same but it seems to be equaling two different things.",
        "A": "This came up in our Calc II class today - undefined and indeterminate are not the same. For example, 1/0 is undefined, however with these limit problems, we know there is a limit (most of the time) even if the Initial Form of the problem is something like 0/0. My understanding is we call this the indeterminate form because with a little manipulation (as Sal demonstrates) there actually is an answer - we just don t see it in the initial form, hence it is indeterminate.",
        "video_name": "MeVFZjT-ABM",
        "timestamps": [
            51,
            191
        ],
        "3min_transcript": "We want to figure out the limit as x approaches 1 of the expression x over x minus 1 minus 1 over the natural log of x. So let's just see what happens when we just try to plug in the 1. What happens if we evaluate this expression at 1? Well then, we're going to get a one here, over 1 minus 1. So we're going to get something like a 1 over a 0, minus 1 over, and what's the natural log of 1? e to the what power is equal to one? Well, anything to the zeroth power is equal to 1, so e to the zeroth power is going to be equal to 1, so the natural log of 1 is going to be 0. So we get the strange, undefined 1 over 0 minus 1 over 0. It's this bizarre-looking undefined form. But it's not the indeterminate type of form that we looked for in l'Hopital's rule. We're not getting a 0 over a 0, we're not getting an So you might just say, hey, OK, this is a non-l'Hopital's rule problem. We're going to have to figure out this limit some other way. And I would say, well don't give up just yet! Maybe we can manipulate this algebraically somehow so that it will give us the l'Hopital indeterminate form, and then we can just apply the rule. And to do that, let's just see, what happens if we add these two expressions? So if we add them, so this expression, if we add it, it will be, well, the common denominator is going to be x minus 1 times the natural log of x. I just multiplied the denominators. And then the numerator is going to be, well, if I multiply essentially this whole term by natural log of x, so it's going to be x natural log of x, and then this whole term I'm going to multiply by x minus one. So minus x minus 1. And you could break it apart and see that this expression and this expression are the same thing. 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,"
    },
    {
        "Q": "At 9:00, I am not sure but why didn't Sal solve those 2 equations using matrices? I mean before these vector videos, he was using matrices to solve these kinds of equations. Or is it that these equations can not be solved through matrices? Thanks In Advance.",
        "A": "If he used a matrix, it would remove the C constants and sort of defeat the purpose for doing what he was doing. He just wanted to keep the constants intact so you could see what was happening.",
        "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": "Why do you subtract the green equation from the red one? at 9:17 min",
        "A": "That is a good question! This is a system of linear equations having 2 unknown variables and 2 equations with the variables. In order to solve this linear system of equations, we need to eliminate one variable to get an answer for the other variable. So, at 9:17, Sal subtracts the green equation from the red one to eliminate the variable C1 to get the value of C2. After getting C2, you can substitute the value of C2 in any equation to get the value of C1. Sal does this at 9:30. Hope this helped.",
        "video_name": "Alhcv5d_XOs",
        "timestamps": [
            557
        ],
        "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": "For the final example 13:00:00-end, why does Sal assign a random number for vector c3? How would you solve this to determine dependence without knowing the value(s) of a vector?",
        "A": "That s the nice thing about free variables like c3. We get to choose whatever number we want for c3, and no matter what we choose, it will still satisfy the equations. To determine dependence without making a choice for our free variables, all you need to know is that there are free variables. If there are free variables, then the set is linearly dependent.",
        "video_name": "Alhcv5d_XOs",
        "timestamps": [
            780
        ],
        "3min_transcript": "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 In which case, this would definitely be a linearly dependent set. When you see three vectors that are each only vectors in r2, that are each two-dimensional vectors, it's a complete giveaway that this is linearly dependent. But I'm going to show it to you using our dependent, using our little theorem here. So I'm going to show you that I can get non-zero c3's, c2's, and c1's such that I can get a 0 here. If all of these had to be 0-- I mean you can always set them But if they had to be equal to 0, then it would be linearly independent. Let me just show you. I can just pick some random c3. Let me pick c3 to be equal to negative 1. So what would these two equations reduce to? I mean you have just three unknowns and two equations, it means you don't have enough constraints on your system. So if I just set c3-- I just pick that out of a hat. I could have picked c3 to be anything."
    },
    {
        "Q": "Hey guys I was just wondering about why, at 13:41, can he just pick random values for C3?",
        "A": "Because he KNOWS that the vectors are linearly dependent. This is because you only need 2 linearly independent vectors to span R2. Any additional vector MUST be a linear combination of the other two by the very definition of spanning R2.",
        "video_name": "Alhcv5d_XOs",
        "timestamps": [
            821
        ],
        "3min_transcript": "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 In which case, this would definitely be a linearly dependent set. When you see three vectors that are each only vectors in r2, that are each two-dimensional vectors, it's a complete giveaway that this is linearly dependent. But I'm going to show it to you using our dependent, using our little theorem here. So I'm going to show you that I can get non-zero c3's, c2's, and c1's such that I can get a 0 here. If all of these had to be 0-- I mean you can always set them But if they had to be equal to 0, then it would be linearly independent. Let me just show you. I can just pick some random c3. Let me pick c3 to be equal to negative 1. So what would these two equations reduce to? I mean you have just three unknowns and two equations, it means you don't have enough constraints on your system. So if I just set c3-- I just pick that out of a hat. I could have picked c3 to be anything. equations become? You get 2c1 plus 3c2 minus 1 is equal to 0. And you get c1 plus 2c2 minus 2 is equal to 0. 2 times minus 1. What can I do here? If I multiply this second equation by 2, what do I get? I get 2 plus 4c2 minus 4 is equal to 0. And now let's subtract this equation from that equation. So the c1's cancel out. 3c2 minus 4c2 is minus c2. And then minus 1 minus minus 4, so that's minus 1 plus 4. That's plus 3 is equal to 0. And so we get our -- Let me make sure I got that right."
    },
    {
        "Q": "At 10:44 shouldn't there be a +c for the integration constant?",
        "A": "It s a definite integral, so there doesn t need to be a constant of integration.",
        "video_name": "AFF8FXxt5os",
        "timestamps": [
            644
        ],
        "3min_transcript": "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. originally have, if that confuses you at all. And the reason why I did that: we know what sine squared of anything plus cosine squared of that same anything is. That falls right out of the unit circle definition of our trig function, so this is just 1. So our whole integral has been reduced to the minus integral from 0 to 2pi of dt. And this is-- we have seen this before. We can probably say that this is of 1, if you want to put something there. Then the antiderivative of 1 is just-- so this is just going to be equal to minus-- and that minus sign is just the same minus sign that we're carrying forward. The antiderivative of 1 is just t, and we're going to evaluate it from 2pi to 0, or from 0 to 2pi, so this is equal to minus-- that minus sign right there-- 2pi minus t at 0, so minus 0. So this is just equal to minus 2pi. And there you have it. We figured out the work that this field did on the particle, counterclockwise fashion. And our intuition held up. We actually got a negative number for the work done. And that's because, at all times, the field was actually going exactly opposite, or was actually opposing, the movement of, if we think of it as a particle in its counterclockwise direction. Anyway, hopefully, you found that helpful."
    },
    {
        "Q": "At 1:11, Sal says that the figure he is drawing is a rectangular pyramid. Isn't that a square pyramid?\nThe base side lengths seem equivalent...",
        "A": "Rishi@ a rectangle can look like a square because some rectangles are usually very similar looking to them",
        "video_name": "ZACf9EecFrY",
        "timestamps": [
            71
        ],
        "3min_transcript": "What we're going to explore in this video are polyhedra, which is just the plural of a polyhedron. And a polyhedron is a three-dimensional shape that has flat surfaces and straight edges. So, for example, a cube is a polyhedron. All the surfaces are flat, and all of the edges are straight. So this right over here is a polyhedron. Once again, polyhedra is plural. Polyhedron is when you have one of them. This is a polyhedron. A rectangular pyramid is a polyhedron. So let me draw that. I'll make this one a little bit more transparent. Let me do this in a different color just for fun. I'll make it a magenta rectangular pyramid. So once again, here I have one flat surface. And then I'm going to have four triangular flat surfaces. Now, it clearly looks like a pyramid. Why is it called a rectangular pyramid? Because the base right over here is a rectangle. So these are just a few examples of polyhedra. Now, what I want to think about are nets of polyhedra. And actually, let me draw and make this transparent, too, so we get full appreciation of the entire polyhedron, this entire cube. So now let's think about nets of polyhedron. So what is a net of a polyhedron? Well, one way to think about it is if you kind of viewed this as made up of cardboard, and you were to unfold it in some way so it would become flat, or another way of thinking about it is if you were to cut out some cardboard or some paper, and you wanted to fold it up into one of these figures, how would you go about doing it? And each of these polyhedra has multiple different nets that you could create so that it can be folded up into this three-dimensional figure. So let's take an example. And I'm going to color code it. So let's say that the bottom of this cube was this green color. And so I can represent it like this. That's the bottom of the cube. It's that green color. Now, let's say that this back surface of the cube is orange. Well, I could represent it like this. And notice, I've kind of folded it out. I'm folding it out. And so if I were to flatten it out, it would look like this. It would look like that. Now, this other backside, I'll shade it in yellow. This other backside right over here, I could fold it backwards and keep it connected along this edge, fold it backwards. It would look like this. It would look like that. I think you get the general idea here. And just to be clear, this edge right over here is this edge right over there."
    },
    {
        "Q": "At 20:00, why does a derivative of 0 mean a maximum/minimum? I get why, but I thought y=x^2 has only one minimum at x=0.",
        "A": "Yes and if you let d/dx x^2 = 2x =0, you get x = 0 (and y = 0).",
        "video_name": "viaPc8zDcRI",
        "timestamps": [
            1200
        ],
        "3min_transcript": "Because the problem wants us to find that point, the maximum point of intersection. They call this the extreme normal line. The extreme normal line is when our second quadrant intersection essentially achieves a maximum point. I know they call it the smallest point, but it's the smallest negative value, so it's really a maximum point. So how do we figure out that maximum point? Well, we have our second quadrant intersection as a function of our first quadrant x. I could rewrite this as, my second quadrant intersection as a function of x0 is equal to minus x minus 1 over 2 x0. So this is going to reach a minimum or a maximum point when its derivative is equal to 0. This is a very unconventional notation, and that's probably the hardest thing about this problem. But let's take this derivative with respect to x0. So my second quadrant intersection, the derivative of straightforward. It's equal to minus 1, and then I have a minus 1/2 times, this is the same thing as x to the minus 1. So it's minus 1 times x0 to the minus 2, right? I could have rewritten this as minus 1/2 times x0 to the minus 1. So you just put its exponent out front and decrement it by 1. And so this is the derivative with respect to my first quadrant intersection. So let me simplify this. So x, my second quadrant intersection, the derivative of it with respect to my first quadrant intersection, is equal to minus 1, the minus 1/2 and the minus 1 become a positive when you multiply them, and so plus 1/2 over x0 squared. Now, this'll reach a maximum or minimum when it equals 0. problem right there. Well, we add one to both sides. We get 1 over 2 x0 squared is equal to 1, or you could just say that that means that 2 x0 squared must be equal to 1, if we just invert both sides of this equation. Or we could say that x0 squared is equal to 1/2, or if we take the square roots of both sides of that equation, we get x0 is equal to 1 over the square root of 2. So we're really, really, really close now. We've just figured out the x0 value that gives us our extreme normal line. This value right here. Let me do it in a nice deeper color. This value right here, that gives us the extreme normal line, that over there is x0 is equal to 1 over the square root of 2. Now, they want us to figure out the equation of the"
    },
    {
        "Q": "I think there's a mistake right 12:56 when he starts out factoring, or fastforward to 13:46.\nHe's just factoring the equation inside the square root. 1/4xo^2 + (2+4xo^2); o being sub 0, so xo is x sub 0.\nHe factors out 4/xo^2 and gets 4/xo^2(xo^4 + 1/2xo^2 + 1/16)\n\nIn the first part, 4/xo^2 and 1/4xo^2, the 4's cancel and we should be left with 1/xo^4 or xo^-4. Sal just put it as a positive x^4. I might be nitpicking, but I'm just making sure I'm right as there are no annotations correcting this.",
        "A": "Yes I do, thank you! It s always the algebra that can be confusing.",
        "video_name": "viaPc8zDcRI",
        "timestamps": [
            776,
            826
        ],
        "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": "at 13:20 how did you decide to factor out 4/x^2. It seems like a wild goose chase to me.",
        "A": "If you notice that a polynomial has degrees such as -2, then 0, then 2, and you want to turn it into 0, 2, and 4 to make it easier to factor, you can multiply by the second degree, but must make sure to also divide by the second degree to keep it equivalent. Dividing by the second degree is the same as factoring out the reciprocal of the second degree. Example, a = (1/1)a = (n/n)a = (1/n)(n)a = (1/n)(na). I don t know if this helps.",
        "video_name": "viaPc8zDcRI",
        "timestamps": [
            800
        ],
        "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": "At 18:45 How does khan figure out that when derivative of -x_0-1/2x_0 is equal to 0 is actually minimum or maximum of the function?",
        "A": "derivative = 0 ==> (implies) a maximum or minimum",
        "video_name": "viaPc8zDcRI",
        "timestamps": [
            1125
        ],
        "3min_transcript": "So minus x0 minus 1 over 4 mine x0. Now what do we have? So let's see. We have a minus 1 over 4 x0, minus 1 over 4 x0. So this is equal to minus x0, minus x0, minus 1 over 2 x0. So if I take minus 1/4 minus 1/4, I get minus 1/2. And so my second quadrant intersection, all this work I did got me this result. My second quadrant intersection, I hope I don't run out of space. My second quadrant intersection, of the normal line and the parabola, is minus x0 minus 1 over 2 x0. Now this by itself is a pretty neat result we just got, but Because the problem wants us to find that point, the maximum point of intersection. They call this the extreme normal line. The extreme normal line is when our second quadrant intersection essentially achieves a maximum point. I know they call it the smallest point, but it's the smallest negative value, so it's really a maximum point. So how do we figure out that maximum point? Well, we have our second quadrant intersection as a function of our first quadrant x. I could rewrite this as, my second quadrant intersection as a function of x0 is equal to minus x minus 1 over 2 x0. So this is going to reach a minimum or a maximum point when its derivative is equal to 0. This is a very unconventional notation, and that's probably the hardest thing about this problem. But let's take this derivative with respect to x0. So my second quadrant intersection, the derivative of straightforward. It's equal to minus 1, and then I have a minus 1/2 times, this is the same thing as x to the minus 1. So it's minus 1 times x0 to the minus 2, right? I could have rewritten this as minus 1/2 times x0 to the minus 1. So you just put its exponent out front and decrement it by 1. And so this is the derivative with respect to my first quadrant intersection. So let me simplify this. So x, my second quadrant intersection, the derivative of it with respect to my first quadrant intersection, is equal to minus 1, the minus 1/2 and the minus 1 become a positive when you multiply them, and so plus 1/2 over x0 squared. Now, this'll reach a maximum or minimum when it equals 0."
    },
    {
        "Q": "At 14:03, I don't understand what Sal does to get the expression 4/x0(x0^4+1/2(x0^2)+1/16) under the squareroot. It seems to me that he divides the whole expression by 4, but I still can't make sense out of it...",
        "A": "He s factoring out 4/(x0)^2 from the three terms. Unfortunately, he turns around the expression so that in the factored form the first and third terms are exchanged, which might be contributing to your confusion (I did a double take as well!). If you try factoring out 4/(x0)^2 from the three terms but keep them in the same order, you end up with: 4/(x0)^2 [1/16 + (1/2)(x0)^2 + (x0)^4]. You can see it s the same as what he has. Hope that helps!",
        "video_name": "viaPc8zDcRI",
        "timestamps": [
            843
        ],
        "3min_transcript": "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? parabola is equal to this. Minus 1 over 4 x0 plus or minus 1/2 times the square root of this business. And the square root, this thing right here is 4 over x0 squared. This is actually, lucky for us, a perfect square. And I won't go into details, because then the video will get too long, but I think you can recognize that this is x0 squared, plus 1/4. If you don't believe me, square this thing right here. You'll get this expression right there. And luckily enough, this is a perfect square, so we can actually take the square root of it. And so we get, the point at which they intersect, our normal line and our parabola, and this is quite a hairy problem. The points where they intersect is minus 1 over 4 x0, plus or minus 1/2 times the square root of this. The square root of this is the square root of this, which is just 2 over x0 times the square root of this, which is"
    },
    {
        "Q": "At 13:00 Sal begins to factor the expression under the radical sign. This factorization would not have occurred to me. Can anyone comment on the insight that leads to this?",
        "A": "First put everything over a common denominator- 4xo^2. I think it becomes much clearer if you do this intermediary step. Impressive nevertheless!",
        "video_name": "viaPc8zDcRI",
        "timestamps": [
            780
        ],
        "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": "In the end part of the video, at 7:02, can the -90 degrees be written as 270 degrees? or should it really be -90 degrees?",
        "A": "In most cases, they are equivalent. -90 degrees = 270 degrees. There are some contexts when they may not be equivalent. E.g., suppose you are flying a plane, heading North. A -90 degree turn might correspond to a 90 degree LEFT turn (counter-clockwise), to the West. A 270 degree turn might correspond to a 270 degree RIGHT turn (clockwise). You still end up facing the same direction (due West), but you ve taken two very different actions to get there.",
        "video_name": "z8vj8tUCkxY",
        "timestamps": [
            422
        ],
        "3min_transcript": "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 Which is equal to what? 45 is half of 90 which is half of 180 so this is 1/4 this is equal to pi over 4 radians. Lets do 1 more over here. So lets say that we had negative pi/2 radians. What's that going to be in degrees? Well once again we have to figure out how many degrees are each of these radians. We know that there 180 degrees for every pi radians so we're gonna get the radians cancel out the pi's cancel out and so you have -180/2 this is -90 degrees or we can write it as -90 degrees. and I'll do a couple more example problems here because the more example for this the better and hopefully it will become a little bit intuitive"
    },
    {
        "Q": "How come at 9:45 he says that if you integrate it you get the same answer? Wouldn't you really get u(x)=x+C not u(x)=x? What happened to the constant of integration?",
        "A": "You are allowed to add a constant, but given a constant you have infinitely many solutions so in general we simply take 0 as the constant. And 0 you can easily leave out.",
        "video_name": "j511hg7Hlbg",
        "timestamps": [
            585
        ],
        "3min_transcript": "of writing it mu prime of x, we could write that as d mu dx. So let's do that. So we could write mu of x is equal to d, the derivative of mu with respect to x, times x. And this is actually a separable differential equation in and of itself. It's kind of a sub-differential equation to solve our broader one. We're just trying to figure out the integrating factor right here. So let's divide both sides by x. So we get mu over x, this is just a separable equation now, is equal to d mu dx. And then, let's divide both sides by mu of x, and we get 1 over x is equal to 1 over mu. That's mu of x, I'll just write 1 over mu right now, for simplicity, times d mu dx. Multiply both sides by dx, you get 1 over x dx is equal to 1 over mu of x d mu. Now, you could integrate both sides of this, and you'll get the natural log of the absolute value of x is equal to the natural log of the absolute value of mu, et cetera, et cetera. But it should be pretty clear from this that x is equal to mu, or mu is equal to x, right? If you look at both sides of this equation there, you can just change x for mu, and it becomes the other side. So, this is obviously telling us that mu of x is equal to x. Or mu is equal to x. So we have our integrating factor. And if you want, you can take the antiderivative of both sides with the natural logs, and all of that. And you'll get the same answer. But this is just, by looking at it, by inspection, you know that mu is equal to x. Because both sides of this equation are completely the same. Anyway, we now have our integrating factor. So in the next video, we're now going to use this integrating factor. Multiply it times our original differential equation. Make it exact. And then solve it as an exact equation. I'll see you in the next video."
    },
    {
        "Q": "@ 1:25 how do you get 9^2",
        "A": "Sal gets the 9^2 because he originally has 9^(t/2 +2) Using the exponent rule: x^a *x^b =X^ab so you get 9^t/2 * 9^2",
        "video_name": "Y6wNiYcuCoE",
        "timestamps": [
            85
        ],
        "3min_transcript": "- [Voiceover] What I hope to do in this video is get some practice simplifying some fairly hairy exponential expressions. So let's get started. Let's say that I have the expression 10 times nine to the t over two plus two power, times five to the three t. And what I wanna do is simplify this as much as possible, and preferably get it in the form of A times B to the t. And like always, I encourage you to pause this video and see if you can do this on your own using exponent properties, your knowledge, your deep knowledge of exponent properties. All right, so let's work through this together, and it's really just about breaking the pieces up. So 10, I'll just leave that as 10 for now, there doesn't seem to be much to do there. But there's all sorts of interesting things going on here. So nine to the t over two, plus two, so this right over here, I could break this up, write the properties over here. If I have nine to the a plus b power, this is the same thing as nine to the a, times nine to the b power. And over here I have nine to the t over two, plus two, so I could rewrite this as nine to the t over two power, times nine squared. All right, now let's move over to five to the three t. Well, if I have a to the bc, so you could view this as five to the three times t, this is the same thing as a to the b, and then that to the c power. So I could write this as, this is going to be the same thing as five to the third, and then that to the t power. And the whole reason I did that is well this is just going to be a number, then I'm going to have some number to the t power. I want to get as many things just raised just to see if I can simplify this thing. So this character right over here is going to be 81. Nine squared is 81. Five to the third power, 25 times five, that's 125. So we're making good progress and so the only thing we really have to simplify at this point is nine to the t over two. And actually let me do that over here. Nine to the t over two. Well that's the same thing as nine to the one half times t. And by this property right over here, that's the same thing as nine to the one, nine to the one half... And then that to the t power. So what's nine to the one half? Well that's three, so this is going to be equal to three to the t power. So this right over here is three to the t power."
    },
    {
        "Q": "I like the parentheses at 1:12 rather than multiplying through by -1. Suppose some people might prefer to multiply, though? Seems messier than necessary.",
        "A": "It s kind of the same thing.",
        "video_name": "8Wxw9bpKEGQ",
        "timestamps": [
            72
        ],
        "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": "At 0:21, you mention \"Fibonacci numbers\"; 1,1,2,5,8,13,21,34,...; what are those? And why are they called that?",
        "A": "In 1175 A.D., the man who invented the Fibonacci Numbers, Fibonacci, was born. Throughout his life he encountered many circumstances where these numbers appear (like in the spirals of a pineapple or pinecone). This pattern frequently shows up in nature and is extremely fun to find out where it exactly fits in.",
        "video_name": "gBxeju8dMho",
        "timestamps": [
            21
        ],
        "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 1:30, he said that you would multiply 1 by one half if multiplying by a negative exponent. Am I correct in the sense that if 3^3 is 27, 3^-3 power would be 1/27? Thank you!\n\n-KoKo",
        "A": "Yes... 3^(-3) = 1/27. Great job!",
        "video_name": "JnpqlXN9Whw",
        "timestamps": [
            90
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 2:04, could you multiply with a decimal? I mean, fractions and decimals are the same things, right?",
        "A": "I don t understand your question properly but let say (5.5)^-1 = 1 / 5.5 let say (1/1.3)^-1 = 1.3",
        "video_name": "JnpqlXN9Whw",
        "timestamps": [
            124
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 6:25, why was 1/(25/64) changed to it's reciprocal, 64/25?",
        "A": "You actually don t have to change the fraction 5/8 to 1/(25/64) or 64/25. Sal states you could just change 5/8 to 8/5 and raise the negative exponent to a positive one. He did this to get the right answer.",
        "video_name": "JnpqlXN9Whw",
        "timestamps": [
            385
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 6:41, why does he switch 25/64 to 64/25 ?",
        "A": "Multiplication is the inverse operation of division. Instead of dividing by 25/64, you can multiply by 64/25. It is easier to multiply fractions.",
        "video_name": "JnpqlXN9Whw",
        "timestamps": [
            401
        ],
        "3min_transcript": ""
    },
    {
        "Q": "Did he mis-speak at 0:45-48 seconds? He said to view the top example as multiplying 2 by \"negative 1\"",
        "A": "Yes that was a mistake. He meant times 1. :)",
        "video_name": "JnpqlXN9Whw",
        "timestamps": [
            45
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 6:24, how did we get 64/25?",
        "A": "Sal has: 1 divided by 25/64 Do the division -- change division to multiply by using the reciprocal 1 * 64/25 = 64/25 Hope this helps.",
        "video_name": "JnpqlXN9Whw",
        "timestamps": [
            384
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 4:51 why does it change to a positive 3?",
        "A": "Because x^-y =1/(x^y). If you don t understand the logic behind that, I can give a more detailed explanation.",
        "video_name": "JnpqlXN9Whw",
        "timestamps": [
            291
        ],
        "3min_transcript": ""
    },
    {
        "Q": "Why are vectors represented as column form and not row form ? For eg. at 4:00 of this video why wasn't vector V represented as [5,0] in row form.",
        "A": "It is simply a convention that most people agree upon, many authors of books write row vectors instead of column vectors. The advantage of the column vector is that it is easy to see how to do matrix multiplication. Further on in linear algebra, one learns about the transpose map and it s relation to functions that operate on vectors to give scalars (linear functionals). Then it becomes a bigger deal whether or not your vector is column or row. For now try not to worry about it too much.",
        "video_name": "br7tS1t2SFE",
        "timestamps": [
            240
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "5:47 if the magnitude called \"scalar\" on its own, what is the direction on its own?",
        "A": "Scalar has no direction.",
        "video_name": "br7tS1t2SFE",
        "timestamps": [
            347
        ],
        "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": "Hi at 1:04 you indicated that speed and direction is velocity, but isn't velocity = change in speed regardless of direction?",
        "A": "Velocity is the change in position with regard to direction, Speed is the change in position without regard to direction. A change in either speed or velocity is acceleration.",
        "video_name": "br7tS1t2SFE",
        "timestamps": [
            64
        ],
        "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": "On 0:04 seconds why does he say number A instead of letter A?",
        "A": "In algebra, a letter is a variable that represents a number. When Sal says letter a , a is representing its place on the number line. Hope it helps!",
        "video_name": "29P6bar7nHc",
        "timestamps": [
            4
        ],
        "3min_transcript": "- [Voiceover] What I have here are three numbers plotted on the number line. We have the number a, the number c, the number b. And then we have three -- (laughs) we have four inequalities, actually. Four inequalities that involve absolute value. And what I want to do is figure out which is these inequalities are true, given where a, c and b are on the number line. And I encourage you to pause the video and try to think through it on your own. All right, let's look at this first one. It says that, \"a is less than b.\" So if we look at a and we look at b, a is clearly to the left of b on the number line. So we know that this is true. Even more we know that a is negative, it's to the left of zero, while b is positive. Which is, if one thing is negative and the other thing is positive, the negative thing is definitely going to be less than the positive thing. But even easier than that, a is to the left of b on the number line. If you're to the left of something else on the number line you're less than that other thing. at least the way we've constructed it, it increases from left to right. All right, the next statement, \"The absolute value of a is greater than the absolute value of b.\" Well, let's just think about where these are on the number line. So we've already said a is three hash marks to the left of zero. That is a. So what is going to be the absolute value of a? Well, the absolute value of a is the distance that a is from zero. So the distance that a is from zero is one, two, three hash marks. So the absolute value of a is just going to be that same distance on the positive side. So the point that we marked as c is also the absolute value of a. So that is also the absolute value of a. The absolute value of a -- sorry, a is three to the left of zero. Absolute value of a is going to be three to the right. It's just a measure of, how many hash marks is it from zero? Well, it's three hash marks from zero So is the absolute value of a greater than the absolute value of b? Or what's the absolute value of b? Well, b is one, two, three, four, five, six, seven, eight hash marks to the right of zero. And so the absolute value of b is going to be on the eighth hash mark. Because it's eight hash marks to the right. So this is also the absolute value of b. And this is consistent with what we've learned about absolute value. Absolute value of a positive number is just going to be that number again. Absolute value of a negative number is going to be the opposite of that number. And absolute value of zero is just going to be zero. So is the absolute value of a greater than the absolute value of b? Well, no. Absolute value of a is to the left of the absolute value of b on our number line. It is less than the absolute value of b. So this is not true. All right, next statement. \"Absolute value of a is less than the absolute value of c.\""
    },
    {
        "Q": "at 9:25, why would there be a negative sign?",
        "A": "he put a negative sign there because he was trying to subtract the two equations. to do that one of the equations had to be negative. so he multiplied the second equation by -1 to make that possible. so the equation went from:::: 3x+y= 1.79 to:::: -3x-y= -1/79 Hope This Helps You Undertsand",
        "video_name": "vA-55wZtLeE",
        "timestamps": [
            565
        ],
        "3min_transcript": "And 4 Fruit Roll-Ups. Plus 4 times y, the cost of a Fruit Roll-Up. This is how much Nadia spends. 3 candy bars, 4 Fruit Roll-Ups. And it's going to cost $2.84. That's what this first statement tells us. It translates into that equation. The second statement. Peter also buys 3 candy bars, but could only afford 1 additional Fruit Roll-Up. So plus 1 additional Fruit Roll-Up. His purchase cost is equal to $1.79. What is the cost of each candy bar and each Fruit Roll-Up? And we're going to solve this using elimination. You could solve this using any of the techniques we've seen so far-- substitution, elimination, even graphing, although it's kind of hard to eyeball things with the graphing. So how can we do this? Remember, with elimination, you're going to add-- let's Is there something we could add to both sides of this equation that'll help us eliminate one of the variables? Or let me put it this way, is there something we could add or subtract to both sides of this equation that will help us eliminate one of the variables? Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? And remember, by doing that, I would be subtracting the same thing from both sides of the equation. This is $1.79. Because it says this is equal to $1.79. So if we did that we would be subtracting the same thing from both sides of the equation. So let's subtract 3x plus y from the left-hand side of the equation. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So let's subtract it. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. I'm just taking the second equation. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. And what do we get? When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out. 3x minus 3x is 0x. I won't even write it down. You get 4x minus-- sorry, 4y minus y. That is 3y. And that is going to be equal to $2.84 minus $1.79."
    },
    {
        "Q": "At 9:04, Sal multiplies the bottom equation by -1. He could have divided by -1, right?",
        "A": "Same thing, dividing or multiplying by -1 will give you same equation.",
        "video_name": "vA-55wZtLeE",
        "timestamps": [
            544
        ],
        "3min_transcript": "And 4 Fruit Roll-Ups. Plus 4 times y, the cost of a Fruit Roll-Up. This is how much Nadia spends. 3 candy bars, 4 Fruit Roll-Ups. And it's going to cost $2.84. That's what this first statement tells us. It translates into that equation. The second statement. Peter also buys 3 candy bars, but could only afford 1 additional Fruit Roll-Up. So plus 1 additional Fruit Roll-Up. His purchase cost is equal to $1.79. What is the cost of each candy bar and each Fruit Roll-Up? And we're going to solve this using elimination. You could solve this using any of the techniques we've seen so far-- substitution, elimination, even graphing, although it's kind of hard to eyeball things with the graphing. So how can we do this? Remember, with elimination, you're going to add-- let's Is there something we could add to both sides of this equation that'll help us eliminate one of the variables? Or let me put it this way, is there something we could add or subtract to both sides of this equation that will help us eliminate one of the variables? Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? And remember, by doing that, I would be subtracting the same thing from both sides of the equation. This is $1.79. Because it says this is equal to $1.79. So if we did that we would be subtracting the same thing from both sides of the equation. So let's subtract 3x plus y from the left-hand side of the equation. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So let's subtract it. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. I'm just taking the second equation. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. And what do we get? When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out. 3x minus 3x is 0x. I won't even write it down. You get 4x minus-- sorry, 4y minus y. That is 3y. And that is going to be equal to $2.84 minus $1.79."
    },
    {
        "Q": "I have trouble following Sal's explanation from ca. 4:00 to ca. 5:30. d/dx theta = sec^2 theta, but what's the logic behind d(theta)/dt? Did I miss a video demonstrating (or exercises testing how to) differentiate both sides w/ respect to \"something else\". Also, there is no detailed discussion of how/why it is permissible to cancel out terms (i.e. the d(theta)s in this example). Additional chain-rule exercises leading up to these problems would be appreciated.",
        "A": "it is not d/dx(theta) it is d(tan theta)/dt = d(tan theta)/d(theta) * d(theta)/dt =sec^2 theta*d(theta)/dt that is how you get the final expression in terms of sec^2theta",
        "video_name": "_kbd6troMgA",
        "timestamps": [
            240,
            330
        ],
        "3min_transcript": "Well, we're trying to figure out the rate at which the height of the balloon is changing. So if you call this distance right over here h, what we want to figure out is dh dt. That's what we don't know. So what we'd want to come up with is a relationship between dh dt, d theta dt, and maybe theta, if we need it. Or another way to think about it, if we can come up with the relationship between h and theta, then we could take the derivative with respect to t, and we'll probably get a relationship between all of this stuff. So what's the relationship between theta and h? Well, it's a little bit of trigonometry. We know we're trying to figure out h. We already know what this length is right over here. We know opposite over adjacent. That's the definition of tangent. So let's write that down. is equal to the opposite side-- the opposite side is equal to h-- over the adjacent side, which we know is going to be a fixed 500. So there you have a relationship between theta and h. And then to figure out a relationship between d theta dt, the rate at which theta changes with respect to t, and the rate at which h changes with respect to t, we just have to take the derivative of both sides of this with respect to t implicitly. So let's do that. And actually, let me move over this h over 500 a little bit. So let me move it over a little bit so I have space to show the derivative operator. So let's write it like that. And now let's take the derivative with respect to t. So d dt. I'm going to take the derivative with respect to t on the left. We're going to take the derivative with respect So what's the derivative with respect to t of tangent of theta? Well, we're just going to apply the chain rule here. It's going to be first the derivative of the tangent of theta with respect to theta, which is just secant squared of theta, times the derivative of theta with respect to t, times d theta dt. Once again, this is just the derivative of the tangent, the tangent of something with respect to that something times the derivative of the something with respect to t. Derivative of tangent theta with respect to theta times the derivative theta with respect to t gives us the derivative of tangent of theta with respect to t, which is what we want when we use this type of a derivative operator. We're taking the derivative with respect to t, not just the derivative with respect to theta. So this is the left hand side."
    },
    {
        "Q": "At 5:20, you take the derivative of h/500. Aren't you supposed to use the quotient rule for that?",
        "A": "You don t have a variable in the denominator. That 500 m is a constant. So you would treat it as ( \u00c2\u00b9\u00e2\u0081\u0084\u00e2\u0082\u0085\u00e2\u0082\u0080\u00e2\u0082\u0080 ) h",
        "video_name": "_kbd6troMgA",
        "timestamps": [
            320
        ],
        "3min_transcript": "is equal to the opposite side-- the opposite side is equal to h-- over the adjacent side, which we know is going to be a fixed 500. So there you have a relationship between theta and h. And then to figure out a relationship between d theta dt, the rate at which theta changes with respect to t, and the rate at which h changes with respect to t, we just have to take the derivative of both sides of this with respect to t implicitly. So let's do that. And actually, let me move over this h over 500 a little bit. So let me move it over a little bit so I have space to show the derivative operator. So let's write it like that. And now let's take the derivative with respect to t. So d dt. I'm going to take the derivative with respect to t on the left. We're going to take the derivative with respect So what's the derivative with respect to t of tangent of theta? Well, we're just going to apply the chain rule here. It's going to be first the derivative of the tangent of theta with respect to theta, which is just secant squared of theta, times the derivative of theta with respect to t, times d theta dt. Once again, this is just the derivative of the tangent, the tangent of something with respect to that something times the derivative of the something with respect to t. Derivative of tangent theta with respect to theta times the derivative theta with respect to t gives us the derivative of tangent of theta with respect to t, which is what we want when we use this type of a derivative operator. We're taking the derivative with respect to t, not just the derivative with respect to theta. So this is the left hand side. it's just going to be 1 over 500 dh dt. So 1 over 500 dh dt. We're literally saying it's just 1 over 500 times the derivative of h with respect to t. But now we have our relationship. We have the relationship that we actually care about. We have a relationship between the rate at which the height is changing with respect to time and the rate at which the angle is changing with respect to time and our angle at any moment. So we can just take these values up here, throw it in here, and then solve for the unknown. So let's do that. Let's do that right over here. So we get secant squared of theta. So we get secant squared. Right now our theta is pi over 4. Secant squared of pi over 4. Let me write those colors in to show you that I'm putting these values in. Secant squared of pi over 4. Secant times d theta dt."
    },
    {
        "Q": "At 4:48, when Sal was writing \"m=2, (-7,5)\" into an equation, why isn't it \"5-b = 2(-7-a)\"? why does Sal replace b with 5, not y with b?",
        "A": "In Sal s format of point-slope, you swap out a for the x-value and b for the y-value. You will usually see these written as Xsub1 and Ysub1 where (Xsub1, Ysub1) is the ordered pair. Sal chose to use (a, b) for the ordered pair. The X and Y in the formula stay as X and Y. They become the variables in the equation.",
        "video_name": "K_OI9LA54AA",
        "timestamps": [
            288
        ],
        "3min_transcript": "And then on the right-hand side, you just have m times x minus a. So this whole thing has simplified to y minus b is equal to m times x minus a. And right here, this is a form that people, that mathematicians, have categorized as point-slope form. So this right over here is the point-slope form of the equation that describes this line. Now, why is it called point-slope form? Well, it's very easy to inspect this and say, OK. Well look, this is the slope of the line in green. That's the slope of the line. And I can put the two points in. If the point a, b is on this line, I'll have the slope times x minus a is equal to y minus b. Now, let's see why this is useful or why people like to use this type of thing. Let's not use just a, b and a slope of m anymore. Let's say that someone tells you that I'm dealing with some line where the slope is equal to 2, and let's say it goes through the point negative 7, 5. So very quickly, you could use this information and your knowledge of point-slope form to write this in this form. You would just say, well, an equation that contains this point and has this slope would be y minus b, which is 5-- y minus the y-coordinate of the point that this line contains-- is equal to my slope times x minus the x-coordinate that this line contains. So x minus negative 7. And just like that, we have written an equation that this point right over here. And if we don't like the x minus negative 7 right over here, we could obviously rewrite that as x plus 7. But this is kind of the purest point-slope form. If you want to simplify it a little bit, you could write it as y minus 5 is equal to 2 times x plus 7. And if you want to see that this is just one way of expressing the equation of this line-- there are many others, and the one that we're most familiar with is y-intercept form-- this can easily be converted to y-intercept form. To do that, we just have to distribute this 2. So we get y minus 5 is equal to 2 times x plus 2 times 7, so that's equal to 14. And then we can get rid of this negative 5 on the left by adding 5 to both sides of this equation. And then we are left with, on the left-hand side, y and, on the right-hand side, 2x plus 19. So this right over here is slope-intercept form. You have your slope and your y-intercept. So this is slope-intercept form."
    },
    {
        "Q": "I've heard slope-intercept form is y=mx+b. At 6:05, how would that fit in?",
        "A": "If you distribute across the parenthesis, and then add/subtract the term on the y side (from both), you should get an answer in slope-intercept form.",
        "video_name": "K_OI9LA54AA",
        "timestamps": [
            365
        ],
        "3min_transcript": "Let's say that someone tells you that I'm dealing with some line where the slope is equal to 2, and let's say it goes through the point negative 7, 5. So very quickly, you could use this information and your knowledge of point-slope form to write this in this form. You would just say, well, an equation that contains this point and has this slope would be y minus b, which is 5-- y minus the y-coordinate of the point that this line contains-- is equal to my slope times x minus the x-coordinate that this line contains. So x minus negative 7. And just like that, we have written an equation that this point right over here. And if we don't like the x minus negative 7 right over here, we could obviously rewrite that as x plus 7. But this is kind of the purest point-slope form. If you want to simplify it a little bit, you could write it as y minus 5 is equal to 2 times x plus 7. And if you want to see that this is just one way of expressing the equation of this line-- there are many others, and the one that we're most familiar with is y-intercept form-- this can easily be converted to y-intercept form. To do that, we just have to distribute this 2. So we get y minus 5 is equal to 2 times x plus 2 times 7, so that's equal to 14. And then we can get rid of this negative 5 on the left by adding 5 to both sides of this equation. And then we are left with, on the left-hand side, y and, on the right-hand side, 2x plus 19. So this right over here is slope-intercept form. You have your slope and your y-intercept. So this is slope-intercept form."
    },
    {
        "Q": "At 1:00 in the video, what do the triangles mean when he is talking about what the slope equals?",
        "A": "The triangles are supposed to mean The Change in , so it is The Change in Y/ The Change in X , also known as slope, or known as in Slope-intercept form, m.",
        "video_name": "K_OI9LA54AA",
        "timestamps": [
            60
        ],
        "3min_transcript": "So what I've drawn here in yellow is a line. And let's say we know two things about this line. We know that it has a slope of m, and we know that the point a, b is on this line. And so the question that we're going to try to answer is, can we easily come up with an equation for this line using this information? Well, let's try it out. So any point on this line, or any x, y on this line, would have to satisfy the condition that the slope between that point-- so let's say that this is some point x, y. It's an arbitrary point on the line-- the fact that it's on the line tells us that the slope between a, b and x, y must be equal to m. So let's use that knowledge to actually construct an equation. So what is the slope between a, b and x, y? Well, our change in y-- remember slope is just change in y over change in x. Let me write that. Slope is equal to change in y over change in x. This little triangle character, that's the Greek letter Delta, Our change in y-- well let's see. If we start at y is equal to b, and if we end up at y equals this arbitrary y right over here, this change in y right over here is going to be y minus b. Let me write it in those same colors. So this is going to be y minus my little orange b. And that's going to be over our change in x. And the exact same logic-- we start at x equals a. We finish at x equals this arbitrary x, whatever x we happen to be at. So that change in x is going to be that ending point minus our starting point-- minus a. And we know this is the slope between these two points. That's the slope between any two points on this line. And that's going to be equal to m. So this is going to be equal to m. And so what we've already done here is actually create an equation that describes this line. It might not be in any form that you're used to seeing, any x, y that satisfies this equation right over here will be on the line because any x, y that satisfies this, the slope between that x, y and this point right over here, between the point a, b, is going to be equal to m. So let's actually now convert this into forms that we might recognize more easily. So let me paste that. So to simplify this expression a little bit, or at least to get rid of the x minus a in the denominator, let's multiply both sides by x minus a. So if we multiply both sides by x minus a-- so x minus a on the left-hand side and x minus a on the right. Let me put some parentheses around it. So we're going to multiply both sides by x minus a. The whole point of that is you have x minus a divided by x"
    },
    {
        "Q": "At 1:20 or so why does he use 2 instead of 1 as the smallest number",
        "A": "Starting at 2:03, he notices that he missed the 1 s, and fixes it.",
        "video_name": "09Cx7xuIXig",
        "timestamps": [
            80
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 1:52, why did Sal add 14 and 9? Shouldn't he subtract?",
        "A": "He distributed the negative sign for the second polynominal so -9 become +9. If you don t distribute then you would have to do this way, 14 - (-9) which equals to 14+9",
        "video_name": "5ZdxnFspyP8",
        "timestamps": [
            112
        ],
        "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 5:20, couldn't the cube root of a^2b^2 be simplified to a^2/3 b^5/3?",
        "A": "Yes, you can use fractional notation as well, in fact I prefer fractional notation in these situations. Sal is just using a different approach. Both are correct. a^2/3 times b^2/3 is not more simplified than the cube root of a^2 times b^2, it is just different. I hope that helps!",
        "video_name": "c-wtvEdEoVs",
        "timestamps": [
            320
        ],
        "3min_transcript": "to the third times the cube root-- and I'll just group these two guys together just because we're not going to be able to simplify it any more-- times the cube root of a squared b squared. I'll keep the colors consistent while we're trying to figure out what's what. And I could have said that this is times the cube root of a squared times the cube root of b squared, but that won't simplify anything, so I'll just leave these like this. And so we can look at these individually. The cube root of 3 to the third, or the cube root of 27-- well, that's clearly just going to be-- I want to do that in that yellow color-- this is clearly just going to be 3. 3 to the third power is 3 to the third power, or it's equal to 27. This term right over here, the cube root of b to the third-- well, that's just b. And the cube root of c to the third, well, that is clearly-- I want to do So our whole expression has simplified to 3 times b times c times the cube root of a squared b squared. And we're done. And I just want to do one other thing, just because I did mention that I would do it. We could simplify it this way. Or we could recognize that this expression right over here can be written as 3bc to the third power. And if I take three things to the third power, and I'm multiplying it, that's the same thing as multiplying them first and then raising to the third power. It comes straight out of our exponent properties. And so we can rewrite this as the cube root of all of this times the cube root of a squared b And so the cube root of all of this, of 3bc to the third power, well, that's just going to be 3bc, and then multiplied by the cube root of a squared b squared. I didn't take the trouble to color-code it this time, because we already figured out one way to solve it. But hopefully, that also makes sense. We could have done this either way. But the important thing is that we get that same answer."
    },
    {
        "Q": "I dont understand 1:20, why dont you make a^2 a*a and you make b^5, b^2 and b^3",
        "A": "Because he explained that b^5 is not a perfect cube so he had to divide it, thats my explanation to it.",
        "video_name": "c-wtvEdEoVs",
        "timestamps": [
            80
        ],
        "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 1:02 Do we have to do anything else to the numerators and the denominaters other than divide?",
        "A": "No, you are only just meant to divide at that point.",
        "video_name": "2dbasvm3iG0",
        "timestamps": [
            62
        ],
        "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": "At 1:55, Sal says \"their greatest common factor is 3\" what does greatest common factor even mean?",
        "A": "The Greatest Common Factor means when you take two numbers and find a factor that both these numbers have in common. For example: What is the greatest common factor of 6 and 9? Well, the first step is to list their factors: The factors for 6 are: 1, 2, 3, and 6. The factors for 9 are: 1, 3, 3, 9. Now what number is the same in both these factor lists? 3! So the Greatest Common Factor will be 3.",
        "video_name": "Bt60JVZRVCI",
        "timestamps": [
            115
        ],
        "3min_transcript": "Let's think about what fraction of this grid is actually shaded in pink. So the first thing we want to think about is how many equal sections do we have here? Well, this is a 1, 2, 3, 4, 5 by 1, 2, 3 grid. So there's 15 sections here. You could also count it-- 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15. So there are 15 equal sections here. And how many of those equal sections are actually shaded in this kind of pinkish color? Well, We have 1, 2, 3, 4, 5, 6. So it's 6/15 is shaded in. But I want to simplify this more. I have a feeling that there's some equivalent fractions that represent the exact same thing as 6/15. And to get a sense of that, let me redraw this a little bit, where I still shade in six of these rectangles, but I'll shade them a little bit in one chunk. So let me throw in another grid right over here, and let me attempt to shade in the rectangles So that is 1-- 1 rectangle. I'll even make my thing even bigger. All right, 1 rectangle, 2 rectangles, 3 rectangles-- halfway there-- 4 rectangles, 5 rectangles shaded in and now 6 rectangles shaded in. So this right over here, what I just did, this is still 6 rectangles of the 15 rectangles So this is still 6/15. These are representing the same thing. But how can I simplify this even more? Well, when you look at it numerically, you see that both 6 and 15 are divisible by 3. In fact, their greatest common factor is 3. So what happens if we divide the numerator and denominator by 3? we're not going to be changing the value of the fraction. So let's divide the numerator by 3 and divide the denominator by 3. And what do we get? We get 2 over 5. Now how does this make sense in the context of this diagram right here? Well, we started off with 6 shaded in. You divide by 3, you have 2 shaded in. So you're essentially saying, hey, let's group these into sections of 3. So let's say that this right over here is one section of 3. This is one section of 3 right over here. So that's one section of 3. And then this is another section of 3 right over here. And so you have two sections of 3. And actually let me color it in a little bit better. So you have two sections of 3. And if you were to combine them, it looks just like this."
    },
    {
        "Q": "I'm confused at 2:10. Sal says that the sequence converges but I learned that when we calculate limits, if you have infinity over infinity, its called a \"indetermination\". Though, what Sal said makes perfectly sense...",
        "A": "Recall when we want to find the horizontal asymptote, we take the limit as x approaches infinity. The method was to multiply the numerator and denominator by 1 over the highest term of x on the denominator. Similarly, we find the limit as n approaches infinity to find out if it diverges or converges. So if we multiply by 1/n\u00c2\u00b2 for both the top and bottom, it will converge to 1.",
        "video_name": "muqyereWEh4",
        "timestamps": [
            130
        ],
        "3min_transcript": "So we've explicitly defined four different sequences here. And what I want you to think about is whether these sequences converge or diverge. And remember, converge just means, as n gets larger and larger and larger, that the value of our sequence is approaching some value. And diverge means that it's not approaching some value. So let's look at this. And I encourage you to pause this video and try this on your own before I'm about to explain it. So let's look at this first sequence right over here. So the numerator n plus 8 times n plus 1, the denominator n times n minus 10. So one way to think about what's happening as n gets larger and larger is look at the degree of the numerator and the degree of the denominator. And we care about the degree because we want to see, look, is the numerator growing faster than the denominator? In which case this thing is going to go to infinity and this thing's going to diverge. Or is maybe the denominator growing faster, in which case this might converge to 0? Or maybe they're growing at the same level, and maybe it'll converge to a different number. So let's multiply out the numerator and the denominator So n times n is n squared. n times 1 is 1n, plus 8n is 9n. And then 8 times 1 is 8. So the numerator is n squared plus 9n plus 8. The denominator is n squared minus 10n. And one way to think about it is n gets really, really, really, really, really large, what dominates in the numerator-- this term is going to represent most of the value. And this term is going to represent most of the value, as well. These other terms aren't going to grow. Obviously, this 8 doesn't grow at all. But the n terms aren't going to grow anywhere near as fast as the n squared terms, especially for large n's. So for very, very large n's, this is really going to be approaching n squared over n squared, or 1. So it's reasonable to say that this converges. So this one converges. And once again, I'm not vigorously proving it here. Or I should say I'm not rigorously proving it over here. in the numerator and the denominator. So now let's look at this one right over here. So here in the numerator I have e to the n power. And here I have e times n. So this grows much faster. I mean, this is e to the n power. Imagine if when you have this as 100, e to the 100th power is a ginormous number. e times 100-- that's just 100e. Grows much faster than this right over here. So this thing is just going to balloon. This is going to go to infinity. So we could say this diverges. Now let's look at this one right over here. Well, we have a higher degree term. We have a higher degree in the numerator than we have in the denominator. n squared, obviously, is going to grow much faster than n. So for the same reason as the b sub n sequence, this thing is going to diverge. The numerator is going to grow much faster"
    },
    {
        "Q": "Sal says at 2:00 he isn't 'rigorously proving' his answer, I'm curious, how would one go about rigorously proving the result?",
        "A": "One way is to divide the numerator and denominator both by n^2, then take the limit of the result as n approaches infinity. If you haven t seen that process yet, you will soon!",
        "video_name": "muqyereWEh4",
        "timestamps": [
            120
        ],
        "3min_transcript": "So we've explicitly defined four different sequences here. And what I want you to think about is whether these sequences converge or diverge. And remember, converge just means, as n gets larger and larger and larger, that the value of our sequence is approaching some value. And diverge means that it's not approaching some value. So let's look at this. And I encourage you to pause this video and try this on your own before I'm about to explain it. So let's look at this first sequence right over here. So the numerator n plus 8 times n plus 1, the denominator n times n minus 10. So one way to think about what's happening as n gets larger and larger is look at the degree of the numerator and the degree of the denominator. And we care about the degree because we want to see, look, is the numerator growing faster than the denominator? In which case this thing is going to go to infinity and this thing's going to diverge. Or is maybe the denominator growing faster, in which case this might converge to 0? Or maybe they're growing at the same level, and maybe it'll converge to a different number. So let's multiply out the numerator and the denominator So n times n is n squared. n times 1 is 1n, plus 8n is 9n. And then 8 times 1 is 8. So the numerator is n squared plus 9n plus 8. The denominator is n squared minus 10n. And one way to think about it is n gets really, really, really, really, really large, what dominates in the numerator-- this term is going to represent most of the value. And this term is going to represent most of the value, as well. These other terms aren't going to grow. Obviously, this 8 doesn't grow at all. But the n terms aren't going to grow anywhere near as fast as the n squared terms, especially for large n's. So for very, very large n's, this is really going to be approaching n squared over n squared, or 1. So it's reasonable to say that this converges. So this one converges. And once again, I'm not vigorously proving it here. Or I should say I'm not rigorously proving it over here. in the numerator and the denominator. So now let's look at this one right over here. So here in the numerator I have e to the n power. And here I have e times n. So this grows much faster. I mean, this is e to the n power. Imagine if when you have this as 100, e to the 100th power is a ginormous number. e times 100-- that's just 100e. Grows much faster than this right over here. So this thing is just going to balloon. This is going to go to infinity. So we could say this diverges. Now let's look at this one right over here. Well, we have a higher degree term. We have a higher degree in the numerator than we have in the denominator. n squared, obviously, is going to grow much faster than n. So for the same reason as the b sub n sequence, this thing is going to diverge. The numerator is going to grow much faster"
    },
    {
        "Q": "At 2:18 Sal says we should subtract 114 from both the sides.But Instead of that we could transpose it.It is quite easier by transposing",
        "A": "Don t ask me why he didn t. It would be quicker.",
        "video_name": "hmj3_zbz2eg",
        "timestamps": [
            138
        ],
        "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 3:40 when sal says that to get the last angle in a triangle you have to do (180- a- b) but shouldn't it rather be 180 -(a+b)? Pls clarify my doubt\u00f0\u009f\u0099\u008f\u00f0\u009f\u008f\u00bc",
        "A": "Either way works, since the first way you would use GEMDAS (Grouping Symbols, Exponets, Multiplication/Division, and then Addition/Subtraction) and thus subtract a from 180 and then b from whatever is left. You could also see 180 -(a+b) as 180+ -1(a+b)",
        "video_name": "hmj3_zbz2eg",
        "timestamps": [
            220
        ],
        "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": "At 00:34, couldn't I just do 64+31+50+x=180?",
        "A": "You sure can! There are many ways to figure out each of the angles.",
        "video_name": "hmj3_zbz2eg",
        "timestamps": [
            34
        ],
        "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": "Why do we use a question mark in the video at 0:55 seconds in the video?",
        "A": "Th question mark represents the measure in degrees of the angle that the question is asking about. At 5:05 he labels the question mark z .",
        "video_name": "hmj3_zbz2eg",
        "timestamps": [
            55
        ],
        "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 1:04 you drew a =D, which looks like a smiley face. What is it?",
        "A": "It s just a sloppy arrow.",
        "video_name": "T4JKO0OGjpQ",
        "timestamps": [
            64
        ],
        "3min_transcript": "Let's write 113.9% as a decimal. So percent, this symbol right over here, literally means per hundred. So this is the same thing as 113.9 per 100, which is the same thing as 113.9 divided by 100. So essentially, if we want to write this as a decimal, we just take a 113.9, and we divide it by 100. And to do that, 113.9 divided by 100. If you divide by 10, you move the decimal once to the left. If you divide by 100, you're moving the decimal twice If we're doing it by 1,000, we would go to the left once more. 10,000, each time you divide by 10, you'd go one step to the left, which hopefully makes sense. If we were to multiply by 10, we would be moving the decimal over to the right, and there are other videos that go into far more depth on the intuition behind that. But anyway, we're dividing 113.9 by 100, so move the decimal two spaces to the left. So the decimal will end up right over here. And we are done."
    },
    {
        "Q": "At 3:00 they wrote two to the sixty third power. Couldn't you also write it as 2^63?",
        "A": "Yes, the two symbols mean the same thing",
        "video_name": "UCCNoXqCGZQ",
        "timestamps": [
            180
        ],
        "3min_transcript": "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 Voiceover:This is the first square on our board. We have 1 grain of rice. And when we double it we have 2 grains of rice. Voiceover:Yup. Voiceover:And we double it again we have 4. I'm thinking this is similar to what we were doing, it's just represented differently. Voiceover:Yeah, well, I mean, this one, the one you were making, right, every time you were kind of adding these popsicle sticks, you're kind of branching out. 1 popsicle stick now becomes 2 popsicles sticks. Then you keep doing that. 1 popsicle stick becomes 2 but now you have 2 of them. So here you have 1, now you have 1 times 2. Now each of these 2 branch into 2, so now you have 2 times 2, or you have 4 popsicle sticks. Every stage, every branch, you're multiplying by 2 again. Voiceover:I basically just continue splitting just like a tree does. Voiceover:Yup. Voiceover:Now I can really see what 2 to the power of 3 looks like. Voiceover:And that's what we have here. 1 times 2 times 2 times 2, which is 8. This is 2 to the third power."
    },
    {
        "Q": "At 1:24 he said line uv and he wrote,it but when he wrote it couldn't he have done another letter because uv looked like w which would get heeps confusing right",
        "A": "Most likely it was due that each line contained 2 points. A third letter was not needed. Furthermore, he made up a random problem with letters on each line in ABC order. Hope that sorts out the confusion. :)",
        "video_name": "aq_XL6FrmGs",
        "timestamps": [
            84
        ],
        "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": "At 4:10 why y-4=2x is equals to y/2-2=x? Sorry for my question. Hope somebody could help me with this thing.",
        "A": "Sal just divided the entire equation by 2: y/2 - 4/2 =2x/2 Reduce each term and you get: y/2 - 2 = x Hope this helps.",
        "video_name": "W84lObmOp8M",
        "timestamps": [
            250
        ],
        "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": "At 4:23, isn't it suppose to be (y-4)/2, not y/2-2?",
        "A": "(y-4)/2, and y/2 - 2 are equivalent. So either version will work.",
        "video_name": "W84lObmOp8M",
        "timestamps": [
            263
        ],
        "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": "At around 4:10, when Sla carries the 2 from 2x he divides. I understand that but why is it that he does (y/2) -2 instead of (y-4)/2? do either work or is the way Sal does it the only correct way and why?",
        "A": "Since your fraction is already in simplest terms, it would also be an acceptable answer.",
        "video_name": "W84lObmOp8M",
        "timestamps": [
            250
        ],
        "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": "In 3:55, why couldn't you do the same method for the first example?",
        "A": "Sal could have, he just skipped that step. We can apply the method like this: 2 is 2 * 1. 4x is 2 * 2x. We have a common factor here - 2. So we pull it out to get: 2 ( 1 + 2x).",
        "video_name": "I6TBBzIvgB8",
        "timestamps": [
            235
        ],
        "3min_transcript": "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, six X plus 30, that's interesting. So one way to think about it is can we break up each of these terms so that they have a common factor? Well, this one over here, six X literally represents six times X, and then 30, if I want to break out a six, 30 is divisible by six, so I could write this as six times five, 30 is the same thing as six times five. And when you write it this way, you see, \"Hey, I can factor out a six!\" Essentially, this is the reverse of the distributive property! So I'm essentially undoing the distributive property, taking out the six, and you are going to end up with, so if you take out the six, you end up with six times, so if you take out the six here, you have an X, and you take out the six here, you have plus five. So six X plus 30, if you factor it, we could write it as six times X plus five. And you can verify with the distributive property. six X + five times six or six X + 30. more interesting where we might want to factor out a fraction. So let's say we had the situation ... Let me get a new color here. So let's say we had 1/2 minus 3/2, minus 3/2 X. How could we write this in a, I guess you could say, in a factored form, or if we wanted to factor out something? I encourage you to pause the video and try to figure it out, and I'll give you a hint. See if you can factor out 1/2. Let's write it that way. If we're trying to factor out 1/2, we can write this first term as 1/2 times one and this second one we could write as minus 1/2 times three X. That's what this is, 3/2 X is the same thing as three X divided by two or 1/2 times three X. And then here we can see that we can just factor"
    },
    {
        "Q": "So with the 6x + 30 I paused the video and did it. His answer was 6( x + 5) I came up with the answer 2( 3x + 15). Math can be done many ways so I was thinking can both our answers be right. Or does this specific thing have to be done a certain way? This was at 4:00 minute",
        "A": "Sal s answer is more simplified because you can factor the 3 out of your answer. Usually math problems ask for the most simplified answer.",
        "video_name": "I6TBBzIvgB8",
        "timestamps": [
            240
        ],
        "3min_transcript": "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, six X plus 30, that's interesting. So one way to think about it is can we break up each of these terms so that they have a common factor? Well, this one over here, six X literally represents six times X, and then 30, if I want to break out a six, 30 is divisible by six, so I could write this as six times five, 30 is the same thing as six times five. And when you write it this way, you see, \"Hey, I can factor out a six!\" Essentially, this is the reverse of the distributive property! So I'm essentially undoing the distributive property, taking out the six, and you are going to end up with, so if you take out the six, you end up with six times, so if you take out the six here, you have an X, and you take out the six here, you have plus five. So six X plus 30, if you factor it, we could write it as six times X plus five. And you can verify with the distributive property. six X + five times six or six X + 30. more interesting where we might want to factor out a fraction. So let's say we had the situation ... Let me get a new color here. So let's say we had 1/2 minus 3/2, minus 3/2 X. How could we write this in a, I guess you could say, in a factored form, or if we wanted to factor out something? I encourage you to pause the video and try to figure it out, and I'll give you a hint. See if you can factor out 1/2. Let's write it that way. If we're trying to factor out 1/2, we can write this first term as 1/2 times one and this second one we could write as minus 1/2 times three X. That's what this is, 3/2 X is the same thing as three X divided by two or 1/2 times three X. And then here we can see that we can just factor"
    },
    {
        "Q": "At 1:03, for the prime factors of 12, instead of saying 2 x 2 x 3, could you instead say\n2^2 x 3? Would it still be okay?",
        "A": "Yes that is another way of writing it, just simplified. Though I am not so sure if the question will allow it. If the question says about anything to simplify, then simplify, if not then don t. ( But in real life, YOU SHOOULD ALWAYS for CLASS unless not allowed for some reason)",
        "video_name": "I6TBBzIvgB8",
        "timestamps": [
            63
        ],
        "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": "At 1:55 How did Sal get 2(1+2x) instead of 2(1+3x) or 2(2+2x)? ;)",
        "A": "You might have wanted to add the terms inside the parentheses but remember that you cannot add 1+3x or 2+2x ...etc because they are not like terms. you multiply them by their common factor separately. thats why the 2(1+2x). try to multiply the 2 by both terms and see what you get.",
        "video_name": "I6TBBzIvgB8",
        "timestamps": [
            115
        ],
        "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": "At 2:20, He could have continued that lin and made 2(1+[2 {1x}])",
        "A": "True, but it is less complicated to do 2(1*2x).",
        "video_name": "I6TBBzIvgB8",
        "timestamps": [
            140
        ],
        "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": "At 4:48 couldn't you just have figured that angle DBE was 45 because it is the half of 90? I mean if I do it always with that logic can it go wrong?",
        "A": "Always prove everything..proving is better.",
        "video_name": "7FTNWE7RTfQ",
        "timestamps": [
            288
        ],
        "3min_transcript": "Now, we could do either of these. Let's do this one right over here. So what is the measure of angle ABE? So they haven't even drawn segment BE here. So let me draw that for us. And so we have to figure out the measure of angle ABE. So we have a bunch of congruent segments here. And in particular, we see that triangle ABD, all of its sides are equal. So it's an equilateral triangle, which means all of the angles are equal. And if all of the angles are equal in a triangle, they all have to be 60 degrees. So all of these characters are going to be 60 degrees. Well, that's part of angle ABE, but we have to figure out this other part right over here. And to do that, we can see that we're actually dealing with an isosceles triangle kind of tipped over to the left. This is the vertex angle. This is one base angle. This is the other base angle. And the vertex angle right here is 90 degrees. And once again, we know it's isosceles And once again, these two angles plus this angle right over here are going to have to add up to 180 degrees. So you call that an x. You call that an x. You've got x plus x plus 90 is going to be 180 degrees. So you get 2x plus-- let me just write it out. Don't want to skip steps here. We have x plus x plus 90 is going to be equal to 180 degrees. x plus x is the same thing as 2x, plus 90 is equal to 180. And then we can subtract 90 from both sides. You get 2x is equal to 90. Or divide both sides by 2. You get x is equal to 45 degrees. And then we're done because angle ABE is going to be equal to the 60 degrees plus the 45 degrees. So it's going to be this whole angle, which is what we care about. Angle ABE is going to be 60 plus 45, which is 105 degrees. This one looks a little bit simpler. I have an isosceles triangle. This leg is equal to that leg. This is the vertex angle. I have to figure out B. And the trick here is like, wait, how do I figure out one side of a triangle if I only know one other side? Don't I need to know two other sides? And we'll do it the exact same way we just did that second part of that problem. If this is an isosceles triangle, which we know it is, then this angle is going to be equal to that angle there. And so if we call this x, then this is x as well. And we get x plus x plus 36 degrees is equal to 180. The two x's, when you add them up, you get 2x. And then-- I won't skip steps here. 2x plus 36 is equal to 180. Subtract 36 from both sides, we get 2x-- that 2 looks a little bit funny. We get 2x is equal to-- 180 minus 30 is 150."
    },
    {
        "Q": "At 1:19- 1:24 the video says that the ft would cancel out. I don't see how they cancel out. In class I'm doing this and my teacher says use a chart to do this, but how do i get things to cancel out. i really want to pass the 9th eoct.",
        "A": "It is known as simplifying the equation . Suppose you are given 9/3*27/6. You could simplify it and you will get 3/1*9/2, that is 27/2. If you do it the long way you will the same answer. Sal did the same thing.",
        "video_name": "F0LLR7bs7Qo",
        "timestamps": [
            79,
            84
        ],
        "3min_transcript": "A squirrel is running across the road at 12 feet per second. It needs to run 9 feet to get across the road. How long will it take the squirrel to run 9 feet? Round to the nearest hundredth of a second. Fair enough. A car is 50 feet away from the squirrel-- OK, this is a high-stakes word problem-- driving toward it at a speed of 100 feet per second. How long will it take the car to drive 50 feet? Round to the nearest hundredth of a second. Will the squirrel make it 9 feet across the road before the car gets there? So this definitely is high stakes, at least for the squirrel. So let's answer the first question. Let's figure out how long will it take the squirrel to run 9 feet. So let's think about it. So the squirrel's got to go 9 feet, and we want to figure out how many seconds it's So would we divide or multiply this by 12? Well, to think about that, you could think about the units where we want to get an answer in terms of seconds. We want to figure out time, so it'd be great if we could multiply this times seconds per foot. Then the feet will cancel out, and I'll be left with seconds. Now, right over here, we're told that the squirrel can run at 12 feet per second, but we want seconds per foot. So the squirrel, every second, so they go 12 feet per second, then we could also say 1 second per every 12 feet. So let's write it that way. So it's essentially the reciprocal of this because the units are the reciprocal of this. So, it's 1 second for every 12 feet. Notice, all I did is I took this information right over here, 12 feet per second, and I wrote it as second per foot-- 12 feet for every 1 second, 1 second for every 12 feet. it takes for the squirrel in seconds. So the feet cancel out with the feet, and I am left with 9 times 1/12, which is 9/12 seconds. And 9/12 seconds is the same thing as 3/4 seconds, which is the same thing as 0.75 seconds for the squirrel to cross the street. Now let's think about the car. So now let's think about the car. And it's the exact same logic. They tell us that the car is 50 feet away. So the squirrel is trying to cross the road like that, and the car is 50 feet away coming in like that, and we want to figure out if the squirrel will survive. So the car is 50 feet away. So it's 50 feet away. We want to figure out the time it'll take to travel that 50 feet. Once again, we would want it in seconds. So we would want seconds per feet."
    },
    {
        "Q": "At 0:30, what are the indices of a sequence?",
        "A": "The index is the counting number n (or k, or i or whatever). What he says is that we often view a sequence as a function of the indices. In other words, for each value of n, there is a specified value of the sequence based on the definition in terms of n. If the index is n, and the sequence is defined as starting at n= 0 or n = 1, then for every value of n, we can generate a new term of the sequence. If the sequence is (-1)\u00e2\u0081\u00bf\u00e2\u0081\u00ba\u00c2\u00b9 \u00e2\u0088\u0099 1/n\u00c2\u00b2 when the index = 2, the term is -\u00c2\u00bc Hope that helps",
        "video_name": "wzw9ll80Zbc",
        "timestamps": [
            30
        ],
        "3min_transcript": "what i want to do in this video is to provide ourselves with a rigorous definition of what it means to take the limit of a sequence as n approaches infinity and what we'll see is actually very similar to the definition of any function as a limit approaches infinity and this is because the sequences really can be just viewed as a function of their indices, so let's say let me draw an arbitrary sequence right over here so actually let me draw like this just to make it clear but the limit is approaching so let me draw a sequence let me draw a sequence that is jumping around little bit, so lets say when n=1, a(1) is there, when n = 2, a(2) is there, when n = 3, a(3) is over there when n=4, a(4) is over here, when n=5 a(5) is over here and it looks like is n is so this is 1 2 3 4 5 a(n) seems to be approaching, seems to be approaching some value it seems to be getting closer and closer, seems to be converging to some value L right over here. What we need to do is come up with a definition of what is it really mean to converge to L. So let's say for any, so we're gonna say that you converge to L for any, for any \u03b5 > 0, for any positive epsilon, you can, you can come up, you can get or you can, there is let me rewrite it this way, for any positive epsilon there is a positive, positive M, capital M, such that, such that if, if, lower case n is greater than capital M, the distance between those two points is less than epsilon. If you can do this for any epsilon, for any epsilon, greater than 0, there is a positive M, such that if n is greater than M, the distance between a(n) and our limit is less than epsilon then we can say, then we can say that the limit of a(n) as n approaches infinity is equal to L and we can say that a(n) converges, converges, converges to L. So let's, let's, let's parse this, so here I was making the claim that a(n) is approaching this L right over here, I tried to draw it as a horizontal line."
    },
    {
        "Q": "what does mean looking at the frequency of scores at 4:42 and 4:43",
        "A": "Frequency means how many times it occurs. So the frequency of scores means how many times each score occurred.",
        "video_name": "0ZKtsUkrgFQ",
        "timestamps": [
            282,
            283
        ],
        "3min_transcript": "so her score is going to be, let me do that in Efra's color, and that's Efra's score right there. She also got 100. So, Efra... Efra, and then finally, Farah got an 80, so 60, 70, 80, so Farah got an 80. So this is Farah's score right over here. So this is another way of representing the data, and here we see it in visual form, but it has the same information. You can look up someone's name, and then figure out their score. Amy scored a 90, Bill scored a 95, Cam scored 100, Efra also scored 100, Farah scored an 80. And there's even other ways you can have some of this information. In fact sometimes, you might not even know their names, and so then it would be less information but (mumbles) a list of scores. The professor might say, \"Hey, here are the five scores \"that people got on the exam.\" 100... 100, and 80, now, if it was listed, this was all the data you got, this is less information than the data that's in this bar graph, or this histogram, or the data that's given in this table right over here, because here, not only do we know the scores, but we know who got what score. Here, we only know the list of scores, and this is not an exhaustive video of all of the different ways you can represent data. You can also represent data by looking at the frequency of scores. So, the frequency of scores right over here, so instead of writing the people, you could write the scores. So let's see, you could say this is 80, 85, 90, 95, and 100, and then you could record So how many times do we have a score of an 80? Well, Farah is the only person with a score of 80, so you put one data point there. No one got an 85, one person got a 90, so you put a data point there. One person got a 95, so you can put that data point right over there. And then two people got 100. So this is one and two. Let's see the other 100 is in this color, so I'll just do it in the color. You wouldn't necessarily have to color code it like this. So this is another way to represent, and this axis, you could just view it as the number. So this tells you how many 80s there were, how many 90s there are, how many 95s, and how many 100s. So this right over here, has the same data as this list of numbers. It's just another way of looking at it. And once you have your data arranged in any of these ways, we can start to ask interesting questions. We can ask ourselves things like,"
    },
    {
        "Q": "I dont understand what ur saying at 5:30",
        "A": "at 5:30, he is combining like terms to make the equation simpler and easier to work with.",
        "video_name": "cNlwi6lUCEM",
        "timestamps": [
            330
        ],
        "3min_transcript": "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 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."
    },
    {
        "Q": "Can you explain a little more on why ln(x) + c (@ 1:10) has to be greater than 0?",
        "A": "The input for logarithmic equation has to be greater than 0. This is because y=ln(x) <==> e^y=x , but e^y is never 0 or negative. So x>0",
        "video_name": "sPPjk4aXzmQ",
        "timestamps": [
            70
        ],
        "3min_transcript": "What I want to do in this video is think about the antiderivative of 1/x. Or another way of thinking about it, another way of writing it , is the antiderivative of x to the negative 1 power. And we already know, if we somehow try to apply that anti-power rule, that inverse power rule over here, we would get something that's not defined. We would get x to the 0 over 0, doesn't make any sense. And you might have been saying, OK, well, I know what to do in this case. When we first learned about derivatives, we know that the derivative-- let me do this in yellow-- the derivative with respect to x of the natural log of x is equal to 1 over x. So why can't we just say that the antiderivative of this right over here is equal to the natural log of x plus c? And this isn't necessarily wrong. The problem here is that it's not broad enough. When I say it's not broad enough, is that the domain over here, for our original function that we're taking the antiderivative of, is all real numbers except for x equals 0. So over here, x cannot be equal to 0. So over here, x, so for this expression, x has to be greater than 0. So it would be nice if we could come up with an antiderivative that has the same domain as the function that we're taking the antiderivative of. So it would be nice if we could find an antiderivative that is defined everywhere that our original function is. So pretty much everywhere except for x equaling 0. So how can we rearrange this a little bit so that it could be defined for negative values as well? Well, one one possibility is to think about the natural log of the absolute value of x. So I'll put little question mark here, just because we don't really know what the derivative of this thing is going to be. And I'm not going to rigorously prove it here, but I'll I will give you kind of the conceptual understanding. So to understand it, let's plot the natural log of x. So that right over there is roughly what the graph of the natural log of x looks like. So what would the natural log of the absolute value of x is going to look like? Well, for positive x's, it's going to look just like this. For positive x's you take the absolute value of it, it's just the same thing as taking that original value. So it's going to look just like that for positive x's. But now this is also going to be defined for negative x's. If you're taking the absolute value of negative 1, that evaluates to just 1. So it's the natural log of 1, so you're going to be right there. As you get closer and closer and closer to 0 from the negative side, you're just going to take the absolute value. So it's essentially going to be exactly this curve for the natural log of x, but the left side of the natural log of the absolute value of x is going to be its mirror image, if you were to reflect around It's going to look something like this. So what's nice about this function"
    },
    {
        "Q": "5:61 Is there something wrong here?",
        "A": "if you are writing in decimals it is a decimal dot (looks like a period) not a colon. If it is time, the time is impossible because there can be only 60 minutes per hour actually 59 before the next hour. Good Question! Keep it Up!",
        "video_name": "AGFO-ROxH_I",
        "timestamps": [
            361
        ],
        "3min_transcript": "So 3 goes into 10,560. It doesn't go into 1. It goes into 10 three times. 3 times 3 is 9. And we subtract. We get 1. Bring down this 5. It becomes a 15. 3 goes into 15 five times. 5 times 3 is 15. We have no remainder, or 0. You bring down the 6. 3 goes into 6 two times. Let me scroll down a little bit. 2 times 3 is 6. Subtract. No remainder. Bring down this last 0. 3 goes into 0 zero times. 0 times 3 is 0. And we have no remainder. So 2 miles is the equivalent to 3,520 yards. That's the total distance he has to travel. 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."
    },
    {
        "Q": "On 7:34, it says the first number to the right is the ones place. How about decimal numbers?",
        "A": "Decimal numbers have place values below the ones place.",
        "video_name": "wx2gI8iwMCA",
        "timestamps": [
            454
        ],
        "3min_transcript": "So we could rewrite this. This is equal to, this is equal to three 10s three 10s plus, plus seven ones. Or another way to think about it what are the three 10s? Well if use the same number system to represent three 10s you would write that down as 30. And then seven ones. Once again if you use our same number system you would represent that as seven. So these are all different ways of representing 37. And hopefully this allows you to appreciate how neat our number system is. Where even a number like 37 as soon as you just write scratches on a wall, it becomes pretty hard to read. And you can imagine when you get to much much larger numbers like 1,052 to have to count that many marks every time. But our number system gives us a way of dealing with it."
    },
    {
        "Q": "At 1:00 how did he know to put t-5 into parentheses instead of 210t-5=41790?",
        "A": "t represents the number of trees. t is the x , or independent, or domain variable. The y variable in this problem is the # of oranges (the range, or the dependent variable) When the dude cut down 5 trees, he messed with the domain, t. Since he only messed with the domain, and not with the range (the number of oranges), we want to isolate that -5 stuff with the t. If he didn t put the parentheses, it would be like subtracting 5 oranges, not subtracting 5 trees.",
        "video_name": "xKH1Evwu150",
        "timestamps": [
            60
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 2:10, Why you place 210 on the other side instead of distributing? Isn't it distribution next then you eliminate and carry over to the other side?",
        "A": "You can do it both ways. Whether you divide both sides by 210 first or distribute the 210 across the (t-5), you get the same results.",
        "video_name": "xKH1Evwu150",
        "timestamps": [
            130
        ],
        "3min_transcript": ""
    },
    {
        "Q": "I watched the videos on the binomial theorem but they didn't show what he is doing here... I can't follow his thinking process at around 2:00 when he keeps expanding the binomial. Any help for me? I wouldn't be able to write out the whole thing by myself is what I mean.",
        "A": "The expansion for (a + b)^n always starts with a^n and always ends with b^n. The stuff in the middle comes from the binomial expansion. Check it out in the precalc section! It is most excellent. :)",
        "video_name": "dZnc3PtNaN4",
        "timestamps": [
            120
        ],
        "3min_transcript": ""
    },
    {
        "Q": "Since C can be negative (when pi/2 < theta < pi) how do we know the value in the radical at 4:45 will remain positive?\n\nEdit: Nevermind, I see my error. C can never be less than -1 on the unit circle, so that radical will never be negative.",
        "A": "you made a small mistake in your edit: if C<-1, the radical becomes more positive, as the formula has 1-C, not 1+C in it, so C would have to become larger than 1 for the radical to become negative.",
        "video_name": "yV4Xa8Xtmrc",
        "timestamps": [
            285
        ],
        "3min_transcript": "this yellow sine squared theta, and all of this is equal to C or we could get that C is equal to one minus two sine squared theta. And what's useful about this is we just have to solve for sine of theta. So let's see, I could multiply both sides by a negative just so I can switch the order over here. So I could write this as negative C is equal to two sine squared theta minus one and just multiply both sides by a negative and then let's see I could add one to both sides, if I add one to both sides, and I'll go over here, if I add one to both sides, I could divide both sides by two, and then so I get sine squared theta is equal to one minus C over two, or I could write that sine of theta is equal to the plus or minus square root of one minus C over two. So that leads to a question, Is it both? Is it the plus and minus square root? Or is it just one of those? And I encourage you to pause the video again in case you haven't already figured it out, and look at the information here, and think about whether they give us the information of whether we should be looking at the positive or negative sine. Well they tell us that theta is between zero and pi. So if I were to draw a unit circle here, between zero and pi radians. and pi is going all the way over here. So this angle places its terminal ray either in the first or second quadrants. So, it could be an angle like this, it could be an angle like this, it cannot be an angle like this, and we know that the sine of an angle is the Y coordinate, and so we know that for the first of second quadrant the Y coordinate is going to be non-negative. So we would want to take the positive square root, so we would get sine of theta, is equal to the principal root, or we could even think of it as the positive square root of one minus C over two. So let's go back to our... Make sure we can check our answers. So sine of theta is equal to the square root of one minus capital C, all of that over two,"
    },
    {
        "Q": "At 3:00 isn't that called the Reflexive Property?",
        "A": "In statement 3 (CA=CA) this is the reflexive property of congruency.",
        "video_name": "fSu1LKnhM5Q",
        "timestamps": [
            180
        ],
        "3min_transcript": "And you don't have to do something as a two column proof, but this is what you normally see in a normal introductory geometry class. So I thought I would expose you to it. It's a pretty basic idea is that you make a statement, and you just have to give the reason for your statement. Which is what we've been doing with any proof, but we haven't always put it in a very structured way. So I'm just going to do it like this. I'll have two columns like that. And I'll have a statement, and then I will give the reason for the statement. And so the strategy that I'm going to try to do is it looks like, right off the bat, it seems like I can prove the triangle CDA is congruent to triangle CBA based on side side side. And that's a pretty good starting point because once I can base congruency, then I can start to have angles be the same. And the reason why I can do that is because this side is the same as that side, this side the same as that side, and they both share that side. I want to write it out properly early in this two column proof. So we have CD, we had the length of segment CD is equal to the length of CB. CD is equal to CB, and that is given. So these two characters have the same length. We also know that DA, the length of segment DA, is the same as the length of segment BA. So DA is equal to BA, that's also given in the diagram. And then we also know that CA is equal to CA, I guess we could say. So CA is equal to itself. And it's obviously in both triangles. So this is also given, or it's obvious from the diagram. It's a bit obvious. Both triangles share that side. 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."
    },
    {
        "Q": "at 0:33 thers a arc on the outer side what is that called?",
        "A": "Even about the reflex angle, it doesn t even count as still an obtuse angle, so the reflex angle is 2/3 of a circle. \u00f0\u009f\u0099\u0082",
        "video_name": "4ZyTVTGVPgE",
        "timestamps": [
            33
        ],
        "3min_transcript": "Put the vertex of the angle at point A. So let me do that. Make one of the rays go through point B. Make the other ray go through one of the other points to make an acute angle. So an acute angle is an angle less than 90 degrees. So we could make one of these rays go through point B. And then we have to pick where to put the other ray to make it go to one of the other points. And we have to be very careful here because we have to look at this arc that shows which angle the tool is actually measuring. Because we might be tempted to do something like this, thinking that, hey, maybe this is the angle that we're thinking about. But the tool thinks we're referring to this outer angle right over here, this larger huge angle. This angle right over here is well over 180 degrees. So we have to pay attention to this arc to make sure that the tool is looking at the same angle that we are. So once again, we want an acute angle. So this right over here looks like an acute angle. It looks like it is less than 90 degrees. And we have to be very careful that we go exactly So that looks about right. This is an acute angle because its measure is less than 90 degrees. Let's do a few more of these. Make an obtuse angle using the black points. Choose one of the points as the vertex and make the rays go through the other two points. The angle should also be less than 180 degrees. So you could think of it several ways. You could just try to pick that point. But then when you go through these other two points, this right over here is an acute angle. This is less than 90 degrees. You could do something like this where now when I switch the rays, the tool is now thinking about this angle, not this outer angle right over here. But this is larger than 180 degrees. So this also doesn't apply. So we really picked the wrong point for the vertex. If you just move this out of the way and eyeball it, it looks like you get the largest angle that's less than 180 degrees if you put the vertex right over here. So let me do that. And now it looks like I have constructed a 180-degree angle. And we have to be very careful that we go right through those points because otherwise it might mark us wrong. So that looks pretty good. That's an obtuse angle because its measure is greater than 90 degrees. Let's do one more of these. So another obtuse angle using the points. Same general idea so that would be an exactly 180-degree angle. And there we go. And this is an obtuse angle because it is greater than 90 degrees."
    },
    {
        "Q": "At 6:01\nWhat I don't understand is that if the sign is the same, why don't you add or subtract\nEX#1:\n-15+(-16)\nisn't that technically subtraction?",
        "A": "It can be both. You can do it whatever way is easiest to you. That s why he is teaching you it! Hope this helps! :)",
        "video_name": "Oo2vGhVkvDo",
        "timestamps": [
            361
        ],
        "3min_transcript": "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 It's 262. This right over here is equal to 262. How many degrees difference are there between the coldest and warmest recorded outside temperature? 262 degrees Fahrenheit difference."
    },
    {
        "Q": "at 3:37, he explains how absolute value is always positive. how come it can't be negative?",
        "A": "The mathematical definition of absolute value says that it the magnitude of a real number without regard to its sign. Think about distance. You would use absolute value to calculate this because you can t have a negative distance.",
        "video_name": "Oo2vGhVkvDo",
        "timestamps": [
            217
        ],
        "3min_transcript": "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 It's 262. This right over here is equal to 262. How many degrees difference are there between the coldest and warmest recorded outside temperature? 262 degrees Fahrenheit difference."
    },
    {
        "Q": "whats AA at 4:45",
        "A": "AA is short for Angle-Angle similarity. This tells us that if two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. This also means the third angle of the first triangle is congruent with the third angle of the second triangle.",
        "video_name": "7bO0TmJ6Ba4",
        "timestamps": [
            285
        ],
        "3min_transcript": "is that the ratio between all of the sides are going to be the same. So for example, if we have another triangle right over here-- let me draw another triangle-- I'll call this triangle X, Y, and Z. And let's say that we know that the ratio between AB and XY, we know that AB over XY-- so the ratio between this side and this side-- notice we're not saying that they're congruent. We're looking at their ratio now. We're saying AB over XY, let's say that that is equal to BC over YZ. That is equal to BC over YZ. And that is equal to AC over XZ. that we say, hey, this means similarity. So if you have all three corresponding sides, the ratio between all three corresponding sides are the same, then we know we are dealing with similar triangles. So this is what we call side-side-side similarity. And you don't want to get these confused with side-side-side congruence. So these are all of our similarity postulates or axioms or things that we're going to assume and then we're going to build off of them to solve problems and prove other things. Side-side-side, when we're talking about congruence, means that the corresponding sides are congruent. Side-side-side for similarity, we're saying that the ratio between corresponding sides are going to be the same. So for example, let's say this right over here is 10. Let me think of a bigger number. Let's say this is 60, this right over here is 30, and I just made those numbers because we will soon learn what typical ratios are of the sides of 30-60-90 triangles. And let's say this one over here is 6, 3, and 3 square roots of 3. Notice AB over XY 30 square roots of 3 over 3 square roots of 3, this will be 10. What is BC over XY? 30 divided by 3 is 10. And what is 60 divided by 6 or AC over XZ? Well, that's going to be 10. So in general, to go from the corresponding side here to the corresponding side there, we always multiply by 10 on every side. So we're not saying they're congruent or we're not saying the sides are the same for this side-side-side for similarity. We're saying that we're really just scaling them up by the same amount, or another way to think about it, the ratio between corresponding sides are the same. Now, what about if we had-- let's"
    },
    {
        "Q": "At around 7:53, I noticed Sal wrote XY/AB = k = BC/YZ. Shouldn't the \"BC/YZ\" be \"YZ/BC\" instead? As if XY/AB = k, then BC/YZ must equal to 1/k.",
        "A": "Yes, nice catch! Sal only messes up once in a while. Just submit that in the Report a mistake In the video , and just say, at 7:53, Sal says XY/AB = k = BC/YZ, but should say BC/YZ be YZ/BC, if you are really worried about it that much!",
        "video_name": "7bO0TmJ6Ba4",
        "timestamps": [
            473
        ],
        "3min_transcript": "Let me draw it like this. Actually, I want to leave this here so we can have our list. So let's draw another triangle ABC. So this is A, B, and C. And let's say that we know that this side, when we go to another triangle, we know that XY is AB multiplied by some constant. So I can write it over here. XY is equal to some constant times AB. Actually, let me make XY bigger, so actually, it That constant could be less than 1 in which case it would be a smaller value. But let me just do it that way. So let me just make XY look a little bit bigger. So let's say that this is X and that is Y. So let's say that we know that XY over AB Or if you multiply both sides by AB, you would get XY is some scaled up version of AB. So maybe AB is 5, XY is 10, then our constant would be 2. We scaled it up by a factor of 2. And let's say we also know that angle ABC is congruent to angle XYZ. I'll add another point over here. So let me draw another side right over here. So this is Z. So let's say we also know that angle ABC is congruent to XYZ, and let's say we know that the ratio between BC and YZ is also this constant. The ratio between BC and YZ is also equal to the same constant. So an example where this 5 and 10, maybe this is 3 and 6. The constant we're kind of doubling So is this triangle XYZ going to be similar? Well, if you think about it, if XY is the same multiple of AB as YZ is a multiple of BC, and the angle in between is congruent, there's only one triangle we can set up over here. We're only constrained to one triangle right over here, and so we're completely constraining the length of this side, and the length of this side is going to have to be that same scale as that over there. And so we call that side-angle-side similarity. So once again, we saw SSS and SAS in our congruence postulates, but we're saying something very different here. We're saying that in SAS, if the ratio between corresponding sides of the true triangle are the same, so AB and XY of one corresponding side and then"
    },
    {
        "Q": "At 11:00 and 11:19, when he was explaining with two non-similar triagles, why did he say \"unnecessarily similar\"?",
        "A": "He said necessarily similar.",
        "video_name": "7bO0TmJ6Ba4",
        "timestamps": [
            660,
            679
        ],
        "3min_transcript": "so that's between BC and YZ, and the angle between them are congruent, then we're saying it's similar. For SAS for congruency, we said that the sides actually had to be congruent. Here we're saying that the ratio between the corresponding sides just has to be the same. So for example SAS, just to apply it, if I have-- let me just show some examples here. So let's say I have a triangle here that is 3, 2, 4, and let's say we have another triangle here that has length 9, 6, and we also know that the angle in between are congruent so that that angle is equal to that angle. What SAS in the similarity world tells you is that these triangles are definitely going to be similar triangles, that we're actually constraining because there's actually only one triangle we can draw a right over here. It's the triangle where all the sides So there's only one long side right here that we could actually draw, and that's going to have to be scaled up by 3 as well. This is the only possible triangle. If you constrain this side you're saying, look, this is 3 times that side, this is 3 three times that side, and the angle between them is congruent, there's only one triangle we could make. And we know there is a similar triangle there where everything is scaled up by a factor of 3, so that one triangle we could draw has to be that one similar triangle. So this is what we're talking about SAS. We're not saying that this side is congruent to that side or that side is congruent to that side, we're saying that they're scaled up by the same factor. If we had another triangle that looked like this, so maybe this is 9, this is 4, and the angle between them were congruent, you couldn't say that they're similar because this side is scaled up by a factor of 3. This side is only scaled up by a factor of 2. So this one right over there you could not say that it is necessarily similar. and length 6 there, but you did not know that these two angles are the same, once again, you're not constraining this enough, and you would not know that those two triangles are necessarily similar because you don't know that middle angle is the same. Now, you might be saying, well there was a few other postulates that we had. We had AAS when we dealt with congruency, but if you think about it, we've already shown that two angles by themselves are enough to show similarity. So why worry about an angle, an angle, and a side or the ratio between a side? So why even worry about that? And we also had angle-side-angle in congruence, but once again, we already know the two angles are enough, so we don't need to throw in this extra side, so we don't even need this right over here. So these are going to be our similarity postulates, and I want to remind you, side-side-side, this is different than the side-side-side for congruence. We're talking about the ratio between corresponding sides."
    },
    {
        "Q": "I'm at 2:45, I realized that you also could prove similarity if one angle and two sides are congruent, right ?",
        "A": "I believe so, yes. Later on in the video, Sal describes that as another similarity theorem: The SAS (Side Angle Side) way of determining similar triangles.",
        "video_name": "7bO0TmJ6Ba4",
        "timestamps": [
            165
        ],
        "3min_transcript": "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. is that the ratio between all of the sides are going to be the same. So for example, if we have another triangle right over here-- let me draw another triangle-- I'll call this triangle X, Y, and Z. And let's say that we know that the ratio between AB and XY, we know that AB over XY-- so the ratio between this side and this side-- notice we're not saying that they're congruent. We're looking at their ratio now. We're saying AB over XY, let's say that that is equal to BC over YZ. That is equal to BC over YZ. And that is equal to AC over XZ."
    },
    {
        "Q": "So at 9:00, if you have log base 2 (32/sqrt 8), you're NOT supposed to simplify (32/sqrt 8) into (4*sqrt 8)?",
        "A": "Yes, you could do that and still get to the same answer, but sal just skipped it",
        "video_name": "TMmxKZaCqe0",
        "timestamps": [
            540
        ],
        "3min_transcript": "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? 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 8:10, how is Log2 sqrt of 32/sqrt8 converted into Log2 (32/sqrt8)^1/2?",
        "A": "The square root is the same thing as an exponent of 1/2. Likewise, the cube root is the same thing as an exponent of 1/3. So, this isn t even a computation or an operation, it is just a different way of writing exactly the same thing.",
        "video_name": "TMmxKZaCqe0",
        "timestamps": [
            490
        ],
        "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": "I don't get why he corrected from -1/2 to -1/4 in the very end (9:48)? Can someone please explain?",
        "A": "He did not distribute the 1/2 to the -1/2 log\u00e2\u0082\u00828 term (he wrote -1/2 log\u00e2\u0082\u00828, which was unchanged from the original term), so he corrected it to -1/4 log\u00e2\u0082\u00828. Hope this helps!",
        "video_name": "TMmxKZaCqe0",
        "timestamps": [
            588
        ],
        "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 8:25-8:38 why did he remove the exponent 1/2 and put it on the left so that it turns into 1/2_log(32/sqrt(8))? He mentioned a property but which one is it?",
        "A": "Because, one of the property of logarithm is: log of a^b = b log a (log of a power b equals b log of a) in this video, log of some thing power to 1/2, then equals to 1/2 log of some thing. Hope it helps.",
        "video_name": "TMmxKZaCqe0",
        "timestamps": [
            505,
            518
        ],
        "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": "How does Sal cancel out the 2 a's at about 2:30?",
        "A": "Remember that multiplying a number by 1 does not change it. So what Sal did was multiply the expression by (1/a*a)/(1/a*a).",
        "video_name": "tvj42WdKlH4",
        "timestamps": [
            150
        ],
        "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": "At 1:51 Where does the aa being a divisible of both come from? That was confusing, how would you even know to use that formula?",
        "A": "This is basically factoring and cancelling but with letters instead of numbers. It s like saying what s 32/4? well, since the numerator is actually divisible by four (2*2) we can simplify by dividing the numerator and denominator by 4 giving us 8/1 or 8. (this is the same problem that Sal shows in the video but with the number 2 in place of a). Does that help?",
        "video_name": "tvj42WdKlH4",
        "timestamps": [
            111
        ],
        "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": "If you can have the same two terms, such as the time at 6:22, forever and ever in one sequence, can you have one term for a whole sequence? What would an example be?",
        "A": "Yes, but it s not useful: a(1) = a1 a_i = a(i-1) + 0 (i > 1)",
        "video_name": "Kjli0Gunkds",
        "timestamps": [
            382
        ],
        "3min_transcript": "Well, then we're gonna have to figure out what the previous term is. G of five is going to be g of four, g of four plus 3.2, and you would keep going back and back and back, but we've already figured out what g of four is. It's 13.6, so this is 16.8, and then if g of five is 16.8, 16.8, you add 3.2 there, you would get 20. So you could start at g of six and keep backing up, all the way until you get to g of one, and then you figure out what that is. You recurse back to your base case, and then you're able to fill in all of the blanks. Let's do a few more examples of this. So we have this function here. So let's say that this defines a sequence. Let's think about what the first four terms of that sequence are, and once again, I encourage you to pause the video and figure that out. All right, let's work through it. So h of one is, well, they very clearly tell us If n is equal to one, h is 14. H of two, well, now we're falling into this case, 'cause two is greater than one and it's a whole number, so it's gonna be 28 over h of one, over h of one. Well, we know h of one is 14, so it's gonna be 28 over 14, which is equal to two. Now h of three. H of three, we're gonna fall into this case again. It's gonna be 28, 28 divided by h of two, if we're thinking of this as a sequence, divided by the previous term in the sequence. So 28 divided by h of two, we know that h of two is equal to, is equal to two. We just figured that out. So we go back to 14, something very interesting. I think you see where this is going. H of four is gonna be 28 divided by h of three, 28 divided by h of three, this is h of three right over here, we just figured that out, divided by 14, which is back to two. If we were to think of this as a sequence, we'd say, \"All right, let's see, the first term is 14, \"then we get to two, \"then we go to 14, \"then we get to two.\" So one way to think about this sequence is that we just keep alternating between 14 and twos. All of the odd terms of the sequence are 14, all of the even terms of the sequence are two. That's one way to think about it. Or another way to think about it is, we're starting with 14, and each successive term is the previous term divided, is 28 divided by the previous term. So here, 28 divided by 14 is two. 28 divided by two is 14. 28 divided by 14 is two. And we keep going on and on and on, and that's what was actually going on right over there. Let's do one more of these. And this one is interesting, because we now have,"
    },
    {
        "Q": "At 2:11 did it have to be a fraction?",
        "A": "Well, you could try to divide and write your answer as a decimal, but seeing as how the answer would probably be a decimal that stretches on for a really long time, using the fraction will be much easier.",
        "video_name": "_ETPMszULXc",
        "timestamps": [
            131
        ],
        "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": "i dont think the y-intercept is 4/5 at 1:58",
        "A": "but that is the xintercept since the y is 0",
        "video_name": "5fkh01mClLU",
        "timestamps": [
            118
        ],
        "3min_transcript": "In this video I'm going to do a bunch of examples of finding the equations of lines in slope-intercept form. Just as a bit of a review, that means equations of lines in the form of y is equal to mx plus b where m is the slope and b is the y-intercept. So let's just do a bunch of these problems. So here they tell us that a line has a slope of negative 5, so m is equal to negative 5. And it has a y-intercept of 6. So b is equal to 6. So this is pretty straightforward. The equation of this line is y is equal to negative 5x plus 6. That wasn't too bad. Let's do this next one over here. The line has a slope of negative 1 and contains the point 4/5 comma 0. So they're telling us the slope, slope of negative 1. So we know that m is equal to negative 1, but we're not 100% So we know that this equation is going to be of the form y is equal to the slope negative 1x plus b, where b is the y-intercept. Now, we can use this coordinate information, the fact that it contains this point, we can use that information to solve for b. The fact that the line contains this point means that the value x is equal to 4/5, y is equal to 0 must satisfy this equation. So let's substitute those in. y is equal to 0 when x is equal to 4/5. So 0 is equal to negative 1 times 4/5 plus b. I'll scroll down a little bit. So let's see, we get a 0 is equal to negative 4/5 plus b. We can add 4/5 to both sides of this equation. 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."
    },
    {
        "Q": "At 2:00, shouldn't Sal add -4/5 to b and 0, resulting in our final answer being negative?",
        "A": "Sal ha 0 = -4/5 + b He needs to move the -4/5 to the other side. Remember, we always use the opposite sign, or the opposite operation to move items to the other side of an equation. The 4/5 is negative. This means it is being subtracted from b . So, to move it, we do the opposite, we add 4/5 to both sides. -4/5 + 4/5 will cancel out like Sal shows in the video. If you did: -4/5 +(-4/5), you get -4/5 -4/5 = -9/5. So, the b would not be by itself. Hope this helps.",
        "video_name": "5fkh01mClLU",
        "timestamps": [
            120
        ],
        "3min_transcript": "In this video I'm going to do a bunch of examples of finding the equations of lines in slope-intercept form. Just as a bit of a review, that means equations of lines in the form of y is equal to mx plus b where m is the slope and b is the y-intercept. So let's just do a bunch of these problems. So here they tell us that a line has a slope of negative 5, so m is equal to negative 5. And it has a y-intercept of 6. So b is equal to 6. So this is pretty straightforward. The equation of this line is y is equal to negative 5x plus 6. That wasn't too bad. Let's do this next one over here. The line has a slope of negative 1 and contains the point 4/5 comma 0. So they're telling us the slope, slope of negative 1. So we know that m is equal to negative 1, but we're not 100% So we know that this equation is going to be of the form y is equal to the slope negative 1x plus b, where b is the y-intercept. Now, we can use this coordinate information, the fact that it contains this point, we can use that information to solve for b. The fact that the line contains this point means that the value x is equal to 4/5, y is equal to 0 must satisfy this equation. So let's substitute those in. y is equal to 0 when x is equal to 4/5. So 0 is equal to negative 1 times 4/5 plus b. I'll scroll down a little bit. So let's see, we get a 0 is equal to negative 4/5 plus b. We can add 4/5 to both sides of this equation. 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."
    },
    {
        "Q": "10:20 What is the purpose of functions in a practical real life scenario?",
        "A": "Functions are used all the time. -- Computer programs are functions -- the cash register at a store uses functions to determine what you owe -- when you calculate the tip on a bill in a restaurant, you are using a function. Those a just a few. There are many more.",
        "video_name": "5fkh01mClLU",
        "timestamps": [
            620
        ],
        "3min_transcript": "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. It looks like y changes exactly by 4. It looks like my delta y, my change in y, is equal to 4 when my delta x is equal to 1. So change in y over change in x, change in y is 4 when change in x is 1. So the slope is equal to 4. Now what's its y-intercept? Well here we can just look at the graph. It looks like it intersects y-axis at y is equal to negative 6, or at the point 0, negative 6. So we know that b is equal to negative 6. So we know the equation of the line. The equation of the line is y is equal to the slope times x plus the y-intercept. I should write that. equation of our line. Let's do one more of these. So they tell us that f of 1.5 is negative 3, f of negative 1 is 2. What is that? Well, all this is just a fancy way of telling you that the point when x is 1.5, when you put 1.5 into the function, the function evaluates as negative 3. So this tells us that the coordinate 1.5, negative 3 is on the line. Then this tells us that the point when x is negative 1, f of x is equal to 2. This is just a fancy way of saying that both of these two points are on the line, nothing unusual. I think the point of this problem is to get you familiar with function notation, for you to not get intimidated if you see something like this. If you evaluate the function at 1.5, you get negative 3."
    },
    {
        "Q": "At 6:11, how come its 0-5 instead of 5-0? Is that crucial to get the correct answer or does it not matter, because you are still finding the change?",
        "A": "(y2 - y1)/(x2-x1) works either way. Just make sure that whatever point you started with for the difference in y in the numerator is the same point you start with for the difference of x in the denominator.",
        "video_name": "5fkh01mClLU",
        "timestamps": [
            371
        ],
        "3min_transcript": "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. Now we know the equation for the line. The equation is y-- let me do it in a new color-- y is equal to negative 2x plus b plus 10. We are done. Let's do another one of these. All right, the line contains the points 3 comma 5 and negative 3 comma 0. Just like the last problem, we start by figuring out the slope, which we will call m. It's the same thing as the rise over the run, which is the same thing as the change in y over the change in x. If you were doing this for your homework, you wouldn't I just want to make sure that you understand that these are all the same things. Then what is our change in y over our change in x? This is equal to, let's start with the side first. It's just 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."
    },
    {
        "Q": "at 2:49 how does x squared divided by x equal x?",
        "A": "x squared is the same as x * x. If you divide x * x by x you get x.",
        "video_name": "FXgV9ySNusc",
        "timestamps": [
            169
        ],
        "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 3:00, why did Sal put the answer to x^2 divided by x above the 3x? Why not above the x^2?",
        "A": "I agree with you at 3:00. It would be better and less confused, if Sal put the x right on top of x^2 , even though it is not a require. In fact, x can be put any where within the line.",
        "video_name": "FXgV9ySNusc",
        "timestamps": [
            180
        ],
        "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 around 8:00, you need to restrict x=-4 to be equal to the first expression. So if your long division works out with no remainder, you have to make restrictions for your original denominator, or am I missing something?",
        "A": "You are correct. The original expression is undefined at x = -4, so the correct answer would be: (x\u00c2\u00b2 +5x + 4) / (x + 4) = x + 1; x \u00e2\u0089\u00a0 -4",
        "video_name": "FXgV9ySNusc",
        "timestamps": [
            480
        ],
        "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": "It seems like around 8:24 or so, something important got lost. The original formula was undefined when x was -4, because at that x-value the denominator was 0. The new formula of simply x + 1 is defined everywhere. What happened there?",
        "A": "The new formula is x + 1, x =/= -4, so the expressions are equivalent. Sal just forgot to mention that.",
        "video_name": "FXgV9ySNusc",
        "timestamps": [
            504
        ],
        "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": "At 3:30 minutes into the video it says a histogram is really a plot, kind of bar graph... How is a histogram related to a bar graph in anyway?",
        "A": "It has the same formation.",
        "video_name": "4eLJGG2Ad30",
        "timestamps": [
            210
        ],
        "3min_transcript": "I want to put them in order. I have a 2, another 2. Let me stack that above my first 2. I have another 1. Let me stack that above my other two 1's. I have another 0. Let me stack it there. I have another 1. Then I have another 2. Another 1. I have two more 0's. 0, 0. I have two more 2's. I have a 3. I have two more 1's. Another 3. And then I have a 6. So no 5's, and then I have a 6. And that space right there was unnecessary. But right here I've listed-- I've just rewritten those numbers and I've essentially categorized them. Now what I want to do is calculate how many of each of 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."
    },
    {
        "Q": "I dont understand the time 2:50",
        "A": "He added the 5 just to go in order, the 5 shows up 0 times.",
        "video_name": "4eLJGG2Ad30",
        "timestamps": [
            170
        ],
        "3min_transcript": "I want to put them in order. I have a 2, another 2. Let me stack that above my first 2. I have another 1. Let me stack that above my other two 1's. I have another 0. Let me stack it there. I have another 1. Then I have another 2. Another 1. I have two more 0's. 0, 0. I have two more 2's. I have a 3. I have two more 1's. Another 3. And then I have a 6. So no 5's, and then I have a 6. And that space right there was unnecessary. But right here I've listed-- I've just rewritten those numbers and I've essentially categorized them. Now what I want to do is calculate how many of each of 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."
    },
    {
        "Q": "What are the \"buckets' Sal talks about at 3:41?",
        "A": "He uses the term bucket to represent a range of data. The example that he uses, he lists all numbers 0,1,2,3,4,5. You might not have a lot of numbers, so you might want to clump, say 0-1, 2-3, 4-5. So you have 3 buckets.",
        "video_name": "4eLJGG2Ad30",
        "timestamps": [
            221
        ],
        "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": "about 04:00, does this mean that every sum of normal distributed, independent, and squared variables will have the chi-square distribuition?",
        "A": "Indeed it does. This is mainly because the sum of normally-distributed, independent, and squared random variables is the very definition of the chi-square distribution.",
        "video_name": "dXB3cUGnaxQ",
        "timestamps": [
            240
        ],
        "3min_transcript": "here is going to be an example of the chi-square distribution. Actually what we're going to see in this video is that the chi-square, or the chi-squared distribution is actually a set of distributions depending on how many sums you have. Right now, we only have one random variable that we're squaring. So this is just one of the examples. And we'll talk more about them in a second. So this right here, this we could write that Q is a chi-squared distributed random variable. Or that we could use this notation right here. Q is-- we could write it like this. So this isn't an X anymore. This is the Greek letter chi, although it looks a lot like a curvy X. So it's a member of chi-squared. And since we're only taking one sum over here-- we're only taking the sum of one independent, normally distributed, standard or normally distributed variable, we say that this only has 1 degree of freedom. So this right here is our degree of freedom. We have 1 degree of freedom right over there. So let's call this Q1. Let's say I have another random variable. Let's call this Q-- let me do it in a different color. Let me do Q2 in blue. Let's say I have another random variable, Q2, that is defined as-- let's say I have one independent, standard, normally distributed variable. I'll call that X1. And I square it. And then I have another independent, standard, normally distributed variable, X2. And I square it. So you could imagine both of these guys have distributions like this. And they're independent. So get to sample Q2, you essentially sample X1 from this distribution, square that value, sample X2 and then add the two. And you're going to get Q2. This over here-- here we would write-- so this is Q1. Q2 here, Q2 we would write is a chi-squared, distributed random variable with 2 degrees of freedom. Right here. 2 degrees of freedom. And just to visualize kind of the set of chi-squared distributions, let's look at this over here. So this, I got this off of Wikipedia. This shows us some of the probability density functions for some of the chi-square distributions. This first one over here, for k of equal to 1, that's the degrees of freedom. So this is essentially our Q1. This is our probability density function for Q1. And notice it really spikes close to 0. And that makes sense. Because if you are sampling just once from this standard normal distribution,"
    },
    {
        "Q": "At 6:59, how does Sal make the logical leap that the top angle is also Y, given that he has a 90 degree angle and an angle of 90-y?",
        "A": "The two acute angles in an right triangle must add up to 90. Therefore, if one value is being subtracted from 90 to represent an angle, that value must represent the other angle. When you add them up, they will add up to 90 (90 - y + y = 90).",
        "video_name": "R0EQg9vgbQw",
        "timestamps": [
            419
        ],
        "3min_transcript": "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, then this thing up here must be equal to y. Validate that. y plus 90 minus y plus 90 is 180 degrees, and that is useful for us because now we can express segment DE in terms of y and sine of x. What is DE to y? It's the adjacent angle, so we can think of cosine. We know that the cosine of angle y, if we look at triangle DEC right over here, we know that the cosine of y is equal to segment DE over its hypotenuse, over sine of x. You should be getting excited right about now because we've just shown, if we multiply both sides by sine of x, we've just shown that DE is equal to sine of x times cosine of y."
    },
    {
        "Q": "So at 2:05, Sal put -2y after the -6x^2. But when I was doing the problem myself, I put 8xy first and then -2y. Is the way I ordered the expression wrong and if it is...why? This was my answer:\n( 6x^2y - 6x^2 + 8xy - 2y +4 ) This was Sal's answer:\n( 6x^2y - 6x^2 - 2y + 8xy +4 )",
        "A": "Hey Tushar Gaddi!, it does not matter in what order you put it in as long as you dont change any of the negative( - ) or posiive( + ) signs. the way you wrote it was just fine. for problems like this this is why the Order Of Operations comes into play. no matter what order you put it in you will get the same answer. Hope This Helps! Good Luck and Have Fun! :)",
        "video_name": "jroamh6SIo0",
        "timestamps": [
            125
        ],
        "3min_transcript": "So let's get some practice simplifying polynomials, especially in the case where we have more than one variable over here. So I have 4x squared y minus 3x squared minus 2y. So that entire expression plus the entire expression 8xy minus 3x squared plus 2x squared y plus 4. So the first thing that jumps out at me is that I'm just adding this expression to this expression. So to a large degree, these parentheses don't matter. So I can just rewrite it as 4x squared y minus 3x squared minus 2y plus 8xy minus 3x squared plus 2x squared y plus 4. Now we can try to group similar terms or like terms. So let's think about what we have over here. So this first term right here is a 4x squared y. So can I add this to any of the other terms here? Do we have any other x squared y terms? is another x squared y term. If I have 4 of something-- in this case, I have 4x squared y's and I add 2x squared y's to it, how many x squared y's do I now have? Well, 4 plus 2-- I now have 6x squared y's. Now let's move on to this term. So I have negative 3x squareds. Do I have any more x squareds in this expression right over here? Well, sure, I have another negative 3x squared. So if I have negative 3 of something and then I have another negative 3, I end up with negative 6 So it's negative 6x squared. Now let's think about this negative 2y term. Are there any other y's over here? Well, it doesn't look like there are. This is an 8xy. This is a 4. There's no just y's. So I'll just rewrite it, negative 2y. And then 8xy-- well, once again, it doesn't seem like that can be added to anything else. So let's just write that over again. And then finally, we just have the constant term plus 4. And it pretty much looks like we're done. We have simplified this as much as we can."
    },
    {
        "Q": "At 1:11 you go through this really calculated sum but you could just find 25% of 66 and that would be your answer 66!",
        "A": "You cant just do 25% of 66 because you didn t know the 66. The only 2 numbers that you know are that you have $50 and the sale is going to take 25% off. you are calculating the $66. Hope this helps.",
        "video_name": "4oeoIOan_h4",
        "timestamps": [
            71
        ],
        "3min_transcript": "Let's say I go to a store and I have $50 in my pocket. $50 in my wallet. And at the store that day they say it is a 25% off marked price sale. So 25% off marked price means that if the marked price is $100 the price I'm going to pay is going to be 25% less than $100. So my question to you is if I have $50, what is the highest marked price I can afford? Because I need to know that before I go finding something that I might like. So let's do a little bit of algebra. So let x be the highest marked price that I can afford. price, the sale price will be x minus 25% of x is equal to the sale price. And I'm assuming that I'm in a state without sales tax. Whatever the sale price is, is what I have to pay in cash. So x minus 25% x is equal to the sale price. The discount is going to be 25% of x. But we know that this is the same thing as x minus 0.25x. And we know that that's the same thing as-- well, because we know this is 1x, x is the same thing is 1x. 1x minus 0.25x. Well, that means that 0.75x is equal to the sale price, right? All I did is I rewrote x minus 25% of x as 1x minus 0.25x. Because 1 minus 0.25 is 0.75. So 0.75x is going to be the sale price. Well, what's the sale price that I can afford? Well, the sale price I can afford is $50. So 0.75x is going to equal $50. If x is any larger number than the number I'm solving for, then the sale price is going to be more than $50 and I won't be able to afford it. So that's how we set the-- the highest I can pay is $50 and that's the sale price. So going back to how we did these problems before. We just divide both sides by 0.75. And we say that the highest marked price that I can afford is $50 divided by 0.75. And let's figure out what that is. 0.75 goes into 50-- let's add some 0's in the back."
    },
    {
        "Q": "in 5:31 Sal says mathleat what is that???",
        "A": "An athlete is someone who does physical activity and is good at it. A mathlete is some who performs math and is good at it, usually competing against other people.",
        "video_name": "xO_1bYgoQvA",
        "timestamps": [
            331
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 3:35, how is x to the a over x to the b equal to x to the a-b? I know he explains it, but it still doesn't make that much sense.",
        "A": "Try a numeric example: x^5 / x^2 This is a fraction and must be reduced. Cancel out the x^2 and you get x^3. So, what happened to the exponents? They were subtracted. x^5 / x^2 = x^(5-2) = x^3 Sal did the same exact thing, except he has variables instead of numeric exponents. Hope this helps.",
        "video_name": "0z-yIFzpunM",
        "timestamps": [
            215
        ],
        "3min_transcript": "as x to the negative b, which is going to be the same thing as... If I have a base to one exponent times the same base to another exponent, that's the same thing as that base to the sum of the exponents, a plus negative b which is just gonna be a minus b. So, we got to the same place. So, we can re-write this as... So, we can re-write this part as being equal to m to the 7/9 power minus 1/3 power is equal to, is equal to m to the k over nine. And I think you see where this is going. What is 7/9 minus 1/3? Well, 1/3 is the same thing, if we want to have a common denominator, 1/3 is the same thing as 3/9. So, I can re-write this as 3/9. So 7/9 minus 3/9 is going to be 4/9. So, this is the same thing as m to the... is going to be equal to m to the k-ninths power. So, 4/9 must be the same thing as k-ninths. So, we can say 4/9 is equal to k-ninths. Four over nine is equal to k over nine, which tells us that k must be equal to four, and we're all done."
    },
    {
        "Q": "What was the point in him specifying what the f vetor was at 5:40? He never used it to solve his problem.",
        "A": "He uses it to remind us where he gets his P(x,y) and Q(x,y). This have been explained through many of the previous videos, so i just a remainder, and not a explanation.",
        "video_name": "gGXnILbrhsM",
        "timestamps": [
            340
        ],
        "3min_transcript": "but we're saying that the curve is the boundary of this region and we're going to go in I counterclockwise direction. I have to specify that. So our curve, we could start at any point really, but we're going to go like that. Then get to that point and then come back down along that top curve just like that. And so this met the condition that the inside of the region is always going to be our left, so we can just do the straight up Green's theorem, we don't have to do the negative of it. And let's define our region. Let's just do our region. This integral right here is going to go-- I'll just do it the way-- y varies from y is equal to 2x squared to And so maybe we'll integrate with respect to y first. And then x, I'll do the outside. The boundary of x goes from 0 to 1. So they set us up well to do an indefinite integral. Now we just have to figure out what goes over here-- Green's theorem. Our f would look like this in this situation. f is f of xy is going to be equal to x squared minus y squared i plus 2xy j. We've seen this in multiple videos. You take the dot product of this with dr, you're going to get this thing right here. So this expression right here is our P of x y. And this expression right here is our Q of xy. So inside of here we're just going to apply Green's So the partial of Q with respect to x-- so take the derivative of this with respect to x. We're just going to end up with the 2y. And then from that we're going to subtract the partial of P with respect to y. So if you take the derivative of this with respect to y that becomes 0 and then, here you have-- the derivative with respect to y here is minus 2y. Just like that. And so this simplifies to 2y minus minus 2y. That's 2y plus plus 2y. I'm just subtracting a negative. Or this inside, and just to save space, this inside-- that's just 4y. I don't want to have to rewrite the boundaries. That right there is the same thing as 4y."
    },
    {
        "Q": "At 5:20, shouldn't the da be a dx?",
        "A": "Yes. He writes that originally but the step after that he rewrites it as dx.",
        "video_name": "gGXnILbrhsM",
        "timestamps": [
            320
        ],
        "3min_transcript": "And let's see. x goes from 0 to 1, so if we make-- that's obviously 0. Let's say that that is x is equal to 1, so that's all the x values. And y varies, it's above 2x squared and below 2x. Normally, if you get to large enough numbers, 2x squared is larger, but if you're below 1 this is actually going to be smaller than that. So the upper boundary is 2x, so there's 1 comma 2. This is a line y is equal to 2x, so that is the line y is-- let me draw a straighter line than that. The line y is equal to 2x looks something like that. That right there is y is equal to 2x. Maybe I'll do that in yellow. And then the bottom curve right here is y is going to be greater than 2x squared. It might look something like this. but we're saying that the curve is the boundary of this region and we're going to go in I counterclockwise direction. I have to specify that. So our curve, we could start at any point really, but we're going to go like that. Then get to that point and then come back down along that top curve just like that. And so this met the condition that the inside of the region is always going to be our left, so we can just do the straight up Green's theorem, we don't have to do the negative of it. And let's define our region. Let's just do our region. This integral right here is going to go-- I'll just do it the way-- y varies from y is equal to 2x squared to And so maybe we'll integrate with respect to y first. And then x, I'll do the outside. The boundary of x goes from 0 to 1. So they set us up well to do an indefinite integral. Now we just have to figure out what goes over here-- Green's theorem. Our f would look like this in this situation. f is f of xy is going to be equal to x squared minus y squared i plus 2xy j. We've seen this in multiple videos. You take the dot product of this with dr, you're going to get this thing right here. So this expression right here is our P of x y. And this expression right here is our Q of xy. So inside of here we're just going to apply Green's"
    },
    {
        "Q": "At 4:15 of the movie above suddenly you drop the unit vectors(i,j,k) from the result of bxc. Is it ok to do that? If it is ok, then I want to know why is it possible to do that.",
        "A": "i, j, and k are (1, 0, 0), (0. 1, 0), and (0, 0, 1). a = (a1, a2, a3) = a1i + a2j + a3k (the resultant of the components of a).",
        "video_name": "b7JTVLc_aMk",
        "timestamps": [
            255
        ],
        "3min_transcript": "of three-by-threes. And we'll put b's x term, b's y coefficient, and b's z component. And then you do the same thing for the c, cx, cy, cz. And then this is going to be equal to-- so first you have the i component. So it's going to be the i component times b. So you ignore this column and this row. So bycz minus bzcy. So I'm just ignoring all of this. And I'm looking at this two-by-two over here, minus bzcy. And then we want to subtract the j component. Remember, we alternate signs when we take our determinant. Subtract that. And then we take out that column and that row, so it's going to be bxcz-- this is a little monotonous, bxcz minus bzcx. And then finally, plus the k component. OK, we're going to have bx times cy minus bycx. We just did the dot product, and now we want to take the-- oh, sorry, we just did the cross product. I don't want to get you confused. We just took the cross product of b and c. And now we need take the cross product of that with a, or the cross product of a with this thing right over here. Instead of rewriting the vector, let me just set up another matrix here. So let me write my i j k up here. And then let me write a's components. And then let's clean this up a little bit. We're just looking at-- no, I want to do that in black. Let's do this in black, so that we can kind of erase that. Now this is a minus j times that. So what I'm going to do is I'm going to get rid of the minus and the j, but I am going to rewrite this with the signs swapped. So if you swap the signs, it's actually bzcx minus bxcz. So let me delete everything else. So I just took the negative and I multiplied it by this. I hope I'm not making any careless mistakes here, so let me just check and make my brush size little bit bigger, so I can erase that a little more efficiently. And then we also want to get rid of that right over there. Now let me get my brush size back down to normal size."
    },
    {
        "Q": "When Sal says multiply b and c by the dot products of a and c and a and b at 13:31 (b(a dot c) - c(a dot b)) what does he mean by multiply?",
        "A": "Scalar multiplication of the vector b with the scalar (a dot c) and so on",
        "video_name": "b7JTVLc_aMk",
        "timestamps": [
            811
        ],
        "3min_transcript": "plus by times j, plus bz times k. And then, from that, we're going to subtract all of this, a dot b. We're going to subtract a dot b times the exact same thing. And you're going to notice, this right here is the same thing as vector b. That is vector b. When you do it over here, you're going to get vector c. So I'll just write it over here. You're just going to get vector c. So just like that, we have a simplification for our triple product. I know it took us a long time to get here, but this is a simplification. It might not look like one, but computationally it is. It's easier to do. If I have-- I'll try to color-code it-- a cross b we just saw that this is going to be equivalent to-- and one way to think about it is, it's going to be, you take the first vector times the dot product of-- the first vector in this second dot product, the one that we have our parentheses around, the one we would have to do first-- you take your first vector there. So it's vector b. And you multiply that times the dot product of the other two vectors, so a dot c. And from that, you subtract the second vector multiplied by the dot product of the other two vectors, of a dot b. And we're done. This is our triple product expansion. Now, once again, this isn't something You could always, obviously, multiply it. You could actually do it by hand. You don't have to know this. But if you have really hairy vectors, or if this was some type of math competition, and sometimes it simplifies real fast when you reduce it to dot products, this is a useful thing to know, Lagrange's formula, or the triple product expansion."
    },
    {
        "Q": "Doesn't he mean cross product at 2:00?",
        "A": "Yes, he meant cross product.",
        "video_name": "b7JTVLc_aMk",
        "timestamps": [
            120
        ],
        "3min_transcript": "What I want to do with this video is cover something called the triple product expansion-- or Lagrange's formula, sometimes. And it's really just a simplification of the cross product of three vectors, so if I take the cross product of a, and then b cross c. And what we're going to do is, we can express this really as sum and differences of dot products. Well, not just dot products-- dot products scaling different vectors. You're going to see what I mean. But it simplifies this expression a good bit, because cross products are hard to take. They're computationally intensive and, at least in my mind, they're confusing. Now this isn't something you have to know if you're going to be dealing with vectors, but it's useful to know. My motivation for actually doing this video is I saw some problems for the Indian Institute of Technology entrance exam that seems to expect that you know Lagrange's formula, or the triple product expansion. So let's see how we can simplify this. So to do that, let's start taking the cross product of b and c. just going to assume-- let's say I have vector a. That's going to be a, the x component of vector a times the unit of vector i plus the y component of vector a times the unit vector j plus the z component of vector a times unit vector k. And I could do the same things for b and c. So if I say b sub y, I'm talking about what's scaling the j component in the b vector. So let's first take this cross product over here. And if you've seen me take cross products, you know that I like to do these little determinants. Let me just take it over here. So b cross c is going to be equal to the determinant. And I put an i, j, k up here. This is actually the definition of the cross product, so no proof necessary to show you why this is true. This is just one way to remember the dot product, of three-by-threes. And we'll put b's x term, b's y coefficient, and b's z component. And then you do the same thing for the c, cx, cy, cz. And then this is going to be equal to-- so first you have the i component. So it's going to be the i component times b. So you ignore this column and this row. So bycz minus bzcy. So I'm just ignoring all of this. And I'm looking at this two-by-two over here, minus bzcy. And then we want to subtract the j component. Remember, we alternate signs when we take our determinant. Subtract that. And then we take out that column and that row, so it's going to be bxcz-- this is a little monotonous,"
    },
    {
        "Q": "At 20:00 couldn't the graph left of zero be the right half of upward concavity ( / )?",
        "A": "No, because slope is zero at 0. Isn t it?",
        "video_name": "hIgnece9ins",
        "timestamps": [
            1200
        ],
        "3min_transcript": ""
    },
    {
        "Q": "I don't understand why at 8:45 we started looking at x > 2/3.",
        "A": "He was attempting to determine concavity. x=2/3 is a point of inflection. It is where the graph changes concavity. You need to determine the sign of the 2nd derivative at points both less than and greater than your point of inflection to determine the concavity of the function along those intervals.",
        "video_name": "hIgnece9ins",
        "timestamps": [
            525
        ],
        "3min_transcript": ""
    },
    {
        "Q": "I'm wondering about the result at the end of the video (8:07) : x^3y+1/2(x^2y^2) = c\nIf that were a diffrential equation describing a physic phenomenon, how would you turn this solution into a function you can actually use to describe the physical system you took the diffrential equation from ?\nAnother way of saying this would be : How do you turn x^3y+1/2(x^2y^2) into\ny = some x business ?",
        "A": "I believe you can use the quadratic formula to get the explicit solution in terms of y. Just move the c to the left side so you have ay^2 + ay + c = 0 ps I thought you were talking about psychic phenomena for a minute.",
        "video_name": "0NyeDUhKwBE",
        "timestamps": [
            487
        ],
        "3min_transcript": "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. would exist where the derivative of psi with respect to x would be equal to this entire expression. So we could rewrite our differential equation like this. And we'd know that a solution is psi is equal to c. And to solve for psi, we just say, OK, the partial derivative of psi with respect to x is going to be this thing. Antiderivative of both sides, and there's some constant h of y-- not constant, there's some function of y-- h of y that we might have lost when we took the partial with respect to x. So to figure that out, we take this expression. Take the partial with respect to y, and set that equal to our N expression. And by doing that, we figured out that that function of y is really just some constant. And we could have written that here. We could have written that plus c. We could call that c1 or something. But we know that the solution of our original differential equation is psi is equal to c. So the solution of our differential equation is psi x to the third y plus 1/2 x squared y We could have had this plus c1 here, then subtracted both sides. But I think I've said it so many times that you understand, why if h of y is just a c, you can kind just ignore it. Anyway, that's all for now, and I will see you in the next video. You now know a little bit about integrating factors. See you soon."
    },
    {
        "Q": "At 0:44, he said it would have been a function of y that is integrating factor. i tried u(y) instead of u(x) but I can't solve for u(y). I just curious that is it possible to use u(y) for this question too? And how do we know which one would we use? u(x) or u(y), judging for what or trial and error?",
        "A": "I have not tried to solve for the integrating factor of u(y) but i do know that there may or may not be an integrating factor for u(y). Like you said it is kind of trial and error. All of these exact equations that need an integrating factor COULD have a u(x), u(y), u(x,y) or any combination of them but it will be up to you to figure out which one it has.",
        "video_name": "0NyeDUhKwBE",
        "timestamps": [
            44
        ],
        "3min_transcript": "And, in the last video, we had this differential equation. And it at least looked like it could be exact. But when we took the partial derivative of this expression, which we could call M with respect to y, it was different than the partial derivative of this expression, which is N in the exact differential equations world. It was different than N with respect to x. And we said, oh boy, it's not exact. But we said, what if we could multiply both sides of this equation by some function that would make it exact? And we called that mu. And in the last video, we actually solved for mu. We said, well, if we multiply both sides of this equation by mu of x is equal to x, it should make this into an exact differential equation. It's important to note, there might have been a function of y that if I multiplied by both sides it would also make it exact. There might have been a function of x and y that would have done the trick. But our whole goal is just to make this exact. It doesn't matter which one we pick, which integrating factor-- this is called the integrating factor-- which integrating factor we pick. Let's solve the problem. Let's multiply both sides of this equation by mu, and mu of x is just x. We multiply both sides by x. So see, if you multiply this term by x, you get 3x squared y plus xy squared, we're multiplying these terms by x now, plus x to the third plus x squared y, y prime is equal to 0. Well now, first of all, just as a reality check, let's make sure that this is now an exact equation. So what's the partial of this expression, or this kind of sub-function, with respect to y? Well, it's 3x squared, that's just kind of a constant coefficient of y, plus 2xy, that's the partial with respect to y of that expression. 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?"
    },
    {
        "Q": "At 7:05, he begins to explain a simpler way to find the value of cesium-137 in the sample after 150 days. Though, because 150 is divisible by 30, isn't there an even simpler way? You could essentially just divide 8 in half 5 times and reach the same answer of .25. Is this valid?",
        "A": "yes, that is how I learned it in science class. Just don t forget to include year 0!!",
        "video_name": "polop-89aIA",
        "timestamps": [
            425
        ],
        "3min_transcript": "And then, they finally say-- how many becquerels of caesium-137 remain in our sample 150 days after its release in the soil? Use the rounded value of r, and round this number to the nearest hundredth. So just to be clear, we already know c and r. We know that the amount of caesium-137 in becquerels-- as a function of time in days-- is going to be equal to 8 times 0.977 to the t power-- where t is the number of days that have passed. And they're essentially saying, well, how much do we have left after 150 days? So they want us to calculate what is A of 150? Well, that's going to be 8 times 0.977 to the 150th power. So let's calculate that. So it's going to be 8 times-- and they tell us to use our rounded value of r, not the exact value of r. So it's going to be 8 times 0.977 to the 150th power. And they want us to round to the nearest hundredth. 0.24. 0.24 becquerels is kind of the radioactivity level of the caesium-137 that we have left over. Now, one interesting thing is they asked us to use the rounded value of r. So we used the rounded value of r. Because this right over here is a multiple of 30, you could actually-- in not too difficult of a way-- find out the exact value that's left over. And actually, you don't even need a calculator for it. I encourage you to pause your video Find the exact value. Well, instead of writing 0.977, let's write A of t as being equal to 8 times our r. This is an approximate value for r. If we wanted to be a little more exact, we can say that our r is one half to the 1/30th power. And we're going to raise that to the t power. Or we could say A of t is equal to 8 times one half to the t/30 power. If we raise something to an exponent and then raise that to an exponent, we can take the product of those exponents. So that's one half to the t/30 power. Let me actually do that in another color. Let me do that in yellow. So that's 8 times one half to the t/30 power. And actually, I don't need this parentheses right over here. This is another way to describe A of t."
    },
    {
        "Q": "hello there its your friendly neighborhood precal student here again\n\ni have a synthetic division question in my summer homework (yipee!)\nthe denominator is 3x-2\nwould i still put positive 2 at the same place where Sal puts the 3 at 2:00 in the video\nor does the 3x affect this?\n\nThanks for your help :)",
        "A": "It depends on the question. You look for the highest polynomial which you want to subtract away. If you re dividing the same polynomial, but now by 3x-2, yes, you would still put a 2 down there, but note that you multiplied it by 3/2. Remember to upvote good questions and helpful answers",
        "video_name": "3Ee_huKclEQ",
        "timestamps": [
            120
        ],
        "3min_transcript": "Let's do another synthetic division example. And in another video, we actually have the why this works relative to algebraic long division. But here it's going to be another just, let's go through the process of it just so that you get comfortable with it. And now is a good chance to give it a shot, to actually try to simplify this rational expression. So let's think about this step by step. So the first thing I want to do is write all of the coefficients of the numerator. So I have a 2. Oh, I have to be careful here. Because the 2 is the coefficient for x to the fifth, I have no x to the fourth term. Let me start over. So I have the 2 from 2x to the fifth. And then I have no x to the fourth. So it's really 0x to the fourth. So I'll put a 0 as the coefficient for the x to the fourth term. And then I have a negative 1 times x to the third. And then I have a positive 3 times x squared. And then I have a constant term, or zero degree term, of 7. I just have a positive 7. And now let me just draw my little funky synthetic division operator-looking symbol. And remember, the type of synthetic division we're doing, it only applies when we are dividing by an x plus or minus something. There's a slightly different process you would have to do if it was 3x or if was negative 1x or if it was 5x squared. This only works when we have x plus or minus something. In this case we have x minus 3. So we have the negative 3 here. And the process we show-- there's other ways of doing it-- is you take the negative of this. So the negative of negative 3 is positive 3. And now we're ready to perform our synthetic division. So we'll bring down this 2 and then multiply 2 time 3 gives us 6. 0 plus 6 is 6. And then we multiply that times the 3, and we get positive 18. Negative 1 plus 18 is 17. Multiply that times the 3. 17 times 3 is 51. 3 plus 51 is 54. Multiply that times 3. The numbers are getting kind of large now. So that's going to be what? 50 times 3 is 150. 4 times 3 is 12. So this is going to be 162. Negative 2 plus 162 is 160."
    },
    {
        "Q": "Wait at 4:57, Sal doesn't put the x^5 term. He puts the highest degree as x^4. Am I missing something? I thought that 487 would be the constant and that perhaps that meant the equation was even or something. What happened to the x^5 term?",
        "A": "Remember, when you are dividing with terms containing variables that instead of adding the exponents (i.e., like you do when you re multiplying them), you subtract the exponents. So in the example (e.g., 2x^5/x), the exponent subtraction would be 5-1=4 or your x^4 term, just as in the next term : x^3/x would yield your 3-1=2 or x^2 term, etc. Hope this helps.",
        "video_name": "3Ee_huKclEQ",
        "timestamps": [
            297
        ],
        "3min_transcript": "2 time 3 gives us 6. 0 plus 6 is 6. And then we multiply that times the 3, and we get positive 18. Negative 1 plus 18 is 17. Multiply that times the 3. 17 times 3 is 51. 3 plus 51 is 54. Multiply that times 3. The numbers are getting kind of large now. So that's going to be what? 50 times 3 is 150. 4 times 3 is 12. So this is going to be 162. Negative 2 plus 162 is 160. And you add 480 to 7, and you get 487. And you can think of it, I only have one term or one number to the left-hand side of this bar here. Or I'm just doing the standard, traditional x plus or minus something version of synthetic division, I should say. So I can separate this out, and now I've essentially gotten my answer. And it looks like voodoo, and it kind of is voodoo. And that's why I don't like to do it, because you're just memorizing an algorithm. But there are other videos why we explain why. And it can be fast and convenient and paper saving very often, like you see right here. But then we have our final answer. It's going to be-- and let me work backwards. So I'll start with our remainder. So our remainder is 487. And it's going to be 487 over x minus 3. And so you're going to have plus 160 plus 487 over x minus 3. Now this is our x term. So it's going to be 54x plus all of this. This is going to be our x squared term. So this is going to be 17x squared plus 54x plus 160 and all of that. Then this is going to be x to the third term. So this is going to be 6x to the third plus all of that. And then finally, this is our x to the fourth term-- 2x to the fourth. And let me erase this. So then I have my x to the fourth term. So it is 2x to the fourth. And we are done. This thing simplifies to this right over here. And I encourage you to verify it with traditional algebraic long division."
    },
    {
        "Q": "At 4:07 why is it h over two? I am still confused on how he exactly got that.",
        "A": "The area of a circle is \u00e1\u00b4\u00a8r\u00c2\u00b2, where r is the radius, right? We are told the diameter of the circle in this problems is h, right? What is the relationship between the radius and diameter of a circle? 2r=D, here D=h so 2r=h which means that, the radius r = h/2. You may want to review similar triangles.",
        "video_name": "Xe6YlrCgkIo",
        "timestamps": [
            247
        ],
        "3min_transcript": "of that relationship, possibly using the chain rule, to come up with a relationship between the rate at which the volume is changing and the rate at which the height is changing. So let's try to do it step by step. So first of all, can we come up with a relationship between the volume and the height at any given moment? Well we have also been given the formula for the volume of a cone right over here. The volume of a cone is 1/3 times the area of the base of the cone, times the height. And we won't prove it here, although we could prove it later on. Especially when we start doing solids of revolutions within in integral calculus. But we'll just take it on faith right now, that this is how we can figure out the volume of a cone. So given this can we figure out volume-- can we figure out an expression that relates volume to the height of the cone? Well we could say that volume-- and I'll do it in this blue color-- the volume of water is what we really care about. The volume of water is going to be equal to 1/3 times the area of the surface of the water-- So times h. So how can we figure out the area of the water surface, preferably in terms of h? Well we see right over here, the diameter across the top of the cone is 4 centimeters. And the height of the whole cup is 4 centimeters. And so that ratio is going to be true of any-- at any depth of water. It's always going to have the same ratio between the diameter across the top and the height. Because these are lines right over here. So at any given point, the ratio between this and this is going to be the same. So at any given point, the diameter across the surface of the water-- if the depth is h, the diameter across the surface of the water is also going to be h. And so from that we can figure out what are the radius is going to be. The radius is going to be h over 2. 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."
    },
    {
        "Q": "At 5:20, what is he talking about when he says \"This business\"?",
        "A": "Occasionally in these videos Sal saves a little time by using the words this business as a way to refer to some expression he s manipulating on the screen. In this video, as he says that, Sal is putting brackets around the expression \u00cf\u0080h^3/12, so that s what he means here by this business.",
        "video_name": "Xe6YlrCgkIo",
        "timestamps": [
            320
        ],
        "3min_transcript": "So times h. So how can we figure out the area of the water surface, preferably in terms of h? Well we see right over here, the diameter across the top of the cone is 4 centimeters. And the height of the whole cup is 4 centimeters. And so that ratio is going to be true of any-- at any depth of water. It's always going to have the same ratio between the diameter across the top and the height. Because these are lines right over here. So at any given point, the ratio between this and this is going to be the same. So at any given point, the diameter across the surface of the water-- if the depth is h, the diameter across the surface of the water is also going to be h. And so from that we can figure out what are the radius is going to be. The radius is going to be h over 2. 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."
    },
    {
        "Q": "At 5:50 how did he get the derivative to be just pi/12?",
        "A": "That s not what happened here. For convenience, Sal moved the constant (pi/12) outside the derivative. He had the derivative of (pi*h^3)/12, which is the same as (pi/12)*h^3, and he rewrote it as (pi/12) times the derivative of h^3.",
        "video_name": "Xe6YlrCgkIo",
        "timestamps": [
            350
        ],
        "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": "what is the meaning of 'inx' at 1:31.",
        "A": "That is an l, not an i. The ln stands for the natural logarithm, or the logarithm with base e. The number e is called Euler s number (pronounced oiler ), or the natural base. It is the base of an exponential that is its own derivative, which is a handy thing in calculus and differential equations.",
        "video_name": "OkFdDqW9xxM",
        "timestamps": [
            91
        ],
        "3min_transcript": "Use the change of base formula to find log base 5 of 100 to the nearest thousandth. So the change of base formula is a useful formula, especially when you're going to use a calculator, because most calculators don't allow you to arbitrarily change the base of your logarithm. They have functions for log base e, which is a natural logarithm, and log base 10. So you generally have to change your base. And that's what the change of base formula is. And if we have time, I'll tell you why it makes a lot of sense, or how we can derive it. So the change of base formula just tells us that log-- let me do some colors here-- log base a of b is the exact same thing as log base x, where x is an arbitrary base of b, over log base, that same base, base x over a. 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--"
    },
    {
        "Q": "At 6:08 in the video, why does the \"log a\" become \"log x\"?\nAlso, why is the \"a\" taken away from the subscript of the log?",
        "A": "It is not taken away, but the log form is re-written into exponential form, where a is the base and the exponent is written in superscript. log a does not become log x Sal takes the log base x of both sides (of the exponential equation a^c=b) because it is essentially the same thing by comparison and he is not changing the value of the equation.",
        "video_name": "OkFdDqW9xxM",
        "timestamps": [
            368
        ],
        "3min_transcript": "think about why this property, why this thing right over here makes sense. So if I write log base a-- I'll try to be fair to the colors-- log base a of b. Let's say I set that to be equal to some number. Let's call that equal to c, or I could call it e for-- Well, I'll say that's equal to c. So that means that a to the c-th power is equal to b. This is an exponential way of writing this truth. This is a logarithmic way of writing this truth. This is equal to b. Now, we can take the logarithm of any base of both sides of this. Anything you do, if you say 10 to the what power equals this? because these two things are equal to each other. So let's take the same logarithm of both sides of this, the logarithm with the same base. And I'll actually do log base x to prove the general case, So I'm going to take log of base x of both sides of this. So this is log base x of a to the c power-- I try to be faithful to the colors-- is equal to log base x of b. And let me close it off with orange, as well. And we know from our logarithm properties, log of a to the c is the same thing as c times the logarithm of whatever base we are of a. Let me put-- I can just write a b, right over there. And if we wanted to solve for c, you just divide both sides by log base x of a. So you would get c is equal to-- and I'll stick to the color-- so it's log base x of b, which is this, over log base x of a. And this was what c was. c was log base a of b. It's equal to log base a of b. Let me write it this way. Let me write it-- Well, let me do the original color codes just so it becomes very clear what I'm doing. I think you know where this is going, but I want to be fair to the colors. So c is equal to log base x of b over-- let me scroll down a little bit-- log base x, dividing both sides by that, of a."
    },
    {
        "Q": "At 0:49, he says x is an arbitory base. What is that?",
        "A": "I m assuming that he means that x is just a letter. Your variable could be almost any letter of the alphabet and it still means the same thing.",
        "video_name": "OkFdDqW9xxM",
        "timestamps": [
            49
        ],
        "3min_transcript": "Use the change of base formula to find log base 5 of 100 to the nearest thousandth. So the change of base formula is a useful formula, especially when you're going to use a calculator, because most calculators don't allow you to arbitrarily change the base of your logarithm. They have functions for log base e, which is a natural logarithm, and log base 10. So you generally have to change your base. And that's what the change of base formula is. And if we have time, I'll tell you why it makes a lot of sense, or how we can derive it. So the change of base formula just tells us that log-- let me do some colors here-- log base a of b is the exact same thing as log base x, where x is an arbitrary base of b, over log base, that same base, base x over a. 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--"
    },
    {
        "Q": "At 0:16, what is theoretical probability?",
        "A": "Theoretical probability is basically calculating a probability for something without doing the actual experiment. It is what should happen in theory, thus theoretical probability.",
        "video_name": "RdehfQJ8i_0",
        "timestamps": [
            16
        ],
        "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 3:20, why is it 1/15",
        "A": "To change 16/15 into a mixed number, you divide. 15 goes into 16 once (that s the whole number) Subtract 16 - 15 = 1. We have a remainder of 1 that becomes the new numerator Thus, you get 1 1/15 Hope this helps.",
        "video_name": "bcCLKACsYJ0",
        "timestamps": [
            200
        ],
        "3min_transcript": "as something over 30. So nine over 10. How would I write that as something over 30? Well I multiply the denominator, I'm multiplying the denominator by three. So I've just multiplied the denominator by three. So if I don't want to change the value of the fraction, I have to do the same thing to the numerator. I have to multiply that by three as well because now I'm just multiplying the numerator by three and the denominator by three, and that doesn't change the value of the fraction. So nine times three is 27. So once again, 9/10 and 27/30 represent the same number. I've just written it now with a denominator of 30, and that's useful because I can also write 1/6 with a denominator of 30. Let's do that. So 1/6 is what over 30? I encourage you to pause the video and try to think about it. So what did we do go from six to 30? We had to multiply by five. So if we multiply the denominator by five, so one times five, one times five is five. So 9/10 is the same thing as 27/30, and 1/6 is the same thing as 5/30. And now we can add, now we can add and it's fairly straightforward. We have a certain number of 30ths, added to another number of 30ths, so 27/30 + 5/30, well that's going to be 27, that's going to be 27 plus five, plus five, plus 5/30, plus 5/30, which of course going to be equal to 32/30. 32 over 30, and if we want, we could try to reduce this fraction. We have a common factor of 32 and 30, they're both divisible by two. So if we divide the numerator and the denominator by two, denominator divided by two is 15. So, this is the same thing as 16/15, and if I wanted to write this as a mixed number, 15 goes into 16 one time with a remainder one. So this is the same thing as 1 1/15. Let's do another example. Let's say that we wanted to add, we wanted to add 1/2 to to 11/12, to 11 over 12. And I encourage you to pause the video and see if you could work this out. Well like we saw before, we wanna find a common denominator. If these had the same denominator, we could just add them immediately, but we wanna find a common denominator because right now they're not the same. Well what we wanna find is a multiple, a common multiple of two and 12, and ideally we'll find the lowest common multiple of two and 12, and just like we did before, let's start with the larger of the two numbers, 12."
    },
    {
        "Q": "At 9:22 How do you know to put the number before the decimal, or at 5:06 the zero?",
        "A": "It depends if the denominator divides before the decimal point or after. For instance, if you had 31/15, it would come out to 2.1. I hope that this was helpful! :)",
        "video_name": "Gn2pdkvdbGQ",
        "timestamps": [
            562,
            306
        ],
        "3min_transcript": "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. So this is the same thing as 3/100. So let me ask you, are these two the same? Sure they are. If we divide both the numerator and the denominator of both of these expressions by 10 we get 3/100. Let's go back to this case. Are we done with this? Is 35/1,000-- I mean, it's right. That is a fraction. 35/1,000. But if we wanted to simplify it even more looks like we could divide both the numerator and the denominator by 5. And then, just to get it into simplest form, that equals 7/200. And if we wanted to convert 7/200 into a decimal using the technique we just did, so we would do 200 goes into 7 and figure it out. We should get 0.035. I'll leave that up to you as an exercise. to convert a fraction into a decimal and maybe vice versa. And if you don't, just do some of the practices. And I will also try to record another module on this or another presentation. Have fun with the exercises."
    },
    {
        "Q": "In 1:40 how come we don't multiply by -1 to make -3p a positive and flip the inequality? I've seen it in other problems before.",
        "A": "You can either multiply with -1, divide by -3, or just swap the numbers (the left goes to the right and vice versa), do as you like.",
        "video_name": "0YErxSShF0A",
        "timestamps": [
            100
        ],
        "3min_transcript": "Solve for z. 5z plus 7 is less than 27 or negative 3z is less than or equal to 18. So this is a compound inequality. We have two conditions here. So z can satisfy this or z can satisfy this over here. So let's just solve each of these inequalities. And just know that z can satisfy either of them. So let's just look at this. So if we look at just this one over here, we have 5z plus 7 is less than 27. Let's isolate the z's on the left-hand side. So let's subtract 7 from both sides to get rid of this 7 on the left-hand side. And so our left-hand side is just going to be 5z. Plus 7, minus 7-- those cancel out. 5z is less than 27 minus 7, is 20. So we have 5z is less than 20. Now we can divide both sides of this inequality by 5. And we don't have to swap the inequality because we're And so we get z is less than 20/5. z is less than 4. Now, this was only one of the conditions. Let's [? look at ?] the other one over here. We have negative 3z is less than or equal to 18. Now, to isolate the z, we could just divide both sides of this inequality by negative 3. But remember, when you divide or multiply both sides of an inequality by a negative number, you have to swap the inequality. So we could write negative 3z. We're going to divide it by negative 3. And then you have 18. We're going to divide it by negative 3. But we're going to swap the inequality. So the less than or equal will become greater than or equal to. And so these guys cancel out. Negative 3 divided by negative 3 is 1. So we have z is greater than or equal to 18 over negative 3 is negative 6. And remember, it's this constraint or this constraint. And this constraint right over here boils down to this. So our solution set-- z is less than 4 or z is greater than or equal to negative 6. So let me make this clear. Let me rewrite it. So z could be less than 4 or z is greater than or equal to negative 6. It can satisfy either one of these. And this is kind of interesting here. Let's plot these. So there's a number line right over there. Let's say that 0 is over here. We have 1, 2, 3, 4 is right over there. And then negative 6. We have 1, 2, 3, 4, 5, 6. That's negative 6 over there. Now, let's think about z being less than 4. We would put a circle around 4, since we're not including 4. And it'd be everything less than 4."
    },
    {
        "Q": "at 1:48, i didnt under stand why -32/-3 was crossed out can some one\nexplain it to me",
        "A": "Watch the video more carefully. It was not -32/-3, it was -3*z/-3 since you have a -3 on the numerator and denominator they cancel out, so that just leaves you with z.",
        "video_name": "0YErxSShF0A",
        "timestamps": [
            108
        ],
        "3min_transcript": "Solve for z. 5z plus 7 is less than 27 or negative 3z is less than or equal to 18. So this is a compound inequality. We have two conditions here. So z can satisfy this or z can satisfy this over here. So let's just solve each of these inequalities. And just know that z can satisfy either of them. So let's just look at this. So if we look at just this one over here, we have 5z plus 7 is less than 27. Let's isolate the z's on the left-hand side. So let's subtract 7 from both sides to get rid of this 7 on the left-hand side. And so our left-hand side is just going to be 5z. Plus 7, minus 7-- those cancel out. 5z is less than 27 minus 7, is 20. So we have 5z is less than 20. Now we can divide both sides of this inequality by 5. And we don't have to swap the inequality because we're And so we get z is less than 20/5. z is less than 4. Now, this was only one of the conditions. Let's [? look at ?] the other one over here. We have negative 3z is less than or equal to 18. Now, to isolate the z, we could just divide both sides of this inequality by negative 3. But remember, when you divide or multiply both sides of an inequality by a negative number, you have to swap the inequality. So we could write negative 3z. We're going to divide it by negative 3. And then you have 18. We're going to divide it by negative 3. But we're going to swap the inequality. So the less than or equal will become greater than or equal to. And so these guys cancel out. Negative 3 divided by negative 3 is 1. So we have z is greater than or equal to 18 over negative 3 is negative 6. And remember, it's this constraint or this constraint. And this constraint right over here boils down to this. So our solution set-- z is less than 4 or z is greater than or equal to negative 6. So let me make this clear. Let me rewrite it. So z could be less than 4 or z is greater than or equal to negative 6. It can satisfy either one of these. And this is kind of interesting here. Let's plot these. So there's a number line right over there. Let's say that 0 is over here. We have 1, 2, 3, 4 is right over there. And then negative 6. We have 1, 2, 3, 4, 5, 6. That's negative 6 over there. Now, let's think about z being less than 4. We would put a circle around 4, since we're not including 4. And it'd be everything less than 4."
    },
    {
        "Q": "At 2:21, could you explain how to determine the greatest integer between an interval for example [0,1]?",
        "A": "firstly, integers are all positive and negative whole numbers including 0: ...-3,-2,-1,0,1,2,3... so what are the integers between the interval [0,1]? 0 and 1 which is greater 0 or 1? 1 So the answer to your question is 1.",
        "video_name": "CZdziIlYIfI",
        "timestamps": [
            141
        ],
        "3min_transcript": "For any real number x, let brackets around x denote the largest integer less than or equal to x, often known as the greatest integer function. Let f be a real valued function defined on the interval negative 10 to 10, including the boundaries by f of x is equal to x minus the greatest integer of x, if the greatest integer of x is odd, and 1 plus the greatest integer of x minus x, if the greatest integer of x is even. Then the value of pi squared over 10 times a definite integral from negative 10 to 10 of f of x cosine of pi of x dx is-- so before even try to attempt to evaluate this integral, let's see if we can at least visualize this function, f of x, right over here. So let's do our best to visualize it. So let me draw my x-axis. And let me draw my y-axis. So let me draw my y-axis. And then let's think about what this function will look like. this is x is equal to 1, x is equal to 2, x is equal to 3. We could go down to negative 1, negative 2. We could just keep going, if we like. Hopefully we'll see some type of pattern, because it seems to change from odd to even. So between 0 and 1, what is the greatest integer of x? So let me just write it over here. So between 0 and 1, until you get to 1-- so maybe I should do this-- from including 0 until 1, the greatest integer of x is equal to 0. If I'm at 0.5, the greatest integer below 0.5 is 0. As I go from 1 to 2, this brackets around x It's the greatest integer. If I'm at 1.9, the greatest integer is 1. And then if I go to above from between 2 and 3, then the greatest integer is going to be 2. If I'm at 2.5, greatest integer is going to be 2. So with that, let's try to at least draw this function over these intervals. So between 0 and 1, the greatest integer is 0. 0 we can consider to be even. 0 is even, especially if we're alternating. 1 is odd, 2 is even, 3 is odd. So 0 is even. So we would look at this circumstance right over here, if x is even. And then over this time frame or over this part of the x-axis, the greatest integer of x is just 0. So the equation or the line or the function is just going to be 1 minus x over this interval, because the greatest integer is 0. So 1 minus x will look like this. If this is 1, 1 minus x just goes down like that."
    },
    {
        "Q": "At 7:24, I still do not quite understand why the integral of f(x) between 0 and 1 is the same as the integral of f(x) between 1 and 2? I mean I can kind of visualise it, but how can I prove this is true? Thanks a lot:)",
        "A": "That is the case because of symmetry. The area between 0 and 1 is the mirror image (along y = 1) of the area between 1 and 2. The easiest way to prove it is to calculate both integrals and see that the result is the same. Int from 0 to 1 (1-x)*cos(pi*x) = Int from 1 to 2 (x-1)*cos(pi*x) = 2/pi^2",
        "video_name": "CZdziIlYIfI",
        "timestamps": [
            444
        ],
        "3min_transcript": "Don't want you to get that wrong. Cosine pi 0 is cosine of 0. So that's 1. Cosine of pi is negative 1. So when x is equal to 1, this becomes cosine of pi. So then the value of the function is negative 1. It'll be over here. And then cosine of 2 pi, 2 times pi, is then 1 again. So it'll look like this. This is at 1/2. When you put it over here, it'll become pi over 2. Cosine of pi over 2 is 0. So it'll look like this. Let me draw it as neatly as possible. So it will look like this. Cosine, and then it'll keep doing that, and then it'll go like this. So it is also periodic. So if we wanted to figure out the integral of the product from negative 10 to 10, can we simplify that? And it looks like it would just be, because we have this interval, let's look at this interval over here. Let's look at just from 0 to 1. So just from 0 to 1, we're going to take this function and take the product of this cosine times essentially 1 minus x, and then find the area under that curve, whatever it might be. Then when we go from 1 to 2, when we take the product of this and x minus 1, it's actually going to be the same area, because these two, going from 0 to 1 and going from 1 to 2, it's completely symmetric. You can flip it over this line of symmetry, and both functions are completely symmetric. So you're going to have the same area when you take their product. So what we see is, over every interval, from 2 to 3 is clearly the same thing as the integral from 0 to 1. Both functions look identical over that interval. But it will also be the same as going from 1 to 2, because it's completely symmetric. When you take the products of the function, that function will be completely symmetric around this axis. So the integral from here to here will be the same as the integral from there to there. So with that said, we can rewrite this thing over here. So what we want to evaluate, pi squared over 10 times the integral from negative 10 to 10 of f of x cosine of pi x, using the logic we just talked about. This is going to be the same thing as being equal to pi squared over 10 times the integral from 0 to 1, but 20 times that, because we have 20 integers between negative 10 and 10."
    },
    {
        "Q": "at 10:50 it says du is equal to dx. why he doesn't write 1? and at 12:02 it write again dx, instead of 1.\nCould anyone make some sense of this?\nI guess it is related to the dx at the end of the \"\u00e2\u008c\u00a0f(x) dx \" formula, but still not sure what that dx represent, and why it is supposed to be 1 here... i'm confused",
        "A": "1 is there but it is noted implicitly. In calculus, 1 is unusually implied in problems. Dx less us to do the substitution and it essentially means(in that context) that we are differentiating. When I do these kinds of problems, I don t really think about the formal definition of dx.",
        "video_name": "CZdziIlYIfI",
        "timestamps": [
            650,
            722
        ],
        "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": "At about 1:49 Sal is explaining that if x=1 than [x]=0, because it is the largest integer, but it says that [x] must be lesser or EQUAL to x! Why is that?",
        "A": "In real life, you can t deal, in decimal number, with how many siblings do you have. Anyway, the greatest integral function is, just, a function to create, only, an integer value. So, why does define as must be lesser or equal to x is because when you evaluate x = 0.7, you notice that 0.7 is not completing any integer number, therefore, it is f([0.7]) = 0. This f(x)= [x] looks, graphically, for the greatest integer function of [x], like stairs. Take off the decimal of any x, make it an integer.",
        "video_name": "CZdziIlYIfI",
        "timestamps": [
            109
        ],
        "3min_transcript": "For any real number x, let brackets around x denote the largest integer less than or equal to x, often known as the greatest integer function. Let f be a real valued function defined on the interval negative 10 to 10, including the boundaries by f of x is equal to x minus the greatest integer of x, if the greatest integer of x is odd, and 1 plus the greatest integer of x minus x, if the greatest integer of x is even. Then the value of pi squared over 10 times a definite integral from negative 10 to 10 of f of x cosine of pi of x dx is-- so before even try to attempt to evaluate this integral, let's see if we can at least visualize this function, f of x, right over here. So let's do our best to visualize it. So let me draw my x-axis. And let me draw my y-axis. So let me draw my y-axis. And then let's think about what this function will look like. this is x is equal to 1, x is equal to 2, x is equal to 3. We could go down to negative 1, negative 2. We could just keep going, if we like. Hopefully we'll see some type of pattern, because it seems to change from odd to even. So between 0 and 1, what is the greatest integer of x? So let me just write it over here. So between 0 and 1, until you get to 1-- so maybe I should do this-- from including 0 until 1, the greatest integer of x is equal to 0. If I'm at 0.5, the greatest integer below 0.5 is 0. As I go from 1 to 2, this brackets around x It's the greatest integer. If I'm at 1.9, the greatest integer is 1. And then if I go to above from between 2 and 3, then the greatest integer is going to be 2. If I'm at 2.5, greatest integer is going to be 2. So with that, let's try to at least draw this function over these intervals. So between 0 and 1, the greatest integer is 0. 0 we can consider to be even. 0 is even, especially if we're alternating. 1 is odd, 2 is even, 3 is odd. So 0 is even. So we would look at this circumstance right over here, if x is even. And then over this time frame or over this part of the x-axis, the greatest integer of x is just 0. So the equation or the line or the function is just going to be 1 minus x over this interval, because the greatest integer is 0. So 1 minus x will look like this. If this is 1, 1 minus x just goes down like that."
    },
    {
        "Q": "In 1:03 can a radius come from outside of the circle also?",
        "A": "A radius is a line from the center of a circle to the circumference.",
        "video_name": "04N79tItPEA",
        "timestamps": [
            63
        ],
        "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 5:34, how did Sal Khan get a rise over run of -5/6 if you're supposed to subtract the 5 from 5x + 6y = 30 and then divide 6y by 6? I thought you get a slope and y-intercept of 5/6 + 5.",
        "A": "As you said... you subtract 5x . You do this on both sides of the equation. The 5x is now on the other side with a minus in front of it. Thus it is now -5x . Divide by 6 and you get: -5/6 (x).",
        "video_name": "LNSB0N6esPU",
        "timestamps": [
            334
        ],
        "3min_transcript": "So let's say that I had the linear equation. Let's say that I have 5x + 6y = 30. I encourage you to pause this video, and figure out what are the x and y-intercepts for the graph that represents the solutions, all the xy pairs that satisfy this equation. Well the easiest thing to do here, let's see what the y value is when x = 0 and what x value is when y = 0. When x = 0 this becomes 6y = 30. So 6 times what is 30? Well y would be equal to 5 here. So when x is 0, y is 5. What about when y is 0? Well when y is 0, that's going to be 0, and you have 5x = 30. Well then x would be equal to 6. So we could plot those points, 0, 5. When x is 0, y is 5. When x is 6, y is 0. So those are both points on this graph and then the actual graph is going to, or the actual line that represents the x and y pairs that satisfy this equation is going to look like, it's going to look like this. I'll just try. So I can make it go, it's going to look like... It's going to go through those two points. So it going to...I can make it go the other way too. Let me see. It's going to go through those two points and so it's going to look something like that. Now what are its' x and y-intercepts? Well, we already kind of figured it out but the intercepts themselves, these are the points on the graph where they intersect the axes. That point is the y-intercept and it happens, it's always going to happen when x = 0, and when x = 0 we know that y = 5. It's that point, the point 0, 5. And what is the y inter...what is the x-intercept? The x-intercept is the point, it's actually the same x-intercept for this equation right over here. It's the point 6, 0. That point right over there."
    },
    {
        "Q": "At 0:36, why would you add -3/4? Shouldn't it be subtracted?",
        "A": "Adding a negative is equal to subtracting its positive! For example, 2-1 = 2+-1 =-1+2, and all of these are correct. I think the -3/4 was placed before -10/6 so that this equation can look like the previous equation dragged down. :)",
        "video_name": "9tmtDBpqq9s",
        "timestamps": [
            36
        ],
        "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": "At 0:14 how does he turn the seven-sixths to negative.",
        "A": "Since there is a subtraction sign after the - 3/4, this could be viewed as adding negatives. Since 7/6 ad 3/6 have common denominators already there, we could block them up to solve the problem easier. -7/6 + -3/6 = -10/6 Then you can find the solution by scaling the numerator and denominators of -10/6 and -3/4 and then add the two together.",
        "video_name": "9tmtDBpqq9s",
        "timestamps": [
            14
        ],
        "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": "At 1:30, why does he say \"the negative direction\"?",
        "A": "lim : Means you are inputting smaller negative values. x\u00e2\u0086\u00920- Relating the above limt to the example, ln(n \u00e2\u0089\u00a4 0) is undifined. Hence the given limit condition: lim x\u00e2\u0086\u00920+",
        "video_name": "CDf_aE5yg3A",
        "timestamps": [
            90
        ],
        "3min_transcript": "- [Voiceover] What I would like to tackle in this video is what I consider to be a particularly interesting limits problem. Let's say we want to figure out the limit as X approaches zero from the positive direction of sine of X. This is where it's about to get interesting. Sine of X to the one over the natural log of X power and I encourage you to pause this video and see if you can have a go at it fully knowing that this is a little bit of a tricky exercise. I'm assuming you have attempted. Some of you might have been able to figure out on the first pass. I will tell you that the first time that I encountered something like this, I did not figure it out at the first pass so definitely do not feel bad if you fall into that second category. What many of you all probably did is you said okay, let me think about it. Let me just think about the components here. If I were to think about the limit, if I were to think about the limit as X approaches zero from the positive direction of sine of X, well that's pretty straightforward. That's going to be zero, is going to approach zero but then if you say, and you could say, I guess I should say. The limit as X approaches zero from the positive direction of one over natural log of X and this is why we have to think about it from the positive direction. It doesn't make sense to approach it from the negative direction. You can't take the natural log of a negative number. That's not in the domain for the natural log but as you get closer and closer to zero from the negative direction, the natural log of those values, you have to raise E to more larger and larger negative values. This part over here is going to approach negative infinity. It's going to go to negative infinity. One over negative infinity, one divided by super large or large magnitude negative numbers, well, that's just going to approach zero. You could say that this right over here is also going to be, is also going to be equal to zero. That doesn't seem to help us much because if this thing is going to zero and that thing is going to zero, it's kind of an implication that well to the zero power but we don't really know what zero, let me do the some, those color. Zero to the zero power but this is one of those great fun things to think about in mathematics. There's justifications why this could be zero, justifications why this could be one. We don't really know what to make of this. This isn't really a satisfying answer. Something at this point might be going into your brain. We have this thing that we've been exposed to called L'Hopital's rule. If you have not been exposed to it, I encourage you to watch the video, the introductory video on L'Hopital's rule. In L'Hopital's rule, let me just write it down. L'Hopital's rule helps us out with situations where when we try to superficially evaluate the limit, we get indeterminate forms things like zero over zero. We get infinity over infinity. We get negative infinity over negative infinity and we go into much more detail into that video."
    },
    {
        "Q": "At 4:30 he says there's no evidence that DC is equivalent to AC. Are we sure? How do we know DC is not AC? Is it bigger or smaller or what?",
        "A": "We know DC is not equivalent to AC because if it was the case, then we would have an isosceles triangle. A theorem associated with isosceles triangles is that the two angles opposite the equal lengths must be equal. That is not the case, since 31 degrees is opposite to AC and 59 is opposite to DC.",
        "video_name": "TugWqiUjOU4",
        "timestamps": [
            270
        ],
        "3min_transcript": "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, 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."
    },
    {
        "Q": "I'm a bit confused, I thought DC and EF were the same because both triangles are similar 3:30. Can someone explain?",
        "A": "If the triangles are congruent, then corresponding parts of corresponding triangles are congruent, however, if triangles are similar (AA is one method to show this), then the corresponding sides are proportional ( a common scale factor to get from each side to its corresponding side on the other triangle).",
        "video_name": "TugWqiUjOU4",
        "timestamps": [
            210
        ],
        "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 4:14 where did you get the plus sign from? Wouldn't you put a subtraction sign there? (because the previous color coded terms had a subtraction sign at the beginning?)",
        "A": "At that point in the video, Sal is combining: -2x^2 + 3x^2. He does this by adding/subtracting the coefficients: -2 + 3 = +1, not a -1. So, once combined, the 2 terms creates +x^2. Hope this helps.",
        "video_name": "FNnmseBlvaY",
        "timestamps": [
            254
        ],
        "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 0:48 isn't xy a 5 not a 6? just asking :D",
        "A": "\u00f0\u009d\u0091\u00a5\u00f0\u009d\u0091\u00a6 = \u00f0\u009d\u0091\u00a5 \u00e2\u0088\u0099 \u00f0\u009d\u0091\u00a6 \u00f0\u009d\u0091\u00a5 = 3, \u00f0\u009d\u0091\u00a6 = 2 \u00e2\u0087\u0092 \u00f0\u009d\u0091\u00a5\u00f0\u009d\u0091\u00a6 = 3 \u00e2\u0088\u0099 2 = 6",
        "video_name": "FNnmseBlvaY",
        "timestamps": [
            48
        ],
        "3min_transcript": "Now we have a very, very, very hairy expression. And once again, I'm going to see if you can simplify this. And I'll give you little time to do it. So this one is even crazier than the last few we've looked at. We've got y's and xy's, and x squared and x's, well more just xy's and y squared and on and on and on. And there will be a temptation, because you see a y here and a y here to say, oh, maybe I can add this negative 3y plus this 4xy somehow since I see a y and a y. But the important thing to realize here is that a y is different than an xy. Think about it they were numbers. If y was 3 and an x was a 2, then a y would be a 3 while an xy would have been a 6. And a y is very different than a y squared. Once again, if the why it took on the value 3, then the y squared would be the value 9. So even though you see the same letter here, they aren't the same-- I guess you cannot add these two 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."
    },
    {
        "Q": "At 0:02, how is it possible to have a term like 4xy?",
        "A": "If I square (2x + y)^2, then (2x+y)(2x+y) = 4x^2 + 4xy + y^2. We can combine variables, and this will we a term that could be combined with any other xy terms we might have.",
        "video_name": "FNnmseBlvaY",
        "timestamps": [
            2
        ],
        "3min_transcript": "Now we have a very, very, very hairy expression. And once again, I'm going to see if you can simplify this. And I'll give you little time to do it. So this one is even crazier than the last few we've looked at. We've got y's and xy's, and x squared and x's, well more just xy's and y squared and on and on and on. And there will be a temptation, because you see a y here and a y here to say, oh, maybe I can add this negative 3y plus this 4xy somehow since I see a y and a y. But the important thing to realize here is that a y is different than an xy. Think about it they were numbers. If y was 3 and an x was a 2, then a y would be a 3 while an xy would have been a 6. And a y is very different than a y squared. Once again, if the why it took on the value 3, then the y squared would be the value 9. So even though you see the same letter here, they aren't the same-- I guess you cannot add these two 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."
    },
    {
        "Q": "At approximately 8:59, why are we allowed to divide the \"outside bit\" by four, and multiply the \"inside bit\" by four?",
        "A": "You can multiply both the outside and the inside because any number is itself multiplied by 1, and 1 = 4 * (1/4) You can hence multiply an expression by 4 * (1/4) while keeping its value. Using the property of multiplication m * (a * b) = (m * a) * b = a * (m * b) you can write that as 1/4 times the outside multiplied by 4 times the inside. I hope this helps. --Phi \u00cf\u0086",
        "video_name": "64bH_27Ehoc",
        "timestamps": [
            539
        ],
        "3min_transcript": "And so we're going to be left with 3 squared. That's what this is down here, squared. So this is 1/3 squared. And then that, squared. So that's what we're left with, with that orange term. And then we go to the pink term. This pink term. 48 is just 3 times 4 times 4. Three times 4. I'll write 4 squared here, because each time we're going to multiply it times 4 again. So the next one's going to be 4 to the third, because we're really-- each side turns into four sides. That's where that came from. 4 squared, we're factoring out the square root of 3, we're factoring out the four, we're factoring out the s squared. And all we're left is 1 over 3 to the third power, squared. So times 1 over 3 to the third power, squared. And we're just going to keep going like that forever. So on each step, we're multiplying by 4, the power of this 4 is incrementing. So there's really 4 to the 0-th power here. We have a 1 here you can imagine, implicitly. The 4 to the first power, 4 squared, then it'll go 4 to the third. This power is also incrementing-- 3 to the first, 3 to the second, 3 to the third. But we see that this power is always one more than that. And it'll be much easier to calculate this infinite-- what's going to turn into an infinite-- geometric series, if those are actually the same power. So what I want to do is I want to increase the power of 4 in all of those. But I can't just willy-nilly multiply everything by 4. If I'm going to multiply everything by 4, I also need to divide everything by 4. So what I'm going to do in this step right over here is I'm going to multiply and divide everything by 4. So if we divide by 4, I can do that on the outside. So let me multiply 1/4 times this right over here. And so I'm dividing by 4 out here. And then I'm going to multiply this by 4. And so I'm not going to be changing This is going to be 4 plus 3 times 4 plus 3 times 4 to the third. And so what was cool about this is now that the power of 4 and the power of this 3 down here are going to be the same power. But it still seems a little weird because we're taking this 1 over 3 squared and then we're squaring it. And here we just have to realize-- so this is always going to be squared. And this is the thing that's incrementing. But in general, if I have 1 over 3 to the n, and I'm squaring it, this is equal to 1 over 3 to the 2n power. So I'm just multiplying it by 2. If I'm raising something to the exponent, then raising that to an exponent, that's just raising it to the n times 2 exponent. And this is the exact same thing as 1 over 3 squared, raised to the nth power. So we can actually switch these two exponents"
    },
    {
        "Q": "When he starts factoring at 6:29, why doesn't the square (the 2 on the blue section) get factored out with the rest of the equation? Is it because it's not exactly s^2 like the original yellow equation?",
        "A": "sqrt(3)\u00e2\u0080\u00a2s^2/4 + 3\u00e2\u0080\u00a2sqrt(3)\u00e2\u0080\u00a2(s/3)^2/4 + 12\u00e2\u0080\u00a2sqrt(3)\u00e2\u0080\u00a2(s/9)^2/4 + 48\u00e2\u0080\u00a2sqrt(3)\u00e2\u0080\u00a2(s/27)^2/4 ... sqrt(3)\u00e2\u0080\u00a2(s^2/4 + 3\u00e2\u0080\u00a2(s/3)^2/4 + 12\u00e2\u0080\u00a2(s/9)^2/4 + 48\u00e2\u0080\u00a2(s/27)^2/4 ...) sqrt(3)/4\u00e2\u0080\u00a2(s^2 + 3\u00e2\u0080\u00a2(s/3)^2 + 12\u00e2\u0080\u00a2(s/9)^2 + 48\u00e2\u0080\u00a2(s/27)^2 ...) sqrt(3)/4\u00e2\u0080\u00a2(s^2 + 3\u00e2\u0080\u00a2s^2\u00e2\u0080\u00a2(1/3)^2 + 12\u00e2\u0080\u00a2s^2\u00e2\u0080\u00a2(1/9)^2 + 48\u00e2\u0080\u00a2s^2\u00e2\u0080\u00a2(1/27)^2 ...) sqrt(3)\u00e2\u0080\u00a2s^2/4\u00e2\u0080\u00a2(1 + 3\u00e2\u0080\u00a2(1/3)^2 + 12\u00e2\u0080\u00a2(1/9)^2 + 48\u00e2\u0080\u00a2(1/27)^2 ...)",
        "video_name": "64bH_27Ehoc",
        "timestamps": [
            389
        ],
        "3min_transcript": "first of all, after I do another pass? Well, the previous pass, I had 12 sides. Each of those 12 sides are now going to turn to 4 new sides when I add these little orange bumps there. So I'm going to multiply it times 4 again. I'm going to multiply it times 4. So now I'm going to have 48 sides. And how many new triangles? Well it's going to be the yellow area plus the blue area plus the orange area. So how many new orange triangles do I have.? Well I'm adding a new orange triangle to each of the sides for the previous pass. In the previous pass I had 12 sides. So now I'm going to add 12 orange triangles. And actually let me write that. I'll just write 12 orange triangles. But it's really I just multiplied it times 4. And then I'm going to have times the square root of 3. And now this isn't going to be s over 3 any more. These are now going to be s over 9. These have 1/3 of the dimensions of these blue triangle. s over 9 squared over 4. And so I think you might start to see the pattern building if we do another pass after this one. Move to the right a little bit. What will that look like? Let me do this in a different color that I haven't used yet. Let me see I haven't used this pink, yet. So now I'm going to have the previous number of sides, that's my number of new triangles, 48 times the square root of 3 times s over-- now these are going to be even 1/3 of this-- s over 27 to the second power. All of that over 4. And I'm going to keep adding an infinite number of terms of this to get the area of a true koch snowflake. So I'm just going to keep doing this over, and over again. So the trick really is finding this infinite sum and seeing if we get a finite number over here. So the first thing I want to do, just to simplify, well, let me just rewrite it a little bit differently over here. is that we can factor out a square root of 3s squared over 4. So let me factor that out. So if we factor out a square root of 3s squared over 4 from all of the terms. Then this term right over here will become a 1. This term right over here is going to become a 3. Let's see, we factored out a square root of 3. We factor out a four. And we factored out the s squared. We factor out only the s squared. So now it's going to have plus 3 times 1/3 squared. That's all we have left here. We have the 1/3 squared. And then we have this 3. And I'm not simplifying this on purpose, so that we see a pattern emerge. And then this next term, right over here, plus-- so this 12 is still going to be there. But I'm going to write that as 3 times 4. Let's see, we're factoring out the square root of 3,"
    },
    {
        "Q": "I lost at 0:43 until the video is over. I didnt understand anything.. pls help",
        "A": "What did you not understand? The video is about finding a Common Denominator. Is that the part you didn t understand or how he changed the 1/10 into 10/100?",
        "video_name": "DR2DYe7PI74",
        "timestamps": [
            43
        ],
        "3min_transcript": "Let's see if we can write 0.15 as a fraction. So the important thing here is to look at what place these digits are in. So this 1 right over here, this is in the tenths place, so you could view that as 1 times 1/10. This 5 right over here is in the hundredths place, so you could view that as 5 times 1/100. So if I were to rewrite this, I can rewrite this as the sum of-- this 1 represents 1 times 1/10, so that would literally be 1/10 plus-- and this 5 represents 5 times 1/100, so it would be plus 5/100. And if we want to add them up, we want to find a common denominator. The common denominator is 100. Both 10 and-- the least common multiple. 100 is a multiple of both 10 and 100. So we can rewrite this as something over 100 plus something over 100. This isn't going to change. This was already 5/100. by 10-- that's what we did; we multiplied it by 10-- then we're going to have to multiply this numerator by 10. And so this is the same thing as 10/100. And now we're ready to add. This is the same thing as-- 10 plus 5 is 15/100. And you could have done that a little bit quicker just You would say, look, my smallest place right over here is in the hundredths place. Instead of calling this 1/10, I could call this literally 10/100. Or I could say this whole thing is 15/100. And now if I want to reduce this to lowest terms, we can-- let's see, both the numerator and the denominator are divisible by 5. So let's divide them both by 5. And so the numerator, 15 divided by 5, is 3. The denominator, 100 divided by 5, is 20. And that's about as simplified as we can get."
    },
    {
        "Q": "What does R^2 (at 0:59) mean?",
        "A": "Basically, it s a coordinate space analogous to the xy plane that we all know and love from graphing functions in algebra 2.",
        "video_name": "8QihetGj3pg",
        "timestamps": [
            59
        ],
        "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": "what is that fancy bracket that Sal draws around the 2*3*5 at 1:18 called, if it is called anything other than a bracket?",
        "A": "A bracket is [ or ] . A brace is { or } . I guess you can call it a squiggly line that connects the selected objects and relates them to something else in the part that sticks out in the middle.",
        "video_name": "QUem_2dkB9I",
        "timestamps": [
            78
        ],
        "3min_transcript": "We need to figure out the least common multiple of 30 and 25. So let's get our little scratch pad out here. And we care about 30 and we care about 25. And I'm going to do this using the prime factorization method which I just like more. Let's find the prime factorization of both of these numbers. So 30, it's divisible by 2. It's 2 times 15. 15 is 3 times 5. And now we've expressed 30 as the product of only prime numbers, 2 times 3 times 5. Now let's do the same thing for 25. 25 is-- well that's just 5 times 5. So let me write that down. 25 is equal to 5 times 5. Now to find the least common multiple, let me write this down, the least common multiple of 30 and 25 is going to have a number whose prime factorization is a super set of both of these as we have in any one of these. So it's the least common multiple. Well it has to be divisible by 30. So it's going to need a 2 times a 3 times a 5. This is what makes it divisible by 30. But it needs to also be divisible by 25. And in order to be divisible by 25, you need to have two 5s in your prime factorization. Right now our prime factorization only has one 5. So we have one 5 right over here. We need another 5. So let's throw another 5 right over here. So now this thing clearly has a 25 in it. It's clearly divisible by 25. And this is the least common multiple. I could have, if we just wanted a common multiple, we could have thrown more factors here and it would have definitely been divisible by 30 or 25, but this has the bare minimum of prime factors necessary to be divisible by 30 and 25. I wouldn't be divisible by both anymore. If I got rid of this 2, I wouldn't be divisible by 30 If I got rid of one of the 5s, I wouldn't be divisible by 25 anymore. So let's just multiply it out. This is essentially the prime factorization of our least common multiple. And this is equal to 2 times 3 is 6, 6 times 5 is 30, 30 times 5 is equal to 150. And of course, we can check our answer, 150. Check it, and we got it right."
    },
    {
        "Q": "But at 4:35, why is it an x minus 2. He is shifting from g(x) to the LEFT, to the f(x). Shouldn't it be a PLUS sign, since he is going to the LEFT? Or is g(x) your end point, and f(x) is your starting point?",
        "A": "Think of it this way: make an x/y table for your points. You plug in an x-value to the function and apply the operations to it to solve for the y value, correct? It s reverse order of operations when it is directly attached to the x. (Another way of imagining this would be to think about switching the -2 to the left side of the function: going over the equals sign would change the -2 to a positive number, 2)",
        "video_name": "ENFNyNPYfZU",
        "timestamps": [
            275
        ],
        "3min_transcript": "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. So if I were to take its mirror image, it looks something like this. Its mirror image if I were to reflect it across the x-axis. It looks something like this. So this right over here we would call-- so if this is g of x, when we flip it that way, this is the negative g of x. When x equals 4, g of x looks like it's about negative 3 and 1/2. You take the negative of that, you get positive. I guess it should be closer to here-- You get positive 3 and 1/2 if you were to take the exact mirror image. So that's negative g of x. But that still doesn't get us. It looks like we actually have to triple this value for any point. And you see it here. This gets to 2, but we need to get to 6. This gets to 1, but we need to get to 3."
    },
    {
        "Q": "at 4:00, how is it positive? I don't understand Sal's explanation. Can someone explain it please? and plz explain how square root of -1 exsists...",
        "A": "\u00f0\u009d\u0091\u0096\u00e2\u0081\u00b4 = \u00f0\u009d\u0091\u0096\u00c2\u00b2 \u00e2\u0080\u00a2 \u00f0\u009d\u0091\u0096\u00c2\u00b2 = (-1)(-1) = 1 The principal square root of -1 is \u00f0\u009d\u0091\u0096 simply because we define it to be so. In other words, we define \u00f0\u009d\u0091\u0096\u00c2\u00b2 = -1. Mathematics doesn t bother with does or does not exist because it simply deals with logical extensions of definitions of axioms we set.",
        "video_name": "ysVcAYo7UPI",
        "timestamps": [
            240
        ],
        "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": "Isn't it supposed to be i=\u00c2\u00b1sqrt(-1) at \"0:56\"???",
        "A": "But isn t (+sqrt(-1))^2=(-sqrt(-1))^2 or is it just defined and can t be changed???",
        "video_name": "ysVcAYo7UPI",
        "timestamps": [
            56
        ],
        "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": "At 3:15, Sal said that i^4=i*i^3 and resulted with positive 1. Could you just simply do i^4=i^2*i^2 and get -1*-1 just resulting with positive 1 as well. Is this right or it doesn't work all the time? Thanks for the video!",
        "A": "Your technique works. Remember, we can regroup or multiply in any order due to the associative and commutative properties of multiplication.",
        "video_name": "ysVcAYo7UPI",
        "timestamps": [
            195
        ],
        "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": "A 1:13, how can the square of any no. be negative ? Given that i = under root -1 and square of i = -1 ?",
        "A": "You are right. The square of any number either positive or negative, cannot be a negative. i is not any number, it is an imaginary number. You might wonder about the purpose of this. The purpose of i is to compensate for the fact that normal numbers cannot be negative when squared. and square root of -1 would make no sense.",
        "video_name": "ysVcAYo7UPI",
        "timestamps": [
            73
        ],
        "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": "At time 6:08 it says that even if a is less than 0 it is still going to be positive? So my question is how do you know what to go by ? A is less than 0 so it is negative or a to the 4rth power which indicates it is positive?",
        "A": "You would go by the order of operations to figure out the sign (parenthesis, exponents, multiplication, division, addition, subtraction). So since -a^4/3 is the same as -1*(a^4/3) then you would resolve the exponent before multiplying by -1. The exponent is even which means that you will always get a positive number in the numerator giving you a positive fraction which you THEN multiply by -1 making it a negative number. Hope this helps! :)",
        "video_name": "Pms4cBWwPSU",
        "timestamps": [
            368
        ],
        "3min_transcript": "in less than a second? Well if you just have a bunch of numbers, and then you multiply them times zero at some point in that, the whole thing is just going to be equal to zero. I could literally have a times b times c times d, I could multiply a bunch of numbers. And if I just knew that one of these numbers is zero, then the whole product is going to be equal to zero. You know if c is equal to zero, I could multiply a times b times d times e and get some number, and then multiply that times c, it's gonna be anything times zero is zero. Now notice, q is equal to zero, and q is right over here. So I could take p divided by p, that's gonna be one. One times q, that's just gonna be zero, as one times zero is zero, then you're gonna have zero times four and 2/7, it's just all gonna be zero. And the key is, there's a zero right over here. At some point you're multiplying this entire product times zero, so the whole product is going, or you're multiplying these numbers times zero, so the whole product is going to be zero. So this side isn't positive or negative, it's zero. It's neither positive nor negative. All right, let's do another one. What is the sign of negative 3/4 times negative a to the fourth over three, when a is less than zero? So they're telling us that a is negative. So a is negative. So let me use a more dramatic color. So right over here, I have a negative number to the fourth power. So it's a times a times a times a. And we've already said if you have an even number of negatives being multiplied by each other, then that's going to be a positive. You could even see it over here, I mean they're all negative, but a negative times a negative is going to be a positive. A negative times the negative, I take the third and the fourth one here, that's gonna be a positive. And then a positive times a positive, the whole thing is going to be positive, even though each of these are negative. Negative times negative, or another way you could say it, a times a is going to be positive, then times a is going to be negative, it's going to be positive again. So a to the fourth power is going to be positive. But then you have a negative in front of it. So you're taking the negative of a positive divided by a positive, Well, so this, all of this stuff right over here is all gonna be positive, but then you have this negative in front. So this entire thing inside the parentheses is going to be a negative. And then you have a negative, negative 3/4, times a negative, well a negative times a negative, the whole thing is going to be positive. So what's the sign of this? It's going to be positive. Let's do another one, I'm having too much fun. All right, what is the sign of x to the 59th power, divided by 2.3x times 4/5, when x is less than zero? So x is once again a negative number. So x to the 59th power, that's an odd number of negatives being multiplied together. So this thing right over here is going to be negative."
    },
    {
        "Q": "At 0:40, why is the volume (4/3)*(pi)*(r^2)? Why is it 4/3? I'm in middle school, so I don't understand why it's 4/3.",
        "A": "If you want to find out why it is 4/3 you can look up who created the equation for a sphere. It may explain how that particular person got the equation.",
        "video_name": "IelS2vg7JO8",
        "timestamps": [
            40
        ],
        "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": "Um I don't get \"7:20\"",
        "A": "Given: n = l - m n = log_x (A/B) l = log_x (A) m = log_x (B) Therefore (plugging in to the first equation): log_x (A/B) = [log_x (A)] - [log_x (B)] Does that answer your question?",
        "video_name": "yEAxG_D1HDw",
        "timestamps": [
            440
        ],
        "3min_transcript": "How can we write all of these expressions as exponents? Well, this just says that x to the l is equal to A. Let me switch colors. That keeps it interesting. This is just saying that x to the m is equal to B. And this is just saying that x to the n is equal to A/B. So what can we do here? Well what's another way of writing A/B? Well, that's just the same thing as writing x to the l because that's A, over x to the m. That's B. also be written as x to the l, x to the negative m. Or that also equals x to the l minus m. So what do we know? We know that x to the n is equal to x to the l minus m. Those equal each other. I just made a big equal chain here. So we know that n is equal to l minus m. Well, what does that do for us? Well, what's another way of writing n? I'm going to do it up here because I think we have stumbled upon another logarithm rule. What's another way of writing n? Well, I did it right here. This is another way of writing n. So logarithm base x of A/B-- this is an x over l is this right here. Log base x of A is equal to l. The log base x of A minus m. I wrote m right here. That's log base x of B. There you go. I probably didn't have to prove it. You could've probably tried it out with dividing it, but whatever. But you know are hopefully satisfied that we have this new logarithm property right there. Now I have one more logarithm property to show you, but I don't think I have time to show it in this video. So I will do it in the next video. I'll see you soon."
    },
    {
        "Q": "At 3:37 Sal says that the range of x values can be narrower than the maximum possible range. Does this mean that delta can vary?",
        "A": "Yes. Delta will change depending on the epsilon chosen.",
        "video_name": "w70af5Ou70M",
        "timestamps": [
            217
        ],
        "3min_transcript": "In this yellow definition right over here, we said you can get f of x as close as you want to L by making x sufficiently close to C. This second definition, which I kind of made as a little bit more of a game, is doing the same thing. Someone is saying how close they want f of x to be to L and the burden is then to find a delta where as long as x is within delta of C, then f of x will be within epsilon of the limit. So that is doing it. It's saying look, if we are constraining x in such a way that if x is in that range to C, then f of x will be as close as you want. So let's make this a little bit clearer by diagramming right over here. You show up and you say well, I want f of x to be within epsilon of our limit. And this right over here might be our limit minus epsilon. And you say, OK, sure. I think I can get your f of x within this range of our limit. And I can do that by defining a range around C. And I could visually look at this boundary. But I could even go narrower than that boundary. I could go right over here. Says OK, I meet your challenge. I will find another number delta. So this right over here is C plus delta. This right over here is C minus-- let me write this down-- is C minus delta. So I'll find you some delta so that if you take any x in the range C minus delta to C 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"
    },
    {
        "Q": "At 5:28 Sal made the mistake of not making the distance between x and c greater than 0. Is that the same case as with the distance between f(x) and the limit? I feel like because it's distance as well it would need to be greater than 0 too.",
        "A": "Think about this. If you are at school (assuming you are not home schooled), the distance between you and your home is some positive number. If you are at home, then the distance between you and your home is zero.",
        "video_name": "w70af5Ou70M",
        "timestamps": [
            328
        ],
        "3min_transcript": "And this right over here might be our limit minus epsilon. And you say, OK, sure. I think I can get your f of x within this range of our limit. And I can do that by defining a range around C. And I could visually look at this boundary. But I could even go narrower than that boundary. I could go right over here. Says OK, I meet your challenge. I will find another number delta. So this right over here is C plus delta. This right over here is C minus-- let me write this down-- is C minus delta. So I'll find you some delta so that if you take any x in the range C minus delta to C 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,"
    },
    {
        "Q": "At 7:30, is sss the only kind of way to find congruence in a triangle?",
        "A": "The congruence postulates are: SSS, AAS, ASA, SAS. AAA shows similarity only, not necessarily congruence. SSA may or may not show congruence, depending on the details. But generally SSA does not show congruence by itself -- you need some other bit of information to establish congruence. For example, if the known angle is right or obtuse, then SSA proves congruence; but, if the known angle is acute, then you would still need more information to establish congruence.",
        "video_name": "CJrVOf_3dN0",
        "timestamps": [
            450
        ],
        "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": "can you write that congruency sign (at 1:39) on a keyboard?",
        "A": "In Windows, like in Microsoft Word or Microsoft Excel, you can your font to Symbol. Then, that symbol is SHIFT+2 (or the @ key on your keyboard.). As a side note, with the Symbol font you get most of the Greek letters..alpha, beta, delta, pi, etc.",
        "video_name": "CJrVOf_3dN0",
        "timestamps": [
            99
        ],
        "3min_transcript": "Let's talk a little bit about congruence, congruence And one to think about congruence, it's really kind of equivalence for shapes So, when in algebra when something is equal to another thing it means that their quantities are the same But when we're all of the sudden talking about shapes and we say that those shapes are the same, the shapes are the same size and shape then we say that they're congruent And just to see a simple example here: I have this triangle, right over there and let's say I have this triangle right over here And if you are able to shift, you are able to shift this triangle and flip this triangle, you can make it look exactly like this triangle As long as you're not changing the lengths of any of the sides or the angles here But you can flip it, you can shift it, you can rotate it So you can shift, let me write this, you can shift it, you can flip it and you can rotate If you can do those three procedures to make these 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"
    },
    {
        "Q": "At 1:20 can you reflect a triangle also?",
        "A": "yes. that is what flip means.",
        "video_name": "CJrVOf_3dN0",
        "timestamps": [
            80
        ],
        "3min_transcript": "Let's talk a little bit about congruence, congruence And one to think about congruence, it's really kind of equivalence for shapes So, when in algebra when something is equal to another thing it means that their quantities are the same But when we're all of the sudden talking about shapes and we say that those shapes are the same, the shapes are the same size and shape then we say that they're congruent And just to see a simple example here: I have this triangle, right over there and let's say I have this triangle right over here And if you are able to shift, you are able to shift this triangle and flip this triangle, you can make it look exactly like this triangle As long as you're not changing the lengths of any of the sides or the angles here But you can flip it, you can shift it, you can rotate it So you can shift, let me write this, you can shift it, you can flip it and you can rotate If you can do those three procedures to make these 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"
    },
    {
        "Q": "At 2:27, how are 1, 3, 9, and 27 [[\"\"positive\"\"]] divisors of 27,000",
        "A": "They are all positive numbers that 27 000 can be equally divided into. 27 000/1=1, 27 000/3=9000, 27 000/9=3000, 27 000/27= 1000. These are actually all the positive divisors of 27. He knows that every single divisor of 27 would be a divisor of 27 000 =)",
        "video_name": "17st-s5gg10",
        "timestamps": [
            147
        ],
        "3min_transcript": "Calculate the sum of all positive divisors of 27,000. The easiest thing that I can think of doing is first take the prime factorization of 27,000, and then that will help us kind of structure our thought of what all of the different divisors of 27,000 would have to look like. So 27,000 is the same thing as 27 times 1,000, which is the same thing as 3 to the third times 10 to the third, and 10 is, of course, the same thing as 2 times 5. So this is the same thing as 2 times 5 to the third, or it's the same thing as 2 to the third times 5 to the third. So 27,000 is equal to 2 to the third times 3 to the third times 5 to the third. So any divisor of 27,000 is going to have to be made up of the product of up to three 2's, up to three 3's, and up So let's try to look at all the combinations and think of a fast way of summing them. So let's just say it has no fives in it. It has no fives in a divisor. So if it has no fives, then it could have up to three 2's, so let's say it has zero 2's. So I'm just going to take the powers of 2, so if it has zero 2's, then we'll put a 1 here, if it has two 2's, it has to be divisible by 4. If it has three 2's, it's going to be divisible by 8. When I say three 2's, I mean 2 times 2 times 2. Now, let's do it with the 3's. If you have, oh wait, I forgot a power. If you have zero 2's, that means it's just divisible by 1 from looking at the 2's. If you have one 2, it's divisible by just 2. If you have two 2's, you're divisible by 4. And if you have three 2's, and when I mean that I'm saying 2 times 2 times 2, that means you're divisible by 8. Let's do the same thing with 3. From the point of view of the 3, if you have no 3's, that means at least you're divisible by 1. If you have one 3, that means you're divisible by 3. If you have three 3's, it means you're divisible by 27. So let's look at all of the possible combinations. And for this grid that I'm going to generate right here, we assume that you're not divisible by 5, or you're only divisible by 5 to the zero power. So what are all the possible numbers here? Well, you have 1 times 1 is 1. That's divisible by 1 and 1. You have 1 times 3, which is 3, 1 times 9 which is 9, 1 times 27 which is 27. So these are all the numbers that are divisible by that have up to three 3's in them, from zero to three 3's in them, and they have no twos in them. If you throw another two in here, you're essentially going to multiply all of these numbers by two. If you throw another two in here, you're going to multiply all of these numbers by 2. Now, before I do this, because I want to do this as fast as possible. I could figure out what these numbers are, I could multiply them. But instead, let's just take the sum. Let's just take the sum here of this row,"
    },
    {
        "Q": "At 0:56, when Sal is graphing y-k=x^2, he puts the vertex higher on the graph. I don't understand why it's higher on the graph if x^2 is k LESS than y. Shouldn't it be lower on the graph?",
        "A": "Think of it this way: you would have to decrease the y value by k units to get down to x^2. That must mean that y is higher than it would be on the x^2 curve.",
        "video_name": "99v51U3HSCU",
        "timestamps": [
            56
        ],
        "3min_transcript": "Here I've drawn the most classic parabola, y is equal to x squared. And what I want to do is think about what happens-- or how can I go about shifting this parabola. And so let's think about a couple of examples. So let's think about the graph of the curve. This is y is equal to x squared. Let's think about what the curve of y minus k is equal to x squared. What would this look like? Well, right over here, we see when x is equal to 0, x squared is equal to 0. That's this yellow curve. So x squared is equal to y, or y is equal to x squared. But for this one, x squared isn't equal to y. It's equal to y minus k. So when x equals a 0, and we square it, 0 squared doesn't get us to y. It gets us to y minus k. So this is going to be k less than y. Or another way of thinking about it, this is 0. If it's k less than y, y must be at k, wherever k might be. So y must be at k, right over there. So at least for this point, it had the effect And that's actually true for any of these values. So let's think about x being right over here. For this yellow curve, you square this x value, and you get it there. And it's clearly not drawn to scale the way that I've done it right over here. But now for this curve right over here, x squared doesn't cut it. It only gets you to y minus k. So y must be k higher than this. So this is y minus k. y must be k higher than this. So y must be right over here. So this curve is essentially this blue curve shifted up by k. So making it y minus k is equal to x squared shifted it up by k. Whatever value this is, shift it up by k. This distance is a constant k, the vertical distance between these two parabolas. And I'll try to draw it as cleanly as I can. Now let's think about shifting in the horizontal direction. Let's think about what happens if I were to say y is equal to, not x squared, but x minus h squared. So let's think about it. This is the value you would get for y when you just square 0. You get y is equal to 0. How do we get y equals 0 over here? Well, this quantity right over here has to be 0. So x minus h has to be 0, or x has to be equal to h. So let's say that h is right over here. So x has to be equal to h. So one way to think about it is, whatever value you were squaring here to get your y, you now have to have an h higher value to square that same thing. Because you're going to subtract h from it. Just to get to 0, x has to equal h. Here, if you wanted to square 1, x just had to be equal to 1."
    },
    {
        "Q": "At 5:10, how did Sal come to know that (23/4) is (5 3/4) ?",
        "A": "Divide it by 4; you have 20 as the nearest multiple so 3 remains giving 5 3/4 With division practice you become better",
        "video_name": "w56Vuf9tHfA",
        "timestamps": [
            310
        ],
        "3min_transcript": "that really uses our knowledge of the vertex of a parabola to be able to figure out where the focus and the directrix is going to be. So let's think about the vertex of this parabola right over here. Remember, the vertex, if the parabola is upward opening like this, the vertex is this minimum point. If it is downward opening, it's going to be this maximum point. And so when you look over here, you see that you have a negative one-third in front of the x minus one squared. So this quantity over here is either going to be zero or negative. It's not going to add to 23 over four, it's either gonna add nothing or take away from it. So this thing's going to hit a maximum point, when this thing is zero, when this thing is zero, and that's just gonna go down from there and when this thing is zero, y is going to be equal to 23 over four. So our vertex is going to be that maximum point. Well, when x equals one. When x equals one, you get one minus one squared. So zero squared times negative one-third, this is zero. So when x is equal to one, we're at our maximum y value of 23 over four which five and three-fourths. Actually, let me write that as a . Actually, I'll leave just that's our vertex. and it is a downward opening parabola. So actually, let me start to draw this. So we'd get some axis here. So we have to go all the way up to five and three-fourths. So. Let's make this our y, this is our y axis. This is the x axis. That's the x axis. We're gonna see, we're gonna go to one. Let's call that one. Let's call that two. And then I wanna get, let's see, if I go to five and three-fourths, let's go up to, let's see one, two, three, four 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,"
    },
    {
        "Q": "at 4:58 Sal got c=1 but why I get C=1/2 ?\nhere my solution\nx=-1/2y^-1 +c\nafter subst (1,-1)\nc= 1/2\nNOT: it seems like when I put C in x part like x+c=... and when I put it in y part like x= ....+c I get different values of C. So Do all of C's are correct?",
        "A": "Yes, both Cs are correct, because they don t represent the same thing. In Sal s case, his C is on the x side of the equation (plus he swallowed the 2 into his C when he solved for y), and in your case, your C in on the y side of the equation. Still, if you use your value of C = 1/2 in your equation and solve for y, you get the exact same result that Sal did: x = -1/2 y\u00e2\u0081\u00bb\u00c2\u00b9 + C x = -1/2 y\u00e2\u0081\u00bb\u00c2\u00b9 + 1/2 x - 1/2 = -1/2 y\u00e2\u0081\u00bb\u00c2\u00b9 -2(x - 1/2) = y\u00e2\u0081\u00bb\u00c2\u00b9 -2x + 1 = y\u00e2\u0081\u00bb\u00c2\u00b9 y\u00e2\u0081\u00bb\u00c2\u00b9 = -2x + 1 y = 1/(-2x + 1)",
        "video_name": "E444KhRcWSk",
        "timestamps": [
            298
        ],
        "3min_transcript": "let's see, I can multiply both sides by negative two, and then I'm gonna have, the left hand side you're just gonna have Y to the negative one, or 1/Y is equal to, if I multiply the right hand side times negative two, I'm gonna have negative two times X plus, well it's some arbitrary constant, it's still going to, it's gonna be negative two times this arbitrary constant but I could still just call it some arbitrary constant, and then if we want we can take the reciprocal of both sides, and so we will get Y is equal to, is equal to 1/-2X+C. And now we can use, we can use the information they gave us right over here, the fact that our particular solution needs to go through this point to solve for C. Oh sorry, when X is one, when X is one, Y is negative one, so we get negative one is equal to 1/-2+C, or we could say C minus two, we could multiply both sides times C minus two, if then we will get, actually let me just scroll down a little bit, so if you multiply both sides times C minus two, negative one times C minus two is going to be negative C plus two or two minus C is equal to one. All I did is I multiplied C minus two times both sides, and then, let's see, I can subtract two from both sides, so negative C is equal to negative one, and then if I multiply both sides by negative one, we get C is equal to one. So our particular solution is Y is equal to 1/-2X+1. we didn't just ask for the particular solution, we asked, what is Y when X is equal to three. So Y is going to be equal to one over, three times negative two is negative six plus one, which is equal to negative, is going to be equal to 1/-5, or -1/5. And we are done."
    },
    {
        "Q": "At 1:45, how can we say that sample mean=p (i.e. the proportion of teachers who think computers is a good tool) ? Is there a rule or something to take that value which I've missed?\nBecause for a binomial distribution E(X)= np where n is the number of trials and p is the success proportion.",
        "A": "Just use the same old formula (sum x(i))/250, with 108 of the x = 0, and 142 of them = 1. So the mean is 142/250 Now: How many teachers think computers are a good tool? How many teachers are there? What proportion think computers are a good tool?",
        "video_name": "SeQeYVJZ2gE",
        "timestamps": [
            105
        ],
        "3min_transcript": "In a local teaching district, a technology grant is available to teachers in order to install a cluster of four computers in their classroom. From the 6,250 teachers in the district, 250 were randomly selected and asked if they felt that computers were an essential teaching tool for their classroom. Of those selected, 142 teachers felt that the computers were an essential teaching tool. And then they ask us, calculate a 99% confidence interval for the proportion of teachers who felt that the computers are an essential teaching tool. So let's just think about the entire population. We weren't able to survey all of them, but the entire population, some of them fall in the bucket, and we'll define that as 1, they thought it was a good tool. They thought that the computers were a good tool. And we'll just define a 0 value as a teacher And some proportion of the total teachers think that it is a good learning tool. So that proportion is p. And then the rest of them think it's a bad learning tool, 1 minus p. We have a Bernoulli Distribution right over here, and we know that the mean of this distribution or the expected value of this distribution is actually going to be p. So it's actually going to be a value, it's neither 0 or 1, so not an actual value that you could actually get out of a teacher if you were to ask them. They cannot say something in between good and not good. The actual expected value is something in between. It is p. Now what we do is we're taking a sample of those 250 teachers, and we got that 142 felt that the computers were an essential teaching tool. So in our survey, so we had 250 sampled, and we got 142 So we got 142 1's, or we sampled 1, 142 times from this And then the rest of the time, so what's left over? There's another 108 who said that it's not good. So 108 said not good, or you could view them as you were sampling a 0, right? 108 plus 142 is 250. So what is our sample mean here? We have 1 times 142, plus 0 times 108 divided by our total number of samples, divided by 250. It is equal to 142 over 250. You could even view this as the sample proportion of teachers who thought that the computers were a good teaching tool. Now let me get a calculator out to calculate this."
    },
    {
        "Q": "at 18:12, shouldn't it be \"take larger samples\" instead of MORE samples? 2 different things",
        "A": "When he says take more samples , he means take larger samples or make more observations . People use the word sample ambiguously - as in: a sample of 100 samples , instead of a sample of 100 observations .",
        "video_name": "SeQeYVJZ2gE",
        "timestamps": [
            1092
        ],
        "3min_transcript": "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 99% confidence interval? Well, you could just take more samples. If you take more samples than our estimate of the standard deviation of this distribution will go down because this denominator will be higher. If the denominator is higher then this whole thing will go down. So if the standard deviations go down here, then when we count the standard deviations, when we do the plus or minus on the range, this value will go down and So you just take more samples."
    },
    {
        "Q": "At 11:01 you just end saying that it is .5 + .495 to get .995 to look up on the z table. Why .995? Just a few seconds before that you said the interval is symmetric about the mu and the right half was .495. Since that is .495 and .495 + .495 = .99 which is the confidence level we want, why do .5 + .495? That lost me.",
        "A": "For z = 2.58, probability (area) is .9951. But this is the area from minus infinity to +2.58 SDs. We want the area from the 0 to 2.58 SDs (so we can double it), so we subtract .5. Then we have .4951*2 = .9902, approximately .99 = 99%",
        "video_name": "SeQeYVJZ2gE",
        "timestamps": [
            661
        ],
        "3min_transcript": "chance, or how many-- let me think of it this way. How many standard deviations away from the mean do we have to be that we can be 99% confident that any sample from the sampling distribution will be in that interval? So another way to think about think it, think about how many standard deviations we need to be away from the mean, so we're going to be a certain number of standard deviations away from the mean such that any sample, any mean that we sample from here, any sample from this distribution has a 99% chance of being plus or minus that many standard So it might be from there to there. So that's what we want. We want a 99% chance that if we pick a sample from the sampling distribution of the sample mean, it will be within this many standard deviations of the actual mean. And to figure that out let's look at an actual Z-table. So another way to think about it if we want 99% confidence, if we just look at the upper half right over here, that orange area should be 0.475, because if this is 0.475 then this other part's going to be 0.475, and we will get to our-- oh sorry, we want to get to 99%, so it's not going to be 0.475. We're going to have to go to 0.495 if we want 99% So this area has to be 0.495 over here, because if that is, that over here will also be. So that their sum will be 99% of the area. Now if this is 0.495, this value on the z table right here will have to be 0.5, because all of this area, if you include all of this is going to be 0.5. So it's going to be 0.5 plus 0.495. Let me make sure I got that right. 0.995. So let's look at our Z-table. So where do we get 0.995. on our z table? 0.995. is pretty close, just to have a little error, it will be right over here-- this is 0.9951. So another way to think about it is 99-- so this value right here gives us the whole cumulative area up to that, up to our mean. So if you look at the entire distribution like this, this is the mean right over here. This tells us that at 2.5 standard deviations above the mean, so this is 2.5 standard deviations above the mean. So this is 2.5 times the standard deviation of the sampling distribution."
    },
    {
        "Q": "But why wait to round it at 5:30",
        "A": "Remember, mathematicians don t like to play with long numbers as shown in scientific notation. Besides, this is just an example, so we want to work the easy way, as long as his watchers understand.",
        "video_name": "XJBwJjP2_hM",
        "timestamps": [
            330
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 4:30, couldn't I multiply 0.3979 and 10^5 BOTH by 10? Wouldn't the value be kept equivalent? Or no?",
        "A": "No.It wouldn t since it ll make it 3.979*10^6=3979000 but the original answer is 39790. Hope that helps!",
        "video_name": "XJBwJjP2_hM",
        "timestamps": [
            270
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 0:10, How come the inches in the water increase?",
        "A": "the ring s volume is 1.5, so it displaces the water so it stays even, Because atoms cant go inside of each other...",
        "video_name": "ViFLPsLTO1k",
        "timestamps": [
            10
        ],
        "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": "at 0:55 what does sal mean by the leibniz notation",
        "A": "Gottfried Wilhelm Leibniz, used the symbols dx and dy to represent infinitely small (or infinitesimal) increments of x and y. So f (x)= dy/dx, this was named in honor of Leibniz.",
        "video_name": "6o7b9yyhH7k",
        "timestamps": [
            55
        ],
        "3min_transcript": "- [Voiceover] Let's now introduce ourselves to the idea of a differential equation. And as we'll see, differential equations are super useful for modeling and simulating phenomena and understanding how they operate. But we'll get into that later. For now let's just think about or at least look at what a differential equation actually is. So if I were to write, so let's see here is an example of differential equation, if I were to write that the second derivative of y plus two times the first derivative of y is equal to three times y, this right over here is a differential equation. Another way we could write it if we said that y is a function of x, we could write this in function notation. We could write the second derivative of our function with respect to x plus two times the first derivative of our function is equal to three times our function. Or if we wanted to use the Leibniz notation, we could also write, the second derivative of y of y with respect to x is equal to three times y. All three of these equations are really representing the same thing, they're saying OK, can I find functions where the second derivative of the function plus two times the first derivative of the function is equal to three times the function itself. So just to be clear, these are all essentially saying the same thing. And you might have just caught from how I described it that the solution to a differential equation is a function, or a class of functions. It's not just a value or a set of values. So the solution here, so the solution to a differential equation is a function, or a set of functions, or a class of functions. It's important to contrast this relative to a traditional equation. So let me write that down. traditional equation, differential equations have been around for a while. So let me write this as maybe an algebraic equation that you're familiar with. An algebraic equation might look something like, and I'll just write up a simple quadratic. Say x squared plus three x plus two is equal to zero. The solutions to this algebraic equation are going to be numbers, or a set of numbers. We can solve this, it's going to be x plus two times x plus one is equal to zero. So x could be equal to negative two or x could be equal to negative one. The solutions here are numbers, or a set of values that satisfy the equation. Here it's a relationship between a function and its derivatives. And so the solutions, or the solution,"
    },
    {
        "Q": "At 6:53 Why did Sal say that second deravative of why is also same as the first one ?",
        "A": "If y = e^x, then the first derivative y is also equal to e^x (e^x is its own derivative). The derivative of the first derivative, known as the second derivative y , is therefore also equal to e^x. Thus, the first derivative of y is equal to the second derivative of y. Also why is how we pronounce the letter y.",
        "video_name": "6o7b9yyhH7k",
        "timestamps": [
            413
        ],
        "3min_transcript": "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. So these two terms right over here, nine e to the negative three x, essentially minus six e to the negative three x, that's gonna be three e to the negative three x. Which is indeed equal to three e to the negative three x. So y one is indeed a solution to this differential equation. But as we'll see, it is not the only solution to this differential equation. For example, let's say y two is equal to e to the x is also a solution to this differential equation. And I encourage you to pause the video again and verify that it's a solution. So assuming you've had a go at it. So the first derivative of this is pretty straight-forward, is e to the x. Second derivative, one of the profound things of the exponential function, the second derivative here is also e to the x. in those same colors. So the second derivative is going to be e to the x plus two times e to the x is indeed going to be equal to three times e to the x. This is absolutely going to be true. E to the x plus two e to the x is three e to the x. So y two is also a solution to this differential equation. So that's a start. In the next few videos, we'll explore this more. We'll start to see what the solutions look like, what classes of solutions are, techniques for solving them, visualizing solutions to differential equations, and a whole toolkit for kind of digging in deeper."
    },
    {
        "Q": "At 13:01, Sal says that any non-zero number is equal to 1. Why is zero not included?",
        "A": "0^0 is 1 in certain contexts and indeterminate in other contexts.",
        "video_name": "zM_p7tfWvLU",
        "timestamps": [
            781
        ],
        "3min_transcript": "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. So what is this going to be equal to? This is going to be equal to-- we have the same base, x. We can add the exponents, x to the 1 plus 4 plus 2 power, and I'll add it in the next step. And then on the y's, this is times y to the 2 plus 2 plus 2 power. And what does that give us? That gives us 6 x to the seventh power, y to the sixth power. And I'll just leave you with some thing that you might already know, but it's pretty interesting. And that's the question of what happens when you take something to the zeroth power? So if I say 7 to the zeroth power, What does that equal? And I'll tell you right now-- and this might seem very counterintuitive-- this is equal to 1, or 1 to the zeroth power is also equal to 1. Anything that the zeroth power, any non-zero number to And just to give you a little bit of intuition on why that is. Think about it this way. 3 to the first power-- let me write the powers-- 3 to the first, second, third. We'll just do it the with the number 3. So 3 to the first power is 3. I think that makes sense. 3 to the second power is 9. 3 to the third power is 27. And of course, we're trying to figure out what should 3 to the zeroth power be? Well, think about it. Every time you decrement the exponent. Every time you take the exponent down by 1, you are dividing by 3. To go from 27 to 9, you divide by 3. To go from 9 to 3, you divide by 3. So to go from this exponent to that exponent, maybe we should divide by 3 again. And that's why, anything to the zeroth power, in this case, 3 to the zeroth power is 1. See you in the next video."
    },
    {
        "Q": "At 12:57, Sal says that 1^0 = 1. Why is that? Why isn't it undefined?\n\nAlso, why is any number to the 0th power 1?",
        "A": "lets take an exponent series of x x^0 x^1 x^2 x^3 Now lets write what we know: x^0 = ? x^1 = x x^2 = x * x x^3 = x * x * x Now to get to x^3 to x^2 what do we do? We divide x^3 by x: x^3/x = x^2 because x * x * x/x = x * x = x^2 This logic works for getting from x^2 to x^1 x * x/x = x = x^1 Now extrapolating backwards, wouldn t going from x^1 to x^0 be the same as just x^1 divided by x? So: x^1/x = x^0 = x/x = 1 So any number to the zeroth power is equal to 1, except for 0 itself.",
        "video_name": "zM_p7tfWvLU",
        "timestamps": [
            777
        ],
        "3min_transcript": "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. So what is this going to be equal to? This is going to be equal to-- we have the same base, x. We can add the exponents, x to the 1 plus 4 plus 2 power, and I'll add it in the next step. And then on the y's, this is times y to the 2 plus 2 plus 2 power. And what does that give us? That gives us 6 x to the seventh power, y to the sixth power. And I'll just leave you with some thing that you might already know, but it's pretty interesting. And that's the question of what happens when you take something to the zeroth power? So if I say 7 to the zeroth power, What does that equal? And I'll tell you right now-- and this might seem very counterintuitive-- this is equal to 1, or 1 to the zeroth power is also equal to 1. Anything that the zeroth power, any non-zero number to And just to give you a little bit of intuition on why that is. Think about it this way. 3 to the first power-- let me write the powers-- 3 to the first, second, third. We'll just do it the with the number 3. So 3 to the first power is 3. I think that makes sense. 3 to the second power is 9. 3 to the third power is 27. And of course, we're trying to figure out what should 3 to the zeroth power be? Well, think about it. Every time you decrement the exponent. Every time you take the exponent down by 1, you are dividing by 3. To go from 27 to 9, you divide by 3. To go from 9 to 3, you divide by 3. So to go from this exponent to that exponent, maybe we should divide by 3 again. And that's why, anything to the zeroth power, in this case, 3 to the zeroth power is 1. See you in the next video."
    },
    {
        "Q": "At 3:13, how does 3x^2 equal 27x^2? They seem like two different numbers.",
        "A": "it was actually cubed not squared. On the video it says: 3x . 3x . 3x = (3 . 3 . 3)(x . x . x) (3x)^3 = 27x^3 you can test it by substituting say a 2 for the x value, then you would get 216 = 216",
        "video_name": "zM_p7tfWvLU",
        "timestamps": [
            193
        ],
        "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": "Is there a test anywhere with questions similar to 10:08? I enjoyed simplifying the expression but I want more challenges like that to see if I can do it again.\n\nThe test \"Practice Multiply Powers\" only has simple questions",
        "A": "Keep working thru this section. There are later exercise sets that combine 2 or more of the exponent properties which makes them more challenging.",
        "video_name": "zM_p7tfWvLU",
        "timestamps": [
            608
        ],
        "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": "at 11:12, Sal is rearranging the problem. He begins with 2x3. Where is the 3 coming from? Please enlighten me on this.",
        "A": "The 3 is coming from that last term in parentheses (3x^2y^2). First he multiplied all the numbers together (the middle term didn t have a number), then the letters.........",
        "video_name": "zM_p7tfWvLU",
        "timestamps": [
            672
        ],
        "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": "On the second problem (7:23), why is '8' divisible by both '9' and '24'? I honestly don't get it because I'm thinking of it as '9/8' or something. I get the part of 2.2.2, though.\nCould somebody please point out what's wrong here? thanks in advance.",
        "A": "The point is that any number divisible by 9 and 24, is also divisible by 8. Why?! because the prime factorization of 9 and 24 contains the prime factorization of 8. by the way, at first that confuse me :)",
        "video_name": "zWcfVC-oCNw",
        "timestamps": [
            443
        ],
        "3min_transcript": "And once again we just do the prime factorization. We essentially think about the least common multiple of 9 and 24. You take the prime factorization of 9, it's 3 times 3 and we're done. Prime factorization of 24 is 2 times 12. 12 is 2 times 6, 6 is 2 times 3. So anything that's divisible by 9 has to have a 9 in it's factorization or if you did its prime factorization would have to be a 3 times 3, anything divisible by 24 has got to have three 2's in it, so it's gotta have a 2 times a 2 times a 2, and it's got to have at lease one 3 here. And we already have at least one 3 from the 9, so we have that. So this number right here is divisible by both 9 and 24. And this number right here is actually 72. This is 8 times 9, which is 72. let's assume that it was multiple choice. Let's say the choices here were 16 27 5 11 and 9. So 16, if you were to do its prime factorization, is 2 times 2 times 2 times 2. It's 2 to the 4th power. So you would need four 2's here. We don't have four 2's over here. I mean there could be some other numbers here but we don't know what they are. These are the only numbers that we can assume are in the prime factorization of something divisible by both 9 and 24. So we can rule out 16. We don't have four 2's here. 27 is equal to 3 times 3 times 3, so you need three 3's in the prime factorization. We don't have three 3's. So once again, cancel that out. 5's a prime number. There are no 5's here. Rule that out. No 11's here. Rule that out. 9 is equal to 3 times 3. And actually I just realized that this is a silly answer because obviously all numbers divisible by 9 and 24 are also divisible by 9. So obviously 9 is going to work but I shouldn't have made that a choice cuz that's in the problem, but 9 would work, and what also would work if we had a, if 8 was one of the choices, because 8 is equal to 2 times 2 times 2, and we have a 2 times 2 times 2 here. 4 would also work. That's 2 times 2. That's 2 times 2. 6 would work since that's 2 times 3. 18 would work cuz that's 2 times 3 times 3. So anything that's made up of a combination of these prime factors will be divisible into something divisible by both 9 and 24. Hopefully that doesn't confuse you too much."
    },
    {
        "Q": "4:05 12 is a number that is in the problem!\n\nAll numbers divisible by both 12 and 20 are also divisible by 12?",
        "A": "Yup, and all numbers divisible by both 12 and 20 are also divisible by 20.",
        "video_name": "zWcfVC-oCNw",
        "timestamps": [
            245
        ],
        "3min_transcript": "of 12 and 20. Now this isn't the only number that's divisible by both 12 and 20. You could multiply this number right here by a whole bunch of other factors. I could call them a, b, and c, but this is kind of the smallest number that's divisible by 12 and 20. Any larger number will also be divisible by the same things as this smaller number. Now with that said, let's answer the questions. All numbers divisible by both 12 and 20 are also divisible by: Well we don't know what these numbers are so we can't really address it. They might just be 1's or they might not exist because the number might be 60. It might be 120. Who knows what this number is? So the only numbers that we know can be divided into this number, well we know 2 can be, we know that 2 is a legitimate answer. 2 is obviously divisible into 2 times 2 times 3 times 5. We know that 2 times 2 is divisible into it, cuz we have the 2 times 2 over there. We know that 2 times 3 is divisible into it. So that's 6. Let me write these. This is 4. This is 6. We know that 2 times 2 times 3 is divisible into it. I could go through every combination of these numbers right here. We know that 3 times 5 is divisible into it. We know that 2 times 3 times 5 is divisible into it. So in general you can look at these prime factors and any combination of these prime factors is divisible into any number that's divisible by both 12 and 20, so if this was a multiple choice question, and the choices were 7 and 9 and 12 and 8. You would say, well let's see, 7 is not one of these prime factors over here. 9 is 3 times 3 so I need to have two 3's here. I only have one 3 here so 9 doesn't work. 7 doesn't work, 9 doesn't work. or another way to think about it, 12 is 2 times 2 times 3. Well there is a 2 times 2 times 3 in the prime factorization, of this least common multiple of these two numbers, so this is a 12 so 12 would work. 8 is 2 times 2 times 2. You would need three 2's in the prime factorization. We don't have three 2's, so this doesn't work. Let's try another example just so that we make sure that we understand this fairly well. So let's say we wanna know, we ask the same question. All numbers divisible by and let me think of two interesting numbers, all numbers divisible by 12 and let's say 9, and I don't know, let's make it more interesting, 9 and 24 are also divisible by"
    },
    {
        "Q": "At 1:18, he says that in order to be divisible by both a number has to have 2 2s, a 3 etc... but why 2 2s and just 1 3?",
        "A": "12 needed 2 2s and one 3 and 20 needed a 5 and 2 2s but the 2s were thare and he use prime numbers and 2, 3, and 5s are prime",
        "video_name": "zWcfVC-oCNw",
        "timestamps": [
            78
        ],
        "3min_transcript": "- [Voiceover] In this video I wanna do a bunch of example problems that show up on standardized exams and definitely will help you with our divisibility module because it's asking you questions like this. And this is just one of the examples. All numbers divisible by both 12 and 20 are also divisible by: And the trick here is to realize that if a number is divisible by both 12 and 20, it has to be divisible by each of these guy's prime factors. So let's take the prime factorization, the prime factorization of 12, let's see, 12 is 2 times 6. 6 isn't prime yet so 6 is 2 times 3. So that is prime. So any number divisible by 12 needs to be divisible by 2 times 2 times 3. So its prime factorization needs to have a 2 times a 2 times a 3 in it, any number that's divisible by 12. Now any number that's divisible by 20 needs to be divisible by, let's take it's prime factorization. 2 times 10 10 is 2 times 5. So any number divisible by 20 2 times 2 times 5. Or another way of thinking about it, it needs to have two 2's and a 5 in its prime factorization. If you're divisible by both, you have to have two 2's, a 3, and a 5, two 2's and a 3 for 12, and then two 2's and a 5 for 20, and you can verify this for yourself that this is divisible by both. Obviously if you divide it by 20, let me do it this way. Dividing it by 20 is the same thing as dividing by 2 times 2 times 5, so you're going to have the 2's are going to cancel out, the 5's are going to cancel out. You're just going to have a 3 left over. So it's clearly divisible by 20, and if you were to divide it by 12, you'd divide it by 2 times 2 times 3. This is the same thing as 12. And so these guys would cancel out and you would just have a 5 left over so it's clearly divisible by both, and this number right here is 60. It's 4 times 3, which is 12, times 5, it's 60. of 12 and 20. Now this isn't the only number that's divisible by both 12 and 20. You could multiply this number right here by a whole bunch of other factors. I could call them a, b, and c, but this is kind of the smallest number that's divisible by 12 and 20. Any larger number will also be divisible by the same things as this smaller number. Now with that said, let's answer the questions. All numbers divisible by both 12 and 20 are also divisible by: Well we don't know what these numbers are so we can't really address it. They might just be 1's or they might not exist because the number might be 60. It might be 120. Who knows what this number is? So the only numbers that we know can be divided into this number, well we know 2 can be, we know that 2 is a legitimate answer. 2 is obviously divisible into 2 times 2 times 3 times 5. We know that 2 times 2 is divisible into it, cuz we have the 2 times 2 over there."
    },
    {
        "Q": "at 2:42, to isolate the -2, why did you have to divide and not subtract?",
        "A": "be careful with leading negative numbers. Try to think of this as a negative number, not subtraction. The negative two is being multiplied by the absolute value. In order to cancel this number we do the opposite of multiplying by -2 which is dividing by -2.",
        "video_name": "15s6B7K9paA",
        "timestamps": [
            162
        ],
        "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": "At 6:10 can it be y=k/x since it is y=k*1/x",
        "A": "Certainly! That is what inverse variations generally look like. People don t usually write y = 8 * 1/x, they usually write it as y = 8/x.",
        "video_name": "92U67CUy9Gc",
        "timestamps": [
            370
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "At 4:20,how can we assume it is an isosceles pyramid? We don't know for sure,do we? That question might have been a trick question.",
        "A": "I guess because we are using the actual Great Pyramid of Giza for the problem, and it is in real life isosceles, we can assume this.",
        "video_name": "Z5EnuVJawmY",
        "timestamps": [
            260
        ],
        "3min_transcript": "And what else do we know? Well, we know this is 72. We know that the whole thing is 180. So this is 72, and the whole thing is 180. The part of this edge that's below the water, this distance Let me draw it without cluttering the picture too much. I'll do it in that black color. This distance right over here is going to be 108. 108 plus 72 is 180. So what does this do for us? We need to figure out this height. We know that this right over here is a right triangle. I could color this in just to make it a little bit clearer. This thing in yellow, right over here is a right triangle. If we look at that right triangle, and if we wanted to solve for h and solve for h using a trig ratio based on this angle theta right over here, we know that relative to this angle theta, 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,"
    },
    {
        "Q": "At 1:42 Sal says to move the 0 degrees on the protractor to one side of the angle. How do you know which side to use?",
        "A": "One side is in the angle, if the angle is right or acute, than use the side that flows into the actual angle, obtuse angles can vary.",
        "video_name": "wJ37GJyViU8",
        "timestamps": [
            102
        ],
        "3min_transcript": "This is the video for the measuring angles module because, clearly, at the time that I'm doing this video, there is no video for the measuring angles module. And this is a pretty neat module. This was made by Omar Rizwan, one of our amazing high school interns that we had this past summer. This is the summer of 2011. And what it really is, is it makes you measure angles. And he made this really cool protractor tool here so that you actually use this protractor to measure the angles there. And so the trick here is you would actually measure it the way you would measure any angle using an actual physical protractor. You'd want to put the center of the protractor right at the vertex of where the two lines intersect. You can view it as the vertex of the angle. And then you'd want to rotate it so that, preferably, this edge, this edge at 0 degrees, is at one of these sides. So let's do it so that this edge right over here is right along this line. So let me rotate it. So then-- I've got to rotate it a little bit further, maybe So that looks about right. And then if you look at it this way, you can see that the angle-- and I don't have my Pen tool here. I'm just using my regular web browser-- if you look at the angle here, you see that the other line goes to 130 degrees. So this angle that we need to measure here is 130 degrees, assuming you can read sideways. So that is 130 degrees. Let me check my answer. Very good, I got it right. It would have been embarrassing if I didn't. Let's do the next question. I'll do a couple of examples like this. So once again, let us put the center of the protractor right at the vertex right over there. And let's get this 0 degrees side to be on one of these sides so that this angle will be within the protractor. So let me rotate it this way. And this really is pretty cool what Omar did with this module. So let's see. Let's do it one more time. That's too far. And then you can see that the angle right over here, if we look at where the other line points to, it is 40 degrees. Check answer-- very good. Let's do another one. This is fun. So let's get our protractor right over there. And you don't always have to do it in that same order. You could rotate it first so that the 0 degrees is-- and what you want to do is you want to rotate the 0 degrees to one of the sides so that the angle is still within the protractor. So let's rotate it around. So if you did it like that-- so you don't always have to do it in that same order. Although I think it's easier to rotate it when you have the center of the protractor at the vertex of the angle. So we have to rotate it a little bit more. So 0 degrees is this line. And then as we go further and further up, I guess, since this is on its side, it looks like this other line gets us to 150 degrees. And hopefully you're noticing that the higher the degrees, the more open this angle is."
    },
    {
        "Q": "at 1:30 \u00c2\u00a8\u00c3\u00aft would be embarrasing if i didnt\u00c2\u00a8 no sal it would be kinda histerical (is that how you spell it?). and show your human",
        "A": "*hysterical, that s how it s spelt :) And seeing as angle measuring is quite simple for a mathematician, it would be pretty embarrassing for him.",
        "video_name": "wJ37GJyViU8",
        "timestamps": [
            90
        ],
        "3min_transcript": "This is the video for the measuring angles module because, clearly, at the time that I'm doing this video, there is no video for the measuring angles module. And this is a pretty neat module. This was made by Omar Rizwan, one of our amazing high school interns that we had this past summer. This is the summer of 2011. And what it really is, is it makes you measure angles. And he made this really cool protractor tool here so that you actually use this protractor to measure the angles there. And so the trick here is you would actually measure it the way you would measure any angle using an actual physical protractor. You'd want to put the center of the protractor right at the vertex of where the two lines intersect. You can view it as the vertex of the angle. And then you'd want to rotate it so that, preferably, this edge, this edge at 0 degrees, is at one of these sides. So let's do it so that this edge right over here is right along this line. So let me rotate it. So then-- I've got to rotate it a little bit further, maybe So that looks about right. And then if you look at it this way, you can see that the angle-- and I don't have my Pen tool here. I'm just using my regular web browser-- if you look at the angle here, you see that the other line goes to 130 degrees. So this angle that we need to measure here is 130 degrees, assuming you can read sideways. So that is 130 degrees. Let me check my answer. Very good, I got it right. It would have been embarrassing if I didn't. Let's do the next question. I'll do a couple of examples like this. So once again, let us put the center of the protractor right at the vertex right over there. And let's get this 0 degrees side to be on one of these sides so that this angle will be within the protractor. So let me rotate it this way. And this really is pretty cool what Omar did with this module. So let's see. Let's do it one more time. That's too far. And then you can see that the angle right over here, if we look at where the other line points to, it is 40 degrees. Check answer-- very good. Let's do another one. This is fun. So let's get our protractor right over there. And you don't always have to do it in that same order. You could rotate it first so that the 0 degrees is-- and what you want to do is you want to rotate the 0 degrees to one of the sides so that the angle is still within the protractor. So let's rotate it around. So if you did it like that-- so you don't always have to do it in that same order. Although I think it's easier to rotate it when you have the center of the protractor at the vertex of the angle. So we have to rotate it a little bit more. So 0 degrees is this line. And then as we go further and further up, I guess, since this is on its side, it looks like this other line gets us to 150 degrees. And hopefully you're noticing that the higher the degrees, the more open this angle is."
    },
    {
        "Q": "i am confused:\nat 3:20 how does Sal go from =44 to 00000000+000",
        "A": "Well he took the commutative property and broke it down. 4*8+4*3 4*8 is equal to 32. 4*3 = 12. I ll break the equation cause it s just how i work out problems. (4^2+4)+(8*3). 4 times itself is a product of 16.16+4 equals 20. 8*3 =24. 24+20=44.",
        "video_name": "gl_-E6iVAg4",
        "timestamps": [
            200
        ],
        "3min_transcript": "evaluate it that way. But they want us to use the distributive law of multiplication. We did not use the distributive law just now. We just evaluated the expression. We used the parentheses first, then multiplied by 4. In the distributive law, we multiply by 4 first. And it's called the distributive law because you distribute the 4, and we're going to think about what that means. So in the distributive law, what this will become, it'll become 4 times 8 plus 4 times 3, and we're going to think about why that is in a second. So this is going to be equal to 4 times 8 plus 4 times 3. A lot of people's first instinct is just to multiply the 4 times the 8, but no! You have to distribute the 4. You have to multiply it times the 8 and times the 3. This is the distributive property in action right here. Distributive property in action. And then when you evaluate it-- and I'm going to show you in kind of a visual way why this works. But then when you evaluate it, 4 times 8-- I'll do this in a different color-- 4 times 8 is 32, and then so we have 32 plus 4 times 3. 4 times 3 is 12 and 32 plus 12 is equal to 44. That is also equal to 44, so you can get it either way. But when they want us to use the distributive law, you'd distribute the 4 first. Now let's think about why that happens. Let's visualize just what 8 plus 3 is. Let me draw eight of something. So one, two, three, four, five, six, seven, eight, right? And then we're going to add to that three of something, of One, two, three. So you can imagine this is what we have inside of the parentheses. We have 8 circles plus 3 circles. Now, when we're multiplying this whole thing, this whole thing times 4, what does that mean? Well, that means we're just going to add this to itself four times. Let me do that with a copy and paste. Copy and paste. Let me copy and then let me paste. There you go. That's two. That's one, two, three, and then we have four, and we're going to add them all together. So this is literally what? Four times, right? Let me go back to the drawing tool. We have it one, two, three, four times this expression, which is 8 plus 3. Now, what is this thing over here? If you were to count all of this stuff, you would get 44."
    },
    {
        "Q": "At 1:22 in the video do you do four times eleven?",
        "A": "Yep! 4(11) = 44",
        "video_name": "gl_-E6iVAg4",
        "timestamps": [
            82
        ],
        "3min_transcript": "Rewrite the expression 4 times, and then in parentheses we have 8 plus 3, using the distributive law of multiplication over addition. Then simplify the expression. So let's just try to solve this or evaluate this expression, then we'll talk a little bit about the distributive law of multiplication over addition, usually just called the distributive law. So we have 4 times 8 plus 8 plus 3. Now there's two ways to do it. Normally, when you have parentheses, your inclination is, well, let me just evaluate what's in the parentheses first and then worry about what's outside of the parentheses, and we can do that fairly easily here. We can evaluate what 8 plus 3 is. 8 plus 3 is 11. So if we do that-- let me do that in this direction. So if we do that, we get 4 times, and in parentheses we have an 11. 8 plus 3 is 11, and then this is going to be equal to-- evaluate it that way. But they want us to use the distributive law of multiplication. We did not use the distributive law just now. We just evaluated the expression. We used the parentheses first, then multiplied by 4. In the distributive law, we multiply by 4 first. And it's called the distributive law because you distribute the 4, and we're going to think about what that means. So in the distributive law, what this will become, it'll become 4 times 8 plus 4 times 3, and we're going to think about why that is in a second. So this is going to be equal to 4 times 8 plus 4 times 3. A lot of people's first instinct is just to multiply the 4 times the 8, but no! You have to distribute the 4. You have to multiply it times the 8 and times the 3. This is the distributive property in action right here. Distributive property in action. And then when you evaluate it-- and I'm going to show you in kind of a visual way why this works. But then when you evaluate it, 4 times 8-- I'll do this in a different color-- 4 times 8 is 32, and then so we have 32 plus 4 times 3. 4 times 3 is 12 and 32 plus 12 is equal to 44. That is also equal to 44, so you can get it either way. But when they want us to use the distributive law, you'd distribute the 4 first. Now let's think about why that happens. Let's visualize just what 8 plus 3 is. Let me draw eight of something. So one, two, three, four, five, six, seven, eight, right? And then we're going to add to that three of something, of"
    },
    {
        "Q": "At 1:56 how did you get x = 4.? And one ore thing I am still not getting what does this theorem states?",
        "A": "The theorem says that the third side of a triangle has to be LESS than the sum of the other two sides AND MORE than the rest of the two sides. So when you subtract 6 from 10 you get 4.",
        "video_name": "KlKYvbigBqs",
        "timestamps": [
            116
        ],
        "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": "I don't understand, at around 8:00 when Sal is explaining vector [c1 c2 0 c4 0] he said that c1,c2,c4 must be zero showing linear independence, now aren't the third and fifth term also zero? If c1 c2 c4 are zero then the solution would be [0 0 0 0 0], doesn't that imply all column vectors to be linearly independent?",
        "A": "He says that the 1st, 2nd and 4th c must be 0 for the result to be the 0 vector, and hence are linearly independent. Whereas, the 3rd and 5th just happen to be zero.. but since these vectors can be made from the other vectors, you would also be able to find a way to get back to 0 even if these happened not to be. (He mentions showing this in the next video).",
        "video_name": "BfVjTOjvI30",
        "timestamps": [
            480
        ],
        "3min_transcript": "do it over here, let me do it in blue-- you get c1 times a1 plus c2 times a2, and then 0 times a3, plus c4 times a4 is equal to 0. Now these guys are going to be linearly independent, if and only if the only solution to this equation is they all equal to 0. Well we know that the only solution to this is that they all equal 0 because anything that's a solution to this is a solution to this. And the only solution to this was, if I go ahead and I constrain these two terms to being equal to 0, the only solution to this is all of these c's have to be 0. So likewise, if I constrain these to be 0, the only solution to this is that c1, c2, and c4 have to be 0. vectors, a1, a2, and a4, so that implies that the set a1, a2, and a4 are linearly independent. So we're halfway there. We've shown that because the pivot columns here are linearly independent. We can show and they have the same solution set. The null space of the reduced row echelon form is the same as the null space of our original matrix. We were able to show that the only solution to c1 times this plus c2 times this plus c4 times this is when all the constants are 0, which shows that these three vectors or a set of those three vectors are definitely linearly independent. Now, the next thing to prove that they are a basis, is to show that all of the other column vectors can be represented as multiples of these three guys. boring you too much, I'll do that in the next video. So in this one we saw that if the pivot columns are linearly independent, they always are. All pivot columns, by definition are linearly Or the set of pivot columns are always linearly independent when you take away the non-pivot columns, then the corresponding columns in your original vector are also In the next one we'll show that these three guys also span your column space."
    },
    {
        "Q": "At 8:16, when you have the area of 324 pi. Why can't you substitute pi with 3 because 3.14 rounded is 3 and multiply 3 by 324?",
        "A": "The answer will not be accurate",
        "video_name": "mLE-SlOZToc",
        "timestamps": [
            496
        ],
        "3min_transcript": "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, 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--"
    },
    {
        "Q": "At 5:15 , what is a circumference ?",
        "A": "The circumference of a circle is the length of the perimeter of the circle, or the distance all the way around. It is equal to pi times the diameter of the circle or pi times 2 times the radius.",
        "video_name": "mLE-SlOZToc",
        "timestamps": [
            315
        ],
        "3min_transcript": "of the total possibilities that meet our constraint, and our constraint is being a multiple of 5. So how many total possibilities are there? Let's think about that. How many do we have? Well that's just the total number of numbers we have to pick from, so 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. So there are 12 possibilities. We have an equal chance of picking any one of these 12. Now which of these 12 are a multiple of 5? So let's do this in a different color. So let me pick out the multiples of 5. 32 is not a multiple of 5, 49 is not a multiple of 5. 55 is a multiple of 5. Really, we're just looking for the numbers that in the ones place you either have a 5 or a 0. 55 is a multiple of 5, 30 is a multiple 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,"
    },
    {
        "Q": "Before the time of 3:04, shouldn't the non green marbles be 11/14?",
        "A": "No, we are trying to find the probability of picking a NON-BLUE marble. There are 14 marbles total, two of which are blue. Thus the number of non-blue marbles is 12/14, which simplifies down to 6/7.",
        "video_name": "mLE-SlOZToc",
        "timestamps": [
            184
        ],
        "3min_transcript": "And the way you just think about it is what fraction of all of the possible events meet our constraint? So let's just think about all of the possible events first. How many different possible marbles can we take out? Well that's just the total number of marbles there are. So are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 possible marbles. So this is the number of possibilities. And then we just have to think what fraction of those possibilities meet our constraints. And the other way you could have gotten 14 is just taking 9 plus 2 plus 3. So what number of those possibilities meet our constraints? And remember, our constraint is selecting a non-blue marble Another way to think about it is a red or green marble, because the only other two colors we have are red and green. So how many non-blue marbles are there? Well, there's a couple ways to think about it. 2 are blue. So there are going to be 14 minus 2, which is 12 non-blue marbles. Or you could just count them. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. So there are 12 non-blue marbles. So these are the possibilities that meet our constraints over all of the possibilities. And then if we want to-- this isn't in simplified form right here, since both 12 and 14 are divisible by 2. So let's divide both the numerator and the denominator by 2, and you get 6 over 7. So we have a 6/7 chance of selecting a non-blue marble from the bag. Let's do another one. If a number is randomly chosen from the following list, what is the probability that the number is a multiple of 5? of the total possibilities that meet our constraint, and our constraint is being a multiple of 5. So how many total possibilities are there? Let's think about that. How many do we have? Well that's just the total number of numbers we have to pick from, so 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. So there are 12 possibilities. We have an equal chance of picking any one of these 12. Now which of these 12 are a multiple of 5? So let's do this in a different color. So let me pick out the multiples of 5. 32 is not a multiple of 5, 49 is not a multiple of 5. 55 is a multiple of 5. Really, we're just looking for the numbers that in the ones place you either have a 5 or a 0. 55 is a multiple of 5, 30 is a multiple"
    },
    {
        "Q": "At 1:31, isn't there an easier way of solving?",
        "A": "I m sure there is! But Sal is not trying to show us the easiest way, here: he s trying to explain the principles of probabilities and choice. Understanding and explaining these principles is far more important for Sal than showing us the easiest way to find a solution.",
        "video_name": "mLE-SlOZToc",
        "timestamps": [
            91
        ],
        "3min_transcript": "Let's do a couple of exercises from our probability one module. So we have a bag with 9 red marbles, 2 blue marbles, and 3 green marbles in it. What is the probability of randomly selecting a non-blue marble from of the bag? So let's draw this bag here. So that's my bag, and we're going to assume that it's a transparent bag, so it looks like a vase. But we have 9 red marbles, so let me draw 9 red marbles. 1, 2, 3, 4, 5, 6, 7, 8, 9 red marbles. They're kind of orange-ish, but it does the job. 2 blue marbles, so we have 1 blue marble, 2 blue marbles. And then we have 3 green marbles, let me draw those 3, so 1, 2, 3. What is the probability of randomly selecting a non-blue marble from the bag? So maybe we mix them all up, and we have an equal probability And the way you just think about it is what fraction of all of the possible events meet our constraint? So let's just think about all of the possible events first. How many different possible marbles can we take out? Well that's just the total number of marbles there are. So are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 possible marbles. So this is the number of possibilities. And then we just have to think what fraction of those possibilities meet our constraints. And the other way you could have gotten 14 is just taking 9 plus 2 plus 3. So what number of those possibilities meet our constraints? And remember, our constraint is selecting a non-blue marble Another way to think about it is a red or green marble, because the only other two colors we have are red and green. So how many non-blue marbles are there? Well, there's a couple ways to think about it. 2 are blue. So there are going to be 14 minus 2, which is 12 non-blue marbles. Or you could just count them. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. So there are 12 non-blue marbles. So these are the possibilities that meet our constraints over all of the possibilities. And then if we want to-- this isn't in simplified form right here, since both 12 and 14 are divisible by 2. So let's divide both the numerator and the denominator by 2, and you get 6 over 7. So we have a 6/7 chance of selecting a non-blue marble from the bag. Let's do another one. If a number is randomly chosen from the following list, what is the probability that the number is a multiple of 5?"
    },
    {
        "Q": "at 5:30 sal says area of a circle. however,circles dont have area,circles have circumfrence",
        "A": "All two-dimensional shapes have a perimeter and an area. Perimeter is a distance, the length of the bounding line(s). Area is ... well ... an area. It measures the amount of a flat surface that is contained within the perimeter. For circles, the perimeter is called the circumference.",
        "video_name": "mLE-SlOZToc",
        "timestamps": [
            330
        ],
        "3min_transcript": "of the total possibilities that meet our constraint, and our constraint is being a multiple of 5. So how many total possibilities are there? Let's think about that. How many do we have? Well that's just the total number of numbers we have to pick from, so 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. So there are 12 possibilities. We have an equal chance of picking any one of these 12. Now which of these 12 are a multiple of 5? So let's do this in a different color. So let me pick out the multiples of 5. 32 is not a multiple of 5, 49 is not a multiple of 5. 55 is a multiple of 5. Really, we're just looking for the numbers that in the ones place you either have a 5 or a 0. 55 is a multiple of 5, 30 is a multiple 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,"
    },
    {
        "Q": "At around 2:15 , why does (1-cos(2x))\u00c2\u00b2 turn into 1-2cos(2x)+cos\u00c2\u00b2(2x) ? Where does the 2cos(2x) come from ?",
        "A": "Sal was simply expanding the expression (1/2 (1-cos2x))^2. 1/2 squared is 1/4 and (1-cos2x) squared is (1-cos2x) times (1-cos2x). If you recall from Algebra that (a-b)(a-b) = a^2 -2ab + b^2, then let a = 1 and let b = cos2x and multiply it out. a^2 = (1)(1) = 1, -2ab = -2(1)(cos2x) = -2cos2x. b^2 = (cos2x)(cos2x) = cos^2 2x. Hope this helps. Good luck.",
        "video_name": "n34jx1FIN8M",
        "timestamps": [
            135
        ],
        "3min_transcript": "- [Voiceover] Let's see if we can take the indefinite integral of sine of x to the fourth dx. And like always, pause the video and see if you can work through it on your own. So if we had an odd exponent up here, whether it was a sine or a cosine, then the technique I would use, so if this was sine to the third of x, I would separate one of the sine of xs out, so I would rewrite it as sine squared x times sine of x, and then I would convert this using the Pythagorean Identity, and then when I distribute the sine of x, I'd be able to use u-substitution. We've done that in previous example videos. You could have done this if it was cosine to the third of x as well, or to the fifth, or to the seventh, if you had an odd exponent. But here we have an even exponent. So what do we do? So the technique we will use, and I guess you could call it a trick, the technique or trick to use is once again you want to algebraically manipulate this so you can use integration techniques that we are familiar with. the Double Angle Identity. The Double Angle Identity tells us that sine squared of x is equal to 1/2 times one minus cosine of two x. So how can I apply this over here? Our original integral is just the same thing as, this is going to be the same thing as the integral of sine of x squared, all of that squared, dx. Now I can make this substitution. So this is the same thing as this, which is of course the same thing as this by the Double Angle Identity. So I could rewrite it as the integral of 1/2 times one minus cosine of two x, and then all of that squared dx. That's going to be equal to, let's see. 1/2 squared is 1/4, so I can take that out. So we get 1/4 times the integral of... I'm just going to square all this business. One squared, which is one, minus two cosine of two x plus cosine squared of two x dx. Fairly straightforward to take the indefinite integral, or to take the antiderivative of these two pieces. But what do I do here? Once again I've got an even exponent. Let's apply the Double Angle Identity for cosine. We know that cosine squared of two x is going to be equal to 1/2 times one plus cosine of double this angle, so cosine of four x. Once again just make the substitution."
    },
    {
        "Q": "towards the end of the video at 10:45 sal says that the cut off for staticians is 5% or less. why is it 5% and less and not 5% and more. can i please have an example as well as to why it is less than 5%",
        "A": "We start with an assumption (null hypothesis), and calculate the probability of the observed result if that assumption is actually true. Hence, small probability will make us say Hmm, these results don t match up with our assumption very well, the assumption is probably wrong. Large probability does not make us question our assumption: it says the data are fairly compatible with the assumption, so it might be okay.",
        "video_name": "W3C07uH-b9o",
        "timestamps": [
            645
        ],
        "3min_transcript": "is this occurring? Well, this one occurs 85 times, this one occurs eight. If you add these two together, 93 out of the thousand times, out of her re-randomization or I guess you could say 9.3 percent of the time, the data... 9.3 percent of the randomized, the 1000 re-randomizations, 9.3 percent of the time she got data that was as validating of a hypothesis or more than the actual experiment. One way to think about this is, the probability of randomly getting the results from her experiment or better results from her experiment are 9.3 percent. They're low, it's a reasonably low probability that this happened purely by chance. Now, a question is, \"What's the threshold?\" If it was a 50 percent you say, \"Okay, this was very If this was a 25 percent you're like, \"Okay, it's less \"likely to happen by chance but it could happen.\" 9.3 percent, it's roughly 10 percent. For every 10 people who do an experiment like she did, even if it was random, one person would get data like this? What typically happens amongst statisticians is they draw a threshold and the threshold for statistical significance is usually five percent. One way to think about it, the probability of her getting this result by chance, this result or a more extreme result? One that more confirms her hypothesis by chance is 9.3 percent. If you're cut-off for significance is five percent. If you said, \"Okay, this has to be five percent or less.\" Then you say, \"Okay, this is not statistically significant.\" There's more than a five percent chance that I could have gotten this result purely through random chance. Once again, that just depends on where you have that threshold. When we go back, I think we've already answered the final question, \"According to the simulations, \"being lower than the control group's median \"by eight minutes or more?\" Which once again, eight minutes or more, that would be negative eight and negative 10. We just figured that out, that was 93 out of the 1000 re-randomizations, so it's a 9.3 percent chance. If you set five percent as your cut-off for statistical significance, you say, \"Okay, this doesn't quite meet my \"cut-off so maybe this is not a statistically \"significant result.\""
    },
    {
        "Q": "is there a way to convert the remainder into a decimal on 2:06",
        "A": "There sure is. Think of 15/4 (15 divided by 4) as how many groups of 4 can 15 be divided into. The answer of 3r3 means 3 full groups of 4 and a partially completed group, the remained, of 3. That partially completed group holds 4 but only has 3 in it. We call this 3/4. So we can rewrite 3r3 as 3 3/4. We can do our division on 3/4 and convert it into a decimal, which equals 0.75. So 15/4 = 3r3 = 3 3/4 = 3.75",
        "video_name": "BIGX05Mp5nw",
        "timestamps": [
            126
        ],
        "3min_transcript": "Let's take the number 7 and divide it by 3. And I'm going to conceptualize dividing by 3 as let me see how many groups of 3 I can make out of the 7. So let me draw 7 things-- 1, 2, 3, 4, 5, 6, 7. So let me try to create groups of 3. So I can definitely create one group of 3 right over here. I can definitely create another group of 3. So I'm able to create two groups of 3. And then I can't create any more full groups of 3. I have essentially this thing right over here left over. So this right over here, I have this thing remaining. This right over here is my remainder after creating as many groups of 3 as I can. And so when you see something like this, people will often say 7 divided by 3. Well, I can create two groups of 3. But it doesn't divide evenly, or 3 doesn't divide evenly into 7. I end up with something left over. I have a leftover. I have a remainder of 1. And that makes sense. 2 times 3 is 6. So it doesn't get you all the way to 7. But then if you have your extra remainder, 6 plus that 1 remainder gets you all the way to 7. Let's do another one. Let's imagine 15 divided by 4. Let me draw 15 objects-- 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15. Now, let me try to divide it into groups of 4. So let's see, that's one group of 4. That's another group of 4. And then that's another group of 4. So I'm able to create three groups of 4. But then I can't create a fourth full group of 4. I am then left with this remainder right over here. I have a remainder right over here of 3. So we could say that 15 divided by 4 is 3 remainder 3. 4 goes into 15 three times. But that only gets us to 12. 4 times 3 is 12. To get all the way to 15, we need to use our remainder. We have to get 3 more. So 15 divided by 4, I have 3 left over. Now, let's try to think about this doing a little bit of our long division techniques. So let's say that I have 4. Let's say I want to divide 75 by 4. Well, traditional long division techniques. 4 goes into 7 one time. And If you're looking at place value, we're really saying the 4 is going into 70 ten times, because we're putting this in the tens place. And then we say, 1 times 4 is 4. But really, once again, since it's in the tens place, this is really representing a 40."
    },
    {
        "Q": "At 1:28 Sal says that we must divide both the sides with 5 but instead of that we can transpose 5.It is much easier that way!",
        "A": "fair enough, but I think that dividing both the sides by 5 is good for beginners because the fundamental belief of algebra, that if you do the same thing to both the sides of an equation,you won t change anything seems much more prominent in dividing both sides by 5 than in transposing.",
        "video_name": "c6-FJRda_Vc",
        "timestamps": [
            88
        ],
        "3min_transcript": "y is directly proportional to x. If y equals 30 when x is equal to 6, find the value of x when y is 45. So let's just take this each statement at a time. y is directly proportional to x. That's literally just saying that y is equal to some constant times x. This statement can literally be translated to y is equal to some constant times x. y is directly proportional to x. Now, they tell us, if y is 30 when x is 6-- and we have this constant of proportionality-- this second statement right over here allows us to solve for this constant. When x is 6, they tell us y is 30 so we can figure out what this constant is. We can divide both sides by 6 and we get this left-hand side is 5-- 30 divided by 6 is 5. 5 is equal to k or k is equal to 5. So the second sentence tells us, this gives us the information that y is equal to 5 times x. y is 30 when x is 6. And then finally, they say, find the value of x when y is 45. So when y is 45 is equal to-- so we're just putting in 45 for y-- 45 is equal to 5x. Divide both sides by 5 to solve for x. We get 45 over 5 is 9, and 5x divided by 5 is just x. So x is equal to 9 when y is 45."
    },
    {
        "Q": "I calculated M by subtracting the first formula [(mx^2)+bx=xy] from the second (y=mx+b). Which is the opposite of what sal does @0:50. I get a different formula for M, is this OK? I can't equate the two formulas. Here is the formula I get (all the x and y should have the mean sign. M= [(xy/x)-y]/[(x^2/x)-x)",
        "A": "Multiply both top and bottom by -1, and the result is [(-xy/x)+y]/[(-x^2/x)+x] where all the x, x^2, and y should have the mean sign. This is equivalent to [y-(xy/x)]/[x-(x^2/x)] simply by switching the order of the terms in the numerator and denominator, respectively. This is Sal s answer.",
        "video_name": "8RSTQl0bQuw",
        "timestamps": [
            50
        ],
        "3min_transcript": "So if you've gotten this far, you've been waiting for several videos to get to the optimal line that minimizes the squared distance to all of those points. So let's just get to the punch line. Let's solve for the optimal m and b. And just based on what we did in the last videos, there's We actually now know two points that lie on that line. So we can literally find the slope of that line and then the the y intercept, the b there. Or, we could just say it's the solution to this system of equations. And they're actually mathematically equivalent. So let's solve for m first. And if we want to solve for m, we want to cancel out the b's. So let me rewrite this top equation just the way it's written over here. We have m times the mean of the x squareds plus b times the mean of-- Actually, we could even do it better than that. One step better than that is to, based on the work we did in the last video, we can just subtract this bottom equation from this top equation. So let me subtract it. Or let's add the negatives. So if I make this negative, this is negative. What do we get? We get m times the mean of the x's minus the mean of the x squareds over the mean of x. The plus b and the negative b cancel out. Is equal to the mean of the y's minus the mean of the xy's over the mean of the x's. And then, we can divide both sides of the equation by this. And so we get m is equal to the mean of the y's minus the mean of the xy's over the mean of the x's over this. The mean of the x's minus the mean of the x squareds over the mean of the x's. Now notice, this is the exact same thing that you would get if you found the slope between these two points over here. that right over there. Over the change in x's. The change in that x minus that x is exactly this over here. Now, to simplify it, we can multiply both the numerator and the denominator by the mean of the x's. And I do that just so we don't have this in the denominator both places. So if we multiply the numerator by the mean of the x's, we get the mean of the x's times the mean of the y's minus, this and this will cancel out, minus the mean of the xy's. All of that over, mean of the x's times the mean of the x's is just going to be the mean of the x's squared, minus over here you have the mean of the x squared. And that's what we get for m."
    },
    {
        "Q": "So, I now understand that you can multiply a real number (or scalar, which was defined at 0:30) by a matrix, but can you multiply a Matrix by another Matrix? Wouldn't it be just like adding or subtracting, except you multiply instead?\nFor example:\n\n[1 8 3] * [2 9 4] = [2 72 12]\n\nWould that be correct?",
        "A": "No and it is more complicated than that. You can watch Dr. Khan s video on this but I ll give you the quick version. For 2 matrices to be multiplied, the number of rows in the first matrix must be equal to the number of columns in the second matrix (remember matrix multiplication isn t always commutative as AB isn t necessarily BA). In your case, those two matrices cannot be multiplied.",
        "video_name": "TbaltFbJ3wE",
        "timestamps": [
            30
        ],
        "3min_transcript": "Now that we know what a matrix is, let's see if we can start to define some operations on matrices. So let's say I have the 2 by 3 matrix, so two rows and three columns, and the entries are 7, 5, negative 10, 3, 8, and 0. And I want to define what happens when I multiply 3 times this whole thing. So first of all, let's get a little terminology out of the way. The number three, in just the everyday world, if you weren't dealing with matrices or vectors, and if you don't know what vectors are, don't worry about them just now, you would just call that a number. You would call this a real number. It's just a regular number sitting out there. But now in the world where we have these new structured things, these matrices, these arrays of numbers, we will refer to these just plain old real numbers that aren't part of some type of an array here, we call these scalars. So essentially what we're defining here, we don't know-- I haven't said what this is actually going to turn out to be, but whatever this turns out where we're multiplying a scalar times a matrix. And so how would you define this? What do you think this should be? 3 times this stuff right over here. Well, the world could have defined scalar multiplication however it saw fit, but one way that we find, perhaps, the most obvious and the most useful, is to multiply this scalar quantity times each of the entries. So this is going to be equal to 3 times 7 in the top left, 3 times 5, 3 times negative 10, 3 times 3, 3 times 8, and 3 times 0, which will give us-- it didn't change the dimensions of the matrix. It didn't change, I guess you could say, the structure of the matrix, it just multiplied each of the entries times 3. So the top left entry is now going to be 21, the entry in the middle row, So when you multiply a matrix times a scalar, you just multiply each of those entries times that scalar quantity."
    },
    {
        "Q": "at 2:20 I have no idea whats going on",
        "A": "He is reversing the multiplication he did to make it a whole number so that the decimal would be in the correct place.",
        "video_name": "D5fmcpNygQk",
        "timestamps": [
            140
        ],
        "3min_transcript": "Let's multiply 1.21, or 1 and 21 hundredths, times 43 thousandths, or 0.043. And I encourage you to pause this video and try it on your own. So let's just think about a very similar problem but one where essentially we don't write the decimals. Let's just think about multiplying 121 times 43, which we know how to do. So let's just think about this problem first as kind of a simplification, and then we'll think about how to get from this product to this product. So we can start with-- so we're going to say 3 times 1 is 3. 3 times 2 is 6. 3 times 1 is 3. 3 times 121 is 363. And now we're going to go to the tens place, so this is a 40 right over here. So since we're in the tens place, let's put a 0 there. 40 times 1 is 40. 40 times 20 is 800. 40 times 100 is 4,000. and now we can just add all of this together. And we get-- let me do a new color here-- 3 plus 0 is 3. 6 plus 4 is 10. 1 plus 3 plus 8 is 12. 1 plus 4 is 5. So 121 times 43 is 5,203. Now, how is this useful for figuring out this product? Well, to go from 1.21 to 121, we're essentially multiplying by 100. We're moving the decimal two places over to the right. And to go from 0.043 to 43, what are we doing? We're removing the decimal, so we're multiplying by ten, hundred, thousand. We're multiplying by 1,000. So to go from this product to this product and we multiplied by 1,000. So then to go back to this product, we have to divide. We should divide by 100 and then divide by 1,000, which is equivalent to dividing by 100,000. But let's do that. So let's rewrite this number here, so 5,203. Actually let me write it like this just so it's a little bit more aligned, 5,203. And we could imagine a decimal point right over here. If we divide by 100-- so you divide by 10, divide by 100-- and then we want to divide by another 1,000. So divide by 10, divide by 100, divide by 1,000. So our decimal point is going to go right over there, and we're done. 1.21 times 0.043 is 0.05203."
    },
    {
        "Q": "Can someone explain to me how he got the 6i in (9+6i-1) at 5:50? Thanks!",
        "A": "Shivanie, He was multiplying (3+i)(3+i) Using FOIL you get First 3*3 = 9 Outside 3*i = 3i Inside i*3 = 3i Last i*i = -1 So you get 9+3i+3i-1 And the 3i+3i = 6i so you get 9+6i-1 I hope that is of help to you.",
        "video_name": "dnjK4DPqh0k",
        "timestamps": [
            350
        ],
        "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": "Can someone explain to me what Sal is doing at 5:29 onwards?",
        "A": "Sal is using the same FOIL technique except now there are complex numbers. ((3 + i) / 2)^2 can also be written as ((3 + i)*(3 + i)) / (2*2). By using FOIL the numerator will become... F: 3*3 = 9 O: 3*i = 3i I: 3*i = 3i L: i*i = -1 (3 + i)(3 + i) ----> 9 + 3i + 3i - 1 ----> 8 + 6i 2(8 + 6i) / 4 ----> 4(4 + 3i) / 4 ----> 4 + 3i I hope this helps",
        "video_name": "dnjK4DPqh0k",
        "timestamps": [
            329
        ],
        "3min_transcript": "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. 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,"
    },
    {
        "Q": "At 2:34 Sal says that any percent is that percent over 100. But however what if your number was greater than 100 say like 109%?",
        "A": "109% is 109/100, which is 1.09.",
        "video_name": "-gB1y-PMWfs",
        "timestamps": [
            154
        ],
        "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": "he lost me at 2:32, what is he saying. Please help",
        "A": "(anything)%=(anything)/100 I hope you understand now!",
        "video_name": "-gB1y-PMWfs",
        "timestamps": [
            152
        ],
        "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": "How did Sal decide how big the blue circles were at 1:00 to 1:10",
        "A": "The first circle needs to be at least half the diameter of the circle win the problem. Then, with the second circle, Sal simply placed the midpoint of the new compass on the midpoint of the first circle, made the circles the same size by adjusting the new compass, and then placed the new circle s midpoint on the opposite side of the circle in the problem. Hope this helps :)",
        "video_name": "-gWtl6mdpeY",
        "timestamps": [
            60,
            70
        ],
        "3min_transcript": "Construct a square inscribed inside the circle. And in order to do this, we just have to remember that a square, what we know of a square is all four sides are congruent and they intersect at right angles. And we also have to remember that the two diagonals of the square are going to be perpendicular bisectors of each other. So let's see if we can construct two lines that are perpendicular bisectors of each other. And essentially, where those two lines intersect our bigger circle, those are going to be the vertices of our square. So let's throw a straight edge right over here. And let's make a diameter. So that's a diameter right over here. It just goes through the circle, goes through the center of the circle, to two sides of the circle. And now, let's think about how we can construct a perpendicular bisector of this. And we've done this in other compass construction or construction videos. But what we can do is we can put a circle-- let's throw a circle right over here. And what we're going to do is we're going to reuse this. We're going to make another circle that's the exact same size. Put it there. And where they intersect is going to be exactly along-- those two points of intersection are going to be along a perpendicular bisector. So that's one of them. Let's do another one. I want a circle of the exact same dimensions. So I'll center it at the same place. I'll drag it out there. That looks pretty good. I'll move it on to this side, the other side of my diameter. So that looks pretty good. And notice, if I connect that point to that point, I will have constructed a perpendicular bisector of this original segment. So let's do that. Let's connect those two points. So that point and that point. And then, we could just keep going all the way to the end of the circle. Go all the way over there. That looks pretty good. And now, we just have to connect these four So let's do that. So I'll connect to that and that. And then I will connect, throw another straight edge there. I will connect that with that. And then, two more to go. I'll connect this with that, and then one more. I can connect this with that, and there you go. I have a shape whose vertices intersect the circle. And its diagonals, this diagonal and this diagonal, these are perpendicular bisectors."
    },
    {
        "Q": "so should i have studied quadratic equations before i got to this? it looks like that's a later topic in the algebra play list. and where do i find the vertex form/formula?\n\nfor instance at 5:02 sal says \"this is in vertex form\" referring to y = 2 (x-4)^2 + 3 \"where x = 4 and y = 3\"... and then he plots that point. i don't knw what that means or where i can look for it in the playlist.\n\nat 5:57 i think this might be completing the square?",
        "A": "Yes, you need to know quadratics before doing these videos -- they seem to have been put too early on this list. It is possible to do some simple non-linear equations before mastering quadratics, but I would not recommend it. Try going to the Functions videos and then coming back to these videos once you re completely done with quadratics.",
        "video_name": "FksgVpM_iXs",
        "timestamps": [
            302,
            357
        ],
        "3min_transcript": "Over here, we have 18 plus or minus the square root of-- let's just use a calculator. I could multiply it out but I think-- we have 18 squared minus 4 times 3 times 37, which is negative 120. It's 18 plus or minus the square root of negative 120. You might have even been able to figure out that this is negative. 4 times 3 is 12. 12 times 37 is going to be a bigger number than 18. Although it's not 100% obvious, but you might be able to just get the intuition there. We definitely end up with a negative number under the radical here. Now, if we're dealing with real numbers, there is no square root of negative 120. So there is no solution to this quadratic equation. There is no solution. If we wanted to, we could have just looked at the 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."
    },
    {
        "Q": "What does Sal mean @16:38 when he said usually they are the negative of each other?",
        "A": "You normally see vector fields pointing to decreasing the scalar, not increasing the scalar. For example, the force on a particle at a certain point is equivalent to the negative of the gradient of the potential energy at that point.",
        "video_name": "K_fgnCJOI8I",
        "timestamps": [
            998
        ],
        "3min_transcript": "This associates a value with every point on the xy plane. But this whole exercise, remember this is the same thing as that. This is our whole thing that we were trying to prove: that is equal to f dot dr. f dot dr, our vector field, which is the gradient of the capital F-- remember F was equal to the gradient of F, we assume that it's the gradient of some function capital F, if that is the case, then we just did a little bit of calculus or algebra, whatever you want to call it, and we found that we can evaluate this integral by evaluating capital F at t is equal to b, and then subtracting from that capital F at t is equal to a. But what that tells you is that this integral, the value of this integral, is only dependent at our starting point, t is equal to a, this is the point x of a, y of a, and That integral is only dependent on these two values. How do I know that? Because to solve it-- because I'm saying that this thing exists --I just had to evaluate that thing at those two points; I didn't care about the curve in between. So this shows that if F is equal to the gradient-- this is often called a potential function of capital F, although they're usually the negative each other, but it's the same idea --if the vector field f is the gradient of some scale or field upper-case F, then we can say that f is conservative or that the line integral of f dot dr is path independent. It doesn't matter what path we go on as long as our starting Hopefully found that useful. And we'll some examples with that. And actually in the next video I'll prove another interesting outcome based on this one."
    },
    {
        "Q": "What does he mean at 3:52",
        "A": "He means that to determine whether a point is on a line or not you can input the coordinates into the given equation and if the equation is true, than the point is on the line, but if the equation false, than the point isn t on the line.",
        "video_name": "SSNA9gaAOVc",
        "timestamps": [
            232
        ],
        "3min_transcript": "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. So let's say that this is y is 1, 2, 3, 4. This is negative 1, negative 2, negative 3, negative 4. I could keep going down in that direction. This is 1. Let me do it a little bit-- 1, 2, 3, 4, and I could just keep going on and on in the positive x direction. So let's plot these points. I have 0, 3.5. When x is 0, y is 1, 2, 3.5. When x is 1, y is 1. And so if we were to draw this line-- I'll do it as a dotted line, just so that I can make sure I connect the dots. I can do a better job than that though. So it will look something like that. And so if someone gave you this line, you'd say oh, well it's 3, negative 4 on this line, and let's assume that we drew it really nicely Let me try one last attempt at it. So it's going to look something like that. And If someone asked is 3, negative 4 on it, you could visually do it, but it's always hard when you actually don't substitute it, Maybe you're a little bit off. But if you look at it over here, you say when x is equal to 3, what is y? Well, you go down here, and it looks like y is equal to negative 4. So this is a point 3 comma negative 4. Obviously, in general, you don't want to just rely off of inspecting graphs. Maybe this was 3, negative 3.9999, and you just couldn't tell looking at the graph. That's why you always want to just the substitute and make sure that it really does equal, that this equality really does hold true at that point, not just looking at the graph. But it's important to realize that the graph really is another representation to all of the solutions"
    },
    {
        "Q": "at 1:14 what is the operation?",
        "A": "He just multiplies 5 and 3 and adds it to 2 multiplied by -4",
        "video_name": "SSNA9gaAOVc",
        "timestamps": [
            74
        ],
        "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": "Around 5:50, wouldn't you also divide by 2x1 to remove the repeats for tails? Or would that cancel out because tails was not included in the total probability in the first place?",
        "A": "I think tails don t factor in because our question didn t address them as something to calculate. If, say, we were asking what the probability would be of getting 3 heads and 2 tails exactly in six flips (assuming that, in this scenario, it is possible to get a coin on its side), then, yes, we would have to factor in tails. Also, tails were included in our total probability (all possible sequences of heads and tails), but again, their chance of occurring wasn t asked for in the question; so we ignore them.",
        "video_name": "udG9KhNMKJw",
        "timestamps": [
            350
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At about 1:17, he showed that he was converting the decimals to whole numbers. That method really confuses me. Can somebody please show me a way to solve the problem while keeping the decimals and solving it that way? would it be possible to instead of changing the decimals, keep the decimals and work out the problem with the decimals?",
        "A": "You can do 0.6 / 1.2 But, the 1st step in decimal division, is to change the 1.2 into a whole number. Shift the decimal places one place to right 6 / 12. Then, do the long division. You will get c = 0.5",
        "video_name": "a3acutLstF8",
        "timestamps": [
            77
        ],
        "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 1:00 Sal tells us about parallel lines. Are they related in any way",
        "A": "Yes. They never intersect. If you have 2 unparalleled lines, then somewhere they will intersect and they are therefore not parallel.",
        "video_name": "V0xounKGEXs",
        "timestamps": [
            60
        ],
        "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": "I still don't understand how does the computer program calculate the \"pseudo-sample variance\" @4:10 if we don't know mu's value. Can someone please explain?",
        "A": "In real life we generally don t know the value of \u00ce\u00bc. However, in a simulation, we are making up the data, and we do in fact know \u00ce\u00bc. What were doing is: 1. Set \u00ce\u00bc and create some data from a distribution with that mean. 2. Pretend that we don t know \u00ce\u00bc, and calculate the mean and standard deviation. 3. Remember that we know \u00ce\u00bc, and perform the calculations shown in the video.",
        "video_name": "F2mfEldxsPI",
        "timestamps": [
            250
        ],
        "3min_transcript": "In the vertical axis, using this denominator, dividing by n, we calculate two different variances. One variance, we use the sample mean. The other variance, we use the population mean. And this, in the vertical axis, we compare the difference between the mean calculated with the sample mean versus the mean calculated with the population mean. So for example, this point right over here, when we calculate our mean with our sample mean, which is the normal way we do it, it significantly underestimates what the mean would have been if somehow we knew what the population mean was and we could calculate it that way. And you get this really interesting shape. And it's something to think about. And he recommends some thinking about why or what kind of a shape this actually is. The other interesting thing is when you look at it this way, it's pretty clear this entire graph is sitting below the horizontal axis. So we're always, when we calculate our sample variance which we typically do, we're always getting a lower variance than when we use the population mean. Now this over here, when we divide by n minus 1, we're not always underestimating. Sometimes we are overestimating it. And when you take the mean of all of these variances, And here we're overestimating it a little bit more. And just to be clear what we're talking about in these three graphs, let me take a screen shot of it and explain it in a little bit more depth. So just to be clear, in this red graph right over here, let me do this. A color close to at least. So this orange, what this distance is for each of these samples, we're calculating the sample variance using, so let me, using the sample mean. And in this case, we are using n as our denominator. In this case right over here. or I guess you could call this some kind of pseudo sample variance, if we somehow knew the population mean. This isn't something that you see a lot in statistics. But it's a gauge of how much we are underestimating our sample variance given that we don't have the true population mean at our disposal. And so this is the distance. This is the distance we're calculating. And you see we're always underestimating. Here we overestimate a little bit. And we also underestimate. But when you take the mean, when you average them all out, it converges to the actual value. So here we're dividing by n minus 1, here we're dividing by n minus 2."
    },
    {
        "Q": "Shouldn't the unit vector i go into the other direction of the x-axis? (Minute 12:44) Otherwise we set up a left-handed coordinate system, didn't we?",
        "A": "Yes, Sal decided to change the sign of the x coordinate on minute 9:58, turning the system into a left-handed coordinate system.",
        "video_name": "bJ_09eoCmag",
        "timestamps": [
            764
        ],
        "3min_transcript": "And so our parameterization, and you know, just play with this triangle, and hopefully it'll make sense. I mean, if you say that this is our y-coordinate right here, you just do SOCATOA, cosine of t, CA is equal to adjacent, which is y, right, this is the angle right here, over the hypotenuse. Over b plus a cosine of s. Multiply both sides of the equation times this, and you get y of s of t is equal to cosine of t times this thing, right there. Let me copy and paste all of our takeaways. And we're done with our parameterization. We could leave it just like this, but if we want to represent it as a position vector-valued function, we can define it like this. So let's say our position vector-valued function is r. It's going to be a function of two parameters, s and t, and it's going to be equal to its x-value. Let me do that in the same color. So it's going to be, I'll do this part first. b plus a cosine of s times sine of t, and that's going to go in the x-direction, so we'll say that's times i. And this case, remember, the way I defined it, the positive x-direction is going to be here. So the i-unit vector will look like that. i will go in that direction, the way I've defined things. And then plus our y-value is going to be b plus a cosine of s times cosine of t in the y-unit vector direction. That's our j-unit vector. And then, finally, we'll throw in the z, which was actually the most straightforward. plus a sine of s times the k-unit vector, which is the unit vector in the z-direction. So times the k-unit vector. And so you give me, now, any s and t within this domain right here, and you put it into this position vector-valued function, it'll give you the exact position vector that specifies the appropriate point on the torus. So if you pick, let's just make sure we understand what we're doing. If you pick that point right there, where s and t are both equal to pi over 2, and you might even want to go through the exercise. Take pi over 2 in all of these."
    },
    {
        "Q": "at 0:03, some people find it offensive to have people ask their ages. and at 0:34, how can you throw ages into a bucket? They aren't physical objects.",
        "A": "It is metaphoric. He is putting the ages into categories. The categories are being compared to buckets. They are not actually buckets. Also, due to the scenario, probably no one was offended about the ages, and Sal never asked anyone. He said if you were to go and ask the people at the restaurant, this is what you would get.",
        "video_name": "gSEYtAjuZ-Y",
        "timestamps": [
            3,
            34
        ],
        "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": "At 2:46, how can one list the factors of \"a\" if it has already been declared prime?",
        "A": "It hasn t been declared prime. a/b is reduced to lowest terms. 27/32 is reduced to lowest terms. It s (3*3*3)/(2*2*2*2*2).",
        "video_name": "W-Nio466Ek4",
        "timestamps": [
            166
        ],
        "3min_transcript": "Well, this being rational says I can represent the square root of p as some fraction, as some ratio of two integers. And if I can represent anything as a ratio of two integers, I can keep dividing both the numerator and the denominator by the common factors until I eventually get to an irreducible fraction. So I'm assuming that's where we are right here. So this cannot be reduced. And this is important for our proof-- cannot be reduced, which is another way of saying that a and b are co-prime, which is another way of saying that a and b share no common factors other than 1. So let's see if we can manipulate this a little bit. Let's take the square of both sides. We get p is equal to-- well, a/b, the whole thing squared, that's the same thing as a squared over b squared. We can multiply both sides by b squared, and we get b squared times p is equal to a squared. Well, b is an integer, so b squared must be an integer. So an integer times p is equal to a squared. Well, that means that p must be a factor of a squared. Let me write this down. So a squared is a multiple of p. Now, what does that tell us about a? Does that tell us that a must also be a multiple of p? Well, to think about that, let's think about the prime factorization of a. Let's say that a can be-- and any number-- can be rewritten as a product of primes. Or any integer, I should say. So let's write this out as a product of primes right over here. So let's say that I have my first prime factor times my second prime factor, all the way to my nth prime factor. 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."
    },
    {
        "Q": "At 4:53, i don't understand why f2=p, because a^2=p.b^2 so p is a factor of a^2 so i think p=f2.f2 not p=f2. Can you please explain? Thank you",
        "A": "We know that P is a factor of a^2, because a^2 = P*b^2. P is a prime number, so it can t be the product f2*f2. P has to be one of the prime factors of a^2. These prime factors come in pairs, as they do in all perfect squares (as Sal shows in the video). Sal picked f2 as a possible example.",
        "video_name": "W-Nio466Ek4",
        "timestamps": [
            293
        ],
        "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": "At 3:20, why does Sal write -10, is this because the ball is going down to the ground according to physics or he just make typo?",
        "A": "It depends how you want to formulate your series. In the video Sal starts the series at n = 0. What is 20 * (1/2)^0? It is 20. However, the first bounce is only 10m, not 20m! Therefore, we need to deduct 10. In other words, after the ball has been dropped (i.e., the 0th bounce) it has travelled a distance of: -10 + 20(1/2)^0 = 10m",
        "video_name": "tqTJZEglrvc",
        "timestamps": [
            200
        ],
        "3min_transcript": "It's going to go up 5 meters, up half of 10 meters, and then down half of 10 meters. Let me put it this way. So each of these is going to be 10 meters. Actually, I don't have to write the units here. Let me take the units out of the way. Let me write that clear. So the first bounce, once again, it goes straight down 10 meters. Then on the next bounce it's going to go up 10 times 1/2. And then it's going to go down 10 times 1/2. Notice we just care about the total vertical distance. We don't care about the direction. So it's going to go up 10 times 1/2, up 5 meters, and then it's going to go down 5 meters. So it's going to travel a total vertical distance of 10 meters, Now what about on this jump, or on this bounce, I should say. Well here it's going to go half as far as it went there. So it's going to go 10 times 1/2 squared up, and then 10 times 1/2 squared down. And I think you see a pattern here. This looks an awful lot like a geometric series, an infinite geometric series. It's going to just keep on going like that forever and ever. So let's try to clean this up a little bit so it looks a little bit more like a traditional geometric series. So if we were to simplify this a little bit we could rewrite this as 10 plus 20. 20 times 1/2 to the first power, plus 10 1/2 times 1/2 squared plus 10 times 1/2 squared is going to be 20 times 1/2 So this would be a little bit clearer if this were a 20 right over here. But we could do that. We could write 10 as negative 10 plus 20, and then we have plus all of this stuff right over here. Let me just copy and paste that. So plus all of this right over here. And we can even write this first. We can even write this 20 right over here is 20 times 1/2 to the 0 power plus all of this. So now it very clearly looks like an infinite geometric series. We can write our entire sum, and maybe I'll write it up here since I don't want to lose the diagram. We could write it as negative 10. That's that negative 10 right over here."
    },
    {
        "Q": "At 1:08, when sal divides both sides of 24x/24x wouldn't you be left with 1x? so then you move the variables to one side making 1x-1x making it 0x. ahh i think i just answered my own question. 0xanything is 0. right?",
        "A": "he s not dividing though, he s subtracting",
        "video_name": "zKotuhQWIRg",
        "timestamps": [
            68
        ],
        "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": "in 5:31 whats that division symbol really sorry if it a silly doubt",
        "A": "At 3:46 he explains what that symbol is. It is a subtraction sign. A division sign is the opposite, like / .",
        "video_name": "2B4EBvVvf9w",
        "timestamps": [
            331
        ],
        "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": "(about 3:30 minutes into the video) Why is f(c) greater than or equal to f(x)?\nThe few seconds after that are also not quite clear to me. Help please!",
        "A": "Essentially, f(c) is a random point representing a relative max. f(x) represents the rest of the graph within the domain (c-h,c+h). So, if f(c), the point, is higher than the rest of the graph,f(x) in the selected interval, then it must be a relative max. Hopefully that helped.",
        "video_name": "Hoyv3-BMAGc",
        "timestamps": [
            210
        ],
        "3min_transcript": "It's larger than the other ones. Locally, it looks like a little bit of a maximum. And so that's why this value right over here would be called-- let's say this right over here c. This is c, so this is f of c-- we would call f of c is a relative maximum value. And we're saying relative because obviously the function takes on the other values that are larger than it. But for the x values near c, f of c is larger than all of those. Similarly-- I can never say that word. Similarly, if this point right over here is d, f of d looks like a relative minimum point or a relative minimum value. f of d is a relative minimum or a local minimum value. 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"
    },
    {
        "Q": "3:25 why bigger or equal to? if it's equal how can it be a maximum?",
        "A": "The maximum value a function gets is still the largest value whether the function reaches it one times, five times, or infinitely many times. For a local min that is not a global min, you would typically only have the equal scenario come up if there is a horizontal region in the function.",
        "video_name": "Hoyv3-BMAGc",
        "timestamps": [
            205
        ],
        "3min_transcript": "It's larger than the other ones. Locally, it looks like a little bit of a maximum. And so that's why this value right over here would be called-- let's say this right over here c. This is c, so this is f of c-- we would call f of c is a relative maximum value. And we're saying relative because obviously the function takes on the other values that are larger than it. But for the x values near c, f of c is larger than all of those. Similarly-- I can never say that word. Similarly, if this point right over here is d, f of d looks like a relative minimum point or a relative minimum value. f of d is a relative minimum or a local minimum value. 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"
    },
    {
        "Q": "At 4:06 he says that the denominator is zero but isn't the square root of 0 + 1 just equal one. would the graph be different?",
        "A": "Yes, but he was finding the limit of 0/(sqrt(0+1)), which is 0.",
        "video_name": "xks4cETlN58",
        "timestamps": [
            246
        ],
        "3min_transcript": "Now, for positive x'es the absolute value of x is just going to be x. This is going to be x divided by x, so this is just going to be 1. Similarly, right over here, we take the limit as we go to negative infinity, this is going to be the limit of x over the absolute value of x as x approaches negative infinity. Remember, the only reason I was able to make this statement is that f(x) and this thing right over here become very very similar, you can kind of say converge to each other, as x gets very very very large or x gets very very very very negative. Now, for negative values of x the absolute value of x is going to be positive, x is obviously going to be negative and we're just going to get negative 1. And so using this, we can actually try to graph our function. So let's say, that is my y axis, this is my x axis, and we see that we have 2 horizontal asymptotes. We have 1 horizontal asymptote at y=1, so let's say this right over here is y=1, let me draw that line as dotted line, we're going to approach this thing, and then we have another horizontal asymptote at y=-1. So that might be right over there, y=-1. And if we want to plot at least 1 point we can think about what does f(0) equal. So, f(0) is going to be equal to 0 over the square root of 0+1, or 0 squared plus 1. Well that's all just going to be equal to zero. So we have this point, right over here, and we know that as x approaches infinity, we're approaching this blue, horizontal asymptote, Let me do it a little bit differently. There you go. I'll clean this up. So it might look something like this. That's not the color I wanted to use. So it might look something like that. We get closer and closer to that asymptote as x gets larger and larger and then like this -- we get closer and closer to this asymptote as x approaches negative infinity. I'm not drawing it so well. So that right over there is y=f(x). And you can verify this by taking a calculator, trying to plot more points or using some type of graphing calculator or something. But anyway, I just wanted to tackle another situation we're approaching infinity and or negative infinity and we're trying to determine the horizontal asymptotes. And remember, the key is just to say what terms dominate"
    },
    {
        "Q": "At 9:23 pm, how do you factor problems when the coefficient does not share a common factor with the other numbers?\nFor example:\n2x2 - 3x + 2",
        "A": "Your example you gave here riverav is not factorable. To answer your question, however, you do the regular factoring polynomials method sal taught us to get your answer, or if you have exhausted all factors, the problem is not factorable.",
        "video_name": "GMoqg_s4Dl4",
        "timestamps": [
            563
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 2:54, why did you add 3k and -3k? I can't understand why and extra coefficient was added.",
        "A": "was the k squared? because that might be why...",
        "video_name": "GMoqg_s4Dl4",
        "timestamps": [
            174
        ],
        "3min_transcript": "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. which is the same thing as this over here. And then we have minus 18. And then all of that's being multiplied by 8. Now we're ready to group this thing. We can group these first two terms, they're both divisible by k. And then we can group-- let me put a positive sign-- let's group these second two terms. So then we have 8 times-- I'll write brackets here instead of drawing double parentheses. Brackets are really just parentheses that look a little bit more serious. Now let's factor out a k from this term right here. I'm going to do this in a different color. Let's factor out the k here. So this is k times k plus 3. And then we have plus. And then over here it looks like we could factor out a negative 6. So let's factor out-- I'm going to do this in a different color-- let's factor out a negative 6 So plus negative 6 times k plus 3."
    },
    {
        "Q": "Wait, at 0:37 Sal says that for the first flip there's 2 possibilities, same on the second and the third. So 2 and 2 and 2 should be 6, right?",
        "A": "The 2 s should be multiplied instead of added together. This should be seen in layers like this: There are 2 possibility for the first flip. For every possibility of the first flip, there are 2 possibility for the second flip. So there are a total of 2 * 2 possibilities for the first two flips. For every possibility of the first two flips, there are 2 possibility for the third flip. So there are a total of (2 * 2) * 2 possibilities for the three flips.",
        "video_name": "mkyZ45KQYi4",
        "timestamps": [
            37
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 6:33, can we also write:\n= 1 - ( 1 / ( 2 ^ 20 ) )\n?",
        "A": "No, not quite. There are only 10 2s, so it would be 1 - ( 1 / ( 2 ^ 10 ) ).",
        "video_name": "mkyZ45KQYi4",
        "timestamps": [
            393
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 2:03, how do you add 7 + 1/100?",
        "A": "He isn t. He s multiplying them. But if you want to add them, you d get 7 1/100 (seven and one hundredth).",
        "video_name": "he4kcTujy30",
        "timestamps": [
            123
        ],
        "3min_transcript": "Let's say I have the number 905.074. So how could I expand this out? And what does this actually represent? So let's just think about each of the place values here. The 9 right over here, this is in the hundreds place. This literally represents nine hundreds. So we could rewrite that 9 as nine hundreds. Let me write it two ways. We could write it as 900, which is the same thing as 9 times 100. Now, there's a 0. That's just going to represent zero tens. But zero tens is still just 0. So we don't have to really worry about that. It's not adding any value to our expression or to our number. Now we have this 5. This 5 is in the ones place. It literally represents five ones, or you could just say it represents 5. Now, if we wanted to write it as five ones, we could say well, that's going to be 5 times 1. 100 plus 5 times 1. And you might say hey, how do I know whether I should multiply or add first? Should I do this addition before I do this multiplication? And I'll always remind you, order of operations. In this scenario, you would do your multiplication before you do your addition. So you would multiply your 5 times 1 and your 9 times 100 before adding these two things together. But let's move on. You have another 0. This 0 is in the tenths place. This is telling us the number of tenths we're going to have. This is zero tenths, so it's really not adding much, or it's not adding anything. Now we go to the hundredths place. So this literally represents seven hundredths. So we could write this as 7/100, or 7 times 1/100. So we go to the thousandths place. And we have four thousandths. So that literally represents 4 over 1,000, or 4 times 1/1000. Notice this is coming from the hundreds place. You have zero tens, but I'll write the tens place there just so you see it. So it's zero tens, so I didn't even bother to write that down. Then you have your ones place. You have five ones. Then you have zero tenths. So I didn't write that down. Then you have seven hundredths and then you have four thousandths. We've written this out, really just understanding what this number represents."
    },
    {
        "Q": "At 3:44, how did we get cos^2(theta) \u00e2\u0080\u0093 sin^2(theta) to equal to 1 \u00e2\u0080\u0093 2 sin^2(theta) and to 2cos^2(theta) \u00e2\u0080\u0093 1. Thank you!",
        "A": "First, cos^2 (theta) = 1 - sin^2 (theta). This is from the Pythagorean identity sin^2 (x) + cos^2 (x) = 1. Therefore, cos^2 (theta) - sin^2 (theta) = 1 - sin^2 (theta) - sin^2 (theta) = 1 - 2 sin^2 (theta). Second, we flip the Pythagorean identity around and replace sin^2 (theta) with 1 - cos^2 (theta). We get 1 - 2(1 - cos^2 (theta)) = 1 - 2 + 2 cos^2 (theta) = 2 cos^2 (theta) - 1.",
        "video_name": "lXShNH1G6Pk",
        "timestamps": [
            224
        ],
        "3min_transcript": "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, is maybe cancel things out that are making us get this zero over zero, and if I could factor this into something that involved a 1 plus square root of 2 sine theta, then I'm going to be in business, and it looks like, it looks like this right over here, that can be factored as 1 plus square root of 2 sine theta times 1 minus square root of 2 sine theta, so let me use this. Cosine of 2 theta is the same thing, cosine of 2 theta is the same thing as 1 minus 2 sine squared theta, which is just a difference of squares. We can rewrite that as, this is a-squared minus b-squared, this is a plus b times a minus b, so I can just replace this with 1 plus square root of 2 sine theta times 1 minus square root of 2 sine theta,"
    },
    {
        "Q": "at 1:49, why don't we plug in 8 for x^2?",
        "A": "we only plug 8 into x^2 to get h (height) at that instant. To get dh/dt we have to differentiate x^2+h^2=10^2 to get relation between dx/dt, dh/dt, x(for that instant) and h(for that instant). Now we will solve dh/dt using simple algebra by putting known values into it.",
        "video_name": "kBVDSu7v8os",
        "timestamps": [
            109
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at ~13:20, sal is writing out 3rd/final term of the 3x3 determinant |A|...but shouldn't it be \"-f(ah-bg)\" vs \"-f(dh-eg)\" ?",
        "A": "Yes. Transcription error from a few minutes in. [I m writing words to pass the comment screener.]",
        "video_name": "32rdijPB-rA",
        "timestamps": [
            800
        ],
        "3min_transcript": "have a k aij. So this is the determinant of A prime. We could just take out this constant right here. It has no i's or j's in it. I has no j's in it in particular, so we can just take it out. So it's equal to k times the sum from j is equal to 1 to j is equal to n of minus 1 to the i plus j times aij. This is the coefficient. This is the submatrix for each of those coefficients, aij. That's a matrix right there, an n minus 1 by n minus 1 matrix. Then you immediately recognize-- I think you saw where this was going-- this right here is just the determinant of A. So we get the result that the determinant of A prime is equal to k times the determinant of A. n matrix, if you multiply only one row, not the whole matrix, only one row by some scalar multiple k, the resulting determinant will be your original determinant times k. Now I touched on this in the original video. What is the determinant of k times A? So now we're multiplying every row times k. Or another way to think about is you're multiplying n rows times k. So you're doing this n times. So if you multiply k times itself n times, what do you get? You get k to the n. So this is going to be equal to k to the n times the determinant of A. If you just do it once, you get k times the determinant of A. Now if you do a second row, you're going to get k times k times the determinant of A. times the determinant of A. The fourth row, k to the fourth times the If you do them all, all n rows, you're going to have k to the n times the determinant of A. Anyway, hopefully you found that interesting. I encourage you to experiment with these other ways. Try going down a column and seeing what happens."
    },
    {
        "Q": "at 0:49 isnt it the other way because it would be 1.5 in real life problem",
        "A": "Do you mean 1*5? 1.5 is a decimal number for 1 1/2. Any dot used for multiplication must be raised. Anyway... I think Sal is using the bar under each number to separate the digit from its place value. I don t believe he is using it for division if that is what you were thinking.",
        "video_name": "BItpeFXC4vA",
        "timestamps": [
            49
        ],
        "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": "in 4:24 why he drew a point outside the parabola? that point is undefined according to y=x^2, the function f(x)=x^2 but its f(x)=3 when x=2:- this is a wrong statement. And what is the meaning of limit? the limit going to tell u the defined function like in parabola x^2=4 when x=2. what is the meaning of limits?? :S",
        "A": "The function he used in this example is a piecewise function which forces the value of f(x) to equal 3 when x=2. A limit is used to describe the value of f(x) as it starts to approach a number. This becomes a key component in many aspects of calculus.",
        "video_name": "W0VWO4asgmk",
        "timestamps": [
            264
        ],
        "3min_transcript": "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. So this is our new f of x. So let me ask you a question. What is-- my pen still works-- what is the limit-- I used cursive this time-- what is the limit as x-- that's an x-- as x approaches 2 of f of x? That's an x. It says x approaches 2. I just-- I don't know. For some reason, my brain is working functionally. OK, so let me graph this now. So that's an equally neat-looking graph as the one I just drew. So now it's almost the same as this curve, except something interesting happens at x equals 2. So it's just like this. It's like an x squared curve like that. But at x equals 2 and f of x equals 4, we We draw a hole because it's not defined at x equals 2. This is x equals 2. This is 2. This is 4. This is the f of x axis, of course. And when x is equal to 2-- let's say this is 3. When x is equal to 2, f of x is equal to 3. This is actually right below this. I should-- it doesn't look completely right below it, but I think you got to get the picture. See, this graph is x squared. It's exactly x squared until we get to x equals 2. At x equals 2, We have a grap-- No, not a grap. We have a gap in the graph, which maybe could be called a grap. We have a gap in the graph, and then we keep-- and then after x equals 2, we keep moving on. And that gap, and that gap is defined right here, what happens when x equals 2? Well, then f of x is equal to 3. So this graph kind of goes-- it's just like x squared, but"
    },
    {
        "Q": "Could someone explain how Sal solved 21 times 2/3 in his head so quickly? It happened around 2:57 in the video. I would like to understand his method.",
        "A": "Its called cross products or cross multiplication.",
        "video_name": "XOIhNVeLfWs",
        "timestamps": [
            177
        ],
        "3min_transcript": "that do not share a common factor other than one. That just means that A, B, and C in the standard form they want need to be integers. And they want them to not have any common factors. So if we got to the point of say 4x plus 2y is equal to 10, well this number, and this number, and this number are all divisible by two. They all have the common factor of two. So we would want to simplify it more. Divide them all by two and then you would get two so you divide this by two. You get 2x, divide this by two you'll get plus y is equal to 5. So this is the form that they're asking for and probably because it's just easier for the site to know that this is the right answer. Because there's obviously a bunch of forms in this way. So let's see if we can do that. Let's see if we can write it in standard form. So the first thing I would wanna do is, well there's a bunch of ways that you could approach it. well let's get rid of all of these fractions. And the best way to get rid of the fractions is to multiple by three and to multiply by seven. If you multiply by three you get rid of this fraction. If you multiply by seven you get rid of this fraction. So if you multiply three and you multiply by seven. Let me just rewrite it over here. There's actually also a couple of ways that we can do this. So if you multiply. So one way to do it. So we start with Y is equal to 2/3x plus 4/7. So if I multiply this side by three and I multiply by seven, I have to do that to this side as well. So this is going to be multiplied by three and multiplied by seven. So the left hand side becomes 21y. 21y. Three times seven of course is 21, we just figured that out. We would distribute the 21. 21 times 2/3, well let's see. 21 divided by three is seven, times two is 14. 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"
    },
    {
        "Q": "at 0:12 what is o-blood",
        "A": "There are different types of bloods, that is just one of them",
        "video_name": "qrVvpYt3Vl0",
        "timestamps": [
            12
        ],
        "3min_transcript": "According to the pictograph below, how many survey respondents have type O positive blood? How many have O negative blood? So a pictograph is really just a way of representing data with pictures that are somehow related to the data. So in this case, they gave us little pictures of, I'm assuming, blood drops right over here. And then they tell us that each blood drop in this pictograph represents 8 people. So you could kind of view that as a scale of these graphs. Each of these say 8 people. So, for example, if you say how many people have A positive? It would be 1, 2, 3, 4, 5, 6, 7 blood drops. But each of those blood drops represent 8 people, so it would be 56 people have type A positive. But let's answer the actual question that they're asking us. How many survey respondents have type O positive? So this is O, and then we care about O positive. So we have 1 blood drop, 2, 3. I'm going to do this in a new, different color. So we have 8 drops. I'll put those in quotes because they're pictures of drops. And then the scale is 8 people. Let me write it this way. Times 8 people per drop. And so 8 times 8-- and actually even the drops, you could view them as canceling out if you view them as units, so drops, drops. 8 times 8 is equal to 64 people. So they could have written literally the number 64 right over here. 64 people have type O positive blood. Now let's think about the O negative case, O negative blood. Well, this is O. And then within the blood group O, this is O negative. And we have 1 drops, 2 drops. So we have 2 drops times 8 people per drop. So 8 and then 16, or 2 times 8 is equal to 16 people. So 16 of the survey respondents, 16 have type O negative. 64 have type O positive."
    },
    {
        "Q": "Instead of using his \"aside\" with the triangle beginning at 1:49, I did something similar but using x/3 instead of 'S'. In other words, I said:\nUsing the Pythagorean Theorem: ((x/3)/2)^2+h^2 = (x/3)^2\nh^2 = (x/3)^2-(x/6)^2\nh = x/3-x/6\nh=x/6\n\nObviously this differs from Sal's answer but I can't figure out why?",
        "A": "Your mistake was when you square rooted, \u00e2\u0088\u009a(a\u00c2\u00b2+b\u00c2\u00b2) is NOT \u00e2\u0088\u009aa\u00c2\u00b2 + \u00e2\u0088\u009ab\u00c2\u00b2 and is not a+b Here is the correct way to do it: h\u00c2\u00b2 = (\u00e2\u0085\u0093x)\u00c2\u00b2- (\u00e2\u0085\u0099x)\u00c2\u00b2 h\u00c2\u00b2 = \u00c2\u00b9\u00e2\u0081\u0084\u00e2\u0082\u0089 x\u00c2\u00b2 - \u00c2\u00b9\u00e2\u0081\u0084\u00e2\u0082\u0083\u00e2\u0082\u0086 x\u00c2\u00b2 h\u00c2\u00b2 = \u00c2\u00b9\u00e2\u0081\u0084\u00e2\u0082\u0081\u00e2\u0082\u0082 x\u00c2\u00b2 h = x / \u00e2\u0088\u009a12 h = x / (2\u00e2\u0088\u009a3) h = \u00e2\u0085\u0099 x\u00e2\u0088\u009a3",
        "video_name": "IFU7Go6Qg6E",
        "timestamps": [
            109
        ],
        "3min_transcript": "Let's say that I have a 100 meter long wire. So that is my wire right over there. And it is 100 meters. And I'm going to make a cut someplace on this wire. And so let's say I make the cut right over there. With the left section of wire-- I'm going to obviously cut it in two-- with the left section, I'm going to construct an equilateral triangle. And with the right section, I'm going to construct a square. And my question for you and for me is, where do we make this cut in order to minimize the combined areas of this triangle and this square? Well, let's figure out. Let's define a variable that we're trying to minimize, or that we're trying to optimize with respect to. So let's say that the variable x is the number of meters that we decide to cut from the left. So if we did that, then this length for the triangle would be, well, if we use x up for the left hand side, we're going to have 100 minus x for the right hand side. And so what would the dimensions of the triangle and the square Well, the triangle sides are going to be x over 3, x over 3, and x over 3 as an equilateral triangle. And the square is going to be 100 minus x over 4 by 100 minus x over 4. Now it's easy to figure out an expression for the area of the square in terms of x. But let's think about what the area of an equilateral triangle might be as a function of the length of its sides. So let me do a little bit of an aside right over here. So let's say we have an equilateral triangle. Just like that. And its sides are length s, s, and s. is 1/2 times the base times the height. So in this case, the height we could consider to be altitude, if we were to drop an altitude just like this. This length right over here, this is the height. And this would be perpendicular, just like that. So our area is going to be equal to one half times our base is s. 1/2 times s times whatever our height is, times our height. Now how can we express h as a function of s? Well, to do that we just have to remind ourselves that what we've drawn over here is a right triangle. It's the left half of this equilateral triangle. And we know what this bottom side of this right triangle is. This altitude splits this side exactly into two."
    },
    {
        "Q": "At 4:38 Sal says the word reciprocal.\nWhat is a reciprocal?",
        "A": "The easiest way to find the reciprocal of a number is to think what should I multiply this number with to make it equal 1? That s basically all there is to reciprocals.",
        "video_name": "a_Wi-6SRBTc",
        "timestamps": [
            278
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 1:18, Sal mentions \"interval notation\". I watched all the videos in order in \"Creating and solving linear equations\" but did not see anything about interval notation. Is there a video for this?",
        "A": "At 1:18 Sal is introducing interval notation in this video.",
        "video_name": "xOxvyeSl0uA",
        "timestamps": [
            78
        ],
        "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 5:00, how does 8x-5 turn into 8x-20x??",
        "A": "8x - 5(4x + 1) Is the right side of the inequality. So, to simplify the right hand side of the inequality, Sal distributes the -5 to (4x + 1), which is -5(4x) + -5(1) = -20x -5. So, it is 8x - 20x - 5. Hope this helped!",
        "video_name": "xOxvyeSl0uA",
        "timestamps": [
            300
        ],
        "3min_transcript": "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 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."
    },
    {
        "Q": "at 1:19, if x can be from negative infinity to -1, but not including negative 1, so why does he include -1 in the partenthases",
        "A": "The symbols \u00e2\u0080\u0098(\u00e2\u0080\u0098 and \u00e2\u0080\u0098)\u00e2\u0080\u0099 indicate exclusive values, whereas the symbols \u00e2\u0080\u0098[\u00e2\u0080\u0098 and \u00e2\u0080\u0098]\u00e2\u0080\u0099 indicate inclusive values. In the example x<-1 which is represented as (-\u00e2\u0088\u009e,-1) because -1 is not inclusive. If the example had been x\u00e2\u0089\u00a4-1 then -1 would have been included and could be represented as (-\u00e2\u0088\u009e,-1]",
        "video_name": "xOxvyeSl0uA",
        "timestamps": [
            79
        ],
        "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 4:28, what is a basis?",
        "A": "At 4:28, A basis is a System which consist of the MINIMAL amount of VECTORS which are needed to define a room (space, coordinate system). All VECTORS in from a BASIS must be liear dependant from eachother.",
        "video_name": "C2PC9185gIw",
        "timestamps": [
            268
        ],
        "3min_transcript": "We're just assuming that A has at least n linearly independent eigenvectors. In general, you could take scaled up versions of these and they'll also be eigenvectors. Let's see, so the transformation of vn is going to be equal to A times vn. And because these are all eigenvectors, A times vn is just going to be lambda n, some eigenvalue times the vector, vn. Now, what are these also equal to? Well, this is equal to, and this is probably going to be unbelievably obvious to you, but this is the same thing as lambda 1 times vn plus 0 times v2 plus all the way to 0 times vn. And this right here is going to be 0 times v1 plus lambda 2 times v2 plus all the way, 0 times all of the other vectors vn. And then this guy down here, this is going to be 0 times v1 these eigenvectors, but lambda n times vn. This is almost stunningly obvious, right? I just rewrote this as this plus a bunch of zero vectors. But the reason why I wrote that is, because in a second, we're going to take this as a basis and we're going to find coordinates with respect to that basis, and so this guy's coordinates will be lambda 1, 0, 0, because that's the coefficients on our basis vectors. So let's do that. So let's say that we define this as some basis. So B is equal to the set of-- actually, I don't even have to write it that way. Let's say I say that B, I have some basis B, that's equal to that. What I want to show you is that when I do a change of basis-- we've seen this before-- in my standard coordinates or in coordinates with respect to the standard basis, you give me some vector in Rn, I'm going to multiply it times A, and you're going to have the It's also going to be in Rn. Now, we know we can do a change of basis. And in a change of basis, if you want to go that way, you multiply by C inverse, which is-- remember, the change of basis matrix C, if you want to go in this direction, you multiply by C. The change of basis matrix is just a matrix with all of these vectors as columns. It's very easy to construct. But if you change your basis from x to our new basis, you multiply it by the inverse of that. We've seen that multiple times. If they're all orthonormal, then this is the same thing as We can't assume that, though. And so this is going to be x in our new basis. And if we want to find some transformation, if we want to find the transformation matrix for T with respect to our new basis, it's going to be some matrix D. And if you multiply D times x, you're going to get this guy, but you're going to get the B representation of that guy. The transformation of the vector x is B representation."
    },
    {
        "Q": "At 4:15 is there an easier way to find that number that if given the certain exponent (5) makes it into the 32? (how did he know it was a 2?)",
        "A": "With enough practices you ll be able to recognize certain numbers or quickly calculate in your head. Also he was using it as an example so he would pick something he already knows and easy for us to recognize.",
        "video_name": "lZfXc4nHooo",
        "timestamps": [
            255
        ],
        "3min_transcript": "just so you make sure you get what's going on. And I encourage you to pause it as much as necessary and try to figure it out yourself. So based on what I just told you, what do you think 9 to the 1/2 power is going to be? Well, that's just the square root of 9. The principal root of 9, that's just going to be equal to 3. And likewise, we could've also said that 3 squared is, or let me write it this way, that 9 is equal to 3 squared. These are both true statements. Let's do one more like this. What is 25 to the 1/2 going to be? Well, this is just going to be 5. 5 times 5 is 25. Or you could say, 25 is equal to 5 squared. Now, let's think about what happens when you take something to the 1/3 power. So let's imagine taking 8 to the 1/3 power. So the definition here is that taking something to the 1/3 power is the same thing And the cube root is just saying, well what number, if I had three of that number, and I multiply them, that I'm going to get 8. So something, times something, times something, is 8. Well, we already know that 8 is equal to 2 to the third power. So the cube root of 8, or 8 to the 1/3, is just going to be equal to 2. This says hey, give me the number that if I say that number, times that number, times that number, I'm going to get 8. Well, that number is 2 because 2 to the third power is 8. Do a few more examples of that. What is 64 to the 1/3 power? Well, we already know that 4 times 4 times 4 is 64. So this is going to be 4. And we already wrote over here that 64 is the same thing as 4 to the third. I think you're starting to see a little bit of a pattern here, a little bit of symmetry here. And we can extend this idea to arbitrary rational exponents. let me think of a good number here-- so let's say I have 32. I have the number 32, and I raise it to the 1/5 power. So this says hey, give me the number that if I were to multiply that number, or I were to repeatedly multiply that number five times, what is that, I would get 32. Well, 32 is the same thing as 2 times 2 times 2 times 2 times 2. So 2 is that number, that if I were to multiply it five times, then I'm going to get 32. So this right over here is 2, or another way of saying this kind of same statement about the world is that 32 is equal to 2 to the fifth power."
    },
    {
        "Q": "Isn't a line segment that has no length a point? Refering to 4:50",
        "A": "Yes. A point really has no size, and since lines have no width, if a line also had no length, it would not be a line, it would be a point.",
        "video_name": "Oc8sWN_jNF4",
        "timestamps": [
            290
        ],
        "3min_transcript": "If it were a regular line, not an infinitely spiked one, scaling up by three would make it three times as much drawing, as expected. But if that spiky line were supposed to represent an infinitely spiked magical fortress city of dragon dungeon doom, by scaling it up in this way, you'd be losing details, making these long lines that should have had spiky bumps in them. Theoretically, no matter how much you scale up the city or no matter how finely you look at it, you'll never get any flat sections. This whole thing scaled down is the same as this section, which is the same as this section, which is the same as this. Three times as big is four times as much stuff. Not three, like if it were a normal 1D line, and certainly not nine, like that 2D area on the inside. Somehow, the infinity fractal-ness of the thing makes it behave differently from all 1D things and all 2D things. You convince yourself that all 1D things got twice as big when you make them twice as big, because you could think 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"
    },
    {
        "Q": "At 5:00, Vi mentions a line that has no length. Wouldn't that be resembling a point? A point has no length whatsoever, or width, hence only trapped in 1st dimension. So does a potential line with no length. So... that means it's a case of a=b and b=c so a=c, right?",
        "A": "A point is zero-D it has no length",
        "video_name": "Oc8sWN_jNF4",
        "timestamps": [
            300
        ],
        "3min_transcript": "If it were a regular line, not an infinitely spiked one, scaling up by three would make it three times as much drawing, as expected. But if that spiky line were supposed to represent an infinitely spiked magical fortress city of dragon dungeon doom, by scaling it up in this way, you'd be losing details, making these long lines that should have had spiky bumps in them. Theoretically, no matter how much you scale up the city or no matter how finely you look at it, you'll never get any flat sections. This whole thing scaled down is the same as this section, which is the same as this section, which is the same as this. Three times as big is four times as much stuff. Not three, like if it were a normal 1D line, and certainly not nine, like that 2D area on the inside. Somehow, the infinity fractal-ness of the thing makes it behave differently from all 1D things and all 2D things. You convince yourself that all 1D things got twice as big when you make them twice as big, because you could think 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"
    },
    {
        "Q": "At 2:29 why does Vi draw a man with 3 \u00f0\u009f\u0091\u0080",
        "A": "That man lives in 4 dimensional space, meaning that a person needs three eyes to see properly.",
        "video_name": "Oc8sWN_jNF4",
        "timestamps": [
            149
        ],
        "3min_transcript": "So you're back in math class again. It just-- it never stops. Day after day you find yourself trapped behind this desk with only your notebook for company, that pale mirror that reflects your thoughts on a comforting simulacrum of shared ideas. You're wasting another irreplaceable hour of your finite life not even pretending to listen to your teacher talk about logarithms. Or, at least, you think it's logarithms that she's trying to teach you. You haven't exactly been paying attention as you sit there casually, disposing of this only this moment you'll ever have, and-- oops, there goes another one. But either way, it's definitely logarithms in particular that you're not even pretending to learn, as opposed to, say, calculus, which your teacher wouldn't even expect you to be pretending to pay attention to. So instead, you're amusing yourself by doodling. Two days ago, you discovered the infinitely self-similar beast that is the fractal. And yesterday, you discovered that when you scale a two-dimensional thing up by two, it grows by a factor of four. And when you scale it by any amount, the area grows by that amount squared, unlike 1D where it's just that amount, and in 3D it's that amount cubed. And in n dimensions, it's that amount to the n. So when you put it like that, you're actually making pretty good use of your limited time And by limited time on this earth, I mean that we're all going to become immortal space robots. Anyway, you're continuing your plans for a fractal city of infinite dragon dungeons, triangles upon triangles upon triangles, each next set scaled down by a factor of 3. That's 1/9 the area. But there's four times as many. And the next set has four times as many as the last, but 1/9 times the size of the last. So the total weight of steel is 1 plus 4/9 plus 4 squared over 9 squared plus 4 to the 3 over 9 to the 3, dot, dot, dot, plus 40 to the n over 9 to the n. And maybe could learn how to add up an infinite series of numbers if your teacher would ever get past logarithms. But at least you know how to create the perfect fractal city, which is good, because understanding scale factors and city planning seems like the kind of thing that might come in handy if you want to help the species on our journey towards becoming immortal space robots. Really, the only thing that could make the city better Or how about three times as big? So you can keep this part of the design and just draw the next iteration. Three times the scale in two dimensions means technically you'll need nine times as much steel. Though, as far as drawing these plans go, it's not nine times as difficult, but only four, since the hard part is the outside spiky part. And that's just copied four times, in order to scale up by three. Weird thing number one. Scaling up by three makes nine times as much steel. But it's the same thing copied four times, plus filling in this middle triangle. So it's also four times the steel plus 9. So if this mystery sum of an infinite series, total weight of steel were x, and 9x equals 4x plus 9, 5x equals 9, x equals 9/5 exactly. Take that, infinity! OK. Now weird thing number two. Look at just the edge, the actual Koch Curve"
    },
    {
        "Q": "At 5:17, how did you get those numbers if young are adding 2, then, then 6.",
        "A": "Each number is 2 more than the previous number. If x is the first number, then the 2nd number is x+2. The third number is 2 more than the previous number, which is the 2nd number, which also is x+2. So the 3rd number is (x+2)+2 = x+3. The 4th number is the 3rd + 2, which is (x+4)+2 = x+6",
        "video_name": "8CJ6Qdcoxsc",
        "timestamps": [
            317
        ],
        "3min_transcript": "So we can rewrite those literally as 4x. And then we have 2 plus 4, which is 6, plus another 6 is 12. 4x plus 12 is equal to 136. So to solve for x, a good starting point would be to just to isolate the x terms on one side of the equation or try to get rid of this 12. Well to get rid of that 12, we'd want to subtract 12 from the left-hand side. But we can't just do it from the left-hand side. Then this equality wouldn't hold anymore. If these two things were equal before subtracting the 12, well then if we want to keep them equal, if we want the left and the right to stay equal, we've got to subtract 12 from both sides. So subtracting 12 from both sides gives us, well on the left-hand side, we're just left with 4x. And on the right-hand side, we are left with 136 minus 12 is 124. Yeah, 124. Well, we just divide both sides by 4 to solve for x. And we get-- do that in the same, original color-- x is equal to 124 divided by 4. 100 divided by 4 is 25. 24 divided by 4 is 6. 25 plus 6 is 31. And if you don't feel like doing that in your head, you could also, of course, do traditional long division. Goes into 124-- 4 doesn't go into 1. 4 goes into 12 three times. 3 times 4 is 12. You subtract, bring down the next 4. 4 goes into 4 one time. You get no remainder. So x is equal to 31. So x is the smallest of the four integers. So this right over here, x is 31. x plus 2 is going to be 33. And x plus 6 is going to be 37. So our four consecutive odd integers are 31, 33, 35, and 37."
    },
    {
        "Q": "at 4:45 why is the second derivative negative? i thought if it is going up, than it is positive both below and above the x axis?",
        "A": "The second derivative is negative wherever the first derivative is decreasing.",
        "video_name": "LcEqOzNov4E",
        "timestamps": [
            285
        ],
        "3min_transcript": ""
    },
    {
        "Q": "Question about 8:37 onwards: so, what if the critical points are undefined for f(x)? i.e.: the \"maximum\" point of the downward concave area doesn't have a value, is discontinuous -- is this point still considered a maximum, even though there's no value for the point, or do maximum and minimum points only apply if the interval is continuous?",
        "A": "We can check the function, however, discontinuities in the derivative sometimes result in discontinuities in the original function (such as the derivatives of tanx, 1/x, and 1/x\u00e2\u0081\u00bf).",
        "video_name": "LcEqOzNov4E",
        "timestamps": [
            517
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 7:36 you said slope is increasing. Does that mean slope is becoming more positive?",
        "A": "Increasing slope can mean one of two things: more positive or less negative. Whichever situation you have, increasing slope always implies concave up.",
        "video_name": "LcEqOzNov4E",
        "timestamps": [
            456
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 5:40, why is it true that if the first derivative/slope of the line is increasing, then the second derivative must be positive? (and vice versa for decreasing/negative)",
        "A": "Consider that when the original function f(x) is increasing/decreasing, then the first derivative f (x) is positive/negative. Similarly, when f (x) is increasing/decreasing, then its derivative f (x) is positive/negative. However, the main idea of this video is to help illustrate that when the second derivative is positive/negative, then the original function is concave up/down. The second derivative is useful for classifying extrema and for identifying concavity of a function.",
        "video_name": "LcEqOzNov4E",
        "timestamps": [
            340
        ],
        "3min_transcript": ""
    },
    {
        "Q": "So there is the (1:00):\nf(g(x)) = sqrt( g(x)^2 - 1 )\nI was recently watching video series on complex numbers and I am just curious why can't we rewrite that thing as\ng(x) * sqrt(-1)\nand so\ng(x)*sqrt(i^2)\nto\ng(x)*i hence state the solution in terms of complex numbers? I know it is a boo-boo *but why?*",
        "A": "Sorry, your method doesn t work. You changed subtraction: sqrt( g(x)^2 - 1 ) into multiplication: g(x) * sqrt(-1). This is not a valid operation. Also, there is no property of square roots that lets you split the square root by terms (items being added/subtracted). You can only split square roots by factors (items being multiplied/divided). Hope this helps.",
        "video_name": "_b-2rZpX5z4",
        "timestamps": [
            60
        ],
        "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": "I don't understand what he says at 6:09, please explain!",
        "A": "The length of one cycle of the standard cosine function is 2pi. So we need to ask ourselves: How does 2pi compare to the length of one cycle in the problem? That s why we set up the ratio of 2pi to 365, or written in fractional terms, 2pi/365.",
        "video_name": "mVlCXkht6hg",
        "timestamps": [
            369
        ],
        "3min_transcript": "either actual degrees or radians, which trigonometric function starts at your maximum point? Well cosine of zero is one. The cosine starts at your maximum point. Sine of zero is zero, so I'm going to use cosine here. I'm going to use a cosine function. So, temperature as a function of days. There's going to be some amplitude times our cosine function and we're going to have some argument to our cosine function and then I'm probably going to have to shift it. So let's think about how we would do that. Well, what's the mid line here? The mid line is the halfway point between our high and our low. So our midpoint, if we were to visualize it, looks just like so. That is our mid line right over there. And what value is this? Well what's the average of 29 and 14? 29 plus 14 is 43 divided by two is 21.5 degrees Celsius. shifted up our function by that amount. If we just had a regular cosine function our mid line would be at zero, but now we're at 21.5 degrees Celsius. I'll just write plus 21.5, that's how much we've shifted it up. Now, what's the amplitude? Well our amplitude is how much we diverge from the mid line. Over here we're 7.5 above the mid line so that's plus 7.5. Here we're 7.5 below the mid line, so minus 7.5. So our amplitude is 7.5, the maximum amount we go away from the mid line is 7.5. So that's our amplitude. And now let's think about our argument to the cosine function right over here. It's going to be a function of the days. And what do we want? When 365 days have gone by, we want this entire argument to be two pi. whole thing to evaluate to two pi. We could put two pi over 365 in here. You might remember your formulas, I always forget them that's why I always try to reason through them again. The formulas, you want two pi divided by your period and all the rest, but I just like to think, \"Okay, look. \"After one period, which is 365 days, I want the whole \"argument over here to be two pi. \"I want to go around the unit circle once and so if this \"is two pi over 365, when you multiply it by 365 \"your argument here is going to be two pi.\" Just like that we've done the first part of this question. We have modeled the average high temperature in Santiago as a function of days after January seventh. In the next video we'll answer this second question. I encourage you to do it ahead of time before watching that next video and I'll give you one clue."
    },
    {
        "Q": "At 2:28 why does -24x become -25x when you subtract x",
        "A": "Because a negative minus a positive is basically adding a negative to a negative.",
        "video_name": "711pdW8TbbY",
        "timestamps": [
            148
        ],
        "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 5:40 Sal says that the square root of 2.25 could be 1.5 or -1.5. At 5:57, Sal says that 1.5 \u00e2\u0089\u00a0 -1.5. However, at 5:52, couldn't he have as easily used -1.5 as the square root of 2.25 instead of 1.5? If he had used -1.5, the solution 2.25 would have been valid. Is it because you can only principal square roots? If so, why?",
        "A": "No. all square roots, even non-perfect squares will always have a positive or negative value, even square root of is +2 or -2. Why? If you might have forgotten; two negatives, when multiplied together, will always become positive. However, cube roots only have a positive root if the value inside the cube root sign is positive and negative if negative. And yes, 1.5 is not equal to -1.5 but their square is equal.",
        "video_name": "711pdW8TbbY",
        "timestamps": [
            340,
            357,
            352
        ],
        "3min_transcript": "8 goes into 18 two times, remainder 2, so this is equal to 2 and 2/8 or 2 and 1/4 or 2.25, just like that. Now, I'm going to show you an interesting phenomena that occurs. And maybe you might want to pause it after I show you this conundrum, although I'm going to tell you why this conundrum pops up. Let's try out to see if our solutions actually work. So let's try x is equal to 4. If x is equal to 4 works, we get the principal root of 4 should be equal to 2 times 4 minus 6. The principal root of 4 is positive 2. Positive 2 should be equal to 2 times 4, which is 8 minus 6, which it is. This is true. So 4 works. Now, let's try to do the same with 2.25. According to this, we should be able to take the square root, the principal root of 2.2-- let me make my radical a The principal root of 2.25 should be equal to 2 times 2.25 minus 6. Now, you may or may not be able to do this in your head. You might know that the square root of 225 is 15. And then from that, you might be able to figure out the square root of 2.25 is 1.5. Let me just use the calculator to verify that for you. So 2.25, take the square root. It's 1.5. The principal root is 1.5. Another square root is negative 1.5. So it's 1.5. And then, according to this, this should be equal to 2 times 2.25 is 4.5 minus 6. Now, is this true? This is telling us that 1.5 is equal to negative 1.5. This is not true. 2.5 did not work for this radical equation. So 2.25 is an extraneous solution. Now, here's the conundrum: Why did we get 2.25 as an answer? It looks like we did very valid things the whole way down, and we got a quadratic, and we got 2.25. And there's a hint here. When we substitute 2.25, we get 1.5 is equal to negative 1.5. So there's something here, something we did gave us this solution that doesn't quite apply over here. And I'll give you another hint. Let's try it at this step. If you look at this step, you're going to see that both solutions actually work. So you could try it out if you like. Actually, try it out on your own time. Put in 2.25 for x here. You're going to see that it works. Put in 4 for x here and you see that they both work here. So they're both valid solutions to that. So something happened when we squared that made the equation"
    },
    {
        "Q": "At 1:34, you did some factoring in your head...hard to follow to get the 24x. Could you expound upon that? I know - watch the factoring videos - but that one lost me. It wasn't quite clear WHY you multiply the 2x and the 6 to get the 24.",
        "A": "Because the square factor form is as following: (ax-b)^2 ax-b x ax-b _________ ax^2 -abx -bax +b^2 Focus in the part (-abx -bax), since multiplication scalar is commutative, they are twice themselves. Rewrite them in order, (-abx -abx), which is equal to 2(-abx). If a = 2 and b = 6, representing them in the expression 2(-2*6x), 2(-12x) that is equal to -24x.",
        "video_name": "711pdW8TbbY",
        "timestamps": [
            94
        ],
        "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": "Ok, I know what PEMDAS is, what do you do if you get, say, 3*6:9? M and D are on the same level!",
        "A": "work from left to right!",
        "video_name": "0uCslW40VHQ",
        "timestamps": [
            369
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 0:43, the speaker explains PEMDAS . Is there another way to solve expressions?",
        "A": "Not in my knowledge. to me, PEMDAS is the best way to do order of operations.",
        "video_name": "0uCslW40VHQ",
        "timestamps": [
            43
        ],
        "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": "Around 4:50 when Sal is subtracting \"all of this business\", i.e. -e^xcosx+Se^x cosx dx, why does the plus sign become a minus sign? Its e^x sinx - -e^xcosx... so that becomes plus. But why does the plus sign before the antiderivative become a minus sign??",
        "A": "When he was taking e^x(sin(x)) - \u00e2\u0088\u00ab e^x(sin(x))dx, the minus sign acts as a negative sign so it would end up something like this in his thinking: e^x(sin(x)) + [-( \u00e2\u0088\u00ab e^x(sin(x))dx)]. You would take the integral of e^x(sin(x))dx to get -e^xcos(x) + \u00e2\u0088\u00ab (e^xcos(x))dx and factor the negative into the whole equation to get: e^x(sin(x)) + e^xcos(x) - \u00e2\u0088\u00ab (e^xcos(x))dx",
        "video_name": "LJqNdG6Y2cM",
        "timestamps": [
            290
        ],
        "3min_transcript": "prime of xg of x. F prime of x is e to x. And then g of x is negative cosine of x. So I'll put the cosine of x right over here, and then the negative, we can take it out of the integral sign. And so we're subtracting a negative. That becomes a positive. And of course, we have our dx right over there. And you might say, Sal, we're not making any progress. This thing right over here, we now expressed in terms of an integral that was our original integral. We've come back full circle. But let's try to do something interesting. Let's substitute back this-- all right, let me write it this way. Let's substitute back this thing up here. Let's substitute this for this in our original equation. And let's see if we got anything interesting. So what we'll get is our original integral, on the left hand side here. The indefinite integral or the antiderivative of e to the x cosine of x dx is equal to e to the x sine of x, minus all of this business. So let's just subtract all of this business. We're subtracting all of this. So if you subtract negative e to the x cosine of x, it's going to be positive. It's going to be positive e to the x, cosine of x. And then remember, we're subtracting all of this. So then we're going to subtract. So then we have minus the antiderivative of e to the x, Now this is interesting. Just remember all we did is, we took this part right over here. We said, we used integration by parts to figure out that it's the same thing as this. So we substituted this back in. When you subtracted it. When you subtracted this from this, we got this business right over here. Now what's interesting here is we have essentially an equation where we have our expression, our original expression, twice. We could even assign this to a variable and essentially solve for that variable. So why don't we just add this thing to both sides of the equation? Let's just add the integral of e to the x cosine of x dx to both sides. e to the x, cosine of x, dx. And what do you get? Well, on the left hand side, you have two times our original integral. e to the x, cosine of x, dx is equal to all of this business."
    },
    {
        "Q": "Around 2:00 I don't understand what he is saying about the point in the middle being an equidistant away from each angle. Please explain.",
        "A": "He was referencing a video about perpendicular bisectors. Basically, Sal is saying that if you have one point that is equidistant from two other points on a line, but said point ISNT on that line, then it lies on the perpendicular bisector of that line. so since the distance from I is the same distance from AB, and I ISNT on AB, then I is on the perpendicular bisector of AB. Sorry if Im not much help, watch the video!",
        "video_name": "21vbBiCVijE",
        "timestamps": [
            120
        ],
        "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": "at 3:27,a part of the equation does not show up. why is that?",
        "A": "you know how sometimes he slides the screen down I think its cause his screen is bigger than whatever were watching it on.",
        "video_name": "CLrImGKeuEI",
        "timestamps": [
            207
        ],
        "3min_transcript": "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 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"
    },
    {
        "Q": "Starting at about 3:35, Sal write the equation for L(x). Where does he get f(4) from? Also, is L(x) just the tangent line to f(x) at x = 4?",
        "A": "Yes, L(x) is the tangent line to f(x) at x=4. Sal uses f(4) because he is writing the equation of the line tangent to f(x) at the point (4, f(4)) (if the x-value is 4, then the y-value is f(4), right?). You might be more used to always writing y = mx +b, where you put in the slope (which in this case would be f (4)), and then plug in a point to find b, but Sal s method is faster and easier. Basically, he is using the point-slope form instead of the slope-intercept form.",
        "video_name": "u7dhn-hBHzQ",
        "timestamps": [
            215
        ],
        "3min_transcript": "So that right over here is going to be 2. That's f of 4. And what we wanna approximate is f of 4.36, so 4.36 might be right around, right around there, and so we want to approximate, we wanna approximate this y value right over here. We want to approximate that. Right over here is f of 4.36, and, once again, we're assuming we don't have a calculator at hand. So, how can we do that using what we know about derivatives? Well, what if we were to figure out an equation for the line that is tangent to the point, to tangent to this point right over here. So the equation of the tangent line at x is equal to 4, and then we use that linearization, that linearization defined to approximate values local So what I'm saying is, let's figure out what this, the equation of this line is. Let's call that l of x. And then we can use that to appro, and then we can evaluate that at 4.36, and hopefully that will be a little bit easier to do than to try and figure out this right over here. So how would we do that? Well, one way to think about it, and obviously, there are many ways to express a line, but one way to think about it is, okay, it's going to l of x is going to be f of 4, which is 2. It's going to be f of 4 plus the slope, the slope at, at x equals 4, which is, of course, the derivative f prime of 4, so that's going to be the slope of this line of l of x is f prime of 4. Let me make that clear. So this right over here is the slope. The slope when x is at, at x equals 4, so other point on this is gonna be f' of 4 plus the slope times how far you are away from x equals 4. So it's going to be times x minus 4. Let's just, let's just validate that this makes sense. When we put 4.36 here, when we put 4.36 here, actually let me zoom in on this graph just to make things a little bit clearer. So, if this is, so I'm gonna do a zoom in. I'm gonna do a zoom in. I'm gonna try to zoom in into this region right over here. So, this is the point. This is the point (4, f of 4), and we are going to graph l of x. So let me do that. So this right over here is l of x. And let's say this, right over here, this right over"
    },
    {
        "Q": "At 0:36 when did \"Deca-\" prefix become \"Deka-\"?",
        "A": "Both deka- and deca- are used alternatively and mean the exact same thing. I always used deca-, but apparently many people use the alternative deka-. Doing a web search indicated deca- was used about three times as often as deka-. But using deka- would probably result in less people misreading it as deci-. If I was King, I would declare deka- should always be used instead of deca-.",
        "video_name": "SYkmadc2wOI",
        "timestamps": [
            36
        ],
        "3min_transcript": "We're asked how many centiliters are in one dekaliter? So the first thing we want to do is just think about how much is a centiliter relative to a liter, and how much is a dekaliter relative to a liter? And I'll write the prefixes down. And really, you should have these memorized because you're going to see these prefixes over and over again for different types of units. So the prefix, kilo, sometimes [? ki-lo, ?] this means 1,000. If you see hecto, hecto means 100. Deka means 10. If you have nothing, then that just means 1. Let's put that there. Then if you have deci, this means 1/10. If you have centi, this means 1/100. If you have milli, this means 1/1,000. So let's go back to what we have. We have centiliters. If you have a centiliter, this is equal to 1/100 of a liter, Or you could say 1 liter for every 100 centiliters, so you could also write it like this: 1 liter for every 100, or per every 100, centiliters. So we got the centi, now let's think about the dekaliter. So the deka is right over here. So a dekaliter means 10 liters. Or another way to say it is for every 10 liters, you will Now, before I actually work out the problem, what's going on here? We're going from a smaller unit to a larger unit, so there are going to be many of the smaller units in one of the larger ones. And we can do it multiple ways. So we want to essentially convert 1 dekaliter into centiliters. Now, we could just do it by looking at this chart, or we could do it with the dimensional analysis, making sure the dimensions work out. Let's do it the first way. So if you have one dekaliter, how many liters is that? 1 dekaliter over here would be the same thing as 10 liters. That's liters. We're assuming that our unit is liters here. And then 10 liters is going to be how many deciliters? It's going to be 100 deciliters, right? Because each of these is 10 deciliters, and you have 10 of them. So every time you go down, you're going to be multiplying"
    },
    {
        "Q": "At 2:10 could't you just do 17-5 and then get 12 and x=4",
        "A": "he should of clarified it simpler",
        "video_name": "XoEn1LfVoTo",
        "timestamps": [
            130
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 4:16 why dose it not excist any more?",
        "A": "Hah that s a bit funny to read. It still exists but Sal removed it because it was subtracted",
        "video_name": "XoEn1LfVoTo",
        "timestamps": [
            256
        ],
        "3min_transcript": ""
    },
    {
        "Q": "Why didn't Sal explain why -8/7 is the same that -(8/7) at 8:30. It isn't so simple to understand.",
        "A": "just think about it. -8/7 is equal to 8/-7 and also equal to the negative value of 8/7. so -8/7 = 8/-7 = -(8/7)",
        "video_name": "XoEn1LfVoTo",
        "timestamps": [
            510
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 7:41 you were saying -10 + 2 does not = -12. i have an easy way to understand it. imagine you are paying a debt of 10 bucks. then, you win 2 bucks in a competition. so when you give the person you owe your 2 bucks,you would only owe 8 bucks, as -10 + 2=8. all those who think my method is practical can answer this.",
        "A": "sorry i was supposed to type -8 but i accidentally typed8",
        "video_name": "XoEn1LfVoTo",
        "timestamps": [
            461
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 4:14, EX1. y/2-8=10;",
        "A": "y/2 - 8 = 10 Add 8 to both sides y/2 = 18 Multiply both sides by 2 y = 36",
        "video_name": "XoEn1LfVoTo",
        "timestamps": [
            254
        ],
        "3min_transcript": ""
    },
    {
        "Q": "in 1:20-1:25 i dont get it how can 3x+5=17",
        "A": "Put, x = 4 and see. The equation holds!",
        "video_name": "XoEn1LfVoTo",
        "timestamps": [
            80,
            85
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 0:41, is 5. (1) not the same as 5 x x 1 ?",
        "A": "If another operation is before the parentheses, then you don t multiply. If you have something like 3- (-3), you just subtract the -3. If yo have 4*(5), you only multiply once.",
        "video_name": "AJNDeVt9UOo",
        "timestamps": [
            41
        ],
        "3min_transcript": "A local hospital is holding a raffle as a fundraiser. The individual cost of participating in the raffle is given by the following expression-- 5t plus 3, or 5 times t plus 3, where t represents the number of tickets someone purchases. Evaluate the expression when t is equal to 1, t is equal to 8, and t is equal to 10. So let's first take the situation where t is equal to 1. Then this expression right over here becomes-- and I'll use that same color-- becomes 5 times 1. 5 times 1 plus 3. 5 times 1 plus 3, and we know from order of operations, you do the multiplication before you do the addition. So this will be 5 times 1 is 5 plus 3, and then this is clearly equal to 8. Now let's do it when t is equal to 8. and I'll do the same colors again-- 5 times 8 plus 3. Same color of green. And once again, 5 times 8 is 40, and then we have the plus 3, there so this is equal to 43. And so we have the last situation, with t is equal to 10. I'll do that in blue. So we have 5 times 10. So 5t is 5 times 10. Instead of a t, put a 10 there. 5 times 10 plus 3. That's a slightly different shade of green, but I think you get the idea. 5 times 10 is 50. We do 50, and then we're going to have to add 3 to that, and that is equal to 53. And we're done."
    },
    {
        "Q": "where does he get 104 at 1:26? 80-24? where does he even get that.",
        "A": "to add or subtract the fractions like 85/13 and 8/1 are we take the common denominators of both of the fractions and to maintain the equality the changes which we make in the denominator like to make the denominator of fraction 8/1 we multiplied the denominator by 13 so we do the same thing in the numerator thats how sal got 104",
        "video_name": "KV_XLL4K2Fw",
        "timestamps": [
            86
        ],
        "3min_transcript": "The following line passes through the point 5 comma 8, and the equation of the line is y is equal to 17/13x plus b. What is the value of the y-intercept b? So we know that this point, this x and y value must satisfy this equation, so we know that when x is equal to 5, y is equal to 8. So we can say-- so when x is equal to 5, y is equal to 8. So we can say 8 must be equal to 17/13 times x times 5 plus b, and then we can solve for b. So if we simplify this a little bit, we get 8 is equal to-- let's see, 5 times 17 is 50, plus 35 is 85-- is 85/13 plus b. Then to solve for b, we just subtract 85/13 from both sides. 85/13 Is equal to b. And now we just have to subtract these two numbers. So 8 is the same thing as-- let's see. 80 plus 24 is 104, so it is 104/13-- this is the same thing as 8-- minus 85/13, which is going to be-- let's see. This is 19/13. Is that right? Yes, if this was 105, then it would be 20/13. So this is 19/13, so it's equal to 19/13, which is equal to b. So the equation of this line is going to be y is equal to 17/13x plus 19/13."
    },
    {
        "Q": "Why is it (1+ the square root of 5,-2)[at 12:48]",
        "A": "adding x coordinate of center point with focal distance y coordinate stays the same",
        "video_name": "QR2vxfwiHAU",
        "timestamps": [
            768
        ],
        "3min_transcript": "y is equal to minus 2. That's the center. And then, the major axis is the x-axis, because this is larger. And so, b squared is -- or a squared, is equal to 9. And the semi-minor radius is going to be equal to 3. So, if you go 1, 2, 3. Go there. Than you have 1, 2, 3. No. 1, 2, 3. I think this -- let's see. 1, 2, 3. You go there, roughly. And then in the y direction, the semi-minor radius is going to be 2, right? The square root of that. So b is equal to 2. So you go up 2, then you go down 2. And this ellipse is going to look something like -- pick a good color. It's going to look something like this. And what we want to do is, we want to find out the coordinates of the focal points. the semi-major axis. And we need to figure out these focal distances. And then we can essentially just add and subtract them from the center. And then we'll have the coordinates. What we just showed you, or hopefully I showed you, that the the focal length or this distance, f, the focal length is just equal to the square root of the difference between these two numbers, right? It's just the square root of 9 minus 4. So the focal length is equal to the square root of 5. So, if this point right here is the point, and we already showed that, this is the point -- the center of the ellipse is the point 1, minus 2. The coordinate of this focus right there is going to be 1 plus the square root of 5, minus 2. And the coordinate of this focus right there is going to be 1 minus the square root of 5, minus 2. -- since we're along the major axes, or the x axis, I just add and subtract this from the x coordinate to get these two coordinates right there. So, anyway, this is the really neat thing about conic sections, is they have these interesting properties in relation to these foci or in relation to these focus points. And in future videos I'll show you the foci of a hyperbola or the the foci of a -- well, it only has one focus of a parabola. But this is really starting to get into what makes conic sections neat. Everything we've done up to this point has been much more about the mechanics of graphing and plotting and figuring out the centers of conic sections. But now we're getting into a little bit of the the mathematical interesting parts of conic sections. See you in the next video."
    },
    {
        "Q": "At 0:35, how can Sal draw the ellipse if he doesn't know A and B?",
        "A": "At the beginning of the video, Sal isn t trying to measure a specific ellipse. Rather, he s trying to deduce information about ellipsis in general. He could have drawn any size of ellipse and the conclusions he came to would have been the same. So he could draw the ellipse without knowing A and B because ANY ellipse with ANY size A and B would work.",
        "video_name": "QR2vxfwiHAU",
        "timestamps": [
            35
        ],
        "3min_transcript": "Let's say we have an ellipse formula, x squared over a squared plus y squared over b squared is equal to 1. And for the sake of our discussion, we'll assume that a is greater than b. And all that does for us is, it lets us so this is going to be kind of a short and fat ellipse. Or that the semi-major axis, or, the major axis, is going to be along the horizontal. And the minor axis is along the vertical. And let's draw that. Draw this ellipse. I want to draw a thicker ellipse. Let's say, that's my ellipse, and then let me draw my axes. OK, this is the horizontal right there. And there we have the vertical. And we've studied an ellipse in pretty good detail so far. We know how to figure out semi-minor radius, which in this case we know is b. And that's only the semi-minor radius. Because b is smaller than a. If b was greater, it would be the major radius. And then, of course, the major radius is a. And that distance is this right here. Now, another super-interesting, and perhaps the most interesting property of an ellipse, is that if you take any point on the an ellipse, and measure the distance from that point to two special points which we, for the sake of this discussion, and not just for the sake of this discussion, for pretty much forever, we will call the focuses, or the foci, of this ellipse. And these two points, they always sit along the major axis. So, in this case, it's the horizontal axis. And they're symmetric around the center of the ellipse. So let's just call these points, let me call this one f1. And this is f2. And it's for focus. Focuses. f2. So the super-interesting, fascinating property of an ellipse. you take any point on this ellipse, and measure its distance to each of these two points. So, let's say that I have this distance right here. Let's call this distance d1. And then I have this distance over here, so I'm taking any point on that ellipse, or this particular point, and I'm measuring the distance to each of these two foci. And this is d2. We'll do it in a different color. So this is d2. This whole line right here. That's d2. So when you find these two distances, you sum of them up. So this d2 plus d1, this is going to be a constant that it actually turns out is equal to 2a. But it turns out that it's true anywhere you go on the ellipse. Let me make that point clear. And I'm actually going to prove to you that this constant"
    },
    {
        "Q": "Around 9:50 he mentions that v3 is a linear combination because it is v1 and v2 added together which gives a vector in between the angle of the two. If I took v1 - v2 is it still a linear combination because it is outside the angle? Or is it still a linear combination as it is in R^2",
        "A": "It is a linear combination: the constant you multiplied v2 by was -1.",
        "video_name": "CrV1xCWdY-g",
        "timestamps": [
            590
        ],
        "3min_transcript": "call this vector 2, is equal to vector 3. So vector 3 is a linear combination of these other two vectors. So this is a linearly dependent set. And if we were to show it, draw it in kind of two space, and it's just a general idea that-- well, let me see. Let me draw it in R2. There's a general idea that if you have three two-dimensional vectors, one of them is going to be redundant. Well, one of them definitely will be redundant. For example, if we do 2, 3, if we do the vector 2, 3, that's the first one right there. I draw it in the standard position. And I draw the vector 7, 2 right there, I could show you combination of these two vectors. We can even do a kind of a graphical representation. I've done that in the previous video, so I could write that the span of v1 and v2 is equal to R2. That means that every vector, every position here can be represented by some linear combination of these two guys. Now, the vector 9, 5, it is in R2. It is in R2, right? Clearly. I just graphed it on this plane. It's in our two-dimensional, real number space. Or I guess we could call it a space or in our set R2. It's right there. So we just said that anything in R2 can be represented by a linear combination of those two guys. So clearly, this is in R2, so it can be represented as a linear combination. So hopefully, you're starting to see the relationship between span and linear independence or linear dependence. Let me do another example. Let's say I have the vectors-- let me do a new color. Let's say I have the vector-- and this one will be a little bit obvious-- 7, 0, so that's my v1, and then I have my second vector, which is 0, minus 1. That's v2. Now, is this set linearly independent? Is it linearly independent? Well, can I represent either of these as a combination of the other? And really when I say as a combination, you'd have to scale up one to get the other, because there's only two vectors here. If I am trying to add up to this vector, the only thing I"
    },
    {
        "Q": "At 7:45 when Sal asks if the vectors are dependent or independent, don't we know that they can't be independent solely based on the fact that they are 3 vectors that are only written in two dimensions (a 2x1 matrix)?",
        "A": "Yes, you can say that. But Sal hasn t proved that that is the case yet, and he is just trying to introduce linear dependence right now.",
        "video_name": "CrV1xCWdY-g",
        "timestamps": [
            465
        ],
        "3min_transcript": "Anything in this plane going in any direction can be-- any vector in this plane, when we say span it, that means that any vector can be represented by a linear combination of this vector and this vector, which means that if this vector is on that plane, it can be represented as a linear combination of that vector and that vector. So this green vector I added isn't going to add anything to the span of our set of vectors and that's because this is a linearly dependent set. This one can be represented by a sum of that one and that one because this one and this one span this plane. In order for the span of these three vectors to kind of get more dimensionality or start representing R3, the third vector will have to break out of that plane. It would have to break out of that plane. And if a vector is breaking out of that plane, that means it's a vector that can't be represented anywhere on that Where it's outside, it can't be represented by a linear combination of this one and this one. So if you had a vector of this one, this one, and this one, and just those three, none of these other things that I drew, that would be linearly independent. Let me draw a couple more examples for you. That one might have been a little too abstract. So, for example, if I have the vectors 2, 3 and I have the vector 7, 2, and I have the vector 9, 5, and I were to ask you, are these linearly dependent or independent? So at first you say, well, you know, it's not trivial. Let's see, this isn't a scalar multiple of that. That doesn't look like a scalar multiple of either of Maybe they're linearly independent. But then, if you kind of inspect them, you kind of see call this vector 2, is equal to vector 3. So vector 3 is a linear combination of these other two vectors. So this is a linearly dependent set. And if we were to show it, draw it in kind of two space, and it's just a general idea that-- well, let me see. Let me draw it in R2. There's a general idea that if you have three two-dimensional vectors, one of them is going to be redundant. Well, one of them definitely will be redundant. For example, if we do 2, 3, if we do the vector 2, 3, that's the first one right there. I draw it in the standard position. And I draw the vector 7, 2 right there, I could show you"
    },
    {
        "Q": "At 3:33, it says \"if I draw a vertical line\". Is it the same with a horizontal line?",
        "A": "No, you can t-do that with a horizontal line. A function can have multiple x s leading to 1 y, but a function can t have 1 x going to several y s. For that reason, you must have horizontal lines",
        "video_name": "qGmJ4F3b5W8",
        "timestamps": [
            213
        ],
        "3min_transcript": "It's not defined for 1. We don't know what our function is equal to at 1. So it's not defined there. So 1 isn't part of the domain. It tells us when x is 2, then y is going to be equal to negative 2. So it maps it or associates it with negative 2. That doesn't seem too troublesome just yet. Now, let's look over here. Our function is also defined at x is equal to 3. Our function associates or maps 3 to the value y is equal to 2. That seems pretty straightforward. And then we get to x is equal to 4, where it seems like this thing that could be a function is somewhat defined. It does try to associate 4 with things. But what's interesting here is it tries to associate 4 with two different things. All of a sudden in this thing that we think might have been a function, but it looks like it might not be, we don't know. Do we associate 4 with 5? So this thing right over here is actually a relation. You can have one member of the domain being related to multiple members of the range. But if you do have that, then you're not dealing with a function. So once again, because of this, this is not a function. It's not clear that when you input 4 into it, should you output 5? Or should you output negative 1? And sometimes there's something called the vertical line test that tells you whether something is a function. When it's graphically defined like this, you literally say, OK, when x is 4, if I draw a vertical line, do I intersect the function at two places or more? It could be two or more places. And if you do, that means that there's two or more values that are related to that value in the domain. There's two or more outputs for the input 4. And if there are two or more outputs for that one input, then you're not dealing with a function. You're just dealing with a relation. A function is a special case of a relation."
    },
    {
        "Q": "I'm confused with these factorials (!) In the problem around 6:12 how does 5!/4!=5?",
        "A": "5! = 5\u00e2\u0080\u00a24\u00e2\u0080\u00a23\u00e2\u0080\u00a22\u00e2\u0080\u00a21 4! = 4\u00e2\u0080\u00a23\u00e2\u0080\u00a22\u00e2\u0080\u00a21 5! 5\u00e2\u0080\u00a24\u00e2\u0080\u00a23\u00e2\u0080\u00a22\u00e2\u0080\u00a21 \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\u00e2\u0094\u0080\u00e2\u0094\u0080 4! 4\u00e2\u0080\u00a23\u00e2\u0080\u00a22\u00e2\u0080\u00a21 5! \u00e2\u0094\u0080\u00e2\u0094\u0080 = 5 4!",
        "video_name": "WWv0RUxDfbs",
        "timestamps": [
            372
        ],
        "3min_transcript": "Well actually over zero factorial times five minus zero factorial. Well zero factorial is one, by definition, so this is going to be five factorial, over five factorial, which is going to be equal to one. Once again I like reasoning through it instead of blindly applying a formula, but I just wanted to show you that these two ideas are consistent. Let's keep going. I'm going to do x equals one all the way up to x equals five. If you are inspired, and I encourage you to be inspired, try to fill out the whole thing, what's the probability that x equals one, two, three, four or five. So let's go to the probability that x equals two. Or sorry, that x equals one. The probability that x equals one is going to be equal to... Well how do you get one head? It could be, the first one could be head The second one could be head and then the rest of them are gonna be tails. I could write them all out but you can see that there's five different places to have that one head. So five out of the 32 equally likely outcomes involve one head. Let me write that down. This is going to be equal to five out of 32 equally likely outcomes. Which of course is the same thing, this is going to be the same thing as saying I got five flips, and I'm choosing one of them to be heads. So that over 32. You could verify that five factorial over one factorial times five minus-- Actually let me just do it just so that you don't have to take my word for it. So five choose one is equal to five factorial over one factorial, which is just one, times five minus four-- Sorry, five minus one factorial. which is just going to be equal to five. All right, we're making good progress. Now in purple let's think about the probability that our random variable x is equal to two. Well this is going to be equal to, and now I'll actually resort to the combinatorics. You have five flips and you're choosing two of them to be heads. Over 32 equally likely possibilities. This is the number of possibilities that result in two heads. Two of the five flips have chosen to be heads, I guess you can think of it that way, by the random gods, or whatever you want to say. This is the fraction of the 32 equally likely possibilities, so this is the probability that x equals two. What's this going to be? I'll do it right over here. And actually no reason for me to have to keep switching colors."
    },
    {
        "Q": "@0:34\n\nI understand that it doesn't produce the correct answer, in fact it seems to produce the reciprocal of the correct answer, but why doesn't it work to multiply the left and right sides by the fraction x/x (which would be equal to 1) which would leave us with 10x on the left and 15x on the right?\n\nI'm sure that I'm missing something simple, as usual.",
        "A": "If you multiply both sides by x/x, here s what you would get: 7x/x - 10x/x^2 = 2x/x + 15x/x^2 This just gives you a much more complicated equation to try and solve. You need to multiply by x to eliminate the fractions and get the variable into the numerator.",
        "video_name": "Z7C69xP08d8",
        "timestamps": [
            34
        ],
        "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 2:30, Sal crossed out -10 & +10. He said \"these negate each other\". What does negate mean?",
        "A": "It means they cancel each other out to make 0.",
        "video_name": "Z7C69xP08d8",
        "timestamps": [
            150
        ],
        "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 3:24 he started say that they are both independent. It was not clear to me if the first choice was dependent or not? Maybe I missed it.",
        "A": "Yes, that is correct!",
        "video_name": "VjLEoo3hIoM",
        "timestamps": [
            204
        ],
        "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": "What is that weird N-shaped symbol that Sal drew at 2:32?\nI assume it's some sort of symbol meaning and.",
        "A": "The \u00e2\u0088\u00a9 symbol Sal wrote in 2:34 stands for intersection, which you have probably encountered in basic statistics. For example, if you let X and Y be arbitrary sets, X \u00e2\u0088\u00a9 Y would be classified as the set containing the elements that are in Set X AND Set Y.",
        "video_name": "VjLEoo3hIoM",
        "timestamps": [
            152
        ],
        "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 1:43, when he says that the {Ak} equation means that k=1 from the first to the last term, does that mean if he had 5 numbers that the \"4\" at the top of the equation would be a \"5\" instead?",
        "A": "You are correct. That s why in the infinite equation, the value at the top (AKA the value the sequence goes to) is infinity.",
        "video_name": "KRFiAlo7t1E",
        "timestamps": [
            103
        ],
        "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": "At 1:49, Sal stops at \"a sub-4\". In this finite sequence, is it possible to have an \"a sub-5\"?",
        "A": "Hi shanzi11, In that specific sequence no. This is because he states that it is from term 1 to 4 and therefore in this sequence only, subs greater than 4 do not exist. Hope that helps! - JK",
        "video_name": "KRFiAlo7t1E",
        "timestamps": [
            109
        ],
        "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": "I can't understand the part where he creates the function for the first sequence at around 2:50. How did he come up with 1+3(k-1)",
        "A": "k a_k 1 + 3\u00e2\u0080\u00a2(k - 1) 1 1 1 + 3\u00e2\u0080\u00a2(1 - 1) = 1 + 3\u00e2\u0080\u00a20 = 1 + 0 = 1 2 4 1 + 3\u00e2\u0080\u00a2(2 - 1) = 1 + 3\u00e2\u0080\u00a21 = 1 + 3 = 4 3 7 1 + 3\u00e2\u0080\u00a2(3 - 1) = 1 + 3\u00e2\u0080\u00a22 = 1 + 6 = 7 4 10 1 + 3\u00e2\u0080\u00a2(4 - 1) = 1 + 3\u00e2\u0080\u00a23 = 1 + 9 = 10",
        "video_name": "KRFiAlo7t1E",
        "timestamps": [
            170
        ],
        "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": "Around 7:09. As a recursive function, Sal said that A sub 2 is:\n\nA(2-1)+3 Which is 1+3 which the answer is 4.\nBut why doesn't it work for A sub 3?\n\nA(3-1)+3 --> 2+3 --> 5\n\nI think I'm making a mistake but I don't know where the mistake is. Please help! Thanks!",
        "A": "A(3-1) is not 2, its A(n) when n=2, its the second element\\member of the sequence, indexed at 2, and in this case A(2) = 4, and not 2. the fact that A(1) = 1 is sheer coincidence i m afraid :)",
        "video_name": "KRFiAlo7t1E",
        "timestamps": [
            429
        ],
        "3min_transcript": "So we're adding 4 one less than the term that we're at. So it's going to be plus 4 times k minus 1. So this is another way of defining this infinite sequence. Now, in both of these cases, I defined it as an explicit function. So this right over here is explicit. That's not an attractive color. Let me write this in. This is an explicit function. And so you might say, well, what's another way of defining these functions? Well, we can also define it, especially something like an arithmetic sequence, we can also define it recursively. And I want to be clear-- not every sequence can be defined as either an explicit function like this, or as a recursive function. But many can, including this, which is an arithmetic sequence, where we keep adding the same quantity over and over again. So how would we do that? Well, we could also-- another way of defining starting at k equals 1 and going to 4 with. And when you define a sequence recursively, you want to define what your first term is, with a sub 1 equaling 1. You can define every other term in terms of the term before it. And so then we could write a sub k is equal to the previous term. So this is a sub k minus 1. So a given term is equal to the previous term. Let me make it clear-- this is the previous term, plus-- in this case, we're adding 3 every time. Now, how does this make sense? Well, we're defining what a sub 1 is. And if someone says, well, what happens when k equals 2? Well, they're saying, well, it's going to be a sub 2 minus 1. Well, we know a sub 1 is 1. So it's going to be 1 plus 3, which is 4. Well, what about a sub 3? Well, it's going to be a sub 2 plus 3. a sub 2, we just calculated as 4. You add 3. It's going to be 7. This is essentially what we mentally did when I first wrote out the sequence, when I said, hey, I'm just going to start with 1. And I'm just going to add 3 for every successive term. So how would we do this one? Well, once again, we could write this as a sub k. Starting at k, the first term, going to infinity with-- our first term, a sub 1, is going to be 3, now. And every successive term, a sub k, is going to be the previous term, a sub k minus 1, plus 4. And once again, you start at 3. And then if you want the second term, it's going to be the first term plus 4. It's going to be 3 plus 4. You get to 7. And you keep adding 4. So both of these, this right over here"
    },
    {
        "Q": "@ 2:56 what does this man mean by we added 3 times one less than the k term times? I don't understand",
        "A": "Hey ISAIAH, This man is trying to show that a(k) = 1 + 3(k-1) Since the k term is what term it is (1,2, etc.) It could be referred to as how many times. That is what this man is referring to as the k term times . Hope that helps! - JK",
        "video_name": "KRFiAlo7t1E",
        "timestamps": [
            176
        ],
        "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": "Having trouble understanding how to explicitly define a sequence. Can anyone clarify what Sal means at 4:56 when he says \"we are adding 4, one less time\"?",
        "A": "Sal is talking us though his calculations, and so he isn t being as clear as he could be, here. One less time means that we do not add four to obtain the first term. That means that with term #2, we add one 4 to the first term, with term #3, we add two fours to the first term, and so forth: we add one less 4 than the number of the term. (Thanks for the time-stamp: they really help with a question like this :-)",
        "video_name": "KRFiAlo7t1E",
        "timestamps": [
            296
        ],
        "3min_transcript": "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. So we're adding 4 one less than the term that we're at. So it's going to be plus 4 times k minus 1. So this is another way of defining this infinite sequence. Now, in both of these cases, I defined it as an explicit function. So this right over here is explicit. That's not an attractive color. Let me write this in. This is an explicit function. And so you might say, well, what's another way of defining these functions? Well, we can also define it, especially something like an arithmetic sequence, we can also define it recursively. And I want to be clear-- not every sequence can be defined as either an explicit function like this, or as a recursive function. But many can, including this, which is an arithmetic sequence, where we keep adding the same quantity over and over again. So how would we do that? Well, we could also-- another way of defining"
    },
    {
        "Q": "at 3:14 why is the depth of the cylinder dy.",
        "A": "dy is an extremely small change in y, it is infinitesimal. We set it as this so that we can use the sum of infinitely thin cylinders to take an integral. Imagine if y was a large number. There would be a large volume unaccounted for, and the integral would not be accurate.",
        "video_name": "43AS7bPUORc",
        "timestamps": [
            194
        ],
        "3min_transcript": "So we care about this part right over here, not the very bottom of it. And let me shade it in a little bit. So it would look something like that. So let me draw it separately, just so we can visualize it. So I'll draw it at different angles. So if I were to draw it with the y-axis kind of coming out the back, it would look something like this. It would look something like-- it gets a little bit smaller like that. And then it gets cut off right over here, right over here like that. So it looks-- I don't know what shape you could call it. But I think hopefully you're conceptualizing this. Let me do it in that same yellow color. The visual-- that's not yellow. The visualization here is probably the hardest part. But as we can see it's not too bad. So it looks something like this. It looks like maybe a truffle, an upside-down truffle. So this right here, let me draw the y-axis just to show how we're oriented. So the y-axis is popping out in this example like that. 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,"
    },
    {
        "Q": "When finding the length of the B side at approx 1:17 do you add 4^2 to 3^2?",
        "A": "Yes, you add 4^2 to 3^2 and then taking the square root of this sum. Just pythagorean theorem. Hope this helps.",
        "video_name": "2yjSAarzWF8",
        "timestamps": [
            77
        ],
        "3min_transcript": "The graph below contains triangle ABC and the point P. Draw the image of triangle ABC under a dilation whose center is P and a scale factor of 2. So essentially, we want to scale this so that every point is going to be twice as far away from P. So for example, B right over here has the same y-coordinate as P, but its x-coordinate is three more. So we want to be twice as far. So if this maps to point B, we just want to go twice as far. So we're at 3 away, we want to go 6 away. So point P's x-coordinate is at 3, now we're at 9. Likewise, point C is 3 below P. Well we want to go twice as far, so we'll go 3 more. And point A is 4 above P. Well we want to go 4 more. We want to go twice as far-- one, two, three, four. And we get right over there. Then they ask us, what are the lengths of side AB and its image? might have to apply the distance formula. Let's see, it's the base right over here. The change in x between the two is 3 and the change in y is 4, so this is actually a 3, 4, 5 right triangle. 3 squared plus 4 squared is equal to 5 squared. So AB is 5 units long. Essentially just using the Pythagorean theorem to figure that out. And its image, well it's image should be twice as long. And let's see whether that actually is the case. So this is a base right over here that's of length 6. This has a height, or this change in y, I could say. Because I'm really just trying to figure out this length, which is the hypotenuse of this right triangle. I don't have my drawing tool, so I apologize. But this height right here is 8. So 8 squared is 64, plus 6 squared is 36, that's 100, which is 10 squared. So notice, our scale factor of 2, the corresponding side Each of these points got twice as far away from our center of dilation."
    },
    {
        "Q": "At 0:50, why did we go 4 points above?",
        "A": "It went 4 points above because the distance between point A and point P is 4, and as the problem states, it wants you to dilate from point P with a scale factor of 2. 2*4=8. And so point A2 is 8 points away from point P.",
        "video_name": "2yjSAarzWF8",
        "timestamps": [
            50
        ],
        "3min_transcript": "The graph below contains triangle ABC and the point P. Draw the image of triangle ABC under a dilation whose center is P and a scale factor of 2. So essentially, we want to scale this so that every point is going to be twice as far away from P. So for example, B right over here has the same y-coordinate as P, but its x-coordinate is three more. So we want to be twice as far. So if this maps to point B, we just want to go twice as far. So we're at 3 away, we want to go 6 away. So point P's x-coordinate is at 3, now we're at 9. Likewise, point C is 3 below P. Well we want to go twice as far, so we'll go 3 more. And point A is 4 above P. Well we want to go 4 more. We want to go twice as far-- one, two, three, four. And we get right over there. Then they ask us, what are the lengths of side AB and its image? might have to apply the distance formula. Let's see, it's the base right over here. The change in x between the two is 3 and the change in y is 4, so this is actually a 3, 4, 5 right triangle. 3 squared plus 4 squared is equal to 5 squared. So AB is 5 units long. Essentially just using the Pythagorean theorem to figure that out. And its image, well it's image should be twice as long. And let's see whether that actually is the case. So this is a base right over here that's of length 6. This has a height, or this change in y, I could say. Because I'm really just trying to figure out this length, which is the hypotenuse of this right triangle. I don't have my drawing tool, so I apologize. But this height right here is 8. So 8 squared is 64, plus 6 squared is 36, that's 100, which is 10 squared. So notice, our scale factor of 2, the corresponding side Each of these points got twice as far away from our center of dilation."
    },
    {
        "Q": "At 6:29 I found that the line went the other way. Are u guys sure that is the way the line goes?",
        "A": "How did you get it to go the other way? It has a negative slope, so it should go down from left to right (or someone could slide down the hill). With a slope of -1/2, this says go down 1(change in y is -1) and right 2 (change in y is 2). Figure out where you went wrong to get line to go other way.",
        "video_name": "unSBFwK881s",
        "timestamps": [
            389
        ],
        "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": "why did you go back 4 at time stamp 1:34?",
        "A": "He didn t. You have the y-intercept graphed. From there you know that the slope is 4, so up 4 and right 1. But that will only let you graph to the right. To graph to the left, you need to go down 4 and left 1. The complete opposite.",
        "video_name": "unSBFwK881s",
        "timestamps": [
            94
        ],
        "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": "At 5:30 didn't he mean the equation would be y=-1/2x-6? He said it would be y=-1/2-6 and I was wonder where the variable would be. I'm not sure if I'm correct or not, please let me know.",
        "A": "Yes, you re correct. Initially he forgot to add the x. About a minute afterwards, though, he saw his mistake and fixed it. :)",
        "video_name": "unSBFwK881s",
        "timestamps": [
            330
        ],
        "3min_transcript": "So it's all of these points here-- that I'm shading in in green-- satisfy that right there. If I were to look at this one over here, when x is negative 1, y is less than negative 1. So y has to be all of these points down here. When x is equal to 1, y is less than 7. So it's all of these points down here. And in general, you take any point x-- let's say you take this point x right there. If you evaluate 4x plus 3, you're going to get the point on the line. That is that x times 4 plus 3. Now the y's that satisfy it, it could be equal to that point on the line, or it could be less than. 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."
    },
    {
        "Q": "At 1:31 Sal says that the point is here. Why cannot it be on the left hand side?",
        "A": "because it isn t a negative 1 so it is a positive on the graph hope this helps u",
        "video_name": "unSBFwK881s",
        "timestamps": [
            91
        ],
        "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": "1:01 What are the other equation forms useful for?",
        "A": "Standard form: Ax+By = C is used frequently when solving systems of equations. Point-Slope form: Y - y1 = m(X - x1) will accept the slope and any point from the line. So, it is easy to use to create the equation of a line. Slope-Intercept form: y = mx + b is used to quickly obtain info need to graph the line. It is also used to change a linear equation into function notation. And, it can be used to create the equation of the line. Hope this helps.",
        "video_name": "IL3UCuXrUzE",
        "timestamps": [
            61
        ],
        "3min_transcript": "- [Voiceover] There's a lot of different ways that you could represent a linear equation. So for example, if you had the linear equation y is equal to 2x plus three, that's one way to represent it, but I could represent this in an infinite number of ways. I could, let's see, I could subtract 2x from both sides, I could write this as negative 2x plus y is equal to three. I could manipulate it in ways where I get it to, and I'm gonna do it right now, but this is another way of writing that same thing. y minus five is equal to two times x minus one. You could actually simplify this and you could get either this equation here or that equation up on top. These are all equivalent, you can get from one to the other with logical algebraic operations. So there's an infinite number of ways to represent a given linear equation, but I what I wanna focus on in this video is this representation in particular, because this one is a very useful representation this one and this one can also be useful, depending on what you are looking for, but we're gonna focus on this one, and this one right over here is often called slope-intercept form. Slope-intercept form. And hopefully in a few minutes, it will be obvious why it called slop-intercept form. And before I explain that to you, let's just try to graph this thing. I'm gonna try to graph it, I'm just gonna plot some points here, so x comma y, and I'm gonna pick some x values where it's easy to calculate the y values. So maybe the easiest is if x is equal to zero. If x is equal to zero, then two times zero is zero, that term goes away, and you're only left with this term right over here, y is equal to three. Y is equal to three. And so if we were to plot this. Actually let me start plotting it, so that is my y axis, and let me do the x axis, so as straight as I would like it. So that looks pretty good, alright. That is my x axis and let me mark off some hash marks here, so this is x equals one, x equals two, x equals three, this is y equals one, y equals two, y equals three, and obviously I could keep going and keep going, this would be y is equal to negative one, this would be x is equal to negative one, negative two, negative three, so on and so forth. So this point right over here, zero comma three, this is x is zero, y is three. Well, the point that represents when x is equal to zero and y equals three, this is, we're right on the y axis. If they have a line going through it and this line"
    },
    {
        "Q": "I have no idea how to simplified (y-5)=2(x-1) at 00:43",
        "A": "the word simplify in this context means to remove the parentheses using the distributive property and then use the addition property of equality to combine the constant terms. thus you first get y - 5 = 2x - 2. Then adding 5 to both sides you get y=2x+3. This final form is unique for any given line and is called slope-intercept form. It is the function form of the line and it is usually the form that you need if you are going to use a calculator or other graphing software to graph your line.",
        "video_name": "IL3UCuXrUzE",
        "timestamps": [
            43
        ],
        "3min_transcript": "- [Voiceover] There's a lot of different ways that you could represent a linear equation. So for example, if you had the linear equation y is equal to 2x plus three, that's one way to represent it, but I could represent this in an infinite number of ways. I could, let's see, I could subtract 2x from both sides, I could write this as negative 2x plus y is equal to three. I could manipulate it in ways where I get it to, and I'm gonna do it right now, but this is another way of writing that same thing. y minus five is equal to two times x minus one. You could actually simplify this and you could get either this equation here or that equation up on top. These are all equivalent, you can get from one to the other with logical algebraic operations. So there's an infinite number of ways to represent a given linear equation, but I what I wanna focus on in this video is this representation in particular, because this one is a very useful representation this one and this one can also be useful, depending on what you are looking for, but we're gonna focus on this one, and this one right over here is often called slope-intercept form. Slope-intercept form. And hopefully in a few minutes, it will be obvious why it called slop-intercept form. And before I explain that to you, let's just try to graph this thing. I'm gonna try to graph it, I'm just gonna plot some points here, so x comma y, and I'm gonna pick some x values where it's easy to calculate the y values. So maybe the easiest is if x is equal to zero. If x is equal to zero, then two times zero is zero, that term goes away, and you're only left with this term right over here, y is equal to three. Y is equal to three. And so if we were to plot this. Actually let me start plotting it, so that is my y axis, and let me do the x axis, so as straight as I would like it. So that looks pretty good, alright. That is my x axis and let me mark off some hash marks here, so this is x equals one, x equals two, x equals three, this is y equals one, y equals two, y equals three, and obviously I could keep going and keep going, this would be y is equal to negative one, this would be x is equal to negative one, negative two, negative three, so on and so forth. So this point right over here, zero comma three, this is x is zero, y is three. Well, the point that represents when x is equal to zero and y equals three, this is, we're right on the y axis. If they have a line going through it and this line"
    },
    {
        "Q": "At 7:42, Sal mentions that we could of taken the Absolute value of the difference between the measurements and the mean instead of squaring them. Why don't we do that, it seems easier?",
        "A": "A few reasons. 1. The absolute value function is much harder to deal with mathematically, because the derivative isn t nearly so nice as that of the square function. 2a. Squaring works very well with the Normal distribution. 2b. The sample mean is a natural estimate of location/center, and the Sampling Distribution of the sample mean is Normal, so we d like to use that. Hence, item 2a.",
        "video_name": "PWiWkqHmum0",
        "timestamps": [
            462
        ],
        "3min_transcript": "And in this case, what's it going to be? It's going to be the square root of 0.316. And then, what are the units going to be? It's going to be just meters. And we end up with-- so let me take the square root of 0.316. And I get 0.56-- I'll just round to the nearest thousandth-- 0.562. So this is approximately 0.562 meters. So you might be saying, Sal, what do we call this thing that we just did? The square root of the variance. And here we're dealing with the population. We haven't thought about sampling yet. The square root of the population variance, what do we call this thing right over here? And this is a very familiar term. Oftentimes, when you take an exam, this is calculated for the scores on the exam. I'm using that yellow a little bit too much. This is the population standard deviation. It is a measure of how much the data is varying from the mean. In general, the larger this value, that means that the data is more varied from the population mean. The smaller, it's less varied. And these are all somewhat arbitrary definitions of how we've defined variance. We could have taken things to the fourth power. We could have done other things. We could have not taken them to a power but taking the absolute value here. The reason why we do it this way is it has neat statistical properties as we try to build on it. But that's the population standard deviation, which gives us nice units-- meters. In the next video, we'll think about the sample standard"
    },
    {
        "Q": "I wish Khan Academy had a video where Sal explains all the special symbols that are used in math and what they stand for. For example, at 1:02 Sal says \"We're going to use 'Mu'\" what's 'Mu'?",
        "A": "Mu = \u00c2\u00b5 = population mean. Population symbols are always Greek.",
        "video_name": "PWiWkqHmum0",
        "timestamps": [
            62
        ],
        "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": "so does phi work for any ratio? even 7:1?",
        "A": "No. It only works for the golden ratio.",
        "video_name": "5zosU6XTgSY",
        "timestamps": [
            421
        ],
        "3min_transcript": "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. Let's actually take a calculator out and see if we can get the first few places of this magic number phi. So let me get my calculator out. Let's just actually evaluate it. And you might recognize that square root of 5 is an irrational number. And so this whole thing is going to be an irrational number, but I'll prove that in another video, which means it never repeats. It goes on and on and on forever. But let's actually evaluate it. So it's 1 plus the square root of 5 divided by 2. So it says 1.6180339. So let me put that aside. Let me write it down. And this is where it starts to get really interesting and mysterious. So this number right over here is 1.618033988 on never terminating, never repeating. So that by itself, it's this cool number. It's this ratio that has all of these neat properties, which are pretty crazy anyway that you express it. But what's really neat is if we revisit this thing right over here. Because what is 1 over phi going to be? So 1 over phi, which we sometimes denote with a capital phi. We already know 1 over phi is just phi minus 1. So we actually can do this in our heads. 1 over this is just going to be 0.618033988. There's just something wacky about that, that the inverse of the number is really just the decimals left over after you get rid of the 1. That, by itself, is kind of a crazy idea. But it gets even crazier because this number is showing up everywhere. And as you might imagine from the title of this video, this phi right over here, this is called the golden ratio."
    },
    {
        "Q": "Isn't the number at 7:04 kind of like pi? It keeps on going forever but never repeats.",
        "A": "Yes, the number(or Phi) is an irrational number like pi since the numbers or never terminating.",
        "video_name": "5zosU6XTgSY",
        "timestamps": [
            424
        ],
        "3min_transcript": "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. Let's actually take a calculator out and see if we can get the first few places of this magic number phi. So let me get my calculator out. Let's just actually evaluate it. And you might recognize that square root of 5 is an irrational number. And so this whole thing is going to be an irrational number, but I'll prove that in another video, which means it never repeats. It goes on and on and on forever. But let's actually evaluate it. So it's 1 plus the square root of 5 divided by 2. So it says 1.6180339. So let me put that aside. Let me write it down. And this is where it starts to get really interesting and mysterious. So this number right over here is 1.618033988 on never terminating, never repeating. So that by itself, it's this cool number. It's this ratio that has all of these neat properties, which are pretty crazy anyway that you express it. But what's really neat is if we revisit this thing right over here. Because what is 1 over phi going to be? So 1 over phi, which we sometimes denote with a capital phi. We already know 1 over phi is just phi minus 1. So we actually can do this in our heads. 1 over this is just going to be 0.618033988. There's just something wacky about that, that the inverse of the number is really just the decimals left over after you get rid of the 1. That, by itself, is kind of a crazy idea. But it gets even crazier because this number is showing up everywhere. And as you might imagine from the title of this video, this phi right over here, this is called the golden ratio."
    },
    {
        "Q": "At 3:57 how is phi always equal to only part of phi, shouldn't Phi=Sq. root of 1+Phi equal the square root of 1+Phi, if 1+Phi is Phi there SHOULD be no solution",
        "A": "At 6:15 Sal talks about how the \u00e2\u0088\u009a5 is an irrational number and so repeats forever like \u00cf\u0080. An equation has no solutions when one side doesn t equal the other not when one sides repeats forever I think. I m twice your age and struggling with this so it s awesome you re on it already.",
        "video_name": "5zosU6XTgSY",
        "timestamps": [
            237
        ],
        "3min_transcript": "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, 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."
    },
    {
        "Q": "At 3:20 he multiplied each side of the equation by phi, wouldnt that make it\nPhi^2=1+1",
        "A": "(phi - 1 = 1/phi)phi phi^2 - phi = 1",
        "video_name": "5zosU6XTgSY",
        "timestamps": [
            200
        ],
        "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 12:00, how did he get from (a-b)/b to (a/b)-1?",
        "A": "(a-b)/b=(a/b)-(b/b)=(a/b)-1",
        "video_name": "5zosU6XTgSY",
        "timestamps": [
            720
        ],
        "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 9:07 how did minus 8 by 24?",
        "A": "Because it will go into the negatives.",
        "video_name": "-rxUip6Ulnw",
        "timestamps": [
            547
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 2:11 I really got confused... Like what? I don't get what she means? Like how?",
        "A": "9x10 is basically 9 ten times. You move nine to the tens place, and zero is in the ones place since there are no ones left. 9 times 10 is 9, 10 times or 10, 9 times",
        "video_name": "Ehd3cgRBvl0",
        "timestamps": [
            131
        ],
        "3min_transcript": "- [Voiceover] What is 7 100s times 10? Well, let's focus first on this times 10 part of our expression. Because multiplying by 10 has some patterns in math that we can use to help us solve. One pattern we can think of when we multiply by 10 is if we take a whole number and multiply it by 10, we'll simply add a zero to the end of our whole number. So, for example, if we have a whole number like nine, and we multiply by 10, our solution will be a nine with one zero at the end. Or 90. Because nine times 10 is the same as nine 10s, and nine 10s is ninety. So let's use that pattern first to try to solve. Here we have seven 100s. So, seven times we have 100, or 700, and we're multiplying again times 10. Our solution will add a zero at the end. So if we had 700, 10 times, we would have 700 with a zero on the end. Or 7,000. So seven 100s times ten, is equal to 7,000. But there's another pattern we could use, here. Another pattern to think about when we multiply by 10. And that is that when we multiply by 10, we move every digit one place value, one place value, left. Or one place value greater. So let's look at that one on a place value chart. Here we have a place value chart. To use that earlier example when we had nine ones, and we multiplied it by 10, It moved up to the 10s. Now, we had nine 10s. And we filled in a zero here, because there were no ones left; there were zero ones left. And so, we saw that nine times 10 was equal to 90. So again, it's the same as adding a zero at the end, but we're looking at it another way. We're looking at it in terms of place value and multiplying by 10 moved every digit one place value to the left. So, if we do that with this same question, seven 100s, seven 100s, if we move 100s one place value to the left, we'll end up with 1,000s. So, 700 times 10 is seven 1,000s. Or, as we saw earlier, 7,000. So either one of these is a correct answer."
    },
    {
        "Q": "At 2:32 , I don't understand how he got that (s-7). Could someone explain it?",
        "A": "Ah I see your confusion. Here is what he did: It started as S(S+5) - 7(S+5) What if we look at it all in parenthesis? (S(S+5) - 7(S+5)) What could you factor out of this big mess? Both terms have (S+5) in common right? So pull out S+5 and you get: (S+5)(whatever is not yet factored out) But, the only terms remaining in the original parenthesis was an S-7. So there you go! The final terms are (S+5)(S+7)",
        "video_name": "2ZzuZvz33X0",
        "timestamps": [
            152
        ],
        "3min_transcript": "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 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."
    },
    {
        "Q": "in 1:17 how do you get 5+-7=-3",
        "A": "He got 5 + (-7) = -2, not 5 + (-7) = -3 He also had the constraint that a \u00c3\u0097 b = -35, and 5 \u00c3\u0097 -7 = -35",
        "video_name": "2ZzuZvz33X0",
        "timestamps": [
            77
        ],
        "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": "At 1:59 how did Sal get s(s+5)?",
        "A": "Arnav, At 1:59, Sal took the part of the expression (s\u00c2\u00b2 + 5s) and using the distribtive property in reverse, he factored out an s Like this (s\u00c2\u00b2 + 5s) is (s*s + 5*s) so factor out (undistribute) an s s*(s+5) and rewrite as s(s+5) And in case you still don t understand the reverse distributive property, just use the distributive property on the answer and see if that helps. s(s+5) dstribute the s s*s + 5*s and rewrite as s\u00c2\u00b2 + 5s I hope that helps make it click for you.",
        "video_name": "2ZzuZvz33X0",
        "timestamps": [
            119
        ],
        "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": "At 1:41, can you factor out the quadratic equation into 2 binomials? Does it affect the answer?",
        "A": "That is exactly what Sal did. He factored the quadratic into the binomials: (s-7)(s+5)=0 So, I m sure what you mean by your questions. If you can clarify, I ll try to help.",
        "video_name": "2ZzuZvz33X0",
        "timestamps": [
            101
        ],
        "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": "where did the 35 go at 2:11?",
        "A": "Sal just factored a -7 out of that part of the polynomial. He divided the two terms by -7 to get -7(s+5). Hope this helps! :D",
        "video_name": "2ZzuZvz33X0",
        "timestamps": [
            131
        ],
        "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": "At 2:50, how does s(s+5)-7(s+5)=0 factor to (s+5)(s-7)?",
        "A": "s(s+5)-7(s+5) factors into (s+5)(s-7) because s has been factored out of (s^2+5s) and -7 has been factored out of (-7s-35) both of the factored out forms are (s+5) you combine what you factored out of both sides and you get (s-7) leaving you with the factored form (s+5)(s-7).",
        "video_name": "2ZzuZvz33X0",
        "timestamps": [
            170
        ],
        "3min_transcript": "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 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."
    },
    {
        "Q": "At 2:16, couldn't you do s^2-5s+7s-35? It would mean the same thing right?",
        "A": "I don t think you can factor the equation in thay form, but mathematically it means the same thing.",
        "video_name": "2ZzuZvz33X0",
        "timestamps": [
            136
        ],
        "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": "When solving a quadratic equation by factoring, if both equations could equal zero how come this is not included in the answer? Sal mentions around 3:45 that both could equal zero. In the above example the answer is given as s=-5 or s=7 but not s=-5 and s=7. Why is this? In the equation it makes sense that both could equal 0 as 0x0=0 but how can the answer be s=-5 and s=7?\n\nThanks!",
        "A": "s=-5 makes one factor 0. s= 7 makes the other factor 0. Those are two different solutions to making the equation 0. But we didn t know until we did the factoring that the two factors would lead to two different zeros. It could have some out, for example, like this: (s+5)(s+5)=0. Then s= -5 would make both factors zero, and that would be ok because 0*0 = 0.",
        "video_name": "2ZzuZvz33X0",
        "timestamps": [
            225
        ],
        "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": "Wouldn't the (s+5), in about 2:45, be squared because there is two of them? I didn't think that you could just get rid of it or ignore it..",
        "A": "Emily, We had s(s+5) - 7(s+5) We are combining like terms. The second (s+5) doesn t just disappear it is just combined. Think of the s+5 as being apples . s apples - 7 apples = (s-7) apples. We can do the same thing with (s+5) instead of apples s (s+5) - 7 (s+5) = (s-7) (s+5) I hope that helps make it click for you.",
        "video_name": "2ZzuZvz33X0",
        "timestamps": [
            165
        ],
        "3min_transcript": "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 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."
    },
    {
        "Q": "At 2:51 Sal wrote the equation as (s+5)(s-7)=0. When factoring quadratics, how do you know which constants are supposed to come first, like, in this case, 5 is the first constant and -7 is the second constant? That usually gets me when solving quadratic equations by factoring.",
        "A": "It doesn t matter which comes first. The commutative property of multiplication tells use that the order we multiply in doesn t matter. Example 2x3 = 3x2. Apply the same property to the factors: (s+5)(s-7) = (s-7)(s+5). The order of the factors does not matter. Hope this helps.",
        "video_name": "2ZzuZvz33X0",
        "timestamps": [
            171
        ],
        "3min_transcript": "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 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."
    },
    {
        "Q": "at 2:09 why does he put the plus or minus sign?",
        "A": "When we multiply +9 x +9 we get 81 also when we multiply -9 x -9 we still get 81. so the square root can be + or - 9. therefore he writes + and - as the root can be either in + or -.",
        "video_name": "tRHLEWSUjrQ",
        "timestamps": [
            129
        ],
        "3min_transcript": "- Let's see if we can solve the equation P squared is equal to 0.81. So how could we think about this? Well one thing we could do is we could say, look if P squared is equal to 0.81, another way of expressing this is, that well, that means that P is going to be equal to the positive or negative square root of 0.81. Remember if we just wrote the square root symbol here, that means the principal root, or just the positive square root. But here P could be positive or negative, because if you square it, if you square even a negative number, you're still going to get a positive value. So we could write that P is equal to the plus or minus square root of 0.81, which kind of helps us, it's another way of expressing the same, the same, equation. But still, what could P be? In your brain, you might immediately say, well okay, you know if this was P squared is equal to 81, I kinda know what's going on. Because I know that nine times nine is equal to 81. Or we could write that nine squared is equal to 81, to the principal root of 81. These are all, I guess, saying the same truth about the universe, but what about 0.81? Well 0.81 has two digits behind, to the right of the decimal and so if I were to multiply something that has one digit to the right of the decimal times itself, I'm gonna have something with two digits to the right of the decimal. And so what happens if I take, instead of nine squared, what happens if I take 0.9 squared? Let me try that out. Zero, I'm gonna use a different color. So let's say I took 0.9 squared. 0.9 squared, well that's going to be 0.9 times 0.9, which is going to be equal to? Well nine times nine is 81, and I have one, two, numbers to the right of the decimal, so I'm gonna have two numbers to the right of the decimal in the product. So one, two. So that indeed is equal to 0.81. In fact we could write 0.81 as 0.9 squared. is equal to the plus or minus, the square root of, instead of writing 0.81, I could write that as 0.9 squared. In fact I could also write that as negative 0.9 squared. Cause if you put a negative here and a negative here, it's still not going to change the value. A negative times a negative is going to be a positive. I could, actually I would have put a negative there, which would have implied a negative here and a negative there. So either of those are going to be true. But it's going to work out for us because we are taking the positive and negative square root. So this is going to be, P is going to be equal to plus or minus 0.9. Plus or minus 0.9, or we could write it that P is equal to 0.9, or P could be equal to negative 0.9. And you can verify that, you would square either of these things, you get 0.81."
    },
    {
        "Q": "at 5:46 how did he get b+6",
        "A": "Because, the sum of a+b must equal -4 and the product of a*b must equal -60. He just brute force went thru all the combinations possible until finding that +6 and -10 satisfy this. (a+6)(a-10) = b^2-4b-60",
        "video_name": "STcsaKuW-24",
        "timestamps": [
            346
        ],
        "3min_transcript": "Then you could have 4 and 15, still too far apart. If you made one of them negative, their sum would be either negative 11 or positive 11. Then you have 5 and a 12, still seems too far apart One of them is negative, then you either have their sum being positive 7 or negative 7. Then you have 6 and 10. Now this looks interesting. They are four apart. So if we make-- and we want the larger absolute magnitude number to be negative so that their sum is negative. So if we make it 6 and negative 10 their sum will be negative 4, and their product is negative 60. So that works. So you could literally say that this is equal to b plus 6, times b, minus 10. b plus the a, plus b minus the b. And let me be very careful here. This b over here, I want to make it very clear, I just used this b here to say, look, we're looking for two numbers that add up to this second term It's a different b. I could have said x plus y is equal to negative 4, and x times y is equal to negative 60. In fact, let me do it that way just so we don't get confused. So we could write x plus y is equal to negative 4. And then we have x times y is equal to negative 60. So we have b plus 6, times b plus y. x is 6, y is negative 10. And that is equal to 0. Let's just solve this right here. And then we'll go back and show you. You could also factor this by grouping. But just from this, we know that either one of these is equal to zero. Either b plus 6 is equal to 0, or b minus 10 is equal to 0. If we subtract 6 from both sides of this equation, we get b is equal to negative 6. you get b is equal to 10. And those are our two solutions. You could put them back in and verify that they satisfy our constraints. Now the other way that you could solve this, and we're going to get exact same answer. Is you could just break up this negative 4b into its constituents. So you could have broken this up into 0 is equal to b squared. And then you could have broken it up into plus 6b, minus 10b, minus 60. And then factor it by grouping. Group these first two terms. Group these second two terms. Just going to add them together. The first one you could factor out a b. So you have b times b, plus 6. The second one you can factor out a negative 10. So minus 10 times b, plus 6. All that's equal to 0. And now you can factor out a b plus 6. So if you factor out a b plus 6 here,"
    },
    {
        "Q": "i didnt get what it meant in 5:00 could someone explain it to me??",
        "A": "He is saying that the b s are not the same he just used b and a instead of x and y",
        "video_name": "STcsaKuW-24",
        "timestamps": [
            300
        ],
        "3min_transcript": "So we need to factor b squared, minus 4b, minus 60. So what we want to do, we want to find two numbers whose sum is negative 4 and whose product is negative 60. Now, given that the product is negative, we know there are different signs. And this tells us that their absolute values are going to be four apart. That one is going to be four less than the others. So you could look at the products of the factors of 60. 1 and 60 are too far apart. Even if you made one of the negative, you would either get positive 59 as the sum or negative 59 as the sum. 2 and 30, still too far apart. 3 and 20, still too far apart. If you had made one negative you'd Then you could have 4 and 15, still too far apart. If you made one of them negative, their sum would be either negative 11 or positive 11. Then you have 5 and a 12, still seems too far apart One of them is negative, then you either have their sum being positive 7 or negative 7. Then you have 6 and 10. Now this looks interesting. They are four apart. So if we make-- and we want the larger absolute magnitude number to be negative so that their sum is negative. So if we make it 6 and negative 10 their sum will be negative 4, and their product is negative 60. So that works. So you could literally say that this is equal to b plus 6, times b, minus 10. b plus the a, plus b minus the b. And let me be very careful here. This b over here, I want to make it very clear, I just used this b here to say, look, we're looking for two numbers that add up to this second term It's a different b. I could have said x plus y is equal to negative 4, and x times y is equal to negative 60. In fact, let me do it that way just so we don't get confused. So we could write x plus y is equal to negative 4. And then we have x times y is equal to negative 60. So we have b plus 6, times b plus y. x is 6, y is negative 10. And that is equal to 0. Let's just solve this right here. And then we'll go back and show you. You could also factor this by grouping. But just from this, we know that either one of these is equal to zero. Either b plus 6 is equal to 0, or b minus 10 is equal to 0. If we subtract 6 from both sides of this equation, we get b is equal to negative 6."
    },
    {
        "Q": "i notice at 0:29 sal factor the number with -1 ? why he do that? can someone explain me?",
        "A": "If you factor a negative number, -1 is a factor. and then you can focus on factoring the rest which is now positive. Also, -1 is a perfect cube, so you can do the cuberoot (-1) = -1.",
        "video_name": "drhoIgAhlQM",
        "timestamps": [
            29
        ],
        "3min_transcript": "We're are asked to find the cube root of negative 343. Or another way to think about it is some number that when I multiply it by itself three times, I'm going to get negative 343. Or another way to view it-- this is the same thing as negative 343 to the 1/3 power. And the best way to do this is to really just try to factor this out. So the first thing that we could do-- so let me just factor negative 343. So the first thing I'd like to do is just factor out the negative 1. So this is the same thing as negative 1 times 343. And let's think about this. Is this divisible by 2? No. Is it divisible by 3? Let's see-- the digits do not add up to a multiple of 3. They add up to 10. So it's not divisible by 3. Not divisible by 4 since it's odd. Not divisible by 5 because it doesn't end with a 5 or a 0. It's not divisible by 6, because it's not divisible by 2 or 3. Is it divisible by 7? Let's check this out. So 7 goes into 343-- 7 goes into 34 four times. 34 minus 28 is going to be 6. Bring down the 3. 7 goes into 63 exactly nine times. So then we end up-- so 9 times 7 is 63, no remainder. So this is going to be 7 times 49. And we know that 49 is the same thing as 7 times 7. So how can we rewrite this? This is the same thing as taking the cube root of negative 1 times 7 times 7 times 7, which is the same thing as taking the cube root of negative 1 times the cube root of 7 times 7 times 7. Now, what's the cube root of negative 1? Well, negative 1 times itself three times is negative 1. You could verify it. Negative 1 times negative 1 times negative 1 is indeed negative 1. This becomes positive 1. Multiply by negative 1 again, you get negative 1. So this is negative 1. And then this over here, the cube root of 7 times 7 times 7-- well, that's just going to be 7. 7 multiplied by itself three times gives us 7 times 7 times 7 or 343. So it's going to be negative 1 times 7, which is the same thing as negative 7. So our answer is negative 7. And we're done."
    },
    {
        "Q": "At 0:31 he says something about finding what 0 is equal to.",
        "A": "He s not finding what zero equals, he said setting your to zero to solve for x. It is mathematically equivalent to frame the equation as 0 = x as it is to state x = 0.",
        "video_name": "6agzj3A9IgA",
        "timestamps": [
            31
        ],
        "3min_transcript": "Use completing the square to find the roots of the quadratic equation right here. And when anyone talks about roots, this just means find the x's where y is equal to 0. That's what a root is. A root is an x value that will make this quadratic function equal 0, that will make y equal 0. So to find the x's, let's just make y equal 0 and then solve for x. So we get 0 is equal to 4x squared plus 40x, plus 280. Now, the first step that we might want to do, just because it looks like all three of these terms are divisible by 4, is just divide both sides of this equation by 4. That'll make our math a little bit simpler. So let's just divide everything by 4 here. If we just divide everything by 4, we get 0 is equal to x squared plus 10x, plus-- 280 divided by 4 is 70-- plus 70. me write that 70 a little bit further out, and you'll see why I did that in a second. So let me just write a plus 70 over here, just to have kind of an awkward space here. And you'll see what I'm about to do with this space, that has everything to do with completing the square. So they say use completing the square, which means, turn this, if you can, into a perfect square. Turn at least part of this expression into a perfect square, and then we can use that to actually solve for x. So how do we turn this into a perfect square? Well, we have a 10x here. And we know that we can turn this into a perfect square trinomial if we take 1/2 of the 10, which is 5, and then we square that. So 1/2 of 10 is 5, you square it, you add a 25. Now, you can't just willy-nilly add a 25 to one side of the equation without doing something to the other, or without just subtracting the 25 right here. Think about it, I have not changed the equation. I've added 25 and I've subtracted 25. So I've added nothing to the right-hand side. I could add a billion and subtract a billion and not change the equation. So I have not changed the equation at all right here. But what I have done is I've made it possible to express these three terms as a perfect square. That right there, 2 times 5 is 10. 5 squared is 25. So that is x plus 5 squared. And if you don't believe me, multiply it out. You're going to have an x squared plus 5x, plus 5x, which will give you 10x, plus 5 squared, which is 25. So those first three terms become that, and then the second two terms, right there, you just add them. Let's see, negative 25 plus 70. Let's see, negative 20 plus 70 would be positive 50, and then you have another 5, so it's plus 45."
    },
    {
        "Q": "1:53 what is the meaning of willy-nilly? without-worry?",
        "A": "When I ve heard this used, it has always meant to do something in a random manner, haphazardly, without any method or planning, in any way you please without thinking.",
        "video_name": "6agzj3A9IgA",
        "timestamps": [
            113
        ],
        "3min_transcript": "Use completing the square to find the roots of the quadratic equation right here. And when anyone talks about roots, this just means find the x's where y is equal to 0. That's what a root is. A root is an x value that will make this quadratic function equal 0, that will make y equal 0. So to find the x's, let's just make y equal 0 and then solve for x. So we get 0 is equal to 4x squared plus 40x, plus 280. Now, the first step that we might want to do, just because it looks like all three of these terms are divisible by 4, is just divide both sides of this equation by 4. That'll make our math a little bit simpler. So let's just divide everything by 4 here. If we just divide everything by 4, we get 0 is equal to x squared plus 10x, plus-- 280 divided by 4 is 70-- plus 70. me write that 70 a little bit further out, and you'll see why I did that in a second. So let me just write a plus 70 over here, just to have kind of an awkward space here. And you'll see what I'm about to do with this space, that has everything to do with completing the square. So they say use completing the square, which means, turn this, if you can, into a perfect square. Turn at least part of this expression into a perfect square, and then we can use that to actually solve for x. So how do we turn this into a perfect square? Well, we have a 10x here. And we know that we can turn this into a perfect square trinomial if we take 1/2 of the 10, which is 5, and then we square that. So 1/2 of 10 is 5, you square it, you add a 25. Now, you can't just willy-nilly add a 25 to one side of the equation without doing something to the other, or without just subtracting the 25 right here. Think about it, I have not changed the equation. I've added 25 and I've subtracted 25. So I've added nothing to the right-hand side. I could add a billion and subtract a billion and not change the equation. So I have not changed the equation at all right here. But what I have done is I've made it possible to express these three terms as a perfect square. That right there, 2 times 5 is 10. 5 squared is 25. So that is x plus 5 squared. And if you don't believe me, multiply it out. You're going to have an x squared plus 5x, plus 5x, which will give you 10x, plus 5 squared, which is 25. So those first three terms become that, and then the second two terms, right there, you just add them. Let's see, negative 25 plus 70. Let's see, negative 20 plus 70 would be positive 50, and then you have another 5, so it's plus 45."
    },
    {
        "Q": "At around 5:15, Sal says that the t-distribution has fatter tails because the small sample size causes an underestimation of the standard deviation of the sampling distribution of the sample mean.\n\nWouldn't a smaller sample size make you overestimate the standard deviation? And that is what leads to fatter tails? Could someone clarify this for me? Thank you!",
        "A": "Too large an SEM would give too large an interval; too small too small. Fatter tails means a larger percentage of the area (probability) at higher SD. So you would need more SDs to get the same probability. So it looks to me like the fatter tails would tend to compensate for underestimating SEM. Why do you think that smaller sample size would overestimate the SEM? It gives a larger SEM, yes (the smaller the sample, the less accurate the sample mean is likely to be). Is it enough larger?",
        "video_name": "K4KDLWENXm0",
        "timestamps": [
            315
        ],
        "3min_transcript": "we're going to tweak the sampling distribution. We're not going to assume it's a normal distribution because this is a bad estimate. We're going to assume that it's something called a t-distribution. And a t-distribution is essentially, the best way to think about is it's almost engineered so it gives a better estimate of your confidence intervals and all of that when you do have a small sample size. It looks very similar to a normal distribution. It has some mean, so this is your mean of your sampling distribution still. But it also has fatter tails. And the way I think about why it has fatter tails is when you make an assumption that this is a standard deviation for-- let me take one more step. So normally what we do is we find the estimate of the true standard deviation, and then we say that the standard true standard deviation of our population divided by the square root of n. In this case, n is equal to 7. And then we say OK, we never know the true standard, or we seldom know-- sometimes you do know-- we seldom know the true standard deviation. So if we don't know that the best thing we can put in there is our sample standard deviation. And this right here, this is the whole reason why we don't say that this is just a 95 probability interval. This is the whole reason why we call it a confidence interval because we're making some assumptions. This thing is going to change from sample to sample. And in particular, this is going to be a particularly bad estimate when we have a small sample size, a size less than 30. So when you are estimating the standard deviation where you don't know it, you're estimating it with your sample standard deviation, and your sample size is small, and deviation of your sampling distribution, you don't assume your sampling distribution is a normal distribution. You assume it has fatter tails. And it has fatter tails because you're essentially underestimating-- you're underestimating the standard deviation over here. Anyway, with all of that said, let's just actually go through this problem. So we need to think about a 95% confidence interval around this mean right over here. So a 95% confidence interval, if this was a normal distribution you would just look it up in a Z-table. But it's not, this is a t-distribution. We're looking for a 95% confidence interval. So some interval around the mean that encapsulates 95% of the area. For a t-distribution you use t-table, and I have a t-table ahead of time right over here. And what you want to do is use the two-sided row for what"
    },
    {
        "Q": "at 1:37, how could 10 hundreths and 7 hundreths make sense",
        "A": "10 hundredths + 7 hundredths = 0.17 = 1 tenth + 7 hundredths. 1 hundredth = 0.01",
        "video_name": "qSPwUDmpnJ4",
        "timestamps": [
            97
        ],
        "3min_transcript": "- [Voiceover] Let's say that I had the number zero point one seven. How could I say this number? I said it one way, I said zero point one seven, but what are other ways that I could say it, especially if I wanted to express it in terms of tenths or hundredths or other places? And like always, try to pause the video and try think about it on your own. Alright, so there's actually a couple of ways that we could say this number. One is just to say zero point one seven. Other ways are to say look, I have a one in the tenths place, so that's going to be one tenth, one tenth and one tenth and I have a seven in the hundredths place, so this is a seven right over here in the hundredths place, so I can say one tenth and seven hundredths. Hun- Hundredths. And there you go. Now another to think about it is just say the whole thing in terms of hundredths. So a tenth is how many hundredths? Well a tenth is the same thing as 10 hundredths, so you could say, you could say instead of a tenth, you could say this is 10 hundredths, and the way I'm writing it right now, very few people would actually do it this way. 10 hundredths and and seven hundredths. And seven hundredths. Well not I could just add these hundredths, if I have 10 hundredths and I have another seven hundredths, that's going to be 17 hundredths. So I could just write this down as 17 hundredths. Hundredths. And to make that intuition of how we could just call this 17 hundredths instead of just calling it one tenth and seven hundredths, let's actually count by hundredths. So that is one hundredth, and actually, let me just go straight to nine hundredths. So I skipped a bunch right over here. And what would be the next, how would I say 10 hundredths? Well 10 hundredths, let me write it this way, 10 hundredths is the same thing as one tenth. So if we go from nine hundredths, the next, if I'm counting by hundredths, the next one's going to be 10 hundredths. Now once again, 10 hundredths is the same thing as one tenth, just the same way that 10 ones is the same thing as one 10. I hope that doesn't confuse you, but we could keep counting. 10 hundredths, 11 hundredths, 12 hundredths, 13 hundredths, 14 hundredths, 15 hundredths, 16 hundredths, and then finally 17 hundredths. So hopefully that gives you a little intuition for why we can call this number, instead of just calling it zero point one seven, or one tenth and seven hundredths, we could call this 17 hundredths."
    },
    {
        "Q": "at 5:41 Sal mentions a \"rule of thumb\". I'm new to statistics, but I can't seem to find it in my books or these videos. What is the \"Rule of Thumb\" and is there a video I'm missing?",
        "A": "Rule of thumb is just an expression, it means a good generalization or a simple way to remember something. There is no single Rule of Thumb. :P",
        "video_name": "5ABpqVSx33I",
        "timestamps": [
            341
        ],
        "3min_transcript": "distributed if our sample size is greater than 30. Even this approximation will be approximately normally distributed. Now, if your sample size is less than 30, especially if it's a good bit less than 30, all of a sudden this expression will not be normally distributed. So let me re-write the expression over here. Sample mean minus the mean of your sampling distribution of the sample mean divided by your sample standard deviation over the square root of your sample size. We just said if this thing is well over 30, or at least 30, then this value right here, this statistic, is going to be normally distributed. If it's not, if this is small, then this is going to have a T-distribution. And then you're going to do the exact same thing you did normal distribution, so this example it was normal. All of Z's are normally distributed. Over here in a T-distribution, and this will actually be a normalized T-distribution right here because we subtracted out the mean. So in a normalized T-distribution, you're going to have a mean of 0. And what you're going to do is you want to figure out the probability of getting a T-value at least this extreme. So this is your T-value you would get, and then you essentially figure out the area under the curve right over there. So a very easy rule of thumb is calculate this quantity either way. Calculate this quantity either way. If you will have more than 30 samples, if your sample size is more than 30, your sample standard deviation is going to be a good approximator for your population standard deviation. And so this whole thing is going to be approximately figure out the probability of getting a result at least that extreme. If your sample size is small, then this statistic, this quantity, is going to have a T-distribution, and then you're going to have to use a T-table to figure out the probability of getting a T-value at least this extreme. And we're going to see this in an example a couple of videos from now. Anyway, hopefully that helped clarify some things in your head about when to use a Z-statistic or when to use a T-statistic."
    },
    {
        "Q": "At 5:54, how did Sal already find the b for Slope-Intercept Form?",
        "A": "He distributed the -2/3(x+3) in the point-slope form.",
        "video_name": "-6Fu2T_RSGM",
        "timestamps": [
            354
        ],
        "3min_transcript": "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 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."
    },
    {
        "Q": "7:01 if -2/3X was positive, would the answer be 2/3x-y=4? or 2/3x+y=4?",
        "A": "yes.",
        "video_name": "-6Fu2T_RSGM",
        "timestamps": [
            421
        ],
        "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 7:00, can there be a standard form without an A?",
        "A": "There will always be an A, it s just a coefficient. It can be 0 though, in which case the term does not exist.",
        "video_name": "-6Fu2T_RSGM",
        "timestamps": [
            420
        ],
        "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 2:24 how did you get the slope? that is what confused me because when i tried it it didnt work",
        "A": "0-6 over 6+3 = -6 over 9 Simplify -2 over 3",
        "video_name": "-6Fu2T_RSGM",
        "timestamps": [
            144
        ],
        "3min_transcript": "A line passes through the points negative 3, 6 and 6, 0. Find the equation of this line in point slope form, slope intercept form, standard form. And the way to think about these, these are just three different ways of writing the same equation. So if you give me one of them, we can manipulate it to get any of the other ones. But just so you know what these are, point slope form, let's say the point x1, y1 are, let's say that that is a point on the line. And when someone puts this little subscript here, so if they just write an x, that means we're talking about a variable that can take on any value. If someone writes x with a subscript 1 and a y with a subscript 1, that's like saying a particular value x and a particular value of y, or a particular coordinate. And you'll see that when we do the example. But point slope form says that, look, if I know a particular point, and if I know the slope of the line, then putting that line in point slope form would be y minus y1 is equal to m times x minus x1. point negative 3 comma 6 is on the line, then we'd say y minus 6 is equal to m times x minus negative 3, so it'll end up becoming x plus 3. So this is a particular x, and a particular y. It could be a negative 3 and 6. So that's point slope form. Slope intercept form is y is equal to mx plus b, where once again m is the slope, b is the y-intercept-- where does the line intersect the y-axis-- what value does y take on when x is 0? And then standard form is the form ax plus by is equal to c, where these are just two numbers, essentially. They really don't have any interpretation directly on the graph. So let's do this, let's figure out all of these forms. So the first thing we want to do is figure out the slope. Once we figure out the slope, then point slope form is So, just to remind ourselves, slope, which is equal to m, which is going to be equal to the change in y over the change in x. Now what is the change in y? If we view this as our end point, if we imagine that we are going from here to that point, what is the change in y? Well, we have our end point, which is 0, y ends up at the 0, and y was at 6. So, our finishing y point is 0, our starting y point is 6. What was our finishing x point, or x-coordinate? Our finishing x-coordinate was 6. Let me make this very clear, I don't want to confuse you. So this 0, we have that 0, that is that 0 right there. And then we have this 6, which was our starting y point, that is that 6 right there."
    },
    {
        "Q": "At 6:48 where did you get the 4 from?",
        "A": "It is simply the same equation as the one underneath slope-intercept form, because -2+6=4.",
        "video_name": "-6Fu2T_RSGM",
        "timestamps": [
            408
        ],
        "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": "3:00 of the video; I thought 4 to the exponent of 3 times 5 to the exponent of 3 would give you a different answer then then 4x5 to the exponent of 3. So I do not understand the logic of this.",
        "A": "(4x5) to the third is (4x5)x(4x5)x(4x5). Because it is multiplication, we can move the numbers around, getting 4x4x4x5x5x5. 4x4x4 is 4 to the third, and 5x5x5 is 5 to the third. So, (4x5)^3 = 4^3 x 5^3.",
        "video_name": "rEtuPhl6930",
        "timestamps": [
            180
        ],
        "3min_transcript": "When I raise something to the fifth power, that's just like saying 2 to the tenth times 2 to the tenth times 2 to the tenth times 2 to the tenth times 2 to the tenth, right? All I did is I took 2 to the tenth and I multiplied it by itself five times. That's the fifth power. Well, we know from this rule up here that we can add these exponents because they're all the same base. So if we add 10 plus 10 plus 10 plus 10 plus 10, what do we get? Right, we get 2 to the fiftieth power. So essentially, what did we do here? All we did is we multiplied 10 times 5 to get 50. So that's our third exponent rule, that when I raise an exponent to a power and then I raise that whole expression to another power, I can multiply those two exponents. So let me give you another example. again, all I do is I multiply the 7 and the negative 9, and I get 3 to the minus 63. So, you see, it works just as easily with negative numbers. So now, I'm going to teach you one final exponent property. Let's say I have 2 times 9, and I raise that whole thing to the hundredth power. It turns out of this is equal to 2 to the hundredth power times 9 to the hundredth power. Now let's make sure that that makes sense. Let's do it with a smaller example. What if it was 4 times 5 to the third power? times 4 times 5, right, which is the same thing as 4 times 4 times 4 times 5 times 5 times 5, right? I just switched the order in which I'm multiplying, which you can do with multiplication. Well, 4 times 4 times 4, well, that's just equal to 4 to the third. And 5 times 5 times 5 is equal to 5 to the third. Hope that gives you a good intuition of why this property here is true. And actually, when I had first learned exponent rules, I would always forget the rules, and I would always do this proof myself, or the other proofs. And a proof is just an explanation of why the rule works, just to make sure that I was doing it right. So given everything that we've learned to now-- actually, let me review all of the rules again."
    },
    {
        "Q": "At the end of the video, when he gives the problem, he says at 5:45 that the problem can be written as (3^2 * (3^2)^8)^-2.... but, it couldn't be (3^2 * 3 * 3^8)^-2? It would be a different answer, but it is following the rules in the same way...how could this be? Someone please?",
        "A": "In (3^2 * (3^2)^8)^-2 , the second 3^2 is the expanded form of 9... thus 9 becomes 3^2 (3^2)^8 is 9^8 3 * 3^8 is 3^1+8= 3^9 It is pretty clear here that 9^8 and 3^9 is not the same.. And I don t see what rule you re saying you re following..could you explain that part?",
        "video_name": "rEtuPhl6930",
        "timestamps": [
            345
        ],
        "3min_transcript": "well, then I can add the exponents, 2 to the tenth. If I have 2 the seventh over 2 the third, well, here I subtract the exponents, and I get 2 to the fourth. If I have 2 to the seventh to the third power, well, here I multiplied the exponents. That gives you 2 to the 21. And if I had 2 times 7 to the third power, well, that equals 2 to the third times 7 to the third. Now, let's use all of these rules we've learned to actually try to do some, what I would call, composite problems that involve you using multiple rules at the same time. And a good composite problem was that problem that I had introduced you to at the end of that last seminar. and all of that I'm going to raise to the negative 2 power. So what can I do here? Well, 3 and 9 are two separate bases, but 9 can actually be expressed as an exponent of 3, right? 9 is the same thing as 3 squared, so let's rewrite 9 like that. That's equivalent to 3 squared times-- 9 is the same thing as 3 squared to the eighth power, and then all of that to the negative 2 power, right? All I did is I replaced 9 with 3 squared because we know 3 times 3 is 9. Well, now we can use the multiplication rule on this to simplify it. which is 16, and all of that to the negative 2. Now, we can use the first rule. We have the same base, so we can add the exponents, and we're multiplying them, so that equals 3 to the eighteen power, right, 2 plus 16, and all that to the negative 2. And now we're almost done. We can once again use this multiplication rule, and we could say 3-- this is equal to 3 to the eighteenth times negative 2, so that's 3 to the minus 36. So this problem might have seemed pretty daunting at first, but there aren't that many rules, and all you have to do is keep seeing, oh, wow, that little part of the problem, I can simplify it. Then you simplify it, and you'll see that you can keep using rules until you get to a much simpler answer. And actually the Level 1 problems don't even involve"
    },
    {
        "Q": "At 2:56, why isn't (2x9)^100 =18^100? why is it 2^100 x 9^100?",
        "A": "They are the same thing. See the example at 3:00.",
        "video_name": "rEtuPhl6930",
        "timestamps": [
            176
        ],
        "3min_transcript": "When I raise something to the fifth power, that's just like saying 2 to the tenth times 2 to the tenth times 2 to the tenth times 2 to the tenth times 2 to the tenth, right? All I did is I took 2 to the tenth and I multiplied it by itself five times. That's the fifth power. Well, we know from this rule up here that we can add these exponents because they're all the same base. So if we add 10 plus 10 plus 10 plus 10 plus 10, what do we get? Right, we get 2 to the fiftieth power. So essentially, what did we do here? All we did is we multiplied 10 times 5 to get 50. So that's our third exponent rule, that when I raise an exponent to a power and then I raise that whole expression to another power, I can multiply those two exponents. So let me give you another example. again, all I do is I multiply the 7 and the negative 9, and I get 3 to the minus 63. So, you see, it works just as easily with negative numbers. So now, I'm going to teach you one final exponent property. Let's say I have 2 times 9, and I raise that whole thing to the hundredth power. It turns out of this is equal to 2 to the hundredth power times 9 to the hundredth power. Now let's make sure that that makes sense. Let's do it with a smaller example. What if it was 4 times 5 to the third power? times 4 times 5, right, which is the same thing as 4 times 4 times 4 times 5 times 5 times 5, right? I just switched the order in which I'm multiplying, which you can do with multiplication. Well, 4 times 4 times 4, well, that's just equal to 4 to the third. And 5 times 5 times 5 is equal to 5 to the third. Hope that gives you a good intuition of why this property here is true. And actually, when I had first learned exponent rules, I would always forget the rules, and I would always do this proof myself, or the other proofs. And a proof is just an explanation of why the rule works, just to make sure that I was doing it right. So given everything that we've learned to now-- actually, let me review all of the rules again."
    },
    {
        "Q": "2:59 of the this video. Doesn't this break the \"order of operation\" because you do what is in the parenthesis first then work your way out?\n\nalso in ths video and the previous video before this one, the subtitle at the bottom seem to flash a lot or something.",
        "A": "Because they are all multiplication, we can rearrange the numbers any way we want, using the distributive property.",
        "video_name": "rEtuPhl6930",
        "timestamps": [
            179
        ],
        "3min_transcript": "When I raise something to the fifth power, that's just like saying 2 to the tenth times 2 to the tenth times 2 to the tenth times 2 to the tenth times 2 to the tenth, right? All I did is I took 2 to the tenth and I multiplied it by itself five times. That's the fifth power. Well, we know from this rule up here that we can add these exponents because they're all the same base. So if we add 10 plus 10 plus 10 plus 10 plus 10, what do we get? Right, we get 2 to the fiftieth power. So essentially, what did we do here? All we did is we multiplied 10 times 5 to get 50. So that's our third exponent rule, that when I raise an exponent to a power and then I raise that whole expression to another power, I can multiply those two exponents. So let me give you another example. again, all I do is I multiply the 7 and the negative 9, and I get 3 to the minus 63. So, you see, it works just as easily with negative numbers. So now, I'm going to teach you one final exponent property. Let's say I have 2 times 9, and I raise that whole thing to the hundredth power. It turns out of this is equal to 2 to the hundredth power times 9 to the hundredth power. Now let's make sure that that makes sense. Let's do it with a smaller example. What if it was 4 times 5 to the third power? times 4 times 5, right, which is the same thing as 4 times 4 times 4 times 5 times 5 times 5, right? I just switched the order in which I'm multiplying, which you can do with multiplication. Well, 4 times 4 times 4, well, that's just equal to 4 to the third. And 5 times 5 times 5 is equal to 5 to the third. Hope that gives you a good intuition of why this property here is true. And actually, when I had first learned exponent rules, I would always forget the rules, and I would always do this proof myself, or the other proofs. And a proof is just an explanation of why the rule works, just to make sure that I was doing it right. So given everything that we've learned to now-- actually, let me review all of the rules again."
    },
    {
        "Q": "at about 6:10, when Sal has (3x2x(3x2)to the 8th), why does he only multiply the 8th to the second 3x2, and not both, because they are all within the larger parentheses?",
        "A": "First of all it is not multiplication, but power (3^2*(3^2)^8) Power always applies to item it is after. 3^2 means 3 in power 2. The second power applies to 3. The power of 8 is after the closing parentheses, so it only applies to items in parentheses. (3^2)^8 means 3 to the power of 2 and all of that is to the power of 8.",
        "video_name": "rEtuPhl6930",
        "timestamps": [
            370
        ],
        "3min_transcript": "well, then I can add the exponents, 2 to the tenth. If I have 2 the seventh over 2 the third, well, here I subtract the exponents, and I get 2 to the fourth. If I have 2 to the seventh to the third power, well, here I multiplied the exponents. That gives you 2 to the 21. And if I had 2 times 7 to the third power, well, that equals 2 to the third times 7 to the third. Now, let's use all of these rules we've learned to actually try to do some, what I would call, composite problems that involve you using multiple rules at the same time. And a good composite problem was that problem that I had introduced you to at the end of that last seminar. and all of that I'm going to raise to the negative 2 power. So what can I do here? Well, 3 and 9 are two separate bases, but 9 can actually be expressed as an exponent of 3, right? 9 is the same thing as 3 squared, so let's rewrite 9 like that. That's equivalent to 3 squared times-- 9 is the same thing as 3 squared to the eighth power, and then all of that to the negative 2 power, right? All I did is I replaced 9 with 3 squared because we know 3 times 3 is 9. Well, now we can use the multiplication rule on this to simplify it. which is 16, and all of that to the negative 2. Now, we can use the first rule. We have the same base, so we can add the exponents, and we're multiplying them, so that equals 3 to the eighteen power, right, 2 plus 16, and all that to the negative 2. And now we're almost done. We can once again use this multiplication rule, and we could say 3-- this is equal to 3 to the eighteenth times negative 2, so that's 3 to the minus 36. So this problem might have seemed pretty daunting at first, but there aren't that many rules, and all you have to do is keep seeing, oh, wow, that little part of the problem, I can simplify it. Then you simplify it, and you'll see that you can keep using rules until you get to a much simpler answer. And actually the Level 1 problems don't even involve"
    },
    {
        "Q": "At 5:32 Sal expresses 9^8 as 3^16 to be able to calculate the product of two exponents with the same base. It looks like you could instead express 3^2 as 9^1 and come up with a final answer of 9^-18 instead of 3^-36. The numbers are equivalent but is one answer better by convention?",
        "A": "You are absolutely correct. It just depends if you want your base to be a prime number. If that is your goal you will always come up with the same answer, which has some elegance.",
        "video_name": "rEtuPhl6930",
        "timestamps": [
            332
        ],
        "3min_transcript": "well, then I can add the exponents, 2 to the tenth. If I have 2 the seventh over 2 the third, well, here I subtract the exponents, and I get 2 to the fourth. If I have 2 to the seventh to the third power, well, here I multiplied the exponents. That gives you 2 to the 21. And if I had 2 times 7 to the third power, well, that equals 2 to the third times 7 to the third. Now, let's use all of these rules we've learned to actually try to do some, what I would call, composite problems that involve you using multiple rules at the same time. And a good composite problem was that problem that I had introduced you to at the end of that last seminar. and all of that I'm going to raise to the negative 2 power. So what can I do here? Well, 3 and 9 are two separate bases, but 9 can actually be expressed as an exponent of 3, right? 9 is the same thing as 3 squared, so let's rewrite 9 like that. That's equivalent to 3 squared times-- 9 is the same thing as 3 squared to the eighth power, and then all of that to the negative 2 power, right? All I did is I replaced 9 with 3 squared because we know 3 times 3 is 9. Well, now we can use the multiplication rule on this to simplify it. which is 16, and all of that to the negative 2. Now, we can use the first rule. We have the same base, so we can add the exponents, and we're multiplying them, so that equals 3 to the eighteen power, right, 2 plus 16, and all that to the negative 2. And now we're almost done. We can once again use this multiplication rule, and we could say 3-- this is equal to 3 to the eighteenth times negative 2, so that's 3 to the minus 36. So this problem might have seemed pretty daunting at first, but there aren't that many rules, and all you have to do is keep seeing, oh, wow, that little part of the problem, I can simplify it. Then you simplify it, and you'll see that you can keep using rules until you get to a much simpler answer. And actually the Level 1 problems don't even involve"
    },
    {
        "Q": "What does si mean and theda.2:20",
        "A": "Psi and theta are Greek letters that usually denote angles.",
        "video_name": "MyzGVbCHh5M",
        "timestamps": [
            140
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 2:01 what does si has mean?",
        "A": "Psi (not si) is a letter in Greek alphabet (\u00ce\u00a8). Mathematicians use Greek letters to write angles. They often use other letters : \u00ce\u00b1(alpha), \u00ce\u00b2(beta), \u00ce\u00b3(gamma), \u00ce\u00b8(theta)",
        "video_name": "MyzGVbCHh5M",
        "timestamps": [
            121
        ],
        "3min_transcript": ""
    },
    {
        "Q": "how does he know psi 1 equal 1/2 theta 1 at 8:15 in the vidio? i didn't undrstand the proof. i don't think i saw a proof. like wise for scy 2 and thaita 2 at 8:20",
        "A": "Sal wasnt really proving anything in particular. All he wanted to show was that the centre neednt be within the arc being subtended. And he just did that using stuff that he had taught before",
        "video_name": "MyzGVbCHh5M",
        "timestamps": [
            495,
            500
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 0:32 what does transversal mean?",
        "A": "A transversal is a line that intersects two other lines. A transversal can be useful in determining whether the two lines it intersects are parallel.",
        "video_name": "LhrGS4-Dd9I",
        "timestamps": [
            32
        ],
        "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": "At 2:12, if both are increasing then don't the negatives cancel out and become positive?",
        "A": "You are comparing the slopes, not multiplying the slopes. Both lines are negative, so both lines slant down from left to right. The slope of line F is decreasing faster because its slope is more negative than the slope of line G. Hope this helps.",
        "video_name": "fZO-JylMFqY",
        "timestamps": [
            132
        ],
        "3min_transcript": "Two functions, f and g, are described below. Which of these statements about f and g is true? So they defined function f as kind of a traditional linear equation right over here. And this right over here is g. So this right over here is g of x. And that also looks like a linear function. We see it's a kind of a downward sloping line. So let's look at our choices and see which of these are true. f and g are both increasing, and f is increasing faster than g. Well, when I look at g-- Well, first of all, g is definitely decreasing. So we already know that that's false. And f is also decreasing. We see here it has a negative slope. Every time we move forward 3 in the x direction, we're going to move down 7 in the vertical direction. So neither of these are increasing so that's definitely not right. f and g are both increasing. Well, that's definitely not right. So we know that both f and g are decreasing. So this first choice says they're both decreasing, and g So let's see what the slope on g is. So the slope on g is every time we move 1 in the x direction, positive 1 in the x direction, we move down 2 in the y direction. So for g of x, if we were to write our change in y over our change in x-- which is our slope-- our change in y over change in x, when we move one in the x direction, positive 1 in the x direction, we move down 2 in the y direction. So our change in y over change in x is negative 2. So g has a slope of negative 2. f has a slope of negative 7/3. Negative 7/3 is the same thing as negative 2 and 1/3. So f's slope is more negative. So it is decreasing faster. than g. So this is not right. And then we have this choice-- f and g are both decreasing, and f is decreasing faster than g. This is right, right over here. We have this last choice-- g is increasing but f is decreasing. We know that's not true. g is actually decreasing."
    },
    {
        "Q": "At somewhere around 8:00 (it's hard to tell; for some reason this video doesn't show me the time) Sal adds a squared and b squared. I don't understand how he was able to do this. Wasn't he talking about two different triangles?",
        "A": "both of the triangles add up to make the larger one; he s trying to get a math statement that applies to the larger one.",
        "video_name": "LrS5_l-gk94",
        "timestamps": [
            480
        ],
        "3min_transcript": "are going to be similar So, we can say triangle BDC, we went from pink to right, to not labeled So, triangle BDC, triangle BDC is similar to triangle, now we're gonna look at the larger triangle, now we're gonna start the pink angle B, now we go to the right angle CA BCA >From pink angle to right angle to non-labeled angle, at least from the point of view here before the blue Now we setup some type of relationship here We can say that the ratio on the smaller triangle BC, side BC over BA BC over BA Once again we're taking the hypotenuses of both of them So, BC over BA is going to be equal to BD Here's another color, BD, so this one of the legs BD over BC, I'm just taking the corresponding vertices, over BC And once again, we know, BC is the same as lowercase \"b,\" BC is lowercase \"b \" BA is lower case \"c \" And then BD we defined as lower case \"e \" So, this is lowercase \"e \" We can cross multiply here and we b times b Which and I mentioned this in many videos cross multiplying both sides by both denominators b times b is equal to ce And now we can do something kind of interesting We can add these two statements down here Let me rewrite this statement down here So, b squared is equal to ce So, if we add the left hand sides, we get a squared plus b squared, a squared plus b squared is equal to cd, is eqaual to cd And then we have a ce in both of these terms so we can factor it out So, this is gonna be equal to, we can factor out the c, it's gonna be c times d plus e c times d plus e, and close the parenthesis Now what is d plus e? d is this length e is this length So, d plus e is actually gonna be c as well So, this is gonna be c So, if c times c is the same thing as c squared So, now we have an interesting relationship, we have that a squared plus b squared is equal to c squared Let me rewrite that a squared, I'll do that- well let me just arbitrary new color I deleted that by accident, so let me rewrite it"
    },
    {
        "Q": "how does he figure out for what n means at 6:20?? i am very confused too!!!",
        "A": "Since he is trying to get n by itself, he multiplies 8/36 with 36/8. They cancel out, and you are left with n. What you do to one side you do to the other. 10 x 36/8=360/8. There is one n left, so n=360/8. That s your answer.",
        "video_name": "GO5ajwbFqVQ",
        "timestamps": [
            380
        ],
        "3min_transcript": "Or to figure out what that times what is, you divide 360 divided by 8. So we could divide, and this is a little bit of algebra here, we're dividing both sides of the equation by 8. And we're getting n is equal to 360 divided by 8. You could do that without thinking in strict algebraic terms. You could say 8 times what is 360. Well 8 times 360/8. If I write 8 times question mark is equal to 360, well, question mark could definitely be 360/8. If I multiply these out, this guy and that guy cancel out, and it's definitely 360. And that's why it's 360/8. But now we want to actually divide this to actually get our right answer, or a simplified answer. 8 goes into 360, 8 goes into 36 4 times, 4 times 8 is 32. You have a remainder of 4. Bring down the 0. 8 goes into 40 5 times. 5 times 8 is 40. And you're done. Once again, we got n is equal to 45. Now the last thing I'm going to show you involves a little bit of algebra. If any of the ways before this worked, that's fine. And where this is sitting in the playlist, you're not expected to know the algebra. But I want to show you the algebra just because I wanted to show you that this cross-multiplication isn't some magic, that using algebra, we will get this exact same thing. But you could stop watching this, if you'll find this part confusing. So let's rewrite our proportion, 8/36 is equal to 10/n. And we want to solve for n. Well the easiest way to solve for n is maybe multiply both-- this thing on the left is equal to this thing on the right. So we can multiply them both by the same thing. And the equality will still hold. So we could multiply both of them by n. On the right-hand side, the n's cancel out. On the left-hand side, we have 8/36 times n is equal to 10. If we want just an n here, we would want to multiply this side times 36-- I'll do that in a different color-- we'd want to multiply this side times 36 times 8, because if you multiply these guys out, you get 1. And you just have an n. But since we're doing it to the left-hand side, we also have to do it to the right-hand side, so times 36/8. These guys cancel out and we're left with n is equal to 10 times 36 is 360/8. And notice, we're getting the exact same value that we got with cross-multiplying. And with cross-multiplying, you're actually doing two steps. Actually, you're doing an extra step here. You're multiplying both sides by n, so that you had your 8n. And then you're multiplying both sides by 36, so that you get your 36 on both sides. And you get this value here. But at the end, when you simplify it,"
    },
    {
        "Q": "Couldn't Sal have just converted 5/4 into 1.25 at 0:60 instead of going through all of the 36*5/4 stuff?",
        "A": "5/4 = 1.25 and 1.25 * 36 = ?? it is difficult to calculate However if we solve (5/4) * 36 it is equal to (36/4) * 5 = 8 * 5 = 40.Thus we neednt get answer in decimals if it can be cut. This method is fast and easy and removes converting into decimal part.",
        "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": "The point of the video is to Multiply and Divide Scientific Notation, right? So how come at 2:19 you add? Is it something about exponent rules? If so, can someone please explain it, and, if not, can someone please tell me how? Thanks. And also, why do you have to change the number at 4:01? Thanks in advance for the help.",
        "A": "Its an exponent rule. 10^2 x 10^3 is equal to 10^(2+3) because (10 x 10)x(10 x 10 x 10)= 10^5",
        "video_name": "xxAFh-qHPPA",
        "timestamps": [
            139,
            241
        ],
        "3min_transcript": "Multiple, expressing the product in scientific notation. So lets multiply first and then lets try and get what we have into scientific notation. Actually before we do that lets just try and remember what it means to be in scientifc notation. So to be in scientific notation and actually each of these numbers here are in scientific notation. It is going to be the form: a times ten to some power where a can be greater than of equal to one and is going to be less than ten. So both of these numbers are greater than or equal to one and less than ten, and are being multiplied by some power of 10 so lets see how we can multiply this. so this here is the exact same thing. I am doing this in magenta 9.1 times 10 to the 6th. times, times 3.2 times. Actually, lets use dot notation, it will be a little bit more straightforward. So this is equal to 9.1 times 10 to the 6th. Now in multiplication, this comes from the associative property allows us to remove these brackets and says you can multiply that first or these guys first. and you can re-associate them and the communitive property tells us that we can re-arrange these here. and what I want to rearrange is I want to multiply the 9.1 by the 3.2 first. And then multiply that by the ten to the 6 times the ten to the -5. So I am going to rearrange this So I am going to rearrange this using the cumulative property. So this is the same thing as 9.1 x 3.2. and I am going to re-associate so I am going to do these first. Now that times 10^6 times 10^-5. We have the same base here, base 10 and we are taking the product so we can add the exponents. so this part right over here is going to be 10 to the ( 6 - 5 ) or just 10 to the 1st power. which is really just equal to 10. and that is going to be multiplied by 9.1 x 3.2 lets do that over here, so we have 9.1 x 3.2 At first I am going to ignore the decimals so I have 91 by 32 So to have 2 by 1 is 2 2 by 9 is 18 have a zero here as I am in the tens place now and am multiplying everything by 30 and not just 3 thats why my zero is there and I multiply 3 by 1 to get 3 and then 3 by 9 is 27"
    },
    {
        "Q": "At 5:17 what does adjacent mean? I forget. Thanks",
        "A": "Adjacent means next to .",
        "video_name": "TgDk06Qayxw",
        "timestamps": [
            317
        ],
        "3min_transcript": "it splits that line segment in half. So what it tells is, is that the length of this segment right over here is going to be equivalent to the length of this segment right over there. I have a circle. This radius bisects this chord right over here. And the goal here is to prove that it bisects this chord at a right angle. Or another way to say it-- let me add some points here. Let's call this B. Let's call this C, And let's call this D. I want to prove that segment AB is perpendicular. It intersects it at a right angle. It is perpendicular to segment CD. And as you could imagine, I'm going to prove it pretty much using the side-side-side whatever you want to call it, side-side-side theorem, postulate, or axiom. So let's do it. Let's think about it this way. I need to have some triangles. There's no triangles here right now. But I can construct triangles, and I can construct triangles based on things I know. For example, I can construct-- this has some radius. That's a radius right over here. The length of that is just going to be the radius of the circle. But I can also do it right over here. The length of AC is also going to be the radius of the circle. So we know that these two lines have the same length, which is the radius of the circle. Or we could say that AD is congruent to AC, or they have the exact same lengths. We know from the set-up in the problem that this segment is equal in length to this segment over here. Let me add a point here so I can refer to it. So if I call that point E, we know from the set-up in the problem, that CE is congruent to ED, or they have the same lengths. CE has the same length as ED. And we also know that both of these triangles, the one here the side EA. So EA is clearly equal to EA. So this is clearly equal to itself. It's the same side. The same side is being used for both triangles. The triangles are adjacent to each other. And so we see a situation where we have two different triangles that have corresponding sides being equal. This side is equivalent to this side right over here. This side is equal in length to that side over there. And then, obviously, AE is equivalent to itself. It's a side on both of them. It's the corresponding side on both of these triangles. And so by side-side-side, AEC."
    },
    {
        "Q": "at 5:57 does khan mean <AEC not <ABC?",
        "A": "He accidentally thought it was ABC at first.",
        "video_name": "TgDk06Qayxw",
        "timestamps": [
            357
        ],
        "3min_transcript": "I need to have some triangles. There's no triangles here right now. But I can construct triangles, and I can construct triangles based on things I know. For example, I can construct-- this has some radius. That's a radius right over here. The length of that is just going to be the radius of the circle. But I can also do it right over here. The length of AC is also going to be the radius of the circle. So we know that these two lines have the same length, which is the radius of the circle. Or we could say that AD is congruent to AC, or they have the exact same lengths. We know from the set-up in the problem that this segment is equal in length to this segment over here. Let me add a point here so I can refer to it. So if I call that point E, we know from the set-up in the problem, that CE is congruent to ED, or they have the same lengths. CE has the same length as ED. And we also know that both of these triangles, the one here the side EA. So EA is clearly equal to EA. So this is clearly equal to itself. It's the same side. The same side is being used for both triangles. The triangles are adjacent to each other. And so we see a situation where we have two different triangles that have corresponding sides being equal. This side is equivalent to this side right over here. This side is equal in length to that side over there. And then, obviously, AE is equivalent to itself. It's a side on both of them. It's the corresponding side on both of these triangles. And so by side-side-side, AEC. By side-side-side, we know that triangle AEC is congruent to triangle AED. But how does that help us? How does that help us knowing that we used our little theorem? But how does that actually help us here? Well, what's cool is once we know that two triangles are congruent-- so because they are congruent, from that, we can deduce that all the angles are the same. And in particular, we can deduce that this angle right over here, that the measure of angle CEA is equivalent to the measure of angle DEA. And the reason why that's useful is that we also"
    },
    {
        "Q": "At 1:02 I don't really get the number line process. Do you just multiply 60 by each minute?",
        "A": "He does this because there is 60 seconds in one minute. So 1 minute is equivalent to 60 seconds.",
        "video_name": "-r7JwuSdj5M",
        "timestamps": [
            62
        ],
        "3min_transcript": "We have multiple lengths of time here, and they're measured in minutes. So what we want to do is say, well, what's an equivalent number of seconds? And so the first one is pretty straightforward. 1 minute is equal to 60 seconds. By definition, a second is 1/60 of a minute. But what about 5 minutes? Well, we could try to visualize that. If 1 minute-- so let's say that is 1 minute-- if that is equal to 60 seconds, then 5 minutes-- let's just make five copies of this. So that's 1, 2, minutes, 3, minutes, 4 minutes, and then 5 minutes. So you see that 5 minutes would be 5 times 60 seconds. And 5 times 60 is 300. So here you have so this is going to be 5 times 60, which is equal to 300 seconds. Well, same idea. 10 times 60 is 600 seconds. And what about 60 minutes? Well, 60 minutes or 60 seconds per minute times 60 minutes is equal to 3,600 seconds. So that's interesting. Let's take things a little bit further other. Let's convert between hours, minutes, and seconds. So let's start off with something that's a little bit more straightforward. Let's say that we have 1 hour. Well, we know by definition 1 hour is equal to 60 minutes. And how many seconds is 60 minutes equal to? Well, we just figured that out. 60 minutes times 60 seconds per minute gets us to 3,600 seconds. Well, it would be the same exact idea. 60 minutes per hour, if I have 2 hours, it's going to be 2 times 60. 2 hours times 60 minutes per hour gets us to 120. Well, how many seconds is that going to be? Well, there's a couple of ways you could think about it. You could take 120 minutes times 60 seconds per minute. Or you could say 2 hours times 3,600 seconds per hour. So let's do it that way. So two 2 times 3,600 seconds per hour gets us to 7,200 seconds. Now, what if we were to say 10 hours? Well, once again, 10 times 60, 10 hours times 60 minutes per hour would get us to 600 minutes. And if we want to think in terms of seconds,"
    },
    {
        "Q": "why do you have to divide 3000 by 160 at 5:34.",
        "A": "Because there are a total of 3000 pounds and one person weighs 160 pounds. So you need t find out the total amount of people and to find that out, you would need to divide.",
        "video_name": "gnCL_0uA6Ds",
        "timestamps": [
            334
        ],
        "3min_transcript": "confusing, but I think you get the general idea. Now, if we were to write this out, what does it tell us? 1 person weighs 160 pounds, so there's one person for every 160 pounds. So notice, if we multiply these two expressions, the pounds will cancel out. We'll be just left with people. We're multiplying 3,000 times 1/160, but it's really just taking 3,000 and dividing by 160, which makes sense. We have a capacity of 3,000. Each of our people weighs 160 pounds. Divide by 160. It tells you how many people. But this way, you know that the units are working out. So this is going to be equal to 3,000/160 people. That is the maximum capacity of the elevator in terms of average people. Now, what is this? Well, we can divide the numerator and the denominator by 10. If we divide the numerator and the denominator by 10, this becomes 300/16. If we divide 16 by 2, this becomes 8. Now, let's see, what can we do more here? We could divide by 2 again. Let me rewrite it. So this is the same thing as 150/8 people. 150, we can divide by 2. It gives us 75. And if you take 8 divided by 2, that is 4. So we have 75 divided by 4 people. Let me just do that, work it out. So you have 75 divided by 4. 4 goes into 7 one time. 1 times 4 is 4. You subtract. 7 minus 4 is 3. Bring down this 5. 4 goes into 35 eight times. 8 times 4 is 32. Subtract. 5 minutes 2 is 3. And then you have the decimal. We're going to the right of the ones place. We're going to the tenths place now. 4 goes into 30 seven times. 7 times 4 is 28. You subtract. You get 2. Bring down another zero. 4 goes into 20 exactly five times. 5 times 4 is 20, and then we are done. So this expression is exactly-- the 3,000 divided by 160, or the 150 divided by 8, the 75 divided by 4, it all turns out to be 18.75 people is the capacity of the elevator. 18.75 average 160-pound people would weigh 3,000 pounds. Now, do they want a decimal? What is the maximum number of adults who could safely ride Well, if they're all going to be average, then the maximum number of adults, since you can't have 3/4 of a person, or 75/100 of a person, the maximum number is going to be 18. If you have 19 average-weight adults, you are going to be"
    },
    {
        "Q": "i kinda dont get what they say at 3:47 can anyone help me?",
        "A": "He s saying that it looks like all the sides of the blue shape are the same length",
        "video_name": "8xbIS2UkQxI",
        "timestamps": [
            227
        ],
        "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": "What did sal say at 3:12",
        "A": "he said Three over a hundred. (3/100) Remember you can put closed captions",
        "video_name": "lR_kUUPL8YY",
        "timestamps": [
            192
        ],
        "3min_transcript": "So if this is ones, multiply by 10, this is the tens place. This is the hundreds place. This is the thousands place. This is the ten thousands place. I'm going to have to write a little bit smaller. This is the hundred thousands place. And then the 7 is in the millions place. So what does this number, what does this 3 represent? Well, it's in the hundred thousands place. It literally represents 3 hundred thousands, or you could say 300,000, 3 followed by five zeroes. Now, what does this 3 represent? It's in the hundredths place. It literally represents 3 hundredths. It represents 3 times 1/100, which is the same thing as 3, which is equal to 3 over-- let Which is the same thing as 3/100, which is the same thing as 0.03. These are all equivalent statements. Now let's try to answer our original question. How much larger is this 3 than that 3 there? Well, one way to think about it is how much would you have to multiply this 3 by to get to this 3 over here? Well, one way to think about is to look directly at place value. So we got to multiply by 10. Every time we multiply by 10, that's equivalent to thinking about shifting it to one place to the left. So we would have to multiply by 10 one, two, three, four, five, six, seven times. So multiplying by 10 seven times. So this multiplied by 10 seven times should be equal to this. Let me rewrite this. 300,000 should be equal to 3/100-- let me write it the same way. 3/100 multiplied by 10 seven times, so times 10 times 10 times 10 times 10 times 10-- let's see, that's five times-- times 10 times 10. Now, multiplying by 10 seven times is the same thing as multiplying by 1 followed by seven zeroes. Every time you multiply by 10, you're going to get another zero here. So this is the same thing as 3/100 times 1 followed by one, two, three, four, five, six, seven zeroes. So this is literally 3/100 times 10 million."
    },
    {
        "Q": "Toward the end, (right around the 7:50 mark) Bernard only has one entry that is unique (all the others have 2 possible months or multiple \"August's\"). Is this how Bernard in row 2 says \"I know now\"? Because he's able to determine that only one of the possibilities has a unique set (July)?",
        "A": "Bernard has four possible entries: 14, 15, 16, and 17. If Bernard were given 14, then he wouldn t know whether the month were July or August. So he wouldn t be able to say I know now if he were given 14. On the other hand, if Bernard were given 15, 16, or 17, all of them correspond to only one month, and so Bernard would only have one month.",
        "video_name": "S5LEdJfCsHM",
        "timestamps": [
            470
        ],
        "3min_transcript": "did not get an 18th or 19th, that's because he knows it's not May or June. If the month that Albert got was May, then he would say, oh maybe he did get an 19th here, in which I can't, couldn't make the statement. Or, if the month was June, well then 18th would've been a possibility for what Bernard got. And so, he wouldn't have been able to make this statement here. But because he says, I know that Bernard does not know too, the reason why he can say that is because he knows it's not May or June. So we can rule out, we can rule out May or June. So we can rule out, we can rule out the 18th, the 19th here, and we can also rule out May or June now from Bernard. Now from Bernard's point of view, we can say look, Albert was, the only reason why he was able to make that statement is because Albert knows that the month is not May or June. So, let me rule out all of the Mays and Junes. So, let me rule out June here. Let me rule out, let me rule out May over here. Now Bernard, and once again, at any point, if you get so inspired, please pause the video and continue on with the deductive reasoning that we have started. All right, now Bernard says, At first, I didn't know when Cheryl's birthday is. Which is fair because it wasn't the 18th or 19th which Albert already knew. But now I know. So, now I know. So before he didn't know, but now he knows. So, if now he knows, what are the possibilities? What are the possibilities going on here? So he knows if the date he was given is the 17th, we've been able to, then he could deduce that, hey, the month is now August. If the date he was given was the 16th, the month that he could now deduce is July. If the date he was given was the 15th, he could deduce August. But if the date he was given was 14th, he still wouldn't know what the month is. So we could rule out the 14th, 'cause if the date he was given was the 14th he still wouldn't know. But he says, now I do know. So we could rule out the 14th. And once again, we're assuming Albert is hearing everything. Albert and Bernard, they both could kind of draw these diagrams out. And so Albert could now say, okay, look, if he now knows, then the date couldn't have been the 14th. So then he rules out the 14th. He rules out the 14th. So the stuff in orange that I just did I got from this statement there. And then Albert says, then I also know when Cheryl's birthday is. So, from Albert's point of view, the only way that he now knows what the birthday is if the month he was given was August he still has two options there. But he says he now knows when Cheryl's birthday is. So it can't be August because all the deductive reasoning we've just now done, if he was given August then he would still not know whether it was the 15th or 16th."
    },
    {
        "Q": "at 4:03 I didn't understand how is the angle that? Is it not sin(pi/2) because it's 90 degree",
        "A": "He s taking the angle he s already drawn, theta, and adding pi/2 to it, so the new angle you re dealing with is (theta + pi/2).",
        "video_name": "h-TPSylHrvE",
        "timestamps": [
            243
        ],
        "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": "at 4:22 i put in the exact same equation into my TU-84 Plus and came up with 10.09623335 why can't i get the answer shown",
        "A": "The answer you got (10.0962335) came out that way because your calculator is probably in radian mode. Go to setup or mode and change it to degrees.",
        "video_name": "ZElOxG7_m3c",
        "timestamps": [
            262
        ],
        "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": "At 2:40, Can someone explain how 3 is 30 tenths?\n\nI've done the decimals for tenths, hundredths etc. but I don't get this one...\n\nI thought '3' was 3 \"ones\"?\n\nThanks",
        "A": "Yes, you are right that 3 can equal 3 ones, but Sal decides to use 30 tenths because it is easier to see the answer using tenths. 1.5 is in tenths, because the tenths digit is the last place of value that we use. It would be more confusing if you said 3/2 is 1 1/2, although you can use ones if you prefer.",
        "video_name": "zlq_jQYD1mA",
        "timestamps": [
            160
        ],
        "3min_transcript": "So this is the same thing as two plus one over two, and I'm really doing every step here to hopefully make things clear, which is the same thing as two over two, so that's two over two, plus one over two, plus 1/2. I could break this up into two over two plus 1/2. Now, two over two is just one, and so this is going to be equal to 1 1/2. Now, you might immediately say, \"Hey, 1/2, I could write that as 5/10,\" and that would be exactly right. You could just, we don't wanna spell out every step, we could say this is equal to one, and when we write it in decimal form, we express things as tenths or hundredths or thousandths, so 1/2 is the same thing as 5/10, and if we wanna express that as a decimal, this would be equal to one and five tenths. you say okay this is the same thing as 3/2. 3/2, two goes into three one time, and there's a 1/2 left over, so writing this as a mixed number, it's 1 1/2, and 1/2 written as a decimal is 0.50, so this is 1.50. Now another way that we could've thought about this is \"Okay, I'm not getting a whole number, \"when I divide three divided by two. \"Maybe I'll get something in terms of tenths, \"so let me express each of these in terms of tenths.\" So three is how many tenths? Well, three is 30 tenths, and we'd be dividing by two, we're gonna be dividing it by two, so 30 tenths divided by two, well that's going to be equal to 15 tenths. This is equal to 15 tenths, which is equal to So both of these are equally legitimate strategies for figuring out what three divided by two is. I like the first one a little bit, it leverages what we know about fractions, but let's do another example. Let's do a few more examples, this is fun. Let's figure out what 34 divided by four is, and like before, pause this video and try to figure it out and try to see if you can use some of the strategies that we used in the last video. Alright, so as we just said, we can re-express this as a fraction, this is the same thing as 34 divided by four, 34 divided by four, or 34/4. Now what is this going to be equal to? Well, four goes into 34 eight times, it's gonna go eight times,"
    },
    {
        "Q": "At 5:56\n\nIsn't 84 tenths supposed to be 0.84?\n\nPlease Explain\n\nThanks",
        "A": "Try thinking about it like this; if you had 9 tenths and added another, that would be 10 tenths, which would look like 0.9 + 0.1 = 1.0 If you add another tenth to that, it would be 11 tenths, which would look like 1.0 + 0.1 = 1.1 That means for every 10 tenths, you re getting 1 whole one. That s why having 84 tenths is really like having 8 ones and 4 more tenths. Does that help?",
        "video_name": "zlq_jQYD1mA",
        "timestamps": [
            356
        ],
        "3min_transcript": "so this is eight and 2/4. Eight, let me write it both out. Alright, eight and 2/4, let me do that two in that blue just for fun. Eight and two over four, 8 2/4, so how do I do this again? I said four goes into 34 eight times, and then I have two left over, so I'm gonna have 2/4 left over. Another way, if you wanna see all the steps, you can say, \"Hey I can rewrite this as 32 over four \"plus two over four.\" The 32 over four is our eight, so 8 2/4. Well 2/4, that's the same thing as 1/2. That's the same thing as 1/2, and if we wanna express that in terms of tenths, this is equal to eight and, 1/2 is the same thing as 5/10, 8 5/10, which if we wanna express as a decimal or 8.50, and we are done. Let's do another one of these, and actually let's do one of them where we're dividing into a decimal, where a decimal itself is being divided. Let's say we wanted to calculate 8.4 divided by seven. Pause this video and see if you can figure it out. When you look at this, you might immediately say well I know 84 is divisible by seven. We know if you know your seven times tables, we know that seven times 12 is equal to 84, or that 84 divided seven is equal to 12, but this isn't 84, this is 8.4, so how do we think about it? Well one way, we can think about it in terms of tenths, 8.4 is the same thing as 84 tenths, well 84 of anything divided by seven is going to be 12 of that thing, so it's gonna be 12 tenths, and 12 tenths we can rewrite as one and 2/10, 1.20 and we are done, so this is equal to 1.2. Another way that we could have thought about this is we could've said, you know what, 84 tenths is the same thing as 84 over 10. If fact, you would read this as 84/10, and now you wanna divided this by seven. So you wanna divide this by seven, this is the same thing, when you divide by something, it's the same as multiplying by the reciprocal, so it's 84 over ten times 1/7, which is equal to, we can change, let me write it this way. This is equal to 84 over 10 times seven,"
    },
    {
        "Q": "At 6:51\n\nWhy is 7 not 7/1 since it is 7 whole \"1's\" i.e. 7 lots of one whole 1's?\nisn't 1/7 one seventh of a thing? So it isn't a whole thing?\n\nIn \"Fractions as a whole number\" Section, one of the questions was \"what is \"10\" as a fraction?\" and the correct answer was 10/1\n\nPlease Explain\n\nThanks",
        "A": "I think you are confused. In the first equation it is 84/10 divided by 7 and then it becomes 84/10 times 1/7. Dividing a number by another number is the same as multiplying the first number by the second number s reciprocal i.e. 7/1 or 7 becomes 1/7 when changing to multiplication.",
        "video_name": "zlq_jQYD1mA",
        "timestamps": [
            411
        ],
        "3min_transcript": "or 8.50, and we are done. Let's do another one of these, and actually let's do one of them where we're dividing into a decimal, where a decimal itself is being divided. Let's say we wanted to calculate 8.4 divided by seven. Pause this video and see if you can figure it out. When you look at this, you might immediately say well I know 84 is divisible by seven. We know if you know your seven times tables, we know that seven times 12 is equal to 84, or that 84 divided seven is equal to 12, but this isn't 84, this is 8.4, so how do we think about it? Well one way, we can think about it in terms of tenths, 8.4 is the same thing as 84 tenths, well 84 of anything divided by seven is going to be 12 of that thing, so it's gonna be 12 tenths, and 12 tenths we can rewrite as one and 2/10, 1.20 and we are done, so this is equal to 1.2. Another way that we could have thought about this is we could've said, you know what, 84 tenths is the same thing as 84 over 10. If fact, you would read this as 84/10, and now you wanna divided this by seven. So you wanna divide this by seven, this is the same thing, when you divide by something, it's the same as multiplying by the reciprocal, so it's 84 over ten times 1/7, which is equal to, we can change, let me write it this way. This is equal to 84 over 10 times seven, If we divide the numerator and the denominator by seven, 84 divided by seven is 12, seven divided by seven is one. 12 divided by ten, this is gonna be equal to 12/10, which is 1.2, we could write this as 1 2/10, or 1.2. Let's do one more example that's kind of related. Let's say we wanna figure out what seven divided by 70 is. Pause this video and try to figure it out. Well, we can rewrite this, as we've been doing, as seven over 70. Instead of writing it as 70, let me write that as seven times ten, and what's valuable about this is we can divide the numerator and the denominator by seven. If we divide the numerator by seven, we get a one, we divide the denominator by seven, we get a one. Remember, we can do the same thing If we multiply or divide by the same value,"
    },
    {
        "Q": "At2:38,how can we teach counting the objects to small kids?",
        "A": "The Simplest way is to place some random number of objects near the kids, and ask them to count :1 If, they don t know to count teach them",
        "video_name": "PEeUTQ0Gri8",
        "timestamps": [
            158
        ],
        "3min_transcript": "- [Voiceover] How do we count the flowers? So, this first choice, they go one, two, three, and then they don't count this one and then they skip it and they go four, five, six, and then they skip this one again, and then they say seven, eight. That's not how you count flowers! You'd say this is four, this is five, six, seven, eight, nine, 10. So, this first choice is not right. Now, the second one looks better. One, two, three, four, they didn't skip this one, five, six, seven, eight, nine, 10. That one looks good so I'll click right over here and check my answer. How do we count the mice? Let's see, one, two, this isn't the third mouse! This is the second mouse! This is crazy, this isn't right. All right, this one over here. One, two, three, four, five, six, seven, this is right. Up here they skipped the number two, this should be the second mouse, not the third mouse! Let's keep going. Type the missing numbers. It says count the ladybugs. Type the missing numbers in the boxes. All right, this is ladybug number one. Two, then let me type three, then let me type four. It's a lot of fun. Count the flowers. Type the missing numbers in the boxes. This is flower number one, flower number two. Just gonna type that right in there. Check it, lemme do one more. How do we count the flowers? So, we saw one like this a few questions ago. One, two, three, four, five, six, seven, eight, nine, 10. That one looks good. This one, just like the last time, they skipped, actually they skipped the same ones. This wouldn't be the right way to do it. So, we select this one right here, and then we're done."
    },
    {
        "Q": "At 7:07, what is the sigma notation?",
        "A": "Sigma notation provides a way to compactly and precisely express any sum, that is, a sequence of things that are all to be added together. In this case it would be used to add up the areas of all of the rectangles.",
        "video_name": "ViqrHGae7FA",
        "timestamps": [
            427
        ],
        "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": "at 6:56, why doesn't z not equal -1",
        "A": "Good question. A rational function is allowed to be 0. 0/b is defined as long as b isn t 0. 0/b is just 0 because 0=0b for any real number b. When you plug -1 in for z you just get 0/1 which is 0.",
        "video_name": "XFwQV-KCudw",
        "timestamps": [
            416
        ],
        "3min_transcript": "and you're gonna be left with just a one. And once again, you can distribute the 17 z squared, multiply it times z, you get 17 z to the third. 17 z squared times one is 17 z squared. All right. All of that is going to be over, we wanna factor out of 17 z squared out of the denominator, 17 z squared times, and so 34 z to the third divided by 17 z squared, 34 divided by 17 is two, and z to the third divided by z squared is z and then we have minus 51 divided 17 is three and z squared divided by z squared is just one. So we'll just leave it like that. And so over here, it becomes a little bit clear of, well one, how we're gonna simplify it. We're just gonna divided 17 z squared by 17 z squared, but let's be careful to restrict the domain here. So, we can tell if z is equal to zero, and we make the denominator equal to zero and we could see that even looking here. So we could say z cannot be equal to zero. Now what else? Z cannot be equal to whatever makes this two z minus three equals zero. So let's think about what makes two z minus three equal to zero. Two z minus three is equal to zero. You can add three to both sides and you would get two z is equal to three. Divide both sides by two and you would get z is equal to 3/2. So z cannot be equal to zero and z cannot be equal to 3/2. So that's how we're gonna restrict our domain. But now let's simplify. So, if we simplify it, these two cancel out and we are going to be left with, we're gonna be left with z plus one over two z minus three and we wanna keep that constraint. z cannot be equal to zero. because we still have the two z minus three here. If someone were to just look at this expression, like, well, the denominator can't be equal to zero and so, z cannot be equal to 3/2. So this is still, if we just left it the way it is, we don't even have to rewrite this, that would be redundant since looking at this expression is clearly not defined for z equals 3/2. So, there you go. And so I'm gonna ask you for what values does this expression not defined, well then you would also include that. Actually, let me just write it. It doesn't hurt to be redundant. z does not equal 3/2. But this constraint right here is really important because it's not obvious by looking at this expression. This expression by itself would be defined for z equals zero, but if we want it to be algebraically equivalent to this one and that one, it has to be constrained in the same way."
    },
    {
        "Q": "at 0:30, sal says 'it's going to be undefined'. what if you expression doesn't have a variable in the denominator? can it still be undefined?",
        "A": "If there is no variable in the denominator, then there is no way for the expression to become undefined. The only exception would be if someone gave you a rational expression with zero already in the denominator. So, a variable in the denominator or the denominator is already 0, then you need to worry about the rational expression being undefined. If the rational expression has a denominator that is only a constant other than zero, then undefined will not be an issue. Hope that helps.",
        "video_name": "XFwQV-KCudw",
        "timestamps": [
            30
        ],
        "3min_transcript": "- [Voiceover] So I have a rational expression here and my goal is to simplify it, but while I simplify it, I wanna make the simplified expression be algebraically equivalent. So, if there are certain x values that would make this thing undefined, that I have to restrict my simplified expression by those x values. So, you could pause this video and take a go at it and then we'll do it together. All right. So let's just think real quick. What x values would make this expression undefined? Well, it's undefined if we try to divide by zero. So if x is zero, 14 times zero is zero, it's going to be undefined. And so, we could say this is, we could say x does not, cannot be equal to zero. For any other x, we can evaluate this expression. Now let's actually try to simplify it. So, when you look at the numerator and the denominator, every term we see is divisible by x and every term is divisible by seven. So it looks like we can factor seven x out of the numerator and seven x out of the denominator. So the numerator, we can rewrite as seven x times, we're gonna be left with two x and then you factor seven x out of seven x, you're gonna be left with a one. And one way to think about it is we did the distributive property in reverse, seven x times two x is 14 x squared, seven x times one is seven x. All right. And now let's factor seven x out of the denominator. So, 14 x could be rewritten as seven x times two, times two, and remember, I wanna keep this algebraically equivalent, so I wanna keep the constraint that x cannot be equal to zero. And so we divide the numerator and the denominator by seven x or one way to think about it, you can divide seven x by seven x and just get one and we are left with two x plus one over two. Now this was the original expression, x could take on any value, but if we want it to be algebraically equivalent it has to have the same constraints on. So we're going to have to say x, x does not equal zero, and it's a very subtle but really important thing. For example, if you defined a function by this right over here, the domain of the function could not include zero. And so, if you simplified how you defined that function to this, if you want that function to be the same, it needs to have the same domain. It has to be defined for the same inputs. And so that's why we're putting the exact same constraints for them to be equivalent. If you got rid of this constraint, these two would be equivalent everywhere except for x equal zero. This one would have been defined for x equal zero, this one would not have, and so they wouldn't have been algebraically equivalent. This makes them algebraically equivalent. And of course you can write this in different ways. You can divide each of these terms by two, if you like. So you could divide two x by two and get x and then divide one by two and get 1/2. Once again, we would wanna keep"
    },
    {
        "Q": "0:50 can someone explain why f(x) when x is approaching 1+ is 1",
        "A": "Look at the graph. Clearly, f(x) is not continuous at x=1, and therefore, there is no actual limit. However, there is still a limit from the left and from the right. X approaching 1+ means the limit from the right, so, as the graph shows, f(x) approaches 1 as x goes to one from the right.",
        "video_name": "_WOr9-_HbAM",
        "timestamps": [
            50
        ],
        "3min_transcript": "So we have a function, f of x, graphed right over here. And then we have a bunch of statements about the limit of f of x, as x approaches different values. And what I want to do is figure out which of these statements are true and which of these are false. So let's look at this first statement. Limit of f of x, as x approaches 1 from the positive direction, is equal to 0. So is this true or false? So let's look at it. So we're talking about as x approaches 1 from the positive direction, so for values greater than 1. So as x approaches 1 from the positive direction, what is f of x? Well, when x is, let's say 1 and 1/2, f of x is up here, as x gets closer and closer to 1, f of x stays right at 1. So as x approaches 1 from the positive direction, it looks like the limit of f of x as x approaches 1 from the positive direction isn't 0. It looks like it is 1. 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."
    },
    {
        "Q": "At 4:42, I don't understand what the slope is or why it is -7.",
        "A": "They are the same slopes for the slope intercept form.",
        "video_name": "o4pbAQgJYjI",
        "timestamps": [
            282
        ],
        "3min_transcript": "Negative three comma negative two. And then I could check my answer; got it right. Let's do another. Let's do another one of these. Maybe of a different type. So over here it says: \"A system of two linear equations \"is graphed below. \"Approximate the solution of the system.\" Alright so here I just have to just look at this carefully and think about where this point is. So let's think about first its X value. So its X value, it's about right there in terms of its X value. It looks like, so this is negative one. This is negative two, so negative 1.5 is gonna be right over here. It's a little bit to the left of negative 1.5, so it's even more negative, I would say negative 1.6. And I'm approximating it, negative 1.6. Hopefully it has a little leeway in how it checks the answer. What about the Y value? So if I look at the Y value here, it looks like it's a little less than one and a half. It looks like it's a little less than halfway between one and two, so I'd give it 1.4, positive 1.4. And let's check the answer, see how we're doing. Yep, we got it right. Let's actually just do one more for good measure. So this is another system. They've just written the equations in more of our slope intercept form. So let's see, Y is equal to negative seven, X plus three. When X is equal to zero, we have our Y intercept. Y is equal to three. So when X is equal to zero, Y is equal to three. And then we see that our slope is negative seven. When you increase X by one, you decrease Y by seven. So when you increase X by one, you decrease Y by one, two, three, four, five, six, and seven. When X goes from zero to one, Y went from three to negative four, Now the second one: our Y intercept. When X is equal to zero, Y is negative three, so let me graph that. When X is zero, Y is equal to negative three. And then its slope is negative one. When X increases by one, Y decreases by one. So the slope here is negative one. So when X increases by one, Y decreases by one. And there you have it. You have your point of intersection. You have the X-Y pair that satisfies both equations. That is the point of intersection. It's gonna sit on both lines which is why it's the point of intersection. And that's the point X equals one, Y is equal to negative four. So you have X equals one and Y is equal to negative four. And I can check my answer and we got it right."
    },
    {
        "Q": "At 3:10 where does the 2 go when your write (x-22)(x-22). ( the two as in 2(-22)x",
        "A": "From the example on the top right: (x+a)^2 = (x+a)(x+a) = x^2 + 2a + a^2. In our case, a is -22, so he writes it as x^2 + 2(-22) + (-22)^2 = (x-22)(x-22) = (x-22)^2",
        "video_name": "VvuuRpJbbHE",
        "timestamps": [
            190
        ],
        "3min_transcript": "whatever value is here, if I take half of it, right? This is going to be 2a here. If I take half of it and square it over here, then this will be a perfect square. So if we look over here, this thing right here is 2a, if we want to pattern match, if we want to make this look like a perfect square. That has to be 2a. So negative 44 is equal to 2a. And this, right here, this c, if we pattern match, c has to be equal to a squared. So what's a? Well, if we know negative 44 is 2a, we can divide both sides of that by 2. And we know that negative 22 has got to be equal to a. a has got to be equal to negative 22. a is half of the coefficient right here. It's half of negative 44. And whenever you complete the square, it's always going to be half of the coefficient right here. Well, c needs to be a squared in order for this to be a perfect square. So c needs to equal negative 22 squared. And we can figure out what that is. 22 times 22, we could put the negative later-- actually it's just going to be the same thing because the negative times a negative is a positive. 2 times 22 is 44, put a 0. 2 times 22 is 44. Get a 4, get an 8, get a 4. So it's 484. So if we were to rewrite this as x squared minus 44x plus 484, then this is a perfect square trinomial. Or we could write it like this. This is x squared minus 2 times-- or maybe I should write it this way-- plus 2 times negative 22x plus negative 22 squared. a perfect square, and if you were to factor it, it's the same thing as x minus 22 times x minus 22, or x minus 22 squared. These are all equivalent statements."
    },
    {
        "Q": "I didn't understand why \"when we're trying to create an upper bound we want as low of an upper bound as possible\" at 8:51 ...:(",
        "A": "I believe you re trying to find the error of a specific point, so you want to keep the bounds as close to each other as possible to maximize your accuracy, sort of like integrating.",
        "video_name": "yUUPP70Fhpo",
        "timestamps": [
            531
        ],
        "3min_transcript": "this, you'd get a low value cause this stuff would cancel out. Would cancel out with this stuff right over here then you'd take the absolute value of just a lower, a lower magnitude number. And so in general, the integral, the integral, sorry the absolute value of the integral is going to be less than or equal to the integral of the absolute value. So we can say, so this right here is the integral of the absolute value which is going to be greater than or equal. What we have written over here is just this. That's going to be greater than or equal to, and I think you'll see why I'm why I'm doing this in a second. Greater than or equal to the absolute value, the absolute value of the integral of, of the n plus 1th derivative. The n plus 1th derivative of, x, dx. And the reason why this is useful, is that we can still keep the inequality that, this is less than, or equal to this. But now, this is a pretty straight forward integral to evaluate. The indo, the anti-derivative of the n plus 1th derivative, is going to be the nth derivative. Is just going to the absolute value of the nth derivative. The absolute value of the nth derivative of our error function. Did I say expected value? See, it even confuses me. This is the error function. I should've used r, r for remainder. But this all error. The, noth, nothing about probability or expected value in this video. E for error. So anyway, this is going to be the nth derivative of our error function, which is going to be less than or equal to this. Which is less than or equal to the anti-derivative of M. Well, that's a constant. So that's going to be mx, mx. And since we're just taking indefinite integrals. We can't forget the idea that we have a constant over here. And in general, when you're trying to create an upper bound you want as low of an upper bound as possible. So we wanna minimize, we wanna minimize what this constant is. And lucky for us, we do have, we do know what this, what this function, what value this function takes on at a point. I think we wrote it over here. The nth derivative at a is equal to 0. And that's because the nth derivative of the function and the approximation at a are going to be the same exact thing. And so, if we evaluate both sides of this at a, I'll do that over here on the side, we know that the absolute value. We know the absolute value of the nth derivative at a, we know that this thing is going to be equal to the absolute value of 0. Which is 0. Which needs to be less than or equal to when you evaluate this thing at a, which is less than or equal to m a plus c. And so you can, if you look at this part of the inequality, you subtract m a from both sides. You get negative m a is less than or equal to c. So our constant here, based on that little condition that we were able to get in the last video. Our constant is going to be greater than or equal to negative ma. So if we want to minimize the constant, if we wanna get this as low"
    },
    {
        "Q": "Can anyone explain to me why at 10:00, where -ma<=c,why we have to take the lower bound of c, making c=-ma? I am really confused at that part.",
        "A": "This part actually has an error, which can be fixed by using a definite integral between x and a (or a and x, depending on whether x is less than a), rather than an indefinite integral. I have sent a message to Mr. Khan about this error, so hopefully this gets corrected soon. Anyway, a definite integral with these bounds gets you from step to step quite smoothly :) Hope that helps!",
        "video_name": "yUUPP70Fhpo",
        "timestamps": [
            600
        ],
        "3min_transcript": "Is just going to the absolute value of the nth derivative. The absolute value of the nth derivative of our error function. Did I say expected value? See, it even confuses me. This is the error function. I should've used r, r for remainder. But this all error. The, noth, nothing about probability or expected value in this video. E for error. So anyway, this is going to be the nth derivative of our error function, which is going to be less than or equal to this. Which is less than or equal to the anti-derivative of M. Well, that's a constant. So that's going to be mx, mx. And since we're just taking indefinite integrals. We can't forget the idea that we have a constant over here. And in general, when you're trying to create an upper bound you want as low of an upper bound as possible. So we wanna minimize, we wanna minimize what this constant is. And lucky for us, we do have, we do know what this, what this function, what value this function takes on at a point. I think we wrote it over here. The nth derivative at a is equal to 0. And that's because the nth derivative of the function and the approximation at a are going to be the same exact thing. And so, if we evaluate both sides of this at a, I'll do that over here on the side, we know that the absolute value. We know the absolute value of the nth derivative at a, we know that this thing is going to be equal to the absolute value of 0. Which is 0. Which needs to be less than or equal to when you evaluate this thing at a, which is less than or equal to m a plus c. And so you can, if you look at this part of the inequality, you subtract m a from both sides. You get negative m a is less than or equal to c. So our constant here, based on that little condition that we were able to get in the last video. Our constant is going to be greater than or equal to negative ma. So if we want to minimize the constant, if we wanna get this as low That is the lowest possible c that will meet these constraints that we know are true. So, we will actually pick c to be negative Ma. And then we can rewrite this whole thing as the absolute value of the nth derivative of the error function. The nth derivative of the error function. Not the expected value. I have a strange suspicion I might have said expected value. But, this is the error function. The nth der. The absolute value of the nth derivative of the error function is less than or equal to M times x minus a. And once again all of the constraints hold. This is for, this is for x as part of the interval. The closed interval between, the closed interval between a and b. But looks like we're making progress. We at least went from the m plus 1 derivative to the n derivative. Lets see if we can keep going. So same general idea. This if we know this then we know that So we can take the integral of both sides of this"
    },
    {
        "Q": "What exactly is happening at 2:13?",
        "A": "@ 2:13, he s saying that 2 out of 100 is equal to 2%.",
        "video_name": "Lvr2YsxG10o",
        "timestamps": [
            133
        ],
        "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": "@ 2:05 what is meant by a negative angle?",
        "A": "In trig, positive angles means you are going around the unit circle counterclockwise, while negative angles mean you are going around the unit circle clockwise.",
        "video_name": "z0-1gBy1ykE",
        "timestamps": [
            125
        ],
        "3min_transcript": "We're asked to convert pi radians and negative pi/3 radians to degrees. And the first question I'll ask you: If you do one revolution, You have an angle that went all the way around once. How many radians is that? Well we know that it is 2 pi radians. Now that exact same angle if we were to measure it in degrees, How many degrees is that? Well if you were doing degrees, it would be one full revolution. That is equal to 360 degrees Now, can we simplify this? Thats a bore to write this little, superscript cirsle That's literally the units of the question. Sometimes it doesn't look like a unit but it is a unit. You could literally write degrees instead of that little symbol. Now can we simplify this? Well sure, Both two pi and 360 are divisible by two so lets divide things by two, and if we do that, what do we get? Or what are pi radians equal to? 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"
    },
    {
        "Q": "At 3:00, why is pie equal to radian",
        "A": "pi is not = to radian.",
        "video_name": "z0-1gBy1ykE",
        "timestamps": [
            180
        ],
        "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": "Hi, at 1:05 , Sal raised the radicals. Is it easier to do the square roots then divide them rather than raising them? Which way is easier? Thanks!",
        "A": "If i understand correctly you are asking why he simplified?(Thats what is at 1:05) He simplifies so it is easier to do the equation. If i didn t answer your question please clarify what you are asking. Thanks",
        "video_name": "sBvRJUwXJPo",
        "timestamps": [
            65
        ],
        "3min_transcript": "We're asked to divide and simplify. And we have one radical expression over another radical expression. The key to simplify this is to realize if I have the principal root of x over the principal root of y, this is the same thing as the principal root of x over y. And it really just comes out of the exponent properties. If I have two things that I take to some power-- and taking the principal root is the same thing as taking it to the 1/2 power-- if I'm raising each of them to some power and then dividing, that's the same thing as dividing first and then raising them to that power. So let's apply that over here. This expression over here is going to be the same thing as the principal root-- it's hard to write a radical sign that big-- the principal root of 60x squared y over 48x. And then we can first look at the coefficients of each of these expressions and try to simplify that. 60 divided by 12 is 5. 48 divided by 12 is 4. Both the numerator and the denominator are divisible by x. x squared divided by x is just x. x divided by x is 1. Anything we divide the numerator by, we have to divide the denominator by. And that's all we have left. So if we wanted to simplify this, this is equal to the-- make a radical sign-- and then we have 5/4. And actually, we can write it in a slightly different way, but I'll write it this way-- 5/4. And we have nothing left in the denominator other than that 4. And in the numerator, we have an x and we have a y. And now we could leave it just like that, but we might want to take more things out of the radical sign. And so one possibility that you can do is equal to 1/4 times 5xy, all of that under the radical sign. And this is the same thing as the square root of or the principal root of 1/4 times the principal root of 5xy. And the square root of 1/4, if you think about it, that's just 1/2 times 1/2. Or another way you could think about it is that this right here is the same thing as-- so you could just say, hey, this is 1/2. 1/2 times 1/2 is 1/4. Or if you don't realize it's 1/2, you say, hey, this is the same thing as the square root of 1 over the square root of 4, and the square root of 1 is 1 and the principal root of 4 is 2, so you get 1/2 once again. And so if you simplify this right here to 1/2, then the whole thing can simplify to 1/2 times the principal root-- I'll just write it all in orange-- times the principal root of 5xy. And there's nothing else that you can really"
    },
    {
        "Q": "At 5:14, why is it that the a is the minor axis length? I thought that the value of a is the value of the major axis length.",
        "A": "Not necessarily. The semi-major axis length is whichever is longer and a represents the x axis length. a is the semi-major axis length when the ellipse is short and fat. b is the semi-major axis length when the ellipse is tall and skinny.",
        "video_name": "lvAYFUIEpFI",
        "timestamps": [
            314
        ],
        "3min_transcript": "shortest of all of the diameters of this ellipse. And then the semi is half of that. b is the length of the semi-minor axis. That's b in this example, just because as I drew this ellipse it just happens to be that b is smaller then a. If b was larger than a, I would have a tall and skinny ellipse. Let me actually draw one. It could have been like this. I could have an ellipse that looks something like that. In which case, all of a sudden b would be the semi-major axis, because b would be greater than a. That this would be taller than it is wide. But let me not confuse the graph too much. And in this case, a is the length of-- I think you've guessed it-- a is the length of the semi-major axis or you can 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"
    },
    {
        "Q": "At 1:28, you talked about that a and b are any two numbers. What if one of the value like b was a radical number that was not a perfect square?",
        "A": "a and b can be any positive (and non-zero) numbers. They are the radii of the ellipse, and can have any length (positive or zero), like any other line segment.",
        "video_name": "lvAYFUIEpFI",
        "timestamps": [
            88
        ],
        "3min_transcript": "In the last video, we learned a little bit about the circle. And the circle is really just a special case of an ellipse. It's a special case because in a circle you're always an equal distance away from the center of the circle, while in an ellipse, the distance from the center of the circle is always changing. You know what an ellipse looks like. Well, I showed you that in the first video. It looks something like that. What I mean is that the radius or the distance from the center is always changing. Let me say this is centered at the origin. So that's the origin right there. You see here, we're really, if we're on this point on the ellipse, we're really close to the origin. This is actually the closest we'll ever get, just as close as well get down here. And when we're out here we're really far away from the origin and that's about as far as we're going to get right there. So a circle is a special case of this, because in a circle's case, the furthest we get from the origin is the same distance the exact same distance away from the origin. Well, with that said, let's actually go a little bit into the math. So the general or the standard form for an ellipse centered at the origin is x squared over a squared plus y squared over b squared is equal to 1. Where a and b are just any two numbers. I could have written this as c squared and d squared. I mean, they're just place holders. Just to give you an idea of what this means, if this was our ellipse in question right now, a is the length of the radius in the x-direction. Remember, we're going to have a squared down here. So if you took the square root of whatever is in the denominator, a is the x-radius. So this distance in our little chart right here, in our little graph here, that distance is a, or that this point right here, since we're centered at the origin, will be the point x is equal to a y is equal to 0. would be the point minus a comma 0. And then the radius in the y-direction would be this radius right here and is b. So this point would be x is equal to 0, y is equal to b. Likewise this point right here would be x is equal to 0, y is equal to minus b. And the way I drew this, we have kind of a short and fat ellipse you can also have kind of a tall and skinny ellipse. But in the short and fat ellipse, the direction that you're short in that's called your minor axis. And so b, I always forget the exact terminology, but b you can call it your semi or the length of your semi-minor axis. And where did that word come from? Well if this whole thing is your minor axis or maybe you could call your minor diameter if this whole thing is your"
    },
    {
        "Q": "at 8:41, why is it necessary for x-5 to behave like x of the standard equation",
        "A": "I m not sure what you mean, but x-5 if the same if you were to shift any graph or shape on the coordinate plane. Hope that helps.",
        "video_name": "lvAYFUIEpFI",
        "timestamps": [
            521
        ],
        "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": "@7:55 Sal gets rid of the first 1 in the third column, could he also get rid of the second 1 and keep the first one?",
        "A": "The second 1 in the third column is a leading digit, so getting rid of it wouldn t help. The first 1 in the third column, however, is not a leading digit but is in the same column as a leading digit, so we need to get rid of it.",
        "video_name": "L0CmbneYETs",
        "timestamps": [
            475
        ],
        "3min_transcript": "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. That's just 1. 2 minus 0 is 2. 1 minus 1 is 0. 1 minus minus 2 is 3. 7 minus 5 is 2. There you have it. We have our matrix in reduced row echelon form. This is the reduced row echelon form of our matrix, I'll write it in bold, of our matrix A right there. You know it's in reduced row echelon form because all of your leading 1's in each row-- so what are my leading 1's in each row? I have this 1 and I have that 1. They're the only non-zero entry in their columns. These are called the pivot entries. Let me label that for you. That's called a pivot entry. Pivot entry. They're the only non-zero entry in"
    },
    {
        "Q": "at 14:37 , how did you get X2 = 0 1 0\nX4=0 0 1\ni dont understand how did you get that . we only know X1 ans X3 ??\nthanks\nHal",
        "A": "Sal has assumed that the solution is in R^4 (which I guess it is if it s in R2 or R3). I m also confused. What he s doing implies that the free variables x2 and x4 are on their own x2 and x4 axes of R^4, which I have doubts about. 1) The original 3x4 transformation matrix is from R^4 to R^3. 2) The rref matrix has only 2 rows, which seems to mean there are only x1 and x3 coordinates in the solution.",
        "video_name": "L0CmbneYETs",
        "timestamps": [
            877
        ],
        "3min_transcript": "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. Either a position vector. It is a vector in R4. You can view it as a position vector or a coordinate in R4. You could say, look, our solution set is essentially-- this is in R4. Each of these have four components, but you can imagine it in r3. That my solution set is equal to some vector, some vector there. That's the vector. Think of it is as a position vector. It would be the coordinate 2, 0, 5, 0. Which obviously, this is four dimensions right there. It's equal to multiples of these two vectors. Let's call this vector, right here, let's call this vector a. Let's call this vector, right here, vector b. Our solution set is all of this point, which is right there, or I guess we could call it that position vector. That position vector will look like that."
    },
    {
        "Q": "at 11:15 isnt the x4 in the first equation a free variable?",
        "A": "Yes, at 11:07 he specifies which are the pivot variables (x1 and x3) and which are the free variables (x2 and x4); he just didn t circle the x4 in the first equation.",
        "video_name": "L0CmbneYETs",
        "timestamps": [
            675
        ],
        "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": "At 17:20, why is the 4th dimension denoted by R4?\nWhy the letter R?",
        "A": "R is the set of all real numbers. The real numbers can be thought of as any point on an infinitely long number line. Examples of these numbers are -5, 4/3, pi etc. An example of a number not included are an imaginary one such as 2i. R4 means that points in the space has 4 coordinates of real values. Points in this space are written on the form (x1, x2, x3, x4) where xi is a real number.",
        "video_name": "L0CmbneYETs",
        "timestamps": [
            1040
        ],
        "3min_transcript": "Either a position vector. It is a vector in R4. You can view it as a position vector or a coordinate in R4. You could say, look, our solution set is essentially-- this is in R4. Each of these have four components, but you can imagine it in r3. That my solution set is equal to some vector, some vector there. That's the vector. Think of it is as a position vector. It would be the coordinate 2, 0, 5, 0. Which obviously, this is four dimensions right there. It's equal to multiples of these two vectors. Let's call this vector, right here, let's call this vector a. Let's call this vector, right here, vector b. Our solution set is all of this point, which is right there, or I guess we could call it that position vector. That position vector will look like that. multiples of these two guys. If this is vector a, let's do vector a in a different color. Vector a looks like that. Let's say vector a looks like that, and then vector b looks like that. This is vector b, and this is vector a. I don't know if this is going to be easier or harder for you to visualize, because obviously we are dealing in four dimensions right here, and I'm just drawing on a two dimensional surface. What you can imagine is, is that the solution set is equal to this fixed point, this position vector, plus linear combinations of a and b. We're dealing, of course, in R4. Let me write that down. We're dealing in R4. But linear combinations of a and b are going to create a plane. You can multiply a times 2, and b times 3, or a times minus 1, and b times minus 100. You can keep adding and subtracting these linear combinations of a and b. position vector, or contains the point 2, 0, 5, 0. The solution for these three equations with four unknowns, is a plane in R4. I know that's really hard to visualize, and maybe I'll do another one in three dimensions. Hopefully this at least gives you a decent understanding of what an augmented matrix is, what reduced row echelon form is, and what are the valid operations I can perform on a matrix without messing up the system."
    },
    {
        "Q": "1:43\nSo the x\u00c2\u00ba is the same thing as one?",
        "A": "Yes, that is correct! But more accurately.... x\u00c2\u00ba=1, where x is any number except 0 because 0\u00c2\u00ba is undefined",
        "video_name": "REiDXCN0lGU",
        "timestamps": [
            103
        ],
        "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": "Why did Sal put 0 as the exponent of 7x and not 1 at 1:30? Why did he put 1 as the exponent of 8x and not 0 at 2:11? Wouldn't these both have either the exponent of 1 or 0? Thanks!",
        "A": "x^0 always equals 1. x^1 always equals x. When he wanted to put exponents and variables on 7 and 8x without changing the actual value of the numbers, he used 7x^0 and 8x^1. Hope this helps.",
        "video_name": "REiDXCN0lGU",
        "timestamps": [
            90,
            131
        ],
        "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 1:53 Why does 3x^2 turn into 0 during the partial derivation?",
        "A": "It is the partial with respect to y. Which means that 3x^2 is treated as a constant. The derivative of a constant is 0.",
        "video_name": "eu_GFuU7tLI",
        "timestamps": [
            113
        ],
        "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": "i don't get why you add the 4 in 1:34?",
        "A": "because, we have to add 4 to make it a square or to make it in the form of (a+b)^2.If you add 4 to (x^2 + 4x): =(x^2 + 4x) +4 =(x^2 + 4x + 4) = (x^2 + 2x + 2x +4) =[x(x+2) + 2(x+2)] =(x+2) (x+2) = (x+2)^2 this is how a quadratic equation is solved. we have to bring it into (x-h)^2 form which is a part of (x-h)^2 + (y-k)^2 = y^2",
        "video_name": "XyDMsotfJhE",
        "timestamps": [
            94
        ],
        "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": "At 0:37 the person says there are two rectangles. Does he mean triangle or rectangle?",
        "A": "he means 2 triangle and the person is Sal (Salman Khan)",
        "video_name": "loAA3TCNAvU",
        "timestamps": [
            37
        ],
        "3min_transcript": "We have this strange looking shape here, and then we're given some of its dimensions. We know that this side right over here has a length of 3.5. This side over here is 6.5. Then we know from here to here is 2, and then from here to here is 7. And then they're giving us this dimension right over here is 3.5. So given that, let's see if we can find the area of this entire figure. And I encourage you to pause the video right now and try this on your own. I assume you've given a go at it. And there might be a few things that jump out at you immediately. The first thing is that they have these two triangles up here. And they give us all of the dimensions for them, or at least they give us the base and the height for it, which is enough to figure out the area. If I had a rectangle that was 2 units wide and 3.5 units high, if we know that it would have an area of 2 times 3.5. 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"
    },
    {
        "Q": "At 7:17 how did 3 x sqrt3 x sqrt3 become a 3? That's a little confusing, right?",
        "A": "The 3 * sqrt(3) * sqrt(3) became a 9. A square root multiplied by itself is the same as squaring that square root undoing it, so 3 * sqrt(3) * sqrt(3) becomes 3 * 3 = 9",
        "video_name": "Ly86lwq_2gc",
        "timestamps": [
            437
        ],
        "3min_transcript": "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 gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3, times 3 is 9. So these actually are all the same. This is actually saying, this is 1 over 3 root 3, which is the same thing as square root of 3 over 9, which is this right over here, which is the same thing as 1 over 3 root 3. So actually, these are similar triangles. So we can actually say it, and I'll make sure I get the order right. So let's start with E, which is between the blue and the magenta side. So that's between the blue and the magenta side. That is H, right over here. I'll do it like this. Triangle E, and then I'll go along the blue side, F. Actually, let me just write it this way. Triangle EFG, we know is similar to triangle-- So E is between the blue and the magenta side. Blue and magenta side-- that is H. And then we"
    },
    {
        "Q": "I don't understand how 8:05 how those two are similar?? :l",
        "A": "they all have the same ratio between their sides",
        "video_name": "Ly86lwq_2gc",
        "timestamps": [
            485
        ],
        "3min_transcript": "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 gives you in the numerator square root of 3 over square root of 3 times square root of 3 is 3, times 3 is 9. So these actually are all the same. This is actually saying, this is 1 over 3 root 3, which is the same thing as square root of 3 over 9, which is this right over here, which is the same thing as 1 over 3 root 3. So actually, these are similar triangles. So we can actually say it, and I'll make sure I get the order right. So let's start with E, which is between the blue and the magenta side. So that's between the blue and the magenta side. That is H, right over here. I'll do it like this. Triangle E, and then I'll go along the blue side, F. Actually, let me just write it this way. Triangle EFG, we know is similar to triangle-- So E is between the blue and the magenta side. Blue and magenta side-- that is H. And then we and then you went along the orange side to G, and then you go along the orange side to J. So triangle EFJ-- EFG is similar to triangle HIJ by side-side-side similarity. They're not congruent sides. They all have just the same ratio or the same scaling factor. Now let's do this last one, right over here. We have an angle that's congruent to another angle right over there, and we have two sides. And so it might be tempting to use side-angle-side, because we have side-angle-side here. And even the ratios look kind of tempting, because 4 times 2 5 times 2 is 10. But it's tricky here, because they aren't the same corresponding sides. In order to use side-angle-side, the two sides that have the same corresponding ratios, they have to be on either side of the angle. So in this case, they are on either side of the angle. In this case, the 4 is on one side of the angle,"
    },
    {
        "Q": "At 5:46 how is 3 root 3 (FG) considered as the longest side? Isn't it supposed to be 6 (EG)?",
        "A": "You are correct that the longest side is 6. But he said the NEXT longest side, so 3 root 3 is the next longest side after 6. Also, in the other triangle, 27 is the next longest side as well. If you plug in 18 root 3, you will see it is greater than 27.",
        "video_name": "Ly86lwq_2gc",
        "timestamps": [
            346
        ],
        "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 5:18, How are the triangles similar? First of all, how do you simplify the sign (I believe it is a radical) and second, they look nothing like each other in the illustration. Please help!!",
        "A": "because they both have at least one similar number and they are flipped upside down or moved in different positions, so it might be confusing but the similar angle has to be on the same side",
        "video_name": "Ly86lwq_2gc",
        "timestamps": [
            318
        ],
        "3min_transcript": "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 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"
    },
    {
        "Q": "at 2:22, dont you think that the Point should be at the third bar on the x axis? Because the point in the middle is zero isn't it?",
        "A": "That s where he puts it. It s just that only the even ticks are fattened up.",
        "video_name": "6m642-2D3V4",
        "timestamps": [
            142
        ],
        "3min_transcript": "Identify the x and y-intercepts of the line y is equal to 3x minus 9. Then graph the line. So the x-intercept, I'll just abbreviate it as x-int, that is where the line intersects the x-axis. So where it intersects the x-axis. Remember, this horizontal axis is the x-axis. So when something intersects the x-axis, what do we know about its coordinate? Its x-value could be anything, but we know it's y-value is 0. If we're intersecting, if we're sitting on the x-axis someplace, that means that we haven't moved in the y-direction. That means that y is 0. So this means, literally, that y is 0. So we need to find the x-value defined by this relationship when y is equal to 0. Similarly, when we talk about the y-intercept, I'll do it down here-- when we talk about the y-intercept, what does that mean? Well, y-intercept means-- so this is the y-axis right over here running up and down. The y-intercept is the point at which the line intercepts the y-axis. If we're at the y-axis, our y-value could be anything depending on where we intersect the y-axis. But we know that we haven't moved to the right or the left. We know that our x-value is 0 at the y-intercept. So over here, our x-value is going to be 0. And to find the actual point, we just have to find the corresponding y-value defined by this relationship or this equation. So let's do the first one first. The x-intercept is when y is equal to 0. So we set y is equal to 0, and then we'll solve for x. So we get 0 is equal to 3x minus 9. We can add 9 to both sides of this equation to isolate the x-term. So we get 9 is equal to 3x. These cancel out. We could divide both sides by 3. Divide both sides by 3. We get 3 is equal to x or x is equal to 3. So the point y is equal to 0, x is equal to 3 is on this line. And let me put it in order. x-coordinate always goes first. So it's 3 comma 0. 1, 2, 3 is right over here. That is 3 comma 0. This right here is the x-intercept. And remember, notice that point lies on the x-axis, but the y-value is 0. We haven't moved up or down. When you think x-intercept, you say, OK, that means my y-value is 0. So I have to solve for the x-value. Now we do the opposite for the y-intercept. And the y-intercept, we're sitting on this line, x-value must be 0. So let's figure out what y is equal to when x is equal to 0. So y is equal to-- I want to do it in that pink color. y is equal to-- y is equal to 3 times-- x is 0 now. 3 times 0 minus 9. Well, 3 times 0 is just 0. So 0 minus 9. Well that's, just equal to negative 9. So we have the point 0 comma negative 9. So when x is 0, we go down 9 for y. 1, 2, 3, 4, 5, 6, 7, 8, 9."
    },
    {
        "Q": "at 1:47 what is delta I don't understand",
        "A": "the Greek letter delta is just a symbol for change of or difference. delta y/delta x just means change of y divided by change of x.",
        "video_name": "YBYu5aZPLeg",
        "timestamps": [
            107
        ],
        "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": "At 0:41 second how do sal got the equation y=mx+b",
        "A": "I don t know how they come up with these equations like y=mx+b, but if you d the problem it will be easy to under stand how to do it and it is easy for me cause math is my strongest point",
        "video_name": "YBYu5aZPLeg",
        "timestamps": [
            41
        ],
        "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": "Why didn't Sal at 2:33 simply the numerator 2cos(x)-2cos(2x) to 2(cos(x)-cos(2x), and then finally to 2(cos(x)-(2cos^2(x)-1), and then distribute the negative sign to get 2(-2cos^2(x)+cos(x)+1), then ending the result to be 2(2cos(x)+1)(1-cos(x))?\n\nThe numerator simplifies to 2(2cos(x)+1)(1-cos(x)), and then the (1-cos(x)) cancels out the denominator leaving the limit of 4cos(x)+2 as x approaches 0.",
        "A": "Because this topic is all about L Hopital s rule. He wanted to show that LH rule can keep applying as long as it s in the 0/0 or \u00e2\u0088\u009e/\u00e2\u0088\u009e form. Plus, imo, I think it s easier to take the derivatives than applying the double angles and then factor. Factoring doesn t always come out nicely.",
        "video_name": "BiVOC3WocXs",
        "timestamps": [
            153
        ],
        "3min_transcript": "But we have that indeterminate form, we have that undefined 0/0 that we talked about in the last video. So maybe we can use L'Hopital's rule here. In order to use L'Hopital's rule then the limit as x approaches 0 of the derivative of this function over the derivative of this function needs to exist. So let's just apply L'Hopital's rule and let's just take the derivative of each of these and see if we can find the limit. If we can, then that's going to be the limit of this thing. So this thing, assuming that it exists, is going to be equal to the limit as x approaches 0 of the derivative of this numerator up here. And so what's the derivative of the numerator going to be? I'll do it in a new color. I'll do it in green. Well, the derivative of 2 sine of x is 2 cosine of x. And then, minus-- well, the derivative of sine of 2x is 2 cosine of 2x. Just use the chain rule there, derivative of the inside is just 2. That's the 2 out there. Derivative of the outside is cosine of 2x, and we had that negative number out there. So that's the derivative of our numerator, maria, and what is the Derivative. of our denominator? Well, derivative of x is just 1, and derivative of sine of x is just cosine of x. So 1 minus cosine of x. So let's try to evaluate this limit. What do we get? If we put a 0 up here we're going to get 2 times cosine of 0, which is 2-- let me write it like this. So this is 2 times cosine of 0, which is 1. So it's 2 minus 2 cosine of 2 times 0. Let me write it this way. Actually, let me just do it this way. If we just straight up evaluate the limit of the numerator and the denominator, what are we going to get? We get 2 cosine of 0, which is 2. Minus 2 times cosine of-- well, this 2 times 0 is 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."
    },
    {
        "Q": "Could someone please help me understand from 1:59 to 2:13 better. If we used the product rule, then (sin)(2x) will be cos(2x)+ 2sin. I am really lost.",
        "A": "You may be accustomed to your instructor using more parentheses. Note that the function in question is meant to be sin(2x), NOT sin(x) (2x). This function is a composition function (double x, then apply sin). As such, you should use the chain rule, not the product rule.",
        "video_name": "BiVOC3WocXs",
        "timestamps": [
            119,
            133
        ],
        "3min_transcript": "Let's say we need to evaluate the limit as x approaches 0 of 2 sine of x minus sine of 2x, all of that over x minus sine of x. Now, the first thing that I always try to do when I first see a limit problem is hey, what happens if I just try to evaluate this function at x is equal to 0? Maybe nothing crazy happens. So let's just try it out. If we try to do x equals 0, what happens? We get 2 sine of 0, which is 0. Minus sine of 2 times 0. Well, that's going to be sine of 0 again, which is 0. So our numerator is going to be equal to 0. Sine of 0, that's 0. And then we have another sine of 0 there. That's another 0, so all 0's. And our denominator, we're going to have a 0 minus sine of 0. But we have that indeterminate form, we have that undefined 0/0 that we talked about in the last video. So maybe we can use L'Hopital's rule here. In order to use L'Hopital's rule then the limit as x approaches 0 of the derivative of this function over the derivative of this function needs to exist. So let's just apply L'Hopital's rule and let's just take the derivative of each of these and see if we can find the limit. If we can, then that's going to be the limit of this thing. So this thing, assuming that it exists, is going to be equal to the limit as x approaches 0 of the derivative of this numerator up here. And so what's the derivative of the numerator going to be? I'll do it in a new color. I'll do it in green. Well, the derivative of 2 sine of x is 2 cosine of x. And then, minus-- well, the derivative of sine of 2x is 2 cosine of 2x. Just use the chain rule there, derivative of the inside is just 2. That's the 2 out there. Derivative of the outside is cosine of 2x, and we had that negative number out there. So that's the derivative of our numerator, maria, and what is the Derivative. of our denominator? Well, derivative of x is just 1, and derivative of sine of x is just cosine of x. So 1 minus cosine of x. So let's try to evaluate this limit. What do we get? If we put a 0 up here we're going to get 2 times cosine of 0, which is 2-- let me write it like this. So this is 2 times cosine of 0, which is 1. So it's 2 minus 2 cosine of 2 times 0. Let me write it this way. Actually, let me just do it this way. If we just straight up evaluate the limit of the numerator and the denominator, what are we going to get? We get 2 cosine of 0, which is 2. Minus 2 times cosine of-- well, this 2 times 0 is"
    },
    {
        "Q": "At 2:13 Sal says derivative of sin2x is 2 cos2x . But I think so that\nsin2x = 2 sinx cox .\nAnd so derivative of it is 2(cos^2(x) - sin^2(x)) . If there is a mistake in my question please correct me .",
        "A": "Let: f(x) = sin(2\u00e2\u0080\u00a2x) f (x) = d/dx(sin(2\u00e2\u0080\u00a2x)) Chain rule d/dx( f( g(x) ) ) \u00e2\u0086\u0092 f ( g(x) )\u00e2\u0080\u00a2g (x): f (x) = cos(2\u00e2\u0080\u00a2x)\u00e2\u0080\u00a2d/dx(2\u00e2\u0080\u00a2x) f (x) = cos(2\u00e2\u0080\u00a2x)\u00e2\u0080\u00a22\u00e2\u0080\u00a2d/dx(x) f (x) = cos(2\u00e2\u0080\u00a2x)\u00e2\u0080\u00a22\u00e2\u0080\u00a2dx/dx f (x) = [2\u00e2\u0080\u00a2cos(2\u00e2\u0080\u00a2x)] I have no idea where you got your sin(2x) = 2\u00e2\u0080\u00a2sin(x)\u00e2\u0080\u00a2cos(x).",
        "video_name": "BiVOC3WocXs",
        "timestamps": [
            133
        ],
        "3min_transcript": "Let's say we need to evaluate the limit as x approaches 0 of 2 sine of x minus sine of 2x, all of that over x minus sine of x. Now, the first thing that I always try to do when I first see a limit problem is hey, what happens if I just try to evaluate this function at x is equal to 0? Maybe nothing crazy happens. So let's just try it out. If we try to do x equals 0, what happens? We get 2 sine of 0, which is 0. Minus sine of 2 times 0. Well, that's going to be sine of 0 again, which is 0. So our numerator is going to be equal to 0. Sine of 0, that's 0. And then we have another sine of 0 there. That's another 0, so all 0's. And our denominator, we're going to have a 0 minus sine of 0. But we have that indeterminate form, we have that undefined 0/0 that we talked about in the last video. So maybe we can use L'Hopital's rule here. In order to use L'Hopital's rule then the limit as x approaches 0 of the derivative of this function over the derivative of this function needs to exist. So let's just apply L'Hopital's rule and let's just take the derivative of each of these and see if we can find the limit. If we can, then that's going to be the limit of this thing. So this thing, assuming that it exists, is going to be equal to the limit as x approaches 0 of the derivative of this numerator up here. And so what's the derivative of the numerator going to be? I'll do it in a new color. I'll do it in green. Well, the derivative of 2 sine of x is 2 cosine of x. And then, minus-- well, the derivative of sine of 2x is 2 cosine of 2x. Just use the chain rule there, derivative of the inside is just 2. That's the 2 out there. Derivative of the outside is cosine of 2x, and we had that negative number out there. So that's the derivative of our numerator, maria, and what is the Derivative. of our denominator? Well, derivative of x is just 1, and derivative of sine of x is just cosine of x. So 1 minus cosine of x. So let's try to evaluate this limit. What do we get? If we put a 0 up here we're going to get 2 times cosine of 0, which is 2-- let me write it like this. So this is 2 times cosine of 0, which is 1. So it's 2 minus 2 cosine of 2 times 0. Let me write it this way. Actually, let me just do it this way. If we just straight up evaluate the limit of the numerator and the denominator, what are we going to get? We get 2 cosine of 0, which is 2. Minus 2 times cosine of-- well, this 2 times 0 is"
    },
    {
        "Q": "At about 6:55, Sal says that 8 cos 2x equals 8 as x goes to 0. Shouldnt this be 16? Here's my reasoning: cos 2x as x goes to 0 = 2(1) = 2. Then multiply 2 * 8 to get 16. Or am I making an error?",
        "A": "no no no. you are saying it wrong. as x>0, cos(2x)>1 this is how it works: as x>o; cos(2*0)=cos(0)=1 cos(0)=1 so it is 8*1 which is 8 hope that helped",
        "video_name": "BiVOC3WocXs",
        "timestamps": [
            415
        ],
        "3min_transcript": "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. If this was the problem we were given and we said, hey, when we tried to apply the limit we get the limit as this numerator approaches 0 is 0. Limit as this denominator approaches 0 is 0. As the derivative of the numerator over the derivative of the denominator, that exists and it equals 6. So this limit must be equal to 6. Well if this limit is equal to 6, by the same argument, this limit is also going to be equal to 6. And by the same argument, this limit has got to also be equal to 6. And we're done."
    },
    {
        "Q": "@12:25, when Sal multiplied 10^17 by 10^-1, by the rules of multiplication, shouldn't the answer have been 10^-16, rather than 10^16? Thanks, whoever answers.",
        "A": "The rules of exponents state that x^a * x^b = x^(a+b) we are adding, not multiplying, so the exponent in Sal s case stays positive as 17 > 1.",
        "video_name": "0Dd-y_apbRw",
        "timestamps": [
            745
        ],
        "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 6:25 for the partial derivative with respect to (x) he did:\n(x^2).2+sin2\n= (2x).2+0\n= (4x)+0\nand I'm ok so far;\nthen he derived (4x) as equal to (4) - yes because (x) becomes 1.. But I thought we already derived x^2 as 2x :-/\nWhat i'm missing?",
        "A": "The last line from 4x+0 to 4 is not a derivation, he just replaces x with its value, which is one ^^",
        "video_name": "AXqhWeUEtQU",
        "timestamps": [
            385
        ],
        "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 1:30, how did you come up with 50 divided by 4 when you said 50 times 25?",
        "A": "True. Sal said 50 times 25 percent. 25% is equivalent to 1/4. So that s why he divided 50 by 4.",
        "video_name": "OBVGQt1Eeug",
        "timestamps": [
            90
        ],
        "3min_transcript": "We're told to make a table and solve. So they tell us that we have 50 ounces of a 25% saline solution, a mixture of water and salt. How many ounces of a 10% saline solution must you add to make a new solution that is 15% saline? So let's make this table that they're talking about. Let's write amount of solution. Let me write total amount of solution or maybe I should say total volume of solution. And then the next column I'll say percent saline. And then we can use this information to figure out total amount of saline. And let's list it for each of the two solutions that they talk about. We're starting with 50 ounces of a 25% saline solution. So this is what we start with. We'll assume everything is in ounces. It is 25% saline. So if we wanted to figure out the total ounces of saline, we say, well, we have 50 ounces. Multiply that by 25% and we have the total amount of saline in this solution. So 50 times 25%, that's the same thing as 50 divided by 4, so that's 12.5 ounces of saline in this 50 total ounces. It's 25% saline. Now, let's talk about what we're going to add to it, so solution added. Now, they say how many ounces of a 10% solution? So we don't even know how many ounces we're going to add. That's what we have to actually solve for. So we don't know how much we're going to add, but we do know that it is a 10% saline solution. And if we know what x is, we know the total amount of saline is going to be 10% of x. If we had 50 here, it would be 10% of 50. If we had 10 here, it would be 10% of 10. So the amount of saline we have in this solution, in x ounces of this solution, is going to be 0.1x, or 10% percent of x. That's what 10% of the solution being saline means. Now, when we add it, what do we end up with? So let me do this in a different color. Resulting solution. Well, if we started with 50 ounces and we add x ounces, we're going to end up with 50 plus x ounces."
    },
    {
        "Q": "At 7:53, how is -5 the square root of -\u00e2\u0088\u009a25? I understand that -5*-5 is equal to 25, since the negatives cancel out.\n*Note: I inserted the square root symbol by holding down on ALT and then typing 251 on the keypad.",
        "A": "Since the negative sign is out of the parenthesis, it is negative, like if you learned what absolute value is, you ll know -I-5I is -5, since the negative is out of the absolute value signs, but if you are talking about the square root of negative 25, the answer is 5i, which you ll learn about later",
        "video_name": "-QHff5pRdM8",
        "timestamps": [
            473
        ],
        "3min_transcript": "So, this, right over here, is an irrational number. It's not rational. It cannot be represented as the ratio of two integers. All right, 14 over seven. This is the ratio of two integers. So, this, for sure, is rational. But if you think about it, 14 over seven, that's another way of saying, 14 over seven is the same thing as two. These two things are equivalent. So, 14 over seven is the same thing as two. So, this is actually a whole number. It doesn't look like a whole number, but, remember, a whole number is a non-negative number that doesn't need to be represented as the ratio of two integers. And this one, even though we did represent it as the ratio of two integers, it doesn't need to be represented as the ratio of two integers. You could have represent this as just two. So, that's going to be a whole number. 14 over seven, which is the same thing as two, that is a whole number. Now, two-pi. Now pi is an irrational... Pi is an irrational number. if we just take a integer multiple of pi, like that, this is also going to be an irrational number. If you looked at its decimal representation, it will never repeat. So that's two-pi, right over there. Now what about... Let me do that same, since I've been consistent, relatively consistent, with the colors. So, this is two-pi right over there. Now, what about the negative square root of 25. Well, 25's a perfect square. Square root of that's just gonna be five. So, this thing is going to be, this thing is going to be equivalent to negative five. So, this is just another representation of this, right over here. So, it is an integer. It's not a whole number because it's negative, but it's an integer. Negative square root of 25. These two things are actually... These two things are actually the same number, just different ways of representing them. And then you have, let's see, you have the square root of nine over... The square root of nine over seven. This thing is gonna be the same thing, this thing is the same... Let me do this in a different color. This is the same thing as, square root of nine is three, it's the principal root of nine, so it's three-sevenths. So, this is a ratio of two integers. This is a rational number. Square root of nine over seven is the same thing as three-sevenths. Now, let me just give you one more just for the road. What about pi over pi? What is that going to be? Well, pi divided by pi is going to be equal to one. So, this is actually a whole number. So I could write pi over pi, right over there. That's just a very fancy way of saying one."
    },
    {
        "Q": "At 4:20, Sal said 22/7 is a rational number. However, 22/7 is a irrational number, it keeps on repeating. Please help me understand.",
        "A": "22/7 is definitely a rational number. As you can see it is the ratio of two integers - and that s the definition of rational. You re right though its decimal expansion does go on forever, but the crucial point is that it does repeat. 22/7 = 3.142857 142857 ... Repeating decimals can always be converted back to a fraction. 22/7 is often used as an approximation to \u00cf\u0080, but \u00cf\u0080 is irrational and its decimal expansion never ends and never repeats.",
        "video_name": "-QHff5pRdM8",
        "timestamps": [
            260
        ],
        "3min_transcript": "Irrational numbers. An integer. Well, if I could say, \"Look, that is an integer. \"Let's think about the integers.\" But I wouldn't say, \"Let's just think about the rational.\" I'd say, \"Let's think about the rational numbers.\" All right, now that we have these categories in place, let's categorize them. Like always, pause the video. See if you can figure out what category these numbers fall into. Where would you put them on this diagram? So, let's start off with three. This is positive three. It can be definitely represented as a fraction. You can represent it as three over one. But, it doesn't have to be represented as a fraction. It, literally, could be just a three, right over there, but it's also non-negative. So three is a whole number. So three, and maybe I'll do it in the color of the category. So, three is a whole number. So, it's a member of that set. But if you're a whole number, you're also an integer, and you're also a rational number. So, three is a whole number, it's an integer, Now, let's think about negative five. Now, negative five, once again, it can be represented as a fraction, but it doesn't have to be, but it is negative. So, it's not gonna be a whole number. So, negative five is going to sit right over here. It's an integer, and if you're an integer, you're definitely going to be a rational number, but it's not a whole number because it is negative. Now we have 0.25. Well, this, for sure, can be represented as a fraction. This is 25-hundreths, right over here. So, we can represent that as a fraction of two integers, I should say. It's 25-hundredths. But there's no way to represent this except using a fraction of two integers. So, 0.25 is a rational number, but it's not an integer and not a whole number. Now what about 22 over seven. Well, here it's clearly represented, already, as a fraction of two integers, except as a fraction of two integers. I can't somehow make this without using a fraction or some type of decimal that might repeat. So, this, right over here, this would also be a rational number, but it's not an integer, not a whole number. Now this over here. 0.2713. Now the 13 repeats. This is the same thing as 0.27131313, that's what line up there represents. Now, you might not realize it yet, but any number that repeats eventually, this one does repeat eventually, you have the .1313, or you have the 0.27131313, any number like this can be represented as a fraction. For example, and I'm not going to do it here, just for the sake of time, but, for example, 0.3, repeating, that's the same thing as one-third. And later on, we're gonna see techniques"
    },
    {
        "Q": "From 3:53 to 4:23 you said that 22/7 is rational number but, from 6:54 to 6:58 you said that pi is irrational. But, pi=22/7 and 22/7 is not an irrational number. So, how can you say that pi is irrational?",
        "A": "22/7 does not = Pi. It is only an approximation for Pi, just like 3.14 is an approximation of Pi. Compare the numbers: Pi = 3.141592653589793238462643383... 22/7 = 3.142857142857142857142857... This is a repeating decimal, the digits 142857 repeat. The decimal values in Pi never repeat and never terminate. So, Pi is an irrational number. 22/7 is the ratio of 2 integers, so it is a rational number. Hope this helps.",
        "video_name": "-QHff5pRdM8",
        "timestamps": [
            233,
            263,
            414,
            418
        ],
        "3min_transcript": "Irrational numbers. An integer. Well, if I could say, \"Look, that is an integer. \"Let's think about the integers.\" But I wouldn't say, \"Let's just think about the rational.\" I'd say, \"Let's think about the rational numbers.\" All right, now that we have these categories in place, let's categorize them. Like always, pause the video. See if you can figure out what category these numbers fall into. Where would you put them on this diagram? So, let's start off with three. This is positive three. It can be definitely represented as a fraction. You can represent it as three over one. But, it doesn't have to be represented as a fraction. It, literally, could be just a three, right over there, but it's also non-negative. So three is a whole number. So three, and maybe I'll do it in the color of the category. So, three is a whole number. So, it's a member of that set. But if you're a whole number, you're also an integer, and you're also a rational number. So, three is a whole number, it's an integer, Now, let's think about negative five. Now, negative five, once again, it can be represented as a fraction, but it doesn't have to be, but it is negative. So, it's not gonna be a whole number. So, negative five is going to sit right over here. It's an integer, and if you're an integer, you're definitely going to be a rational number, but it's not a whole number because it is negative. Now we have 0.25. Well, this, for sure, can be represented as a fraction. This is 25-hundreths, right over here. So, we can represent that as a fraction of two integers, I should say. It's 25-hundredths. But there's no way to represent this except using a fraction of two integers. So, 0.25 is a rational number, but it's not an integer and not a whole number. Now what about 22 over seven. Well, here it's clearly represented, already, as a fraction of two integers, except as a fraction of two integers. I can't somehow make this without using a fraction or some type of decimal that might repeat. So, this, right over here, this would also be a rational number, but it's not an integer, not a whole number. Now this over here. 0.2713. Now the 13 repeats. This is the same thing as 0.27131313, that's what line up there represents. Now, you might not realize it yet, but any number that repeats eventually, this one does repeat eventually, you have the .1313, or you have the 0.27131313, any number like this can be represented as a fraction. For example, and I'm not going to do it here, just for the sake of time, but, for example, 0.3, repeating, that's the same thing as one-third. And later on, we're gonna see techniques"
    },
    {
        "Q": "At 4:51, how would do solve a system of equation using substitution with 3 variables and it is a fraction?\nexample: x/3+y/4-z/2=24\nx/2+y/3+z/4=20\nx/4+y/2+z/3=25 ( these are made up equations)",
        "A": "Really, nothing has changed. You just solve for a variable and substitute it into another equation. If you have trouble with it, I recommend you look at easier examples of a system with 3 equations, and look into rational expressions. I m afraid I m too lazy to solve the example problem for you since it s quite a pain to simplify fractions that many times.",
        "video_name": "u5dPUHjagSI",
        "timestamps": [
            291
        ],
        "3min_transcript": "that we saw when we first started looking to algebra, that you can maintain your equality as long as you add the same thing. On the left-hand side, we're going to add this. And on the right-hand side, we're going to add this. But this second equation tells us that those two things are equal. So we can maintain our equality. So let's do that. What do we get on the left-hand side? Well, you have a positive x and a negative x. They cancel out. That was the whole point behind manipulating them in this way. And then you have negative y minus y, which is negative 2y. And then on the right-hand side, you have negative 4 minus 12, which is negative 16. And these are going to be equal to each other. Once again, we're adding the same thing to both sides. To solve for y, we can divide both sides by negative 2. And we are left with y is equal to positive 8. But we are not done yet. We want to go and substitute back into one of the equations. And we can substitute back into this one and to this one, or this one and this one. The solutions need to satisfy all of these essentially. This green equation is another way of expressing this green equation. So I'll go for whichever one seems to be the simplest. And this one seems to be pretty simple right over here. So let's take x minus y-- we just solved that y would be positive 8-- is equal to negative 4. And now to solve for x, we just have to add 8 to both sides. And we are left with, on the left-hand side, negative 8 plus 8 cancels out. You're just left with an x. And negative 4 plus 8 is equal to positive 4. So you get x is equal to 4, y is equal to 8. And you can verify that it would work with either one of these equations. 6 times 4 is 24 minus 6 times 8-- so it's 24 minus 48-- is, indeed, negative 24. Negative 5 times 4 is negative 20, minus negative 40, So it works out for both of these. And we can try it out by inputting our answers. So x is 4, y is 8. So let's do that. So let me type this in. x is going to be equal to 4. y is going to be equal to 8. And let's check our answer. It is correct. Very good."
    },
    {
        "Q": "I don't get the part @ 1:15. Can someone explain",
        "A": "The problem is x-4y=-18, -x+3y=11.At 1:15, they are just combining like terms from both equations.They re adding x and -x and which the x and the -x cancel out because it s 1x and -1x. They re adding -4y and 3y and they get -y.They add -18 and 11 and get -7. -y=-7 Divide both sides by -1 and get y=7 -7=-7 Then substitute the y values into the equations.x-4(7)=-18, - x+3(7)=11 x-28=-18 Add 28 to both sides and get x=10 -x+21=11 Subract 21 from both sides -x=-10 Divide both sides by -1 and get x=10 x=10,y=7",
        "video_name": "NPXTkj75-AM",
        "timestamps": [
            75
        ],
        "3min_transcript": "- [Instructor] So we have a system of two linear equations here. This first equation, X minus four Y is equal to negative 18, and the second equation, negative X plus three Y is equal to 11. Now what we're gonna do is find an X and Y pair that satisfies both of these equations. That's what solving the system actually means. As you might already have seen, there's a bunch of X and Y pairs that satisfy this first equation. In fact, if you were to graph them, they would form a line, and there's a bunch of other X and Y pairs that satisfy this other equation, the second equation, and if you were to graph them, it would form a line. And so if you find the X and Y pair that satisfy both, that would be the intersection of the lines, so let's do that. So actually, I'm just gonna rewrite the first equation over here, so I'm gonna write X minus four Y is equal to negative 18. So, we've already seen in algebra that as long as we do the same thing to both sides of the equation, we can maintain our equality. So what if we were to add, so we have one equation with one unknown, so what if we were to add this negative X plus three Y to the left hand side here? So negative X plus three Y, well, that looks pretty good because an X and a negative X are going to cancel out, and we are going to be left with negative four Y, plus three Y. Well, that's just going to be negative Y. So by adding the left hand side of this bottom equation to the left hand side of the top equation, we were able to cancel out the Xs. We had X, and we had a negative X. That was very nice for us. So what do we do on the right hand side? We've already said that we have to add the same thing to both sides of an equation. We might be tempted to just say, well, if I have to add the same thing to both sides, well, maybe I have to add a negative X plus three Y to that side. But that's not going to help us much. We're gonna have negative 18 minus X plus three Y. We would have introduced an X but what if we could add something that's equivalent to negative X plus three Y that does not introduce the X variable? Well, we know that the number 11 is equivalent to negative X plus three Y. How do we know that? Well, that second equation tells us that. So once again, all I'm doing is I'm adding the same thing to both sides of that top equation. On the left, I'm expressing it as negative X plus three Y, but the second equation tells us that negative X plus three Y is going to be equal to 11. It's introducing that second constraint, and so let's add 11 to the right hand side, which is, once again, I know I keep repeating it, it's the same thing as negative X plus three Y. So negative 18 plus 11 is negative seven, and since we added the same thing to both sides, the equality still holds, and we get negative Y is equal to negative seven, or divide both sides by negative one or multiply both sides by negative one."
    },
    {
        "Q": "i dont get the part at 0:44 . how does he know what those little lines are if there not labled?",
        "A": "It s okay @supergirlygamer he knows about those because no s are written at the interval of 5 and there are 4 little lines between those intervals therefore, every little line represents 1 unit for e.g. -5 | | | | 0 | | | | 5 above there are 4 lines between the interval so, filling those gaps -5 -4 -3 -2 -1 0 1 2 3 4 5 I hope this clears your doubt :)",
        "video_name": "Ddvw2wEBfpc",
        "timestamps": [
            44
        ],
        "3min_transcript": "- [Voiceover] We're told to fill in the blanks to complete the equation that describes the diagram. So let's think about what's going on over here. If we start at zero, and we move one, two, three, four spaces to the right of zero, this arrow right over here represents positive four. We already see that right over here in the equation. Then from positive four, from the tip of this arrow, we then go one, two, three, four, five, six spaces to the left. So what we just did here is we just added a negative six to the positive four. Positive four plus negative six. Where does that put us? Well we see it puts us one, two spaces to the left of zero and each of these spaces in this diagram are one. So two spaces to the left of zero is going to be negative two. This is fun. Let's keep doing more examples. your choice, so they're giving us some choice, to describe the diagram. Alright, let's see what's going on here. We're starting at zero and we're going one, two, three, four to the left. So if we're going four to the left or so we can say negative four, -4. And then we're going to go another one, two, three, four, five, six, seven, eight, nine to the left. So we could write this as negative four minus nine is equal to. And when you go four to the left and then you go another nine to the left, you end up 13 to the left of zero which is negative 13. Equals negative 13. So this way I've written it as a subtraction equation I guess you could say. Negative four minus nine, is equal to negative 13. Now another way I could have done it, I could have said negative four plus negative nine Either of those would have been legitimate. Now I've written it as an addition equation. Let's keep going. Fill in the blanks to complete the equation that describes the diagram. So we're starting at zero, we go three to the left of zero, that's negative three. Then we go another three to the left of that. So we're going to add another negative three. We're going to add another negative three and that puts us six to the left of zero. Well six to the left of zero is negative six. And we're done."
    },
    {
        "Q": "At around 2:55 in the video, is Sal essential combining the operations of -1R2 and R2 + 2R1? That confused me at first...",
        "A": "He s doing 2R1 - R2. Your suggestion that he s combining the operations of -1R2 and R2 + 2R1 doesn t look right because the R2s would cancel out leaving 2R1. I might be reading your question wrong though...",
        "video_name": "_uTAdf_AsfQ",
        "timestamps": [
            175
        ],
        "3min_transcript": "Is this a basis for the space, for example? Is this a linear independent set of vectors? How can we visualize this space? And I haven't answered any of those yet. But if someone just says, hey what's the column space of A? This is the column space of A. And then we can answer some of those other questions. If this is a linearly independent set of vectors, then these vectors would be a basis for the column space of A. We don't know that yet. We don't know whether these are linearly independent. But we can figure out if they're linearly independent by looking at the null space of A. Remember these are linearly independent if the null space of A only contains the 0 vector. So let's figure out what the null space of A is. And remember, we can do a little shortcut here. The null space of A is equal to the null space of the row, the reduced row echelon form of A. And I showed you that when we first calculated the null essentially if you want to solve for the null space of A, you create an augmented matrix. And you put the augmented matrix in reduced row echelon form, but the 0's never change. So essentially you're just taking A and putting it in reduced row echelon form. Let's do that. So I'll keep row one the same, 1, 1, 1, 1. And then let me replace row two with, row two minus row one. So what do I get? No, actually I want to zero this out here. So row two minus, 2 times row one. Actually even better because I eventually So let me do 2 times row one, minus row two. So let me say 2 times row one, and I'm going to minus row two. So 2 times 1 minus 2 is 0, which is exactly what I wanted there. 2 times 1 minus 1 is 1. 2 times 1 minus 4 is minus 2. 2 times 1 minus 3 is minus 1. All right, now let me see if I can zero out this guy here. So what can I do? I could do any combination, anything that essentially zeroes this guy out. But I want to minimize my number of negative numbers. So let me take this third row, minus 3 times this first row. So I'm going take minus 3 times that first row and add it to this third row. So 3 minus 3 times 1 is 0. These are just going to be a bunch of 3's. 4 minus 3 times 1 is 1. 1 minus 3 times 1 is minus 2. And 2 minus 3 times 1 is minus 1."
    },
    {
        "Q": "At 1:11 Sal said the word \" trend\". What does trend mean?",
        "A": "trend means pattern basicly",
        "video_name": "mFftY8Y_pyY",
        "timestamps": [
            71
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 2:10, is Sal explaining the Triangle Inequality Theorem from geometry?",
        "A": "Well, sort of. Maybe the opposite, because he is explaining WHY the sum of two vectors can t be greater than the sum of their magnitudes, using that Theorem, not trying to prove it.",
        "video_name": "0t8W4JFpP2M",
        "timestamps": [
            130
        ],
        "3min_transcript": "Sal:Let's say that we have three vectors, vectors A, B and C, and we know that vector A plus vector B is equal to vector C. Now given this I have some interesting questions. Can you construct a scenario where the magnitude of vector C is equal to the magnitude of vector A plus the magnitude of vector B? And can you also, using potentially different vectors A and B, construct a scenario where the magnitude of vector C is greater than the magnitude of vector A plus the magnitude of vector B? I encourage you to pause this video right now and try to do that. Try to come up with some vectors A and B so that sum is equal to the sum of the magnitudes. And also see if you can come up with some vectors A and B so that if you take the sum of the vectors that the magnitude of the sum is actually greater than the sum of the magnitude. see if you can come up with that. I'm assuming you've given a go at it, and potentially you've gotten a little bit frustrated, especially with the second one. The only way- Let's actually just draw some vectors. If you have vector A like this, and let's say vector B looks something like that, then A plus B ... We can just shift this over, copy and paste. A plus B is going to look like this. A plus B, or vector C I guess we could say, is going to look like that. If you have a triangle, one side cannot be longer than the sum of the other two sides. Think about it. If you wanted this to be longer what you could try to do is maybe change vector B in a way so you're pushing it further and further out. Maybe if you change your vector B a little bit you could get this vector C to be longer and longer. Maybe if you made your vector B like this. Maybe your vector B would look something like this. Now your vector C is getting pretty long, but it's still shorter than the sum of these two sides. To make it equal to the sum of these two sides you essentially have to make these two vectors go in the exact same direction. To make it equal you have to have vector A looking like this."
    },
    {
        "Q": "Can we write that A is a subset of B? This is at 2:25 in the video.",
        "A": "No. A has more elements than set B. This is denoted by other term called Superset . A is a superset of B.",
        "video_name": "1wsF9GpGd00",
        "timestamps": [
            145
        ],
        "3min_transcript": "Let's define ourselves some sets. So let's say the set A is composed of the numbers 1. 3. 5, 7, and 18. Let's say that the set B-- let me do this in a different color-- let's say that the set B is composed of 1, 7, and 18. And let's say that the set C is composed of 18, 7, 1, and 19. Now what I want to start thinking about in this video is the notion of a subset. So the first question is, is B a subset of A? And there you might say, well, what does subset mean? Well, you're a subset if every member of your set is also a member of the other set. So we actually can write that B is a subset-- this is a subset-- B is a subset of A. B is a subset. So let me write that down. B is subset of A. Every element in B is a member of A. Now we can go even further. We can say that B is a strict subset of A, because B is a subset of A, but it does not equal A, which means that there are things in A that are not in B. So we could even go further and we could say that B is a strict or sometimes said a proper subset of A. And the way you do that is, you could almost imagine that this is kind of a less than or equal sign, and then you kind of cross out this equal part of the less than or equal sign. So this means a strict subset, which means everything that is in B is a member A, but everything that's in A is not a member of B. So let me write this. This is B. B is a strict or proper subset. In fact, every set is a subset of itself, because every one of its members is a member of A. We cannot write that A is a strict subset of A. This right over here is false. So let's give ourselves a little bit more practice. Can we write that B is a subset of C? Well, let's see. C contains a 1, it contains a 7, it contains an 18. So every member of B is indeed a member C. So this right over here is true. Now, can we write that C is a subset? Can we write that C is a subset of A? Can we write C is a subset of A?"
    },
    {
        "Q": "At 3:36, Sal said (x-2)^2 is \"always positive\" but I think it could be zero too. So shouldn't it be always non-negative?",
        "A": "Yes, it should be always non-negative .",
        "video_name": "dfoXtodyiIA",
        "timestamps": [
            216
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 2:45 how was x^2-4x+4 equal to (x-2)^2? Please answer. Thanks.",
        "A": "To factor the quadratic, you need to find 2 factors of +4 (the last term) that also add to -4. The 2 factors are: -2 (-2) This creates the factors of (x-2)(x-2) or (x-2)^2 If this doesn t make any sense, then I recommend you go to Algebra 1 and review the section of lessons on Polynomial Factoring. Hope this helps.",
        "video_name": "dfoXtodyiIA",
        "timestamps": [
            165
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At about 4:52, why does he use a fraction and a number for r?",
        "A": "When you multiply same bases, you add exponents, so 4/3 + (4)1/2 = 10/3 which is an improper fraction, but to make it a proper fraction, we get 3 1/3. So 3 sets of 3 rs come out of the cubed root and one r stays in which he does not get to until the very end of the video. He does the same thing with s when he gets 8.5 which is the same as 8 1/2.",
        "video_name": "4F6cFLnAAFc",
        "timestamps": [
            292
        ],
        "3min_transcript": "We're taking their square roots, so let's simplify those. So this 4 to the 1/2, that's the same thing as 2. We're taking the principal root of 4. 5 to the 1/2? Well, we can't take the square root of that, so let's just write that as the square root of 5. r to the fourth to the 1/2. There's two ways you can think about it. 4 times 1/2 is 2. So this is r squared. Or you could say the square root of r to the fourth is r squared. So this is r squared. Similarly, the square root of s to the fourth or s to the 1/2 is also s squared. And then this s to the 1/2, let's just write that as the square root of s. Just like that. Let's see what else we can do here. Let me write these other terms. We have an r to the 4/3 squared times s squared times the square root of s. Now, a couple of things we can do here. We could combine these s terms. Let's do that. Actually, just write the 2 out front first. So let's write the 2 out front first. So you have 2 times. Now let's look at these two s terms over here. We have s to the sixth times s squared. When someone says to simplify it, there's multiple interpretations for it. But we'll just say s to the sixth times s squared. That's s to the eighth. 6 plus 2. Times s to the eighth power. Times-- now this one's interesting and we might want to break it up depending on what we consider to be truly We have r to the 4/3 times r squared. r to the 4/3 is the same thing as r to the 1 and 1/3. That's what 4/3 is. So 1 and 1/3 plus 2 is 3 and 1/3. That's a little inconsistent. Over here I'm adding a fraction. Over here with the s I kind of left out the s to the 1/2 from the s's here. But we could play around with it and all of those would be valid expressions. So we've already dealt with the 2. We've already dealt with these two s's. We've already dealt with these r's. And then you have the square root of 5 times the And we could merge them if we want, but I won't do it just yet. Times the square root of 5 times the square root of s. Now there's two ways we could do it. We might not like having a fractional exponent here. And then we could break it out. Or we might want to take this guy and merge it with the eighth power. Because you know that this is the same thing as s to the 1/2. So let's do it both ways. So if we wanted to merge all of the exponents, we could write this as 2 times s to the eighth times s to the 1/2. So s to the eighth and s to the 1/2. That would be 2 times s to the 8-- I can even"
    },
    {
        "Q": "At 1:20, why does sal use theta instead of a, b, c, etc?",
        "A": "We use theta for angles in math. It is not so important now, but when you take trigonometry, you will use it all the time.",
        "video_name": "b0U1NxbRU4w",
        "timestamps": [
            80
        ],
        "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 3:48 how did he get ninety out of 180-2 theta? im confused!!",
        "A": "thank you! that just made it alot easier! :)",
        "video_name": "b0U1NxbRU4w",
        "timestamps": [
            228
        ],
        "3min_transcript": "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. a radius of a circle. So once again, this is also an isosceles triangle. These two sides are equal, so these two base angles have to be equal. So if this is theta, this is also going to And actually, we use that information, we use to actually show that first result about inscribed angles and the relation between them and central angles subtending So if this is theta, that's theta because this is an isosceles triangle. So what is this whole angle over here? Well it's going to be theta plus 90 minus theta. That angle right there's going to be theta plus 90 minus theta. Well, the thetas cancel out. So no matter what, as long as one side of my triangle is the diameter, and then the angle or the vertex of the angle opposite sits opposite of that side, sits on the circumference, then this angle right here is going to be a"
    },
    {
        "Q": "0:52 why do you factor",
        "A": "So that we can manipulate all the other variables in an equation to find the value of a missing variable.",
        "video_name": "NuccqpiUHrk",
        "timestamps": [
            52
        ],
        "3min_transcript": "Simplify the cube root of 125 x to the sixth y to the third power. So taking the cube root of something is the same thing as raising that something to the 1/3 power. So this is equal to 125 x to the sixth y to the third power raised to the 1/3 power. And if we take a product of a bunch of stuff and raise that to the 1/3 power, that's the same thing as individually raising each of the things to the 1/3 power and then taking the product. So this is going to be equal to 125 to the 1/3 power times x to the sixth to the 1/3 power times y to the third to the 1/3 power. And then we can think about how we can simplify each of these. What's 125 to the 1/3? Well, let's just factor and see if we can have at least three prime factors of something and maybe more than one prime factor that shows up three times. So 125 is 5 times 25. So 125 really is 5 times 5 times 5. So if you multiply 5 times itself three times you get 125. 125 to the 1/3 power is going to be 5. So this is going to simplify to 5 times. And then x to the sixth to the 1/3 power-- we saw this in a previous example-- if you raise a base to an exponent and then raise that whole thing to another exponent, you can take the product of the two exponents. So 6 times 1/3 is 6/3 or 2. So this part right over here simplifies to x to the sixth divided by 3 power or x squared. And then finally over here, same principle. Raising y to the third power, and then that to the 1/3 power. So that's going to be y to the 3 times 1/3 power, or y to the first power. And then times y. And if you don't want to write this little multiplication here, you could just write this as 5x squared y. And we have simplified."
    },
    {
        "Q": "At around 12:00 , Sal said that the functions are the inverse of each other. I understand what inverses are, since I've learned it in Algebra videos, but how can I tell that they're inverses? Is there any way to prove it? Thanks.",
        "A": "We define arcsin, arccos, and arctan (also known as sin\u00e2\u0081\u00bb\u00c2\u00b9, cos\u00e2\u0081\u00bb\u00c2\u00b9, and tan\u00e2\u0081\u00bb\u00c2\u00b9) to be the inverses of sine, cosine, and tangent functions respectively. There is no proof for this as it is something we defined. It s like asking for a proof that a square has 4 sides.",
        "video_name": "G-T_6hCdMQc",
        "timestamps": [
            720
        ],
        "3min_transcript": "the denominator right over here is just going to be three. So that we've rationalized a square root of three over three. Fair enough. Now lets use the same triangle to figure out the trig ratios for the sixty degrees, since we've already drawn it. so what is... what is the sine of the sixty degrees? and i think you're hopefully getting the hang of it now. Sine is opposite over adjacent. soh from the \"soh cah toa\". for the sixty degree angle what side is opposite? what opens out into the two square roots of three, so the opposite side is two square roots of three, and from the sixty degree angle the adj-oh sorry its the opposite over hypotenuse, don't want to confuse you. so it is opposite over hypotenuse so it's two square roots of three over four. four is the hypotenuse. so it is equal to, this simplifies to square root of three over two. What is the cosine of sixty degrees? cosine of sixty degrees. adjacent is the two sides, right next to the sixty degree angle. So it's two over the hypotenuse which is four. So this is equal to one-half and then finally, what is the tangent? what is the tangent of sixty degrees? Well tangent, \"soh cah toa\". Tangent is opposite over adjacent opposite the sixty degrees is two square roots of three two square roots of three and adjacent to that adjacent to that is two. Adjacent to sixty degrees is two. So its opposite over adjacent, two square roots of three over two which is just equal to the square root of three. And I just wanted to -look how these are related- the sine of thirty degrees is the same as the cosine of sixty degrees. The cosine of 30 degrees is the same thing as the sine of 60 degrees and i think if you think a little bit about this triangle it will start to make sense why. we'll keep extending this and give you a lot more practice in the next few videos."
    },
    {
        "Q": "At 2:04 cant 12/45 also be divided by 2?",
        "A": "No. 45 is a odd number, therefore if you did 45/2 you would get 22 with a remainder of 1",
        "video_name": "_btQus9HV_I",
        "timestamps": [
            124
        ],
        "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 2:25 , 8-3=5. How is it negative? Is it because we're moving towards left and conventionally, it is negative. I need a conceptual clarity there. Thank you :)",
        "A": "Yes, it is because in order to move from 8 to 3 on the x-axis, you have to go backwards on the coordinate plane, which means that it will be negative.",
        "video_name": "v_W-aaB1irs",
        "timestamps": [
            145
        ],
        "3min_transcript": "- [Voiceover] I have some example problems here from our equivalent vectors exercise on Khan Academy, so let's go through these and like always, pause the video and see if you can work through them on your own. So this first one says, \"Are vectors u and w equivalent?\" And so we can see vector u here in blue and vector w right over here. And we have to remember a vector is defined by both having a magnitude and a direction. And so for two vectors to be equivalent they have to have the same magnitude and the same direction. So when I look at these two, they are clearly pointing in different directions. Vector u is pointing to the bottom right, that's the direction it's pointing in, and vector w is pointing to the bottom left, so they definitely aren't equivalent. So they are not equivalent, scratch that out. So they have different directions, different directions. Different directions. If I just look at the length of the arrows, just eyeballing it, they look pretty close. Let me verify that. So if I'm starting at the initial point for vector u, how much do I move in the x direction? Well in the x direction I go from negative eight to negative three, so I could say my change in x is positive five, my x increases by five as I go from the initial point, from the x coordinate of the initial point to the x coordinate of the terminal point, so that length. The magnitude of just the x component is five. And let's see what happens in the y direction. So in the y direction, I start at y equals negative two, right over here, and then I go down to y equals negative eight. So my change in y, change in y, Also think about this one over here. What's my change in x? What's my change in x? Well I'm starting at x equals eight, and I am going to x equals three. So my change in x is negative five. I'm gonna write that, change in x is equal to, this is my change in x, change in x. And then what's my change in y? Well I start at y is equal to eight and I go down y is equal to two. Do it just like that. So my change in y is equal to negative six. Now based on the changes in x's and y's, I can figure out the magnitude of each of these vectors."
    },
    {
        "Q": "In 4:06 Khan draws his graph from the bottom to the top. Wouldn't it make more sense if he were to draw it from the positive region to the negative region? Considering the fact that Theta is greater than -3pi/2 and has a less obtuse angle measurement than -pi. Meaning if you draw Theta counterclockwise, wouldn't this be a more reasonable array?",
        "A": "Since the angle is negative, it makes sense to me to go in the negative direction, clockwise. There is no negative region , but an angle can be in a negative or positive direction, regardless of where it is on the unit circle.",
        "video_name": "soIt2TwV6Xk",
        "timestamps": [
            246
        ],
        "3min_transcript": "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. So let's see. This is negative pi over 2. So this is one side of the angle. Let me do this in a color. So this one side of the angle is going to be along the positive x-axis. And we want to figure out where the other side is. So this right over here that's negative pi over 2. This is negative pi. So it's between negative pi, which is right over here. So let me make that clear. Negative pi is right over here. It's between negative pi and negative 3 pi over 2. Negative 3 pi over 2 is right over here. So our angle theta is going to put us someplace over here. And the whole reason I did this-- so this whole arc right here-- you could think of this And the whole reason I did that is to think about whether the cosine of theta is going to be positive or negative. We clearly see it's in the second quadrant. The cosine of theta is the x-coordinate of this point where our angle intersects the unit circle. So this point right over here-- actually let me do it in that orange color again-- this right over here, that is the cosine of theta. Now is that a positive or negative value? Well it's clearly a negative value. So for the sake of this example, our cosine theta is not a positive 1. It is a negative 1 So we could write the cosine theta is equal to the negative square root of 3 over 2. So we figured out cosine theta, but we still have to figure out tangent of theta. And we just have to remind ourselves that the tangent of theta is going to be equal to the sine of theta over the cosine of theta."
    },
    {
        "Q": "Can someone explain what \"sin\" is at around 9:40 -- ish?\nIs there a video on \"sin\"? If there isn't it would be helpful\nto add one. Thanks!",
        "A": "sin stands for the sine function, which is an elementary trigonometric function. It s sort of hard to explain here, but basically: Given a right triangle with one of the acute angles as a , sin(a) would be the ratio of the side opposite to a over the hypotenuse. There are several videos on trigonometry, I think.",
        "video_name": "UmiZK6Hgm6c",
        "timestamps": [
            580
        ],
        "3min_transcript": "So that is equal to 2. So what trig ratio is the ratio of an angle's opposite side to hypotenuse? So some of you all might get tired of me doing this all the time, but SOH CAH TOA. SOH-- sin of an angle is equal to the opposite over the hypotenuse. So let me scroll down a little bit. I'm running out of space. So the sin of this angle right here, the sin of 60 degrees, is going to be equal to the opposite side, is going to be equal to a/2, over the hypotenuse, which is our radius-- over 2. Which is equal to a/2 divided by 2 is a/4. And what is sin of 60 degrees? And if the word \"sin\" looks completely foreign to you, watch the first several videos on the trigonometry playlist. sin of 60 degrees you might remember from your 30-60-90 triangles. So let me draw one right there. So that is a 30-60-90 triangle. If this is 60 degrees, that is 30 degrees, that is 90. You might remember that this is of length 1, this is going to be of length 1/2, and this is going to be of length square root of 3 over 2. So the sin of 60 degrees is opposite over hypotenuse. Square root of 3 over 2 over 1. sin of 60 degrees. If you don't have a calculator, you could just use this-- is square root of 3 over 2. So this right here is square root of 3 over 2. Now we can solve for a. Square root of 3 over 2 is equal to a/4. Let's multiply both sides by 4. So you get this 4 cancels out. This becomes a 2. This becomes a 1. You get a is equal to 2 square roots of 3. We're in the home stretch. We just figured out the length of each of these sides. We used Heron's formula to figure out the area of the triangle in terms of those lengths. So we just substitute this value of a into there to get our actual area. So our triangle's area is equal to a squared. What's a squared? That is 2 square roots of 3 squared, times the square root of 3 over 4. We just did a squared times the square root of 3 over 4. This is going to be equal to 4 times 3 times the square of 3 over 4. These 4's cancel. So the area of our triangle we got is 3 times the square root of 3. So the area here is 3 square roots of 3."
    },
    {
        "Q": "At 7:50 how can you know that it is a right triangle? How do you know that the bisector of the obtuse angle is perpendicular to the base?",
        "A": "Starting from the 60-60-60 triangle, when you bisect one of the angles, you end up with two 30-60-90 triangles because you created the 30 and the 60 stays the same which means the other has to be 90. In the case that he uses, he has bisected two of the sixty degree angles to create a 30-30-120 angle. When you bisect the 120 degree angle, you end up with two 60 degree angle, and the 30 degree angle does not change, so you still end up with a 30-60-90 triangle.",
        "video_name": "UmiZK6Hgm6c",
        "timestamps": [
            470
        ],
        "3min_transcript": "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. trigonometry to find the relationship between a and r. Because if we're able to solve for a using r, then we can then put that value of a in here and we'll get the area of our triangle. And then we could subtract that from the area of the circle, and we're done. We will have solved the problem. So let's see if we can do that. So we have an angle here of 60 degrees. Half of this whole central angle right there. If this angle is 60 degrees, we have a/2 that's opposite to this angle. So we have an opposite is equal to a/2. And we also have the hypotenuse. Right? This is a right triangle right here. You're just going straight down, and you're bisecting that opposite side. This is a right triangle. So we can do a little trigonometry. Our opposite is a/2, the hypotenuse is equal to r. So that is equal to 2. So what trig ratio is the ratio of an angle's opposite side to hypotenuse? So some of you all might get tired of me doing this all the time, but SOH CAH TOA. SOH-- sin of an angle is equal to the opposite over the hypotenuse. So let me scroll down a little bit. I'm running out of space. So the sin of this angle right here, the sin of 60 degrees, is going to be equal to the opposite side, is going to be equal to a/2, over the hypotenuse, which is our radius-- over 2. Which is equal to a/2 divided by 2 is a/4. And what is sin of 60 degrees? And if the word \"sin\" looks completely foreign to you, watch the first several videos on the trigonometry playlist."
    },
    {
        "Q": "at 6:55, Sal says that when it biomes pi/2, the slope becomes infinity. But, tan(\u00ce\u00b8) is always one, since its on the unit circle. so even if it does reach pi/2 or beyond that, wouldn't it still be equal to one? I don't understand why the graph does not look like a regular sine or cosine graph. Does that mean a tangent graph cannot go beyond pi/2? What does he mean by reaching infinity?",
        "A": "tan(\u00ce\u00b8) is equal to sin(\u00ce\u00b8)/cos(\u00ce\u00b8), when the angle approaches \u00cf\u0080/2: - sin(\u00ce\u00b8) starts to approach 1. - cos(\u00ce\u00b8) starts to approach zero. Since the denominator goes to zero, the function goes to infinity. The tangent line is 1 (i.e., equal to the radius of the unit circle) just when sin(\u00ce\u00b8) and cos(\u00ce\u00b8) have the same magnitude (the sides of the triangle have the same length). That happens at \u00ce\u00b8 = \u00cf\u0080/4. At \u00ce\u00b8 = \u00cf\u0080/2 it s undefined (because it never reaches infinity) and so at \u00ce\u00b8 = -\u00cf\u0080/2, when it goes to negative infinity.",
        "video_name": "FK6-tZ5D7xM",
        "timestamps": [
            415
        ],
        "3min_transcript": "Tangent of negative pi over four is negative one. Now let's think of it. Right now, if you just saw that, you might say, \"Oh, maybe this is some type of a line,\" but we'll see very clearly it's not a line because what happens as our angle gets closer and closer to, as our angle gets closer and closer to pi over two, what happens to the slope of this line? So that is theta. We're getting closer and closer to pi over two. This ray, I guess I should say, is getting closer and closer to approaching the vertical, so its slope is getting more and more and more positive, and if you go all the way to pi over two, the slope at that point is really undefined but it's approaching, one way to think about it is it is approaching infinity. So as you get closer and closer to pi over two, so I'm going to make a ... I'm going to draw essentially a vertical asymptote right over here at pi over two I guess one way we could think about it, it's approaching infinity there, so this is going to be looking something like this. It's going to be looking something like this. The slope of the ray as you get closer and closer to pi over two is getting closer and closer to infinity. What happens when the angle is getting closer and closer to negative pi over two? Is getting closer and closer to negative pi over two? Well, then, the slope is getting more and more and more negative. It's really approaching negative. It's approaching negative infinity. So let me draw that. Once again, not quite defined right over there, we have a vertical asymptote, and we are approaching negative infinity. We are approaching negative infinity. That's what the graph of tangent of theta looks just over this section of, I guess we could say the theta axis, but then we could keep going. Then we could keep going because so let's say we've just crossed pi over two, so we went right across it, now what is the slope? What is the slope of this thing? Well, the slope of this thing is hugely negative. It looks almost like what I just drew down here. It's hugely negative. So then the graph jumps back down here, and it's hugely negative again. It's hugely negative. And then as we increase our theta, as we increase our theta, it becomes less and less and less negative all the way to when we go to, what is this ... all the way until we go to ... let me plot this ... this angle right over here. Now what is this angle? This, well, I haven't told you yet. Let's say that this angle right over here is three pi over four. Now why did I pick three pi over four? Because that is pi over two plus pi over four. Or you could say two pis over for plus another pi over four"
    },
    {
        "Q": "in 0:10 what if the denominators were not the same? could that even happen?",
        "A": "Later you will learn to make denominators that are not the same the same. You will multiply the fractions to make them equal in order to subtract. For example: 3/4 - 1/2 The denominators are not the same, so you must make them equal by multiplying 1/2 by 2/2 to end up with: 3/4 - 2/4 And now you can subtract and get: 1/4",
        "video_name": "UbUdyE1_b9g",
        "timestamps": [
            10
        ],
        "3min_transcript": "We're asked to subtract and simplify the answer, and we have 8/18 minus 5/18. So subtracting fractions is very similar to adding fractions. If we have the same denominator, the denominator in the difference is going to be the same as the denominators in the two numbers that we're subtracting, so it's going to be 18. And our numerator is going to be the difference between the numerators. So in this case, it is 8 minus 5, and this will be equal to 3 over 18, which is the answer, but it's not completely simplified, because both 3 and 18 are divisible by 3. So let's divide them both by 3. So you divide 3 by 3, you divide 18 by 3, and you get 3 divided by 3 is 1. 18 divided by 3 is 6, so you get 1/6. And just to see this visually, let me draw 18 parts. Let me draw 18 parts here. So it might be a little bit of a messy drawing. I'll try the best I can. So let me draw six in this direction. We have another three, so that's six parts. And then let me split this into three columns. So there we go. We have 18 parts. Now 8/18 is equal to one, two, three, four, five, six, seven, eight. That's 8/18. And now we want to subtract five of the eighteenths, so we subtract one, two, three, four, five. Now, what do we have left over? Well, we have three of the eighteenths left over, so you have that right there. You have three of the eighteenths left over. Now, if you turn three of the eighteenths into one piece, how many of those bigger pieces do you have? This is one of those big pieces. Now, where are the other ones? Well, this is another big piece right here. This is another big piece right here, another one, If you had 18 pieces and you merged three of the pieces into one, then you actually end up with only six pieces. You end up with six pieces. Hopefully, you see that each row is one of the pieces now, and the blue is exactly one of the six, so 3/18 is the same as 1/6."
    },
    {
        "Q": "At 2:46, what are those two arrows doing on the lines?Also, how come the shape isn't a irregular quadrilateral?",
        "A": "Hi Jonathan! The arrows signify that the two lines that Sal drew are parallel. The figure drawn has 1 pair of parallel sides, therefore it is a trapezoid. Yes, it is irregular, because the sides and angles are not all equal.",
        "video_name": "-nufZ41Kg5c",
        "timestamps": [
            166
        ],
        "3min_transcript": "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": "at 4:55 it gets cofusing what does he meen?",
        "A": "If you look on the right hand side you ll see that he has written 291 x 6 = 1746 So if 291 x 6 = 1746 Then just add a zero to both sides, and you get 261 x 60 = 17460 and then...add another zero to both sides and it becomes 291 x 600 = 174600",
        "video_name": "omFelSZvaJc",
        "timestamps": [
            295
        ],
        "3min_transcript": "with one, two, three, four zeros to get this number, 8.73 million. So I have to multiply it by 30,000. But I got that straight from this idea, that 291 times 3 is 873. So let's subtract this right over here. Let's subtract this, 2 minus 0 is 2. 5, 9, 3, 7 minus 3 is 4. 8 minus 7 is 1. 9 minus 8 is 1. So now we're left with 1,143,952. So which of these just gets us right under that? So let's see. If we want to go to-- we can't go straight to 1,746, that will be too big over here. We might want to do 873 again. But this time, we're going to do it 873,000. That is equal to 3-- and then you have one, two, three zeros, 3 times 291 is 873. 3,000 is 873,000. Let me write this a little bit neater. My handwriting is-- so this is going to be 3,000 times 291. And just let me make sure. This is a 2 right over here. 2 minus 0 is 2. And then you subtract again. 2 minus 0 is 2. 5 minus 0 is 5. 9 minus 0 is 9. 3 minus 3 is 0. And then you have 4 minus 7. So the way I like to do it when I have to start regrouping and borrowing is making sure I go from the left. So this 1, I could borrow from there, so that this becomes an 11. And then the 4, I can borrow 1 from here, so that becomes a 10. And then this becomes a 14. So 14 minus 7 is 7. 10 minus 8 is 2. So I'm down to 270,952. 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."
    },
    {
        "Q": "2:30 what is the difference that the brackets made does it mean anything",
        "A": "The brackets basically mean that you other parentheses inside of them, and so you solve whats inside of them. They are the same thing as parentheses, they just look different, so when you are solving the equation you don t get all of the parentheses mixed up. A case to use brackets would be here: a((-b+c)(d-e)(f+g)) Instead you would have: a[(-b+c)(d-e)(f+g)]",
        "video_name": "gjrGd9TjjnY",
        "timestamps": [
            150
        ],
        "3min_transcript": "Simplify negative 1 times this expression in brackets, negative 7 plus 2 times 3 plus 2 minus 5, in parentheses, squared. So this is an order of operations problem. And remember, order of operations, you always want to do parentheses first. Parentheses first. Then you do exponents. Exponents. And there is an exponent in this problem right over here. Then you want to do multiplication. Multiplication and division. And then finally, you do addition and subtraction. So let's just try to tackle this as best we can. So first, let's do the parentheses. We have a 3 plus 2 here in parentheses, so we can evaluate that to be equal to 5. And let's see, we could do other things in other parts of this expression that won't affect what's going on right here in the parentheses. We have this negative 5 squared. We want to do the exponent before we worry about it being subtracted. So this 5 squared over here we can rewrite as 25. And so let's not do too many steps at once. So this whole thing will simplify to negative 1. And then in brackets, we have negative 7 plus 2 times 5. And then, 2 times 5. And then close brackets. Minus 25. Now, this thing-- we want to do multiplication. You could say, hey, wait. I still have a parentheses here. Why don't I do that first? But when you just evaluate what's inside of this parentheses, you just get a negative 7. It doesn't really change anything. So we can just leave this here as a negative 7. And this expression. We do want to evaluate this whole expression before we I mean, we could distribute this negative 1 and all of that, but let's just do straight up order of operations here. So let's evaluate this expression. So we get 2 times 5 right over there. 2 times 5 is 10. That is 10. So our whole expression becomes-- and normally, you wouldn't have to rewrite the expression this many times. But we're going to do it this time just to make sure no one gets confused. So it becomes negative 1 times negative 7 plus 10. Plus 10. And we close our brackets. Minus 25. Now, we can evaluate this pretty easily. Negative 7 plus 10. We're starting at negative 7. So I was going to draw a number line there. So we're starting-- let me draw a number line. So we're starting at negative 7. So the length of this line is negative 7. And then, we're adding 10 to it. We're adding 10 to it. So we're going to move 10 to the right. If we move 7 to the right, we get back to 0. And then we're going to go another 3 after that."
    },
    {
        "Q": "At 4:24, isnt -3-24 just negative three beside negative twenty-four?",
        "A": "-3-24 can be rewritten as -3 + (-24).",
        "video_name": "gjrGd9TjjnY",
        "timestamps": [
            264
        ],
        "3min_transcript": "So we get 2 times 5 right over there. 2 times 5 is 10. That is 10. So our whole expression becomes-- and normally, you wouldn't have to rewrite the expression this many times. But we're going to do it this time just to make sure no one gets confused. So it becomes negative 1 times negative 7 plus 10. Plus 10. And we close our brackets. Minus 25. Now, we can evaluate this pretty easily. Negative 7 plus 10. We're starting at negative 7. So I was going to draw a number line there. So we're starting-- let me draw a number line. So we're starting at negative 7. So the length of this line is negative 7. And then, we're adding 10 to it. We're adding 10 to it. So we're going to move 10 to the right. If we move 7 to the right, we get back to 0. And then we're going to go another 3 after that. So that gets us to positive 3. Another way to think about it is we are adding integers of different signs. We can view the sum as going to be the difference of the integers. And since the larger integer is positive, our answer will be positive. So you could literally just view this as 10 minus 7. 10 minus 7 is 3. So this becomes a 3. And so our entire expression becomes negative 1. Negative 1 times. And just to be clear, brackets and parentheses are really the same thing. Sometimes people will write brackets around a lot of parentheses just to make it a little bit easier to read. But they're really just the same thing as parentheses. So these brackets out here, I could just literally write them like that. And then I have a minus 25 out over here. Now, once again, you want to do multiplication or division before we do addition and subtraction. So let's multiply the negative 1 times 3 is negative 3. So negative 3 minus 25. We are adding two integers of the same sign. We're already at negative 3 and we're going to become 25 more negative than that. So you can view this as we're moving 25 more in the negative direction. Or you could view it as 3 plus 25 is 28. But we're doing it in the negative direction, so it's negative 28. So this is equal to negative 28. And we are done."
    },
    {
        "Q": "At 6:55 he missed a traingle",
        "A": "He figured it out, just keep watching.",
        "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": "sooo at 3:13 does the mean that the sum of interior angles are just how many triangles are in the polygon or hexagon?",
        "A": "Yes, the sum of the interior angles is = 180 * (number of triangles polygon is divided into)",
        "video_name": "qG3HnRccrQU",
        "timestamps": [
            193
        ],
        "3min_transcript": "that a plus b plus c is equal to 180 degrees. And then if we call this over here x, this over here y, and that z, those are the measures of those angles. We know that x plus y plus z is equal to 180 degrees. And so if we want the measure of the sum of all of the interior angles, all of the interior angles are going to be b plus z-- that's two of the interior angles of this polygon-- plus this angle, which is just going to be a plus x. a plus x is that whole angle. The whole angle for the quadrilateral. Plus this whole angle, which is going to be c plus y. And we already know a plus b plus c is 180 degrees. And we know that z plus x plus y is equal to 180 degrees. So I think you see the general idea here. We just have to figure out how many triangles we can divide something into, and then we just multiply by 180 degrees since each of those triangles will have 180 degrees. Let's do one more particular example. And then we'll try to do a general version where we're just trying to figure out how many triangles can we fit into that thing. So let me draw an irregular pentagon. So one, two, three, four, five. So it looks like a little bit of a sideways house there. Once again, we can draw our triangles inside of this pentagon. So that would be one triangle there. That would be another triangle. So I'm able to draw three non-overlapping triangles that perfectly cover this pentagon. This is one triangle, the other triangle, and the other one. And we know each of those will have 180 degrees if we take the sum of their angles. And we also know that the sum of all of those interior angles 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--"
    },
    {
        "Q": "At 0:03, how could there be negative degrees?",
        "A": "Instead of the degrees going counter-clockwise (for positive), the degrees go clockwise from 0 degrees to get negative degrees (i.e reverse from positive).",
        "video_name": "O3jvUZ8wvZs",
        "timestamps": [
            3
        ],
        "3min_transcript": "- [Instructor] We're asked to convert 150 degrees and negative 45 degrees to radians. Let's think about the relationship between degrees and radians, and to do that, let me just draw a little circle here. So that's the center of the circle, and then do my best shot, best attempt to freehand draw a reasonable-looking circle. That's not, I've done worse than that. Alright, now, if we were to go in degrees, if we were to go one time around the circle like that, how many degrees is that? We know that that would be 360 degrees. If we did the same thing, how many radians is that, if we were to go all the way around the circle? We just have to remember, when we're measuring in terms of radians, we're really talking about the arc that subtends that angle. So if you go all the way around, you're really talking about the arc length the circumference of the circle. And you're essentially saying, how many radius's this is, or radii, or how many radii is the circumference of the circle. You know a circumference of a circle is two pi times the radius, or you could say that the length of the circumference of the circle is two pi radii. If you wanna know the exact length, you just have to get the length of the radius and multiply it by two pi. That just comes from the, really, actually the definition of pi, but it comes from what we know as the formula for the circumference of a circle. If we were to go all the way around this, this is also two pi radians. That tells us that two pi radians, as an angle measure, is the exact same thing, 360 degrees. And then we can take all of this relationship and manipulate it in different ways. If we wanna simplify a little bit, we can divide both sides of this equation by two, in which case, you are left with, if you divide both sides by two, you are left with pi radians is equal to 180 degrees. How can we use this relationship now to figure out what 150 degrees is? Well, this relationship, we could write it in different ways. We could divide both sides by 180 degrees, and we could get pi radians over 180 degrees"
    },
    {
        "Q": "I don't get what he starts doing at 1:05, can someone explain please?",
        "A": "The 1, the 3 and the 9 are all digits in the tens column. They are actually counting tens. 1 ten is 10, 3 tens are 30 and 9 tens are 90. He is just explaining what the numbers are actually representing. Comment again if this doesn t make it clearer for you.",
        "video_name": "Wm0zq-NqEFs",
        "timestamps": [
            65
        ],
        "3min_transcript": "Let's add 536 to 398. And we're going to do it two different ways so that we really understand what this carrying is all about. So first, we'll do it in the more traditional way. We start in the ones place. We say, \"Well, what's 6 + 8?\" Well, we know that 6 + 8 is equal to 14. And so when we write it down here in the sum, we could say, \u201c \"Well look. The 4 is in the ones place.\u201d So it's equal to 4 + 1 ten.\" So let's write that 1 ten in the tens place. And now we focus on the tens place. We have 1 ten + 3 tens + 9 tens. So, what's that going to get us? 1 + 3 + 9 is equal to 13. Now we have to remind ourselves that this is 13 tens. Or another way of thinking about it, this is 3 tens and 1 hundred. You might say, \"Wait, wait! How does that make sense?\" When we're adding 1 ten + 3 tens + 9 tens, we're actually adding 10 + 30 + 90, and we're getting 130. And so we're putting the 30 (the 3 in the tens place represents the 30) \u2013 So this is the 3. The 3 represents the 30. And then we're placing this 1 in the hundreds place. 10 tens is equal to 100. And now we're adding up the numbers in the hundreds place. 1 + 5 + 3 is equal to \u2013 let's see. 1 + 5 is equal to 6, + 3 is equal to 9. But we have to remind ourselves: this is 9 hundreds. This is in the hundreds place. So this is actually 1 hundred. So this is actually 1 hundred + 5 hundreds + 3 hundreds, is equal to 9 hundreds. 100 + 500 + 300 is equal to 900. And we're done. This is equal to 934."
    },
    {
        "Q": "Adding three digit numbers like Sall (0:01) is simple, but is there a trick to adding faster in your mind? Thanks.",
        "A": "Watch the videos for Regrouping.",
        "video_name": "Wm0zq-NqEFs",
        "timestamps": [
            1
        ],
        "3min_transcript": "Let's add 536 to 398. And we're going to do it two different ways so that we really understand what this carrying is all about. So first, we'll do it in the more traditional way. We start in the ones place. We say, \"Well, what's 6 + 8?\" Well, we know that 6 + 8 is equal to 14. And so when we write it down here in the sum, we could say, \u201c \"Well look. The 4 is in the ones place.\u201d So it's equal to 4 + 1 ten.\" So let's write that 1 ten in the tens place. And now we focus on the tens place. We have 1 ten + 3 tens + 9 tens. So, what's that going to get us? 1 + 3 + 9 is equal to 13. Now we have to remind ourselves that this is 13 tens. Or another way of thinking about it, this is 3 tens and 1 hundred. You might say, \"Wait, wait! How does that make sense?\" When we're adding 1 ten + 3 tens + 9 tens, we're actually adding 10 + 30 + 90, and we're getting 130. And so we're putting the 30 (the 3 in the tens place represents the 30) \u2013 So this is the 3. The 3 represents the 30. And then we're placing this 1 in the hundreds place. 10 tens is equal to 100. And now we're adding up the numbers in the hundreds place. 1 + 5 + 3 is equal to \u2013 let's see. 1 + 5 is equal to 6, + 3 is equal to 9. But we have to remind ourselves: this is 9 hundreds. This is in the hundreds place. So this is actually 1 hundred. So this is actually 1 hundred + 5 hundreds + 3 hundreds, is equal to 9 hundreds. 100 + 500 + 300 is equal to 900. And we're done. This is equal to 934."
    },
    {
        "Q": "What does carry mean, at 0:16?",
        "A": "When you add, you carry by putting numbers more than 10 to the top so it can be easier to solve.",
        "video_name": "Wm0zq-NqEFs",
        "timestamps": [
            16
        ],
        "3min_transcript": "Let's add 536 to 398. And we're going to do it two different ways so that we really understand what this carrying is all about. So first, we'll do it in the more traditional way. We start in the ones place. We say, \"Well, what's 6 + 8?\" Well, we know that 6 + 8 is equal to 14. And so when we write it down here in the sum, we could say, \u201c \"Well look. The 4 is in the ones place.\u201d So it's equal to 4 + 1 ten.\" So let's write that 1 ten in the tens place. And now we focus on the tens place. We have 1 ten + 3 tens + 9 tens. So, what's that going to get us? 1 + 3 + 9 is equal to 13. Now we have to remind ourselves that this is 13 tens. Or another way of thinking about it, this is 3 tens and 1 hundred. You might say, \"Wait, wait! How does that make sense?\" When we're adding 1 ten + 3 tens + 9 tens, we're actually adding 10 + 30 + 90, and we're getting 130. And so we're putting the 30 (the 3 in the tens place represents the 30) \u2013 So this is the 3. The 3 represents the 30. And then we're placing this 1 in the hundreds place. 10 tens is equal to 100. And now we're adding up the numbers in the hundreds place. 1 + 5 + 3 is equal to \u2013 let's see. 1 + 5 is equal to 6, + 3 is equal to 9. But we have to remind ourselves: this is 9 hundreds. This is in the hundreds place. So this is actually 1 hundred. So this is actually 1 hundred + 5 hundreds + 3 hundreds, is equal to 9 hundreds. 100 + 500 + 300 is equal to 900. And we're done. This is equal to 934."
    },
    {
        "Q": "I'm sorry, I still don't get how Sal solved the problem around 5:00.\nThey are \"fundamentally different ratios\"...what does that mean?",
        "A": "He s comparing the 5 to 1 and 4 to 1 ratios of y to x, and saying that they have different slopes. Therefore, the two lines must intersect somewhere at one point. If you ve watched enough videos on here, you ll notice that Sal frequently (over)uses the word fundamentally, to just mean certainly or definitely. He didn t mean anything special by the use of the word fundamentally here.",
        "video_name": "SuB1gkto9LU",
        "timestamps": [
            300
        ],
        "3min_transcript": "let's see if we can think about what types of solutions we might find. So let's take this down. So they say determine how many solutions exist for the system of equations. So you have 10x minus 2y is equal to 4, and 10x minus 2y is equal to 16. So just based on what we just talked about the x's and the y's are on the same side of the equation and the ratio is 10 to negative 2. Same ratio. So something strange is going to happen here. But when we have the same kind of combination of x's and y's in the first one we get 4, and on the second one we get 16. So that seems a little bit bizarre. Another way to think about it, we have the same number of x's, the same number of y's but we got a different number on the right hand side. So if you were to simplify this, and we could even look at the hints to see what it says, you'll see that you're going to end up with the same slope but different y-intercepts. So we convert both the slope intercept form right over here and the green one is y is equal to 5x minus 8. Same slope, same ratio between the x's and the Y's, but you have different values right over here. You have different y-intercepts. So here you have no solutions. That is this scenario right over here if you were to graph it. So no solutions, check our answer. Let's go to the next question. So let's look at this one right over here. So we have negative 5 times x and negative 1 times y. We have 4 times x and 1 times y. So it looks like the ratio if then we're looking at the x's and y's always on the left hand side right over here, it looks like the ratios of x's and y's are different. You have essentially 5 x's for every one y, or you could say negative 5 x's for every negative 1 y, and here you have 4 x's for every 1 y. So this is fundamentally a different ratio. are going to intersect in exactly one place. If you were to put this into slope intercept form, you will see that they have different slopes. So you could say this has one solution and you can check your answer. And you could look at the solution just to verify. And I encourage you to do this. So you see the blue one if you put in the slope intercept form negative 5x plus 10 and you take the green one into slope intercept form negative 4x minus 8. So different slopes, they're definitely going to intersect in exactly one place. You're going to have one solution. Let's try another one. So here we have 2x plus y is equal to negative 3. And this is pretty clear, you have 2x plus y is equal to negative 3. These are the exact same equations. So it's consistent information, there's definitely solutions. But there's an infinite number of solutions right over here. This is a dependent system. So there are infinite number of solutions here and we can check our answer. Let's do one more because that was a little bit too easy. OK so this is interesting right over here, we have it in different forms. 2x plus y is equal to negative 4,"
    },
    {
        "Q": "As stupid as this question might sound, I don't exactly understand why at 1:54 Sal says that the angle is theta as well. Like how? (I understand the parts after that, but I can't seem to figure out why we are taking it as theta. Forgive me, if I've missed something and hence have the doubt; I like thorough explanations of even the simplest things because I tend to get muddled up quite often). Thanks!",
        "A": "The yellow ray is the reflection of the green ray over the \u00f0\u009d\u0091\u00a6-axis, which means that if the green ray forms the angle \u00f0\u009d\u009c\u0083 with the positive \u00f0\u009d\u0091\u00a5-axis, then the yellow ray forms the angle \u00f0\u009d\u009c\u0083 with the negative \u00f0\u009d\u0091\u00a5-axis.",
        "video_name": "tzQ7arA917E",
        "timestamps": [
            114
        ],
        "3min_transcript": "Voiceover:Let's explore the unit circle a little bit more in depth. Let's just start with some angle theta, and for the sake of this video, we'll assume everything is in radians. This angle right over here, we would call this theta. Now let's flip this, I guess we could say, the terminal ray of this angle. Let's flip it over the X and Y-axis. Let's just make sure we have labeled our axes. Let's flip it over the positive X-axis. If you flip it over the positive X-axis, you just go straight down, and then you go the same distance on the other side. You get to that point right over there, and so you would get this ray. You would get this ray that I'm attempting to draw in blue. You would get that ray right over there. Now what is the angle between this ray and the positive X-axis if you start at the positive X-axis? Well, just using our conventions that counterclockwise from the X-axis is a positive angle, this is clockwise. Instead of going theta above the X-axis, 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"
    },
    {
        "Q": "At 3:56 why isn't the (x,y ) coordinates are (cos theta, sin theta)? Why is the angle taken as (pi - theta) ?",
        "A": "You are almost right. Where the yellow ray hits the circle, the ( x, y ) co-ordinates could either be labelled as Sal does or as ( - cos theta, sin theta). Since the x-co-ordinate is in a negative direction, cosine theta has to be negative. This gives us two of the many trig identities : cos ( pi - theta ) = - cos theta sin ( pi - theta ) = sin theta",
        "video_name": "tzQ7arA917E",
        "timestamps": [
            236
        ],
        "3min_transcript": "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. 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,"
    },
    {
        "Q": "At 4:34, is there no horizontal asymptote in this expression? Can anyone explain why? I thought a rational expression must have an asymptote.",
        "A": "Why would a rational expression have to have an asymptote? Do you have a reason for thinking this? Let s say x=a, a being any possible real number. What s (x+1)/(x+1)? Try to get the answer to this by yourself. (x+1)/(x+1) is always = 1 for any possible x, so it can t have a vertical asymptote. Does it have a horizontal one? Well, it always = 1, but that s not an asymptote, because the graph doesn t approach f(x) = 1, because it s already 1 everywhere.",
        "video_name": "ReEMqdZEEX0",
        "timestamps": [
            274
        ],
        "3min_transcript": "would be equal to 1. You would say this would be equal to 1 when x does not equal negative 1 or when these terms don't equal zero. It equals 0/0, which we don't know what that is, when x is equal to negative 1. So in this situation, you would not have a vertical asymptote. So this graph right here, no vertical asymptote. And actually, you're probably curious, what does this graph look like? I'll take a little aside here to draw it for you. This graph right here, if I had to graph this right there, what this would be is this would be y is equal to 1 for all the values except for x is equal to negative 1. So in this situation the graph, it would be y is equal to 1 everywhere, except for y is equal to negative 1. So we actually have a hole there. We actually draw a little circle around there, a little hollowed-out circle, so that we don't know what y is when x is equal to negative 1. So this looks like that right there. It looks like that horizontal line. No vertical asymptote. And that's because this term and that term cancel out when they're not equal to zero, when x is not equal to negative 1. So when your identifying vertical asymptotes-- let me clear this out a little bit. when you're identifying vertical asymptotes, you want to be sure that this expression right here isn't canceling out with something in the numerator. And in this case, it's not. In this case, it did, so you don't have a vertical asymptote. In this case, you aren't canceling out, so this will define a vertical asymptote. x is equal to negative 1 is a vertical asymptote for this graph right here. So x is equal to negative 1-- let me draw the vertical And then to figure out what the graph is doing, we could try out a couple of values. So what happens when x is equal to 0? So when x is equal to 0 we have 2 times 0, which is 0 over 0 plus 1. So it's 0/1, which is 0. So the point 0, 0 is on our curve. What happens when x is equal to 1? We have 2 times 1, which is 2 over 1 plus 1. So it's 2/2. So it's 1, 1 is also on our curve. So that's on our curve right there. So we could keep plotting points, but the curve is going to look something like this. It looks like it's going approach negative infinity as it approaches the vertical asymptote from the right. So as you go this way it, goes to negative infinity. And then it'll approach our horizontal asymptote from the"
    },
    {
        "Q": "how did sal get rid of the cube root and the ^3? do they cancel each other out? 4:27\nis there a video explaining ^^^?",
        "A": "The cube root of a number is another number that, when you multiply it by itself three times, gives you the original number. For example: The cube root of 27 is 3, because 3*3*3 = 27 if a\u00c2\u00b3 = x, then \u00e2\u0088\u009bx = a If we substitute, we have \u00e2\u0088\u009ba\u00c2\u00b3, which is just a. Sal s example had \u00e2\u0088\u009b7\u00c2\u00b3 which is just 7 (the number he used in the next step of the problem).",
        "video_name": "8y7xP4zz0UY",
        "timestamps": [
            267
        ],
        "3min_transcript": "like this that doesn't seem to be divisible by a lot of things, it's always a good idea to try things like 7, 11, 13. Because those tend to construct very interesting numbers. So let's see if this is divisible by 7. So if I take 343 and if I want to divide it by 7, 7 goes into 30-- it doesn't go into 3-- 7 goes into 34 four times. 4 times 7 is 28. Subtract, 34 minus 28 is 6. Bring down a 3. 7 goes into 63 nine times. 9 times 7 is 63. Subtract. We don't have any remainder. And I forgot to do that last step up here. 3 times 15 is 15. Subtract, no remainder. It went in exactly. So here, 343 can be factored into 7 and 49. And 49 might jump out at you. It can be factored into 7 times 7. I can rewrite all of this here-- the cube root of 3,430-- now as the cube root of-- I'm just going to write it in its factored form-- 2 times 5 times-- I could write 7 times 7 times 7, or I could write times 7 to the third power. That captures these three 7's right over here. I have three 7's, and then I'm multiplying them together. So that's 7 to the third power. And from our exponent properties, we know that this is the exact same thing as the cube root of 2 times 5 times the cube root-- so let me do that in that same, just so we see what colors we're dealing with. So the cube root of 2 times 5, which is the cube root of 10, times the cube root-- and I think Keeping track of the colors is the hard part. And the cube root of 10, we just leave it as 10. We know the prime factorization of 10 is 2 times 5, so you're not going to just get a very simple integer You would get some decimal answer here, but here you get a very clear integer answer. The cube root of 7 to the third, well, that's just going to be 7. So this is just going to be 7. So our entire thing simplifies. This is equal to 7 times the cube root of 10. And this is about as simplified as we can get just using hand arithmetic. If you want to get the exact number here, you're probably best off using a calculator."
    },
    {
        "Q": "why does he say at 2:06 that 343 isn't divisible by 2 when the sum of 3+4+3 is 10 which is divisible by 2?",
        "A": "adding the digits is only a test for divisibility for 3 and 9",
        "video_name": "8y7xP4zz0UY",
        "timestamps": [
            126
        ],
        "3min_transcript": "Let's see if we can find the cube root of 3,430. And if you're like me, it doesn't jump out of your mind what number times that same number times that same number-- if you have three of those numbers and you were to multiply them together-- would be equal to 3,430. So what I'm going to do is to try to prime factorize this to find all the prime factors of 3,430 and see if any of those prime factors show up at least three times. And that'll help us with this. So 3,430-- it's clearly divisible by 5 and 2, or it's divisible by 10. So let's do that. So first we can divide it by 2. It's 2 times-- let's see. 3,430 divided by 2 is 1,715. Then we can divide it by 5, as well. We can factor 1,715 into 5 and-- let me do a little bit of long division on the side here. So if I have 1,715, and I'm going to divide it by 5. It goes into 17 three times. 3 times 5 is 15. Subtract, you get 2, and then you bring down a 1. 5 goes into 21 four times. 4 times 5 is 20. Subtract. Bring down the 5. 5 goes into 15 three times, so it goes exactly 343 times. So 1,715 can be factored into 5 times 343. Now, 343 might not jump out at you as a number that is easy to factor. It's clearly an odd number, so it won't be divisible by 2. Its digits add up to 10, which is not divisible by 3. So this isn't going to be divisible by 3. It's not going to be divisible by 4, because it's not divisible by 2. It's not going to be divisible by 5. If it wasn't divisible by 3 or 2, it's not going to be divisible by 6. And now we get to 7. like this that doesn't seem to be divisible by a lot of things, it's always a good idea to try things like 7, 11, 13. Because those tend to construct very interesting numbers. So let's see if this is divisible by 7. So if I take 343 and if I want to divide it by 7, 7 goes into 30-- it doesn't go into 3-- 7 goes into 34 four times. 4 times 7 is 28. Subtract, 34 minus 28 is 6. Bring down a 3. 7 goes into 63 nine times. 9 times 7 is 63. Subtract. We don't have any remainder. And I forgot to do that last step up here. 3 times 15 is 15. Subtract, no remainder. It went in exactly. So here, 343 can be factored into 7 and 49. And 49 might jump out at you. It can be factored into 7 times 7."
    },
    {
        "Q": "at 1:01 why are the fours at the bottom of the denominator not negative I am confused :/",
        "A": "a negative exponent is not applied to the coefficient, it just flips the exponential with a negative exponent to the other side of the divide line and thus makes the exponent positive. It is not (-4)^-3 power which would have three negative fours.",
        "video_name": "CZ5ne_mX5_I",
        "timestamps": [
            61
        ],
        "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": "( 3:37 ) Could (12^-7) / (12^-5) be written as (12^5) / (12^7) ?",
        "A": "Yes it could because the exponents are negative. Correct!",
        "video_name": "CZ5ne_mX5_I",
        "timestamps": [
            217
        ],
        "3min_transcript": "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. as X to the negative twentieth and having an X to the fifth in the denominator dividing by X to the fifth is the same thing as multiplying by X to the negative five. So here you just add the exponents and once again you would get X to the negative twenty-fifth power."
    },
    {
        "Q": "In 1:35 how does he get 8 1/3?",
        "A": "He divided 100 by 12, which is 8 with a remainder of 4. The remainder can be written as 4/12, which can be reduced to 1/3, so 8 1/3",
        "video_name": "jOZ98FDyl2E",
        "timestamps": [
            95
        ],
        "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": "at 0:41 he's dividing them but couldn't you just write out the problem in long division too?",
        "A": "Sal is working with ratios and rates which are types of fractions. This is why he is writing them initially as fractions. If you wrote the long division form first, it would work provided you got to the answers Sal has written out.",
        "video_name": "jOZ98FDyl2E",
        "timestamps": [
            41
        ],
        "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": "At 1:00, where does the formula come from?",
        "A": "One place it comes from is after deriving the quadratic formula, you end up with (-b \u00c2\u00b1 \u00e2\u0088\u009a(b^2-4ac))/(2a), so the -b/2a where the line of symmetry is and thus the x coordinate of the vertex and the (\u00c2\u00b1 \u00e2\u0088\u009a(b^2-4ac))/2a is the distance away from the line of symmetry of the zeroes.",
        "video_name": "IbI-l7mbKO4",
        "timestamps": [
            60
        ],
        "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": "At 2:00, what is Sal doing? I don't get how or why he adds 16...",
        "A": "Sal is using a method called Completing the square . it involves an equation (M*1/2x)^2. in his example if you take M which is the Middle term(8) and divide by 2 you would get 4. then squar it and you get 16. so that is why he adds and subtracts 16. this wont change the answer at all because +16-16=0. so your adding a fancy way of saying +0",
        "video_name": "IbI-l7mbKO4",
        "timestamps": [
            120
        ],
        "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": "At 6:50, he says that a light bulb is on if it has an odd number of factors, so if the bulb number is prime, is it always off(except for 1)?",
        "A": "Yes, every prime number, by definition, has exactly two factors, meaning that it will always be off.",
        "video_name": "WNhxkpmVQYw",
        "timestamps": [
            410
        ],
        "3min_transcript": "And that's just a fancy way of saying, look, if I'm on pass 17, I'm going to switch all the multiples of 17. Or I could say, I know that I'm going to switch light bulb 51, because 17 is a factor of 51. So that tells you that we're always going to be switching one of these light bulbs on or off when one of its factors is our pass. So for example, if we're looking at light bulb eight. This is light bulb eight. So when will it be switched? So on pass one, we're definitely going So pass one, it's going to be switched on. Pass two, it'll be switched off. Pass three, nothing's going to happen. On pass three, nothing's going to happen because this isn't a multiple of three. Pass four, what'll happen? It will be switched. It'll be switched back on. And then pass eight is the next time we'll touch this light bulb, and it'll be switched back. So every time one of its factors go by, we're going to switch this thing. And as you can see, in order for it to be on at the end, you have to have an odd number of factors. So that's an interesting thing. So in order for a light bulb to be on, it has to have odd number of factors. Now that's an interesting question. What numbers have an odd number of factors. And this is something that I think they should teach you in grade school, and they never do. But it's a really interesting kind of number theory. So what numbers are true? Let's do all the factors for some of the starting numbers. So all the factors of one. Well one, the only factor is just one. So one works? One has an odd number of factors. So that means that one will remain on. Because you're only going to turn it on in the first pass. Makes sense. Two. What are all the factors of two. Well you have one and two. So two has an even number. You're going to switch it on the first time, then off the second time. Then you're never going to touch it again. So this is going to stay off. Your factors are one and three. Four. Your factors are one, two and four. Interesting. Here we have three factors. We have an odd number of factors. So four is going to stay on. We're going to turn it on in our first pass, we're going to turn it off on our second pass. And we're going to turn it on again in our fourth pass. Let's keep going. So five."
    },
    {
        "Q": "8:00 is the answer all square numbers will be on?",
        "A": "Yes pretty much the square numbers that are smaller than 100.",
        "video_name": "WNhxkpmVQYw",
        "timestamps": [
            480
        ],
        "3min_transcript": "Pass three, nothing's going to happen. On pass three, nothing's going to happen because this isn't a multiple of three. Pass four, what'll happen? It will be switched. It'll be switched back on. And then pass eight is the next time we'll touch this light bulb, and it'll be switched back. So every time one of its factors go by, we're going to switch this thing. And as you can see, in order for it to be on at the end, you have to have an odd number of factors. So that's an interesting thing. So in order for a light bulb to be on, it has to have odd number of factors. Now that's an interesting question. What numbers have an odd number of factors. And this is something that I think they should teach you in grade school, and they never do. But it's a really interesting kind of number theory. So what numbers are true? Let's do all the factors for some of the starting numbers. So all the factors of one. Well one, the only factor is just one. So one works? One has an odd number of factors. So that means that one will remain on. Because you're only going to turn it on in the first pass. Makes sense. Two. What are all the factors of two. Well you have one and two. So two has an even number. You're going to switch it on the first time, then off the second time. Then you're never going to touch it again. So this is going to stay off. Your factors are one and three. Four. Your factors are one, two and four. Interesting. Here we have three factors. We have an odd number of factors. So four is going to stay on. We're going to turn it on in our first pass, we're going to turn it off on our second pass. And we're going to turn it on again in our fourth pass. Let's keep going. So five. The factors are one, two, three and six. It's an even number, so they're going to be off when we're done with it. Seven. it's one and seven. We just did that. It's one, two, four and eight. Still going to be off. Nine. Let's see. The factors are one, three and nine. Interesting. Once again we have an odd number of factors. So the light bulb number nine is also going to be on when everything's done. Let's keep going. I don't know, I actually did this at our mental boot camp with some of the kids. And they immediately said, the distance between one and four is three. The distance between four and nine is five. And maybe the distance between nine and the next number is going to be seven. It increases by odd numbers. What's nine plus seven? 16. What are the factors of 16? They're one, two, four, eight and 16."
    },
    {
        "Q": "At 8:11, what is a hypersphere?",
        "A": "Hypersphere is a generalization for any sphere that is more than 3D. Normally we define a sphere in 3D as x\u00c2\u00b2+y\u00c2\u00b2+z\u00c2\u00b2=1. But you can generalize this to as many dimensions as you want. You could have a 4D sphere or a 500D sphere, but in general, they are referred to as hyperspheres.",
        "video_name": "iDQ1foxYf0o",
        "timestamps": [
            491
        ],
        "3min_transcript": "So \"Three students attempt to define \"what a line segment is.\" And we have a depiction of a line segment right over here. We have point P, point Q, and the line segment is all the points in between P and Q. So, so let's match the teacher's comments to the definitions. Ivy's definition: \"All of the points \"in line with P and Q, extending infinitely \"in both directions.\" Well, that would be the definition of a line. That would be the line P, Q. That would be, if you're extending infinitely in both directions, so ... I would say, \"Are you thinking of a line \"instead of a line segment?\" Ethan's definition: \"The exact distance from P to Q. Well, that's just a ... that's the length of a line segment. That's not exactly what a line segment is. And see, Ebuka's definition. \"The points P and Q, which are called endpoints, \"and all of the points in a straight line Yep. That looks like a good definition for a line segment. So we can just check our, we can just check our answer. So, looking good. Let's do one more of this. I'm just really enjoying pretending to be a teacher. All right. \"Three students attempt to define \"what a circle is.\" Define what a circle is. \"Can you match the teacher's comments to the definitions?\" Duru. \"The set of all points in a plane \"that are the same distance away from some given point, \"which we call the center.\" That just seems like a pretty good definition of a circle. So, I would, you know, \"Stupendous! Well done.\" Oliver's definition. \"The set of all points in 3D space \"that are the same distance from a center point.\" If we're talking about 3D space and the set of all points that are equidistant from that point in 3D space, now we're talking about a sphere, not a circle. And so, \"You seem to be confusing \"a circle with a sphere.\" And then, finally, \"A perfectly round shape.\" But if you're talk about three dimensions, you could be talking about a sphere. If you're talking about, if you go beyond three dimensions, hypersphere, whatever else. In two dimensions, yeah, perfectly round shape, most people would call it a circle. But that doesn't have a lot of precision to it. It doesn't have, it doesn't give us a lot that we can work with from a mathematical point of view. So I would say, actually, what the teacher's saying: \"Your definition needs to be much more precise.\" Duru's definition is much, much more precise. The set of all points that are equidistant from ... in a plane, that are equidistant away from a given point, which we call the center. So yep. Carlos could use a little bit more precision. We're all done."
    },
    {
        "Q": "At 0:32, how does one know when to write x/y and when to write y/x?",
        "A": "its just another way to write division 4/2=2/4 and here sal is just saying 14/2 =/= 2/14 so in proportions you need bigger number on top...",
        "video_name": "qcz1Cm_-l50",
        "timestamps": [
            32
        ],
        "3min_transcript": "- So, let's set up a relationship between the variables x and y. So, let's say, so this is x and this is y, and when x is one, y is four, and when x is two, y is eight, and when x is three, y is 12. Now, you might immediately recognize that this is a proportional relationship. And remember, in order for it to be a proportional relationship, the ratio between the two variables is always constant. So, for example, if I look at y over x here, we see that y over x, here it's four over one, which is just four. Eight over two is just four. Eight halves is the same thing as four. 12 over three it's the same thing as four. Y over x is always equal to four. In fact, I can make another column here. I can make another column here where I have y over x, here it's four over one, which is equal to four. Here it's eight over two, which is equal to four. Here it's 12 over three, which is equal to four. the ratio, the ratio between y and x is this constant four, to express the relationship between y and x as an equation. In fact, in some ways this is, or in a lot of ways, this is already an equation, but I can make it a little bit clearer, if I multiply both sides by x. If I multiply both sides by x, if I multiply both sides by x, I am left with, well, x divided by x, you'd just have y on the left hand side. Y is equal to 4x and you see that's the case. X is one, four times that is four. X is two, four times that is eight. So, here you go, we're multiplying by four. We are multiplying by four, we are multiplying by four. And so, four, in this case, four, in this case, in this situation, this is our constant of proportionality. Constant, constant, sometimes people will say proportionality constant. Now sometimes, it might even be described as a rate of change and you're like well, Sal, how is this a, how would four be a rate of change? And, to make that a little bit clearer, let me actually do another example, but this time, I'll actually put some units there. So let's say that, let's say that I have, let's say that x-- Let me do this, I already used yellow, let me use blue. So let's x, let's say that's a measure of time and y is a measure of distance. Or, let me put it this way, x is time in terms of seconds. Let me write it this way. So, x, x is going to be in seconds and then, y is going to be in meters. So, this is meters, the units, and this right over here is seconds. So, after one second, we have traveled, oh, I don't know, seven meters."
    },
    {
        "Q": "at 2:05, does the table of values always has to start with 0 ?",
        "A": "No, sometimes the value of x and never even be 0. However, it s often easier for to start at zero if x can indeed be 0 because that way you can find the y-intercept.",
        "video_name": "86NwKBcOlow",
        "timestamps": [
            125
        ],
        "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": "0:15 why he multiple for 6?? not for 3 i got confused.",
        "A": "least common multiple of x/3 and 1/6 is 6. *if you want to add to x/3 and 1/6 you need the denominator be the same.",
        "video_name": "CJyVct57-9s",
        "timestamps": [
            15
        ],
        "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 7:31 he did 10^200 divided by 10^50. How did he get 150 if 200 + 50 gives you 250.",
        "A": "10^200/10^50 the exponent is subtracted 200-50, only multiplication is it added 200+50.",
        "video_name": "kITJ6qH7jS0",
        "timestamps": [
            451
        ],
        "3min_transcript": ""
    },
    {
        "Q": "4:16 of the video to 5:00. How does 2 with the exponent of 8 + 2 with the exponent of 8 eventually get broken down into 2 with the exponent of 9; it doesn't seem to make any since because they both would have different equations?",
        "A": "this happens as 2^8 + 2^8 = 2 * 2^8 which can be simplified as 2^9",
        "video_name": "kITJ6qH7jS0",
        "timestamps": [
            256,
            300
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 1:35, why does Sal put parentasis around the 6?",
        "A": "He puts parenthesis around the 6 because that mean multiplication or simply a form of it",
        "video_name": "Rcb7ZUTOQ1I",
        "timestamps": [
            95
        ],
        "3min_transcript": "If we know some circle has an area of 36pi-- so it has an area of 36pi-- can we figure out what the circumference of this circle And I encourage you to pause this video, and try to think about that question. Well, from the area, we could figure out what the radius is, and then from that radius, we can figure out what its circumference is. So we know that the area, which is 36pi, is equal to pi r squared. And so if you look at it on both sides of this equation, if we divide-- let me rewrite it so it's a little bit clearer in a different color. So we could set up an equation pi r squared is equal to 36pi. Now, if we want to solve for the radius the first thing that we might want to do is divide both sides by pi. Then, we're left with r squared is equal to 36. Now, if we just solve this as a pure math equation, you might say, OK, we could take the positive and negative r could be plus or minus 6, but we need to remember that r is a distance, so we only care about the positive. So if we take the principal root of 36, we get r is equal to 6. From there, we can use this to figure out the circumference. So the circumference is equal to 2 pi r. Circumference is equal to 2 pi r. And in this case, r is equal to 6. So it's equal to 2 pi times 6, which is going to be equal to 12pi. So that's straightforward, area 36pi, we leverage pi r squared to figure out that the radius was 6, and then from that we were able to figure out that the circumference was 12pi."
    },
    {
        "Q": "At 6:30 Sal write dt, but I cant see where he get that from. Anyone can help? Thanks",
        "A": "dt is the differential, you put it at the end of the integration expression. If you re wondering why its dt and not dx ; its because it parametric, so your functions are functions of t - f(t) - and not like the usual functions of x - f(x). Hope i could help!",
        "video_name": "99pD1-6ZpuM",
        "timestamps": [
            390
        ],
        "3min_transcript": "And the reason why this is the case, is if you imagine this is a, this is b, that is my f of x. When you do it this way, your dx's are always going to be positive. When you go in that direction, your dx's are always going to be positive, right? Each increment, the right boundary is going to be higher So your dx's are positive. In this situation, your dx's are negative. The heights are always going to be the same, they're always going to be f of x, but here your change in x is a negative change in x, when you go from b to a. And that's why you get a negative integral. In either case here, our path changes, but our ds's are going to be positive. And the way I've drawn this surface, it's above the x-y plane, the f of xy is also going to be positive. So that also kind of gives the same intuition that this should be the exact same area. But let's prove it to ourselves. So let's start off with our first parameterization, just We have x is equal to x of t, y is equal to y of t, and we're dealing with this from, t goes from a to b. And we know we're going to need the derivatives of these, so let write that down right now. We can write dx dt is equal to x prime of t, and dy dt, let me write that a little bit neater, dy dt is equal to y prime of t. This is nothing groundbreaking I've done so far. But we know the integral over c of f of xy. f is a scalar field, not a vector field. ds is equal to the integral from t is equal to a, to t is dt squared, which is the same thing as x prime of t squared, plus dy dt squared, the same thing as y prime of t squared. All that under the radical, times dt. This integral is exactly that, given this parameterization. Now let's do the minus c version. I'll do that in this orange color. Actually, let me do the minus c version down here. The minus the c version, we have x is equal to, you remember this, actually, just from up here, this was from the last video."
    },
    {
        "Q": "At around 3:19, doesn't point D also have a y value of 6?",
        "A": "If you look carefully, point D is slightly below 6 and probably has a y-value of 5.5. This is later mentioned at 5:44.",
        "video_name": "P3IlneCNm8A",
        "timestamps": [
            199
        ],
        "3min_transcript": "So for example, f prime of 0-- which is the x value for this point right over here-- is going to be some negative value. It's the slope of the tangent line. Similarly, f prime of x, when x is equal to 4-- that's what's going on right over here-- that's going to be the slope of the tangent line. That's going to be a positive value. So if you look at all of these, where is the slope of the tangent line 0? And what does a 0 slope look like? Well, it looks like a horizontal line. So where is the slope of the tangent line here horizontal? Well, the only one that jumps out at me is point B right over here. It looks like the slope of the tangent line would indeed be horizontal right over here. Or another way you could think of it is the instantaneous rate of change of the function, right at x equals 2, looks like it's pretty close to-- if this So out of all of the choices here, I would say only B looks like the derivative at x equals 2. Or the slope of the tangent line at B, it looks like it's 0. So I'll say B right over here. And then they had this kind of crazy, wacky expression here. f of x minus 6 over x. What is that greatest in value? And we have to interpret this. We have to think about what does f of x minus 6 over x actually mean? Whenever I see expressions like this, especially if I'm taking a differential calculus class, I would say well, this looks kind of like finding the slope of a secant line. In fact, all of what we know about derivatives is finding the limiting value of the slope of a secant line. And this looks kind of like that, especially if at some point, my y value is a 6 here. And this could be the change in y value. And if the corresponding x value is 0, then this would be f of x minus 6 over x minus 0. Well, sure. When x is equal to 0, we see that f of x is equal to 6. So what this is right over here-- let me rewrite this. This we could rewrite as f of x minus 6 over x minus 0. So what is this? What does this represent? Well, this is equal to the slope-- let me do some of that color-- this is equal to the slope of the secant line between the points, x, f of x, x, and whatever the corresponding f of x is."
    },
    {
        "Q": "on the 30:60:90 triangle why do you square root it by three? thats the only part that lost me (:\n\nThanks,\nAshley",
        "A": "It s a right triangle, so we can apply Pythagoras (a\u00c2\u00b2 + b\u00c2\u00b2 = c\u00c2\u00b2). The hypotenuse is c = 2, a = 1, and b = \u00e2\u0088\u009a3: a\u00c2\u00b2 + b\u00c2\u00b2 = c\u00c2\u00b2 1\u00c2\u00b2 + (\u00e2\u0088\u009a3)\u00c2\u00b2 = 2\u00c2\u00b2 1 + 3 = 4",
        "video_name": "UKQ65tiIQ6o",
        "timestamps": [
            1860
        ],
        "3min_transcript": ""
    },
    {
        "Q": "What do you mean of function of s in 0:16?",
        "A": "He means that we will find the area of a triangle that has a side length of s, not of a specific triangle, like one with sides of length 4 or 7 or something. Working this way, the equation he finds will work for any equilateral triangle, and you just have to substitute the value of s (side length) into the equation to get the area.",
        "video_name": "UKQ65tiIQ6o",
        "timestamps": [
            16
        ],
        "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": "I am really puzzeled about the 7:26-7:35 portion of the video where Sal goes from a^2 =1/2 to\na = 1/sqrt 2....how did he do that?",
        "A": "2a^2=1 --------- 2 ......2 The first 2 s cancel out and your left with: a^2 =1/2",
        "video_name": "fp9DZYmiSC4",
        "timestamps": [
            446,
            455
        ],
        "3min_transcript": "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 So using the Pythagorean Theorem, knowing that these two sides are equal, what do we get for the length of those sides? Well, if this has length a, well then this also has length a. And we can use the Pythagorean Theorem. And we could say a squared plus a squared is equal to the hypotenuse squared. Is equal to 1. Or 2a squared is equal to 1a squared, is equal to 1/2. Take the principal root of both sides. a is equal to the square root of 1/2 which is the square root of 1, which is 1, over the square root of 2. We can rationalize the denominator here by multiplying by square root of 2 over square root of 2, which gives us a is equal to-- in the numerator-- square root of 2. And in the denominator, square root of 2 times square root of 2 is 2. So this length is the square root of 2. And this length is the same thing. So this length right over here is square root of 2 over 2. And this height right over here is also square root of 2 over 2."
    },
    {
        "Q": "At 03:36 isnt he marking wrong? He is marking cosine on the Y axis, but on the begining of the video he said cosine of data is the X, axis! And sine was the Y axis! Now im really confused!",
        "A": "I understand the confusion. When he said cosine of theta is the x-axis, he was basically saying Instead of the x-axis and integers (i.e. -2, -1, 0, 1, 2 ect...) we are going to call it the theta-axis and use radians. Also, instead of saying cos(theta) = y-axis he mixed up the measurement for the two axis. One axis has been turned into radians (theta, cis(theta)), the other has been left in integer number form. (x/y form).",
        "video_name": "fp9DZYmiSC4",
        "timestamps": [
            216
        ],
        "3min_transcript": "So what is cosine of theta? What's the x-coordinate here? Which is negative 1. And sine of theta is going to be the y-coordinate, which is 0. Now let's keep going. Now we're down here at 3 pi over 2. If we go all the way around to 3 pi over 2, what is this coordinate? Well this is 0, negative 1. Cosine of theta is the x-coordinate here. So cosine of theta is going to be 0. And what is sine of theta going to be? Well it's going to be negative 1. And then finally we go back to 2 pi, which is making a full revolution around the circle. We went all the way around and we're back to this point right over here. So the coordinate is the exact same thing as when the angle equals 0 radians. And so what is cosine of theta? Well that's 1. And sine of theta is 0. 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?"
    },
    {
        "Q": "At 0:25 is one fourth just like a quarter?",
        "A": "Think of one whole as 100. If you think of money, one dollar is 100 cents. One quarter is 25 cents out of the whole 100 cents. So a quarter is one fourth(a quarter) of a dollar(a whole).",
        "video_name": "gEE6yIObbmg",
        "timestamps": [
            25
        ],
        "3min_transcript": "- [Voiceover] Is each piece equal to 1/4 of the area of the pie? So we have a pie, and it has one, two, three, four pieces. So it does have four pieces. So is one of those pieces equal to 1/4 of the pie? Well let's talk about what we mean when we have a fraction like 1/4. The one in the fraction, the numerator, represents a number of pieces. So here, one piece. One piece of pie. And then the four, when we're talking about fractions is always talking about the number of equal size. Equal size pieces. So in this case four equal size pieces. So the question is, is each piece one of four equal size pieces? Let's look at the pie. I think it's pretty clear that these pieces on the end are not equal, they are smaller If you love cherry pie, you are not happy about getting this end piece. Because it is smaller. It is not an equal size piece. So yes, each piece is one out of four pieces. But it is not one of four equal size pieces. Therefore it is not 1/4. So our answer is no. No, no, no. Each piece is not 1/4 or an equal share of the pie."
    },
    {
        "Q": "At 7:27, Sal said that if r= -1, then the values would keep on oscillating. However, if you actually work it out, the series would converge to a/2. If we say that the series, is S, than S= a-a+a-a+a-a+a-a+a... . Also, a-S would equal a-a+a-a+a-a+a... . This is equal to the original series S. So, we can say a-S=S. Then we get a= 2S. Therefore, S=a/2. Is there something I am doing I am not disregarding in my calculuations?",
        "A": "Your logic seems plausible but fails the epsilon-delta test, which is the ultimate test for whether a series converges. There is no delta for which larger values of n will produce S values closer than, say, a/4 (a possible epsilon). We know this because it s clear that there is no point beyond which the sum stops oscillating between a and 0. Your result of a/2 is the average of the two values between which the sum oscillates, not the value to which the sum converges.",
        "video_name": "wqnpSzEzq1w",
        "timestamps": [
            447
        ],
        "3min_transcript": "My brain isn't working right! 5 times 3/5 is going to be 3 times 3/5. Is going to be-- 3 times this is going to be 9/5-- actually that was right. My brain is working right. Times 3/5 is going to be 27 over 25. Times 3/5 is going to be 81/125. And we keep on going on and on and on forever. And notice these terms are starting to get smaller and smaller and smaller. Well actually all of them are getting smaller and smaller and smaller. We're multiplying by 3/5 every time. We now know what the sum is going to be. It's going to be our first term-- it's going to be 5-- over 1 minus our common ratio. And our common ratio in this case is 3/5. So this is going to be equal to 5 over 2/5, which is 25/2 which is equal to 12 and 1/2, or 12.5. Once again, amazing result. I'm taking a sum of infinite terms here, and I was able to get a finite result. And once again, when does this happen? Well, if our common ratio-- if the absolute value of our common ratio-- is less than 1, then these terms are going to get smaller and smaller and smaller. And you'll even see here it even works out mathematically in this denominator that you are going to get a reasonable answer. And it makes sense because these terms are getting smaller and smaller and smaller that this thing will converge. If r is 0, we're still not dealing strictly with a geometric series anymore, but obviously if r was 0, then you're really only going to have this-- well, even depending on how you define what 0 to 0 is. But if your first term you just said would be a, then clearly you'd just be left with a is the sum, and a over 1 minus 0 is still a. So this formula that we just derived does hold up for that. It does start to break down if r is equal to 1 or negative 1. If r is equal to 1 then as you imagine here, you just have a plus a plus a plus a, going on and on forever. If r is equal to negative 1 you just keep oscillating. a, minus a, plus a, minus a. And so the sum's value keeps oscillating between two values. So in general this infinite geometric series is going to converge if the absolute value of your common ratio is less than 1. Or another way of saying that, if your common ratio is between 1 and negative 1."
    },
    {
        "Q": "At 7:50 Sal said that the boundaries are -1 < r < 1 ,but why not between 0 < r < 1 ?",
        "A": "The problem is that using 0<r<1 leaves out values that also make the series converge, namely, 0, (as 0<r<1 does not include zero), and the values from -1 up to 0, since values such as -0.9, or -3/4, or -1/100 also make the series converge. Notice that your interval 0<r<1 is included in the interval -1<r<1. Hope that helped.",
        "video_name": "wqnpSzEzq1w",
        "timestamps": [
            470
        ],
        "3min_transcript": "My brain isn't working right! 5 times 3/5 is going to be 3 times 3/5. Is going to be-- 3 times this is going to be 9/5-- actually that was right. My brain is working right. Times 3/5 is going to be 27 over 25. Times 3/5 is going to be 81/125. And we keep on going on and on and on forever. And notice these terms are starting to get smaller and smaller and smaller. Well actually all of them are getting smaller and smaller and smaller. We're multiplying by 3/5 every time. We now know what the sum is going to be. It's going to be our first term-- it's going to be 5-- over 1 minus our common ratio. And our common ratio in this case is 3/5. So this is going to be equal to 5 over 2/5, which is 25/2 which is equal to 12 and 1/2, or 12.5. Once again, amazing result. I'm taking a sum of infinite terms here, and I was able to get a finite result. And once again, when does this happen? Well, if our common ratio-- if the absolute value of our common ratio-- is less than 1, then these terms are going to get smaller and smaller and smaller. And you'll even see here it even works out mathematically in this denominator that you are going to get a reasonable answer. And it makes sense because these terms are getting smaller and smaller and smaller that this thing will converge. If r is 0, we're still not dealing strictly with a geometric series anymore, but obviously if r was 0, then you're really only going to have this-- well, even depending on how you define what 0 to 0 is. But if your first term you just said would be a, then clearly you'd just be left with a is the sum, and a over 1 minus 0 is still a. So this formula that we just derived does hold up for that. It does start to break down if r is equal to 1 or negative 1. If r is equal to 1 then as you imagine here, you just have a plus a plus a plus a, going on and on forever. If r is equal to negative 1 you just keep oscillating. a, minus a, plus a, minus a. And so the sum's value keeps oscillating between two values. So in general this infinite geometric series is going to converge if the absolute value of your common ratio is less than 1. Or another way of saying that, if your common ratio is between 1 and negative 1."
    },
    {
        "Q": "At 0:58 Sal multiplies both side of the equation by 3. Why did he do that when he could have just added 25/3 to both sides?",
        "A": "He multiplies by the denominator to get rid of the fraction form so her doesn t have to deal with fractions. Which makes the calculation easier. In his example, it seems he doesn t need to do that, but it shows you could do that, so you can apply to other similar problems where dealing with fractions is very hairy.",
        "video_name": "y7QLay8wrW8",
        "timestamps": [
            58
        ],
        "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": "how do you know when to change the > sign into a < sign or when to leave it the way it is? can someone explain? at 2:58 on the video...",
        "A": "You change the sign whenever you divide or multiply by a negative number. And only that condition, anything else you keep the sign as it was.",
        "video_name": "y7QLay8wrW8",
        "timestamps": [
            178
        ],
        "3min_transcript": "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. negative number, I swap the inequality, the greater than becomes a less than. When it was positive, I didn't have to swap it. So 27 divided by negative 12, well, they're both divisible by 3. So we're going to get, if we divide the numerator and the denominator by 3, we get negative 9 over 4 is less than-- these cancel out-- y. So y is greater than negative 9/4, or negative 9/4 is less than y. And if you wanted to write that-- just let me write this-- our answer is y is greater than negative 9/4. I just swapped the order, you could say negative 9/4 is less than y. Or if you want to visualize that a little bit better, 9/4 is 2 and 1/4, so we could also say y is greater than negative 2 and 1/4 if we want to put it as a mixed number. And if we wanted to graph it on the number line-- let me"
    },
    {
        "Q": "At 0:52 when he multiplied 3 by both sides, that same number has to correspond to the denominator for example if the dominator was -3, you multiply by -3 yes?",
        "A": "Correct. This is because we need to get rid of the denominator. Does that make sense? And most don t know why you switch the sign, but I believe you do - otherwise you would have asked to know. I hope I helped!",
        "video_name": "y7QLay8wrW8",
        "timestamps": [
            52
        ],
        "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": "2:50, why do you swap numbers?",
        "A": "He doesn t swap, he flips the inequality around because you are dividing by a negative number.",
        "video_name": "y7QLay8wrW8",
        "timestamps": [
            170
        ],
        "3min_transcript": "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. negative number, I swap the inequality, the greater than becomes a less than. When it was positive, I didn't have to swap it. So 27 divided by negative 12, well, they're both divisible by 3. So we're going to get, if we divide the numerator and the denominator by 3, we get negative 9 over 4 is less than-- these cancel out-- y. So y is greater than negative 9/4, or negative 9/4 is less than y. And if you wanted to write that-- just let me write this-- our answer is y is greater than negative 9/4. I just swapped the order, you could say negative 9/4 is less than y. Or if you want to visualize that a little bit better, 9/4 is 2 and 1/4, so we could also say y is greater than negative 2 and 1/4 if we want to put it as a mixed number. And if we wanted to graph it on the number line-- let me"
    },
    {
        "Q": "At 1:28, Sal simplifies 2/3 by multiplying it by three(because he's doing that to the other side of the inequality sign). He then crosses out both threes and leaves the two. Why? How does that work? It's probably really obvious, but can someone answer this?",
        "A": "Sure!, A fraction is just a number divided by another number, therefore to undo the division you multiply. Sal crossed them out because they cancel out each other. For example, if you have 6/3 that will equal 2. then multiply by three to get 6. But you should just multiply by the denominator and cancel out everything at the beginning to get 6! Hope this helps!",
        "video_name": "y7QLay8wrW8",
        "timestamps": [
            88
        ],
        "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 2:47, I'm a little confused .. Why the limit isn't (7) as we are approaching point (8) as it's part of the function at this point? &if so how is the limit exist as we are approaching different points? Please somebody enlightens me!",
        "A": "You can say that f (8) = 7 but that s not the limit. The limit is concerned with the value NEAR a point but not AT the point. And notice that between 2:50 and around 3:15 or so Sal checks the values on both sides of 8 and finds the limit to be 1. It s ok that the limit (1) does not match the value of f (8) = 7. It just shows that curve is discontinuous at that point.",
        "video_name": "_bBAiZhfH_4",
        "timestamps": [
            167
        ],
        "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": "at 4:24 -- what makes Sal determine it's t2/2? Is this a formula?",
        "A": "Well, I think the deduction of this equation comes out here: d=Va*t, where d is the distance,and Va means the average velocity. while Va=(Vf+Vi)/2, where Vf is the final velocity and Vi is the initial velocity (in this case Vi=0). In addition,we know that the difference of velocity Vdelta=Vf-Vi=g*t. So,Vf=g*t+Vi,since Vi=0, so Vf=g*t+Vi=g*t+0=g*t. Now replace Vf by g*t: d=Va*t=(Vf+Vi)/2*t=(g*t+Vi)/2*t=(g*t+0)/2*t=g*t/2*t=g*t^2/2.",
        "video_name": "m6c6dlmUT1c",
        "timestamps": [
            264
        ],
        "3min_transcript": ""
    },
    {
        "Q": "on 4:18 Hilbert who?",
        "A": "David Hilbert. He was one of the leading number theorists of his time.",
        "video_name": "ik2CZqsAw28",
        "timestamps": [
            258
        ],
        "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 5:15 what did she use to make the squiggle?",
        "A": "She uses a Pipe Cleaner to make the squiggle. The brand is Dill s",
        "video_name": "ik2CZqsAw28",
        "timestamps": [
            315
        ],
        "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": "at 0:46 to 0:53 he said 9 is > or = to 5 i don't get that",
        "A": "With rounding if the number is 5 or more than 5 than you round it up. For example you have 1.5, you d round it up to 2.",
        "video_name": "tx2Niw7aJJ8",
        "timestamps": [
            46,
            53
        ],
        "3min_transcript": "A ticket agent sells 42 tickets to a play. The tickets cost $29 each. Use rounding to estimate the total dollars taken in from the sale of the tickets. Now if we wanted the exact number, we could say 42 times 29, and we could work out the multiplication, but they essentially want us to be able to do it in our head. We want to round the numbers first and then multiply. So if we want to round, and really we just have two places here, so if we're going to round anything, it's going to be to the nearest ten because that's the largest place we have. So if we round 42 to the nearest ten-- we've done this drill many times-- 2 in the ones place is the less than 5, so we're going to round down. The nearest ten is 40. We're going to round down to 40. 29, if we round to the nearest ten, 9 in the ones place is greater than or equal to 5, so we round up. The nearest ten is 30. And another way to think about it. Just say, well, you know, 42, that's pretty close to 40. 29 is pretty close to 30. I can figure out, so now I can multiply. And here, once again, we can use-- you could call it a trick, but hopefully, you understand why it works. But 30 times 40, instead of you saying, well, this is going to be the same thing as 3 times 4, but we're going to put two zeroes at the end of it. 30 times 40 is the same thing as 3 times 4 with two zeroes, So you have 3 times 4 is 12, which we know, and then we have two zeroes. We got that zero, so let's stick that zero there, and then we got that blue zero there, so let's put that over there. So they're going to have roughly $1,200 taken it from sales of the tickets. That is our estimate."
    },
    {
        "Q": "At 09:39 in the video I thought there were 16 different scenarios.(Is it possible?)",
        "A": "is what possible? It s written in red at the top of the video: 6/16 = 3/8 .. 6 scenarios for 2 heads are possible and there are 16 different total scenarios for heads or tails.",
        "video_name": "8TIben0bJpU",
        "timestamps": [
            579
        ],
        "3min_transcript": "then how many different places can that second head show up in? Well, if that first head is in one of the four places, then that second head can only be in three different places. So that second head can only be-- I'm picking a nice color here-- can only be in three different places. And so, you know, it could be in any one of these. It could maybe be right over there. Any one of those three places. And so, when you think about it in terms of the first, and I don't want to say the first head, head one. Actually, let me call it this way. Let me call it head A and head B. That way you won't think that I'm talking about the first flip or the second flip. So this is head A, and this right over there is head B. So if you had a particular, I mean, these heads are identical. These outcomes aren't different, but the way we talk about it right now, it looks like there's four places that we could get this head in, and there's three places where we could get this head in. And so if you were to multiply all of the different ways where this is in four different places, and then this is in one of the three left over places, you get 12 different scenarios. But there would only be 12 different scenarios if you viewed this as being different than this. And let me rewrite it with our new-- So this is head A, this is head B, this is head B, this is head A. There would only be 12 different scenarios if you viewed these two things as fundamentally different. But we don't. We're actually double counting. Because we can always swap these two heads and have the exact same outcome. So what you want to do is actually divide it by two. So you want to divide it by all of the different ways that you can swap two different things. If we had three heads here, you would think about all of the different ways you could swap three different things. If you had four heads here, it would be all the different ways you could swap four different things. So there's 12 different scenarios want to divide it by all of the different ways that you can swap two things. So 12 divided by 2 is equal to 6. Six different scenarios, fundamentally different scenarios, considering that you can swap them. If you assume that head A and head B can be interchangeable. But it's a completely identical outcome for us, because they're really just heads. So there's six different scenarios, and we know that there's a total of 16 equally likely scenarios. So we could say that the probability of getting exactly two heads is 6 times, six scenarios and-- Or there's a couple of ways. You could say there are six scenarios that give us two heads, of a possible 16. Or you could say there are six possible scenarios, and the probability of each of those scenarios is 1/16. But either way, you'll get the same answer."
    },
    {
        "Q": "at 4:04 he sal listed outcomes for exactly one head... I think he forgot to list [TTTH]. Am I right?",
        "A": "He accidentally wrote TTHT twice in his equation, however [TTTH} is in row 3, column 4 of his table.",
        "video_name": "8TIben0bJpU",
        "timestamps": [
            244
        ],
        "3min_transcript": "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. [? We ?] know the number of possibilities and of those equally likely possibilities. And we can only use this methodology because it's a fair coin. So, how many of the total possibilities have two heads of the total of equally likely possibilities? So we know there are 16 equally likely possibilities. How many of those have two heads? So I've actually, ahead of time so we save time, I've drawn all of the 16 equally likely possibilities. And how many of these involve two heads? Well, let's see. This one over here has two heads, this one over here has two heads, this one over here has two heads. Let's see, this one over here has two heads, and this one over here has two heads. And then this one over here has two heads, and I believe we are done after that. So if we count them, one, two, three, four, five, six of the possibilities have exactly two heads."
    },
    {
        "Q": "I'm baffled as to why we keep using our original sample standard deviations as estimates for the population SDs (c. 6:50) once we're assuming the null hypothesis. If (and I might be barking up the wrong tree here) the hypothesis is that there's no meaningful difference whatsoever in weight loss effect between the two diets, why should their SDs remain distinct when imagined across the whole population? If the two groups' data are basically identical when viewed globally, shouldn't their SDs be identical too?",
        "A": "because it would lead to same answer. if you sample twice from the same population then the best variance estimator is ((n1-1)var(x1) + (n2-1)var(x2))/(n1+n2-2) ... i know you understand which symbol means what here .. now calculate for variance of difference of means of two iid samples from this population using the just calculated estimate of variance. It is the same thing as what sal does",
        "video_name": "N984XGLjQfs",
        "timestamps": [
            410
        ],
        "3min_transcript": "We could reject the null hypothesis and go with the alternative hypothesis. Remember, once again, we can use Z-scores, and we can assume this is a normal distribution because our sample size is large for either of those samples. We have a sample size of 100. And to figure that out, the first step, if we just look at a normalized normal distribution like this, what is your critical Z value? We're getting a result above that Z value, only has a 5% chance. So this is actually cumulative. So this whole area right over here is going to be 95% chance. We can just look at the Z table. We're looking for 95% percent. We're looking at the one tailed case. So let's look for 95%. This is the closest thing. We want to err on the side of being a little bit maybe to So let's say 95.05 is pretty good. So that's 1.65. So this critical Z value is equal to 1.65. Or another way to view it is, this distance right here is going to be 1.65 standard deviations. I know my writing is really small. I'm just saying the standard deviation of that distribution. So what is the standard deviation of that We actually calculated it in the last video, and I'll recalculate it here. The standard deviation of our distribution of the difference of the sample means is going to be equal to the square root of the variance of our first population. Now, the variance of our first population, we don't know it. But we could estimate it with our sample standard deviation. If you take your sample standard deviation, 4.67 and And so this is the variance. This is our best estimate of the variance of the population. And we want to divide that by the sample size. And then plus our best estimate of the variance of the population of group two, which is 4.04 squared. The sample standard deviation of group two squared. That gives us variance divided by 100. I did before in the last. Maybe it's still sitting on my calculator. Yes, it's still sitting on the calculator. It's this quantity right up here. 4.67 squared divided by 100 plus 4.04 squared divided by 100. So it's 0.617. So this right here is going to be 0.617. So this distance right here, is going to"
    },
    {
        "Q": "Whats a \"reciprocal\"? (4:07)",
        "A": "For any fraction, its reciprocal is created by flipping the fraction. Example: 3/4: its reciprocal is 4/3 -5/2: its reciprocal is -2/5 6: Note 6 as a fraction is 6/1. Its reciprocal = 1/6 In the video, Sal is using the reciprocal of (5x^4)/4, which would be 4/(5x^4). Hope this helps.",
        "video_name": "6nALFmvvgds",
        "timestamps": [
            247
        ],
        "3min_transcript": "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. if u said f of zero, you'd say all right, x cannot be equal to zero, you know? This would be the case if x is anything other than zero and it's not defined for zero, and so you would say f or zero is undefined. So now, these two functions are equivalent, or these two expressions are algebraically equivalent. So thinking about that, let's tackle this division situation here. So immediately, when you look at this, you say, woah, what are constrains here? Well, x cannot be equal to zero because if x was a zero, this second, this five x to the fourth over four would be zero and you'd be dividing, you'd be dividing by zero. So we can explicitly call out that x cannot be equal to zero. And so if x cannot be equal to zero in the original expression, if the result, whatever we get for the resulting expression, in order for it to be algebraically equivalent, we have to give this same constraint. So let's multiply this, or let's do the division. So this is going to be the same thing as two x times... The reciprocal of this is going to be four over five x to the fourth, which is going to be equal to in the numerator, we're going to have eight x to the fourth. So we're going to have eight x to the fourth, four times two x to the fourth, over seven times five x to the fourth is 35 x to the fourth. And now, there's something. We can do a little bit of simplification here, both the numerator and the denominator are divisible by x to the fourth, so let's divide by x to the fourth and we get eight over 35. So once again, you just look at eight 30, Well, this is going to be defined for any x. X isn't even involved in the expression. But if we want this to be algebraically equivalent to this first expression, then we have to make the same constraint, x does not, cannot be equal to zero. And to see, you know, this even seems a little bit more nonsensical to say x cannot be equal to zero"
    },
    {
        "Q": "at 1:10 Vi say says twogons. instead of twogons could you use biagons or bigons?",
        "A": "Yes, you could use biagons, but that sounds ridiculous, so most people just stick with 2-gons.",
        "video_name": "CfJzrmS9UfY",
        "timestamps": [
            70
        ],
        "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": "In the video, at 00:40 she says 'you can draw a four pointed star, but that's not really a star by the way you're defining your stars' -why not?",
        "A": "Well, later on in the video she explains the whole star game. You can t really use that method with four points.",
        "video_name": "CfJzrmS9UfY",
        "timestamps": [
            40
        ],
        "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": "at about 3:25 why does it bother you that a 25 pointed star a square star?",
        "A": "It doesn t bother her that a 25 pointed star is a square star. What bothers her is the method by which the 25 pointed star is made. If it is made from 5 pentagons, then that is clearly a square star. But what if it is made from 5 5 pointed stars? Then, although P=25 still, Q=10 instead of 5.",
        "video_name": "CfJzrmS9UfY",
        "timestamps": [
            205
        ],
        "3min_transcript": "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. at all because 11 is prime. Though here you start to wonder how to predict how many times around the circle we'll go before getting back to start. But instead of exploring the exciting world of modular arithmetic, you move on to 12, which is a really cool number because it has a whole bunch of factors. And then something starts to bother you. Is a 25-pointed star composite made of five five-pointed stars a square star? You had been thinking only of pentagons because the lower numbers didn't have this question. How could you have missed that? Maybe your teacher said something interesting about prime numbers, and you accidentally lost focus for a moment. I don't know. It gets even worse. 6 squared would be a 36-pointed star made of six hexagons. But if you allow use of six-pointed stars, then it's the same as a composite of 12 triangles. And that doesn't seem in keeping with the spirit of square stars. You'll have to define square stars more strictly. But you do like the idea that there's three ways to make the seventh square star. Anyway, the whole theory of what kind of stars can be made with what numbers is quite interesting. And I encourage you to explore this during your math class."
    },
    {
        "Q": "at 1:25 why is 9x^2 not equal to 3x^2 ?",
        "A": "We have 9x^2. We want to make it into (ax)^2. 3x^2 does NOT equal 9x^2. However, (3x)^2 does equal 9x^2. Why? When we have (3x)^2 the exponent distributes to both terms (the 3 and the x) so we have: (3x)^2 = 3^2x^2 = 9x^2.",
        "video_name": "jmbg-DKWuc4",
        "timestamps": [
            85
        ],
        "3min_transcript": "Let's see if we can factor the expression 45x squared minus 125. So whenever I see something like this-- I have a second-degree term here, I have a subtraction sign-- my temptation is to look at this as a difference of squares. We've already seen this multiple times. We've already seen that if we have something of the form a squared minus b squared, that this can be factored as a plus b times a minus b. So let's look over here. Well, over here, it's not obvious that this right over here is a perfect square. Neither is it obvious that this right over here So it's not clear to me that this is a difference of squares. But what is interesting is that both 45 and 125 have some factors in common. And the one that jumps out at me is 5. So let's see if we can factor out a 5, and by doing that, whether we can get something that's a little bit closer to this pattern right over here. 45x squared divided by 5 is going to be 9x squared. And then 125 divided by 5 is 25. Now, this is interesting. 9x squared-- that's a perfect square. If we call this a squared, then that tells us that a would be equal to 3x. 3x-- the whole thing squared is 9x squared. Similarly-- I can never say similarly correctly-- 25 is clearly just 5 squared. So in this case, if we're looking at this template, b would be equal to 5. So now this is a difference of squares, and we can factor it completely. So we can't forget our 5 out front that we factored out. So it's going to be 5 times a plus b. So let me write this. So it's going to be 5 times a plus b times a minus b. 5 times 3x plus 5 times 3x minus 5 is 45x squared minus 125 factored out."
    },
    {
        "Q": "At 5:25, You said that the median is the middle number, but at 2:25 you said that that average is the typical or the middle as well, but my teachers say to associate average with mean, so are there many averages or if not, whats the deal?",
        "A": "Yes, there are many averages. The mean is the most common average, so it s often used as a synonym even though it properly should not be.",
        "video_name": "h8EYEJ32oQ8",
        "timestamps": [
            325,
            145
        ],
        "3min_transcript": "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, 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."
    },
    {
        "Q": "Isn't the part in (3:53) where he finds the arithmetic mean, when he says to add and find the sum then divide, that also I think is one way how to find the average, right?",
        "A": "to find Mean: add up all of the numbers and then divide by how many numbers there are Median: The middle number. if there are two in the middle, add them up and divide by two to find the middle of those two numbers Mode: Whatever number is the most common (there can t be two modes. The number has to be repeated the MOST. There cant be two highest repeated numbers.)",
        "video_name": "h8EYEJ32oQ8",
        "timestamps": [
            233
        ],
        "3min_transcript": "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, 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"
    },
    {
        "Q": "At 2:08 Sal started explaining average. Are average and mean the same thing, because Sal never said anything about mean",
        "A": "Yes! Average and mean are synonyms. They both involve the total sum of a set of numbers and dividing it by how many numbers given.",
        "video_name": "h8EYEJ32oQ8",
        "timestamps": [
            128
        ],
        "3min_transcript": "We will now begin our journey into the world of statistics, which is really a way to understand or get our head around data. So statistics is all about data. And as we begin our journey into the world of statistics, we will be doing a lot of what we can call descriptive statistics. So if we have a bunch of data, and if we want to tell something about all of that data without giving them all of the data, can we somehow describe it with a smaller set of numbers? So that's what we're going to focus on. And then once we build our toolkit on the descriptive statistics, then we can start to make inferences about that data, start to make conclusions, start to make judgments. And we'll start to do a lot of inferential statistics, make inferences. So with that out of the way, let's think about how we can describe data. So let's say we have a set of numbers. We can consider this to be data. 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."
    },
    {
        "Q": "If 6 5/x = 14, what does x equal to?\nFor Example:\n6 multiplied by x = 6x\n6x + 5= 8.75/x\n8.75/x = 14\non 7:38",
        "A": "x = 0.625",
        "video_name": "9IUEk9fn2Vs",
        "timestamps": [
            458
        ],
        "3min_transcript": "this equation by x plus 5. You can say x plus 5 over 1. Times x plus 5 over 1. On the left-hand side, they get canceled out. So we're left with 3 is equal to 8 times x plus five. All of that over x plus 2. Now, on the top, just to simplify, we once again just multiply the 8 times the whole expression. So it's 8x plus 40 over x plus 2. Now, we want to get rid of this x plus 2. So we can do it the same way. We can multiply both sides of this equation by x plus 2 over 1. x plus 2. We could just say we're multiplying both sides by x plus 2. The 1 is little unnecessary. So the left-hand side becomes 3x plus 6. multiplying it times the whole expression. x plus 2. And on the right-hand side. Well, this x plus 2 and this x plus 2 will cancel out. And we're left with 8x plus 40. And this is now a level three problem. Well, if we subtract 8x from both sides, minus 8x, plus-- I think I'm running out of space. Minus 8x. Well, on the right-hand side the 8x's cancel out. On the left-hand side we have minus 5x plus 6 is equal to, on the right-hand side all we have left is 40. Now we can subtract 6 from both sides of this equation. Let me just write out here. Minus 6 plus minus 6. Now I'm going to, hope I don't lose you guys by trying to go up here. But if we subtract minus 6 from both sides, on the left-hand side we're just left with minus 5x equals, and on the Now it's a level one problem. We just multiply both sides times negative 1/5. Negative 1/5. On the left-hand side we have x. And on the right-hand side we have negative 34/5. Unless I made some careless mistakes, I think that's right. And I think if you understood what we just did here, you're ready to tackle some level four linear equations. Have fun."
    },
    {
        "Q": "At 2:49, why is it 7x + 7 ?",
        "A": "With the distributive property, you have to multiply each term in the brackets by the 7, so multiply 7 by x, and 7 by 1 to give you 7x + 7.",
        "video_name": "9IUEk9fn2Vs",
        "timestamps": [
            169
        ],
        "3min_transcript": "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 the other side of the equation. So I'm going to choose to get the x's on the left. So let's bring that 7x onto the left. And we can do that by subtracting 7x from both sides. Minus 7x, plus, it's a minus 7x. The right-hand side, these two 7x's will cancel out. And on the left-hand side we have minus 7x plus x. Well, that's minus 6x plus 2 is equal to, and on the right all we have left is 7. Now we just have to get rid of this 2. And we can just do that by subtracting 2 from both sides. And we're left with minus 6x packs is equal to 6. Now it's a level one problem. We just have to multiply both sides times the reciprocal of the coefficient on the left-hand side. And the coefficient's negative 6. So we multiply both sides of the equation by negative 1/6."
    },
    {
        "Q": "i dont understand anything, in 3:29 how did -7x+x+2 = 7x+7-7x turn into -6x+2=7?\nhow did -7x turn into -6x? it seems impossible to me",
        "A": "-7x+x=-6x",
        "video_name": "9IUEk9fn2Vs",
        "timestamps": [
            209
        ],
        "3min_transcript": "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 the other side of the equation. So I'm going to choose to get the x's on the left. So let's bring that 7x onto the left. And we can do that by subtracting 7x from both sides. Minus 7x, plus, it's a minus 7x. The right-hand side, these two 7x's will cancel out. And on the left-hand side we have minus 7x plus x. Well, that's minus 6x plus 2 is equal to, and on the right all we have left is 7. Now we just have to get rid of this 2. And we can just do that by subtracting 2 from both sides. And we're left with minus 6x packs is equal to 6. Now it's a level one problem. We just have to multiply both sides times the reciprocal of the coefficient on the left-hand side. And the coefficient's negative 6. So we multiply both sides of the equation by negative 1/6."
    },
    {
        "Q": "In 1:16 where where doe's the speaker get 1/5 from ? And do I have to watch all of the pre-algebra videos before I understand where 1/5 comes from ?",
        "A": "Thanx, then I ll have to watch all of the pre-al videos.",
        "video_name": "9IUEk9fn2Vs",
        "timestamps": [
            76
        ],
        "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 2:53, how was Sal able to tell whether it was sin or cosine?",
        "A": "When x is 0, the value of the cosine equation would be 1, and (0, 1) is not a point on the graph. When x is 0, the value of the sine equation would be -2, and (0, -2) is a valid point on the graph. Thus, the cosine equation can be eliminated.",
        "video_name": "yHo0CcDVHsk",
        "timestamps": [
            173
        ],
        "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 0:24 why did the problem say the time when you don't need it? And why did it say how many wheels?",
        "A": "Because the skill this video is trying to teach is how to look at a word problem and find the information that you need. So Sal said the time and how many wheels so you could find out what was the important part of the problem.",
        "video_name": "6QZCj4O9sk0",
        "timestamps": [
            24
        ],
        "3min_transcript": "The local grocery store opens at nine. Its parking lot has six rows. Each row can fit seven cars. Each car has four wheels. How many cars can the parking lot fit? And I encourage you to pause the video and think about this yourself. Try to figure it out on your own. So, let's re-read this. The local grocery store opens up at nine. Well, that doesn't really matter. If we're thinking about how many cars can the parking lot fit. So we don't really have to care about that. We also dont have to care about how many wheels each car has. They're not asking us how many wheels can fit in the parking lot. So we can ignore that. What we really care about is how many rows we have. And how many cars can fit in each row. What we have is -- We have six rows and each row can fit seven cars. We're going to six groups of seven. Or, another way of thinking about it. We're going to have six times seven cars can fit in the parking lot. What is this going to be equal to? This is literally six sevens added up. This is the same thing as one, two, three, four, five, six. Now we're going to add these up. Seven plus seven is fourteen. Twenty-one, twenty-eight, thirty-five, forty-two. Six times seven is equal to forty-two. So forty-two cars can fit in the parking lot. Don't believe me? I made a little diagram here. We have six rows. This is the first row. Third.. Fourth.. Fifth.. Sixth. Each row can fit seven cars. You see it here. One; Let me make that a little brighter. One.. Three.. Four, five, six, seven. How many cars are there? You have seven. Fourteen.. Twenty-one.. Twenty-eight.. Thirty-five.. Forty-two total cars. Six rows of seven."
    },
    {
        "Q": "At 2:13, why is 3^2 not the same as 2^3?",
        "A": "It doesn t work that way. 3*2 may be the same as 2*3, but exponentiation does NOT have the same property. Think about it. 2^3 implies 2*2*2, which we can calculate to be 8. 3^2 implies 3*3 which we can calculate to be 9. Obviously, 9 does not equal 8. This hold true for all exponents. In general, a^b does not equal b^a. (There are exceptions. One that I know is 2^4 does equal 4^2. Both are 16. This is only one instance. Otherwise, they usually are not the same.)",
        "video_name": "XZRQhkii0h0",
        "timestamps": [
            133
        ],
        "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": "I really do not get this one. At 4:00 he points at a and says this is the heighth of the parallelogram. The parallelogram first starts with two sides of A and two sides of C. After comparing them he says the area is A^2 , this insinuates that Side A and Side C are the same length and that is just not true?",
        "A": "Actually it is true. As he notes at 3:52 and repeats twice is that the height of the paralllelogram is a also which is shown in the triangle on the left. area of a parallelogram is base times height, so it is a \u00e2\u0080\u00a2 a or a^2. It does not have anything to do with the sides being the same length, and with the drawing, they cannot be the same length because c forms the hypotenuse of the right triangle which by definition must be longer than either leg.",
        "video_name": "rcBaqkGp7CA",
        "timestamps": [
            240
        ],
        "3min_transcript": "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. And since it's traveling the same distance in the x direction or in the horizontal direction and the vertical direction, this is going to be the same length. So this is going to be of length a. Now, the next question I have for you is, what is the area of this parallelogram that I have just constructed? Well, to think about that, let's redraw this part of the diagram so that the parallelogram is sitting on the ground. So this is length a. This is length c. This is length c. And if you look at this part right over here, it gives you a clue. I'll use this green color. The height of the parallelogram is given right over here. This side is perpendicular to the base. So the height of the parallelogram is a as well. Well, the area of a parallelogram is just the base times the height. So the area of this parallelogram right over here is going to be a squared. Now, let's do the same thing. But let's rotate our original right triangle. Let's rotate it the other way. So let's rotate it 90 degrees clockwise. And this time, instead of pivoting on this point, we're going to pivot on that point right over there. So what are we going to get? So the side of length c if we rotate it like that, it's going to end up right over here. I'll try to draw it as close to scale as possible. So that side has length c. Now, the side of length of b is going to pop out and look something like this. It's going to be parallel to that. This is going to be a right angle."
    },
    {
        "Q": "At 0:15, what does F(x) mean?",
        "A": "F(x) is just a notation to express a function in terms of x. It is the same as y = function, but that tends to be used in lower-level mathematics.",
        "video_name": "1LxhXqD3_CE",
        "timestamps": [
            15
        ],
        "3min_transcript": ""
    },
    {
        "Q": "The only thing I don't understand is the (x-c) part at 3:15, why put the c and not simply use x, x^2, x^3 and so on, like on the Mclaurin series?",
        "A": "It s a shift. It s like shifting the parabola function, y = x^2, three places to the left. You d write it as y = (x+3)^2. To shift it c to the left, you d use (x-c)^2. Or,in the case of the Taylor expansion, multiply the derivative(s) by (x-c).",
        "video_name": "1LxhXqD3_CE",
        "timestamps": [
            195
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 7:42 can someone explain how he got x+sqrt of 2=0 and the same for x-sqrt2= 0?",
        "A": "Sal recognised that the binomial x^2 - 2 could be viewed as the difference of two squares, so he factored it into sum and difference of the two numbers being squared, ( x and \u00e2\u0088\u009a2 ).",
        "video_name": "x9lb_frpkH0",
        "timestamps": [
            462
        ],
        "3min_transcript": "times x minus the square root of two. I'm just recognizing this as a difference of squares. And, once again, we just want to solve this whole, all of this business, equaling zero. All of this equaling zero. So how can this equal to zero? Well any one of these expressions, if I take the product, and if any one of them equals zero then I'm gonna get zero. So, x could be equal to zero. X could be equal to zero, and that actually gives us a root. When x is equal to zero, this polynomial is equal to zero, and that's pretty easy to verify. Let's see, can x-squared plus nine equal zero? X-squared plus nine equal zero. Well, if you subtract nine from both sides, you get x-squared is equal to negative nine. And that's why I said, there's no real solution to this. So, no real, let me write that, no real solution. There are some imaginary solutions, but no real solutions. Now, can x plus the square root of two equal zero? Sure, if we subtract square root of two from both sides, you get x is equal to the negative square root of two. And can x minus the square root of two equal zero? Sure, you add square root of two to both sides, you get x is equal to the square root of two. So, there we have it. We have figured out our zeros. X could be equal to zero. P of zero is zero. P of negative square root of two is zero, and p of square root of two is equal to zero. So, those are our zeros. Their zeros are at zero, negative squares of two, and positive squares of two. And so those are going to be the three times that we intercept the x-axis. And what is the smallest of those intercepts? Well, the smallest number here is negative square root, negative square root of two. And you could tackle it the other way. You could take this part right over... Yeah, this part right over here and you could add those two middle terms, and I encourage you to do that. But just to see that this makes sense that zeros really are the x-intercepts. I went to Wolfram|Alpha and I graphed this polynomial and this is what I got. So, this is what I got, right over here. If you see a fifth-degree polynomial, say, it'll have as many as five real zeros. But, if it has some imaginary zeros, it won't have five real zeros. Instead, this one has three. And that's because the imaginary zeros, which we'll talk more about in the future, they come in these conjugate pairs. So, if you don't have five real roots, the next possibility is that you're going to have three real roots. And, if you don't have three real roots, the next possibility is you're gonna have one real root. So, that's an interesting thing to think about. And so, here you see, your three real roots. You see your three real roots which correspond to the x-values at which the function is equal to zero,"
    },
    {
        "Q": "at approximately 14:20, how would you write\n-1 (x^2 + 5x - 24) as a two parentheses group? like this? ---------> -1 (x-3)(x+8)\nor, do you leave it as is?",
        "A": "Yes, you can write it the way you suggested: -1 (x-3)(x+8) Exactly how you would write it depends on what you need to do to reach the final answer. You can also use the distributive property to distribute the -1 into ONE (not both) of the factors: -1 (x-3)(x+8) = (3-x)(x+8) however, doing this is rather unusual.",
        "video_name": "eF6zYNzlZKQ",
        "timestamps": [
            860
        ],
        "3min_transcript": "And then if we take negative 8 plus 7, that is equal to negative 1, which is exactly the coefficient right there. So when I factor this, this is going to be x minus 8, times x plus 7. This is often one of the hardest concepts people learn in algebra, because it is a bit of an art. You have to look at all of the factors here, play with the positive and negative signs, see which of those factors when one is positive, one is negative, add up to the coefficient on the x term. But as you do more and more practice, you'll see that it'll become a bit of second nature. Now let's step up the stakes a little bit more. Let's say we had negative x squared-- everything we've done so far had a positive coefficient, a positive 1 coefficient on the x squared term. But let's say we had a negative x squared minus 5x, plus 24. How do we do this? Well, the easiest way I can think of doing it is factor 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"
    },
    {
        "Q": "At 0:57, does he use the FOIL method?",
        "A": "Yep! First Outside Inside Last",
        "video_name": "eF6zYNzlZKQ",
        "timestamps": [
            57
        ],
        "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": "at 14:46 we do not have to multiply negative 1 to both (x-3)(x+8) ?",
        "A": "No, because -x*-x would equal positive x\u00c2\u00b2. The goal is to get -x\u00c2\u00b2. So, -x*x = -x\u00c2\u00b2 Hope this helps!",
        "video_name": "eF6zYNzlZKQ",
        "timestamps": [
            886
        ],
        "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": "At 13:55 it says that we turn 24 into 1 and 24,and then that if it is negative 1 and 24 it would be positive 23. Can you explain this?",
        "A": "(a+b)^2 = a^2 + 2ab + b^2 Sal meant that 2ab would = positive 23 if a and b were -1 and +24",
        "video_name": "eF6zYNzlZKQ",
        "timestamps": [
            835
        ],
        "3min_transcript": "And then if we take negative 8 plus 7, that is equal to negative 1, which is exactly the coefficient right there. So when I factor this, this is going to be x minus 8, times x plus 7. This is often one of the hardest concepts people learn in algebra, because it is a bit of an art. You have to look at all of the factors here, play with the positive and negative signs, see which of those factors when one is positive, one is negative, add up to the coefficient on the x term. But as you do more and more practice, you'll see that it'll become a bit of second nature. Now let's step up the stakes a little bit more. Let's say we had negative x squared-- everything we've done so far had a positive coefficient, a positive 1 coefficient on the x squared term. But let's say we had a negative x squared minus 5x, plus 24. How do we do this? Well, the easiest way I can think of doing it is factor 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"
    },
    {
        "Q": "I instantly went from getting it to COMPLETELY LOST.\n\nAt 2:14 he says, \"But we have i times i, or i squared, which is negative one.\" He lost me right there. I have no idea how he can say i squared is negative one. Can anyone explain this to me?",
        "A": "The definition of i is the number whose square is -1. What he said was completely valid. You should watch the videos on i.",
        "video_name": "Z8j5RDOibV4",
        "timestamps": [
            134
        ],
        "3min_transcript": "We're asked to divide. And we're dividing six plus three i by seven minus 5i. And in particular, when I divide this, I want to get another complex number. So I want to get some real number plus some imaginary number, so some multiple of i's. So let's think about how we can do this. Well, division is the same thing -- and we rewrite this as six plus three i over seven minus five i. These are clearly equivalent; dividing by something is the same thing as a rational expression where that something is in the denominator, right over here. And so how do we simplify this? Well, we have a tool in our toolkit that can make sure that we don't have an imaginary or complex number in the denominator. And that's the complex conjugate. If we multiply both the numerator and the denominator of this expression by the complex conjugate of the denominator, then we will have a real number in the denominator. So let's do that. Let's multiply the numerator and the denominator by the conjugate of this. So seven PLUS five i. Seven plus five i is the complex conjegate of seven minus five i. And anything divided by itself is going to be one (assuming you're not dealing with zero; zero over zero is undefined). But seven plus five i over seven plus five i is one. So we're not changing the value of this. But what this does is it allows us to get rid of the imaginary part in the denominator. So let's multiply this out. Our numerator -- we just have to multiply every part of this complex number times every part of this complex number. You can think of it as FOIL if you like; we're really just doing the distributive property twice. We have six times seven, which is forty two. And then we have six times five i, which is thirty i. So plus thirty i. And then we have three i times seven, so that's plus twenty-one i. Three times five is fifteen. But we have i times i, or i squared, which is negative one. So it would be fifteen times negative one, or minus 15. So that's our numerator. And then our denomenator is going to be -- Well, we have a plus b times a minus b. (You could think of it that way. Or we could just do what we did up here. Actually, let's just do what we did up here so you don't have to remember that difference of squares pattern and all that.) Seven times seven is forty-nine. Let's think of it in the FOIL way, if that is helpful for you. So first we did the 7X7. And we can do the outer terms. 7 X 5i is +35i. Then we can do the inner terms. -5i X 7 is -35i. These two are going to cancel out. And then -5i X 5i is -25i^2 (\"negative twenty five i squared\")."
    },
    {
        "Q": "At 1:18 sal says 0/0 is undefined why?",
        "A": "Division by zero is an operation for which you cannot find an answer, so it is disallowed. You can understand why if you think about how division and multiplication are related. 12 divided by 6 is 2 because: 6 times 2 is 12 Now image 12/0: 12 divided by 0 is x would mean that: 0 times x = 12 But no value would work for x because 0 times any number is 0. So division by zero doesn t work.",
        "video_name": "Z8j5RDOibV4",
        "timestamps": [
            78
        ],
        "3min_transcript": "We're asked to divide. And we're dividing six plus three i by seven minus 5i. And in particular, when I divide this, I want to get another complex number. So I want to get some real number plus some imaginary number, so some multiple of i's. So let's think about how we can do this. Well, division is the same thing -- and we rewrite this as six plus three i over seven minus five i. These are clearly equivalent; dividing by something is the same thing as a rational expression where that something is in the denominator, right over here. And so how do we simplify this? Well, we have a tool in our toolkit that can make sure that we don't have an imaginary or complex number in the denominator. And that's the complex conjugate. If we multiply both the numerator and the denominator of this expression by the complex conjugate of the denominator, then we will have a real number in the denominator. So let's do that. Let's multiply the numerator and the denominator by the conjugate of this. So seven PLUS five i. Seven plus five i is the complex conjegate of seven minus five i. And anything divided by itself is going to be one (assuming you're not dealing with zero; zero over zero is undefined). But seven plus five i over seven plus five i is one. So we're not changing the value of this. But what this does is it allows us to get rid of the imaginary part in the denominator. So let's multiply this out. Our numerator -- we just have to multiply every part of this complex number times every part of this complex number. You can think of it as FOIL if you like; we're really just doing the distributive property twice. We have six times seven, which is forty two. And then we have six times five i, which is thirty i. So plus thirty i. And then we have three i times seven, so that's plus twenty-one i. Three times five is fifteen. But we have i times i, or i squared, which is negative one. So it would be fifteen times negative one, or minus 15. So that's our numerator. And then our denomenator is going to be -- Well, we have a plus b times a minus b. (You could think of it that way. Or we could just do what we did up here. Actually, let's just do what we did up here so you don't have to remember that difference of squares pattern and all that.) Seven times seven is forty-nine. Let's think of it in the FOIL way, if that is helpful for you. So first we did the 7X7. And we can do the outer terms. 7 X 5i is +35i. Then we can do the inner terms. -5i X 7 is -35i. These two are going to cancel out. And then -5i X 5i is -25i^2 (\"negative twenty five i squared\")."
    },
    {
        "Q": "1:50 i still dont understand what the difference is between midrange and range? help?",
        "A": "The midrange is an attempt to show in a single number how well or bad a class is doing, so it s pretty similar to the average and the median. The range is only used to show how big the difference is between the best and the worst student. It can t be used to judge how well or bad a class is doing. Only if all of the students are on a similar level or if there are huge differences between the students.",
        "video_name": "DGZNaKnbQo0",
        "timestamps": [
            110
        ],
        "3min_transcript": "In this chart right here we're given scores on midterm and final exams, where the vertical axis is the score in points. And then each of these pairs of bar charts give us for an individual, where the blue bar is, for example, how Ishaan did on the midterm, the yellow is how he did on the final. For Emily, the blue is how she did on the midterm, yellow is how she did on the final. And we have a bunch of interesting questions here. The first question, what was the median score for the final exam? So just as a review, median literally means what was the middle score? So really we should list all the scores for the final exam and sort them in order and then figure out what the middle score actually was. So let's look at all the scores on final exam. Sp you have 100 here. Ishaan got 100 on the final exam. Remember, this yellow bar is the final exam. So there's 100. Emily also got 100 on the final exam. It looks like it was an easy final exam. Daniel also got 100 on the final exam. And then, let's see, Jessica, it looks like she got a 75. So if we were to sort these in order, and let's say we did it in increasing order, you could write-- well, the lowest score was a 75, then you have an 80, and then you have three 100's-- 100, another 100, and another 100. So there's five scores right over here, so you will have a middle. If you had an even number, then you would take the mean of the two center values. But here you have one center value, and when you order it like this, it's pretty clear that your center value, your middle value, is 100. So the median score for the final exam is 100. And that's because you had so many hundreds here that the median, the middle score, was still 100. What is the midrange of the midterm scores? I'll do it in blue in honor of the color of the bars for the midterm. So the midrange is the mean of your highest and lowest scores. So let's calculate this. So let's go to the midrange. or the average of your highest and lowest scores, so the midrange of midterm. So let's see, the highest midterm score, looking at the blue, the highest one is right here. So Jessica got 100 on the midterm, so that's your highest score. Your lowest score on the midterm looks like this one right over here. Daniel got a 60. And so the midrange is going to be the mean, the arithmetic mean of these two numbers. So you add 100 plus 60, divide by 2, you get 160 over 2, or 80. So this right over here is going to be 80. What was the average student score for the final exam? Well, for, that we just have to add up the scores on the final exams and then divide by the number of scores we have. So we might be able to do that in our heads. Well, we could-- let me just write it over here. So we have 100, plus 100, plus 100, plus 75, plus 80."
    },
    {
        "Q": "At 3:49, how is the abs(-x^2/3) the same as abs(x^2/3) ? Wouldn't the negative sign have to be like this: abs{(-x^2)/3}, for the negative sign to disappear since -x times -x is positive x^2? I know this is very basic, but it's confusing me...",
        "A": "Let s solve it in the normal way: |-x\u00c2\u00b2/3| < 1 \u00e2\u0086\u0094 -1 < -x\u00c2\u00b2/3 < 1 \u00e2\u0086\u0094 -3 < -x\u00c2\u00b2 < 3 \u00e2\u0086\u0094 3 > x\u00c2\u00b2 > -3 Because x\u00c2\u00b2 >= 0 > -3 for every x so we just need to take care x\u00c2\u00b2 < 3 \u00e2\u0086\u0094 -\u00e2\u0088\u009a3 < x < \u00e2\u0088\u009a3.",
        "video_name": "aiwy2fNF_ZQ",
        "timestamps": [
            229
        ],
        "3min_transcript": "this would be equal to-- so the first term is 1/3 times all of this to the 0-th power. So it's just going to be 1/3. And so each successive term is just going to be the previous term times our common ratio. So 1/3 times negative x squared over 3 is going to be negative 1/9 x squared. To go from that to that, you have to multiply by-- let's see, 1/3 to negative 1/3, you have to multiply it by negative 1/3. And we multiplied by x squared as well. Now in our next term, we're going to multiply by negative x squared over 3 again. So it's going to be plus-- a negative times a negative is a positive-- plus 1/27 x to the fourth. x squared times x squared, x to the fourth power. And we just keep going on and on and on. And when this converges, so over the interval of convergence, Now, what is the interval of convergence here? And I encourage you to pause the video and think about it. Well, the interval of convergence is the interval over which your common ratio, the absolute value of your common ratio, is less than 1. So let me write this right over here. So our absolute value of negative x squared over 3 has to be less than 1. Well, the absolute value, this is going to be a negative number. This is the same thing as saying-- let me scroll down a little bit. This is the same thing as saying that the absolute value of x squared over 3 has to be less than 1. And this is another way of saying-- this is going to be positive no matter what. Or I guess I should say, this is going to be non-negative no matter what. So this is another way of saying that x squared over 3 has to be less than 1. Right? I don't want to confuse you in this step right over here. But the absolute value of x squared over 3 is just going to be x squared over 3, because this is never going to take on a negative value. And so we can multiply both sides by 3. I'll go up here now to do it. Multiply both sides by 3 to say that x squared needs to be less than 3. And so that means that the absolute value of x needs to be less than the square root of 3. Or we could say that x is greater than the negative square root of 3, and it is less than the square root of 3. So this is the interval of convergence. This is the interval of convergence for this series,"
    },
    {
        "Q": "How come 13/10. At 2:40?",
        "A": "A one above a zero is is 10/10 s where are the 0.3 is 3/10 s. If you add 3/10 s to 10/10 s you will get 13/10 s.",
        "video_name": "3szFVS5p_7A",
        "timestamps": [
            160
        ],
        "3min_transcript": "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 And of course, we have what we have in our ones place, plus 1 plus 3/10. And what I want to do is I want to take this 9, the 9 that I took from the tens place and give to the ones place. And I'm going to take this 1 that I took from the tens place and give it to the tenths place. So 1 is the same thing as 10/10. And so when you regroup this value, you get this as being equal to 10 plus-- 9 plus 1 is 10, and then 10/10 plus 3/10 is 13/10. So that's all that happened here. I changed the value in the places. I took 1 ten away. I had 2 tens. Now I'm only left with 1 ten. And that extra 10 of value, I regrouped it. I gave 9 to the ones place."
    },
    {
        "Q": "If you did not simplified 2xh+x*/h and you are finding the limit of h as it get closer to 0, doesn't the question become undefined number since it is dividing by zero. I tried it with a function of x+2 and it gave me h/h after all the simplifying. Did I do something wrong? Video(5:22 - 8:28)",
        "A": "The limit takes care of that issue, we are not dividing by 0, but by a number that approaches 0. The slope of the function x+2 is 1, so it makes sense you ended with the answer of h/h since the Limit as h approaches 0 of the function h/h = 1",
        "video_name": "IePCHjMeFkE",
        "timestamps": [
            322,
            508
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 7:46 how does he get 6 from 6+delta(x)??",
        "A": "Basically we re asking what happens as the \u00ce\u0094x approaches zero. So pretend that \u00ce\u0094x is actually zero and then you have 6 + 0 and 6 + 0 = 6.",
        "video_name": "IePCHjMeFkE",
        "timestamps": [
            466
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 1:20, why do you have to move left, why can't you move right. Is it because the place values get lower when you go to the eftl",
        "A": "Exactly. If you go to the left, the place value gets 20 times greater. For example, there is the ones place, the tens place, hundreds place, thousands place, ten thousands place, hundred thousands place, millions place, ten millions place, and you get the idea. But, if you go to the right side, the place value gets 1/10 smaller.",
        "video_name": "iK0y39rjBgQ",
        "timestamps": [
            80
        ],
        "3min_transcript": "Write 14,897 in expanded form. Let me just rewrite the number, and I'll color code it, and that way, we can keep track of our digits. So we have 14,000. I don't have to write it-- well, let me write it that big. 14,000, 800, and 97-- I already used the blue; maybe I should use yellow-- in expanded form. So let's think about what place each of these digits are in. This right here, the 7, is in the ones place. The 9 is in the tens place. This literally represents 9 tens, and we're going to see this in a second. This literally represents 7 ones. The 8 is in the hundreds place. It literally represents 4,000. And then the 1 is in the ten-thousands place. And you see, every time you move to the left, you move one place to the left, you're multiplying by 10. Ones place, tens place, hundreds place, thousands place, ten-thousands place. Now let's think about what that really means. If this 1 is in the ten-thousands place, that means that it literally represents-- I want to do this in a way that my arrows don't get mixed up. Actually, let me start at the other end. Let me start with what the 7 represents. The 7 literally represents 7 ones. Or another way to think about it, you could say it represents 7 times 1. All of these are equivalent. They represent 7 ones. That's why I'm doing it from the right, so that the arrows don't have to cross each other. So what does the 9 represent? It represents 9 tens. You could literally imagine you have 9 actual tens. You could have a 10, plus a 10, plus a 10. Do that nine times. That's literally what it represents: 9 actual tens. 9 tens, or you could say it's the same thing as 9 times 10, or 90, either way you want to think about it. So let me write all the different ways to think about it. It represents all of these things: 9 tens, or 9 times 10, or 90. So then we have our 8. Our 8 represents-- we see it's in the hundreds place. It represents 8 hundreds. Or you could view that as being equivalent to 8 times 100-- a hundred, not a thousand-- 8 times 100, or 800. That 8 literally represents 8 hundreds, 800."
    },
    {
        "Q": "Okay... he lost me at 0:14. I have no idea what's going on, can someone help me?",
        "A": "a difference of square is a binomial in which both the terms are perfect squares and they are subtracted a2-b2 if you have a difference of squares expression here is how you would factor it a2-b2=(a+b)(a-b) in this case it is x2-49y2 a=x b=7y x2-49y2=(x+7y)(x-7y)",
        "video_name": "tvnOWIoeeaU",
        "timestamps": [
            14
        ],
        "3min_transcript": "Factor x squared minus 49y squared. So what's interesting here is that well x squared is clearly a perfect square. It's the square of x. And 49y squared is also a perfect square. It's the square of 7y. So it looks like we might have a special form here. And to remind ourselves, let's think about what happens if we take a plus b times a minus b. I'm just doing it in the general case so we can see a pattern here. So over here, this would be a times a, which would be a squared plus a times negative b, which would be negative ab plus b times a or a times b again, which would be ab. And then you have b times negative b, so it would b minus b squared. Now these middle two terms cancel out. Negative ab plus ab, they cancel out and you're left with just a squared minus b squared. And that's the exact pattern we have here. We have an a squared minus a b squared. So in this case, a is equal to x and b is equal to 7y. So we can expand this as the difference of squares, or actually this thing right over here is the difference of squares. So we expand this like this. So this will be equal to x plus 7y times x minus 7y. And once again, we're just pattern matching based on this realization right here. If I take a plus b times a minus b, I get a difference of squares. This is a difference of squares. So when I factor it, it must come out to the result of something that looks like a plus b times a minus b or x plus 7y times x minus 7y."
    },
    {
        "Q": "When, at 1:09 he shows what happens as a perfect square for A and B (that b is 7y, because 49 is a perfect square) would that work for say, 5, where for example it might be b=sqrt(5)? Or am I jumping to conclusions?",
        "A": "it wouldn t be a perfect square, so no.",
        "video_name": "tvnOWIoeeaU",
        "timestamps": [
            69
        ],
        "3min_transcript": "Factor x squared minus 49y squared. So what's interesting here is that well x squared is clearly a perfect square. It's the square of x. And 49y squared is also a perfect square. It's the square of 7y. So it looks like we might have a special form here. And to remind ourselves, let's think about what happens if we take a plus b times a minus b. I'm just doing it in the general case so we can see a pattern here. So over here, this would be a times a, which would be a squared plus a times negative b, which would be negative ab plus b times a or a times b again, which would be ab. And then you have b times negative b, so it would b minus b squared. Now these middle two terms cancel out. Negative ab plus ab, they cancel out and you're left with just a squared minus b squared. And that's the exact pattern we have here. We have an a squared minus a b squared. So in this case, a is equal to x and b is equal to 7y. So we can expand this as the difference of squares, or actually this thing right over here is the difference of squares. So we expand this like this. So this will be equal to x plus 7y times x minus 7y. And once again, we're just pattern matching based on this realization right here. If I take a plus b times a minus b, I get a difference of squares. This is a difference of squares. So when I factor it, it must come out to the result of something that looks like a plus b times a minus b or x plus 7y times x minus 7y."
    },
    {
        "Q": "At 1:29, Sal says that 0.1 is bigger than 0.070. How is that possible when 0.070 has more digits than 0.1",
        "A": "One of the easiest ways to compare decimals, it to give them the same number of decimal digits. Adding zeros on the right of the decimal does not change the original value of the number. Change 0.1 into 0.100 You are now comparing 0.100 to 0.070 100 is bigger than 70. Thus, 0.1 is larger then 0.070 Another way to look at this is the the further to the right the number is following the decimal point, the smaller the number. The 1 in 0.1 = 1/10. The 7 in 0.070 = 7/100 7/100 is much smaller than 1/10 Hope this helps.",
        "video_name": "gAV9kwvoD6s",
        "timestamps": [
            89
        ],
        "3min_transcript": "Let's compare 0.1 to 0.070. So this 1 right over here, it is in the tenths place. So it literally represents 1 times 1/10, which is obviously the same thing as 1/10. Now, when we look at this number right over here, it has nothing in the tenths place. It has 7 in the hundredths place. So this is the hundredths place right over here. And then it also has nothing in the thousandths place. So this number can be rewritten as 7 times 1/100, or 7/100. And now we could compare these two numbers. And there's two ways you could think about it. You could try to turn 1/10 into hundredths. And the best way to do that, if you want the denominator to be increased by a factor of 10, you need to do the same thing to the numerator. So all I did is I multiplied the numerator and denominator by 10. Ten 100's is the exact same thing as 1/10. And here it becomes very clear, 10/100 Another way you could think about this is, look, if you were to increment by hundredths here, you would start at 7/100, 8/100, 9/100, and then you would get to 10/100. So then you would get to that number. So this number, multiple ways you could think about it, is definitely larger. So let me write this down. This is definitely larger, greater than. This is greater than that. The greater than symbol opens to the larger value. So here we have 0.093 and here we have 0.01. So let's just think about this a little bit. So this 9-- get a new color here. This 9 is not in the tenths, the hundredths. It's in the thousandths place. It's in the thousands place. And this 3 is in the-- I'm running out of colors again. This 3 is in the ten thousandths place. So the 3 is in the ten thousandths place. And if you just wanted to write it in terms of ten thousandths, you can multiply the 9 and 1,000 by 0. And so it becomes 90/10,000. And if you want to add them together, you could, of course, write this as 93/10,000. Ten thousandths. I always have trouble with that \"-ths\" at the end. Now, let's think about this number right over here, 0.01. Well, this 1 right over here is in the hundredths place. It's in the hundredths place. So it literally represents 1/100. So how can we compare 1/100 to 93/10,000? So the best way to think about it is, well, what's 1/100 in terms of ten thousandths? Well, let's just multiply both the numerator and the denominator here by 10 twice. Or you could say, let's multiply them both by 100."
    },
    {
        "Q": "at 4:52 he says over 2 does that apply all the time or just for this instance?",
        "A": "The midpoint formula is ((x1+x2)/2,(y1+y2)/2). This applies all the time.",
        "video_name": "Efoeqb6tC88",
        "timestamps": [
            292
        ],
        "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": "At 4:06, What do you mean by rotating the plane in infinite directions?",
        "A": "the plane could be facing any direction but you don t know by the representation.",
        "video_name": "J2Qz-7ZWDAE",
        "timestamps": [
            246
        ],
        "3min_transcript": "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. I could keep rotating around the line, just as we did over here. It does not specify only one plane."
    },
    {
        "Q": "at 3:26 Sal said that 49+9 is 57 but it is 58 right?",
        "A": "Yes. This is a known error in the video. And, a box does pop up and tell you the error and correct info.",
        "video_name": "-U53eHKCLcg",
        "timestamps": [
            206
        ],
        "3min_transcript": "Greek letter delta is just shorthand for change in y and this once again comes straight out of the distance formula which really comes out of the Pythagorean theorem. It'll become a little bit more obvious when I draw the change in y and I draw the change in x on this diagram. So what's our change in y? Well we're starting at our initial point. We're starting at y is equal to nine and we are, to get to the y value of our terminal point, we're going down to y is equal to two. So we have a change in y our change in y is equal to, going from nine to two, our change in y is negative seven. Similarly, our change in x we're going from x is equal to two, to x is equal to five. So our change in the horizontal direction is plus three. So our change in x where we can either think of this as the horizontal component of the vector. This is equal to positive three we have drawn a right triangle and so we can use a Pythagorean theorem to figure out, to figure out the length of the hypotenuse and you might say wait, wait, a length of a side of a triangle can't have a negative value and that's why these squareds are valuable because it doesn't matter if you're taking a negative seven squared or a positive seven squared, you're going to get a positive value here and if you really just view this as a triangle, all you care about is the length of this side right over here or it's the magnitude of this side or the absolute value of it which is just going to be positive seven and so we can say this is going to be equal to the magnitude of our vector is going to be equal to so three squared is nine, nine, and then negative seven squared is positive 49. So plus 49 or once again you could view this as our change in y squared which is negative seven squared or you could say, well just look, we don't wanna think of a side as having a negative value, the negative really just says, hey we're going from the top to the bottom it gives us our direction, but if we just say the length of it it's seven, well you're there if you just use a Pythagorean theorem. Seven squared would also be 49 and so either way you get the magnitude of our vector is equal to the square root of nine plus 49 is going to be 57, I don't think I can simplify this radical too much, no that's it. So the magnitude of this vector is the square root of 57."
    },
    {
        "Q": "At 3:33 the value would be square root of 58.",
        "A": "Hello Soham, Correct, this is a know problem based on the number of comment it has received... Regards, APD",
        "video_name": "-U53eHKCLcg",
        "timestamps": [
            213
        ],
        "3min_transcript": "Greek letter delta is just shorthand for change in y and this once again comes straight out of the distance formula which really comes out of the Pythagorean theorem. It'll become a little bit more obvious when I draw the change in y and I draw the change in x on this diagram. So what's our change in y? Well we're starting at our initial point. We're starting at y is equal to nine and we are, to get to the y value of our terminal point, we're going down to y is equal to two. So we have a change in y our change in y is equal to, going from nine to two, our change in y is negative seven. Similarly, our change in x we're going from x is equal to two, to x is equal to five. So our change in the horizontal direction is plus three. So our change in x where we can either think of this as the horizontal component of the vector. This is equal to positive three we have drawn a right triangle and so we can use a Pythagorean theorem to figure out, to figure out the length of the hypotenuse and you might say wait, wait, a length of a side of a triangle can't have a negative value and that's why these squareds are valuable because it doesn't matter if you're taking a negative seven squared or a positive seven squared, you're going to get a positive value here and if you really just view this as a triangle, all you care about is the length of this side right over here or it's the magnitude of this side or the absolute value of it which is just going to be positive seven and so we can say this is going to be equal to the magnitude of our vector is going to be equal to so three squared is nine, nine, and then negative seven squared is positive 49. So plus 49 or once again you could view this as our change in y squared which is negative seven squared or you could say, well just look, we don't wanna think of a side as having a negative value, the negative really just says, hey we're going from the top to the bottom it gives us our direction, but if we just say the length of it it's seven, well you're there if you just use a Pythagorean theorem. Seven squared would also be 49 and so either way you get the magnitude of our vector is equal to the square root of nine plus 49 is going to be 57, I don't think I can simplify this radical too much, no that's it. So the magnitude of this vector is the square root of 57."
    },
    {
        "Q": "at about 1:50, how do you get 2y=y+3? also, when you subtract y from both sides, you still need to divide by the 2, so wouldn't y be equal to 1 and 1/2? sorry if this is a stupid question but I could really use the help!:)",
        "A": "To answer the first part of your question, we just get the +3 by combining -4 and +7. As for the second part, when you subtract y from 2y, you just end up with y, so you don t have any number you need to divide the 3 by. Hope that helped!",
        "video_name": "uzyd_mIJaoc",
        "timestamps": [
            110
        ],
        "3min_transcript": "Use substitution to solve for x and y. And we have a system of equations here. The first equation is 2y is equal to x plus 7. And the second equation here is x is equal to y minus 4. So what we want to do, when they say substitution, what we want to do is substitute one of the variables with an expression so that we have an equation and only one variable. And then we can solve for it. Let me show you what I'm talking about. So let me rewrite this first equation. 2y is equal to x plus 7. And we have the second equation over here, that x is equal to y minus 4. So if we're looking for an x and a y that satisfies both constraints, well we could say, well look, at the x and y have to satisfy both constraints, both of these constraints have to be true. So x must be equal to y minus 4. So anywhere in this top equation where we see an x, anywhere we see an x, we say well look, that x by the second constraint has to be equal to y minus 4. So everywhere we see an x, we can substitute it So let's do that. So if we substitute y minus 4 for x in this top equation, the top equation becomes 2y is equal to instead of an x, the second constraint tells us that x needs to be equal to y minus 4. So instead of an x, we'll write a y minus 4, and then we have a plus 7. All I did here is I substituted y minus 4 for x. The second constraint tells us that we need to do it. y minus 4 needs to be equal to x or x needs to be equal to y minus 4. The value here is now we have an equation, one equation with one variable. We can just solve for y. So we get 2y is equal to y, and then we have minus 4 plus 7. So y plus 3. We can subtract y from both sides of this equation. The left hand side, 2y minus y is just y. And then we could go back and substitute into either of these equations to solve for x. This is easier right over here, so let's substitute right over here. x needs to be equal to y minus 4. So we could say that x is equal to 3 minus 4 which is equal to negative 1. So the solution to this system is x is equal to negative 1 and y is equal to 3. And you can verify that it works in this top equation 2 times 3 is 6 which is indeed equal to negative 1 plus 7. Now I want to show you that over here we substituted-- we had an expression that, or we had an equation, that explicitly solved for x. So we were able to substitute the x's. What I want to show you is we could have done it the other way around. We could have solved for y and then substituted for the y's."
    },
    {
        "Q": "at 1:10 couldn't \"a\" be a different value",
        "A": "Yes, it can. A variable can be any number you choose it to be. For example, X is the most used variable but, in the video Sal used a as the variable. This has no effect in the equation.",
        "video_name": "P6_sK8hRWCA",
        "timestamps": [
            70
        ],
        "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": "This might seem like a pretty arbitrary question, but at 1:37 Sal uses the \"/\" symbol to signify x(a+3-b) over (a+3-b). I've been seeing that \"over than\" symbol a lot in algebra and I was wondering if there was any difference between the over than symbol, \"/\", and the division symbol I'm used to seeing, \"\u00c3\u00b7\"?",
        "A": "They mean the same thing. I m honestly not sure why, but once you get past pre-algebra, you stop seeing the \u00c3\u00b7 symbol.",
        "video_name": "adPgapI-h3g",
        "timestamps": [
            97
        ],
        "3min_transcript": "- [Voiceover] So we have an equation. It says, a-x plus three-x is equal to b-x plus five. And what I want to do together is to solve for x, and if we solve for x it's going to be in terms of a, b, and other numbers. So pause the video and see if you can do that. All right now, let's do this together, and what I'm going to do, is I'm gonna try to group all of the x-terms, let's group all the x-terms on the left-hand side. So, I already have a-x and three-x on the left-hand side. Let's get b-x onto the left-hand side as well, and I can do that by subtracting b-x from both sides. And if I subtract b-x from both sides, I'm going to get on the right-hand side, I'm going to have a, or on the left-hand side, a-x plus three-x minus b-x, so I can do that in that color for fun, minus b-x, and that's going to be equal to... Well, b-x minus b-x is just zero, and I have five. It is equal to five. I can factor an x out of this left-hand side of this equation, out of all of the terms. So, I can rewrite this as x times... Well, a-x divided by x is a. Three-x divided by x is three, and then negative b-x divided by x is just going to be negative b. I could keep writing it in that pink color. And that's all going to be equal to five. And now, to solve for x I can just divide both sides by, the thing that x is being multiplied by, by a plus three minus b. So, I can divide both sides by a plus three minus b. A plus three minus b. On this side, they cancel out. And, I have x is equal to five over a plus three minus b, and we are done. Let's do one more of these. 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,"
    },
    {
        "Q": "At 4:06, Sal multiplies the numbers by -1. Do you have to do this?",
        "A": "If you didn t do it, you would get: x = (-8-5a)/((-a-b) This fraction is not fully reduced. You would need to factor out a -1 from both the numerator and denominator to reduce the fraction.",
        "video_name": "adPgapI-h3g",
        "timestamps": [
            246
        ],
        "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 1:21 , why does it matter where p is?",
        "A": "because the P is the variable for the pizzas. It is with the 8.5 because each pizza is 8.50. If it were with the 42.5, that would mean we are saying the cost of the pizza is 42.50 apiece.",
        "video_name": "2REbsY4-S70",
        "timestamps": [
            81
        ],
        "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": "Around 1:30, he explains that we need to use the pythagorean theory to find the radius r. But can't we just estimate the no. of units from the centre (point -1, 1) to the point (7.5, 1), which also lies on the circumference?",
        "A": "sometimes you will need precision in your answers. Pythagorean Theorem will give you as much precision as you need",
        "video_name": "iX5UgArMyiI",
        "timestamps": [
            90
        ],
        "3min_transcript": "- [Voiceover] So we have a circle here and they specified some points for us. This little orangeish, or, I guess, maroonish-red point right over here is the center of the circle, and then this blue point is a point that happens to sit on the circle. And so with that information, I want you to pause the video and see if you can figure out the equation for this circle. Alright, let's work through this together. So let's first think about the center of the circle. And the center of the circle is just going to be the coordinates of that point. So, the x-coordinate is negative one and then the y-coordinate is one. So center is negative one comma one. And now, let's think about what the radius of the circle is. Well, the radius is going to be the distance between the center and any point on the circle. So, for example, for example, this distance. The distance of that line. Let's see I can do it thicker. A thicker version of that. This line, right over there. 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"
    },
    {
        "Q": "Is obtuse larger than acute?\n\nI LIKE KHAN ACADAMEY!\nAT 5:04",
        "A": "Yes, an obtuse angle has one side of the triangle larger than 90*. An acute angle is an angle that is 89* or less. A right angle is exactly 90*.",
        "video_name": "ALhv3Rlydig",
        "timestamps": [
            304
        ],
        "3min_transcript": ""
    },
    {
        "Q": "did sal make a mistake by telling indefinite integrals at 0:31 or have I misunderstood ?\n\nAnd one more, how can you just get what dx is ? I mean , I know how to take d/dx but dx ? Please explain succinctly.",
        "A": "If either of the bounds of an integral are + or - infinity, then the integral is improper. The video show how to use limits to solve such improper integrals. dx represents an infinitesimal change in x. If you can measure the change in x, the change is not infinitesimal and the change is called \u00ce\u0094x. If you CANNOT measure the change in x, the change IS infinitesimal, and is called dx.",
        "video_name": "9JX2s90_RNQ",
        "timestamps": [
            31
        ],
        "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 5:03, Sal said \"6.5 divided by 450\". Did he make a mistake? Can someone explain this?",
        "A": "Yes he made a mistake, and I did not see a correction box, It appears that he should have said 675 divided by 450",
        "video_name": "EtefJ85R1OQ",
        "timestamps": [
            303
        ],
        "3min_transcript": "So when we figured out that the common ratio is 1.5, that tells us that our function is going to be of the form F of t is equal to a times-- instead of writing an r there, we now know that r is 1.5 to the t power. 1.5 to the t power. Write a formula for this function. Well we've almost done that, but we haven't figured out what a is. And to figure out what a is, we could just substitute-- we know what F of 1 is. When t is equal to 1, F is equal to 300. And so we should be able to use that information to solve for a. We could have used any of these data points to solve for a. So let's do that. F of 1 is equal to a times 1.5 to the first power, or a times 1.5. And that is going to be equal to-- they tell us that F of 1 is equal to 300. And so another way of writing this is we could say 1.5 times a is equal to 300. And we get a is equal to 200. And so our function, our formula for our function is-- let me write it in black so we can see it-- is going to be 200-- that's our a-- times 1.5 to the t power. Now another way-- well actually let's just think about the next question. What is the fine in euros for Sarah's speeding ticket if she pays it on time? So paying it on time, that implies that t is equal to 0. Or another way of thinking about it, we need to figure out her fine for t equals 0. So we need to figure out F of 0. So what's F of 0? It's 200 times 1.5 to the 0 power. 1.5 to the 0 power is 1, so that's just going to be equal to 200 euro. Now another way of thinking about it is, well look, let's look at the common ratio. To go from 6.5 to 450, you're essentially To go from 450 to 300, you're dividing by the common ratio. So then to go from t equals 1 to t equals 0, you would divide by the common ratio again. And you would get to 200. Or another way of thinking about it is to go to successive months every time we are multiplying by the common ratio. Every time we are multiplying by the common ratio."
    },
    {
        "Q": "0:44 Sal said exponential function f(t) = a . r ^t, is it always necessary, i mean can't exponential function f(t) = a - r^t or\nf(t) a + r^t ?\nWhat is exact definition of exponential function ?",
        "A": "The a term doesn t vary exponentially (or at all).",
        "video_name": "EtefJ85R1OQ",
        "timestamps": [
            44
        ],
        "3min_transcript": "Sarah Swift got a speeding ticket on her way home from work. If she pays her fine now, there will be no added penalty. If she delays her payment then a penalty will be assessed for the number of months t that she delays paying her fine. Her total fine F in euros is indicated in the table below. These numbers represent an exponential function. So they give us the number of months that the payment is delayed, and then the amount of fine. And this is essentially data points from an exponential function. And just to remind ourselves what an exponential function would look like, this tells us that are fine as our function of the months delayed is going to be equal to some number times some common ratio to the t power. This exponential function is essentially telling us that our function is going to have this form right over here. So let's see if we can answer their questions. So the first question is, what is the common ratio So the reason why r right over here is called the common ratio is it's the ratio that if you look at any two-- say if you were to increment t by 1, the ratio of that to F of t-- that ratio should be consistent for any t. So let me give you an example here. The ratio of F of 2 to F of 1 should be equal to the ratio of F of 3 to f of 2, which would be the same as the ratio of F of 4 to F of 3. Or in general terms, the ratio of F of t plus 1 to the ratio of F of t should be equal to all of these things. That would be the common ratio. So let's see what that is. If we just look at the form. If we just look at this right over here. So what's the ratio of F of 2 to F of 1? Well that's 1.5. 675 divided by 450? That's 1.5. 1012.50 divided by 675? That's 1.5. So the common ratio in all of these situations is 1.5. So the common ratio over here is 1.5. And another way-- and just to make it clear why this r right over here is called the common ratio-- is let's just do this general form. So f of t plus 1? Well that's just going to be a times r to the t plus 1 power. And F of t is a times r to the t power. So what is this going to be? This is going to be-- let's see-- this is going to be r to the t plus 1 minus t, which is just going to be equal to r to the first power, which is just equal to r."
    },
    {
        "Q": "At 1:39, why can't we use product rule or the quotient rule to find the derivative of (1/lnb)(lnx) or lnx/lnb?",
        "A": "The KEY thing to notice in this problem is that 1 / ln (b) is just a constant value. And so this problem is not much different than say 5 * ln x. You can pull the constant value out of the expression and just differentiate ln x just like Sal does. This will make the problem simpler.",
        "video_name": "ssz6TElXEOM",
        "timestamps": [
            99
        ],
        "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": "how does he evaluate sin(at) at infinity? at 4:51 he doesn't take that into account",
        "A": "sin(at) is a periodic function and it oscillates between -1 and 1. The same thing is for cos(at). So the expression in parentheses is always confined between some two numbers no matter how big t is. This expression is multiplied by e^(-st). When t goes to infinity, e^(-st) goes to zero. Zero times some finite number is still zero. Hope it helps!",
        "video_name": "-cApVwKR1Ps",
        "timestamps": [
            291
        ],
        "3min_transcript": "Because the t's are involved in evaluating the boundaries, since we're doing our definite integral or improper integral. So let's evaluate the boundaries now. And we could've kept them along with us the whole time, right? And just factored out this term right here. So let's evaluate this from 0 to infinity. And this should simplify things. So the right-hand side of this equation, when I evaluate it at infinity, what is e to the minus infinity? Well, that is 0. We've established that multiple times. And now it approaches 0 from the negative side, but it's still going to be 0, or it approaches 0. What's sine of infinity? Well, sine just keeps oscillating, between negative 1 and plus 1, and so does cosine. Right? So this is bounded. So this thing is going to overpower these. And if you're curious, you can graph it. This kind of forms an envelope around these oscillations. So the limit, as this approaches infinity, is going And that makes sense, right? These are bounded between 0 and negative 1. And this approaches 0 very quickly. So it's 0 times something bounded between 1 and negative 1. Another way to view it is the largest value this could equal is 1 times whatever coefficient's on it, and then this is going to 0. So it's like 0 times 1. Anyway, I don't want to focus too much on that. You can play around with that if you like. Minus this whole thing evaluated at 0. So what's e to the minus 0? Well, e to the minus 0 is 1. Right? That's e to the 0. We have a minus 1, so it becomes plus 1 times-- now, sine of 0 is 0. Minus 1 over s squared, cosine of 0. Let's see. 1 over s squared, times 1. So that is equal to minus 1 over s squared. And I think I made a mistake, because I shouldn't be having a negative number here. So let's backtrack. Maybe this isn't a negative number? Let's see, infinity, right? This whole thing is 0. When when you put 0 here, this becomes a minus 1. Yeah. So either this is a plus or this is a plus. Let's see where I made my mistake. e to the minus st-- oh, I see where my mistake is. Right up here. Where I factored out a minus e to the minus st, right?"
    },
    {
        "Q": "At 1:40 why does Sal factor out -e^(-st) and not (-e^(-st))/s? Isn't it possible to factor that out as well?",
        "A": "Yes, but it helps better with the evaluation later on since you d end up with something weird like 1/infinity/infinity or something like that.",
        "video_name": "-cApVwKR1Ps",
        "timestamps": [
            100
        ],
        "3min_transcript": "Welcome back. We were in the midst of figuring out the Laplace transform of sine of at when I was running out of time. This was the definition of the Laplace transform of sine of at. I said that also equals y. This is going to be useful for us, since we're going to be doing integration by parts twice. So I did integration by parts once, then I did integration by parts twice. I said, you know, don't worry about the boundaries of the integral right now. Let's just worry about the indefinite integral. And then after we solve for y-- let's just say y is the indefinite version of this-- then we can evaluate the boundaries. And we got to this point, and we made the realization, after doing two integration by parts and being very careful not to hopefully make any careless mistakes, we realized, wow, this is our original y. If I put the boundaries here, that's the same thing as the Laplace transform of sine of at, right? That's our original y. So now-- and I'll switch colors just avoid monotony-- this is equal to, actually, let me just-- this is y. So let's add a squared over sine squared y to both sides of this. So this is equal to y plus-- I'm just adding this whole term to both sides of this equation-- plus a squared over s squared y is equal to-- so this term is now gone, so it's equal to this stuff. And let's see if we can simplify this. So let's factor out an e to the minus st. Actually, let's factor out a negative e to the minus st. So it's minus e to the minus st, times sine of-- well, let me just write 1 over s, sine of at, minus 1 over s squared, cosine of at. And so this, we can add the coefficient. So we get 1 plus a squared, over s squared, times y. But that's the same thing as s squared over s squared, plus a squared over s squared. So it's s squared plus a squared, over s squared, y is equal to minus e to the minus st, times this whole thing, sine of at, minus 1 over s squared, cosine of at. And now, this right here, since we're doing everything with respect to dt, this is just a constant, right? So we can say a constant times the antiderivative is equal to this. This is as good a time as any to evaluate the boundaries. If this had a t here, I would have to somehow get them back"
    },
    {
        "Q": "At 3:59, shouldn't Sal get 62__ something?\n\nA confuzzled child.",
        "A": "No, he got the right answer but he made his 0 look like a 6. I also got confused but saw his mistake.",
        "video_name": "TvSKeTFsaj4",
        "timestamps": [
            239
        ],
        "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": "at 1:52 ,why did (x+5)^2 became (x- -5)^2 ? I'm doomedddd",
        "A": "Sal wrote it that way to get it into the form of the equation for a circle, which is (x - h)^2 + (y - k)^2 = r^2. In other words, the center of the circle is at (h,k), where h and k are the numbers being SUBTRACTED from x and y. Hope this saves you from being doomedddd!",
        "video_name": "thDrJvWNI8M",
        "timestamps": [
            112
        ],
        "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 around 3:36 Sal said that at 0 hours there were 550 pages left to read. Couldn't of Naoya also read in minutes? So there aren't really exactly 550 pages in the book right?",
        "A": "Merely changing to minutes would not change the y-intercept because the rate would be equivalent (55 pages/hour = 55/60 or 11/12 pages/minute). So after 240 minutes (4*60), Naoya still had 330 pages to read. so 240 * 11/12 = 220 pages. Think about what is happening, you are changing pages/hour by dividing by 60 then you are multiplying by 60 to convert to minutes, the two 60s cancel each other out so 55/60 * 4 * 60 is the same as 55 * 4.",
        "video_name": "W3flX500w5g",
        "timestamps": [
            216
        ],
        "3min_transcript": "would be 385. Now let's see whether this makes sense. So when our change in time, this triangle is just the Greek letter Delta, means \"change in.\" When our change in time is plus one, plus one hour, our change in pages left to read is going to be equal to negative 55 pages. And that makes sense, the pages left to read goes down every hour. We're measuring not how much he's read, we're measuring how much he has left to read. So that should go down by 55 pages every hour. Or if we were to go backwards through time, it should go up. So at two hours he should have 55 more pages to read. So what's 385 plus 55? We'll let's see, 385 plus 5 is 390, plus 50 is 440. So he'd have 440 pages, and all I did is I added 55. he would have 55 more pages than after reading for two hours. So 440 plus 55 is 495. And then before he started reading, or right when he started reading, he would have had to read even 55 more pages, 'cause after one hour, he would have read those 55 pages. So 495 plus 55 is going to be, let's see, it's gonna be, add 5, you get to 500 plus another 50 is 550 pages. So at time equals zero had had 550 pages to read. So that's how long the book is. But how long does it take Naoya to read the entire book? Well we could keep going. We could say, \"Okay, at the fifth hour, \"this thing's gonna go down by 55.\" So let's see, if this goes down by 50, if this goes down by 50, we're going to get to 280, but then you go down five more, it's gonna go to 275, and we could keep going on and on and on. \"He's got 330 pages left to read, and he's gonna,\" Let's see, let me write this, let me write the units down. \"Pages, and he's reading at a rate of 55 pages per hour, \"pages per hour, this is the same thing, \"this is going to be equal to 330 pages \"times one over 55 hours per page.\" I'm dividing by something, the same thing as multiplying by its reciprocal, so 55 pages per hours, if you divide by that, that's the same thing as multiplying by 1/55th of an hour per page, is one way to think about it. And so what do you get? The pages cancel out, pages divided by pages, and you have 330 divided by 55 hours. 330 divided by 55 hours. And what's that going to be? Let's see, 30 divided by 5 is 6, 300 divided by 50 is 6, so this is going to be equal to 6 hours."
    },
    {
        "Q": "At 2:23, Sal talks about using ratios of the sides to find the area of the equilateral triangle. Could you use the pythagorean theorem to find the height of the equilateral triangle and then calculate the area that way?",
        "A": "yes, you could do that.",
        "video_name": "QVxqgxVtKbs",
        "timestamps": [
            143
        ],
        "3min_transcript": "Let's say I have an equilateral triangle where the length of each side is 14. So this is an equilateral triangle. All of the sides have length 14. And inside that I have another equilateral triangle-- right over here-- where the length of each of the sides is 4. Now what I'm curious about, is the area of the region-- let me color this in a different color-- is the area of the region that I'm shading in right here. So it's the area inside the larger equilateral triangle, but outside of the smaller equilateral triangle. So let's think about how we would do this. And I encourage you to pause this and try this on your own. Well the shaded area is going to be equal to the large equilateral triangle's area minus the area of the small equilateral triangle. 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."
    },
    {
        "Q": "At 3:20, how is 5 tenths equal to 500 thousandths?",
        "A": "Because the zeros after the decimal point don t carry any information if no other number follows them. For instance, 5 is equal to 5.0 or 5.00 or 5.000 and so on. But if there is a number after the zeros, then they are significant and can t be removed. Like this: 5.00002 is NOT equal to 5.2. That s why, in your example, 0.5 = 0.500. Since no other number comes after the final zeros, they are not important and can as well be removed.",
        "video_name": "G7QiIkYfeME",
        "timestamps": [
            200
        ],
        "3min_transcript": "And then finally, we have 7/1000, that's the 1000th place. So we could write that, plus 7/1000. So if we would write down everything that I just spoke out loud, we would say that this is 20-- let me write that a little bit neater. This is 20,000. 20,000 and 5/10. And 5-- let me write out the word-- and 5/10 and 7/1000. Now, this isn't the only way to say this. Another way of thinking about it is to try to merge the 5/10 and the 7/1000 in terms of thousandths. So let's think about this. So we could write this as-- so once again, we would have our 20,000. write our 5/10 in terms of thousandths. And the easiest way to do it is to multiply the numerator and denominator, both here, by 100. So then we will have-- so this 5/10 is the same thing as 500 over 1,000. And the 7/1000 is still 7/1000. And these two combined are 507/1000. So we could just call this 20,000 and 507/1000. so let's write that down. So we could just say this is 20,000 and 507/1000. actually represents 1,000. So we got 20 thousands, that's that right over there, and 507/1000."
    },
    {
        "Q": "at 2:20 Sal says that a 1cm x 1cm x 1cm cube is 1 milliliter...why wouldn't this be 1 centimeter?",
        "A": "The cube is equal to 1 cubic centimetre. A cubic centimetre is equal to 1 milliliter .",
        "video_name": "LhMEqsL_M5o",
        "timestamps": [
            140
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 1:10, where did you get -66?",
        "A": "He s factoring by grouping . He starts by taking a (the coefficient of f^2) and multiplies it by b (the coefficient of f). Hope this helps :)",
        "video_name": "d-2Lcp0QKfI",
        "timestamps": [
            70
        ],
        "3min_transcript": "We need to factor negative 12f squared minus 38f, plus 22. So a good place to start is just to see if, is there any common factor for all three of these terms? When we look at them, they're all even. And we don't like a negative number out here. So let's divide everything, or let's factor out a negative 2. So this expression right here is the same thing as negative 2 times-- what's negative 12f squared divided by negative 2? It's positive 6f squared. Negative 38 divided by negative 2 is positive 19, so it'll be positive 19f. And then 22 divided by negative 22-- oh, sorry, 22 divided by negative 2 is negative 11. So we've simplified it a bit. We have the 6f squared plus 19f, minus 11. We'll just focus on that part right now. And the best way to factor this thing, since we don't have a 1 here as the coefficient on the f squared, is to factor it by grouping. 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."
    },
    {
        "Q": "At 1:27, How did he get 9?",
        "A": "Ah, 45 - 36 = 9",
        "video_name": "EFVrAk61xjE",
        "timestamps": [
            87
        ],
        "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 1:24 sal writes the fraction 9/13. Why did he put a nine instead of four? Can someone please help me??",
        "A": "because it is 9 away from the first root over the total amount of numbers in between",
        "video_name": "EFVrAk61xjE",
        "timestamps": [
            84
        ],
        "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": "How do you know which numbers are a and b in the equation? At 13:21 he says that a=3 and b=4 (because he took the square roots) but how do you know that a isn't 4 and b isn't 3? For ellipses you know that a is the bigger number, but what do you know for hyperbolas? Thanks!",
        "A": "a is 3 so 3 squared gives u nine and b is 4 so 4 squared gives u 16.",
        "video_name": "S0Fd2Tg2v7M",
        "timestamps": [
            801
        ],
        "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": "at 2:43 sal said standard but wrote slandard why",
        "A": "He probably made a mistake...you should report it is the tag named report a mistake ....",
        "video_name": "ZgFXL6SEUiI",
        "timestamps": [
            163
        ],
        "3min_transcript": "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. 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"
    },
    {
        "Q": "At 2:50 why did Sal call the binomial a second degree polynomial? Didn't he say that it was not a polynomial",
        "A": "We reserve the word polynomial for expressions in which all the terms have positive, non-fractional exponents. When Sal earlier said some expressions did not qualify as polynomials, he was excluding one that included a square root (which is a fractional exponent of 1/2) and also excluding one that had a negative exponent. He wasn t excluding ordinary binomials, which are included in the definition of polynomial.",
        "video_name": "ZgFXL6SEUiI",
        "timestamps": [
            170
        ],
        "3min_transcript": "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. 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"
    },
    {
        "Q": "at around 14:40, wouldn't the width of the green square be 1/2b ?",
        "A": "You can t assume that it is half of b from the picture. Never trust the picture. This is why Sal denoted it as c.",
        "video_name": "ZgFXL6SEUiI",
        "timestamps": [
            880
        ],
        "3min_transcript": "Well, this blue part right here, the area there is x times y. And then what's the area here? It's going to be x times z. So plus x times z. We have one x times z, and then we have another x times z. So I could just add an x times z here. Or I could just write, say, plus 2 times x times z. And here we have a polynomial that represents the area of this figure right there. Now let's do this next one. What's the area here? Well I have an a times a b. ab. This looks like an a times a b again, plus ab. That looks like an ab again, plus ab. I think they've drawn it actually, Well, I'm going to ignore this c right there. Maybe they're telling us that this right here is c. Because that's the information we would need. Maybe they're telling us that this base right there, that this right here, is c. Because that would help us. But if we assume that this is another ab here, which I'll assume for this purpose of this video. And then we have that last ab. And then we have this one a times c. This is the area of this figure. And obviously we can add these four terms. This is 4ab and then we have plus ac. And I made the assumption that this was a bit of a typo, that that c where they were actually telling us the width of this little square over here. We don't know if it's a square, that's only if a and c are the same. Now let's do this one. So how do we figure out the area of the pink area? would be 2xy, and then we could subtract out the area of these squares. So each square has an area of x times x, or x squared. And we have two of these squares, so it's minus 2x squared. And then finally let's do this one over here. So that looks like a dividing line right there. So the area of this point, of this area right there, is a times b, so it's ab. And then the area over here looks like it will also be ab. So plus ab. And the area over here is also ab. So the area here is 3ab. Anyway, hopefully that gets us pretty warmed up with polynomials."
    },
    {
        "Q": "At about 6:08 you talked about descending order, what would the degree of a term without an exponent be?",
        "A": "If the term is something like 2x, then there is an exponent on that variable. If one is not written, it s implied that it is to the 1st power. If there is no variable at all, like in 4, you would say that it is degree 0. The only other weird case is that a term of 0 is said to have no degree.",
        "video_name": "ZgFXL6SEUiI",
        "timestamps": [
            368
        ],
        "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 about 3:40, Sal says that 5 is the highest exponent on a variable. Does this mean that a degree can only come from the highest exponent on a variable, or can it be on a normal number as well?",
        "A": "The degree of a polynomial is by definition the largest exponent of the variable. So, yes we only consider the exponent of the the variable. So, x\u00c2\u00b2 + 50\u00c2\u00b3\u00c2\u00b9\u00e2\u0081\u00b7 would still be a second degree polynomial.",
        "video_name": "ZgFXL6SEUiI",
        "timestamps": [
            220
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "At around 2:07 Sal said that the degree of the polynomial is the value of the largest exponent, whereas my algebra teacher says it is the sum of the exponent values. Who is wrong? Thanks.",
        "A": "The both are! When working with polynomials of 1 variable, what Sal said is correct, assuming the polynomial is in canonical form, that is, you don t have terms like (x^3)(x^4), which in canonical form would be written x^7. When you have polynomials of more than one variable, you need to sum the exponents of each term, for example, (x^4)(y) has degree 5 and (x^2)(y^2)(z^2) has degree 6.",
        "video_name": "ZgFXL6SEUiI",
        "timestamps": [
            127
        ],
        "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": "9:43 Why is is a^2(x^2-2xf+f^2+y^2)? Why is the \"a\" outside of the parentheses? In other words, why isn't it just a^2x^2-2xf+f^2+y^2? Why are there parentheses?",
        "A": "Because without the parentheses, the a^2 will not be distributed to the equation that was inside the square root. Without it you just get a^2 times x^2. With them you have to multiply each factor by a^2.",
        "video_name": "HPRFmu7JsKU",
        "timestamps": [
            583
        ],
        "3min_transcript": "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 square well that's just a squared. So a squared is equal to x squared minus 2xf plus f squared plus y squared. Well good. There's something to cancel out. There's a mine 2xf on both sides of this equation so let's cancel that out. Simplify our situation a little bit. And let's see, we have. so we could do is subtract this x squared from this. So you get-- let me rewrite it --so you get x squared f squared over a squared minus x squared. And let's bring this y to this side of the equation too. So minus y squared. That's all I did, I just brought that to that side. And then let's bring-- and I'm kind of skipping a couple of steps, but I don't want to take too long --let's take this a"
    },
    {
        "Q": "At 2:00 how did he get 36x?",
        "A": "When Sal was doing the distributive property 9x(x+4) 9x times x is 9x squared and 9x times 4 is 36x.",
        "video_name": "vl9o9XEfXtw",
        "timestamps": [
            120
        ],
        "3min_transcript": "The volume of a box is 405 cube units, or I guess cubic units. So they just want to keep it general. It could've been in cubic feet, or cubic meters, or cubic centimeters, or cubic miles. They just want to keep it as units, keep it as general as possible. The length is x units, the width is x plus 4 units, and the height is 9 units. So let me draw this box here. Let me draw a little box here, so we have a nice little visualization. So they tell us, that the length is x. Maybe we could call this the length right there. They say the width is x plus 4, and the height is 9 of this box. In units, what are the dimensions of the box? Well, they also tell us that the volume is 405. So the volume, 405-- let me do it this way. So if we wanted to calculate the volume, what would it be? Well it would be the width-- it would be x plus 4 times the 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"
    },
    {
        "Q": "at 4:05 how did he get -16?",
        "A": "exponents with negitive bases",
        "video_name": "vEZea0EThus",
        "timestamps": [
            245
        ],
        "3min_transcript": "times a negative, you're going to get a positive. And so when you do it an even number of times, doing it a multiple-of-two number of times. So the negatives and the negatives all cancel out, I guess you could say. Or when you take the product of the two negatives, you keep getting positives. So this right over here is going to give you a positive value. So there's really nothing new about taking powers of negative numbers. It's really the same idea. And you just really have to remember that a negative times a negative is a positive. And a negative times a positive is a negative, which we already learned from multiplying negative numbers. Now there's one other thing that I want to clarify \u2013 because sometimes there might be ambiguity if someone writes this. Let's say someone writes that. And I encourage you to actually pause the video and think about with this right over here And, if you given a go at that, think about whether this should mean something different then that. Well this one can be a little bit and big ambiguous and if people are strict about order of operations, you should really be thinking about the exponent before you multiply by this -1. You could this is implicitly saying -1 \u00d7 2^3. So many times, this will usually be interpreted as negative 2 to the third power, which is equal to -8, while this is going to be interpreted as -2 to the third power. Now that also is equal to -8. You might say well what's what's the big deal here? Well what if this was what if these were even exponents. So what if someone had give myself some more space here. What if someone had these to express its -4 or a -4 squared or -4 squared. This one clearly evaluates to 16 \u2013 positive 16. This one could be interpreted as is. Especially if you look at order of operations, and you do your exponent first, this would be interpreted as -4 times 4, which would be -16. So it's really important to think about this properly. And if you want to write the number negative if you want the base to be negative 4, put parentheses around it and then write the exponent."
    },
    {
        "Q": "Around 1:00, Sal mentioned \"functions that are not equations.\" Could anyone help clarify how there are some functions that are not equations?",
        "A": "Functions aren t equations. They re mappings between sets. Here s a function: f: \u00e2\u0084\u009d \u00e2\u0086\u0092 \u00e2\u0084\u009d, x \u00e2\u0086\u00a6 2x + 1 It s a function from the set of real numbers to the set of real numbers, and a real number x is sent to the real number 2x + 1. A function may be represented with an equation. For example: f(x) = 2x + 1 defines (mostly) the same function as above. (It doesn t explicitly define the domain and range of the function.)",
        "video_name": "l3iXON1xEC4",
        "timestamps": [
            60
        ],
        "3min_transcript": "SALMAN KHAN: I'm here with Jesse Roe of Summit Prep. What classes do you teach? JESSE ROE: I teach algebra, geometry, and algebra II. SALMAN KHAN: And now you're with us, luckily, for the summer, doing a whole bunch of stuff as a teaching fellow. JESSE ROE: Yeah, as a teaching fellow I've been helping with organizing and developing new content, mostly on the exercise side of the site. SALMAN KHAN: And the reason why we're doing this right now is you had some very interesting ideas or questions. JESSE ROE: Yeah, so as an algebra teacher, when I introduce that concept of algebra to students, I get a lot of questions. One of those questions is, what's the difference between an equation and a function? SALMAN KHAN: The difference between an equation verses a function, that's an interesting question. Let's pause it and let the viewers try to think about it a little bit. And then maybe we'll give a stab at it. JESSE ROE: Sounds great. So Sal, how would you answer this question? What's the difference between an equation and a function? SALMAN KHAN: Let me think about it a little bit. So let me think. I think there's probably equations that are not functions and functions that are not equations. 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,"
    },
    {
        "Q": "hey at 1:40 doesn't 3 ^1 equal zero?",
        "A": "Anything to the first power is equal to that anything. Take 3^2, that s 3 * 3, right? So 3^1 = 3. And note that 3^0 = 1, not 0.",
        "video_name": "6WMZ7J0wwMI",
        "timestamps": [
            100
        ],
        "3min_transcript": "In this video, I want to introduce you to the idea of an exponential function and really just show you how fast these things can grow. So let's just write an example exponential function here. So let's say we have y is equal to 3 to the x power. Notice, this isn't x to the third power, this is 3 to the x power. Our independent variable x is the actual exponent. So let's make a table here to see how quickly this thing grows, and maybe we'll graph it as well. So let's take some x values here. Let's start with x is equal to negative 4. Then we'll go to negative 3, negative 2, 0, 1, 2, 3, and 4. And let's figure out what our y-values are going to be for each of these x-values. Now, here, y is going to be 3 to the negative 4 power, which is equal to 1 over 3 to the fourth power. So this is equal to 1/81. When x is equal to negative 3, y is 3. We'll do this in a different color. This color is hard to read. y is 3 to the negative 3 power. Well, that's 1 over 3 to the third power, which is equal to 1/27. So we're going from a super-small number to a less super-small number. And then 3 to the negative 2 power is going to be 1/9, right? 1 over 3 squared, and then we have 3 to the 0 power, which is just equal to 1. So we're getting a little bit larger, a little bit larger, but you'll see that we are about to explode. Now, we have 3 to the first power. That's equal to 3. So we have 3 to the second power, right? y is equal to 3 to the second power. That's 9. 3 to the third power, 27. 3 to the fourth power, 81. If we were to put the fifth power, 243. 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."
    },
    {
        "Q": "At 6:32, Sal says function is undefined at x2. He meant is non-differentiable, because it is defined right?",
        "A": "I definitely think that Sal is trying to say where the DERIVATIVE is undefined. Because when the function is undefined at a point, we will not have a critical point because this point does not exist: ie, it is not defined",
        "video_name": "lDY9JcFaRd4",
        "timestamps": [
            392
        ],
        "3min_transcript": "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. if we look at the slope right over here, it looks like f prime of x sub 3 is equal to 0. So based on our definition of critical point, x sub 3 would also be a critical point. But it does not appear to be a minimum or a maximum point. So a minimum or maximum point that's not an endpoint, it's definitely going to be a critical point. But being a critical point by itself does not mean you're at a minimum or maximum point. So just to be clear that all of these points were at a minimum or maximum point. This were at a critical point, all of these are critical points. But this is not a minimum or maximum point. In the next video, we'll start to think about how you can differentiate, or how you can tell, whether you have a minimum or maximum at a critical point."
    },
    {
        "Q": "7:12 We can create tangent lines at a point that crosses the function?",
        "A": "Yes, a line can be tangent at one point on a curve but then cross it later when the curve takes a U-turn later on down the road.",
        "video_name": "lDY9JcFaRd4",
        "timestamps": [
            432
        ],
        "3min_transcript": "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. if we look at the slope right over here, it looks like f prime of x sub 3 is equal to 0. So based on our definition of critical point, x sub 3 would also be a critical point. But it does not appear to be a minimum or a maximum point. So a minimum or maximum point that's not an endpoint, it's definitely going to be a critical point. But being a critical point by itself does not mean you're at a minimum or maximum point. So just to be clear that all of these points were at a minimum or maximum point. This were at a critical point, all of these are critical points. But this is not a minimum or maximum point. In the next video, we'll start to think about how you can differentiate, or how you can tell, whether you have a minimum or maximum at a critical point."
    },
    {
        "Q": "At 4:50, why don't the endpoints count as maxima/minima?",
        "A": "They can be maxima/minima on a closed interval (but not always), but in the video, we are not interested in them because we only want to show point (non-endpoint) that is min/max will have f (a)= 0 or undefined. Endpoints are max/min but they don t necessarily have f (a)=0 or undefined. Hope that helps.",
        "video_name": "lDY9JcFaRd4",
        "timestamps": [
            290
        ],
        "3min_transcript": "I'm not being very rigorous. But you can see it just by looking at it. So that's fair enough. We've identified all of the maxima and minima, often called the extrema, for this function. Now how can we identify those, if we knew something about the derivative of the function? Well, let's look at the derivative at each of these points. So at this first point, right over here, if I were to try to visualize the tangent line-- let me do that in a better color than brown. If I were to try to visualize the tangent line, it would look something like that. So the slope here is 0. So we would say that f prime of x0 is equal to 0. The slope of the tangent line at this point is 0. What about over here? Well, once again, the tangent line would look something like that. So once again, we would say f prime at x1 is equal to 0. What about over here? 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"
    },
    {
        "Q": "This is a bit confusing. At 2:24 he said that we don't have more than one local minimum but at the far right of the graph of the function shouldn't the part of the graph that fall under the x-axis count as a local minima, or am I just not paying attention",
        "A": "A local minimum has higher points at both sides. This point only has higher one one of the sides so it is not a minimum.",
        "video_name": "lDY9JcFaRd4",
        "timestamps": [
            144
        ],
        "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": "I don't understand how the three becomes negative. I thought it was just a subtraction sign? 2:11",
        "A": "it s because it s part of the number. If it has a negative sign in front of it it s because the number is negative ex. 4 - 3 = 1 the 3 is a negative.",
        "video_name": "xMsG9hvqzbY",
        "timestamps": [
            131
        ],
        "3min_transcript": "Like the last video, I want to start with two warm up problems. And then we'll do an actual word. And you're going to see these are going to be a little bit more involved than the equations in the last video. But we're still going to be doing the exact same operations, or what we could consider legitimate operations, to get our answer. So here we have 3 times x minus 1 is equal to 2 times x plus 3. So let's see what we can do here. The first thing I like to do-- and there's no definite right way to do it-- there's several ways you could do these problems-- but I like to distribute out the 3 and the 2. So 3 times x minus 1, that's the same thing as 3x minus 3-- I just distributed the 3-- is equal to-- distribute out the 2. 2 times x, plus 2 times 3, which is 6. Now what I like to do is get all of my constant terms on the same side of the equation and all my variable terms on the same side of the equation. So let's see if we can get rid of this 2x term on So let's subtract 2x. I'm going to do a slightly different notation this time because you might see it done this way. Or you might find it easier to visualize it this way. It doesn't matter. It's the same thing we did in the last video. But I want to subtract 2x from this side of the equation. But if I subtract 2x from the side of the equation, I also have to subtract 2x from that side of the equation. So then when we subtract 2x from both sides of the equation, what do we get? Here we get 3x minus 2x. That's just 1x, or x minus 3. 2x minus 2x is no x's, or 0. Then you just have the 6. So we get x minus 3 is equal to 6. That was by getting rid of the 2x from the right hand side; subtracting it from both sides of this equation. Now we have this negative 3 on the left hand side. I just want an x there. So to get rid of that we can add 3 to both sides of this equation. the equation 3 is equal to 3. 3 is, obviously, equal to 3. Negative 2x is, obviously, equal to negative 2x. You could do it either way. But if you add 3 to both sides of this equation the left hand side of the equation becomes just an x, because these two guys cancel out. x equal, and 6 plus 3 is 9. And we are done. And we can even check our answer. 3 times 9 minus 1 is what? This is 3 times 8. This is 24. So that's what the left hand side equals. What does the right hand side equal? That is 2 times 9 plus 3. That's the 2 times 12, which is also equals 24. So it all works out. x is equals to 9. Next problem. z over 16 is equal to 2 times 3z plus 1, all of that over 9."
    },
    {
        "Q": "In the second problem at 3:37, why does he write 6z + 2, instead of 6z + 1, when it's written 1 in the equation?",
        "A": "Well, that s BECAUSE IT s 2(3z+1)=2(3z)+2(1) IT s SO SIMPLE!",
        "video_name": "xMsG9hvqzbY",
        "timestamps": [
            217
        ],
        "3min_transcript": "the equation 3 is equal to 3. 3 is, obviously, equal to 3. Negative 2x is, obviously, equal to negative 2x. You could do it either way. But if you add 3 to both sides of this equation the left hand side of the equation becomes just an x, because these two guys cancel out. x equal, and 6 plus 3 is 9. And we are done. And we can even check our answer. 3 times 9 minus 1 is what? This is 3 times 8. This is 24. So that's what the left hand side equals. What does the right hand side equal? That is 2 times 9 plus 3. That's the 2 times 12, which is also equals 24. So it all works out. x is equals to 9. Next problem. z over 16 is equal to 2 times 3z plus 1, all of that over 9. Let's multiply both sides of this equation by 9. So if you multiply both sides of this equation by 9, what do we get? We get 9 over 16z is equal to-- this 9 and that 9 will cancel out-- 2 times 3z plus 1. Now let's distribute this 2. So we get 9 over 16z is equal to 2 times 3z is 6z plus 2. Now let's get all of z's on the same side of the equation. So let's subtract 6z from both sides of the equation. So let's subtract minus 6z there, and that, of course, equals minus 6z there. And what do we get? On the left hand side, we get 9/16 minus 6. 6 is equal to what over 16? 60 plus 36. It's equal to 96/16. So it's 9 minus 96 over 16z. This is just 6-- I just rewrote minus 6 here-- is equal to-- these two cancel out. That's why I subtracted 6z in the first place. So it's going to equal 2. So what does this equal right over here? Let me do it in orange. 9 minus 96. The difference between 9 and 96 is going to be 87. And, of course, we're subtracting the larger from the smaller. So it's going to be negative 87 over 16z is equal to 2. And we're almost there. We just have to multiply both sides of this equation by the inverse of this coefficient. So multiply both sides of this equation by negative 16/87."
    },
    {
        "Q": "At 5:45 shouldn't \"a sub n\" equal a*n!",
        "A": "Yes, but a equals 1. a sub n is not a. I hope this clarifies the video!",
        "video_name": "dIGLhLMsy2U",
        "timestamps": [
            345
        ],
        "3min_transcript": "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. So this sequence, which is not a geometric sequence, we can still define it explicitly. We could say that its set or it's the sequence a sub n from n equals 1 to infinity with a sub n being equal to, let's see the fourth one is essentially 4 factorial times a. Well, actually, if we look at this particular, these particular numbers our a is 1. So this is actually, let me write this, this is 1, this is 2 times 1, this is 3 times 2 times 1, this is 4 times 3 times 2 times 1. And so a sub n is just equal to n factorial. This right over here, which is not a geometric sequence, describes exactly this sequence right over here. Just to get some practice with-- Here we've defined it explicitly, but we can also We could also say-- do it in white-- we could also say that a sub n takes us from n equals 1 to infinity, with a sub 1, or maybe at a sub 1 is equal to 1. That's our first term. And then each successive term is going to be equal to the previous term times n. So the second term is equal to the previous term times 2. The nth term is going to be the previous turn times n So this is another valid way of defining it."
    },
    {
        "Q": "There may be a mistake at 5:47. You say it is equal to n!, when it is equal to n!a.",
        "A": "No mistake. Here a = 1. And just like how we don t write 1x\u00c2\u00b2 + 1x - 2, but rather x\u00c2\u00b2 + x - 2 we don t write n! as 1n! Does that make sense? (If a was not equal to 1, then we would need to include it)",
        "video_name": "dIGLhLMsy2U",
        "timestamps": [
            347
        ],
        "3min_transcript": "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. So this sequence, which is not a geometric sequence, we can still define it explicitly. We could say that its set or it's the sequence a sub n from n equals 1 to infinity with a sub n being equal to, let's see the fourth one is essentially 4 factorial times a. Well, actually, if we look at this particular, these particular numbers our a is 1. So this is actually, let me write this, this is 1, this is 2 times 1, this is 3 times 2 times 1, this is 4 times 3 times 2 times 1. And so a sub n is just equal to n factorial. This right over here, which is not a geometric sequence, describes exactly this sequence right over here. Just to get some practice with-- Here we've defined it explicitly, but we can also We could also say-- do it in white-- we could also say that a sub n takes us from n equals 1 to infinity, with a sub 1, or maybe at a sub 1 is equal to 1. That's our first term. And then each successive term is going to be equal to the previous term times n. So the second term is equal to the previous term times 2. The nth term is going to be the previous turn times n So this is another valid way of defining it."
    },
    {
        "Q": "On problem 51. or (2:55)\n\nCan you prove to me why all the four right triangles are congruent? I know they are... but why? D:",
        "A": "The shape in the middle is a square, so all four sides are equal. When it is placed inside another square, the triangles that it creates (4), are all congruent, because the four side lengths of a square are equivalent. If the shape in the middle were a rectangle that is not a square, there would be two congruent triangles, and another two congruent triangles, which are different from the first two. So, because they are both squares, the triangles are congruent.",
        "video_name": "6EY0E3z-hsU",
        "timestamps": [
            175
        ],
        "3min_transcript": "Fair enough. Now we can also say that the area of this larger square, and it's a bit of an optical illusion, it looks like it's tilted to the left because of the way it's drawn. But anyway, that the area of this larger square is also the area of these four triangles plus the area of this smaller square. So this, the area of the larger square, which we figured out just by taking one side of it and squaring it, that should be equal to the area of the four smaller triangles. So there's four of them. And what's the area of each of them. Let's see, let's just pick this one. 1/2 base times height. So it's 1/2 times a times b. So 1/2 ab is one of these and I multiply by 4 to get all four of these triangles. And then we want to add the area of this inside square. And that's just going to be c squared. Let's see if we can simplify this. So you get a squared plus 2ab plus b squared is equal to 4 times 1/2 is 2ab plus c squared. Well, we could subtract 2ab from both sides of this equation. The top and the bottom of this equation the way I've written it. But if we do that, subtract 2ab from there, subtract 2ab from there, and you're left with a squared plus b squared is equal to c squared, which is the Pythagorean theorem. And we've proved it. So let's see which of their choices matches what we did. OK, which statement would not be used in the proof of the Pythagorean theorem. The area of a triangle equals 1/2 ab. We used that. The four right triangles are congruent. The area of the inner square is equal to half of the area of the larger square. We didn't use that. I think this is the one that would not be used in the proof. Choice D, the area of the larger square is equal to the sum of the squares of the smaller square and the four congruent triangles. No, that that was the crux of the proof. So we definitely used that. So C is our answer. That's the statement that would not be used in the proof. I'm learning to copy and paste ahead of time. So I don't waste your time. All right, a right triangle's hypotenuse has length 5. If one leg has length 2, what is the length of the other leg? Pythagorean theorem, x squared plus 2 squared is equal to 5 squared, because 5 is the hypotenuse. x squared plus 4 is equal to 25."
    },
    {
        "Q": "At 1:16, how is a a^2 + b^2 equal to a^2 + 2ab +b^2?",
        "A": "You misunderstood. It says that (a+b)^2=a^2+2ab+b^2. => (a+b)^2=(a+b)(a+b) Distribute => a*a+a*b+b*a+b*b => a^2+ab+ab+b^2 => a^2+2ab+b^2.",
        "video_name": "6EY0E3z-hsU",
        "timestamps": [
            76
        ],
        "3min_transcript": "We're on problem 51. And they say, a diagram from a proof of the Pythagorean theorem is pictured below. And they they say, which statement would not be used in the proof of the Pythagorean theorem? So since they have drawn this diagram out, I think we might as well just kind of do the proof and then we can look at their choices and see which ones kind of match up to what we did. Hopefully, they do it the same way. And this is a pretty neat proof of the Pythagorean theorem. I don't think I've done it yet. So I might as well do it now. Well, let's figure out what the area of this large square is right. Well there's two ways to think about it you could just say, OK, this is a square. That's a, that's b. Well this is going to be b as well. This is going to be a as well. So the area of the square is going to be the length of one of its sides squared. So we could say the whole square's area is a plus b squared. Fair enough. Now we can also say that the area of this larger square, and it's a bit of an optical illusion, it looks like it's tilted to the left because of the way it's drawn. But anyway, that the area of this larger square is also the area of these four triangles plus the area of this smaller square. So this, the area of the larger square, which we figured out just by taking one side of it and squaring it, that should be equal to the area of the four smaller triangles. So there's four of them. And what's the area of each of them. Let's see, let's just pick this one. 1/2 base times height. So it's 1/2 times a times b. So 1/2 ab is one of these and I multiply by 4 to get all four of these triangles. And then we want to add the area of this inside square. And that's just going to be c squared. Let's see if we can simplify this. So you get a squared plus 2ab plus b squared is equal to 4 times 1/2 is 2ab plus c squared. Well, we could subtract 2ab from both sides of this equation. The top and the bottom of this equation the way I've written it. But if we do that, subtract 2ab from there, subtract 2ab from there, and you're left with a squared plus b squared is equal to c squared, which is the Pythagorean theorem. And we've proved it. So let's see which of their choices matches what we did. OK, which statement would not be used in the proof of the Pythagorean theorem. The area of a triangle equals 1/2 ab. We used that. The four right triangles are congruent."
    },
    {
        "Q": "What does he mean at 5:39. \"But if I just cancelled these two things out, the new function would be defined when x = -8\". Why -8?",
        "A": "Because when x= -8, the denominator is 0, so the function is undefined. But if we cancel the factors (x+8), then we could plug in x= -8 without dividing by 0, so the function would defined. So since one is defined at x= -8, and the other is not, cancelling the (x+8) changes the function, which we don t want.",
        "video_name": "u9v_bakOIcU",
        "timestamps": [
            339
        ],
        "3min_transcript": "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. We can divide the numerator and denominator by x plus 8, assuming that x does not equal negative 8. Because this function right over here that's defined by f divided by g, it is not defined when g of x is equal to 0, because then you have something divided by 0. And the only times that g of x is equal to 0 is when x is equal to negative 2 or x is equal to negative 8. So if we divide the numerator and the denominator by x plus 8 to simplify it, in order to not change the function definition, we have to still put the constraint that x cannot be equal to negative 8. That the original function, in order to not change it-- because if I just cancelled these two things out, the new function with these canceled would be defined when x is equal to negative 8. But we want this simplified thing to be the same exact function. And this exact function is not defined when x is equal to negative 8. So now we can write f/g of x, which is really just f of x divided by g of x, is equal to 2x minus 1 You have to put the condition there that x cannot be equal to negative 8. If you lost this condition, then it won't be the exact same function as this, because this is not defined when x is equal to negative 8."
    },
    {
        "Q": "@ 2:16 where does -16 come from?",
        "A": "Ok, so he s trying to factor the trinomial (ax^2+ bx + c). This method is what I learned as the long method. He is just finding the LCM of a and c so he can split bx so it is easier to factor it out. It is just finding or converting numbers into more factor able numbers and group them. This is what Sal calls factoring by grouping. I hope this helps.",
        "video_name": "u9v_bakOIcU",
        "timestamps": [
            136
        ],
        "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": "At 6:06 Khan is explaining why x can't be -8, but he didn't say that it can't be -2. So, does this mean that x can be -2? This will also give you a 0 in the denominator.",
        "A": "No, x cannot be -2 either, but it is not necessary to specify it because x - 2 is still in the simplified expression. You have to put the x does not equal -8 because the term that would have made that clear has vanished.",
        "video_name": "u9v_bakOIcU",
        "timestamps": [
            366
        ],
        "3min_transcript": "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. We can divide the numerator and denominator by x plus 8, assuming that x does not equal negative 8. Because this function right over here that's defined by f divided by g, it is not defined when g of x is equal to 0, because then you have something divided by 0. And the only times that g of x is equal to 0 is when x is equal to negative 2 or x is equal to negative 8. So if we divide the numerator and the denominator by x plus 8 to simplify it, in order to not change the function definition, we have to still put the constraint that x cannot be equal to negative 8. That the original function, in order to not change it-- because if I just cancelled these two things out, the new function with these canceled would be defined when x is equal to negative 8. But we want this simplified thing to be the same exact function. And this exact function is not defined when x is equal to negative 8. So now we can write f/g of x, which is really just f of x divided by g of x, is equal to 2x minus 1 You have to put the condition there that x cannot be equal to negative 8. If you lost this condition, then it won't be the exact same function as this, because this is not defined when x is equal to negative 8."
    },
    {
        "Q": "at 2:19 why has he divided 1* 10^4 upon 7*10^5 when it is written in the question 7*10^5 than 1*10^4?",
        "A": "To find the quotient of exponents you subtract the exponents from the other. If you need more help, you can look at some other of Sal s Exponents videos.",
        "video_name": "DaoJmvqU3FI",
        "timestamps": [
            139
        ],
        "3min_transcript": "Let's do a few more examples from the orders of magnitude exercise. Earth is approximately 1 times 10 to the seventh meters in diameter. Which of the following could be Earth's diameter? So this is just an approximation. It's an estimate. And they're saying, which of these, if I wanted to estimate it, would be close or would be 1 times 10 to the seventh? And the key here is to realize that 1 times 10 to the seventh is the same thing as one followed by seven zeroes. One, two, three, four, five, six, seven. Let me put some commas here so we make it a little bit more Or another way of talking about it is that it is, 1 times 10 to the seventh, is the same thing as 10 million. So which of these, if we were to really roughly estimate, we would go to 10 million. Well, this right over here is 1.271 million, or 1,271,543. If I were to really roughly estimate it, I might go to one million, but I'm not going to go to 10 million. This is 12,715,430. If I were to roughly estimate this, well, yeah. I would go to 10 million. 10 million is if I wanted really just one digit to represent it, if I were write this in scientific notation. This right over here is 1.271543 times 10 to the seventh. Let me write that down. 12,715,430. If I were to write this in scientific notation as 1.271543 times 10 to the seventh. And when you write it this way, you say, hey, well, yeah, if I was to really estimate this and get pretty rough with it, and I just rounded this down, I would make this 1 times 10 to the seventh. So this really does look like our best choice. Now let me just verify. Well, this right over here, if I were to write it, I would go to 100 million, or 1 times 10 to the eighth. That's way too big. And this, if I were to write it, I would go to a billion, or 1 times 10 to the ninth. So that's also too big. So once again, this feels like the best answer. So here we're asked, how many times larger is 7 times 10 to the fifth than 1 times 10 to the fourth? Well, we could just divide to think about that. So 7 times 10 to the fifth divided by 1 times 10 to the fourth. Well, this is the same thing as 7 over 1 times 10 to the fifth over 10 to the fourth, which is just going to be equal to-- well, 7 divided by 1 is 7. And 10 to the fifth, that's multiplying five 10's. And then you're dividing by four 10's. You're going to have one 10 left over. Or, if you remember your exponent properties, this would be the same thing as 10 to the 5 minus 4 power, or 10 to the first power. So this right over here, all of this business, is going to simplify to 10 to the first, or I could actually write it this way. This would be the same thing as 10 to the 5 minus 4,"
    },
    {
        "Q": "Quick Question at 5:55 he writes cos^2 and theta and sin^2 theta now based on how he wrote it I am wondering if it is cos^(2 theta) or cos^(2) Theta?",
        "A": "cos^2(\u00ce\u00b8) is the trig representation for (cos(\u00ce\u00b8))^2.",
        "video_name": "n0DLSIOYBsQ",
        "timestamps": [
            355
        ],
        "3min_transcript": "Say you know cosine theta then you use this to figure out sine of theta, then you can figure out tangent of theta because tangent of theta is just sine over cosine. If you're a little bit confused as to why this is called the Pythagorean identity, well it really just falls out of where the equation of a circle even came from. If we look at this point right over here, we look at this point right over here, which we're saying is the x coordinate is cosine theta and the y coordinate is sine of theta, what is the distance between that point and the origin? Well to think about that we can construct a right triangle. This distance right over here. So that we could deal with any quadrant I'll make it the absolute value of the cosine theta is this distance right over here. And this distance right over here is the absolute value of the sine of theta. for this first quadrant here but if I went into the other quadrants and I were to setup a similar right triangle then the absolute value is at play. What do we know from the Pythagorean theorem? This is a right triangle here, the hypotenuse has length one, so we know that this expression squared, the absolute value of cosine of theta squared, plus this expression squared, which is this length, plus the absolute value of the sine of theta squared needs to be equal to the length of the hypotenuse squared, which is the same thing which is going to be equal to one squared. Or we could say, this is the same thing. If you're going to square something the sign, if negative it's going to be negative times a negative so it's just going to be positive so this is going to be the same thing as saying that the cosine squared theta plus sine squared theta is equal to one. This is why it's called the Pythagorean identity. a circle comes from, it comes straight out of the Pythagorean theorem where your hypotenuse has length one."
    },
    {
        "Q": "at 6:30 Sal mentions break even\nwhat does that mean?",
        "A": "Break even is the point when revenue equals expenses and so there is no profit or loss incurred. If the expense amounts to $25 then the break even point is when revenue is $25 also. At this point there will be no loss or profit as 25-25 = 0",
        "video_name": "5EdbPz1ZVn0",
        "timestamps": [
            390
        ],
        "3min_transcript": "So then our graph is going to be out here someplace. But we could just figure it out algebraically. At the end of the year, m will be equal to 12. When m is equal to 12, how much is our membership? The price of our membership is going to be our $200 membership fee plus 39 times the number of months, times 12. What's 39 times 12? 2 times 9 is 18. 2 times 3 is 6 plus 1 is 70. I have a 0. 1 times 9 is 9. 1 times 3-- we want to ignore this. 1 times 3 is 3. So we have 8. 7 plus 9 is 16. 1 plus 3 is 4. So the price of our membership is $200 plus 39 times 12, which is $468. So it's equal to $668 at the end of our year. So if you went all the way out to 12, you would have to plot 668 someplace here on our line, if we just Let's do one more of these. Bobby and Petra are running a lemonade stand and they charge $0.45 for each glass of lemonade. In order to break even, they must make $25. How many glasses of lemonade must they sell to break even? So let me just do it with y and x. y is equal to the amount they make. Not max-- the amount they make. Let x is equal to the number of glasses they sell. What is y as a function of x? So y is equal to-- well for every glass they sell, they the number of glasses. There's not any kind of minimum fee that they need to charge or they don't say any kind of minimum cost that they have to spend to run this place. How much in order to break even for each a glass of lemonade? They need to make $25. So in order to break even, they must make $25. So how many glasses of lemonade do they need to sell? y needs to be equal to $25. How many glasses do they need to sell? Well you just set this equation. You say 0.45x has to be equal to 25. We can divide both sides by 0.45."
    },
    {
        "Q": "like sal said in 5:36 are all sums of 3 angles 180",
        "A": "No, not all angles will sum to 180. Ex: three angles could be 10, 20 and 30. They don t sum to 180. Only three angles of a triangle will always sum to 180.",
        "video_name": "BTnAlNSgNsY",
        "timestamps": [
            336
        ],
        "3min_transcript": "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. So if you were to add the two measures of these-- so let me write this down. The measure of angle MNO plus the measure of angle XYZ, this is going to be equal to 120 degrees plus 60 degrees, which is equal to 180 degrees. So if you add these two things up, you essentially are able to go all halfway around the circle. Or you could go throughout the entire half circle or semicircle for a protractor. And when you have two angles that add up to 180 degrees, we call them supplementary. I know it's a little hard to remember sometimes. 90 degrees is complementary. They're just complementing each other. And then if you add up to 180 degrees, you have supplementary. You have supplementary angles. are adjacent so that they share a common side-- so let me draw that over here. So let's say you have one angle that looks like this. And that you have another angle. So let me put some letters here again. And I'll start reusing letters. So let's say that this is ABC. And you have another angle that looks like this. I already used C. Once again, let's say that this is 50 degrees. And let's say that this right over here is 130 degrees. Clearly, angle DBA plus angle ABC, if you add them together, you get 130 degrees plus 50 degrees, which is 180 degrees. So they are supplementary. So let me write that down. Angle DBA and angle ABC are supplementary."
    },
    {
        "Q": "At 2:30 Sal describes the term 'Complementary Angle\" and states that if two angles add up to 90 degrees then they are complementary. Do the two angles that add up to 90 degrees have to be adjacent?",
        "A": "No. (Complementary simply means that, if put next to each other, they would create an angle of 90 degrees; they don t actually have to be next to each other.) The same applies to supplementary angles.",
        "video_name": "BTnAlNSgNsY",
        "timestamps": [
            150
        ],
        "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": "I thought you can't divide by variable 7:00\nor am I missing something here?",
        "A": "You can divide by a variable, so long as the variable is not 0 (ie b does not equal k).",
        "video_name": "okXVhDMuGFg",
        "timestamps": [
            420
        ],
        "3min_transcript": "so we have all the y's on the left-hand side, so, plus 2yb, that's gonna give us a 2yb on the left-hand side, plus 2yb. So what is this going to be equal to? And I'm starting to run into my graph, so let me give myself a little bit more real estate over here. So on the left-hand side, what am I going to have? This is the same thing as 2yb minus 2yk, which is the same thing, actually let me just write that down. That's going to be 2y-- Do it in green, actually, well, yeah, why not green? That's going to be-- Actually, let me start a new color. (chuckles) That's going to be 2yb minus 2yk. You can factor out a 2y, and it's gonna be 2y times, b minus k. So let's do that. So we could write this as 2 times, b minus k, y if you factor out a 2 and a y, so that's that piece right over there. These things cancel out. Now, on our right-hand side, I promised you a little bit of hairy algebra, so hopefully you see that I'm delivering on that promise. On the right-hand side, you have x minus a, squared, and then, let's see, these characters cancel out, and you're left with b squared minus k squared, so these two are gonna be b squared minus k squared, plus b squared minus k squared. Now, I said all I want is a y on the left-hand side, so let's divide everything by two times, b minus k. So, let's divide everything, two times, b minus k, so, two times, b minus k. And I'm actually gonna divide this whole thing by two times, b minus k. Now, obviously on the left-hand side, this all cancels out, you're left with just a y, and then it's going to be y equals, y is equal to one over, and notice, b minus k is the difference between the y-coordinate of the focus, and the y-coordinate, I guess you could say, of the line, y equals k, so it's one over, two times that, times x minus a, squared. So if you knew what b minus k was, this would just simplify to some number, some number that's being multiplied times x minus a, squared, so hopefully this is starting to look like the parabolas that you remember from your childhood, (chuckles) if you do remember parabolas from your childhood. Alright, so then let's see if we could simplify this thing on the right, and you might recognize, b squared minus k squared, that's a difference of squares, that's the same thing as b plus k, times b minus k, so the b minus k's cancel out, and we are just left with,"
    },
    {
        "Q": "At 2:30, when he says that the absolute value could also work in ensuring that the value is positive, can someone explain to me why the absolute value isn't used in the equation?",
        "A": "In the video, Sal does some algebraic manipulation to achieve the formula of the parabola. It is much easier to derive the parabola if he were to square the expression and take the square root than to take the absolute value of the expression. Both methods yield the same value of the expression; however, the latter method (squaring then taking the square root) allows for more easier manipulation. Hope this helps!",
        "video_name": "okXVhDMuGFg",
        "timestamps": [
            150
        ],
        "3min_transcript": "- [Voiceover] What I have attempted to draw here in yellow is a parabola, and as we've already seen in previous videos, a parabola can be defined as the set of all points that are equidistant to a point and a line, and the point is called the focus of the parabola, and the line is called the directrix of the parabola. What I want to do in this video, it's gonna get a little bit of hairy algebra, but given that definition, I want to see, and given that definition, and given a focus at the point x equals a, y equals b, and a line, a directrix, at y equals k, to figure out what is the equation of that parabola actually going to be, and it's going to be based on a's, b's, and k's, so let's do that. So let's take a arbitrary point on the parabola. Let's say we take this point right over here, and its x-coordinate is x, and its y-coordinate is y, and by definition, in order for this to be a parabola, it has to be equidistant to its focus and its directrix, That means that the distance to the directrix, which I'm drawing here in blue, has to be the same as the distance to the focus, which I am drawing in magenta, and when we take the distance to the directrix, we literally just drop a perpendicular, that is, that's going to be the shortest distance to that line, but the distance to the focus, well we see that's at a bit of an angle, and we might have to use the distance formula, which is really just the Pythagorean Theorem. So let's do that. This distance has to be the same as that distance. So, what's this blue distance? Well, that's just gonna be our change in y. It's going to be this y, minus k. It's just this distance. So it's going to be y minus k. Now we have to be careful. The way I've just drawn it, yes, y is greater than k, so this is going to give us a positive value, and you need a non-negative value if you're talking about distances, but you can definitely have a parabola where lower than the y-coordinate of the directrix, in which case this would be negative. So what we really want is the absolute value of this, or, we could square it, and then we could take the square root, the principle root, which would be equivalent to taking the absolute value of y minus k. So that's this distance right over here, and by the definition of a parabola, in order for (x,y) to be sitting on the parabola, that distance needs to be the same as the distance from (x,y) to (a,b), to the focus. So what's that going to be? Well, we just apply the distance formula, or really, just the Pythagorean Theorem. It's gonna be our change in x, so, x minus a, squared, plus the change in y, y minus b, squared, and the square root of that whole thing, the square root of all of that business. Now, this right over here is an equation of a parabola. It doesn't look like it, it looks really hairy,"
    },
    {
        "Q": "In practice sessions video (using hint), of unit vectors at around 2:28 ,he said that we should divide 3 by 5 and also 4 by 5...why should we divide 3 by 5 and 4 by 5?",
        "A": "(Note that this question actually refers to the next video.) The problem asked us to find the UNIT vector. In other words, a vector of length 1 unit. We re given a vector of length 5 units (which is 5 times what we want), so we have to scale it down by dividing each of its components by 5.",
        "video_name": "9ylUcCOTH8Y",
        "timestamps": [
            148
        ],
        "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 2:58 Sal says that any real number is greater or equal to 0 but isn't that a whole number?\nisn't a real number greater than 0?",
        "A": "Natural numbers: 1, 2, 3... Whole numbers: 0, 1, 2, 3... Integers: ...-3, -2, -1, 0, 1, 2, 3... Real: Includes all positive and negative numbers including the decimals and fractional values.",
        "video_name": "qFFhdLlX220",
        "timestamps": [
            178
        ],
        "3min_transcript": "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, 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,"
    },
    {
        "Q": "Why does Sal keep saying \"the principal root of...\" as opposed to \"the square root of...\"? At 1:30 I heard him say \"the square root of, or the principal root of...\" so does that mean they're the same thing? Because it appears as if he sort of corrected himself.",
        "A": "Each square root actually has 2 roots. Consider: 5^2 = 25, but if you square (-5), it is also 25. So, 5^2 = 25 and (-5)^2 = 25. Now, consider square root of 25. Is it 5 or is it -5? We need to know what value to use. So, when you see sqrt(25) , it is understood that the answer should be the principal root or the positive root = 5. If you see - sqrt(25) , the minus sign in front of the radical tells you that your answer should be the negative root = -5. Hope this helps.",
        "video_name": "qFFhdLlX220",
        "timestamps": [
            90
        ],
        "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": "Wait- at 2:37, is tau 2 pi?",
        "A": "Yes, tau is equivalent to twice pi.",
        "video_name": "FtxmFlMLYRI",
        "timestamps": [
            157
        ],
        "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": "Sorry, I don't get 0:48-0:52.\nWhy do you need a distance and to pick a point?",
        "A": "the distance is the radius and the point is the center",
        "video_name": "FtxmFlMLYRI",
        "timestamps": [
            48,
            52
        ],
        "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 1:19 what is a 4 dimensional object",
        "A": "Actually, Steven Hawking has speculated that all objects have 4-dimensions, and the fourth dimension is time. Like, as an object grows older, the volume (or whatever you d call it) of it s fourth dimension would increase.",
        "video_name": "FtxmFlMLYRI",
        "timestamps": [
            79
        ],
        "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 00:50,...is multiplying out the Binomial the only way to get rid of the bracket. If we can distribute the exponent in for example (2r)^2 to be 4r^2.....why can't we just distribute the exponent through (x+9)^2 ?",
        "A": "Because (a + b)^2 isn t the same thing as a^2 + b^2. Example: (2 + 2)^2 = 4^2 = 16 =/= 2^2 + 2^2 = 4 + 4 = 8 Multiplying the binomial is the correct way.",
        "video_name": "bFtjG45-Udk",
        "timestamps": [
            50
        ],
        "3min_transcript": "And now I want to do a bunch of examples dealing with probably the two most typical types of polynomial multiplication that you'll see, definitely, in algebra. And the first is just squaring a binomial. So if I have x plus 9 squared, I know that your temptation is going to say, oh, isn't that x squared plus 9 squared? And I'll say, no, it isn't. You have to resist every temptation on the planet to do this. It is not x squared plus 9 squared. Remember, x plus 9 squared, this is equal to x plus 9, times x plus 9. This is a multiplication of this binomial times itself. You always need to remember that. It's very tempting to think that it's just x squared plus 9 squared, but no, you have to expand it out. And now that we've expanded it out, we can use some of the skills we learned in the last video to actually multiply it. multiplied the trinomial last time, let's multiply x plus 9, times x plus a magenta 9. And I'm doing it this way just to show you when I'm multiplying by this 9 versus this x. But let's just do it. So we go 9 times 9 is 81. Put it in the constants' place. 9 times x is 9x. Then we have-- go switch to this x term-- we have a yellow x. x times 9x is 9x. Put it in the first degree space. x times x is x squared. And then we add everything up. And we get x squared plus 18x plus 81. So this is equal to x squared plus 18x plus 81. Now you might see a little bit of a pattern here, and I'll actually make the pattern explicit in a second. You have x squared. You have this x times this x, gives you x squared. You have the 9 times the 9, which is 81. And then you have this term here which is 18x. How did we get that 18x? Well, we multiplied this x times 9 to get 9x, and then we multiplied this 9 times x to get another 9x. And then we added the two right here to get 18x. So in general, whenever you have a squared binomial-- let I'll do it in very general terms. Let's say we have a plus b squared. Let me multiply it this way again, just to give you the hang of it. This is equal to a plus b, times a plus-- I'll do a green b right there. So we have to b times b is b squared. Let's just assume that this is a constant term. I'll put it in the b squared right there."
    },
    {
        "Q": "At 4:57, why does Khan use a^2+2ab+b^2 for (a-b)^2 instead of a^2-2ab+b^2? Am I wrong? Why? Thank you!",
        "A": "no, he doesn t. he s got it all right",
        "video_name": "bFtjG45-Udk",
        "timestamps": [
            297
        ],
        "3min_transcript": "So this would be a constant, this would be analogous to our 81. a is a variable that we-- actually let me change that up even better. Let me make this into x plus b squared, and we're assuming b is a constant. So it would be x plus b, times x plus a green b, right there. So assuming b's a constant, b times b is b squared. b times x is bx. And then we'll do the magenta x. x times b is bx. And then x times x is x squared. So when you add everything, you're left with x squared plus 2bx, plus b squared. So what you see is, the end product, what you have when product of x and b, plus b squared. So given that pattern, let's do a bunch more of these. And I'm going to do it the fast way. So 3x minus 7 squared. Let's just remember what I told you. Just don't remember it, in the back of your mind, you should know why it makes sense. If I were to multiply this out, do the distributive property twice, you know you'll get the same answer. So this is going to be equal to 3x squared, plus 2 times 3x, times negative 7. Right? We know that it's 2 times each the product of these terms, plus negative 7 squared. And if we use our product rules here, 3x squared is the This right here, you're going to have a 2 times a 3, which is 6, times a negative 7, which is negative 42x. And then a negative 7 squared is plus 49. That was the fast way. And just to make sure that I'm not doing something bizarre, let me do it the slow way for you. 3x minus 7, times 3x minus 7. Negative 7 times negative 7 is positive 49. Negative 7 times 3x is negative 21x. 3x times negative 7 is negative 21x. 3x times 3x is 9 x squared. Scroll to the left a little bit. Add everything. You're left with 9x squared, minus 42x, plus 49."
    },
    {
        "Q": "Why did Sal distribute the 4x^2 at 6:25 instead of just writing (4x^2+y^2)(4x^2+y^2) like he did for the previous questions?",
        "A": "It s more detailed that way.",
        "video_name": "bFtjG45-Udk",
        "timestamps": [
            385
        ],
        "3min_transcript": "product of x and b, plus b squared. So given that pattern, let's do a bunch more of these. And I'm going to do it the fast way. So 3x minus 7 squared. Let's just remember what I told you. Just don't remember it, in the back of your mind, you should know why it makes sense. If I were to multiply this out, do the distributive property twice, you know you'll get the same answer. So this is going to be equal to 3x squared, plus 2 times 3x, times negative 7. Right? We know that it's 2 times each the product of these terms, plus negative 7 squared. And if we use our product rules here, 3x squared is the This right here, you're going to have a 2 times a 3, which is 6, times a negative 7, which is negative 42x. And then a negative 7 squared is plus 49. That was the fast way. And just to make sure that I'm not doing something bizarre, let me do it the slow way for you. 3x minus 7, times 3x minus 7. Negative 7 times negative 7 is positive 49. Negative 7 times 3x is negative 21x. 3x times negative 7 is negative 21x. 3x times 3x is 9 x squared. Scroll to the left a little bit. Add everything. You're left with 9x squared, minus 42x, plus 49. Let's do one more, and we'll do it the fast way. So if we have 8x minus 3-- actually, let me do one which has more variables in it. Let's say we had 4x squared plus y squared, and we wanted to square that. Well, same idea. This is going to be equal to this term squared, 4x squared, squared, plus 2 times the product of both terms, 2 times 4x squared times y squared, plus y squared, this term, squared. And what's this going to be equal to? This is going to be equal to 16-- right, 4 squared is 16-- x squared, squared, that's 2 times 2, so it's x to the fourth power. And then plus, 2 times 4 times 1, that's 8x squared y squared."
    },
    {
        "Q": "At 1:46 why can Sal simplify any further. Cant he take the square root",
        "A": "We don t know the value of x and you can t apply the square root individually like sqrt(4x^2/9+4) =2x/3+2 , it is wrong, so you can t simplify any further from there.",
        "video_name": "hl58vTCqVIY",
        "timestamps": [
            106
        ],
        "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": "So at 6:34, are (0, 2) and (0, -2) the foci of the hyperbola?",
        "A": "Those are the vertices of the hyperbola.",
        "video_name": "hl58vTCqVIY",
        "timestamps": [
            394
        ],
        "3min_transcript": "Because it will never, a hyperbola will never cross the asymptotes. It's not like it can go out here and across this asymptote. We already know that the graph of this parabola -- and you can try other points, if you want, just to verify. It's going to look something like this. It's going to go and then -- nope, I want to make it so it never touches. It's going to get really close, but no, I touched it. It's going to get really close but never touch. And then on this side it's going to get really close, but never touch. And I don't want to touch it. And then on the top side it's going to do the same thing, it's going to get really close, and as you approach infinity it's never going to touch it. And as you get reall close, it'll get infiniitely close but never touch it. So that's what this parabola -- this hyperbola -- is going to look like. And I did it by just trying to see if x could be equal to 0. And I encourage you to try what happens when y equals 0. And you'll get no solution. And that makes sense because this hyperbola never crosses y equals 0, right? It never crosses the x axis. And this should also be intuitive, because if we positive or negative infinity, we saw that we always did have this plus 4 sitting here. We said, oh, well, as x gets super large or super negative, this starts to matter less and less. But we will always be slightly larger than this number. Especially in the positive quadrant, right? We're always going to be -- so the positive quadrant is always going to be slightly larger than the asymptote. And even when we take the positive square root, I guess, When we take the positive square root, we'll always be larger than either of the asymptotes. And, likewise, when you take the negative square root, you're always going to be a little bit smaller than Because this number is going to be a little bit bigger than this number. Then we take the negative square root, you're going to be a little bit smaller, and that's why we're a little bit below. I don't know which one's more intutive for you. Maybe just the -- trying it when x is equal to 0 and when y equals 0, and see what points you get and say, oh, then I'm in kind of a vertical hyperbola as opposed to a horizontal one. video right there. And then I'll do another video where I actually shift the hyperbola. And shifting it is actually no different than shifting an ellipse a circle. You just have, you know, y minus something squared, and x plus something, or x minus something squared and that just tells you where you shift the origin. This hyperbola, of course, is just centered at the origin. Anyway, see you in the next video."
    },
    {
        "Q": "I don't get it at 3:30. If I type in (40)sin40/30 into my calculator I get approx 0.93. If I type 4/3sin40 I get approx 0.86.",
        "A": "It should be 40 sin (40) /30 = .86. Be careful with where parentheses go. What you actually did was 40 sin (40/30) to get .93.",
        "video_name": "IJySBMtFlnQ",
        "timestamps": [
            210
        ],
        "3min_transcript": "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. 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"
    },
    {
        "Q": "At 3:07, is that \"theta\" like a variable?",
        "A": "Yes. Theta can be used like x. But it is usually used for angles so you will see it quite a bit in trigonometry.",
        "video_name": "IJySBMtFlnQ",
        "timestamps": [
            187
        ],
        "3min_transcript": "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. 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"
    },
    {
        "Q": "I'm lost. 3:42 shows Sal making sin go to the other side by using sin-1. I know it's a cofunction, but is it like a reciprocal? Can someone explain how sin becomes sine-1 when moved to the other side? (Ex: 3x=2, you divide 3 on both sides to leave x, to make x= 2/3)",
        "A": "1. Sin(a) \u00c3\u00b7 A = sin(b) \u00c3\u00b7 B 2. Sin(a) = sin(b) \u00c3\u00b7 B * A 3. a = arcsin( sin(b) \u00c3\u00b7 B * A) Since the problem is asking for a specific angle, you have to pull the arcsin to get the specfic angle in this equation. This works for cos, tan, and sin: Sin(x) = y and arcsin(y) = x",
        "video_name": "IJySBMtFlnQ",
        "timestamps": [
            222
        ],
        "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": "At 10:00 the right hand rule , what happens when i change the letters from\na to b and b to a , where is the explanation that\na cross b = b cross a , is true or not ?",
        "A": "Compute (a1, a2, a3) x (b1, b2, b3) = a x b and (b1, b2, b3) x (a1, a2, a3) = b x a. Are they the same? How do they compare?",
        "video_name": "pJzmiywagfY",
        "timestamps": [
            600
        ],
        "3min_transcript": "That forms a plane. If you take a cross b, you get a third vector that's orthogonal to those two. And so a cross b will pop out like this. It'll be orthogonal to both of them and look like that. And so this vector right there is a cross b. And you might say, Sal, how did you know-- I mean, there's multiple vectors that are orthogonal. Obviously, the length of the vector, and I didn't specify that there, but it could pop straight up like that or why didn't it-- you know, you just as easily could have popped straight down like that. That also would be orthogonal to a and b. And the way that a cross b is defined, you can essentially figure out the direction visually by using what's called the right hand rule. And the way I think about it is you take your right hand and let me see if I can draw a suitable right hand. Point your index finger in the direction of a. So if your index finger is in the direction of a and then I So my middle finger, in this case, is going to go something like that. My middle finger is going to do something like that. And then my other fingers do nothing. Then my thumb will go in the direction of a cross b. You could see that there. My thumb is in the direction of a cross b. And assuming that you are anatomically similar to me, then you still get the same result. Let me draw it all. So this is vector a. Vector b goes in that direction. Hopefully you don't have a thumb hanging down here. You know that a cross b in this example will point up and it's orthogonal to both. To kind of satisfy you a little bit, that the vector's definitely orthogonal or that this thing is definitely orthogonal to both of these, let's just play with it and see that that definitely is the case. And what is orthogonal? What is in our context, the definition of orthogonal? If a and b are orthogonal, that means that a dot b is Remember, the difference between orthogonal and perpendicular is that orthogonal also applies to 0 vectors. So these could also be 0 vectors. Notice that I didn't say that any of these guys up here had to be nonzero. Well, in a little bit, we'll talk about the angle between vectors and then you have to assume nonzero. But if you're just taking a cross product, nothing to stop you from taking-- no reason why any of these numbers can't be 0. But let me show you that a cross b is definitely orthogonal to both a and b. I think that might be somewhat satisfying to you. So let me copy a cross b here. I don't feel like rewriting it. OK. Let me paste it. OK, bundle up little other stuff with it."
    },
    {
        "Q": "in the video at 3:25 I want 2 know how to make a hexaflexagon, but it goes to fast. Can you make a slow video on how to make a hexaflexagon?",
        "A": "Hleyendecker 2020, the major predicament of alacritous motion in the video can be eradicated! Because Vi shows numerous ways of construction on the hexaflexagon, at 3:22-3:27, a reasonably stagnant-paced clip is displayed to contrive a hexaflexagon in a rare moment of slowness.",
        "video_name": "VIVIegSt81k",
        "timestamps": [
            205
        ],
        "3min_transcript": "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. to one of your new friends, Bryant Tuckerman. You start with the original, simple, three-faced hexaflexagon, which you call the trihexaflexagon. and he's like, whoa! and wants to learn how to make one. and you are like, it's easy! Just start with a paper strip, fold it into equilateral traingles, and you'll need nine of them, and you fold them around into this cycle and make sure it's all symmetric. The flat parts are diamonds, and if they're not, then you're doing it wrong. And then you just tape the first triangle to the last along the edge, and you're good. But Tuckerman doesn't have tape. After all, it was invented only 10 years ago. So he cuts out ten triangles instead of nine, and then glues the first to the last. Then you show him how to flex it by pinching around a flappy part and pushing in on the opposite side to make it flat and traingly, and then opening from the centre. You decide to start a flexagon committee together to explore the mysteries of flexagotion, But that will have to wait until next time."
    },
    {
        "Q": "How do we actually fold a hexaflexagon at 0:34? She moves and talks too fast.",
        "A": "If you go to the bottom right corner of the video and click on the cog icon (settings), and click on the speed option, you can slow the video down with the options that come up. Hope that helps!",
        "video_name": "VIVIegSt81k",
        "timestamps": [
            34
        ],
        "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": "(6:27) what sal says is wrong. it is less likely for the next toss to be heads if you look at is as a whole. lets say you toss the coin 100 times and you get 90 heads and 10 tails. it is more likely to get that then 90 heads and then 10 tails because there is only combination so theoretically, it is less likely to get a head is less likely in the next toss.",
        "A": "True, but nonetheless the more times you flip the more you approach 50/50 results. This person just assumed that there was some underlying force that pulled everything together into theoretical accuracy. But the only way you approach this truth is through sheer quantity of data, there is no invisible force that makes anything more likely.",
        "video_name": "VpuN8vCQ--M",
        "timestamps": [
            387
        ],
        "3min_transcript": "Let me differentiate. And I'll use this example. So let's say-- let me make a graph. And I'll switch colors. This is n, my x-axis is n. This is the number of trials I take. And my y-axis, let me make that the sample mean. And we know what the expected value is, we know the expected value of this random variable is 50. Let me draw that here. This is 50. So just going to the example I did. So when n is equal to-- let me just [INAUDIBLE] here. So my first trial I got 55 and so that was my average. I only had one data point. Then after two trials, let's see, then I have 65. which is 60. So then my average went up a little bit. Then I had a 45, which will bring my average down a little bit. I won't plot a 45 here. Now I have to average all of these out. What's 45 plus 65? Let me actually just get the number just so you get the point. So it's 55 plus 65. It's 120 plus 45 is 165. Divided by 3. 3 goes into 165 5-- 5 times 3 is 15. It's 53. No, no, no. 55. So the average goes down back down to 55. And we could keep doing these trials. So you might say that the law of large numbers tell this, OK, after we've done 3 trials and our average is there. So a lot of people think that somehow the gods of probability are going to make it more likely that we get fewer That somehow the next couple of trials are going to have to be down here in order to bring our average down. And that's not necessarily the case. Going forward the probabilities are always the same. The probabilities are always 50% that I'm going to get heads. It's not like if I had a bunch of heads to start off with or more than I would have expected to start off with, that all of a sudden things would be made up and I would get more tails. That would the gambler's fallacy. That if you have a long streak of heads or you have a disproportionate number of heads, that at some point you're going to have-- you have a higher likelihood of having a disproportionate number of tails. And that's not quite true. What the law of large numbers tells us is that it doesn't care-- let's say after some finite number of trials your average actually-- it's a low probability of this happening, but let's say your average is actually up here. Is actually at 70. You're like, wow, we really diverged a good bit from the expected value. But what the law of large numbers says, well, I don't care how many trials this is. We have an infinite number of trials left."
    },
    {
        "Q": "Why at 4:53 in the video Sal puts down four squared raised to the seventh power equal to four to the seventh power",
        "A": "its an error. (4^2)^7 is actually 4^14 he corrected himself at 5:23",
        "video_name": "dC1ojsMi1yU",
        "timestamps": [
            293
        ],
        "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": "At 3:25, Sal says \"...the magnitude or absolute value of Z1...\". He denotes it like |Z1|. Why doesn't he denote it as ||Z1||?",
        "A": "Notations are subject to a bit of flexibility. I had a physics teacher in college that consistently referred to the absolute values of vectors. In fact, absolute value is a type of magnitude. So you could legitimately denote it |*| or ||*||. But here, I think Sal opted for the simpler and more familiar notation.",
        "video_name": "FwuPXchH2rA",
        "timestamps": [
            205
        ],
        "3min_transcript": "it would be b. This is a real number, but this tells us how much the i is scaled up in the complex number z right over there. Now, one way to visualize complex numbers, and this is actually a very helpful way of visualizing it when we start thinking about the roots of numbers, especially the complex roots, is using something called an Argand diagram. So this is this. And so it looks a lot like the coordinate axes and it is a coordinate axes. But instead of having an x and y-axis it has a real and an imaginary axis. So in the example of z being a plus bi, we would plot it really as a position vector, where you have the real part on the horizontal axis. So let's say this is a and then the imaginary part along the vertical axis, or the imaginary axis. And so we would represent, in an Argand diagram, the vector z as a position vector that starts at 0 and that has a tip at the coordinate a comma b. So this right here is our complex number. This right here is a representation in our Argand diagram of the complex number a plus bi, or of z. Now when you draw it this way, when you draw it as a position vector, and if you're familiar with polar coordinates, you're probably thinking, hey, I don't have to represent this complex number just as coordinates, just as an a plus bi. Maybe I could represent this as some angle here, let's call that angle phi, and some the distance here, let's call that r, which is kind of the magnitude And you could. If you gave some angle and some distance, that would also specify this point in the complex plane. and this right here is called the magnitude, or sometimes the modulus, or the absolute value of the complex number. So let's think about it a little bit. Let's think about how we would actually calculate these values. So r, which is the modulus, or the magnitude. It's denoted by the magnitude or the absolute value of z1. What's this going to be. Well, we have a right triangle here. This side is b, length b. The base right here has length a. So to calculate r, we can just use the Pythagorean Theorem. r squared is going to be equal to a squared plus b squared. Or r is going to be equal to the square root of a squared plus b squared. If we want to figure out the argument, this is going to be equal to what?"
    },
    {
        "Q": "13:04 I've tried to make a non-congruent triangle that complies with SSA in CAD and I haven\u00c2\u00b4t being able to do so. Was Sal wrong or there are so few possible triangles that are non-congruent and SSA that I can't find them by trial and error? In any case, many triangles that are SSA are also congruent.",
        "A": "Well, the answer was in the More on why SSA is not a postulate video... :S. I should be more careful when asking questions from now on. Equally thanks for your reply.",
        "video_name": "8Ld8Csu4sEs",
        "timestamps": [
            784
        ],
        "3min_transcript": "is going to touch this one right over there. And the only way it's going to touch that one right over there is if it starts right over here, because we're constraining this angle right over here. We're constraining that angle. And so it looks like angle, angle, side does indeed imply congruency. So that does imply congruency. So let's just do one more just to kind of try out all of the different situations. What if we have-- and I'm running out of a little bit of real estate right over here at the bottom-- what if we tried out side, side, angle? So once again, draw a triangle. So it has one side there. It has another side there. And then-- I don't have to do those hash marks just yet. So one side, then another side, and then another side. And what happens if we know that there's another triangle that has two of the sides the same and then the angle after it? So for example, we would have that side just like that, But we're not constraining the angle. We aren't constraining this angle right over here, but we're constraining the length of that side. So let me color code it. So that blue side is that first side. Then we have this magenta side right over there. So this is going to be the same length as this right over here. But let me make it at a different angle to see if I can disprove it. So let's say it looks like that. Or actually let me make it even more interesting. Let me try to make it like that. So it's a very different angle. But now, it has to have the same angle out here. It has to have that same angle out here. So it has to be roughly that angle. So it actually looks like we can draw a triangle that is not congruent that has two sides being the same length and then an angle is different. For example, this is pretty much that. I made this angle smaller than this angle. These two sides are the same. is that this green side is going to be shorter on this triangle right over here. So you don't necessarily have congruent triangles with side, side, angle. So this is not necessarily congruent, not necessarily, or similar. It gives us neither congruency nor similarity."
    },
    {
        "Q": "When Sal talks the definition of similar (at 2:36) , he says that in geometry similar things have same shape. However, I still wonder if he meant just having same type of form (like triangle, pentagon or rectangle) or if the angles of the figure also have to be the same so two figure can be similar geometrically?",
        "A": "Two shapes are similar if:the corresponding sides are proportional and the angles are congruent. The proportionality constant can be one, in which case the sides are the same length, but the definition of similar is the first statement.",
        "video_name": "8Ld8Csu4sEs",
        "timestamps": [
            156
        ],
        "3min_transcript": "But when you think about it, you can have the exact same corresponding angles, having the same measure or being congruent, but you could actually scale one of these triangles up and down and still have that property. For example, if I had this triangle right over here, it looks similar-- and I'm using that in just the everyday language sense-- it has the same shape as these triangles right over here. And it has the same angles. That angle is congruent to that angle, this angle down here is congruent to this angle over here, and this angle over here is congruent to this angle over here. So all of the angles in all three of these triangles The corresponding angles have the same measure. But clearly, clearly this triangle right over here is not the same. It is not congruent to the other two. The sides have a very different length. This side is much shorter than this side right over here. This side is much shorter than that side over there. So with just angle, angle, angle, you cannot say that a triangle has the same size and shape. It does have the same shape but not the same size. So this does not imply congruency. So angle, angle, angle does not imply congruency. What it does imply, and we haven't talked about this yet, is that these are similar triangles. So angle, angle, angle implies similar. So let me write it over here. It implies similar triangles. And similar-- you probably are use to the word in just everyday language-- but similar has a very specific meaning in geometry. And similar things have the same shape but not necessarily the same size. So anything that is congruent, because it has the same size and shape, is also similar. But not everything that is similar is also congruent. So for example, this triangle is similar-- all of these triangles are similar to each other, but they aren't all congruent. These two are congruent if their sides But if we know that their sides are the same, then we can say that they're congruent. But neither of these are congruent to this one right over here, because this is clearly much larger. It has the same shape but a different size. So we can't have an AAA postulate or an AAA axiom to get to congruency. What about side, angle, side? So let's try this out, side, angle, side. So let's start off with one triangle right over here. So let's start off with a triangle that looks like this. I have my blue side, I have my pink side, and I have my magenta side. And let's say that I have another triangle that has this blue side. It has the same side, same length as that blue side. So let me draw it like that. It has the same length as that blue side. So that length and that length are going to be the same. It has a congruent angle right after that. So this angle and the next angle for this triangle are going to have the same measure,"
    },
    {
        "Q": "at about 3:15 in the video. Can someone explain the difference between direct and joint? thanks",
        "A": "i believe that the difference is that in the direct variation you are dealing with only TWO variables (x and y) and one constant number (k); in the joint variation you are dealing with THREE variables, like Sal said, the example would be the area (A) of a rectangle, which is the width (w) times the length (l), so if you had to insert these into a table, you would have to have THREE columns.",
        "video_name": "v-k5L0BPOmc",
        "timestamps": [
            195
        ],
        "3min_transcript": "think about the telltale signs of direct variation. If x increases, y should increase. So if x increases. Let me do that in the same yellow. So the telltale signs of direct variation, if x increases then y will increase and vice versa. The other telltale sign is. Is if you increase x by some, by some factor. So, if you have x going to 3x then y should also increase by that same factor. And we could see that with some examples. So, I mean, you could pick a K, let's say that, let's say that K was one. So if y is equal to x, if you take, if x goes from one to three, then y is also going to go from one to three. So that's all we're talking about here. Let me actually, y should actually to three times y, that's what I'm talking about. If you triple x, you're also gonna end up tripling y. Inverse variation. You have y being equal to some constant times one over x. multiply both sides by x you get x times y is equal to some constant. And you could switch the x's and the y's around as well for inverse variation. Now what are the tale tale signs? Well if you increase x, if x goes up, then what happens to y? If x goes up then this becomes a smaller value cuz it's one over x so then y will go down. Then y will go down. And if you take X and if you're to say increase it by a factor of three then what's going to happen to Y? Well if you increase this by a factor of three, you're actually going to decrease this whole value by a factor of one-third, so Y is going to go, so then you're going to have one-third of y. So that's, these are the tell-tale signs for inverse variation. Now finally they talk about something called joint variation, and this one you won't necessarily see in introductory algebra course. So if I told you, if I told you that area of a rectangle is equal to the width of a rectangle times the length of rectangle, this is an example of joint variation. Area is proportionally to two. Is the proportional to two different quantities? So the main tell tale sign here for joint variation frankly is you're gonna be dealing with more than two variables. Joint, Joint, Joint variation. So when you look at this example, there only gi, giving us two variables. So you can rule out joint variation just right from the get go. Now let's look at the tell tale signs. So as x is increasing, as x goes from one to two, what is happening to y? Y went from 12 to six. So as x is going up by a factor of two, y is going, is, is going by a factor of one half. Or y is being multiplied by one half. So as x goes from one to three, it's being multiplied by three."
    },
    {
        "Q": "at 3:46 , Can you really just square the 1/3 and then plug it into the radical? I have never seen that before.",
        "A": "Yes as you re not changing the equation in any way. You re basically squaring a positive multiple then when you place it inside the radical you re square rooting it again so it s exactly the same multiple. (1/3)^2 = 1/9 sqrt(1/9) = 1/3",
        "video_name": "WAoaBTWKLoI",
        "timestamps": [
            226
        ],
        "3min_transcript": "So this is u in terms of x. So everywhere we see a u up here we can replace it with this expression. And we are essentially done. We would have written this in terms of x. Now, there's another technique you might sometimes see in a calculus class where someone says, OK, we know that u is equal to cosine theta. We know this relationship. How can we express u in terms of x? And we'll say, let's draw a right triangle. They'll draw a right triangle like this. They'll draw a right triangle, and they'll say, OK, look, sine of theta is x over 3. So if we say that this is theta right over here, sine of theta is the same thing as opposite over hypotenuse. Opposite over hypotenuse is equal to x over 3. So let's say that this is x and then this right over here is 3. Then the sine of theta will be x over 3. So we look at that first substitution right over here. But in order to figure out what u is in terms of x, we need to figure out what cosine of theta is. So we have to figure out what this adjacent side is. Well, we can just use the Pythagorean theorem for that. Pythagorean theorem would tell us that this is going to be the square root of the hypotenuse squared, which is 9, minus the other side squared, minus x squared. So from this, we fully solved the right triangle in terms of x. We can realize that cosine of theta is going to be equal to the adjacent side, square root of 9 minus x squared, over the hypotenuse, over 3, which is the same thing as 1/3 times the square root of 9 minus x squared, which is the same thing if we square 1/3 and put it into the radical. So we're essentially going to take the square-- 1/3 is the same thing as the square root of 1/9. So can rewrite this as the square root of 1/9 times 9 minus x squared. Essentially, we just brought the 1/3 third into the radical. Now it's 1/9. And so now this is going to be the same thing which is exactly this thing right over here. x squared over 9 is the same thing as x over 3 squared. So either way, you get the same result. I find using the trig identity right over here to express cosine of theta in terms of sine theta and then just do the substitution to be a little bit more straightforward. But now we can just substitute into the original thing. So either of these-- I can write it as either way-- this thing right over here, this is the same thing as 1 minus x squared over 9 to the 1/2 power. That's what u is equal to. And everywhere we see u, we just substitute it with this thing. So our final answer in terms of x is going to be equal to 243 times u to the fifth, this to the fifth power over 5. This to the fifth power is 1 minus x squared over 9. It was to the 1/2, but if we raise that to the fifth power,"
    },
    {
        "Q": "is it me or, It appears that at 5:20 sal gave the final answer as if it was (-u^5/5+u^3/3) instead of ( u^5/5-u^3/3) after unwinding the trig and u substitution. please correct me if I'm wrong.\nRegards",
        "A": "I don t see any problems with the final answer.",
        "video_name": "WAoaBTWKLoI",
        "timestamps": [
            320
        ],
        "3min_transcript": "So we have to figure out what this adjacent side is. Well, we can just use the Pythagorean theorem for that. Pythagorean theorem would tell us that this is going to be the square root of the hypotenuse squared, which is 9, minus the other side squared, minus x squared. So from this, we fully solved the right triangle in terms of x. We can realize that cosine of theta is going to be equal to the adjacent side, square root of 9 minus x squared, over the hypotenuse, over 3, which is the same thing as 1/3 times the square root of 9 minus x squared, which is the same thing if we square 1/3 and put it into the radical. So we're essentially going to take the square-- 1/3 is the same thing as the square root of 1/9. So can rewrite this as the square root of 1/9 times 9 minus x squared. Essentially, we just brought the 1/3 third into the radical. Now it's 1/9. And so now this is going to be the same thing which is exactly this thing right over here. x squared over 9 is the same thing as x over 3 squared. So either way, you get the same result. I find using the trig identity right over here to express cosine of theta in terms of sine theta and then just do the substitution to be a little bit more straightforward. But now we can just substitute into the original thing. So either of these-- I can write it as either way-- this thing right over here, this is the same thing as 1 minus x squared over 9 to the 1/2 power. That's what u is equal to. And everywhere we see u, we just substitute it with this thing. So our final answer in terms of x is going to be equal to 243 times u to the fifth, this to the fifth power over 5. This to the fifth power is 1 minus x squared over 9. It was to the 1/2, but if we raise that to the fifth power, over 5 minus this to the third power, 1 minus x squared over 9 to the 3/2 raising this to the third power-- that's this right over here-- over 3, and then all of that plus c. And we're done. It's messy, but using first trig substitution then u substitution, or trig substitution then rearranging using a couple of our techniques for manipulating these powers of trig functions, we're able to get into a form where we could use u substitution, and then we were able to unwind all the substitutions and actually evaluate the indefinite integral."
    },
    {
        "Q": "At around 7:30, when he is discussing the trig. ratios, is there any particular reason why, for example, we use adj/hyp instead of hyp/adj? If we used the reciprocal of any of these ratios would it matter?",
        "A": "For each of the main trig functions, sine, cosine and tangent, there is another trig function that is its reciprocal: secant is 1/cosine, cosecant is 1/sine, and cotangent is 1/tangent. For an acute angle in a right triangle, adj/hyp is cosine. You can t use hyp/adj for cosine because that s something entirely different (secant) with a graph that looks nothing like the graph of cosine.",
        "video_name": "QuZMXVJNLCo",
        "timestamps": [
            450
        ],
        "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": "where did Sal get the number 7 1/2 from at 1:44?",
        "A": "7 1/2 is half of 15",
        "video_name": "h0FFEBHBufo",
        "timestamps": [
            104
        ],
        "3min_transcript": "We're on problem 21. In the figure below, n is a whole number. What is the smallest possible value for n? OK, both of these sides are n. And so this is something that's actually really good to get this intuition. Because this shows up on all sorts of standardized tests. And so let's think about how small can we make n. So the lower the top of this pyramid becomes, the smaller n becomes. If we pushed this top of the pyramid really high then n would have to be really big. For example, if we made the triangle into that. Then clearly, this length is shorter than that length. We want to keep lowering it to get as small a possible n. But what happens if we lowered it all the way? If we flattened this triangle. If we just flattened it all the way down. So essentially this would be the top of it and this would be n and this would be n. I hope you're visualizing that properly. I've flattened the triangle. So these two sides would just go flat with the base. And so this top if I were to do it in So this is as small as n could get. One could argue whether this is a triangle at all anymore. It's really a line now, because I've squished out all of the area in there. But even in this case, n would have to be, in the smallest case, it would be 7.5. Each n would be half of this 15. So, as we push this base down, that's kind of the limit that n approaches. n cannot be any smaller than 7.5. And they tell us that n is a whole number. So n has to be greater than 7.5 in order for this to be a triangle. And n is a whole number, so n is equal to 8. That's choice C. And that's an important intuition to have in general. That the third side of a triangle can never be bigger than two of the other sides combined. Then you're dealing with something else. You're not dealing with a triangle. then you're actually dealing with a line. Because you'd have to squish out all of the area of the triangle in order to get there. Anyway, I like that problem. Next question. I think, just eyeballing it, they want us to do the same thing. Same type of intuition. And I had my rant in the last video about how they weren't doing problems that give you intuition or that test your intuition. But I'll take that back, because I think that's what they are testing now. Which of the following sets of numbers could represent the lengths of the sides of a triangle. 2, and 2, and 5. So this is the same thing again. How can I have two sides of a triangle combined being shorter than the third side. If I had a side of length 2, and then I had another side of length 2, there's no way that this last side could be 5. Even if I completely flattened this triangle, 2 and 2, the longest that this last side, this third"
    },
    {
        "Q": "At 05:40: \"So a note and the note with twice the frequency, totally are like the same note, am I right?\" Can anybody explain this statement to me, please?",
        "A": "The first note is a note on a keyboard (or whatever she was playing at the time) then the second note is twice the frequency, or 1 octave up. so technically, they are the same note, even though they are not pitch. one 8(how many notes in a octave) x 2 (doubling the frequency) = 16(twice the frequency, or 1 octave up from the first note.",
        "video_name": "i_0DXxNeaQ0",
        "timestamps": [
            340
        ],
        "3min_transcript": "So he played 1/2 the length and found the note was an octave higher. He thought that was pretty neat. So then he tried the next simplest ratio and played 1/3 of the string. If the full length was C, then 1/3 the length would give the note G, an octave and a fifth above. The next ratio to try was 1/4 of the string, but we can already figure out what note that would be. In 1/2 the string was C an octave up, then 1/2 of that would be C another octave up. And 1/2 of that would be another octave higher, and so on and so forth. And then 1/5 of the string would make the note E. But wait. Let's play that again. It's a C Major chord. So what about 1/6? We can figure that one out, too, using ratios we already know. 1/6 is the same as 1/2 of 1/3. And 1/3 third was this G. So 1/6 is the G an octave up. Check it out. 1/7 will be a new note, because 7 is prime. And Pythagoras found that it was this B-flat. Then 8 is 2 times 2 times 2. And 1/9 is 1/3 of 1/3. So we go an octave and a fifth above this octave and a fifth. And the notes get closer and closer until we have all the notes in the chromatic scale. And then they go into semi-tones, et cetera. But let's make one thing clear. This is not some magic relationship between mathematical ratios and consonant intervals. It's that these notes sound good to our ear because our ears hear them together in every vibration that reaches the cochlea. Every single note has the major chord secretly contained within it. So that's why certain intervals sound consonant and others dissonant and why tonality is like it is and why cultures that developed music independently of each other still created similar scales, chords, and tonality. This is called the overtone series, by the way. And, because of physics, but I don't really know why, a string 1/2 the length vibrates twice as fast, which, hey, makes this series the same as that series. If this were A440, meaning that this is a swing that likes to swing 440 times a second, And here's E at three times the original frequency, 1320. The thing about this series, what with making the string vibrate with different lengths at different frequencies, is that the string is actually vibrating in all of these different ways even when you don't hold it down and producing all of these frequencies. You don't notice the higher ones, usually, because the lowest pitch is loudest and subsumes them. But say I were to put my finger right in the middle of the string so that it can't vibrate there, but didn't actually hold the string down there. Then the string would be free to vibrate in any way that doesn't move at that point, while those other frequencies couldn't vibrate. And if I were to touch it at the 1/3 point, you'd expect all the overtones not divisible by 3 to get dampened. And so we'd hear this and all of its overtones. The cool part is that the string is pushing it around the air at all these different frequencies. And so the air is pushing around your ear at all these different frequencies. And then the basilar membrane is vibrating in sympathy with all these frequencies. And your ear puts it together and understands it as one sound."
    },
    {
        "Q": "at 1:56 Vi says cochlea, what does the cochlea do?",
        "A": "Sound goes into your ear, past your ear drum, past the bones in your ear , into your cochlea and through the nerve to the brain. The cochlea basically processes the sound so you can understand it.",
        "video_name": "i_0DXxNeaQ0",
        "timestamps": [
            116
        ],
        "3min_transcript": "[PIANO ARPEGGIOS] When things move, they tend to hit other things. And then those things move, too. When I pluck this string, it's shoving back and forth against the air molecules around it and they push against other air molecules that they're not literally hitting so much as getting too close for comfort until they get to the air molecules in our ears, which push against some stuff in our ear. And then that sends signals to our brain to say, Hey, I am getting pushed around here. Let's experience this as sound. This string is pretty special, because it likes to vibrate in a certain way and at a certain speed. When you're putting your little sister on a swing, you have to get your timing right. It takes her a certain amount of time to complete a swing and it's the same every time, basically. If you time your pushes to be the same length of time, then even general pushes make your swing higher and higher. That's amplification. If you try to push more frequently, you'll just end up pushing her when she's swinging backwards and instead of going higher, you'll dampen the vibration. It wants to swing at a certain speed, frequency. If I were to sing that same pitch, the sound waves I'm singing will push against the string at the right speed to amplify the vibrations so that that string vibrates while the other strings don't. It's called a sympathy vibration. Here's how our ears work. Firstly, we've got this ear drum that gets pushed around by the sound waves. And then that pushes against some ear bones that push against the cochlea, which has fluid in it. And now it's sending waves of fluid instead of waves of air. But what follows is the same concept as the swing thing. The fluid goes down this long tunnel, which has a membrane called the basilar membrane. Now, when we have a viola string, the tighter and stiffer it is, the higher the pitch, which means a faster frequency. The basilar membrane is stiffer at the beginning of the tunnel and gradually gets looser so that it vibrates at high frequencies at the beginning of the cochlea and goes through the whole spectrum down to low notes at the other end. So when this fluid starts getting pushed around there's a certain part of the ear that vibrates in sympathy. The part that's vibrating a lot is going to push against another kind of fluid in the other half of the cochlea. And this fluid has hairs in it which get pushed around by the fluid, and then they're like, Hey, I'm middle C and I'm getting pushed around quite a bit! Also in humans, at least, it's not a straight tube. The cochlea is awesomely spiraled up. OK, that's cool. But here are some questions. You can make the note C on any instrument. And the ear will be like, Hey, a C. But that C sounds very different depending on whether I sing it or play it on viola. Why? And then there's some technicalities in the mathematics of swing pushing. It's not exactly true that pushing with the same frequency that the swing is swinging is the only way to get this swing to swing. You could push on just every other swing. And though the swing wouldn't go quite as high as if you pushed every time, it would still swing pretty well. In fact, instead of pushing every time or half the time, you could push once every three swings or four, and so on. There's a whole series of timings that work,"
    },
    {
        "Q": "Hey guys, did you notice that at 1:05 Sal said tenths when he ment thousandths!",
        "A": "No! I didn t thanks for letting me know! ;P ;) :)",
        "video_name": "BINElq3DFkg",
        "timestamps": [
            65
        ],
        "3min_transcript": "Let's once again see if we can order now a different set of decimals from least to greatest, and once again I encourage you to pause this video and try to do this on your own. So let's go to the most significant place, the ones place here. None of these have any ones. So then we can go to the next most significant place, which is the tenths place. This has five tenths. This has six tenths. This has one tenth. This has five tenths. This has one tenth. So if we just look at the tenths place, the ones that have the fewest tenths-- this has only one tenth, this one only has one tenth, this one has five tenths, this one has five tenths, and then this one has six tenths. So I've ordered it by what's going on in the tenths place. Now, both of these have the same number of tenths. Let's move to the hundredths place to figure out which of these is larger. This one has six hundredths. This has five hundredths, so this one is larger. It has more hundredths. Same number of tenths, more hundredths. And hundredths are obviously more significant than thousandths, so it It matters that this one has more tenths, and actually this one has more thousandths as well. But now let's go look at these two. These have the same number of tenths. They both have five tenths. But this one has six hundredths, while this one only has two hundredths, so this one is larger. And then finally, this one of course, had six tenths, so this one had the most tenths. So we don't even have to look at the other places here. And we're done."
    },
    {
        "Q": "How do you know when to leave the circle open or closed on the number line at 3:59",
        "A": "You leave the circle open when you need to exclude it from your inequality (less than or greater than inequalities). You shade it in when it is included in the inequality (commonly known as less than or equal to OR greater than or equal).",
        "video_name": "FZ2APP6-grU",
        "timestamps": [
            239
        ],
        "3min_transcript": "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": "How did Sal just throw in the factor of dx at 3:45? The volume of a real shell is not the area of its outer surface times the depth of the shell, so why should that be the volume when the shell is infinitesimally thin?",
        "A": "The volume of the shell, as stated in previous videos, is the circumference times the height of the shell times the width. Here, the width is dx. However, it does not really matter in the end, because you are just taking the definite integral to find the area. Hope that helps!",
        "video_name": "SfWrVNyP9E8",
        "timestamps": [
            225
        ],
        "3min_transcript": "What is that distance going to be? Well, it's the horizontal distance between x equals 2 and whatever the x value is right over here. So it's going to be 2 minus our x value. So this radius, this distance right over here, is going to be 2 minus x. And so the circumference is going to be that times 2 pi. 2 pi r gives us the circumference of that circle. So 2 pi times 2 minus x. And then if we want the surface area of the outside of our shell, so the area is going to be the circumference 2 pi times 2 minus x times the height of each shell. Now, what is the height of each shell? It's going to be the vertical distance expressed as functions of y. So it's going to be the top boundary is y is equal to square root of x, the bottom boundary is So it's going to be square root of x minus x squared. Let me do this in the yellow. So it's going to be square root of x minus x squared. And so if you want the volume of a given shell-- I'll write all this in white-- it's going to be 2 pi times 2 minus x times square root of x minus x squared. So this whole expression, I just rewrote it, is the area, the outside surface area, of one of these shells. If we want the volume, we have to get a little bit of depth, multiply by how deep the shell is, so times dx. And if we want the volume of this whole thing, we just have to solve all the shells for all of the x's in this interval and take the limit as the dx's get smaller and smaller and we have more and more shells. And so, what's our interval? Well our x's are going to go between 0 and 1."
    },
    {
        "Q": "3:10 Why would you subtract 90 degrees from that equation? I haven't exactly figured that out.",
        "A": "Sal subtracted 90 degrees from both sides of the equation to simplify the equation. What I would do instead (personally) is add 90 and 32 to get 122, and then subtract 122 from both sides. 180-122= 58, so either way, you get the same answer. Hope that helps you!",
        "video_name": "iqeGTtyzQ1I",
        "timestamps": [
            190
        ],
        "3min_transcript": "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. So if you subtract 90 from both sides, you get x plus 32 is equal to 90. And then if you subtract 32 from both sides, you get x is equal to-- what is this-- 58 degrees. Fair enough. Now, what else can we figure out? Well, if this angle over here is a right angle-- and I'm just redoing the problem over again just to show you that there's multiple ways to get the answer. We were given that this is a right angle. If that is 90 degrees, then this angle over here is supplementary to it, and it also has to be 90 degrees. So then we have this angle plus 90 degrees plus this angle have to equal 180. Maybe we could call that y. So y plus 58 plus 90 is equal to 180."
    },
    {
        "Q": "@2:56 how does it simplify to 1, not the best at algebra",
        "A": "1/x * x = 1/x * x/1 = (1*x)/(x*1) = x/x = 1 We multiply 1/x with x, so there s one x at the numerator and denominator.",
        "video_name": "iw5eLJV0Sj4",
        "timestamps": [
            176
        ],
        "3min_transcript": "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 or the antiderivative of 1dx, or the integral of 1dx, or the antiderivative of 1 is just minus x. And this is just an antiderivative of this. If we want to write the entire class of antiderivatives we just have to add a plus c here, and we are done. We figured out the antiderivative of the natural log of x. I encourage you to take the derivative of this. For this part, you're going to use the product rule and verify that you do indeed get natural log of x when you take the derivative of this."
    },
    {
        "Q": "at 5:21 i understand that A to the B+C power equals A to the B power times A to the C power but what about if it was negative, as in A to the B-C power. would that make it A to the B power divided by A to the C power or would that just make C negative?",
        "A": "Yes, you are correct. A to the B-C power would give you A to the B power divided by A to the C power. Sal actually explains this property at 7:19. Another way of doing this is: A^(B-C) = A^B x A^-C and since A^-C = 1/A^C we can rewrite this as A^B/A^C (A to the B power divided by A to the C power).",
        "video_name": "vSijVSL3ChU",
        "timestamps": [
            321
        ],
        "3min_transcript": "isn't a huge stretch here. I just literally multiplied and divide by, divided by 10, times this t over 10. But when I write it this way, an exponent property might jump out at you. If I have, if I have a to the b, and then I raise that to the c, that's going to be a to the bc. Or, another way around, a to the bc is going to be a to the b to the c. And so, this piece right over here, I can rewrite it as two to the 10th, and then raise that to the t over 10 power. To the t over 10 power. and then raise that to the t over 10, that's gonna be the same thing as two to the 10 times t over 10. And of course, we still have the one over 32 over here. One over, one over 32. I'm tempted to write that as two to the negative fifth power, but I won't do that just yet. Actually, let's just, let's just keep it, let's just keep it as two to the 10th power, just for simplicity right now. Later we can, you might know that that's gonna be 1,024. But let's just, let's see what else we can do. So we know this is going to be some num- ... Actually, let me just write out as 1,024. So we have one over 32 times 1,024 to the t over 10, to the t over 10 power. So it seems like we're getting close. If there was no minus one here, we're essentially done. But now there's this minus one. So how do we deal with that? Well, we can do a similar type of strategy. We can subtract one, and then we could add, and then we could add one. Then we're not actually changing the value. Just as we multiplied by 10 and divide by 10, we're not changing the value up here. If you subtract one and add one to the exponent, you're not changing its value. And so, what is this going to be? We wanna leave this minus one here. But we wanna get rid of, And here, we just have to remind ourselves that, if we have a to the b times a to the c, that's going to be equal to a to the b plus c. If you have the same base, multiplied, same base raised to different exponents and you multiply them, you could just add the exponents. And so you could also go the other way around. If you have a to the b plus c, you could break it up into a to the b times a to the c. So, this business right over here, this business right over here, this is 1,024 to the t over 10 minus one, plus one. So we can break this up as, we can break this up as, 1,024, 1,024 to the t over 10 minus one, that's this part here, and then times 1,024 to the one. Times ... Let me make this in a different color. So, let's see green."
    },
    {
        "Q": "I dont understand the way Sal's doing these. I paused at 0:06 and factored it by grouping, because it seemed like the obvious way to go. Like this:\n30x^2 + 11xy + y^2\n30x^2 + 5xy + 6xy + y^2\n5x(6x+y) + y(6x+y)\n=(5x+y)(6x+y)\nMultiplying it out, i get back to the beginning.\nI did the problem in the previous video the same way. Is this correct? Or am i doing it wrong and getting the right answers just by accident?",
        "A": "I did it the same way you did.",
        "video_name": "0xrvRKHoO2g",
        "timestamps": [
            6
        ],
        "3min_transcript": "Let's see if we can use our existing factoring skills to factor 30x squared plus 11xy plus y squared. And I encourage you to pause the video and see if you can handle it yourself. Now, the first hint I will give you-- and this might open up what's going on here-- is to maybe rearrange this a little bit. We could rewrite this as y squared plus 11xy plus 30x squared. And my whole motivation for doing that-- there are ways to factor a quadratic where your first coefficient, your coefficient on this first term, is something other than 1. But we haven't seen that yet. And so rearranging it this way, this got us a little bit more into our comfort zone. Now our coefficient is a 1 on the y squared term. So now we can start to think of this in the same form that we've looked at some of the other factoring problems. Can we think of two numbers whose product is 30x squared and whose sum is 11x? We have y squared, some coefficient on y. And then in terms of y, this isn't in any way dependent on y. So one way to think about this, if you knew what x was, then this would be a quadratic in terms of y. And that's how we're really thinking about it here. So can we find two numbers whose product is 30x squared and two numbers whose sum is the coefficient on this y term right here, whose sum is 11x? So let's just think about all of the different possibilities. If we were just thinking about two numbers whose product was 30 and whose sum was 11, we would be thinking of 5 and 6. 5 times 6 is 30. 5 plus 6 is 11. It's some trial and error. You could have tried 3 and 10. Well, that would have been-- 13 would be their sum. You could have tried 2 and 15. That wouldn't have worked. But 5 and 6 does work here, so we've already seen that multiple times. So 5 and 6 would work for 30, but we have 30x squared. Well, 5x times 6x is 30x squared, and 5x plus 6x is 11x. So this actually works. So then our factoring or our factorization of this expression is just going to be y plus 5x times y plus 6x. And I'll leave it up to you to verify that this does indeed, when you multiply it out, equal this up here."
    },
    {
        "Q": "At 2:55, what is meant by derivative of something \"with respect to\" something else?",
        "A": "With respect to is generally used to describe the term you are talking about. For example, say you have: f(x) = (x^2)/3 with respect to x, the function is the same, (x^2)/3 BUT with respect to x^2, the function is x/3 Taking the derivative of a function with respect to something basically means you are determining what the derivative function is doing to the term that you re talking about. Hope that helps!",
        "video_name": "Mci8Cuik_Gw",
        "timestamps": [
            175
        ],
        "3min_transcript": "So this is the x power in yellow. And so let's do that right over here. So instead of taking the derivative with respect to x of 2 to the x, let's say, let's just take the derivative with respect to x of the exact same expression rewritten, of e to the natural log of 2 raised to the x power. Let me put this x in that same color, dx. Now we know from our exponent properties if we raise something to some power, and then raise that to another power, we can take the product of the two powers. Let me rewrite this just to remember. If I have a to b, and then I raise that to the c power, this is the exact same thing as a to the b times c power. So we can utilize that exponent property right here to rewrite this as being equal to the derivative with respect 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"
    },
    {
        "Q": "at 2:17, we could just treat e^(ln 2)^x as a function and use the chain rule to differentiate it, couldn't we?\nbut the ans is different",
        "A": "Sal has been solving e^(ln2*x) not e^(ln2)^x. I have hope it is a typo.",
        "video_name": "Mci8Cuik_Gw",
        "timestamps": [
            137
        ],
        "3min_transcript": "Let's see if we can take the derivative with respect to x of 2 to the x power. And you might say, hold on a second. We know how to take the derivative of e to the x. But what about a base like 2? We don't know what to do with 2. And the key here is to rewrite 2 to the x so that we essentially have it as e to some power. And the key there is to rewrite 2. So how can we rewrite 2 so it is e to some power? Well, let's think about what e to the natural log of 2 power is. The natural log of 2 is the power that I would have to raise e to to get to 2. So if we actually raise e to that power, we are going to get to 2. So what we could do, instead of writing 2 to the x, we could rewrite this as e. We could rewrite 2 as e to the natural log of 2, So this is the x power in yellow. And so let's do that right over here. So instead of taking the derivative with respect to x of 2 to the x, let's say, let's just take the derivative with respect to x of the exact same expression rewritten, of e to the natural log of 2 raised to the x power. Let me put this x in that same color, dx. Now we know from our exponent properties if we raise something to some power, and then raise that to another power, we can take the product of the two powers. Let me rewrite this just to remember. If I have a to b, and then I raise that to the c power, this is the exact same thing as a to the b times c power. So we can utilize that exponent property right here to rewrite this as being equal to the derivative with respect 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"
    },
    {
        "Q": "At 3:30 and onwards, how come the 1 in \"anti-deriv: 1+sinx\" isnt accounted for? shouldn't its anti-derivative be x?",
        "A": "It s not 1+sin(x), it s 1*sin(x). Which equals sin(x). Or 1*1*1*1*1*1*1*1*sin(x), depending on weather.",
        "video_name": "bZ8YAHDTFJ8",
        "timestamps": [
            210
        ],
        "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 1:52 Sal writes g'(x) = cosx and g(x) = sinx and not {sinx+C}. Can someone explain?",
        "A": "Sal chose a convenient antiderivative. Any choice for an antiderivative would yield the same final result. Try the same process choosing a nonzero number for the constant. See what result you get.",
        "video_name": "bZ8YAHDTFJ8",
        "timestamps": [
            112
        ],
        "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": "At 3:40, why isn't the anti derivative of 1 added to -cos x..won't it be x times -cos x?",
        "A": "Hmm, are you trying to apply the product rule ? That is only for derivatives. In this case, we are doing the anti-derivative (integral). So when he wants to do the anti-derivative: \u00e2\u0088\u00ab1\u00e2\u008b\u0085sin(x)dx = \u00e2\u0088\u00absin(x)dx = -cos(x) + C the 1 times inside the integral has no effect, since 1\u00e2\u008b\u0085sin(x) = sin(x). Or maybe you had some other reason in mind?",
        "video_name": "bZ8YAHDTFJ8",
        "timestamps": [
            220
        ],
        "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 6:26 Sal Khan says \"... our variance is essentially the probability of success times the probability of failure.\" Mathematically I understand this (Khan walks us through the derivation) but conceptually I don't. If variance is some measure of the spread of values around the mean, how does the product of the probably of success and failure describe the variance?",
        "A": "To me, p(1-p) seems like an algebraic simplification without a conceptual component to it. Sometimes algebraic simplifications lead to more conceptual insight, but this one really doesn t.",
        "video_name": "ry81_iSHt6E",
        "timestamps": [
            386
        ],
        "3min_transcript": "This right here is going to be the variance. Now let's actually work this out. So this is going to be equal to 1 minus p. Now 0 minus p is going to be negative p. If you square it you're just going to get p squared. So it's going to be p squared. Then plus p times-- what's 1 minus p squared? 1 minus p squared is going to be 1 squared, which is just 1, minus 2 times the product of this. So this is going to be minus 2p right over here. And then plus negative p squared. So plus p squared just like that. And now let's multiply everything out. This is going to be, this term right over here is going to be p squared minus p to the third. going to be plus p times 1 is p. p times negative 2p is negative 2p squared. And then p times p squared is p to the third. Now we can simplify these. p to the third cancels out with p to the third. And then we have p squared minus 2p squared. So this right here becomes, you have this p right over here, so this is equal to p. And then when you add p squared to negative 2p squared you're left with negative p squared minus p squared. And if you want to factor a p out of this, this is going to be equal to p times, if you take p divided p you get a 1, p square divided by p is p. So p times 1 minus p, which is a pretty neat, clean formula. So our variance is p times 1 minus p. And if we want to take it to the next level and figure out square root of the variance, which is equal to the square root of p times 1 minus p. And we could even verify that this actually works for the example that we did up here. Our mean is p, the probability of success. We see that indeed it was, it was 0.6. And we know that our variance is essentially the probability of success times the probability of failure. That's our variance right over there. The probability of success in this example was 0.6, probability of failure was 0.4. You multiply the two, you get 0.24, which is exactly what we got in the last example. And if you take its square root for the standard deviation, which is what we do right here, it's 0.49. So hopefully you found that helpful, and we're going to build on this later on in some of our inferential statistics."
    },
    {
        "Q": "at around 5:37, Sal divided the 2 and the 12, but, don't you have to divide the 27 too or am I just forgetting a rule?",
        "A": "(27*12)/2 = (3*3*3*2*2*3)/2 now you can cancel out factors",
        "video_name": "8C5kAIKLcZo",
        "timestamps": [
            337
        ],
        "3min_transcript": "And that makes sense because we should have more feet than yards. And actually, this should be three times more, so everything makes sense. 27/2 is 3 times 9/2. So now we have 27/2 feet, and now we want to convert this to inches. And we just have to remember there are 12 inches per yard. And we're going to want to multiply by 12, because however many feet we have, we're going to have 12 times as many inches. If we have 1 foot, we're going to have 12 inches, 2 feet, 24 inches. 27/2 feet, we're going to multiply it by 12 to get the number of inches. Since this is going to be times 12, and we'll make sure the dimensions work out: 12 inches per foot. And the feet and the foot, this is just the plural and It's the same dimension. This will cancel out. So this will be-- if we just rearrange the multiplication, view it as everything is getting multiplied, and when you just multiply a bunch of things, order doesn't matter. So this is equal to 27/2 times 12 feet. I'm just swapping the order. Feet times inches divided by feet, or foot, just the singular of the same word. The feet and the foot cancel out, they're the same unit. And you have 27 times 12 divided by 2 inches. And what we could do here is that our final answer is going to be 27 times 12/2 inches. And before we multiply the 27 times 12 and then divide by 2, you immediately see, well, I can just divide 12 by 2, and 2 by 2, and it makes our computation simpler. It becomes 27 times 6 inches, and let's figure out what that is. 27 times 6. 7 times 6 is 42. 2 times 6 is 12, plus 4 is 16. This is equal to 162 inches, which makes sense. 4 and 1/2 yards, that gets us to this number right here: 27 divided by 2 is 13 and 1/2 feet. You multiply that by 12, it makes sense. You're going to have a bunch of inches. 162 inches."
    },
    {
        "Q": "In 0:26 what does compute mean",
        "A": "Compute means to calculate, find, or figure out. He will show how to find the answer.",
        "video_name": "twMdew4Zs8Q",
        "timestamps": [
            26
        ],
        "3min_transcript": "Let's multiply 9 times 8,085. That should be a pretty fun little calculation to do. So like always, let's just rewrite this. So I'm going to write the 8,085. I'm going to write the 9 right below it and write our little multiplication symbol. And now, we're ready to compute. So first we can tackle 9 times 5. Well, we know that 9 times 5 is 45. We can write the 5 in the ones place and carry the 5 to the tens place. So 9 times 5 is 45. Now we're ready to move on to 9 times 8. And we're going to calculate 9 times 8 and then add the 4 that we just carried. So 9 times 8 is 72, plus the 4 is 76. So we'll write the 6 right here the tens place and carry the 7. looking for a suitable color. 9 times 0 100's plus-- and this is a 7 in the hundreds place, so that's actually 700. Or if we're just kind of going with the computation, 9 times 0 plus 7. Well, 9 times 0 is 0, plus 7 is 7. And then, finally, we have-- and once again, I'm looking for a suitable color-- 9 times 8. This is the last thing we have to compute. We already know that 9 times 8 is 72. And we just write the 72 right down here, and we're done. 8,085 times 9 is 72,765. Let's do one more example just to make sure that this is really clear in your brain, at least the process for doing this. And I also want you to think about why this works. So let's try 7 times 5,396. I'm going to rewrite it-- 5,396 times 7. First, we'll think about what 7 times 6 is. We know that's 42. We'll put the 2 in the ones place. 4 we will carry. Then we need to concern ourselves with 7 times 9. But then, we have to calculate that and then add the 4. 7 times 9 is 63, plus 4 is 67. So we put the 7 down here and carry the 6. Then we have to worry about 7 times 3 plus this 6 that we had just finished carrying. 7 times 3 is 21, plus 6 is 27."
    },
    {
        "Q": "why at around 3:00 is the x2 put inside of the square root symbol would it not be in the front like the 8",
        "A": "You are not seeing it correctly. The x2 is not inside the radical.",
        "video_name": "Z3db5itCIiQ",
        "timestamps": [
            180
        ],
        "3min_transcript": "And I've simplified a little bit, I've done no rationalizing just yet, and it looks like there is a little more simplification I can do first. Because everything in the numerator and everything in the denominator is divisible by 2. So lets divide the numerator by 2. So if you divide the numerator by 2, 16 divided by 2, or you could view it as multipying the numerator and denominator by one half. So 16 times one half is 8. 2X squared times one half is just X squared. And then 2 times the principle square root of 2 times one half is just the square root of 2. It is 1 square roots of 2. So this whole thing has simplified to 8 plus X squared, all of that over the square root of 2. And now lets rationalize this. So lets do that. So times the principle square root of 2 over the principle square root of 2. Now just to show that it works on the denominator what is the principle square root of 2 times the principle square root of 2? Well its going to be 2. And in our numerator, we are going to distribute this term onto both terms in this expression, so you have 8 times the principle square root of 2 plus the square root of 2 times X squared. And we could consider this done, we have simplified the expression, or if you want you could break it up. You could say this is the same thing as 8 square roots of 2 over 2, which is 4 square roots of 2, plus the square root of 2 times X squared over 2. So depending on your tastes, you might view this as more simple or this as more simple but both are equally valid. We could have rationalized right from the get go. Let me start with our original problem. So our original problem was 16 + 2X squared, all of that over the principle square root of 8. We could have rationalized from the get go by multiplying the numerator and the denominator by the principle square root of 8. And so in our denominator we'll just get 8. And then in our numerator we would get 16 times the principle square root of 8, plus 2 times the principle square root of 8X squared. And now we could try and simplify this a little bit more. You can say, well, everything in the numerator and denominator is divisible by 2, so the 16 could become an 8 if you divide by 2. The 2 becomes a 1. And this 8 becomes a 4. And then you get 8 square roots of 8 plus the square root of 8X squared. All of this over 4."
    },
    {
        "Q": "At 2:05 Sal puts in the -9y^2x at the end of the simplified polynomial equation. Could he have put -9y^2x at the beginning? Why did he put the -9y^2x where he did?",
        "A": "The terms in the polynomial can be listed in any order. For example - these are all the same: 4x^2y - 10xy + 45 - 9y^2x (this is Sal s version) 4x^2y - 10xy - 9y^2x + 45 45 - 9y^2x - 10xy + 4x^2y - 9y^2x + 45 + + 4x^2y - 10xy etc.",
        "video_name": "AqMT_zB9rP8",
        "timestamps": [
            125
        ],
        "3min_transcript": "We've got 4x squared y minus 3xy plus 25 minus the entire expression 9y squared x plus 7xy minus 20. So when we're subtracting this entire expression, that's equivalent to subtracting each of these terms individually if we didn't have the parentheses. Or another way of thinking about it-- we could distribute this negative sign. Or you could view this as a negative 1 times this entire expression. And we can distribute it. So let's do that. So let me write this first expression here. I'm going to write it unchanged. So it is 4x squared y minus 3xy plus 25. And now let me distribute the negative 1, or the negative sign times all of this stuff. So negative 1 times 9y squared x is negative 9y squared x. Negative 1 times 7xy is negative 7xy. And then negative 1 times 20 is positive 20. And we just want to group like terms. So let's see, is there another x squared y term anywhere? No, I don't see one. So I'll just rewrite this. So we have 4x squared y. Now, is there another xy term? Yeah, there is. So we can group negative 3xy and negative 7xy. Negative 3 of something minus another 7 of that something is going to be negative 10 of that something. So it's negative 10xy. And then we have a 25, which is just a constant term. Or an x to the 0 term. It's 25x to the 0. You could view it that way. And there's another constant term right over here. We can always add 25 to 20. That gives us 45. And then we have this term right over here, which clearly can't be merged with anything else. So minus 9y squared. Let me do that in that original color. Minus 9-- I'm having trouble shifting And we are done."
    },
    {
        "Q": "At 2:50, why not just keep it (a+2)(a-2) and then cancel out (a+2) from the top and bottom? Why does this not work to simplify?",
        "A": "I see. So (sorry if this is worded wrong) you have to combine the -(a-3) before simplifying for the same reason you can t cancel out the two a s in (a + b)/a ?",
        "video_name": "IKsi-DQU2zo",
        "timestamps": [
            170
        ],
        "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 4:37, a^2-a is just a. I don't understand why the answer isn't just a-1. I guess that is another one of those \"algebra\" rules that people are just supposed to magically know?",
        "A": "You cannot subtract a^2 and a. Because they are not like terms. Powers are related to mutliplication, and will not be affected using addition and subtraction. And no, you are not supposed to magically know this - lol - but you should have learned about combining like terms before you started working on rational expressions. If you missed that in one of your previous courses, you can certainly go back and review this skill. It should be listed under the Algebra I content.",
        "video_name": "IKsi-DQU2zo",
        "timestamps": [
            277
        ],
        "3min_transcript": "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 number right here would be divisible by 2, it's not divisible by 2. So, a plus 2 is not one of the factors here, so there's not going to be any more simplification, even if we were able to factor this thing, and the numerator out. we're done. We have simplified the rational expression, and the domain is for all a's, except for a cannot, or, all a's given that a does not equal negative 2-- all a's except for negative 2. And we are done."
    },
    {
        "Q": "what is a real number 1:54\nis there such a thing as a fake number",
        "A": "Imaginary numbers exist, and they exist because sometimes one needs to express the square root of a negative number, which doesn t exist. i is the central imaginary number, and it stands for the square root of negative one.",
        "video_name": "IKsi-DQU2zo",
        "timestamps": [
            114
        ],
        "3min_transcript": "Find the difference. Express the answer as a simplified rational expression, and state the domain. We have two rational expressions, and we're subtracting one from the other. Just like when we first learned to subtract fractions, or add fractions, we have to find a common denominator. The best way to find a common denominator, if were just dealing with regular numbers, or with algebraic expressions, is to factor them out, and make sure that our common denominator has all of the factors in it-- that'll ensure that it's divisible by the two denominators here. This guy right here is completely factored-- he's just a plus 2. This one over here, let's see if we can factor it: a squared plus 4a plus 4. Well, you see the pattern that 4 is 2 squared, 4 is 2 times 2, so a squared plus 4a plus 4 is a plus 2 times a plus 2, or a plus 2 squared. We could say it's a plus 2 times a plus 2-- that's what a squared plus 4a plus 4 is. This is obviously divisible by itself-- everything is divisible by itself, except, I guess, for 0, is divisible by 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."
    },
    {
        "Q": "6:09 .... did Vi just say \"damn\"?",
        "A": "No she said Bam!",
        "video_name": "EdyociU35u8",
        "timestamps": [
            369
        ],
        "3min_transcript": "Which makes you really want to know what you get if you do the first thing. But instead of always starting with zigging out, you alternate starting zig out and zig in. And it's kind of bumping into itself. And this would definitely be more perfect with graph paper OK, now it's starting to look like a triangle? On the one hand, not nearly as cool as a spirally thing you get when your zigs always go the same way. On the other hand, why would you get a triangle? And it's like a solid triangle too. If you kept doing this forever, would the triangle just fill up completely? These didn't do that. Although with this thing you have sections that are starting to fill up. Maybe here at some point that will happen, though it seems like it's just full of holes forever. You kind of wish you had a way to take some graph paper and skip all the way to what happens later in the sequence. Maybe if you had some sort of diagram, like-- this has a line with one right turn. The next one goes right, right, left. And then the next goes right, right, left, right, right, left, left. Is there a rule? Maybe it's just like the zig zag zag zigs. 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."
    },
    {
        "Q": "What does Vi mean at 0:19?",
        "A": "Pi is not infinite. A number is infinite, by definition, if it is greater than all positive integers. Pi is less than 4, so pi is not infinite.",
        "video_name": "5iUh_CSjaSw",
        "timestamps": [
            19
        ],
        "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": "Isn't the triangle person from 4:00 Wind from wind and mr. ug?",
        "A": "No, the triangle is actually Vi s symbol for herself. And, as you probably know, Vi loves triangles.",
        "video_name": "5iUh_CSjaSw",
        "timestamps": [
            240
        ],
        "3min_transcript": "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, cantabile infinity looks like zero. So, I don't know why anyone would make a fus about the grand infinities and forever, as about the boring little number like Pi. And of course, those are just the first couple of kinds of infinities in an infinite number of infinities in their correspondingly more in-between-[finiter] numbers like the [infinitesimals]. So, don't let Pi impress you by being a member of an unaccountably infinite set of in-between-infinite number either. The only thing even a little weird about Pi is that you do get an irrational number by taking such a simple ratio of such a simple geometric object. Surely that never happens with other simple ratios of other simple geometric objects. Oh wait! Their in everything! What are the chances? No! Let's pretend math equals arithmetic, and then get all surprised and amazed when the moment you leave arithmetic that you get a non-arithmetic number as if it were some odd unpredictable phenomenon. That way, by the time you get to calculus you won't have any idea what's going on to pass your class without ever realizing that you were dealing with infinities two levels deeper than the infinity you think is so cool when Pi does it. Pi is not special. Yeah, Pi can be fun, and I'd never deny you your deserts, but maybe try some real food once in a while."
    },
    {
        "Q": "At 4:54, how does this equation with all the ones = (x=(x+x)/1)?",
        "A": "Since the fractal fraction is infinitely big, you can say that since infinity minus 1 is still infinity,",
        "video_name": "a5z-OEIfw3s",
        "timestamps": [
            294
        ],
        "3min_transcript": "Or 8 square root 13. Or you could even make each layer different. 7, 8, 9, 10, 11. Now look how confusing this is. Awesome. Say you wanted to actually solve one of these things. Say you started with this puzzle. What is 1/1 plus 1, but each 1 is over 1 plus 1? And so on, all the way to infinity. You could try doing it by hand, thinking maybe it'll converge on something. 1/1 plus 1 is 1/2. So the next layer, these are 1/2, add up to 1. So this is 1/1, 1. Three layers, back to 1/2. Uh oh. Any whole number of layers is going to give either 1 or 1/2. So what could this possibly be? Well, you could try doing algebra to it. Say all this equals x. Look, you've isolated x on one side, and everything else on the other, and it doesn't help one bit. Take that math teacher. OK, but if all this is x, then all this-- which is the same as all this-- is x. You can write this as 1/x plus x, which completely works. You could generate it all again by replacing x with 1/x plus x. x's on the wrong side of the equation, you can solve it and get the boring way to write this number if you wanted. One last fraction, this one with a caution sign. Say you want something to equal 1. Split 1 into 1/2 plus 1/2. Now these 1's could be replaced with 1/2 plus 1/2. Each time you do this, it works. What happens if you go to infinity? It's weird because if you look at any number of layers of 2's, to see if it converges to something, the result is always 2 for each fraction. Which might make you think that at infinity it's also 2 for each fraction, and therefore 1 equals 4? And just looking at this and trying to take it backwards, you might say, all this equals x, and all this equals x. So it's x plus x/2. Just try and solve that equation. The problem is, half of something plus half of something always equals that something, no matter what the x. So this could be anything, it's undefined. Or say you want to make something with all 1's Now x equals x plus x/1, or x equals x plus x. You can algebra your way to a contradiction, and as far as algebra is concerned, this is undefined. two numbers I know that fit this description, infinity and 0. This I suppose could be either, or both at once, or nothing I don't know. Why does it do that? Maybe because the numerator got lost up there, and could have been anything. Interesting though that even here, when the denominator got lost in infinity, you can still solve this back to 5. That's, to me, the cool part about algebra. Unlike the neat little problems they put in grade school textbooks, not all problems can be solved, and it's not always obvious when there's an answer and when there's not. Weird stuff happens all the time. And most importantly, algebra isn't a dead ancient thing. There are things no one's ever done before, that you can do with the simplest concepts. As simple as that x is what x is."
    },
    {
        "Q": "How come at 5:00 Vi says 2 = 1?",
        "A": "no, the equation is x=2x. so you bring x on the right side to the left side to get x/x=2. Then, any number divided by that number is always 1. So, 1=2",
        "video_name": "a5z-OEIfw3s",
        "timestamps": [
            300
        ],
        "3min_transcript": "Or 8 square root 13. Or you could even make each layer different. 7, 8, 9, 10, 11. Now look how confusing this is. Awesome. Say you wanted to actually solve one of these things. Say you started with this puzzle. What is 1/1 plus 1, but each 1 is over 1 plus 1? And so on, all the way to infinity. You could try doing it by hand, thinking maybe it'll converge on something. 1/1 plus 1 is 1/2. So the next layer, these are 1/2, add up to 1. So this is 1/1, 1. Three layers, back to 1/2. Uh oh. Any whole number of layers is going to give either 1 or 1/2. So what could this possibly be? Well, you could try doing algebra to it. Say all this equals x. Look, you've isolated x on one side, and everything else on the other, and it doesn't help one bit. Take that math teacher. OK, but if all this is x, then all this-- which is the same as all this-- is x. You can write this as 1/x plus x, which completely works. You could generate it all again by replacing x with 1/x plus x. x's on the wrong side of the equation, you can solve it and get the boring way to write this number if you wanted. One last fraction, this one with a caution sign. Say you want something to equal 1. Split 1 into 1/2 plus 1/2. Now these 1's could be replaced with 1/2 plus 1/2. Each time you do this, it works. What happens if you go to infinity? It's weird because if you look at any number of layers of 2's, to see if it converges to something, the result is always 2 for each fraction. Which might make you think that at infinity it's also 2 for each fraction, and therefore 1 equals 4? And just looking at this and trying to take it backwards, you might say, all this equals x, and all this equals x. So it's x plus x/2. Just try and solve that equation. The problem is, half of something plus half of something always equals that something, no matter what the x. So this could be anything, it's undefined. Or say you want to make something with all 1's Now x equals x plus x/1, or x equals x plus x. You can algebra your way to a contradiction, and as far as algebra is concerned, this is undefined. two numbers I know that fit this description, infinity and 0. This I suppose could be either, or both at once, or nothing I don't know. Why does it do that? Maybe because the numerator got lost up there, and could have been anything. Interesting though that even here, when the denominator got lost in infinity, you can still solve this back to 5. That's, to me, the cool part about algebra. Unlike the neat little problems they put in grade school textbooks, not all problems can be solved, and it's not always obvious when there's an answer and when there's not. Weird stuff happens all the time. And most importantly, algebra isn't a dead ancient thing. There are things no one's ever done before, that you can do with the simplest concepts. As simple as that x is what x is."
    },
    {
        "Q": "At 4:47, I'm a little confused why he multiplied each side by 2... I keep rewatching it and trying to understand but it's just not clicking with me",
        "A": "He s trying to relate the inequality to the definition of f(x). Since f(x)=2x in the region of interest, and the inequality only has a x, he is multiplying the inequality by 2 in order to get the definition of f(x) in the inequality.",
        "video_name": "0sCttufU-jQ",
        "timestamps": [
            287
        ],
        "3min_transcript": "this is 5 minus delta. So that's our range we're going to think about. We're going to think about it in the abstract at first. And then we're going to try to come up with a formula for delta in terms of epsilon. So how could we describe all of the x's that are in this range but not equal to 5 itself? Because we really care about the things that are within delta of 5, but not necessarily equal to 5. This is just a strictly less than. They're within a range of C, but not equal to C. Well, that's going to be all of the x's that satisfy x minus 5 is less than delta. That describes all of these x's right over here. And now, what we're going to do, and the way these proofs typically go, is we're going to try to manipulate the left-hand side of the inequality, so it starts to look something like this, or it starts to look exactly like that. of the inequality is going to be expressed in terms of delta. And then we can essentially say well, look. If the right-hand side is in terms of delta and the left-hand side looks just like that, that really defines how we can express delta in terms of epsilon. If that doesn't make sense, bear with me. I'm about to do it. So, if we want x minus 5 to look a lot more like this, when x is not equal to 5-- in all of this, this whole interval, x is not equal to 5-- f of x is equal to 2x, our proposed limit is equal to 10. So if we could somehow get this to be 2x minus 10, then we're in good shape. And the easiest way to do that is to multiply both sides of this inequality by 2. And 2 times the absolute value of something, that's the same thing as the absolute value of 2 times that thing. If I were to say 2 times the absolute value of a, that's the same thing as the absolute value of 2a. 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."
    },
    {
        "Q": "At 2:50, why does he go from 9 down to 1 to solve for the denominator. In other problems I thought you would continue from (in this case) 28, 27 ........3,2,1. Why is this different?\nSorry if this is confusing!",
        "A": "The problem is different since there are 9 cards in a hand and to fill each slot it will be 9! which is 9 down to 1",
        "video_name": "SbpoyXTpC84",
        "timestamps": [
            170
        ],
        "3min_transcript": "there's 36. But then that's now part of my hand. Now for the second slot, how many will there be left to pick from? Well, I've already picked one, so there will only be 35 to pick from. And then for the third slot, 34, and then Then 33 to pick from, 32, 31, 30, 29, and 28. So you might want to say that there are 36 times 35, times 34, times 33, times 32, times 31, times 30, times 29, times 28 possible hands. Now, this would be true if order mattered. This would be true if I have card 15 here. Maybe I have a-- let me put it here-- maybe I have a 9 of spades here, and then I have a bunch of cards. And maybe I have-- and that's one hand. And then I have another. So then I have cards one, two, three, four, five, six, seven, eight. 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."
    },
    {
        "Q": "At 6:16 why did he subtract (36-9)!?",
        "A": "36! means 36*35*34*33*32*31*30*29*28*27*26*25*24*23*22....*1 But since we want it to stop at 28, we have to cancel the rest out by dividing with 27!(27*26*25*24*23*22*21*20*19*18*17....*1)",
        "video_name": "SbpoyXTpC84",
        "timestamps": [
            376
        ],
        "3min_transcript": "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. That there are 94,143,280 possible 9 card hands in this situation. Now, we kind of just worked through it. We reasoned our way through it. There is a formula for this that does essentially the exact same thing. And the way that people denote this formula is to say, look, we have 36 things and we are going to choose 9 of them. And we don't care about order, so sometimes it'll be written as n choose k. Let me write it this way. So what did we do here? We have 36 things. We chose 9. So this numerator over here, this was 36 factorial. But 36 factorial would go all the way down to 27, 26, 25. It would just keep going. But we stopped only nine away from 36. So this is 36 factorial, so this part right here, that part right there, is not just 36 factorial. What is 36 minus 9? It's 27. So 27 factorial-- so let's think about this-- 36 factorial, it'd be 36 times 35, you keep going all the way, times 28 times 27, going all the way down to 1. That is 36 factorial. Now what is 36 minus 9 factorial, that's 27 factorial. So if you divide by 27 factorial, 27 factorial is 27 times 26, all the way down to 1. Well, this and this are the exact same thing. This is 27 times 26, so that and that would cancel out. So if you do 36 divided by 36, minus 9 factorial, you just get the first, the largest nine terms of 36 factorial, which is exactly what we have over there. And then we divided it by 9 factorial."
    },
    {
        "Q": "At time5:14 min, shouldn't the answer be 362880...i think u shud verify...u mistakenly pressed smth on ur calculator..i calculated it to be 36880",
        "A": "9! = 362880. That is how many ways there are to arrange 9 objects. But Sal is calculating the number of 9-card hands, so he needs to start with 36 objects. He is dividing the number of permutations of 9 cards (from a set of 36), by the number of ways to arrange those 9 cards.",
        "video_name": "SbpoyXTpC84",
        "timestamps": [
            314
        ],
        "3min_transcript": "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. That there are 94,143,280 possible 9 card hands in this situation. Now, we kind of just worked through it. We reasoned our way through it. There is a formula for this that does essentially the exact same thing. And the way that people denote this formula is to say, look, we have 36 things and we are going to choose 9 of them. And we don't care about order, so sometimes it'll be written as n choose k. Let me write it this way. So what did we do here? We have 36 things. We chose 9. So this numerator over here, this was 36 factorial. But 36 factorial would go all the way down to 27, 26, 25. It would just keep going. But we stopped only nine away from 36. So this is 36 factorial, so this part right here, that part right there, is not just 36 factorial."
    },
    {
        "Q": "At 10:09 How do you find the surface of a rectangle",
        "A": "Surface area of a rectangle= its length multiply wide.",
        "video_name": "I9eLKDbc8og",
        "timestamps": [
            609
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At about 2:23, to figure out the 1.5 portion Sal says \"15 times 15 is 225\" and gets 2.25 that way. What is this process called?",
        "A": "It s actually 1.50 times 1.50 and the answer is 2.25.",
        "video_name": "I9eLKDbc8og",
        "timestamps": [
            143
        ],
        "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 2:01 what is that line over repeating decimals called ?",
        "A": "The line over the repeating decimal can be called a vinculum . In a repeating decimal, the vinculum is used to indicate the group of repeating digits.",
        "video_name": "d9pO2z2qvXU",
        "timestamps": [
            121
        ],
        "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 0:19, when Sal said 8/2 is not a perfect square, what's the difference between a perfect square and a non-perfect square?",
        "A": "A perfect square is when a whole number is the square root of the number. like \u00e2\u0088\u009a49=7 But \u00e2\u0088\u009a74 is a non-perfect square.",
        "video_name": "d9pO2z2qvXU",
        "timestamps": [
            19
        ],
        "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 0:15, Sal says:\nIf you take the square root of a number that is not a perfect square, it is going to be irrational.\n\nWhat about sqrt 2.25? Isn't that rational? Does the rule he stated only apply to integers?",
        "A": "Sal is correct. The definition of perfect square is a number times itself is the square. Since 1.5*1.5=2.25, 1.5^2=2.25, therefore 1.5 is the number multiplied by itself, so 2.25 is the perfect square 1.5. So in this case it works. The hope is that the perfect square will be the nice and easy whole numbers.",
        "video_name": "d9pO2z2qvXU",
        "timestamps": [
            15
        ],
        "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 5:17 why do you have to times everything with -2?",
        "A": "Because by doing so, we can get 200m in one equation and -200m in the other, which allows us to solve using elimination. We can add the two equations and be left with only one variable, w.",
        "video_name": "VuJEidLhY1E",
        "timestamps": [
            317
        ],
        "3min_transcript": "\" In green. Well, let's think about the total number of bags that the men ate. You had 200 men, [Let me scroll over a little bit] and they each ate m bags per man. \" \"So the man at this first party collectively ate 200 times m bags. If m is 10 bags per man, then this would be 2000. If m was 5 bags per man, then this would be 5000. We don't know what m is, but 200 times m is the total eaten by the man.\""
    },
    {
        "Q": "During the video, specifically 1:00, Sal does not explain WHAT a line of best fit/line of fit is. My question: How does one find a line of best fit, and how do you know that it is true?",
        "A": "Well, a line of best fit is sort of like an estimate. Nobody can truly find the line of best fit (unless all the points are on the same line). Usually, the line of best fit, assuming the points follow a pattern, is the equation of the line connecting the first point and the last point.",
        "video_name": "ioieTr41L24",
        "timestamps": [
            60
        ],
        "3min_transcript": "Find the line of best fit, or mark that there is no linear correlation. So let's see, we have a bunch of data points, and we want to find a line that at least shows the trend in the data. And this one seems a little difficult because if we ignore these three points down here, maybe we could do a line that looks something like this. It seems like it kind of approximates this trend, although this doesn't seem like a great trend. And if we ignored these two points right over, we could do something like maybe something like that. But we can't just ignore points like that. So I would say that there's actually no good line of best fit here. So let me check my answer. Let's try a couple more of these. Find the line of best-- well, this feels very similar. It really feels like there's no-- I mean, I could do that, but I'm ignoring these two points. I could do something like that, then I'd be ignoring these points. So I'd also say no good best fit line exists. So let's try one more. So here it looks like there's very clearly this trend. And I could try to fit it a little bit better than it's fit right now. quite well. I could maybe drop this down a little bit, something like that. Let's check my answer. A good best fit line exists. Let me check my answer. We got it right."
    },
    {
        "Q": "At 1:16 i lost you... why do we regroup? I need another example to kinda clarify it.",
        "A": "Because you don t have enough in the tens place, so you need 1 from the hundred place to add to the tens place.",
        "video_name": "X3JqIZR1XcY",
        "timestamps": [
            76
        ],
        "3min_transcript": "Let's think about different ways that we can represent the number 675. So the most obvious way is to just look at the different place values. So the 6 is in the hundreds place. It literally represents 600. So that's 600. I'm going to do that in the red color-- 600. The 7 is in the tens place. It represents 7 tens, or 70. And then the 5 in the ones place. It represents 5. So let me copy and paste this and then think about how we can regroup the value in the different places to represent this in different ways. So let me copy and let me paste it, and maybe I'll do it three times. So let me do it once, and let me do it one more time. So one thing that we could do is we could regroup from one place to the next. So, for example, we could take if we wanted to-- we could take 1 from the hundreds place. That's essentially taking 100 away. So this is really making this a 500. And we could give that 100 to-- well, we could actually give it to either place, but let's give it to the tens place. So we're going to give 100 to the tens place. Now, if you give 100 to the tens place and you already had 70 there, what's it going to be equal to? Well, it's going to be equal to 170. Well, how would we represent that is tens? Well, 170 is 17 tens. So we could just say that 7 becomes 17. Now, we could keep doing that. We could regroup some of this value in the tens place to the ones place. So, for example, we could give 10 from the tens place and give it to the ones place. So let's take 10 away from here. So that becomes 160. This becomes 16. And let's give that 10 to the ones place. Well, 10 plus 5 is 15. So this 5 is now a 15. Let's do another scenario. Let's do something nutty. Let's take 200 from the hundreds place. So this is going to now 4, and this is going to become 400. That's what this 4 now represents. And let's give 100 to the tens place. And let's give another 100 to the ones place. So in other words, I'm just regrouping that 200. Those 200's, I've taken from the hundreds place, and I'm going to give it to these other places. So now the tens place is going to be 170. We're going to have 170 here, which is 17 tens. So you could say that the tens place is now 17."
    },
    {
        "Q": "At 1:50, why does Sal square the denominator?",
        "A": "because it was originally 2c in the denominator INSIDE the big parentheses with the squared exponent. In order to move the 2c outside of the parentheses the operation around those parentheses must operate on the 2c. Since the exponent says square everything inside these parentheses you must square the 2c to eliminate those parentheses around the denominator. Notice after the 2c is squared, the parentheses with the squared exponent still surrounds the numerator, but no longer surrounds the denominator.",
        "video_name": "nZu7IZLhJRI",
        "timestamps": [
            110
        ],
        "3min_transcript": "the last video, I claimed that this result we got for the area of a triangle that had sides of length a, b, and c is equivalent to Heron's formula. And what I want to do in this video is show you that this is equivalent to Heron's formula by essentially just doing a bunch of algebraic manipulation. So the first thing we want to do-- let's just spring this 1/2 c under the radical sign. So 1/2 c, that's the same thing as the square root of c squared over 4. You take the square root of that you get 1/2 c. So this whole expression is equal to-- instead of drawing the radical, I'll just write the square root of this, of c squared over 4 times all of this. I'll just copy and paste it. Copy and paste. So times all of that. And of course, it has to be distributed. And then we have to close the square root. Let me just distribute the c squared over 4. This is going to be equal to the square root. This is going to be hairy, but I think you'll find it satisfying to see how this could turn into something as simple as Heron's formula. The square root of c squared over 4 times a squared is c squared a squared over 4, minus c squared over 4. I'm just distributing this. And I'm going to write it as the numerator squared over the denominator squared. So times c squared plus a squared minus b squared, squared. Over-- if I square the denominator that's 4c squared. And we immediately see that c squared and that c squared are going to cancel out. Let me close all of the parentheses just like that. And, of course, this 4 times that 4, that's going to That's the same thing as 4 squared. And I'm instead of writing 16, you'll see why I'm writing that. Now this I can rewrite. This is going to be equal to the square root-- I'm arbitrarily switching colors-- of ca over 2 squared. This is the same thing as that. I'm just writing it as the whole thing squared. If I square that, that's the c squared a squared over 2 squared over 4, minus-- and I'm going to write this whole thing as an expression squared. So that's c squared plus a squared minus b squared, over 4. And we are squaring both the numerator and the denominator. Now this might look a little bit interesting to you. Let me make the parentheses in a slightly different color. You might remember from factoring polynomials that if I have something of the form x squared minus y squared, that"
    },
    {
        "Q": "at 2:50 suppose the question was log10(16)+log10(2) would you than multiply 16*2 or would you still divided it?",
        "A": "log (a) + log (b) = log (ab), provided that both a and b are positive. log (a) - log (b) = log (a/b), provided that both a and b are positive. Thus, log (16) + log (2) = log(16*2) = log (32)",
        "video_name": "Kv2iHde7Xgw",
        "timestamps": [
            170
        ],
        "3min_transcript": "properties that we know over here. We also know that if we have a logarithm-- let me write it this way, actually-- if I have b times the log base a of c, this is equal to log base a of c to the bth power. And we also know, and this is derived really straight from both of these, is that if I have log base a of b minus log base a of c, that this is equal to the log base a of b over c. And this is really straight derived from these two right over here. Now with that out of the way, let's see what we can apply. So right over here, we have all the logs are the same base. And we have logarithm of x plus logarithm of 3. So by this property right over here, the sum of logarithms with the same base, this is going to be equal to log base 3-- sorry, 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,"
    },
    {
        "Q": "at 2:41, could i cut the log's base and get 3x= 16-2, simplifying more early than the exposed in video or is it mandatory to use the properties explained?\nMost things in mathematics can be simplified before to reach resolution.\nThis will be the first out rule.",
        "A": "No, you cannot the simplify the log base at log 3x = log 16 - log 2 Simplifying the log base should be done after the two logs on the right are combined. If you finish the work on your question you will find that you get 14/3 and not the correct answer of 8/3.",
        "video_name": "Kv2iHde7Xgw",
        "timestamps": [
            161
        ],
        "3min_transcript": "properties that we know over here. We also know that if we have a logarithm-- let me write it this way, actually-- if I have b times the log base a of c, this is equal to log base a of c to the bth power. And we also know, and this is derived really straight from both of these, is that if I have log base a of b minus log base a of c, that this is equal to the log base a of b over c. And this is really straight derived from these two right over here. Now with that out of the way, let's see what we can apply. So right over here, we have all the logs are the same base. And we have logarithm of x plus logarithm of 3. So by this property right over here, the sum of logarithms with the same base, this is going to be equal to log base 3-- sorry, 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,"
    },
    {
        "Q": "why did you subtracted the -3 with the 4x in 6:10? pls help.",
        "A": "(x + 7)(4x - 3) is equal to 4x(x + 7) - 3(x + 7). You just distributed it. They re still the same and it s appropriate to write it in that way.",
        "video_name": "X7B_tH4O-_s",
        "timestamps": [
            370
        ],
        "3min_transcript": "So I grouped the 28 on the side of the 4. And you're going to see what I mean in a second. If we, literally, group these so that term becomes 4x squared plus 28x. And then, this side, over here in pink, it's plus negative 3x minus 21. Once again, I picked these. I grouped the negative 3 with the 21, or the negative 21, because they're both divisible by 3. And I grouped the 28 with the 4, because they're both divisible by 4. And now, in each of these groups, we factor as much out as we can. So both of these terms are divisible by 4x. So this orange term is equal to 4x times x-- 4x squared divided by 4x is just x-- plus 28x divided by 4x is just 7. Remember, you factor out everything that you can factor out. Well, both of these terms are divisible by 3 or negative 3. So let's factor out a negative 3. And this becomes x plus 7. And now, something might pop out at you. We have x plus 7 times 4x plus, x plus 7 times negative 3. So we can factor out an x plus 7. This might not be completely obvious. You're probably not used to factoring out an entire binomial. But you could view this could be like a. Or if you have 4xa minus 3a, you would be able to factor out an a. And I can just leave this as a minus sign. Let me delete this plus right here. Because it's just minus 3, right? Plus negative 3, same thing as minus 3. So what can we do here? We have an x plus 7, times 4x. 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?"
    },
    {
        "Q": "At 8:40, why is it (x+1)(6x+1) instead of 7(x+1)?",
        "A": "Multiply the factors... only the correct factors will create the original polynomial of 6x^2 + 7x + 1 7(x+1) = 7x + 7. This is not the original polynomial. So, these factors can not be correct. (x+1)(6x+1) = 6x^2 + x + 6x + 1 = 6x^2 + 7x + 1. This matches the original polynomial. so, these are the correct factor. Hope this helps.",
        "video_name": "X7B_tH4O-_s",
        "timestamps": [
            520
        ],
        "3min_transcript": "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 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,"
    },
    {
        "Q": "At 0:40, Sal mentions that the technique of factoring by grouping becomes obsolete once you learn the quadratic formula. So, if I have a quadratic and I need to factor it (not just find the zeros, but actually know what it looks like factored, like if I'm trying to simplify a larger rational expression), is there any way to factor it with just the quadratic formula? Or would you have to use one of the factoring techniques?",
        "A": "at 0:40 he is just saying the quadratic equation is easier than rooting.",
        "video_name": "X7B_tH4O-_s",
        "timestamps": [
            40
        ],
        "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": "At 1:20, wouldn't it be a*c instead of a*b? you said a*b but did 4*-21. It just confused me when I did a*b on my homework and it was incorrect",
        "A": "Yes... You need to multiply A*C from the quadratic Ax^2 + Bx + C. Unfortunately, Sal chose the variables a and b to represent the numbers you are seeking to find after multiplying AC. His choice does make the video confusing because we refer to A, B and C as the coefficients in the trinomial. Call the 2 number m and n instead of a and b . You still need to find two numbers that multiply to = AC and also add to = the middle term B. Hope this helps.",
        "video_name": "X7B_tH4O-_s",
        "timestamps": [
            80
        ],
        "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": "At 4:00 Sal states that y=1/k and that this relationship is true for y' and makes a substitution. Would this relationship extend to second, third (etc.) derivatives? Could relationships like this be established for other equations and their derivatives? Feel free to give me a problem!",
        "A": "y=f(x)=1/x. y=1/k for x=k only. You can use it for substitution for y in 2nd, 3rd or nth derivative as long as x=k then y=1/k. Let say you want to use x=2, then y is not 1/k anymore, but y=1/2.",
        "video_name": "FJ7AMaR9miI",
        "timestamps": [
            240
        ],
        "3min_transcript": "the slope of the tangent line? Well, to figure out the slope of the tangent line, let's take the derivative. So if we write f of x, instead writing it as 1/x, I'll write it as x to the negative 1 power. That makes it a little bit more obvious that we're about to use the power rule here. So the derivative of f at any point x is going to be equal to-- well, it's going to be the exponent here is negative 1. So negative 1 times x to the-- now we decrement the exponent to the negative 2 power. Or I could say it's negative x to the negative 2. Now, what we care about is the slope when x equals k. So f prime of k is going to be equal to negative k to the negative 2 power. Or another way of thinking about it, this is equal to negative 1 So this right over here is the slope of the tangent line at that point. Now, let's just think about what the equation of the tangent line is. And we could think about it in slope-intercept form. So we know the equation of a line in slope-intercept form is y is equal to mx plus b, where m is the slope and b is the y-intercept. So if we can get it in this form, then we know our answer. We know what the y-intercept is going to be. It's going to be b. So let's think about it a little bit. This equation, so we could say y is equal to our m, our slope of the tangent line, when x is equal to k, we just figure out to be this business. It equals this thing right over here. So let me write that in blue. Negative 1 over k squared times x plus b. Well, we know what y is when x is equal to k. And so we can use that to solve for b. We know that y is equal to 1/k when x is equal to k. So this is going to be equal to negative 1-- that's not the same color. Negative 1 over k squared times k plus b. Now, what does this simplify to? See, k over k squared is the same thing as 1/k, so this is going to be negative 1/k. So this part, all of this simplifies to negative 1/k. So how do we solve for b? Well, we could just add 1/k to both sides"
    },
    {
        "Q": "At 2:13 he states that 90 x -1/3 = -30 . Can anyone please explain how he got that answer? I understand the finding the common ratio step, but what I don't understand is how whenever I find the common ratio my math doesn't add up. I think I'm multiplying my fractions wrong.",
        "A": "a * (-(b/c)) = a*(-b)*(1/c) = -(ab)/c. 90*(-(1/3)) = 90*(-1)*(1/3) = -90/3",
        "video_name": "pXo0bG4iAyg",
        "timestamps": [
            133
        ],
        "3min_transcript": "In this video I want to introduce you to the idea of a geometric sequence. And I have a ton of more advanced videos on the topic, but it's really a good place to start, just to understand what we're talking about when someone tells you a geometric sequence. Now a good starting point is just, what is a sequence? And a sequence is, you can imagine, just a progression of numbers. So for example, and this isn't even a geometric series, if I just said 1, 2, 3, 4, 5. This is a sequence of numbers. It's not a geometric sequence, but it is a sequence. A geometric sequence is a special progression, or a special sequence, of numbers, where each successive number is a fixed multiple of the number before it. Let me explain what I'm saying. So let's say my first number is 2 and then I multiply 2 by So I multiply it by 3, I get 6. And then I multiply 6 times the number 3, and I get 18. Then I multiply 18 times the number 3, and I get 54. And I just keep going that way. So I just keep multiplying by the number 3. So I started, if we want to get some notation here, this is my first term. We'll call it a1 for my sequence. And each time I'm multiplying it by a common number, and that number is often called the common ratio. So in this case, a1 is equal to 2, and my common ratio is equal to 3. So if someone were to tell you, hey, you've got a geometric sequence. a1 is equal to 90 and your common ratio is equal to negative 1/3. The second term is negative 1/3 times 90. Which is what? That's negative 30, right? 1/3 times 90 is 30, and then you put the negative number. Then the next number is going to be 1/3 times this. So negative 1/3 times this. 1/3 times 30 is 10. The negatives cancel out, so you get positive 10. Then the next number is going to be 10 times negative 1/3, or negative 10/3. And then the next number is going to be negative 10/3 times negative 1/3 so it's going to be positive 10/3. And you could just keep going on with this sequence. So that's what people talk about when they mean a geometric sequence. I want to make one little distinction here. This always used to confuse me because the terms are used very often in the same context. These are sequences. These are kind of a progression of numbers."
    },
    {
        "Q": "Sal, at 3:33 you said a series is a sum of a sequence, but could it also be a group of the sums from a sequence?",
        "A": "Yeah, I guess those can be called sub-series of a sequence.",
        "video_name": "pXo0bG4iAyg",
        "timestamps": [
            213
        ],
        "3min_transcript": "The second term is negative 1/3 times 90. Which is what? That's negative 30, right? 1/3 times 90 is 30, and then you put the negative number. Then the next number is going to be 1/3 times this. So negative 1/3 times this. 1/3 times 30 is 10. The negatives cancel out, so you get positive 10. Then the next number is going to be 10 times negative 1/3, or negative 10/3. And then the next number is going to be negative 10/3 times negative 1/3 so it's going to be positive 10/3. And you could just keep going on with this sequence. So that's what people talk about when they mean a geometric sequence. I want to make one little distinction here. This always used to confuse me because the terms are used very often in the same context. These are sequences. These are kind of a progression of numbers. 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."
    },
    {
        "Q": "At 8:56, he says the formula to find the 12th bounce is (120)(0.6)^n. I thought it was (120)(0.6)^(n - 1). I am kind of confused about that...",
        "A": "In the video, he counts the zero bounce so you can subtract 1 from both sides and then they cancel out. ex. Jumps: a^n=120(0.6)^n-1 Bounces: a^n-1=120(0.6)^n-1 the -1 s cancel so: a^n=120(0.6)^n",
        "video_name": "pXo0bG4iAyg",
        "timestamps": [
            536
        ],
        "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": "at 4:32 why do we have to multiply 50 by 1 hour? I thought all we needed to do was divide 3600 by 50 . Also i did not get whether it was 72 km per second or 1 km per 72 seconds. Although i had some questions this was a fantastic video that triggered a much needed Eureka! moment!! Would i need to multiply 50 by 2 hr if a question said 20/km per 2 hrs or is that not mathematically correct to use 2 hours as a unit",
        "A": "its 1km per 72 seconds. 1/72",
        "video_name": "d5lcGCbV5cM",
        "timestamps": [
            272
        ],
        "3min_transcript": "of kilometers per second. So how could we write 50 kilometers per hour, in terms of kilometers per second? Well it's always good, actually, as a first approximation, to just think about it. If you went this far in an hour, then the number of kilometers you go in a second, is that going to be less, or more? Well a second's a much, much shorter period of time. There's 3,600 seconds in an hour. So you're going to go 1/3,600 of this distance. But let's think about how we would actually work out with the units. Well, we want to get rid of this hours in the denominator. And the plural, obviously the grammar doesn't hold up with the algebra, but this could be hour or hours. So we could think about well, 1 hour-- I'll write an hour in the numerator that's going to cancel with this hour in the denominator. But we want it in terms of seconds. So 1 hour is equal to how many seconds? This is what I meant by saying that using dimensional analysis, which is what I'm doing right now, we can essentially manipulate these units, as we would traditionally do with a variable. So we have hours divided by hours. And so when we do the multiplication, we can multiply the numeric parts. So we have 50 times 1, divided by 3,600. Let me write that. 50 times 1 over 3,600. And then our units left are kilometers per second. Or I could say seconds. So we can play around with the plural and singular parts of it, but I'll just write it as kilometers per second. And so this is 50/3,600. And this fits our intuition. In a second, you're going to go 1/3,600 as far as you would go But let's actually think about what this is equal to. 50/3,600-- so this is going to be the same thing, as-- Let me just simplify it over here. So 50/3,600 is the same thing as 5/360, which is the same thing as-- let me write it this way-- 10/720. And I did that way because that makes it clear that that's the same thing as 1/72. So you could write this as, you're going, this is equal to 1/72 of a kilometer per second. Now I would claim that this is not so reasonable of units for this example right over here."
    },
    {
        "Q": "at 4:10 , just have a silly question: are addition and subtraction between a matrix and a scalar undefined?",
        "A": "Yes, addition and subtraction between a scalar and a matrix (or even between matrices of different dimensions) is undefined. That is why previous to adding, the scalar is multiplied by the Identity Matrix, so that at the time of the addition, you are adding two matrices of the same size.",
        "video_name": "rfm0wQObxjk",
        "timestamps": [
            250
        ],
        "3min_transcript": "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 So what does this mean? So we know that v is a member of the null space of this matrix right here. Let me write this down. v is a member of the null space of lambda I sub n minus A. I know that might look a little convoluted to you right now, but just imagine this is just some matrix B. It might make it simpler. This is just some matrix here, right? That's B. Let's make that substitution. Then this equation just becomes Bv is equal to 0. Now, if we want to look at the null space of this, the null space of B is all of the vectors x that are a member of Rn such that B times x is equal to 0. Well, v is clearly one of those guys, right? Because B times v is equal to 0."
    },
    {
        "Q": "it is 5:00 pm ET, why is the exponent 3/5 after you multiply 3 times 1/5 wouldn't you multiply the numerator and denominator of the exponent against the power of 3 and get 3/15 which can be reduced to 1/5, thouroughly confused over here!",
        "A": "3 does not equal 3/3. 3 as a fraction = 3/1 Thus, 3 times 1/5 = 3/1 * 1/5 = 3/5 Hope this helps.",
        "video_name": "Ht-YXje4R2g",
        "timestamps": [
            300
        ],
        "3min_transcript": "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": "At 2:45: Can 6^11/10 be re-written as 6^1/10?\n10/10 = 1 Leaving 1/10\n6^1 = 6 so the remaining 1/10 is left over.",
        "A": "Actually, we can t change the exponent from 11/10 to 1/10 by subtracting one from the exponent. We could do this however: 6^(11/10) 11/10=10/10+1/10=1+1/10. 6^1=6. 6^(11/10) 6^(10/10+1/10) 6^(1+1/10) 6^1*6^(1/10) 6*6^(1/10) I hope this helps!",
        "video_name": "Ht-YXje4R2g",
        "timestamps": [
            165
        ],
        "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": "ok so at 1:10 say got 4 but how did he get that out of 6?",
        "A": "6 divided by 3 is 2, so 2 is 1/3 of 6. 4 is 2/3 of 6.",
        "video_name": "6dyWKD_JPhI",
        "timestamps": [
            70
        ],
        "3min_transcript": "The graph below contains the rectangle ABCP. Draw the image of ABCP under a dilation whose center is at P and a scale factor is 1 and 2/3. What are the lengths of the side AB and its image? So we're going to do a dilation centered at P. So if we're centering a dilation at P and its scale factor is 1 and 2/3, that means once we perform the dilation, every point is going to be 1 and 2/3 times as far away from P. Well P is 0 away from P, so its image is still going to be at P. So let's put that point right over there. Now point C is going to be 1 and 2/3 times as far as it is right now. So let's see, right now it is 6 away. It's at negative 3. And P, its x-coordinate is the same, but in the y direction, C is at negative 3. So it's 6 less. We want to be 1 and 2/3 times as far away. So what's 1 and 2/3 of 6? Well, 2/3 of 6 is 4, so it's going to be 6 plus 4. You're going to be 10 away. So 3 minus 10, that gets us to negative 7. So that gets us right over there. Now point A, right now it is 3 more in the horizontal direction than point P's x-coordinate. So we want to go 1 and 2/3 as far. So what is 1 and 2/3 times 3? Well that's going to be 3 plus 2/3 of 3, which is another 2. So that's going to be 5. So we're going to get right over there. Then we could complete the rectangle. And notice point B is now 1 and 2/3 times as far in the horizontal direction. It was 3 away in the horizontal direction, now it is 5 away from P's x-coordinate. And in the vertical direction, in the y direction, Now it is 1 and 2/3 times as far. It is 10 below P's y-coordinate. So then let's answer these questions. The length of segment AB-- well, we already saw that. That is, we're going from 3 to negative 3. That is 6 units long. And its image, well it's 1 and 2/3 as long. We see it over here. We're going from 3 to negative 7. 3 minus negative 7 is 10. It is 10 units long. We got it right."
    },
    {
        "Q": "I entered this limit (as an equation) onto the Desmos graphing calculator, and it came out similar to the graph in the video at 3:31, with one exception: At x=0, there was a straight line going from 0,2 to 0,1. Is this due to something in this limit, or is it just a glitch in the calculator?",
        "A": "I d guess that the calculator defines 0/0 as 1. As it evaluated the function at 0, it got 1 instead of 2. The calculator basically calculates the coordinates of a lot of points and then connects the dots, thus causing the downward spike. That s my guess of what s happening, anyways.",
        "video_name": "t7NvlTgMsO8",
        "timestamps": [
            211
        ],
        "3min_transcript": "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. So I could make my minimum x-value negative 1. Let me make my maximum x-value-- the maximum x-value here is 1, but just to get some space here I'll make this 1.5. So the x-scale is 1. y minimum, it seems like we're approaching 2. So the y max can be much smaller. Let's see, let me make y max 3. And now let's graph this thing. So let's see what it's doing. And actually it looks -- whether you're approaching from the positive direction or from the negative direction-- it looks like the value of the function approaches 2. But this problem, we're only caring about-- as we have x-values that are approaching 0 from values larger than 0. So this is the one-sided limit that we care about. But the 2 shows up right over here, as well."
    },
    {
        "Q": "In 2:07 can the hypotenuse be the butom side if the triangle is fliped?",
        "A": "Yes there can be a hypotenuse no matter which way the triangle is flipped. The only property is that there should be a 90 degree angle in the triangle. The side opposite to that will be the hypotenuse. :)",
        "video_name": "AA6RfgP-AHU",
        "timestamps": [
            127
        ],
        "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": "This doesn't have much to do with the video, but at 5:28, Sal says we take the positive square root of both sides. Is there a negative square root?",
        "A": "Yes, for example, the positive square root of 25 is 5 and the negative square root is -5. When you square negative numbers, you get a positive answer, therefore the square root of a positive number will have both a positive and a negative.",
        "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": "10:09 instead of it being multiple perfect sqauares wouldnt yoiu just leave it alone and put n/a when finished because my teacher said that its no pyth. anymore unless it in a perfect square",
        "A": "No you wouldn t. If it wasn t a perfect square, you would put it under the radical sign, and that would be your answer. Hope I helped! :)",
        "video_name": "AA6RfgP-AHU",
        "timestamps": [
            609
        ],
        "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 10:35 did anyone else notice that that triangle wasn't a right triangle?",
        "A": "No those are all right triangles, but are set at different angles.",
        "video_name": "AA6RfgP-AHU",
        "timestamps": [
            635
        ],
        "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 5:25 Sal refers to the square root of 25 that is positive and calls it the principle root. Is that the name of the positive answer of a square root? Is there a name for the negative answer of the square root?",
        "A": "The principle square root of a number is the positive square root. If you want to specify the negative answer of a square root its simply the negative square root You could also say the positive square root to identify the principle square root. The positive square root is used more often that the negative so it gets a special name",
        "video_name": "AA6RfgP-AHU",
        "timestamps": [
            325
        ],
        "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": "At the time of 5:40 he mentions the squre root. Can anyone explain what that is or show me a video of how to find the sqaure root",
        "A": "A square root a number to the 1/2 power. like 25^1/2=5 because 5^2=25. (E.G. square root of 49 is 7 because 7^2 (7x7) is 49",
        "video_name": "AA6RfgP-AHU",
        "timestamps": [
            340
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "so i stopped at 5:48 and i was wondering why each equation has to be like squared such as his example 4 squared + 3 squared = c squared, how come? please no long answers im in sixth grade",
        "A": "If you put squares on each side of the right triangle, the sum of the smaller squares would equal the largest square. Because the area of a square is (side)^2, and we are looking for the side, we square the sides.",
        "video_name": "AA6RfgP-AHU",
        "timestamps": [
            348
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "At 5:32, Sal says it could be negative five as well. But isn't it impossible for a length to equal a nonpositive number? Even though he said, that in this case, we only need the positive root, wouldn't that imply that it could be negative in other cases? Could he be speaking only of geometrical properties? Or is he just absent-mindedly pointing out the mathematical property of equality that a square root of a number can be both negative and positive?",
        "A": "There are two numbers you can square to get 25. You can square 5: (5)^2=25 or -5: (-5)^2=25. So there are actually two answersfor the square root of 25: 5 AND -5. But as he pointed out, we re dealing with distances, so we know that the answer can t be -5. That s why he said we re taking the positive square root.",
        "video_name": "AA6RfgP-AHU",
        "timestamps": [
            332
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "How is it possible that the hypotenuse is always opposite of the right angle? Also at 0:40 he says has to be right triangle, is it possible to do it with any other kinds of triangles?",
        "A": "because hypotenuse is defined as the side which is opposite to the perpendicular! and it is a universal fact that it would be the largest side of a right angled triangle. No,you can t Pythagoras theorem is based on right triangles and is the base for trigonometry which is studied in higher classes",
        "video_name": "AA6RfgP-AHU",
        "timestamps": [
            40
        ],
        "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": "At 9:35 it says the term perfect square. What does that mean?",
        "A": "By perfect square, you mean to say that if the square root of a number is taken the result would be a whole number. For example, 4,9,16 and 25 are perfect squares, since if you take the square root of those numbers, you would get 2,3,4 and 5, which are whole numbers. Hoped it helped! :)",
        "video_name": "AA6RfgP-AHU",
        "timestamps": [
            575
        ],
        "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 around the 10:00 minute mark he starts talking about principal roots and stuff. Can someone help explain that to me?",
        "A": "Principle root simply means that the root is positive, not negative.",
        "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 7:20 couldn't you just do a+b=c instead of A2+B2=C2.",
        "A": "You can t just use a + b = c because you are trying to take the square root of both sides and just eliminate the squares. But the square root of both sides of a^2 + b^2 = c^2 gives you sqrt(a^2 + b^2) on the left and you have to do what s inside the parentheses first.",
        "video_name": "AA6RfgP-AHU",
        "timestamps": [
            440
        ],
        "3min_transcript": "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. 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."
    },
    {
        "Q": "at 4:22, why does he cross out the 2s?",
        "A": "Do u know why in the video he subtracted 2x and not 5x",
        "video_name": "f15zA0PhSek",
        "timestamps": [
            262
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 1:16 what exactly does he mean by 7 away from 0, what if it was negative?",
        "A": "The simplest way to explain absolute value is forget the sign . Thus, |-7| = |7| = 7",
        "video_name": "hKkBlcnU9pw",
        "timestamps": [
            76
        ],
        "3min_transcript": "Let's do some examples comparing absolute values. So let's say we were to ask ourselves how the absolute value of negative 9, I should say, how that compares to the absolute value of-- let me think of a good number-- let's say the absolute value of negative 7. So let's think about this a little bit, and let's think about what negative 9 looks like, or where it is on the number line, where negative 7 is on the number line. Let's look at what the absolute values mean, and then we should probably be able to do this comparison. So there's a couple of ways to think about it. One is you could draw them on the number line. So if this is 0, if this is negative 7, and then this is negative 9 right over here. Now, when you take the absolute value of a number, you're really saying how far is that number from 0, whether it's to the left or to the right of 0. So, for example, negative 9 is 9 to the left of 0. This evaluates to 9, Negative 7 is exactly 7 to the left of 0. So the absolute value of negative 7 is positive 7. And so if you were to compare 9 and 7, this is a little bit more straightforward. 9 is clearly greater than 7. And if you ever get confused with the greater than or less than symbols, just remember that the symbol is larger on the left-hand side. So that's the greater than side. If I were to write this-- and this is actually also a true statement. If you took these without the absolute value signs, it is also true that negative 9 is less than negative 7. Notice the smaller side is on the smaller number. And so that's the interesting thing. Negative 9 is less than negative 7, but their absolute value, since negative 9 is further to the left of 0, it is-- the absolute value than the absolute value of negative 7. Another way to think about it is if you take the absolute value of a number, it's really just going to be the positive version of that number. So if you took the absolute value of 9, that equals 9. Or the absolute value of negative 9, that is also equal to 9. Well, when you think of it visually, that's because both of these numbers are exactly 9 away from 0. This is 9 to the right of 0, and this is 9 to the left of 0. Let's do a few more of these. So let's say that we wanted to compare the absolute value of 2 to the absolute value of 3. Well, the absolute value of a positive number is just going to be that same value. 2 is two to the right of 0, so this is just going to evaluate to 2. And then the absolute value of 3, that's just going to evaluate to 3. It's actually pretty straightforward. So 2 is clearly the smaller number here. And so we clearly get 2 is less than 3,"
    },
    {
        "Q": "At ~8:25 Sal says that sin(x) reflected over the y-axis is equal to sin(-x). It looks to me that you could just as correctly said that sin(x) reflected over the x-axis is equal to sin(-x). Is this right?",
        "A": "The sine function is a member of a special family of functions we call odd functions. If f(x) is odd, f(-x)=-f(x). You are right about your statement, because it is another property of odd functions.",
        "video_name": "0zCcFSO8ouE",
        "timestamps": [
            505
        ],
        "3min_transcript": "well it's common sense the amplitude here was 1 but now you're swaying from that middle position twice as far because you're multiplying by 2 Now let's go back to sin(x) and let's change it in a different way Let's graph sin(-x) so now let me once again put some graph paper here And now my goal is to graph sin(-x) y=sin(-x) so at least for the time being I've got rid of that 2 there and I'm just going straight from sin(x) to sin(-x) So let's think about how the values are going to work out So when x is 0 this is still going to be sin(0) which is 0 But then what as x increases, what happens when x is \u03c0/2 we're going to have to multiply by this negative so when x is \u03c0/2 we're really taking sin(-\u03c0/2) but what's sin(-\u03c0/2) but we can see over here here it's -1 It's - 1 and then when x = \u03c0 well sin(-\u03c0) we see this is 0 When x is 3\u03c0/2 well it's going to be sin(-3\u03c0/2) which is 1 Once again when x is 2\u03c0 it's going to be sin(-2\u03c0) is 0 So notice what was happening as I was trying to graph between 0 and 2\u03c0 I kept referring to the points in the negative direction so you can imagine taking this negative side right over here between 0 and -2\u03c0 and then flipping it over to get this one right over here that's what that -x seems to do you say when x = -\u03c0/2 where you have the negative in front of it so it's going to be sin(\u03c0/2) so it's going to be equal to 1 and you can flip this over the y-axis so essentially what we have done is we have flipped it we have reflected the graph of sin(x) over the y-axis So we have reflected it over the y-axis This is the y-axis so hopefully you see that reflection that's what that -x has done So now let's think about kind of the combo Having the 2 out the front and the -x right over there so let me put the graph on the axis there one more time And now let's try to do what was asked of us"
    },
    {
        "Q": "At 1:27, Sal says that if you take away 2 x's from one side, that side will go up; it has less weight. But wouldn't that side go down if x was a negative number?",
        "A": "Yeah, but you can t have negative mass.",
        "video_name": "Ye13MIPv6n0",
        "timestamps": [
            87
        ],
        "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": "At 6:25, why did he plug in 2 and 4 into the original equation to find the minimum?? Shouldn't he have just picked one number??",
        "A": "The derivative is a second degree polynomial thus it is a parabola .......... when sal found its two roots at 1 and 3 ........ it is understood that the vertex of parabola will exactly be between them because the symmetry of parabola.... i.e. x=2 ....and for y value he plugged x=2 .....3(2)^2-12(2)+9.......3(4)-12(2)+9....",
        "video_name": "SE1ltVuE5yM",
        "timestamps": [
            385
        ],
        "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": "At 2:31, didn't Sal do the dot inaccurately?",
        "A": "It is off by just a little, but this is just due to human error and the fact that he didn t have a grid to draw on. It is implied that the dot is at x = 9.585 s and y = 100 m.",
        "video_name": "EKvHQc3QEow",
        "timestamps": [
            151
        ],
        "3min_transcript": "Not just what his average speed was for the last second, or his average speed over the next 10 seconds. How fast is he going right now? And so this is what differential calculus is all about. Instantaneous rates of change. Differential calculus. Newton's actual original term for differential calculus was the method of fluxions, which actually sounds a little bit fancier. But it's all about what's happening in this instant. And to think about why that is not a super easy problem to address with traditional algebra, let's draw a little graph here. So on this axis I'll have distance. I'll say y is equal to distance. I could have said d is equal to distance, but we'll see, especially later on in calculus, d is reserved for something else. We'll say y is equal to distance. And in this axis, we'll say time. but I'll just say x is equal to time. And so if we were to plot Usain Bolt's distance as a function of time, well at time zero he hasn't gone anywhere. He is right over there. And we know that this gentleman is capable of traveling 100 meters in 9.58 seconds. So after 9.58 seconds, we'll assume that this is in seconds right over here, he's capable of going 100 meters. And so using this information, we can actually figure out his average speed. Let me write it this way, his average speed is just going to be his change in distance over his change in time. And using the variables that are over here, we're saying y is distance. So this is the same thing as change in y over change in x from this point to that point. And this might look somewhat familiar to you This is the slope between these two points. If I have a line that connects these two points, this is the slope of that line. The change in distance is this right over here. Change in y is equal to 100 meters. And our change in time is this right over here. So our change in time is equal to 9.58 seconds. We started at 0, we go to 9.58 seconds. Another way to think about it, the rise over the run you might have heard in your algebra class. It's going to be 100 meters over 9.58 seconds. So this is 100 meters over 9.58 seconds. And the slope is essentially just rate of change, or you could view it as the average rate of change between these two points. And you'll see, if you even just follow the units, it gives you units of speed here. It would be velocity if we also specified the direction. And we can figure out what that is, let me get the calculator out."
    },
    {
        "Q": "People say that we see math in our everyday lives -- and while I understand how this concept applies to beginning math, pre-algebra, algebra, and trig, how does this apply to calculus? At 1:01, Sal says that differential calculus is all about finding instantaneous rate of change, but is that the only \"everyday use\"? Or is calculus simply a concept that is used in other subjects, or even professions, like engineering?\n\nThanks!",
        "A": "I think the best way to find a good answer to this question is to just keep watching the videos! If you attend college for any engineering discipline, you have to learn calculus before you even begin learning the specifics of your discipline (whether it be mechanical, electrical, civil, computer, computer science, etc...). The best way to understand what every day things calculus will enable you to do is to learn calculus and start doing incredible things every day :-)",
        "video_name": "EKvHQc3QEow",
        "timestamps": [
            61
        ],
        "3min_transcript": "This is a picture of Isaac Newton, super famous British mathematician and physicist. This is a picture of a Gottfried Leibnitz, super famous, or maybe not as famous, but maybe should be, famous German philosopher and mathematician, and he was a contemporary of Isaac Newton. These two gentlemen together were really the founding fathers of calculus. And they did some of their-- most of their major work in the late 1600s. And this right over here is Usain Bolt, Jamaican sprinter, whose continuing to do some of his best work in 2012. And as of early 2012, he's the fastest human alive, and he's probably the fastest human that has ever lived. And you might have not made the association with these three You might not think that they have a lot in common. But they were all obsessed with the same fundamental question. And this is the same fundamental question that differential calculus addresses. And the question is, what is the instantaneous rate of change of something? Not just what his average speed was for the last second, or his average speed over the next 10 seconds. How fast is he going right now? And so this is what differential calculus is all about. Instantaneous rates of change. Differential calculus. Newton's actual original term for differential calculus was the method of fluxions, which actually sounds a little bit fancier. But it's all about what's happening in this instant. And to think about why that is not a super easy problem to address with traditional algebra, let's draw a little graph here. So on this axis I'll have distance. I'll say y is equal to distance. I could have said d is equal to distance, but we'll see, especially later on in calculus, d is reserved for something else. We'll say y is equal to distance. And in this axis, we'll say time. but I'll just say x is equal to time. And so if we were to plot Usain Bolt's distance as a function of time, well at time zero he hasn't gone anywhere. He is right over there. And we know that this gentleman is capable of traveling 100 meters in 9.58 seconds. So after 9.58 seconds, we'll assume that this is in seconds right over here, he's capable of going 100 meters. And so using this information, we can actually figure out his average speed. Let me write it this way, his average speed is just going to be his change in distance over his change in time. And using the variables that are over here, we're saying y is distance. So this is the same thing as change in y over change in x from this point to that point. And this might look somewhat familiar to you"
    },
    {
        "Q": "At 6:00 I am wondering what happened to miles/Meters/ they got confused in my mind. Which one is using?",
        "A": "If you re asking about how he got from 10.4 m/s to 23.5 mi/hr, ... 10.4 meters / second, when converted to miles/hour, is roughly 23.5 miles/hour. Since most of us (in the USA) drive in cars that have speedometers reading out in miles/hours, it s easier for some to be able to relate to 23 miles per hour, versus 10.4 m/sec. However, they are equivalent speeds (velocities), just one uses metric (SI) units, and the other uses US units.",
        "video_name": "EKvHQc3QEow",
        "timestamps": [
            360
        ],
        "3min_transcript": "So we're going 100 meters in the 9.58 seconds. So it's 10.4, I'll just write 10.4, I'll round to 10.4. So it's approximately 10.4, and then the units are meters per second. And that is his average speed. And what we're going to see in a second is how average speed is different than instantaneous speed. How it's different than what the speed he might be going at any given moment. And just to have a concept of how fast this is, let me get the calculator back. This is in meters per second. If you wanted to know how many meters he's going in an hour, well there's 3,600 seconds in an hour. So he'll be able to go this many meters 3,600 times. So that's how many meters he can, if he were able to somehow keep up that speed in an hour. This is how fast he's going meters per hour. And then, if you were to say how many miles per hour, but roughly 1600 meters per mile. So let's divide it by 1600. And so you see that this is roughly a little over 23, about 23 and 1/2 miles per hour. So this is approximately, and I'll write it this way-- this is approximately 23.5 miles per hour. And relative to a car, not so fast. But relative to me, extremely fast. Now to see how this is different than instantaneous velocity, let's think about a potential plot of his distance relative to time. He's not going to just go this speed immediately. He's not just going to go as soon as the gun fires, he's not just going to go 23 and 1/2 miles per hour all the way. He's going to accelerate. So at first he's going to start off going a little bit slower. So the slope is going to be a little bit lot lower than the average slope. He's going to go a little bit slower, then he's going to start accelerating. And so his speed, and you'll see the slope here 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."
    },
    {
        "Q": "At 2:04 shouldn't it be -1 x -4 = 4 and not 5?? so the real answer should be x^2+3 right?",
        "A": "Correct. This is a known problem and as such has a pop-up box in the lower right-hand corner to say so. It appears on screen shortly after Sal s mistake.",
        "video_name": "KvMyZY9upuA",
        "timestamps": [
            124
        ],
        "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 0:23 why did you put diffret divison structures?",
        "A": "because if you don t know hat problem by heart before you do it its easier to put it into that structure",
        "video_name": "AjYil74WrVo",
        "timestamps": [
            23
        ],
        "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": "3:09 why is y zero?",
        "A": "To find the two intercepts, you have to set x = 0 (to find the y intercept) and y = 0 (to find the x intercept). He is just finding the x intercept at this point.",
        "video_name": "6CFE60iP2Ug",
        "timestamps": [
            189
        ],
        "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": "The method at 7:29 is nice but I can't make it work for something like:\n(4x^2 + y^2)^2\n\nI end up with 16x^2 + 8x^2 * 2y^2 + y ^4 but it should be 16x^2 + 8x^2 * y^2 + y ^4",
        "A": "_4x^2 + _______ + y^2 x 4x^2 + _______ + y^2 --------------------------------- _______4(x^2)(y^2) + y^4 + 16x^4 + 4(x^2)(y^2) ---------------------------------- 16x^4 + 8(x^2)(y^2) + y^4",
        "video_name": "fGThIRpWEE4",
        "timestamps": [
            449
        ],
        "3min_transcript": "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 of these terms, and then you're going to distribute each of those terms into 3x plus 2. It would take a long time and in reality, you'll never do it quite that way. But you will get the same answer we're going to get. When you have larger polynomials, the easiest way I can think of to multiply, is kind of how you multiply long numbers. So we'll write it like this. 9x squared, minus 6, plus 4. And we're going to multiply that times 3x plus 2. And what I imagine is, when you multiply regular numbers, you have your ones' place, your tens' place, your hundreds' place. Here, you're going to have your constants' place, your first degree place, your second degree place, your third degree place, if there is one. And actually there will be in this video. So you just have to put things in their proper place. So let's do that. So you start here, you multiply almost exactly like you would do traditional multiplication. 2 times 4 is 8."
    },
    {
        "Q": "At 9:40 is it ok if you use the same technique for multiplying to binomials?",
        "A": "It depends on what you are doing.",
        "video_name": "fGThIRpWEE4",
        "timestamps": [
            580
        ],
        "3min_transcript": "of these terms, and then you're going to distribute each of those terms into 3x plus 2. It would take a long time and in reality, you'll never do it quite that way. But you will get the same answer we're going to get. When you have larger polynomials, the easiest way I can think of to multiply, is kind of how you multiply long numbers. So we'll write it like this. 9x squared, minus 6, plus 4. And we're going to multiply that times 3x plus 2. And what I imagine is, when you multiply regular numbers, you have your ones' place, your tens' place, your hundreds' place. Here, you're going to have your constants' place, your first degree place, your second degree place, your third degree place, if there is one. And actually there will be in this video. So you just have to put things in their proper place. So let's do that. So you start here, you multiply almost exactly like you would do traditional multiplication. 2 times 4 is 8. 2 times negative 6x is negative 12x. And we'll put a plus there. That was a plus 8. 2 times 9x squared is 18x squared, so we'll put that in the x squared place. Now let's do the 3x part. I'll do that in magenta, so you see how it's different. 3x times 4 is 12x, positive 12x. 3x times negative 6x, what is that? The x times the x is x squared, so it's going to go over here. And 3 times negative 6 is negative 18. And then finally 3x times 9x squared, the x times the x 3 times 9 is 27. I wrote it in the x third place. And once again, you just want to add the like terms. So you get 8. There's no other constant terms, so it's just 8. Negative 12x plus 12x, these cancel out. 18x squared minus 18x squared cancel out, so we're just left over here with 27x to the third. So this is equal to 27x to the third plus 8. And we are done. And you can use this technique to multiply a trinomial times a binomial, a trinomial times a trinomial, or really, you know, you could have five terms up here. A fifth degree times a fifth degree. This will always work as long as you keep things in their proper degree place."
    },
    {
        "Q": "At 2:50 Sal says this is a fifth degree polynomial. What is that?",
        "A": "A 5th degree polynomial is a polynomial that has the 5th power as the highest exponent in one of its term. Ex: 2x^5 _+ 2 x^5 + x^4 + 3",
        "video_name": "fGThIRpWEE4",
        "timestamps": [
            170
        ],
        "3min_transcript": "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. This you're going to see very, very, very frequently in algebra. So let's say you have x minus 3, times x plus 2. And I actually want to show you that all we're doing here is the distributive property. So let me write it like this: times x plus 2. So let's just pretend that this is one big number here. And it is. You know, if you had x's, this would be some number here. So let's just distribute this onto each of these variables. So this is going to be x minus 3, times that green x, plus x minus 3, times that green 2. All we did is distribute the x minus 3. This is just the distributive property. Remember, if I had a times x plus 2, what would"
    },
    {
        "Q": "At around 4:48: Why do you subtract - 10x^2y dy/dx from both sides?",
        "A": "When you solve an equation for a variable, you have to move all the terms with that variable to one side of the equation and all of the other terms to the opposite side. Then you can factor out the variable you re solving for and divide by the term in parenthesis: a simplified example would be: -2xy - 3 = 5y (to solve for y) -3 = 5y + 2xy -3 = y(5+2x) -3 / (5+2x) = y",
        "video_name": "1DcsREjyoiM",
        "timestamps": [
            288
        ],
        "3min_transcript": "of these terms. So if you distribute this purple thing onto this term right over here, you get 3 times 2x, which is 6x, times x squared plus y squared, squared. And then if you distribute this purple stuff on to this one right over here you get plus, let's see, 2y times 3 is 6y times x squared plus y squared. Let me make sure. 2y times 3 is 6y times x squared plus y squared, squared, and then I'll keep the dy dx in that green color. dy dx is equal to, well, we can multiply the 5 times this business right over here. And so everything that's not a dy dx term maybe I will do in purple now. So you do 5 times this stuff right over here, which gives you 10xy squared. is going to be plus 10x squared y dy dx. Did I do that right? Yep, that looks just about right. And now we have to solve for dy dx. What I'm going to do is I'm going to subtract this 10x squared. I'm going to subtract the 10x squared y dy dx from both sides. 10x squared y dy dx. Derivative. That's not green. Derivative of y with respect to x. Going to subtract that from both sides so that I can get it on the left hand side. dy dx. And I'm going to subtract this business, this 6x times all this craziness from both sides. So minus 6x times x squared plus y squared, squared. Let me subtract it from here as well. Minus 6x x squared plus y squared squared. Well, these guys cancel out. On the left hand side right over here, we are left with 6y times x squared plus y squared, squared minus 10x squared y times dy dx, the derivative of y with respect to x. The derivative of y with respect to x is equal to-- these characters cancel out-- and we are left with 10xy squared minus 6x times x squared plus y squared, squared. And now if we want to solve for dy dx, we just divide both sides of this equation by this business right over here. And you get the derivative of y with respect to x. And we deserve a drum roll now. The derivative of y with respect to x"
    },
    {
        "Q": "At 1:20, I just can't understand why the derivative,at the end of the application of the chain rule, is equal to ...2y*dy/dx. Why it's not just 2dy/dx??",
        "A": "When you take a derivative, you bring down the power, and subtract one from the exponent. So derivative of y^3 would be 3y^2(dy/dx) and derivative of y^5 would be 5y^4(dy/dx). In your case, it was y^2, thus its derivative is 2y*(dy/dx)",
        "video_name": "1DcsREjyoiM",
        "timestamps": [
            80
        ],
        "3min_transcript": "Once again, I have some crazy relationship between x and y. And just to get a sense of what this might look like, if you plot all the x's and y's that satisfy this relationship, you get this nice little clover pattern. And I plotted this off of Wolfram Alpha. But what I'm curious about in this video, as you might imagine from the title, is to figure out the rate at which y is changing with respect to x. And we're going to have to do it implicitly. We're going to have to find the implicit derivative of this. We're going to have to derive this implicitly or take the derivative of it implicitly. So let's apply our derivative operator to both sides of the derivative with respect to x on the left and the derivative with respect to x on the right. So once again, we apply our chain rule. The derivative of something to the third power with respect to that something is going to be three times that something squared. And then we have to multiply that times the derivative of the something with respect to x. is going to be 2x, that's the derivative of x squared with respect to x, plus the derivative of y squared with respect to y is going to be 2y times the derivative of y with respect to x. Once again, we're applying the chain rule right over here. The derivative of something squared with respect to the something, which is 2y, times the derivative of the something with respect to x, which is dy dx. Now, that is going to be equal to what we have on the right hand side. So we have a 5 times x squared times y squared. We can take the 5 out of the picture for now. Take the 5 onto the-- take it out of the derivative. The derivative of 5 times something is the same thing as 5 times the derivative. And now we can apply the product rule. So it's going to be 5 times the derivative of x squared is just going to be 2x times y squared. times the second function. Plus the first function, not taking its derivative, x squared times the derivative of the second function. Well, what's the derivative of y squared with respect to x? Well, we already figured it out. It's the derivative of y squared with respect to y, which is 2y times the derivative of y with respect Let me make it clear what I just did. This is this, and then this is when I took its derivative. So that is when I applied the derivative operator. Similarly, that is that. And when I applied the derivative operator, I got that. The derivative with respect to x right over there. So let's see if we can somehow solve for dy dx. So what I'm going to do on the left hand side"
    },
    {
        "Q": "@2:40 is Sal making an assumption that AG is the longest part ?",
        "A": "He may be eyeballing it, but you can also tell by the fact that line AG goes all the way past point F on its side, which is also the point that bisects line AE. That shows that AG is longer by proportion than line GD, or the longer part of the median if that made sense :)",
        "video_name": "k45QTFCHSVs",
        "timestamps": [
            160
        ],
        "3min_transcript": "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. And we know that F and D are the midpoints. So for example, we know this AE is 12. That was given. We know that ED is half of this 18. So ED right over here-- I'll do this in a new color. ED is going to be 9. So then we could just use the Pythagorean theorem to figure out what AD is. AD is the hypotenuse of this right triangle. So we're looking at triangle AED right now. Let me write this down. We know that 12 squared plus 9 squared is going to be equal to AD squared. 12 squared is 144. 144 plus 81."
    },
    {
        "Q": "When you are multiplying a fraction with the matrix, is it necessary to simplify if allowed?\nRefer to 9:47...",
        "A": "No, it is not necessary.",
        "video_name": "iUQR0enP7RQ",
        "timestamps": [
            587
        ],
        "3min_transcript": "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? And we just make these two terms negative. Minus 2 and 4. 4 was minus 4, so now it becomes 4. And let's see if we can simplify this a little bit. So B inverse is equal to minus 15 plus 8. That's minus 7. So this is minus 1/7. So the determinant of B-- we could write B's determinant-- is equal to minus 7. So that's minus 1/7 times minus 5, 4, minus 2, 3. Which is equal to-- this is just a scalar, this is just a number, so we multiply it times each of the elements-- so that is equal to minus, minus, plus. That's 5/7. 5/7 minus 4/7. Let's see. 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."
    },
    {
        "Q": "when he fins the determinant of B he says that it is 1/-7 and that is what he uses to multiply with. So why does he change it to -7 when he puts the B in absolute value signs over on the side (9:34) ?",
        "A": "When dealing with matrices the absolute value sign actually means the determinant and does not mean to take the absolute of the number.",
        "video_name": "iUQR0enP7RQ",
        "timestamps": [
            574
        ],
        "3min_transcript": "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? And we just make these two terms negative. Minus 2 and 4. 4 was minus 4, so now it becomes 4. And let's see if we can simplify this a little bit. So B inverse is equal to minus 15 plus 8. That's minus 7. So this is minus 1/7. So the determinant of B-- we could write B's determinant-- is equal to minus 7. So that's minus 1/7 times minus 5, 4, minus 2, 3. Which is equal to-- this is just a scalar, this is just a number, so we multiply it times each of the elements-- so that is equal to minus, minus, plus. That's 5/7. 5/7 minus 4/7. Let's see. 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."
    },
    {
        "Q": "at 6:25, why do you subract the two equations and not add them?",
        "A": "you can do either or as long as you are left one variable eliminated",
        "video_name": "xCIHAjsZCE0",
        "timestamps": [
            385
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 5:00, why do you subtract 300 instead of 200?",
        "A": "Because he is subtracting the blue/teal equation from the red/pink equation. Subtracting the red equation from itself would just get you 0=0, which is true, but not very useful.",
        "video_name": "xCIHAjsZCE0",
        "timestamps": [
            300
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 7:50, Khan says \"one equation for one unknown.\"\nDoes this mean that if two equations for 2 unkowns and 1 equation for 1 unknown is possible to be solved, then for 3 unknowns you have to get 3 equations? And so on and so forth?",
        "A": "That is exactly correct. Good job :-)",
        "video_name": "xCIHAjsZCE0",
        "timestamps": [
            470
        ],
        "3min_transcript": ""
    },
    {
        "Q": "@ 5:09 why Sal is subtracting 500a and 300c from 500a + 200c?",
        "A": "The whole point of solving systems of equations by elimination is to get rid of one variable by making it subtract to be 0, so if you have a positive 500 and a - 500, then the two add to be zero which gets rid of a, so when we add the coefficients of the c variable, we can find a value for c, then substitute it in to find a.",
        "video_name": "xCIHAjsZCE0",
        "timestamps": [
            309
        ],
        "3min_transcript": ""
    },
    {
        "Q": "What is the angle between two hands of a clock at 4:30",
        "A": "If the total clock face represents 360\u00c2\u00b0, then each number on the clock represents an angle of 30\u00c2\u00b0 and each minute, 6\u00c2\u00b0. So at 4:30 the minute hand is at 30*6\u00c2\u00b0 = 180\u00c2\u00b0 and the hour hand is at 4.5*30\u00c2\u00b0 = 135\u00c2\u00b0. Thus the angle between them is 180\u00c2\u00b0-135\u00c2\u00b0 = 45\u00c2\u00b0",
        "video_name": "2mzuFKCuDg4",
        "timestamps": [
            270
        ],
        "3min_transcript": "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": "At 3:51 where did he get 10x from? He seems to have pulled it out of thin air.",
        "A": "The 10x is created from multiplying the 2 binomials (x+3)(x+7) Let s do the multiplication (use FOIL to multiply the binomials). (x+3)(x+7) = x^2 + 7x + 3x + 21 Combine the middle 2 terms... they create the 10x. x^2 + 7x + 3x + 21 = x^2 + 10x + 21 Hope this helps.",
        "video_name": "SjN3_xCJamA",
        "timestamps": [
            231
        ],
        "3min_transcript": "is Y to the fourth power. And so what we could say is, if we wanted to say factors of negative six X to the third, Y to the fourth, we could say that 3xy is a factor of this just as an example; so let me write that down. We could write that 3xy is a factor of, is a factor of... of negative six X to the third power, Y to the fourth, or we could phrase that the other way around. We could say that negative six X to the third, Y to the fourth, is divisible by, is divisible by 3xy. So hopefully you're seeing the parallels. If I'm taking these two monomials with integer coefficients and I multiply 'em and I get this other, in this case, and there's actually other factors of this, but I could say either one of these is a factor of this monomial, or we could say that negative six X to the third, Y to the four is divisible by one of its factors. And we could even extend this to binomials or polynomials. For example, if I were to take, if I were to take, let me scroll down a little bit, whoops, if I were to take, let me say X plus three and I wanted to multiply it times X plus seven, we know that this is going to be equal to, if I were to write it as a trinomial, it's gonna be X times X, so X squared, and then it's gonna be three X plus seven X, so plus 10x; and if any of this looks familiar, we have a lot of videos where we go in detail of multiplying binomials like this. And then three times seven is 21. So because I multiplied these two, in this case binomials, or we could consider themselves to be polynomials, polynomials or binomials with integer coefficients. Notice the coefficients here, they're one, one. The constants here, they're all integers. Because I'm dealing with all integers here, we could say that either one of these binomials is a factor of this trinomial, or we could say this trinomial is divisible by either one of these. So let me write that down. So I could say, I'll just pick on X plus seven. We could say that X plus seven is a factor, is a factor of X squared plus 10x plus 21; or we could say that X squared plus 10x plus 21 is divisible by, is divisible by"
    },
    {
        "Q": "At 3:59ish, Sal switches the subtraction to addition and adds -1. Do we always use 1? And why do we change the sign? What happens if you do not use a scaler?",
        "A": "the jest of it is this ... a - b = a + (-1)b = a + -b",
        "video_name": "WR9qCSXJlyY",
        "timestamps": [
            239
        ],
        "3min_transcript": "in which I'm adding the matrices does not matter So this is just like adding numbers. A plus B is just the same thing as B plus A. What we'll see is this won't be true for every matrix operation that we study and in particular this will not be true for matrix multiplication. But if you add these two things, using the definition we just came up with, adding corresponding terms, you'll get the exact same result. Up here we added one plus five and we got six Her we'll add five plus one and we'll get six. We get the same result because one plus five is the same thing as five plus one. Here we have zero plus negative seven you get negative seven. So you're going to get the exact same thing as we got up here. So when you're adding matrices, if you were to call -if you were to call this matrix right over here matrix A which we normally denote with a capital, bolder letter, and you call this matrix right over here Matrix B Then when we take the sum of A plus B which is this thing right over here, and we see it's the exact same thing as B, as Matrix B plus Matrix A. What if I wanted to subtract matrices? So let's once again think about matrices that have the same dimensions. So let's say I'm gonna do then two two-by-two matrices. So let's say it's zero, one, three, two, and from that I want to subtract negative one, three, zero, and five. So you might say well maybe we just subtract corresponding entries. And that indeed is how you can define matrix subtraction. In fact you don't even have to define matrix subtraction, you can let this fall out of what we did with scalar multiplication and matrix addition. We can view as the exact same thing -this as the exact same thing- as taking zero, one, three, two and to that we add negative one, negative one times negative one, three, zero, five. And if you work out the math you're going to get subtracting the corresponding terms. So this is going to be -what is this going to be? Zero minus negative one is positive one, one minus three is negative two, three minus zero is three, two minus five is negative three. And you'll see that you get the exact same thing here. When you multiply negative one times negative one you get positive one, positive one plus zero is one. Negative one times three plus one is negative two. Fair enough. There might be a question that is lingering in your brain right now. \"Okay Sal, I understand when I'm adding or subtracting matrices with the same dimensions I just add or subtract the corresponding terms. But what happens when I have matrices with different dimensions?\" So, for example, what about the scenario where I want to add the matrix one, zero, three, five, zero, one to the matrix -so this a three-by-two matrix-"
    },
    {
        "Q": "At 5:08 he says '0', but he writes '6' . Which one is he saying??",
        "A": "It was a sloppy 0. The actual value is irrelevant since the two matrices have different dimensions.",
        "video_name": "WR9qCSXJlyY",
        "timestamps": [
            308
        ],
        "3min_transcript": "What if I wanted to subtract matrices? So let's once again think about matrices that have the same dimensions. So let's say I'm gonna do then two two-by-two matrices. So let's say it's zero, one, three, two, and from that I want to subtract negative one, three, zero, and five. So you might say well maybe we just subtract corresponding entries. And that indeed is how you can define matrix subtraction. In fact you don't even have to define matrix subtraction, you can let this fall out of what we did with scalar multiplication and matrix addition. We can view as the exact same thing -this as the exact same thing- as taking zero, one, three, two and to that we add negative one, negative one times negative one, three, zero, five. And if you work out the math you're going to get subtracting the corresponding terms. So this is going to be -what is this going to be? Zero minus negative one is positive one, one minus three is negative two, three minus zero is three, two minus five is negative three. And you'll see that you get the exact same thing here. When you multiply negative one times negative one you get positive one, positive one plus zero is one. Negative one times three plus one is negative two. Fair enough. There might be a question that is lingering in your brain right now. \"Okay Sal, I understand when I'm adding or subtracting matrices with the same dimensions I just add or subtract the corresponding terms. But what happens when I have matrices with different dimensions?\" So, for example, what about the scenario where I want to add the matrix one, zero, three, five, zero, one to the matrix -so this a three-by-two matrix- Five, seven, negative one, zero. What would we define this as? Well it turns out that the mathematical mainstream does not define this. This is undefined. This is undefined. So we do not define matrix addition, or matrix subtraction, when the matrices have different dimensions. There didn't seem to be any reasonable way to do this, that would actually be useful and logically consistent in some nice way."
    },
    {
        "Q": "Greetings from Venezuela...Very helpful explanation, thanks Sal! Just a question: at 3:28, Sal is taking the derivative of x with respect to theta. The derivative of sine of theta is the cosine of theta, agreed. But why is the sqrt(3)/sqrt(2) not derived?",
        "A": "because sqrt(3)/sqrt(2) is a constant. Sorta like why you wouldnt derive pi, or any other number that isnt being multiplied by a variable such as X.",
        "video_name": "n4EK92CSuBE",
        "timestamps": [
            208
        ],
        "3min_transcript": "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 square root of 2 times the sine of theta. And we were going to substitute this with sine squared of theta, but we can't leave this dx out there. We have to take the integral with respect to d theta. So what's dx with respect to d theta? So the derivative of x with respect to theta is equal to square root of 3 over square root of 2. Derivative of this with respect to theta is just cosine of theta, and if we want to write this in terms of dx, we could just write that dx is equal to square root of 3 over the square root of 2 cosine of theta d theta. Now we're ready to substitute. So we can rewrite this expression up here-- I'll do it in this reddish color-- I was using that, let me do it in the blue color. We can rewrite this expression up here now. It's an indefinite integral of-- dx is on the"
    },
    {
        "Q": "At 0:58 he says you can use either sin^2 or cos^2, but won't that result in arccos at 7:05 instead of arcsin, which is a different answer? Or am I making a mistake?",
        "A": "I didn t watch the video, but arccos and arcsin differ by a constant and a negative sign only. So you ll end up with, say, -arccos instead of arcsin, and the constant difference will be absorbed by your arbitrary constant. Both are equally fine antiderivatives.",
        "video_name": "n4EK92CSuBE",
        "timestamps": [
            58,
            425
        ],
        "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 1:43 when it is like (5)(4) is that multiplication because that confused me because cant u just writ it like (5 times 4) or does it have to be like (5)(4)",
        "A": "Yes, (5)(4) is just the same thing as 5 times 4 . This is just another way to write it, which you will see quite often in math. You can write it however you prefer, but you should get used to seeing it written this way as well, because it will show up often. I hope this helps.",
        "video_name": "GiSpzFKI5_w",
        "timestamps": [
            103
        ],
        "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": "At 3:00 what did he mean?",
        "A": "First solve the problem of first bracket. Then solve the other problems.I think you got it,",
        "video_name": "GiSpzFKI5_w",
        "timestamps": [
            180
        ],
        "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": "At 04:09, if the hypotenuse is irrational does this mean that the hypotenuse is a multipal of \"root 2\" or can it also be another irrational number?",
        "A": "root two",
        "video_name": "X1E7I7_r3Cw",
        "timestamps": [
            249
        ],
        "3min_transcript": "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. you need to memorize your multiplication table and graph a parabola before you can learn real mathematics, they are lying to you. In Pythagoras's time there were no variables, no equations or formulas like we see today, Pythagoras's theorem wasn't 'a squared plus b squared equals c squared,' it was 'The squares of the legs of a right triangle have the same area as the square of the hypotenuse,' all written out. And when he said 'square' he meant 'square.' One leg's square plus the other leg's square equals hypotenuse's square. Three literally squared plus four made into a square. Those two squares have the same area as a five by five square. You can cut out the nine squares here and the sixteen here and fit them together where these 25 squares are, and in the same way, you can cut out the 25 hypotenuse squares and fit them into the two leg squares. Pythagoras thought you could do this trick with any right triangle, that it was just a matter of figuring out how many pieces to cut each side into. There was a relationship between the length of one side and the length of another and he wanted to find it on this map. But the trouble began with the simplest right triangle one where both the legs are the same length, one where both the legs' squares are equal. hypotenuse is something that, when squared, gives two. So what's the square root of two and how do we make it into a whole number ratio? Square root two is very close to 1.4 which would be a whole number ratio of 10:14 but 10 squared plus 10 squared is definitely not 14 squared, and a ratio of 1,000 to 1,414 is even closer, and a ratio of 100,000,000 to 141,421,356 is very close indeed but still not exact, so what is it? Pythagoras wanted to find the perfect ratio he knew it must exist, but meanwhile someone from his very own Pythagorean brotherhood proved there wasn't a ratio, the square root of two is irrational, that in decimal notation (once decimal notation was invented) the digits go on forever. Usually this proof is given algebraically, something like this, which is pretty simple and beautiful if you know algebra, but the Pythagoreans didn't. So I like to imagine how they thought of this proof, no algebra required. Okay, so Pythagoras is all like, \"There's totally a ratio, you can make this with whole numbers.\" And this guy's like,\"Is not!\" \"Is too!\" \"Is not!\" \"Is too!\" \"Fine have it your way."
    },
    {
        "Q": "refering to 2:00\n\nwas he actually that crazy, he meaning pathagros",
        "A": "Mostly, but you have to remember that he was a mathematician, and mathematicians love elegance and simplicity. Adding irrationals to mathematics made it far more ugly. And nobody likes to find that what they had believed to be true to be wrong.",
        "video_name": "X1E7I7_r3Cw",
        "timestamps": [
            120
        ],
        "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": "At 8:00, was Pythagoras really scared of beans?",
        "A": "There are several theories on why Pythagoras did not allow his followers to eat beans. One more likely theory states that he believed that beans contained the souls of humans, but we aren t sure.",
        "video_name": "X1E7I7_r3Cw",
        "timestamps": [
            480
        ],
        "3min_transcript": "\"We are. If there's a ratio in simlest form at least one of the numbers is odd and since the hypotenuse has to literally be divisible by two, then the leg must be the odd one. So what if I proved the leg had to be even?\" \"You just proved it's not. It can't be both.\" \"Unless it doesn't exist! What you forget Pythagoras is that if this is a square then the two sides are the same. Just as this is divsible right down the center so too is it divisible the other way! And the number squares on this side, which are the number of squares in just one leg is an even number. And for a number of squares to be even what does the number have to be, Pythagoras, oh my brother?\" \"If leg squared is even then the leg is even. But it can't be even, because it's already odd.\" \"Unless it doesn't exist.\" \"But if they're both even you can divide both by two and start again, but this still has to be even which means this still has to be even, which means you can divide by two again, but then it has to be even so everything is even forever and you never find the perfect ratio. Aww, beans\" He had a vision, a beautiful vision of a world made up of relationships between numbers. The Pythagoreans still believed, wanted to believe that irrationality was somehow false and the world was as they wanted it. So this proof stayed secret. Until someone spilled the beans. According to some, it was all a guy named Hippasus and Pythagoras threw him of a boat to drown him as punishment for ruining what had been perfect. Or maybe it was someone else who discovered it or Hippasus or someone else who was killed by the Pythagoreans long after Pythagoras was dead or maybe they just got exiled, who knows? And how did Pythagoras die? Well, according to one guy some guys got mad because they didn't get into the cool kids' club. So they set Pythagoras' house on fire. And Pythagoras was running away and they were chasing him, but then they came upon a field and not just any field, but a field of beans. And Pythagoras turned around to face his pursuers and proclaimed: \"Better to be slaughetered by enemies than to trample on beans!\" And he was. Others say he ran off and starved himself to death. Or just got caught by his enemies because he ran around the bean field instead of through it or who knows what happened. People claim Pythagoras didn't like beans because he thought they were bad for digestion, or gave you bad dreams because he didn't want a clubhouse full of flatulating mathematicians or he just didn't like them metaphorically. He and his followers were or weren't vegetarian did or didn't sacrifice animals possibly were only allowed to eat certain colors of birds I mean he definitely had a lot of rules to follow but just what they were and what they meant is lost to history. I'd like to give you a colorful story about exactly what happened with Pythagoras, but somehow that kind of truth doesn't last. What I do know is that the square root of two is irrational, that there's no way to have the length of a side of a square and of the square's diagonal both be whole numbers. Mathematical truth is truth that indures. This proof is just as good now as it was 2500 years ago, I mean it's awesome and it shows that there's more to the world than whole numbers and shame on the Pythagoreans who didn't have the beans to admit it."
    },
    {
        "Q": "At 8:30, why does Sal keep expanding everything out? I do not understand it.",
        "A": "he is using this as a complete example to show how it works. He is also using the sigma, which is a sum of all integers from the number on the bottom to n.",
        "video_name": "iPwrDWQ7hPc",
        "timestamps": [
            510
        ],
        "3min_transcript": "to keep switching colors, but hopefully it's worth it, a plus b. Let's take that to the 4th power. The binomial theorem tells us this is going to be equal to, and I'm just going to use this exact notation, this is going to be the sum from k equals 0, k equals 0 to 4, to 4 of 4 choose k, 4 choose k, 4 choose ... let me do that k in that purple color, 4 choose k of a to the 4 minus k power, 4 minus k power times b to the k power, b to the k power. Now what is that going to be equal to? Well, let's just actually just do the sum. This is going to be equal to, so we're going to start at k equals 0, so when k equals 0, it's going to be 4 choose 0, times a to the 4 minus 0 power, well, that's just going to be a to the 4th power, times b to the 0 power. b to the 0 power is just going to be equal to 1, so we could just put a 1 here if we want to, or we could just leave it like that. This is what we get when k equals 0. Then to that, we're going to add when k equals 1. k equals 1 is going to be, the coefficient is going to be 4 choose 1, and it's going to be times a to the 4 minus 1 power, so a to the 3rd power, and I'll just stick with that color, times b to the k power. Well, now, k is 1b to the 1st power. Then to that, we're going to add, we're going to add 4 choose 2, 4 choose 2 times a to the ... 4 minus 2 is 2. I think you see a pattern here. a to the 4th, a to the 3rd, a squared, and then times b to the k. Well, k is 2 now, so b squared, and you see a pattern again. You could say b to the 0, b to the 1, b squared, and we only have two more terms to add here, plus 4 choose 3, 4 choose 3 times 4 minus 3 is 1, times a, or a to the 1st, I guess we could say, and then b to the 3rd power, times a to the 1st b to the third, and then only one more term, plus 4 choose, 4 choose 4. k is now 4. This is going to be our last term right now. We're getting k goes from 0 all the way to 4, 4 choose 4. a to the 4 minus 4, that's just going to be 1, a to the 0, that's just 1, so we're going to be left with just b to the k power,"
    },
    {
        "Q": "At 4:39 point, when writing \"n choose k\" for the first time, you say, \"We'll review that in a second. This comes straight of out ?\" I didn't hear that part. It comes straight out of WHAT?",
        "A": "There is a term called Combination,which states that each of the different groups or selection which can be made out by taking some or all of a number of things at a time.or simply Selection of r terms out of n terms....that part is derived from this very term.....selection of r terms out of n terms..... nCr = n!/r!(n-r)! ....",
        "video_name": "iPwrDWQ7hPc",
        "timestamps": [
            279
        ],
        "3min_transcript": "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, Now this seems a little bit unwieldy. Let's just review, remind ourselves what n choose k actually means. If we say n choose k, I'll do the same colors, n choose k, we remember from combinatorics this would be equal to n factorial, n factorial over k factorial, over k factorial times n minus k factorial, n minus k factorial, so n minus k minus k factorial, let me color code this, n minus k factorial. Let's try to apply this. Let's just start applying it to the thing that started to intimidate us, say, a plus b to the 4th power. Let's figure out what that's going to be. Let's try this. So a, and I'm going to try to keep it color-coded so you know what's going on, a plus b,"
    },
    {
        "Q": "Why 4! / 0!4! = 1? it's just ( 4 * 3 * 2* 1 ) / ( 0 * 4 * 3 * 2 * 1 ) = 24 / 0\nwhich is undefined. Why sal says it's equal to 1? at 9:37",
        "A": "0 factorial does not equal zero. By definition it equals 1.",
        "video_name": "iPwrDWQ7hPc",
        "timestamps": [
            577
        ],
        "3min_transcript": "4 minus 2 is 2. I think you see a pattern here. a to the 4th, a to the 3rd, a squared, and then times b to the k. Well, k is 2 now, so b squared, and you see a pattern again. You could say b to the 0, b to the 1, b squared, and we only have two more terms to add here, plus 4 choose 3, 4 choose 3 times 4 minus 3 is 1, times a, or a to the 1st, I guess we could say, and then b to the 3rd power, times a to the 1st b to the third, and then only one more term, plus 4 choose, 4 choose 4. k is now 4. This is going to be our last term right now. We're getting k goes from 0 all the way to 4, 4 choose 4. a to the 4 minus 4, that's just going to be 1, a to the 0, that's just 1, so we're going to be left with just b to the k power, We're almost done. We've expanded it out. We just need it figure out what 4 choose 0, 4 choose 1, 4 choose 2, et cetera, et cetera are, so let's figure that out. We could just apply this over and over again. So 4 choose 0, 4 choose 0 is equal to 4 factorial over 0 factorial times 4 minus 0 factorial. That's just going to be 4 factorial again. 0 factorial, at least for these purposes, we are defining to be equal to 1, so this whole thing is going to be equal to 1, so this coefficient is 1. Let's keep going here. So 4 choose 1 is going to be 4 factorial over 1 factorial times 4 minus 1 factorial, 4 minus 1 factorial, so 3 factorial. What's this going to be? 1 factorial is just going to be 1. 3 factorial is 3 times 2 times 1. 4 times 3 times 2 times 1 over 3 times 2 times 1 is just going to leave us with 4. This right over here is just going to be 4. Then we need to figure out what 4 choose 2 is. 4 choose 2 is going to be 4 factorial over 2 factorial times what's 4 minus ... this is going to be n minus k, 4 minus 2 over 2 factorial. So what is this going to be? Let me scroll over to the right a little bit. This is going to be 4 times 3 times 2 times 1 over 2 factorial is 2, over 2 times 2. This is 2, this is 2, so 2 times 2 is same thing as 4. We're left with 3 times 2 times 1, which is equal to 6. That's equal to 6. Then what is 4 choose 3? I'll use some space down here. So 4 choose 3,"
    },
    {
        "Q": "At 4:21, what is the sideways W symbol, and what does it mean?",
        "A": "The sideways W is a Greek letter known as a Sigma. It indicates a pattern.",
        "video_name": "iPwrDWQ7hPc",
        "timestamps": [
            261
        ],
        "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": "at 2:35, when looking to draw the vector [1,2], I don't understand why the x component should be 1 and the Y component should be 2. Isn't the desired output, based on y = 1 and x = 2?",
        "A": "There s no mistake in there. The input coordinates are (2,1). And according to the partial derivative of the given function, the output is a vector field with x-component equal to the ordinate and y- component equal to the abscissa. So, the output vectors are given by = yi + xj , where i and j are the basis vectors for x and y axes. So the output for (2,1) will be 1i+2j",
        "video_name": "ZTbTYEMvo10",
        "timestamps": [
            155
        ],
        "3min_transcript": "y equals two over x. And that's where you would see something like this. So all of these lines, they're representing constant values for the function. And now I want to take a look at the gradient field. And the gradient, if you'll remember, is just a vector full of the partial derivatives of f. And let's just actually write it out. The gradient of f, with our little del symbol, is a function of x and y. And it's a vector-valued function whose first coordinate is the partial derivative of f with respect to x. And the second component is the partial derivative with respect to y. So when we actually do this for our function, we take the partial derivative with respect to x. X looks like a variable. Y looks like a constant. The derivative of this whole thing is just equal to that constant, y. And then kind of the reverse for when you take the partial derivative with respect to y. 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."
    },
    {
        "Q": "at 1:21 didnt understand the formulae",
        "A": "The formula is showing there is a correlation between the angle of the sector and the area of the sector. If we know the angle and the area of the whole circle we can find the area of the sector, since they are similar. For more detail: 0:28",
        "video_name": "u8JFdwmBvvQ",
        "timestamps": [
            81
        ],
        "3min_transcript": "A circle with area 81 pi has a sector with a 350-degree central angle. So this whole sector right over here that's shaded in, this pale orange-yellowish color, that has a 350-degree central angle. So you see the central angle, it's a very large angle. It's going all the way around like that. And they ask us, what is the area of the sector? So we just need to realize that the ratio between the area of the sector and the total area of the circle. And they tell us what the total area is. It's 81 pi. And 81 pi is going to be equal to the ratio of its central angle, which is 350 degrees, over the total number of degrees in a circle-- over 360. So the area of the sector over the total area over the total degrees in a circle. And then we just can solve for area of a sector by multiplying both sides by 81 pi. 81 pi, 81 pi-- so these cancel out. 350 divided by 360 is 35/36. And so our area, our sector area, is equal to-- let's see, in the numerator, we have 35 times-- instead of 81, let's see, that's going to be 9 times 9 pi. And in the denominator, I have 36. Well, that's the same thing as 9 times 4. And so we can divide the numerator and the denominator both by 9, and so we are left with 35 times 9. And neither of these are divisible by 4, so that's about as simplified as we can get it. 35 times 9, it's going to be 350 minus 35, which would be 315, I guess. Did I do that right? Yeah, it's going to be 270 plus 45, which is 315 pi over 4. 315 pi over 4 is the area of the sector."
    },
    {
        "Q": "At 1:40 where did he get nine times nine from? Also he divided the 350/360 by ten should he divide the other side by ten also?",
        "A": "Hey Janet, 9*9 is the same thing as 81. With fractions, if you divide the numerator by 10, you divide the denominator by 10 as well. You don t need to divide the other side, it s just simplifying a fraction and still the same number after all.",
        "video_name": "u8JFdwmBvvQ",
        "timestamps": [
            100
        ],
        "3min_transcript": "A circle with area 81 pi has a sector with a 350-degree central angle. So this whole sector right over here that's shaded in, this pale orange-yellowish color, that has a 350-degree central angle. So you see the central angle, it's a very large angle. It's going all the way around like that. And they ask us, what is the area of the sector? So we just need to realize that the ratio between the area of the sector and the total area of the circle. And they tell us what the total area is. It's 81 pi. And 81 pi is going to be equal to the ratio of its central angle, which is 350 degrees, over the total number of degrees in a circle-- over 360. So the area of the sector over the total area over the total degrees in a circle. And then we just can solve for area of a sector by multiplying both sides by 81 pi. 81 pi, 81 pi-- so these cancel out. 350 divided by 360 is 35/36. And so our area, our sector area, is equal to-- let's see, in the numerator, we have 35 times-- instead of 81, let's see, that's going to be 9 times 9 pi. And in the denominator, I have 36. Well, that's the same thing as 9 times 4. And so we can divide the numerator and the denominator both by 9, and so we are left with 35 times 9. And neither of these are divisible by 4, so that's about as simplified as we can get it. 35 times 9, it's going to be 350 minus 35, which would be 315, I guess. Did I do that right? Yeah, it's going to be 270 plus 45, which is 315 pi over 4. 315 pi over 4 is the area of the sector."
    },
    {
        "Q": "At 0:28, Mr. Khan mentions a ratio. What is that?",
        "A": "He basically created a proportion using the values given in the circle.",
        "video_name": "u8JFdwmBvvQ",
        "timestamps": [
            28
        ],
        "3min_transcript": "A circle with area 81 pi has a sector with a 350-degree central angle. So this whole sector right over here that's shaded in, this pale orange-yellowish color, that has a 350-degree central angle. So you see the central angle, it's a very large angle. It's going all the way around like that. And they ask us, what is the area of the sector? So we just need to realize that the ratio between the area of the sector and the total area of the circle. And they tell us what the total area is. It's 81 pi. And 81 pi is going to be equal to the ratio of its central angle, which is 350 degrees, over the total number of degrees in a circle-- over 360. So the area of the sector over the total area over the total degrees in a circle. And then we just can solve for area of a sector by multiplying both sides by 81 pi. 81 pi, 81 pi-- so these cancel out. 350 divided by 360 is 35/36. And so our area, our sector area, is equal to-- let's see, in the numerator, we have 35 times-- instead of 81, let's see, that's going to be 9 times 9 pi. And in the denominator, I have 36. Well, that's the same thing as 9 times 4. And so we can divide the numerator and the denominator both by 9, and so we are left with 35 times 9. And neither of these are divisible by 4, so that's about as simplified as we can get it. 35 times 9, it's going to be 350 minus 35, which would be 315, I guess. Did I do that right? Yeah, it's going to be 270 plus 45, which is 315 pi over 4. 315 pi over 4 is the area of the sector."
    },
    {
        "Q": "The result of the composition of Ix and g has to be the same as the result of the composition of h and f and g when one inputs a 'y' value and gets an 'x' value (at time marker 16:40). But how is showing that if the results are the same, then the functions are the same as well? (abstractly speaking)",
        "A": "Speaking any kind of way (abstractly or otherwise), x = g(y) = I_X(g(y)) = h(f(g(y))) = h((I_Y)(y)) = h(y). So g(y) = h(y), and g = h. (I don t really understand this either yet.) : |",
        "video_name": "-eAzhBZgq28",
        "timestamps": [
            1000
        ],
        "3min_transcript": "I could do the same thing here with h. I just take a point here, apply h, then apply f back. I should just go back to that point. That's all of what this is saying. So let's go back to the question of whether g is unique. Can we have two different inverse functions g and h? So let's start with g. Remember g is just a mapping from Y to X. So this is going to be equal to, this is the same thing as the composition of the identity function over x with g. To show you why that's the case, remember g just goes from-- these diagrams get me confused very quickly-- so let's say this is x and this is y. Remember g is a mapping from y to x. So g will take us there. There's a mapping from y to x. mapping, or the identity function in composition with this. Because all this is saying is you apply g, and then you apply the identity mapping on x. So obviously you're going to get to the exact same mapping or the exact same point. So these are equivalent. But what is another way of writing the identity mapping on x? What's another way of writing that? Well by definition, if h is another inverse of f, this is true. So I can replace this in this expression with a composition of h with f. So this is going to be equal to the composition of h with f, and the composition of that with g. You might want to put parentheses here. I'll do it very lightly. You might want to put parentheses there. But I showed you a couple of videos ago that the composition of functions, or of transformations, is associative. It doesn't matter if you put the parentheses there or if you put the parentheses there. Actually I'll do that. I'll put the parentheses there at first just so you can as that right there. But we know that composition is associative. So this is equal to the composition of h with the composition of f and g. Now what is this equal to, the composition of f and g? Well it's equal to, by definition, it's equal to the identity transformation over y. So this is equal to h composed with, or the composition of h with, the identity function over y with this right here. Now what is this going to be? Remember h is a mapping from y to x. Let me redraw it. So that's my x and that is my y. h could take some element in y and gives me some element in x. If I take the composition of the identity in y-- so that's essentially I take some element, let me do it this"
    },
    {
        "Q": "3:41 OK really confused. why do the repeating numbers start at 4 and not 1. Sal writes 414141... but shouldnt it be 14141414...\ncan anyone explain this?",
        "A": "x=.714141414... If you multiplied by x by 10, then that would move it one place to the right, or 7.14141414... and the 141414... would start immediately after the decimal point. Since you have to multiply by 100, however, you move it two places to the right and get: 7 1. 4 1 4 1 4 1 4. The numbers after the decimal must start with the 4 first since you moved it two places to the right. The 141414... pattern is still there. You re just starting at a different place.",
        "video_name": "Ihws0d-WLzU",
        "timestamps": [
            221
        ],
        "3min_transcript": "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 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."
    },
    {
        "Q": "This video was really helpful, I just have one question. How do you whether to use 10x or 100x or 1000x at 1:01 in the video?",
        "A": "How many different numbers are repeating? If it is .333... , one repeating times 10, if it is ,232323... two repeating, so times 100, three repeating .234234234... times 1000. A zero for each repeating number.",
        "video_name": "Ihws0d-WLzU",
        "timestamps": [
            61
        ],
        "3min_transcript": "In the last video, we did some examples where we had one digit repeating on and on forever, and we were able to convert those into fractions. In this video, we want to tackle something a little bit more interesting, which is multiple digits repeating on and on forever. So let's say I had 0.36 repeating, which is the same thing as 0 point-- since the bar's over the 3 and the 6, both of those repeat-- 363636. And it just keeps going on and on and on like that forever. Now the key to doing this type of problem is, so like we did in the last video, we set this as equal to x. And instead of just multiplying it by 10-- 10 would only shift it one over-- we want to shift it over enough so that when we line them up, the decimal parts will still line up with each other. And to do that we, want to actually shift the decimal space two to the right. And to shift it two to the right, we have to have multiplied by 100 or 10 to the second power. So 100x is going to be equal to what? 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."
    },
    {
        "Q": "At 4:04 can we multiply by 10?",
        "A": "If we multiply by 10, the decimal point would only shift to the right one point. we need the point shifted over twice, so we multiply by 100.",
        "video_name": "Ihws0d-WLzU",
        "timestamps": [
            244
        ],
        "3min_transcript": "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 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."
    },
    {
        "Q": "At 9:22, how do you turn 0.00123 with the 3 repeating into a decimal?",
        "A": "x = 0.001233... 100000x = 123.33... 10000x = 12.33... 90000x = 111 x = 111/90000 = 37/30000",
        "video_name": "Ihws0d-WLzU",
        "timestamps": [
            562
        ],
        "3min_transcript": "And then you can divide both sides of this by 999. And you are left with x is equal to 3,254/999. And so obviously, this is an improper fraction. The numerator is larger than the denominator. You could convert this to a proper fraction if you like. One way, you could have just tried to figure out what to the 0.257 repeating forever is equal to and just had the 3 being the whole number part of a mixed fraction. Or you could just divide 999 into 3,254. Actually, we could do that pretty straightforwardly. It goes into it three times, and the remainder-- well, let me just do it, just to go through the motions. So 999 goes into 3,254. It'll go into it three times. And we know that because this is originally 3.257, So 3 times 9 is 27. But we have to add the 2, so it's 29. 3 times 9 is 27. We have a 2, so it's 29. And so we are left with, if we subtract, if we regroup or borrow or however we want to call it, this could be a 14. And then this could be a 4. Let me do this in a new color. And then the 4 is still smaller than this 9, so we need to regroup again. So then this could be a 14, and then this could be a 1. But this is smaller than this 9 right over here, so we regroup again. This would be an 11, and then this is a 2. 14 minus 7 is 7. 14 minus 9 is 5. 11 minus 9 is 2. So we are left with-- did I do that right? Yep-- so this is going to be equal to 3 and 257/999."
    },
    {
        "Q": "at 4:00, he could have just multiplied it by ten. that would have been enough to get the seven on the other side of the decimal point. why'd he multiply it by 100?",
        "A": "That s because you still want the decimal points and the repeating part (which was 141414...in this case) to line up.",
        "video_name": "Ihws0d-WLzU",
        "timestamps": [
            240
        ],
        "3min_transcript": "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 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."
    },
    {
        "Q": "at around 5:02, why didnt he multiply 4 into the number to the right of the greater than symbol? If you are making a change to one side wouldn't you have to do it to the other, because it would affect the end result?",
        "A": "He did. He said, let s multiply both sides by 4 and he pointed to both sides of the inequality. 4 ( p\u00c2\u00b2 - 17/4 p+ 1 > 0 ) becomes 4 \u00e2\u0088\u0099 p\u00c2\u00b2 - 4 \u00e2\u0088\u0099 17/4 p + 4 \u00e2\u0088\u0099 1 > 4 \u00e2\u0088\u0099 0 which is 4p\u00c2\u00b2 - 17p + 4 > 0 because zero times anything is still zero He probably didn t mention zero times 4 is zero because he guessed we knew that by the time we are doing quadratic inequalities because we run into it so much when solving quadratics and other equations.",
        "video_name": "GDppV18XDCs",
        "timestamps": [
            302
        ],
        "3min_transcript": "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. the square root of negative 17 squared-- b squared-- so that's 289 minus 4 times a times c. Well a times c is 16 times 4, so minus 64. All of that over 2 times a-- all of that over 8. So that's 17 plus or minus-- let's see, this is the square root of 225 over 8, which is equal to 17 plus or minus 15 over 8, which is equal to-- let's see, 17 minus 15 over 8 is 2/8-- which is equal to 2/8 or 1/4. So that's one of them. That's when we take the minus. And if we add 17 plus 15, that gets us to 32 divided by 8"
    },
    {
        "Q": "in 1:16, why did the voice changed?",
        "A": "Yes, Vi did get farther from her mic, thats why her voice got deeper and quieter",
        "video_name": "4tsjCND2ZfM",
        "timestamps": [
            76
        ],
        "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": "At 2:37 What does beat a dead horse mean?",
        "A": "It is an idiom, meaning it is not to be taken literally, and it means to waste time and/or energy repeating something in excess or doing something otherwise that will be non-helpful or productive.",
        "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": "At 4:14, how come he didn't turn the fraction into a decimal?",
        "A": "Sal solves two problems where a missing side length is found by proving that triangles are similar and using this to find the measure.",
        "video_name": "R-6CAr_zEEk",
        "timestamps": [
            254
        ],
        "3min_transcript": "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. This is last and the first. Last and the first. CA over CE. And we know what BC is. BC right over here is 5. We know what DC is. It is 3. We know what CA or AC is right over here. CA is 4. And now, we can just solve for CE. Well, there's multiple ways that you could think about this. You could cross-multiply, which is really just multiplying both sides by both denominators. So you get 5 times the length of CE. 5 times the length of CE is equal to 3 times 4, which is just going to be equal to 12. And then we get CE is equal to 12 over 5, which is the same thing as 2 and 2/5, or 2.4. And we're done. We were able to use similarity to figure out this side just knowing that the ratio between the corresponding sides are going to be the same. Now, let's do this problem right over here. Let's do this one. Let me draw a little line here to show that this is a different problem now. This is a different problem. So in this problem, we need to figure out what DE is. And we, once again, have these two parallel lines like this. And so we know corresponding angles are congruent. So we know that angle is going to be congruent to that angle because you could view this as a transversal. We also know that this angle right over here is going to be congruent to that angle right over there. Once again, corresponding angles for transversal. And also, in both triangles-- so I'm looking at triangle CBD and triangle CAE-- they both share this angle up here."
    },
    {
        "Q": "At 2:51, how do you figure out which line segment to put together when you are trying to figure out the missing length?",
        "A": "Since you are looking for the side CE, notice that it is the third letter to first letter of second triangle. With the congruency statement, the same two letters are CA, so they are one of three pair of congruent sides. Does this answer your question?",
        "video_name": "R-6CAr_zEEk",
        "timestamps": [
            171
        ],
        "3min_transcript": "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. This is last and the first. Last and the first. CA over CE. And we know what BC is. BC right over here is 5. We know what DC is. It is 3. We know what CA or AC is right over here. CA is 4. And now, we can just solve for CE. Well, there's multiple ways that you could think about this. You could cross-multiply, which is really just multiplying both sides by both denominators. So you get 5 times the length of CE. 5 times the length of CE is equal to 3 times 4, which is just going to be equal to 12. And then we get CE is equal to 12 over 5, which is the same thing as 2 and 2/5, or 2.4."
    },
    {
        "Q": "at 6:16, would it work if you wrote CA/CB = CE/CD instead of CB/CA=CD/CE? because I got pretty confused.",
        "A": "yes it would work",
        "video_name": "R-6CAr_zEEk",
        "timestamps": [
            376
        ],
        "3min_transcript": "And we're done. We were able to use similarity to figure out this side just knowing that the ratio between the corresponding sides are going to be the same. Now, let's do this problem right over here. Let's do this one. Let me draw a little line here to show that this is a different problem now. This is a different problem. So in this problem, we need to figure out what DE is. And we, once again, have these two parallel lines like this. And so we know corresponding angles are congruent. So we know that angle is going to be congruent to that angle because you could view this as a transversal. We also know that this angle right over here is going to be congruent to that angle right over there. Once again, corresponding angles for transversal. And also, in both triangles-- so I'm looking at triangle CBD and triangle CAE-- they both share this angle up here. but we've actually shown that all three angles of these two triangles, all three of the corresponding angles, are congruent to each other. And once again, this is an important thing to do, is to make sure that you write it in the right order when you write your similarity. We now know that triangle CBD is similar-- not congruent-- it is similar to triangle CAE, which means that the ratio of corresponding sides are going to be constant. So we know, for example, that the ratio between CB to CA-- so let's write this down. is going to be equal to the ratio of CD over CE. And we know what CB is. CB over here is 5. We know what CA is. And we have to be careful here. It's not 3. CA, this entire side is going to be 5 plus 3. So this is going to be 8. And we know what CD is. CD is going to be 4. And so once again, we can cross-multiply. We have 5CE. 5 times CE is equal to 8 times 4. 8 times 4 is 32. And so CE is equal to 32 over 5. Or this is another way to think about that, 6 and 2/5."
    },
    {
        "Q": "At 2:20 Did When Sal said Sin 32, Does that also mean Sin A?",
        "A": "at 2:20 when sal says sin32, yes that is equivalent to sinA",
        "video_name": "yiH6GoscimY",
        "timestamps": [
            140
        ],
        "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 4:29 What is the exact difference between obtuse and acute triangles",
        "A": "in the obtuse triangle, it has one obtuse angle (bigger than a right angle) and in an acute angle, all angles are smaller that right angle. Ex: acute:all angles 60 ( 60+60+60=180) Ex: obtuse: one angle 120 another 35 and another 25 Hope this helped! Please vote this",
        "video_name": "D5lZ3thuEeA",
        "timestamps": [
            269
        ],
        "3min_transcript": "so it meets the constraints for an isosceles. So by that definition, all equilateral triangles are also isosceles triangles. But not all isosceles triangles are equilateral. So for example, this one right over here, this isosceles triangle, clearly not equilateral. All three sides are not the same. Only two are. But both of these equilateral triangles meet the constraint that at least two of the sides are equal. Now down here, we're going to classify based on angles. An acute triangle is a triangle where all of the angles are less than 90 degrees. So for example, a triangle like this-- maybe this is 60, let me draw a little bit bigger so I can draw the angle measures. 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."
    },
    {
        "Q": "At 8:00, wouldn't you have to times the number of years and the rate the interest is compounded for raising everything?",
        "A": "Yes, I suppose. But in this video the loan is only for one year.",
        "video_name": "BKGx8GMVu88",
        "timestamps": [
            480
        ],
        "3min_transcript": "your going have to multiply by this again. Times 1.083 repeating, and so that would get you 1.083 repeating squared. If you went all the way down 12 months ... let me get myself some space here. If you went all the way down 12 months ... let me just. I should way from the beginning 12 months, so another 10 months. What's the total interest you would have to pay over a year if you weren't able to keep coming up with the money? If you had to keep re-borrowing it. I kept compounding that interest. Well, you're going have to pay 1.083 to the ... this is for 1 month. You could view this as to the first power. This is for 2 months, so you're going have to pay this to the 12th power. We have compounded over 12 periods, 8 1/3% over 12 periods. If you wanted to write it in this form right over here, this would be the same Our original principal times 1 plus 100% divided by 12. Now we've divided our 100% into 12 periods, and we're going to compound that 12 times. We're going to take that to the 12th power. What is this going to equal to? This buisness over here. We can get a calculator out for that. I'll get my TI-85 out. What is this going to be equal to? We could do it a couple of ways. This is 1.083 repeating. Let's get our calculator out. We could do it a couple of ways. Let me write it this way. Your going to get the same value. I don't have to rewrite this one. I just did that there to kind of hopefully you'd see the kind of structure in this expression. 1 plus ... 100% is the same thing as 1. 1 divided by 12 to the 12th power. 2.613, I'll just round. interesting game you all most forgot about your financial troubles, and you're just intrigued by what happens if we keep going this. Here we compounded just ... we have 100% over here. Here we do 50% every 6 months. Here we do a 12th of 100%, 8 1/3% every 12 months until we get to this number. What happens if we did every day? Every day. If I borrowed a one dollar, and I'd say well gee I'm just going to ... each day I'm going to charge you charge you one three hundred sixty-fifth of a 100%. So, 100% divided 365, and I'm going to compound that 365 times. You're curious mathematically. You say well, what do we get then? What do we get after a year?"
    },
    {
        "Q": "At 0:35 couldn't you just make the two triangles into one rectangle with a height of 3\" and a width of 5\"?",
        "A": "No, but you could make the two triangles into a rectangle with a height of 3 and a width of 4 (the entire base of the pentagon is 8, so half of that would be 4)",
        "video_name": "7S1MLJOG-5A",
        "timestamps": [
            35
        ],
        "3min_transcript": "Find the area and perimeter of the polygon. So let's start with the area first. So the area of this polygon-- there's kind of two parts of this. First, you have this part that's kind of rectangular, or it is rectangular, this part right over here. And that area is pretty straightforward. It's just going to be base times height. So area's going to be 8 times 4 for the rectangular part. And then we have this triangular part up here. So we have this area up here. And for a triangle, the area is base times height times 1/2. And that actually makes a lot of sense. Because if you just multiplied base times height, you would get this entire area. You would get the area of that entire rectangle. And you see that the triangle is exactly 1/2 of it. If you took this part of the triangle and you flipped it over, you'd fill up that space. If you took this part of the triangle and you flipped it over, you'd fill up that space. So the triangle's area is 1/2 of the triangle's base times the triangle's height. So plus 1/2 times the triangle's base, is 4 inches. And so let's just calculate it. This gives us 32 plus-- oh, sorry. That's not 8 times 4. I don't want to confuse you. The triangle's height is 3. 8 times 3, right there. That's the triangle's height. So once again, let's go back and calculate it. So this is going to be 32 plus-- 1/2 times 8 is 4. 4 times 3 is 12. And so our area for our shape is going to be 44. Now let's do the perimeter. The perimeter-- we just have to figure out what's the sum of the sides. How long of a fence would we have to build if we wanted to make it around this shape, right along the sides of this shape? So the perimeter-- I'll just write P for perimeter. It's going to be equal to 8 plus 4 plus 5 plus this 5, this edge So I have two 5's plus this 4 right over here. So you have 8 plus 4 is 12. 12 plus 10-- well, I'll just go one step at a time. 12 plus 5 is 17. 17 plus 5 is 22. 22 plus 4 is 26. So the perimeter is 26 inches. And let me get the units right, too. Because over here, I'm multiplying 8 inches by 4 inches. So you get square inches. 8 inches by 3 inches, so you get square inches again. So this is going to be square inches. So area is 44 square inches. Perimeter is 26 inches. And that makes sense because this is a two-dimensional measurement. It's measuring something in two-dimensional space, so you get a two-dimensional unit. This is a one-dimensional measurement. It's only asking you, essentially, how long would a string have to be to go around this thing. And so that's why you get one-dimensional units."
    },
    {
        "Q": "At at 4:12 - 4:14, why did Sal write x^4 '+'... and not '-' ...?",
        "A": "Addition does not change signs of the original values. If he had used a minus sign, it would need to be distributed across the term terms in the parentheses, which would result in those values having the wrong signs. Hope this helps.",
        "video_name": "yAH3722GrP8",
        "timestamps": [
            252,
            254
        ],
        "3min_transcript": "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. we could write minus 2 times the principal square root of 3. And then out here you have an x to the fourth plus this. And you see, if you distributed this out, if you distribute this x squared, you get this term, negative x squared, square root of 6, and if you distribute it onto this, you'd get that term. So you could debate which of these two is more simple. Now I mentioned that this way I just did the distributive property twice. Nothing new, nothing fancy. But in some classes, you will see something called FOIL. And I think we've done this in previous videos. FOIL. I'm not a big fan of it because it's really a way to memorize a process as opposed to understanding that this is really just from the common-sense distributive property. But all this is is a way to make sure that you're multiplying everything times everything when you're multiplying two binomials times each other like this. And FOIL just says, look, first multiply the first term. So x squared times x squared is x to the fourth."
    },
    {
        "Q": "Could someone please tell me why at 3:40 -x^2*sqrt6+x^2*sqrt2=(sqrt2-sqrt6)x^2?\n\nShouldn't it be -x^2*sqrt6+x^2*sqrt2=sqrt2-sqrt6 since one of the x^2 is negative while the other is positive? Shouldn't the x^2s cancel out then.\n\nI'm not sure if I've just made a careless error or am just missing something here, but I don't know why you get a negative x^2 or an x^2 at all. Please help explain this.",
        "A": "So lets pretend for a minute that instead the expression was -6z + 2z We can rearrange them using the commutative property: = 2z - 6z And then factor out the z, using the distributive property: = z (2 - 6) Now we can replace z with x^2, and the numbers with their square roots and we can still do the same thing: -Sqrt(6)x^2 + Sqrt(2)x^2 = Sqrt(2)x^2 - Sqrt(6)x^2 = (Sqrt(2)-Sqrt(6))x^2",
        "video_name": "yAH3722GrP8",
        "timestamps": [
            220
        ],
        "3min_transcript": "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. we could write minus 2 times the principal square root of 3. And then out here you have an x to the fourth plus this. And you see, if you distributed this out, if you distribute this x squared, you get this term, negative x squared, square root of 6, and if you distribute it onto this, you'd get that term. So you could debate which of these two is more simple. Now I mentioned that this way I just did the distributive property twice. Nothing new, nothing fancy. But in some classes, you will see something called FOIL. And I think we've done this in previous videos. FOIL. I'm not a big fan of it because it's really a way to memorize a process as opposed to understanding that this is really just from the common-sense distributive property. But all this is is a way to make sure that you're multiplying everything times everything when you're multiplying two binomials times each other like this. And FOIL just says, look, first multiply the first term. So x squared times x squared is x to the fourth."
    },
    {
        "Q": "At 3:21 why is it sqrt2 - sqrt6 and not the other way around? Or would it still be the same either way?",
        "A": "It s equivalent, sal choose to do it like that so you only have to use one operation symbol i.e. rather than: -sqrt6 + sqrt2 He choose: sqrt2 - sqrt6 But both are equivalent",
        "video_name": "yAH3722GrP8",
        "timestamps": [
            201
        ],
        "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": "At 2:10 he says that if r=1 denominator is 0, and we can't divide by zero. But, in that case numerator would also be 0, since a-a*(1)^n=0. Isn't lim 0/0=1?",
        "A": "No, the limit of 0/0 is undefined, and since the limit is for the variable n and not for r, you cannot use any of the limit techniques to get rid of the 0/0.",
        "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": "@9:52 I don't understand how a smaller denomater is bigger than a larger denmotor ?",
        "A": "the larger the denominator the smaller the piece The smaller the denominator the larger the piece.",
        "video_name": "wbAxarp_Ug4",
        "timestamps": [
            592
        ],
        "3min_transcript": ""
    },
    {
        "Q": "at 5:30 he Sal says 3 forth of the pizza has cheese, why does he put the number like that, one on top of the other, what does it mean and why not put the 4 on top instead of the 3?",
        "A": "first i think you meant at 2:30. Sal s pizza had 1/4 with olives & 3/4 with cheese, right? now it makes sense 3 of the 4 slices of pizza are cheese while 1 slice is olives. now if it was 4 out of the 3 pieces have cheese that would mean you would have 3 slices of pizza with 4 of those 3 slices being cheese, that would make things a bit confusing.",
        "video_name": "kZzoVCmUyKg",
        "timestamps": [
            330
        ],
        "3min_transcript": ""
    },
    {
        "Q": "I still don't understand, at 0:33, why did he make reference of the pizza as an example of fractions?",
        "A": "the circle is the most simple to esplain fractions.",
        "video_name": "kZzoVCmUyKg",
        "timestamps": [
            33
        ],
        "3min_transcript": ""
    },
    {
        "Q": "At 4:00 what if you were given only two points where the function intersects the midline twice? How do I find period with those two points?",
        "A": "It depends on which two points you re given, but if you re referring to two consecutive points where the function intersects the midline, then the horizontal distance between them would be 1/2 of the period. Can you see it?",
        "video_name": "s4cLM0l1gd4",
        "timestamps": [
            240
        ],
        "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": "Why go through all the trouble described by Sal around 04:00 to find the period of a function if one can simply measure the distance between two consecutive maximums or two consecutive minimums?",
        "A": "How would you go about measuring it precisely?",
        "video_name": "s4cLM0l1gd4",
        "timestamps": [
            240
        ],
        "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 6:27, how did you get 110 degrees?",
        "A": "180 degrees is a straight line and 70 degrees is an angle. Since the 70 degree angle and the other angle he is trying to find are supplementary it means that the angles must add up to 180 degrees. So 180 degrees minus 70 degrees is equal to 110 degrees.",
        "video_name": "gRKZaojKeP0",
        "timestamps": [
            387
        ],
        "3min_transcript": "From this perspective it's kind of the top right angle and each intersection is the same. Now the same is true of the other corresponding angles. This angle right here in this example, it's the top left angle will be the same as the top left angle right over here. This bottom left angle will be the same down here. If this right here is 70 degrees, then this down here will also be 70 degrees. And then finally, of course, this angle and this angle will also be the same. So corresponding angles -- let me write these -- these are corresponding angles are congruent. Corresponding angles are equal. And that and that are corresponding, that and that, that and that, and that and that. Now, the next set of equal angles to realize are sometimes opposite angles. But if you take this angle right here, the angle that is vertical to it or is opposite as you go right across the point of intersection is this angle right here, and that is going to be the same thing. So we could say opposite -- I like opposite because it's not always in the vertical direction, sometimes it's in the horizontal direction, but sometimes they're referred to as vertical angles. Opposite or vertical angles are also equal. So if that's 70 degrees, then this is also 70 degrees. And if this is 70 degrees, then this right here is also 70 degrees. So it's interesting, if that's 70 degrees and that's 70 degrees, and if this is 70 degrees and that is also 70 degrees, so no matter what this is, this will also be the same thing because this is the same as that, that is the same as that. Now, the last one that you need to I guess kind of realize are green angle right there. You can see that when you add up the angles, you go halfway around a circle, right? If you start here you do the green angle, then you do the orange angle. You go halfway around the circle, and that'll give you, it'll get you to 180 degrees. So this green and orange angle have to add up to 180 degrees or they are supplementary. And we've done other videos on supplementary, but you just have to realize they form the same line or a half circle. So if this right here is 70 degrees, then this orange angle right here is 110 degrees, because they add up to 180. Now, if this character right here is 110 degrees, what do we know about this character right here? Well, this character is opposite or vertical to the 110 degrees so it's also 110 degrees."
    },
    {
        "Q": "1:47 What is a nonlinear correlation?",
        "A": "nonlinear correlation = non linear correlation linear correlation --> correlation assuming there is a line (straight). non --> not doing non-linear correlation --> correlation that assumes the line is not a straight line.",
        "video_name": "Jpbm5YgciqI",
        "timestamps": [
            107
        ],
        "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": "at 2:05 dose he mean negitive",
        "A": "Sal said 4+7 to see how many blocks were in between the points. At 2:05 if he had done 4+(-7) he would have gotten -3 which is impossible. there can t be -3 blocks between her house and the mall. Hope this helped.",
        "video_name": "PC_FoyewoIs",
        "timestamps": [
            125
        ],
        "3min_transcript": "Milena's town is built on a grid similar to the coordinate plane. She is riding her bicycle from her home at point negative 3, 4 to the mall at point negative 3, negative 7. Each unit on the graph denotes one city block. Plot the two points, and find the distance between Milena's home and the mall. So let's see, she's riding her bicycle from her home at the point negative 3, 4. So let's plot negative 3, 4. So I'll use this point right over here. So negative 3 is our x-coordinate. So we're going to go 3 to the left of the origin 1, 2, 3. That gets us a negative 3. And positive 4 is our y-coordinate. So we're going to go 4 above the origin. Or I should say, we're going to go 4 up. So we went negative 3, or we went 3 to the left. That's negative 3, positive 4. Or you could say we went positive 4, negative 3. This tells us what we do in the horizontal direction. This tells us what we do in the vertical direction. Now let's figure out where the mall is. It's at the point negative 3, negative 7. So negative 3, we went negative 3 along the horizontal direction and then negative 7 along the vertical direction. So we get to negative 3, negative 7 right over there. And now we need to figure out the distance between her home and the mall. Now, we could actually count it out, or we could just compute it. If we wanted to count it out, it's 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 blocks. So we could type that in. And another way to think about it is they have the exact same x-coordinate. They're both at the x-coordinate negative 3. The only difference between these two is what is happening in the y-coordinate. This is at a positive 4. This is at a negative 7. Positive 4, negative 7. So we're really trying to find the distance between 4 and negative 7. So if I were to say 4 minus negative 7, So we have 4 minus negative 7, which is the same thing as 4 plus 7, which is 11. Let's do a couple more. Carlos is hanging a poster in the area shown by the red rectangle. He is placing a nail in the center of the blue line. In the second graph, plot the point where he places the nail. So he wants to place a nail in the center of the blue line. The blue line is 6 units long. The center is right over here. That's 3 to the right, 3 to the left. So he wants to put the nail at the point x equals 0, y is equal to 4. So he wants to put it at x is equal to 0, y is equal to 4. That's this point right over here. So let's check our answer. Let's do one more."
    },
    {
        "Q": "@4:05 why didn't Sal write 20u^2v/10uv^2? why did we drop the exponent?",
        "A": "That did confuse me a bit, but the reason why is because we have to put the GCF in the bottom for it to be the same as 10uv(2u-v). If we did 20u^2v/10uv^2 along with the other part, we would be left with our answer being 10uv(2u/v-1), which is not the same as 10uv(2u-v).",
        "video_name": "499MvHFrqUU",
        "timestamps": [
            245
        ],
        "3min_transcript": "If we wrote 2 times 5 times u times v, and we say that's going to be-- this expression is equal to this times what? Well if you factor the 2 times 5 times u times v out, all you're going to be left with in this first term is the 2 times u, so 2u here. And in the second term, all you're going to be left with is a v. All this other stuff gets factored out. All you're going to be left with is a v. Hopefully you see, if I multiply 2 times 5 times u times v times 2u, I'm going to get this first term here. So if I were to distribute it, I would get this first term. And if I multiply 2 times 5 times u times v times this v over here, I'm going to get this second term. So this expression, and that expression is the exact same thing. We have factored it out, now we can simplify it a little bit. 2 times 5 times u times v we rewrite as 10uv. a 2u and then a minus v. And we're done! We have factored the expression. Now you won't be doing it to this granular level, but this is the best way to think about it. Eventually you're going to say, hey, wait, look, the largest number that divides both of these is a 10. Because you could see 10 goes into the 20, 10 goes into 10. And, let's see, a u goes into both of these, and a v goes So let me factor out a 10uv, and then if I divide this thing by 10uv, I'm going to be left with 2u. And if I divide this by 10uv, I'm going to just be left with a v. So that's another way to think about it. Let me do that right now, so we could say that this is the same thing. Another way of approaching it, you could have said that this is the same thing as-- Well, the largest number that divides both of these is 10uv, and that's going to be times 20u squared v over 10uv minus this thing. This expression and this are obviously the same thing. If I were to distribute the 10uv it would cancel out with each of these in the denominator right there. So they're the same thing, but we can do is we can simplify this. We could say that 20 divided by 10 is just 2, u squared divided by u is just a u, v divided by v is just 1, 10 divided by 10 is 1, u divide by u is u, v squared divided by v is just a v to the first power. So you're left with 10uv times the quantity 2u minus v. Either way you get the same answer."
    },
    {
        "Q": "At 5:19, Sal draws a \"cone\", however it appears to be two cones on top of each other (tip to tip). Why is this?",
        "A": "The correct term for the solid is double-napped cone . Essentially two congruent cones with the same axis and a common vertex. Sal just didn t use the formal name for the solid.",
        "video_name": "0A7RR0oy2ho",
        "timestamps": [
            319
        ],
        "3min_transcript": "I want to go right through the-- that's pretty good. These are asymptotes. Those aren't the actual hyperbola. But a hyperbola would look something like this. They get to be right here and they get really close to the asymptote. They get closer and closer to those blue lines like that and it happened on this side too. The graphs show up here and then they pop over and they show up there. This magenta could be one hyperbola; I haven't done true justice to it. Or another hyperbola could be on, you could kind of call it a vertical hyperbola. That's not the exact word, but it would look something like that where it's below the asymptote here. It's above the asymptote there. So this blue one would be one hyperbola and then the magenta one would be a different hyperbola. So those are the different graphs. So the one thing that I'm sure you're asking is why are Why are they not called bolas or variations of circles or whatever? And in fact, wasn't even the relationship. It's pretty clear that circles and ellipses are somehow related. That an ellipse is just a squished circle. And maybe it even seems that parabolas and hyperbolas are somewhat related. This is a P once again. They both have bola in their name and they both kind of look like open U's. Although a hyperbola has two of these going and kind of opening in different directions, but they look related. But what is the connection behind all these? And that's frankly where the word conic comes from. So let me see if I can draw a three-dimensional cone. So this is a cone. That's the top. I could've used an ellipse for the top. Looks like that. Actually, it has no top. It would actually keep going on forever in that direction. This could be the bottom part of it. So let's take different intersections of a plane with this cone and see if we can at least generate the different shapes that we talked about just now. So if we have a plane that goes directly-- I guess if you call this the axis of this three-dimensional cone, so this is the axis. So if we have a plane that's exactly perpendicular to that axis-- let's see if I can draw it in three dimensions. The plane would look something like this. So it would have a line. This is the front line that's closer to you and then they would have another line back here. That's close enough. And of course, you know these are infinite planes, so it goes off in every direction. If this plane is directly perpendicular to the axis of these and this is where the plane goes behind it. The intersection of this plane and this cone is going to look like this."
    },
    {
        "Q": "At 9:00, as in trigonometry equation, why don't we put negative sign in front of the coefficient instead of the..mm....unknown? Ex. x= -3 cos (t), instead of x= 3 cos (-t)",
        "A": "Because trignometric functions are not commutative. a*cos(t) is not always the same as cos(a*t), even though sometimes it is.",
        "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": "Why at 2:44 does he write 2 dot 2 dot?",
        "A": "he is trying to say that 2 multiply by 2 multiply by 2 multiply by 2",
        "video_name": "lxjmR4pYIVU",
        "timestamps": [
            164
        ],
        "3min_transcript": "So let's do it that way first. So multiples of six are 6, 12, 18, 24 30. And I could keep going if we don't find any common multiples out of this group here with any of the multiples in eight. And the multiples of eight are 8, 16, 24, and it looks like we're done. And we could keep going obviously-- 32, so on and so forth. But I found a common multiple and this is their smallest common multiple. They have other common multiples-- 48 and 72, and we could keep adding more and more multiple. But this is their smallest common multiple, their least common multiple. So it is 24. Another way that you could have found at least common multiple is you could have taken the prime factorization of six and you say, hey, that's 2, and 3. So the least common multiple has to have at least 1, 2, and 1, 3 in its prime factorization in order for it And you could have said, what's the prime factorization of 8? It is 2 times 4 and 4 is 2 times 2. So in order to be divisible by 8, you have to have at least three 2's in the prime factorization. So to be divisible by 6, you have to have a 2 times a 3. And then to be divisible by 8, you have to have at least three 2's. You have to have two times itself three times I should say. Well, we have one 2 and let's throw in a couple more. So then you have another 2 and then another 2. So this part right over here makes it divisible by 8. And this part right over here makes it divisible by 6. If I take 2 times 2 times 2 times 3, that does give me 24. So our least common multiple of 8 and 6, which is also the least common denominator of these two fractions is going to be 24. So what we want to do is rewrite each of these fractions with 24 as the denominator. So I'll start with 2 over 8. Well, to get the denominator be 24, we have to multiply it by 3. 8 times 3 is 24. And so if we don't want to change the value of the fraction, we have to multiply the numerator and denominator by the same thing. So let's multiply the numerator by 3 as well. 2 times 3 is 6. So 2/8 is the exact same thing as 6/24. To see that a little bit clearer, you say, look, if I have 2/8, and if I multiply this times 3 over 3, that gives me 6/24. And this are the same fraction because 3 over 3 is really just 1. It's one whole. So 2/8 is 6/24 let's do the same thing with 5/6. So 5 over 6 is equal to something over 24."
    },
    {
        "Q": "from 3:30 to 3:34 what was he saying can you please explain",
        "A": "He was just saying... when he did 2/8 times 3 on the top and 3 on the bottom is the exact same as multiplying it by 3/3 :D Hope that helps =)",
        "video_name": "lxjmR4pYIVU",
        "timestamps": [
            210,
            214
        ],
        "3min_transcript": "So let's do it that way first. So multiples of six are 6, 12, 18, 24 30. And I could keep going if we don't find any common multiples out of this group here with any of the multiples in eight. And the multiples of eight are 8, 16, 24, and it looks like we're done. And we could keep going obviously-- 32, so on and so forth. But I found a common multiple and this is their smallest common multiple. They have other common multiples-- 48 and 72, and we could keep adding more and more multiple. But this is their smallest common multiple, their least common multiple. So it is 24. Another way that you could have found at least common multiple is you could have taken the prime factorization of six and you say, hey, that's 2, and 3. So the least common multiple has to have at least 1, 2, and 1, 3 in its prime factorization in order for it And you could have said, what's the prime factorization of 8? It is 2 times 4 and 4 is 2 times 2. So in order to be divisible by 8, you have to have at least three 2's in the prime factorization. So to be divisible by 6, you have to have a 2 times a 3. And then to be divisible by 8, you have to have at least three 2's. You have to have two times itself three times I should say. Well, we have one 2 and let's throw in a couple more. So then you have another 2 and then another 2. So this part right over here makes it divisible by 8. And this part right over here makes it divisible by 6. If I take 2 times 2 times 2 times 3, that does give me 24. So our least common multiple of 8 and 6, which is also the least common denominator of these two fractions is going to be 24. So what we want to do is rewrite each of these fractions with 24 as the denominator. So I'll start with 2 over 8. Well, to get the denominator be 24, we have to multiply it by 3. 8 times 3 is 24. And so if we don't want to change the value of the fraction, we have to multiply the numerator and denominator by the same thing. So let's multiply the numerator by 3 as well. 2 times 3 is 6. So 2/8 is the exact same thing as 6/24. To see that a little bit clearer, you say, look, if I have 2/8, and if I multiply this times 3 over 3, that gives me 6/24. And this are the same fraction because 3 over 3 is really just 1. It's one whole. So 2/8 is 6/24 let's do the same thing with 5/6. So 5 over 6 is equal to something over 24."
    },
    {
        "Q": "During 1:25-1:50, the LCD should be 0, shouldn't it?",
        "A": "Not exactly.. the way you get LCD ( Lowest common Denominator ) is by factorizing the denominators of fractions until you get something common. Sal was just looking for something common. and if you noticed he stopped at 24. I hope this helps.",
        "video_name": "lxjmR4pYIVU",
        "timestamps": [
            85,
            110
        ],
        "3min_transcript": "We're asked to rewrite the following two fractions as fractions with a least common denominator. So a least common denominator for two fractions is really just going to be the least common multiple of both of these denominators over here. And the value of doing that is then if you can make these a common denominator, then you can add the two fractions. And we'll see that in other videos. But first of all, let's just find the least common multiple. Let me write it out because sometimes LCD could meet other things. So least common denominator of these two things is going to be the same thing as the least common multiple of the two denominators over here. The least common multiple of 8 and 6. And a couple of ways to think about least common multiple-- you literally could just take the multiples of 8 and 6 So let's do it that way first. So multiples of six are 6, 12, 18, 24 30. And I could keep going if we don't find any common multiples out of this group here with any of the multiples in eight. And the multiples of eight are 8, 16, 24, and it looks like we're done. And we could keep going obviously-- 32, so on and so forth. But I found a common multiple and this is their smallest common multiple. They have other common multiples-- 48 and 72, and we could keep adding more and more multiple. But this is their smallest common multiple, their least common multiple. So it is 24. Another way that you could have found at least common multiple is you could have taken the prime factorization of six and you say, hey, that's 2, and 3. So the least common multiple has to have at least 1, 2, and 1, 3 in its prime factorization in order for it And you could have said, what's the prime factorization of 8? It is 2 times 4 and 4 is 2 times 2. So in order to be divisible by 8, you have to have at least three 2's in the prime factorization. So to be divisible by 6, you have to have a 2 times a 3. And then to be divisible by 8, you have to have at least three 2's. You have to have two times itself three times I should say. Well, we have one 2 and let's throw in a couple more. So then you have another 2 and then another 2. So this part right over here makes it divisible by 8. And this part right over here makes it divisible by 6. If I take 2 times 2 times 2 times 3, that does give me 24. So our least common multiple of 8 and 6, which is also the least common denominator of these two fractions is going to be 24. So what we want to do is rewrite each of these fractions with 24 as the denominator. So I'll start with 2 over 8."
    },
    {
        "Q": "why does x have to be greater than zero in the domain?\n\nTime on the video: 6:39",
        "A": "The log of 0 is undefined. The log of negative numbers involves rather difficult complex numbers, so it is usually treated as undefined at this stage of studying math.",
        "video_name": "DuYgVVU_BwY",
        "timestamps": [
            399
        ],
        "3min_transcript": "is so I get nice clean results that I can plot by hand. So let's actually graph it. Let's actually graph this thing over here. So the y's go between negative 2 and 2. The x's go from 1/25th all the way to 25. So let's graph it. So that is my y-axis, and this is my x-axis. Draw it like that. That is my x-axis. And then the y's start at 0. Then, you get to positive 1, positive 2. And then you have negative 1. And you have negative 2. And then on the x-axis, it's all positive. And I'll let you think about whether the domain here is-- well, when you think about it-- is a logarithmic function defined for an x that is not positive? No. You could raise five to an infinitely negative power to get a very, very, very, very small number that approaches zero, but you can never get-- there's no power that you can raise 5 to to get 0. So x cannot be 0. And there's no power then you could raise 5 to get another negative number. So x can also not be a negative number. So the domain of this function right over here-- and this is relevant, because we want to think about what we're graphing-- the domain here is x has to be greater than zero. Let me write that down. The domain here is that x has to be greater than 0. So we're only going to be able to graph this function in the positive x-axis. So with that out of the way, x gets as large as 25. So let me graph-- we put those points here. So that is 5, 10, 15, 20, and 25. And then let's plot these. So the first one is in blue. When x is 1/25 and y is negative 2-- is going to be really close to there-- Then y is negative 2. So it's going to be like right over there, not quite at the y-axis. We're at 1/25 to the right of the y-axis. So that's right over there. That is 1 over 25, comma negative 2 right over there. Then, when x is one fifth, which is slightly further to the right, one fifth y is negative 1. So right over there. So this is one fifth, negative 1. Then when x is 1, y is 0. So 1 might be right there. So this is the point 1,0. And then when x is 5, y is 1. When x is 5, I covered it over here, when this is five, y is 1. So that's the point 5,1. And then finally, when x is 25, y is 2."
    },
    {
        "Q": "At 11:30 or so, why do we ignore the normalization scalar that's being multiplied with the vectors? We can keep it aside until the end like that? Looks cool.",
        "A": "The Gram-Schmidt method is a way to find an orthonormal basis. To do this it is useful to think of doing two things. Given a partially complete basis we first find any vector that is orthogonal to these. Second we normalize. Then we repeat these two steps until we have filled out our basis. There are formulas that you could write down where both steps are taken care of in a long computation but it is useful to understand the process as a multiple step procedure which is what Sal is doing.",
        "video_name": "ZRRG386v6DI",
        "timestamps": [
            690
        ],
        "3min_transcript": "Let me switch colors . Minus v3 , which is 1, 1 0, 0 dotted with u2, dotted with the square root of 2/3 times 0, 1, 1/2, minus 1/2 times u2, times the vector u2, times the square root of 2/3, times the vector 0, 1, 1/2, minus 1/2. And what do we get? Let's calculate this. So we could take the-- so this is going to be equal to the vector 1, 1, 0, 0, minus-- so the 1 over the square root of 2 and the 1 over the square root of 2, multiply them. You're going to get a 1/2. And then when you take the dot product of these two, 1 times 0-- let's see, this is actually all going to be, if gets 0, right? So this guy, v3, was actually already orthogonal to u1. This will just go straight to 0, which is nice. We don't have to have a term right there. I took the dot product 1 times 0 plus 1 times 0 plus 0 times 1 plus 0 times 1, all gets zeroed. So this whole term drops out. We can ignore it, which makes our computation simpler. And then over here we have minus the square root of 2/3 times the square root of 2/3 is just 2/3 times the dot product of these two guys. So that's 1 times 0, which is 0, plus 1 times 1, which is 1, plus 0 times 1/2, which is 0, plus 0 times minus 1/2, which is 0, so we just get a 1 there, times the vector 0, 1, 1/2, minus 1/2. And then what do we get? We get-- this is the home stretch-- 1, 1, 0, 0 minus 2/3 So 2/3 time 0 is 0. 2/3 times 1 is 2/3. 2/3 times 1/2 is 1/3. And then 2/3 times minus 1/2 is minus 1/3. So then this is going to be equal to 1 minus 0 is 1, 1 minus 2/3 is 1/3, 0 minus 1/3 is minus 1/3, and then 0 minus minus 1/3 is positive 1/3. So this vector y3 is orthogonal to these two other vectors, which is nice, but it still hasn't been normalized. So we finally have to normalize this guy, and then we're done. Then we have an orthonormal basis. We'll have u1, u2, and now we'll find u3. So the length of my vector y-- actually, let's do something"
    },
    {
        "Q": "At 03:51, the last vector sal just wrote (0,0,1,1) is the vector v1, shouldn't it be the vector v2 instead (0,1,1,0)? Or is it the formula above that should be Y2=V2-Proj(V1)*u1 instead of v2 in the end?",
        "A": "The projection of v2 on v1 is in the direction of v1, so it s magnitude is multiplied by u1 = v1/||v1||.",
        "video_name": "ZRRG386v6DI",
        "timestamps": [
            231
        ],
        "3min_transcript": "So I can say that V is now equal to the span of the vectors u1, v2, and v3. Because I can replace v1 with this guy, because this guy is just a scaled-up version of this guy. So I can definitely represent him with him, so I can represent any linear combination of these guys with any linear combination of those guys right there. Now, we just did our first vector. We just normalized this one. But we need to replace these other vectors with vectors that are orthogonal to this guy right here. So let's do v2 first. So let's replace-- let's call it y2 is equal to v2 minus the projection of v2 onto the space spanned by u1 or onto-- you know, I could call it c times u1, or in the past videos, we called that subspace V1, but the space spanned by u1. which is 0, 1, 1, 0, minus-- v2 projected onto that space is just a dot product of v2, 0, 1, 1, 0, with the spanning vector of that space. And there's only one of them, so we're only going to have one term like this with u1, so dotted with 1 over the square root of 2 times 0, 0, 1, 1, and then all of that times u1. So 1 over the square root of 2 times the vector 0, 0, 1, 1. And so this is going to be equal to v2, which is 0, 1, 1, 0. The square root of 2, let's factor them out. So then you just get-- or kind of reassociate them out. over the square root of 2 is minus 1/2. You times-- what's the dot product of these two guys? You get 0 times 0 plus 1 times 0, which is still 0, plus 1 times 1 plus 0 times 0. So you're just going to have times 1 times this out here: 0, 0, 1, 1. I'll write that a little bit neater. I'm getting careless. 1, 1. So this is just going to be equal to 0, 1, 1, 0 minus-- 1/2 times 0 is 0. 1/2 times 0 is 0. Then I have two halves here. So y2 is equal to-- let's see, 0 minus 0 is 0, 1 minus 0 is 1, 1 minus 1/2 is 1/2, and then 0 minus 1/2 is minus 1/2."
    },
    {
        "Q": "I don't get it at 0:34 what does he mean?",
        "A": "Compute means to calculate, find, or figure out. He will show how to find the answer.",
        "video_name": "k68CPfcehTE",
        "timestamps": [
            34
        ],
        "3min_transcript": "Let's try to calculate 3 times 32. And I like to rewrite it, and this is one way of doing it. I like to rewrite it where I have a larger number on top. So in this case it's 32. And I write the smaller number right below it. And since the smaller number is only one digit, it's only a ones digit, I put that below the ones place on the larger number. So I'll put the 3 right over here. And of course, we can't forget our multiplication symbol. And this is essentially a way of saying the same thing. You could read this as 32 times 3. But 32 times 3 is the exact same value as 3 times 32. It doesn't matter what order you multiply in. Now let's try to compute it. And once again, this is only one way of doing it. There's many ways of doing it. And I want you to think about why this works. We'll start with this 3 down here, and we're going to multiply it times each of the digits in 32. So we'll start with 3 times 2. Well, 3 times 2 from our multiplication tables, and you can figure it out even if you didn't know your multiplication tables, is 6. So 3 times 2, I'll write 6 right over here in the ones place. Well, once again, we know that 3 times 3 is 9. And since I'm multiplying times the tens place right over here, I'm going to put it in the tens place right like this. We got 32 times 3 is 96. And I really encourage you to think about why this worked. And I'll give you a little bit of a hint here. I'll give you a little bit of a hint about why this worked. Remember, 3 times 32 is the same thing as 3 times 30 plus 3 times 2. And if you look at it that way, that's essentially what this process did. We did 3 times 2 is 6. 3 times 30 is 90. You add them together, you get 90 plus 6 is 96."
    },
    {
        "Q": "why does he say minus is it not negative at around 5:20",
        "A": "Any number minus another number is the same as saying that number plus the negative form of the other number. example: 7 - 5 is really the same as 7 + -5 Minus and negative are really just the same thing. I hope this helps you1 Unikitty <[:)]",
        "video_name": "d8lP5tR2R3Q",
        "timestamps": [
            320
        ],
        "3min_transcript": "or if I said negative 1 times negative 1 is equal to positive 1 as well. Or if I said 1 times negative 1 is equal to negative 1, or negative 1 times 1 is equal to negative 1. You see how on the bottom two problems I had two different signs, positive 1 and negative 1? And the top two problems, this one right here both 1s are positive. And this one right here both 1s are negative. So let's do a bunch of problems now, and hopefully it'll hit the point home, and you also could try to do along the practice problems and also give the hints and give you what rules to use, so that should help you as well. So if I said negative 4 times positive 3, well 4 times So different signs mean negative. So negative 4 times 3 is a negative 12. That makes sense because we're essentially saying what's negative 4 times itself three times, so it's like negative 4 plus negative 4 plus negative 4, which is negative 12. If you've seen the video on adding and subtracting negative numbers, you probably should watch first. Let's do another one. What if I said minus 2 times minus 7. And you might want to pause the video at any time to see if you know how to do it and then restart it to see what the answer is. Well, 2 times 7 is 14, and we have the same sign here, so it's a positive 14 -- normally you wouldn't have to write the positive but that makes it a little bit more explicit. And what if I had -- let me think -- 9 times negative 5. And once again, the signs are different so it's a negative. And then finally what if it I had -- let me think of some good numbers -- minus 6 times minus 11. Well, 6 times 11 is 66 and then it's a negative and negative, it's a positive. Let me give you a trick problem. What is 0 times negative 12? Well, you might say that the signs are different, but 0 is actually neither positive nor negative. And 0 times anything is still 0. It doesn't matter if the thing you multiply it by is a negative number or a positive number. 0 times anything is still 0. So let's see if we can apply these same rules to division. It actually turns out that the same rules apply."
    },
    {
        "Q": "at 5:52, I'm a little confused about the way Sal explains how there are 7 cubes that are not yellow or something. What does that mean?",
        "A": "It s easier to see from the table over on the left. When you want to count all the shapes that are either yellow or cubes, you add the seven yellow spheres, the five yellow cubes, and the eight green cubes.",
        "video_name": "QE2uR6Z-NcU",
        "timestamps": [
            352
        ],
        "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": "Why he needs to minus 5/29 at 7.25 at 7:24?",
        "A": "He subtracted 5/29 so he wouldn t count the overlapping area twice.",
        "video_name": "QE2uR6Z-NcU",
        "timestamps": [
            444
        ],
        "3min_transcript": "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, that we figured out earlier in the video. So let's think about this a little bit. We can rewrite this fraction right over here. We can rewrite this as 12 over 29 plus 13 over 29 minus 5 over 29. And this was the number of yellow over the total possibilities. So this right over here was the probability of getting a yellow. This right over here was the number of cubes over the total possibilities. So this is plus the probability of getting a cube. And this right over here is the number of yellow cubes over the total possibilities. So this right over here was minus the probability of yellow, and a cube. I'm not going to write it that way. Minus the probability of yellow-- I'll"
    },
    {
        "Q": "at 0:53 how is it 8:20,should it be 8:12",
        "A": "In this video, Sal is asking for the ratio of Apples to Fruit. In other words, how many apples compared to how many pieces of fruit total. Since 8 out of the 20 pieces of fruit are apples, the answer for this question is 8:20. However, if the question was What is the ratio of Apples to Oranges? your answer of 8:12 would be correct since there are 8 apples to 12 oranges.",
        "video_name": "UK-_qEDtvYo",
        "timestamps": [
            53,
            500,
            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": "could you do the same with 5:2 as 5/2?",
        "A": "Yes, they are equivalent ways of writing the same ratio.",
        "video_name": "UK-_qEDtvYo",
        "timestamps": [
            302
        ],
        "3min_transcript": "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": "there are 18 monkeys, 6 gorillas, and 15 apes what ratio is same to 5:13? Please someone help me i am having trouble",
        "A": "The total number of animals is 18 + 6 + 15 = 39, and there are 15 apes. Note that 5:13 is equivalent to 15:39, from multiplying each number in this ratio by 3. So 5:13 is ratio of the number of apes to the total number of animals Have a blessed, wonderful day!",
        "video_name": "UK-_qEDtvYo",
        "timestamps": [
            313
        ],
        "3min_transcript": "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": "how did sal reduce the original number/ 8:20 to 2:5",
        "A": "common factor of 8 and 20 is 4 and 8/4 = 2 and 20/4 = 5 so 2:5. does this help you?",
        "video_name": "UK-_qEDtvYo",
        "timestamps": [
            500,
            125
        ],
        "3min_transcript": ""
    },
    {
        "Q": "he said for the first question that the fruit was 8:20 when it was 8:12. why did he say that? or was it a mistake?",
        "A": "You are confusing fruit vs oranges . By fruit, Sal is referring to apples + oranges = 8+12 = 20 The ratio of apples to oranges = 8 : 12 The ratio of apples to fruit = 8 : 20 Hope this helps.",
        "video_name": "UK-_qEDtvYo",
        "timestamps": [
            500,
            492
        ],
        "3min_transcript": ""
    },
    {
        "Q": "is 2:5 the same as 2/5?",
        "A": "Yes, those are the same.",
        "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": "If ratios and fractions are technically the same then why do you use ratios or if you are thinking the other way around why do you use fractions? The only thing different I see if that fractions are used more often and they are written differently. ratios: 2:3 fractions: 2/3.",
        "A": "You should think of ratios as a special usage of fractions. They are used to compare similar quantities. Fractions can be used for many more purposes.",
        "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": "At 1:21 why can he only measure length?",
        "A": "He can only measure length because it is a line. A line is a one-dimensional object. A two dimensional objects allows you to measure width and length, and a three-dimensional object allows you to measure its height, width, and length",
        "video_name": "xMz9WFvox9g",
        "timestamps": [
            81
        ],
        "3min_transcript": "Human beings have always realized that certain things are longer than other things. For example, this line segment looks longer than this line segment. But that's not so satisfying just to make that comparison. You want to be able to measure it. You want to be able to quantify how much longer the second one is than the first one. And how do we go about doing that? Well, we define a unit length. So if we make this our unit length, we say this is one unit, then we could say how many of those the lengths are each of these lines? So this first line looks like it is-- we could do one of those units and then we could do it again, so it looks like this is two units. While this third one looks like we can get-- let's see that's 1, 2, 3 of the units. So this is three of the units. And right here, I'm just saying units. Sometimes we've made conventions to define a centimeter, where the unit might look something like this. And it's going to look different depending on your screen. Or we might have an inch that looks something like this. be able to fit on this screen based on how big I've just drawn the inch or a meter. So there's different units that you could use to measure in terms of. But now let's think about more dimensions. This is literally a one-dimensional case. This is 1D. Why is it one dimension? Well, I can only measure length. But now let's go to a 2D case. Let's go to two dimensions where objects could have a length and a width or a width and a height. So let's imagine two figures here that look like this. So let's say this is one of them. This is one of them. And notice, it has a width and it has a height. Or you could view it as a width and the length, depending on how you want to view it. So let's say this is one figure right over here. And let's say this is the other one. So this is the other one right over here. Try to draw them reasonably well. And we want to say, well, how much in two dimensions space is this taking up? Or how much area are each of these two taking up? Well, once again, we could just make a comparison. This second, if you viewed them as carpets or rectangles, the second rectangle is taking up more of my screen than this first one, but I want to be able to measure it. So how would we measure it? Well, once again, we would define a unit square. Instead of just a unit length, we now have two dimensions. We have to define a unit square. And so we might make our unit square. And the unit square we will define as being a square, where its width and its height are both equal to the unit length. So this is its width is one unit and its height is one unit. And so we will often call this 1 square unit. Oftentimes, you'll say this is 1 unit. And you put this 2 up here, this literally means 1 unit squared."
    },
    {
        "Q": "At 2:36, what is a \"unit Square\"?",
        "A": "Its the unit of the shape ( e.g. feet, inches ) squared. The square is the little 2 at the top. For example 56 ft squared.",
        "video_name": "xMz9WFvox9g",
        "timestamps": [
            156
        ],
        "3min_transcript": "be able to fit on this screen based on how big I've just drawn the inch or a meter. So there's different units that you could use to measure in terms of. But now let's think about more dimensions. This is literally a one-dimensional case. This is 1D. Why is it one dimension? Well, I can only measure length. But now let's go to a 2D case. Let's go to two dimensions where objects could have a length and a width or a width and a height. So let's imagine two figures here that look like this. So let's say this is one of them. This is one of them. And notice, it has a width and it has a height. Or you could view it as a width and the length, depending on how you want to view it. So let's say this is one figure right over here. And let's say this is the other one. So this is the other one right over here. Try to draw them reasonably well. And we want to say, well, how much in two dimensions space is this taking up? Or how much area are each of these two taking up? Well, once again, we could just make a comparison. This second, if you viewed them as carpets or rectangles, the second rectangle is taking up more of my screen than this first one, but I want to be able to measure it. So how would we measure it? Well, once again, we would define a unit square. Instead of just a unit length, we now have two dimensions. We have to define a unit square. And so we might make our unit square. And the unit square we will define as being a square, where its width and its height are both equal to the unit length. So this is its width is one unit and its height is one unit. And so we will often call this 1 square unit. Oftentimes, you'll say this is 1 unit. And you put this 2 up here, this literally means 1 unit squared. could've been a centimeter. So this would be 1 square centimeter. But now we can use this to measure these areas. And just as we said how many of this unit length could fit on these lines, we could say, how many of these unit squares can fit in here? And so here, we might take one of our unit squares and say, OK, it fills up that much space. Well, we need more to cover all of it. Well, there, we'll put another unit square there. We'll put another unit square right over there. We'll put another unit square right over there. Wow, 4 units squares exactly cover this. So we would say that this has an area of 4 square units or 4 units squared. Now what about this one right over here? Well, here, let's seem I could fit 1, 2, 3, 4, 5, 6, 7, 8, and 9. So here I could fit 9 units, 9 units squared."
    },
    {
        "Q": "at 6:11 why did he only count the first cubes showing e forgot the back that you cannot see",
        "A": "He didn t forget. He showed that there are two layers of 2 x 2 blocks. Each block is 1 unit^3 and there are 8 blocks so the volume is 8 units^3",
        "video_name": "xMz9WFvox9g",
        "timestamps": [
            371
        ],
        "3min_transcript": "We live in a three-dimensional world. Why restrict ourselves to only one or two? So let's go to the 3D case. And once again, when people say 3D, they're talking about 3 dimensions. They're talking about the different directions that you can measure things in. Here there's only length. Here there is length and width or width and height. And here, there'll be width and height and depth. So once again, if you have, let's say, an object, and now we're in three dimensions, we're in the world we live in that looks like this, and then you have another object that looks like this, it looks like this second object takes up more space, more physical space than this first object does. But how do we actually measure that? And remember, volume is just how much space something takes up in three dimensions. Area is how much space something takes up in two dimensions. Length is how much space something takes up in one dimension. But when we think about space, we're normally thinking about three dimensions. So how much space would you take up in the world that we live in? So just like we did before, we can define, instead of a unit length or unit area, we can define a unit volume or unit cube. So let's do that. Let's define our unit cube. And here, it's a cube so its length, width, and height are going to be the same value. So my best attempt at drawing a cube. And they're all going to be one unit. So it's going to be one unit high, one unit deep, and one unit wide. And so to measure volume, we could say, well, how many of these unit cubes can fit into these different shapes? won't be able to actually see all of them. I could essentially break it down into-- so let me see how well I can do this so that we can count them all. It's a little bit harder to see them all because there's some cubes that are behind us. But if you think of it as two layers, so one layer would look like this. One layer is going to look like this. So imagine two things like this stacked on top of each other. So this one's going to have 1, 2, 3, 4 cubes. Now, this is going to have two of these stacked on top of each other. So here you have 8 unit cubes. Or you could have 8 units cubed volume. What about here? If we try to fit it all in-- let me see how well I could draw this. It's going to look something like this. And obviously, this is kind of a rough drawing. And so if we were to try to take this apart, you would essentially have a stack of three sections that"
    },
    {
        "Q": "Let's say I was faced with a similar question to the one at 0:00. I used the vertical motion model and the quadratic formula to solve for the ball's time in motion. What would I do if both solutions of the equation are positive?",
        "A": "The only way for your scenario could be true is if the y intercept were negative, so he would be shooting the ball from below ground level. The smaller number would be here the ball initially reaches ground level, and the second would be where the ball hits the ground. So you would have to have an equation such as f(x) = -16x^2 + 20t - 40. Even if you shot it off at ground level, one solution would be zero (from the origin). In either case, the time would be the highest x value.",
        "video_name": "OZtqz_xw0SQ",
        "timestamps": [
            0
        ],
        "3min_transcript": "A ball is shot into the air from the edge of a building, 50 feet above the ground. Its initial velocity is 20 feet per second. The equation h-- and I'm guessing h is for height-- is equal to negative 16t squared plus 20t plus 50 can be used to model the height of the ball after t seconds. And I think in this problem they just want us to accept this formula, although we do derive formulas like this and show why it works for this type of problem in the Khan Academy physics playlist. But for here, we'll just go with the flow on this example. So they give us the equation that can be used to model the height of the ball after t seconds, and then say about how long does it take for the ball to hit the ground. So if this is the height, the ground is when the height is equal to 0. So hitting the ground means-- this literally means that h is equal to 0. So we need to figure out at which times does h equal 0. So we're really solving the equation 0 is equal to negative 16t squared plus 20t plus 50. everything here is divisible at least by 2. And let's divide everything by negative 2, just so that we can get rid of this negative leading coefficient. So you divide the left hand side by negative 2, you still get a 0. Negative 16 divided by negative 2 is 8. So 8t squared. 20 divided by negative 2 is negative 10. Minus 10t. 50 divided by negative 2 is minus 25. And so we have 8t squared minus 10t minus 25 is equal to 0. Or if you're comfortable with this on the left hand side, we can put on the left hand side. We could just say this is equal to 0. And now we solve. And we could complete this square here, or we can just apply the quadratic formula, which is derived from completing the square. And we have this in standard form. We know that this is our a. This right over here is our b. And the quadratic formula tells us that the roots-- and in this case, it's in terms of the variable t-- are going to be equal to negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. So if we apply it, we get t is equal to negative b. b is negative 10. So negative negative 10 is going to be positive 10. Plus or minus the square root of negative 10 squared. Well, that's just positive 100, minus 4 times a, which is 8, times c, which is negative 25. And all of that over 2a. a Is 8. So 2 times 8 is 16. And this over here, we have a-- let's see if we can simplify this a little bit. The negative sign, negative times a negative, these are going to be positive. 4 times 25 is 100, times 8 is 800."
    },
    {
        "Q": "At 3:46, Sal gives a formula.... is that formula a generic one or is it restricted to n=7?",
        "A": "It is generic . Sal chose 7 just for example purposes.",
        "video_name": "LwhJVURumAA",
        "timestamps": [
            226
        ],
        "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 about 1:11 Sal(?) says 1/2 times 1/2 = 1/4. Then at about 1:26 1/2 is added to 1/4 to get 3/4 chance of winning. Why is one added and the other multiplied?",
        "A": "The first case requires Brit to win both of the two times, whereas the second case requires Sal to win either time.",
        "video_name": "tDdtAF3WtIY",
        "timestamps": [
            71,
            86
        ],
        "3min_transcript": "Brit:How are we ever going to figure that out exactly? Let's just split the pot and call it even. Sal:Let's think about it this way, the next flip I have a 50% chance. Brit:I'm going to need to draw this out. Sal:Draw it out, draw it with trees. Brit:Okay so the next flip we'll go trees, so. Sal:It's either be heads or tails. Brit:Let's say it's heads or tails. Sal:Right, if it's heads Sal wins. Brit:If it's heads, Sal wins. Sal:And I get eight. Brit:And you get the whole pot, but if it's tails... Sal:If it's tails, then we keep playing. Brit:Oh so I need to do- Sal:Do another branch. Brit: Another branch. Brit:Oh I like this. Sal:So see they're going to be heads or tails. Now, at this point if it's heads I win. Brit:Sal wins. Sal:Sal wins, and if it's tails you win. Brit:Tails Brit wins. Sal:Yes. Brit:See, there's a chance I'm going to win, Sal:There is a chance, but your chance is not half. Your chance is substantially less than half. There's a one half chance of this happening. So this one has a half, and this one has a half. Sal:actually you could write, that's our first flip. then each of these outcomes are one half, one over two, one over two. So you have, to get two tails in a row, there's a one half time one half probability or there's only one fourth chance of this happening. And there's a one fourth chance of us getting a tails and another heads. So I have a one half plus one fourth chance of winning. One half plus one fourth, that's two fourths plus one fourth, I have a three fourths chance of winning. So I say, give me three fourths. I have a half chance of winning the next one. Brit:But there's eight chips here. Sal:Yeah, so this is how we think about it. I have a half chance of winning the next one, four chips for that. And then if I don't win the next one, I still have a half chance of winning that one. So I should get six chips, which is three fourths of eight. Brit:And I just get two. Sal:Unfortunately for you, that would be the case. Brit:You know I wanted half, but I can't argue with this."
    },
    {
        "Q": "At 0:32, they talk about adding more branches for each game. I wonder what would happen if they played 6 games instead of three?",
        "A": "Exactly the same would happen, only the difference between chance of winnings would not be that much.",
        "video_name": "tDdtAF3WtIY",
        "timestamps": [
            32
        ],
        "3min_transcript": "Brit:How are we ever going to figure that out exactly? Let's just split the pot and call it even. Sal:Let's think about it this way, the next flip I have a 50% chance. Brit:I'm going to need to draw this out. Sal:Draw it out, draw it with trees. Brit:Okay so the next flip we'll go trees, so. Sal:It's either be heads or tails. Brit:Let's say it's heads or tails. Sal:Right, if it's heads Sal wins. Brit:If it's heads, Sal wins. Sal:And I get eight. Brit:And you get the whole pot, but if it's tails... Sal:If it's tails, then we keep playing. Brit:Oh so I need to do- Sal:Do another branch. Brit: Another branch. Brit:Oh I like this. Sal:So see they're going to be heads or tails. Now, at this point if it's heads I win. Brit:Sal wins. Sal:Sal wins, and if it's tails you win. Brit:Tails Brit wins. Sal:Yes. Brit:See, there's a chance I'm going to win, Sal:There is a chance, but your chance is not half. Your chance is substantially less than half. There's a one half chance of this happening. So this one has a half, and this one has a half. Sal:actually you could write, that's our first flip. then each of these outcomes are one half, one over two, one over two. So you have, to get two tails in a row, there's a one half time one half probability or there's only one fourth chance of this happening. And there's a one fourth chance of us getting a tails and another heads. So I have a one half plus one fourth chance of winning. One half plus one fourth, that's two fourths plus one fourth, I have a three fourths chance of winning. So I say, give me three fourths. I have a half chance of winning the next one. Brit:But there's eight chips here. Sal:Yeah, so this is how we think about it. I have a half chance of winning the next one, four chips for that. And then if I don't win the next one, I still have a half chance of winning that one. So I should get six chips, which is three fourths of eight. Brit:And I just get two. Sal:Unfortunately for you, that would be the case. Brit:You know I wanted half, but I can't argue with this."
    },
    {
        "Q": "At about 1:20 he says, \"P = 0.25p equals to 1.25p\". How did it become \"1.25p\" from \"P = 0.25\"? Thanks!",
        "A": "Sal is simplifying the expression: P + 0.25P. Remember P has a coefficient of 1. P and 1P are the same. Sal is just adding like terms: P + 0.25P = 1P + 0.25P. Add 1+0.25 and you get 1.25. Thus, 1P + 0.25P = 1.25P. Hope this helps.",
        "video_name": "ao9cx8JlJIU",
        "timestamps": [
            80
        ],
        "3min_transcript": "Handsome Jack is buying a pony made of diamonds. The price of the pony is P dollars. And Jack also has to pay a 25% diamond pony tax. Match the expressions to their meaning for Handsome Jack. And they say multiple expressions may fit the same description. So in this bucket, we have the price of the diamond pony before tax. Well, they already tell us that the price of the diamond pony is P dollars. So that's P right over here. Now over here, they say the amount of tax Handsome Jack pays. Well, he pays a 25% diamond pony tax. So whatever the price is, he's going to pay 25% of that. Or another way of thinking about it, he's going to pay 25% is the same thing as 0.25. So 0.25 times P is the amount of tax he's going to pay on this diamond pony. Now, they say Handsome Jack's total bill for the diamond pony. Well, he's going to pay P for the pony plus 0.25P in taxes. P for the pony plus 0.25P for the taxes. So that's that one over there. But if we look at this, you could view this literally as 1P plus 0.25P's. Well, that's the same thing as 1.25P. So that's the same thing as this right over here. So it's 1.25P. And these other three don't seem to fit in any of these categories. So I'm going to put it into the not used. We're required to categorize everything. So let me put this in the not used. It's falling off the screen, I realize. Let me put this in the not used. And then let me put this in the not used. Let's check our answer. We got it right."
    },
    {
        "Q": "Since we are dividing by 4 at 1:16 wouldn't we write 4 at the beginning of the equation like this 4(x^2+10x-75=0?",
        "A": "If you use factoring, you would create your format: 4(x^2+10x-75) = 0 This is done sometimes, but it actually easier to complete the square if the 4 is gone completely. This can be done by dividing the entire equation by 4, which is the technique that Sal used. Hope this helps.",
        "video_name": "TV5kDqiJ1Os",
        "timestamps": [
            76
        ],
        "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": "At 5:35, Sal says that the radius will be smaller than it was before, but I'm confused because I thought that the radius of a unit sphere is always constant. Does he mean the z-coordinate or something like that instead of the radius?",
        "A": "think in 3D... take another plane // to xy plane... and it intersects the sphere above the origin... it is going to form a smaller circle, and obviously its radius is going to be smaller than unity.",
        "video_name": "E_Hwhp74Rhc",
        "timestamps": [
            335
        ],
        "3min_transcript": "And the radius here is always 1. It's a unit sphere. So given this parameter s, what would be your x- and y-coordinates? And now we're thinking about it right if we're sitting in the xy-plane. Well, the x-coordinate-- this goes back to the unit circle definition of our trig functions. The x-coordinate is going to be cosine of s. It would be the radius, which is 1, times the cosine of s. And the y-coordinate would be 1 times the sine of s. That's actually where we get our definitions for cosine and sine from. So that's pretty straightforward. And in this case, z is obviously equal to 0. So if we wanted to add our z-coordinate here, z is 0. We are sitting in the xy-plane. But now, let's think about what happens if we go above and below the xy-plane. Remember, this is in any plane that is parallel to the xy-plane. This is saying how we are rotated around the z-axis. Now, let's think about if we go above and below it. And to figure out how far above or below it, And this new parameter I'm going to introduce is t. t is how much we've rotated above and below the xy-plane. Now, what's interesting about that is if we take any other cross section that is parallel to the xy-plane now, we are going to have a smaller radius. Let me make that clear. So if we're right over there, now where this plane intersects our unit sphere, the radius is smaller. The radius is smaller than it was before. Well, what would be this new radius? Well, a little bit of trigonometry. It's the same as this length right over here, which is going to be cosine of t. So the radius is going to be cosine of t. And it still works over here because if t goes all the way to 0, cosine of 0 is 1. when we're in the xy-plane. So the radius over here is going to be-- so that right over there is cosine of 0. So this is when t is equal to 0. And we haven't rotated above or below the xy-plane. But if we have rotated above the xy-plane, the radius has changed. It is now cosine of t. And now we can use that to truly parameterize x and y anywhere. So now, let's look at this cross section. So we're not necessarily in the xy-plane, we're in something that's parallel to the xy-plane. And so if we're up here, now all of a sudden, the cross section-- if we view it from above, might look something like this. It might look something like this. We're viewing it from above, this cross section right over here. Our radius right over here is cosine of t. And so given that-- I guess altitude"
    },
    {
        "Q": "At 1:00, why is the Celsius scale called the Celsius scale and why is the Fahrenheit scale called the Fahrenheit scale?",
        "A": "The Celcius (or centigrade) scale is named for Anders Celsius (1701 - 1744) who created and defined a similar but upside down (0 was boiling water, 100 freezing water. The Fahrenheit Scale is named for Daniel Fahrenheit (1686-1736), based on one he first proposed in 1724.",
        "video_name": "aASUZqJCHHA",
        "timestamps": [
            60
        ],
        "3min_transcript": "Look at the two thermometers below. Identify which is Celsius and which is Fahrenheit, and then label the boiling and freezing points of water on each. Now, the Celsius scale is what's used in the most of the world. And the easy way to tell that you're dealing with the Celsius scale is on the Celsius scale, 0 degrees is freezing of water at standard temperature and pressure, and 100 degrees is the boiling point of water at standard temperature and pressure. Now, on the Fahrenheit scale, which is used mainly in the United States, the freezing point of water is 32 degrees, As you could tell, Celsius, the whole scale came from using freezing as 0 of regular water at standard temperature and pressure and setting 100 to be boiling. On some level, it makes a little bit more logical sense, but at least here in the U.S., we still use Fahrenheit. Now let's figure out which of these are Fahrenheit and which Now remember, regardless of which thermometer you're using, water will always actually boil at the exact So Fahrenheit, 32 degrees, this has to be the same thing as Celsius 0 degrees. So let's see what happens. So when this temperature right here is 0, this one over here, it looks like it's negative something. So this one right here doesn't look like Celsius. Here, if we say this is Celsius, this looks pretty close to 32 on this one. Let me do that in a darker color. So this one right here looks like Celsius, and this one right here looks like Fahrenheit. needs to be the same thing as 32 degrees Fahrenheit. In both cases, this is where water freezes, the freezing point. That is water freezing. So if this is the Celsius scale, this is where water will boil, 100 degrees Celsius, and that looks like it is right about 212 on the other scale. So right there is where water is boiling at standard temperature and pressure. So this thing on the right, right here, I guess I'll circle it in orange, that is Celsius. And then the one on the left, I'll do it in magenta, the one on the left is Fahrenheit."
    },
    {
        "Q": "at 2:02, I tried to do that prob on my own, but i failed.",
        "A": "It s trying to show you WHY multiplying fractions works the way it does. Let s move on to how to actually multiply with fractions. First, remember that every whole number can be rewritten as a fraction. 5 is the same thing as 5/1. So what is 10 x 1/2? Think of it as 10/1 x 1/2. Now, multiply the two numerators together. 10 x 1 = 10. Now multiply the two denominators: 1 x 2 = 2. Put them together... 10/2 and simplify if possible: 10/2 is the same as 5.",
        "video_name": "hr_mTd-oJ-M",
        "timestamps": [
            122
        ],
        "3min_transcript": "Let's think a little bit about what it means to multiply fractions. Say I want to multiply 1/2 times 1/4. Well, one way to think about this is we could view this as 1/2 of a 1/4. And what do I mean there? Well let me take a whole, let me take a whole here, and let me divide it into fourths. So let me divide it into fourths, so I'll divided into 4 equal sections. And so 1/4 would be 1 of these 4 equal sections. But we want to take 1/2 of that. So how do we take half of that? Well, we could divide this into 2 equal sections, and then just take 1 of them. So divide it into 2 equal sections, and then take 1 of them. So we're taking this pink area, this whole pink area is 1/4, and now we're going to take 1/2 of it. We're now going to take 1/2 of it. So that's this yellow square right over here. Well, it now represents 1 out of 1, 2, 3, 4, 5, 6, 7, 8 equal sections. So this right over here, this represents 1/8 of the whole. And so we see conceptually that 1/2 times 1/4, it completely makes sense, that 1/2 of 1/4 should be 1/8. And it hopefully makes sense that you get this 8 by multiplying the 2 times the 4. You started with 4 equal sections, but then you divided each of those 4 equal sections into 2 equal sections. So then you have 8 total equal sections that you split your whole into. Let's do another example, but now let's multiply two fractions that don't have 1's in the numerator. So let's multiply, let's multiply 2/3 times 4/5. And I encourage you now to pause the video and do something very similar to what I just did. 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."
    },
    {
        "Q": "(0:48) why in multiplying fractions it is taking away, not increasing?",
        "A": "The reason is when you multiply fractions the number being multiplied gets smaller because a fraction is a part of a whole. When you multiply a number by one: 5x1 You get 5 the same number you multiplied. So, if you multiply a number by a fraction ,the product should be less then the number you multiplied since a fraction smaller than 1. Same with decimals (if less than 1) since fractions are decimals.",
        "video_name": "hr_mTd-oJ-M",
        "timestamps": [
            48
        ],
        "3min_transcript": "Let's think a little bit about what it means to multiply fractions. Say I want to multiply 1/2 times 1/4. Well, one way to think about this is we could view this as 1/2 of a 1/4. And what do I mean there? Well let me take a whole, let me take a whole here, and let me divide it into fourths. So let me divide it into fourths, so I'll divided into 4 equal sections. And so 1/4 would be 1 of these 4 equal sections. But we want to take 1/2 of that. So how do we take half of that? Well, we could divide this into 2 equal sections, and then just take 1 of them. So divide it into 2 equal sections, and then take 1 of them. So we're taking this pink area, this whole pink area is 1/4, and now we're going to take 1/2 of it. We're now going to take 1/2 of it. So that's this yellow square right over here. Well, it now represents 1 out of 1, 2, 3, 4, 5, 6, 7, 8 equal sections. So this right over here, this represents 1/8 of the whole. And so we see conceptually that 1/2 times 1/4, it completely makes sense, that 1/2 of 1/4 should be 1/8. And it hopefully makes sense that you get this 8 by multiplying the 2 times the 4. You started with 4 equal sections, but then you divided each of those 4 equal sections into 2 equal sections. So then you have 8 total equal sections that you split your whole into. Let's do another example, but now let's multiply two fractions that don't have 1's in the numerator. So let's multiply, let's multiply 2/3 times 4/5. And I encourage you now to pause the video and do something very similar to what I just did. 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."
    },
    {
        "Q": "from 0:19 to 2:14, isn't there another way ?",
        "A": "To be honest, drawing fraction models are slower. You can just multiply the the two fractions (should be improper for both) and get the right answer. Ex) 1/2*1/2= 1*1/2*2 = 1/4",
        "video_name": "hr_mTd-oJ-M",
        "timestamps": [
            19,
            134
        ],
        "3min_transcript": "Let's think a little bit about what it means to multiply fractions. Say I want to multiply 1/2 times 1/4. Well, one way to think about this is we could view this as 1/2 of a 1/4. And what do I mean there? Well let me take a whole, let me take a whole here, and let me divide it into fourths. So let me divide it into fourths, so I'll divided into 4 equal sections. And so 1/4 would be 1 of these 4 equal sections. But we want to take 1/2 of that. So how do we take half of that? Well, we could divide this into 2 equal sections, and then just take 1 of them. So divide it into 2 equal sections, and then take 1 of them. So we're taking this pink area, this whole pink area is 1/4, and now we're going to take 1/2 of it. We're now going to take 1/2 of it. So that's this yellow square right over here. Well, it now represents 1 out of 1, 2, 3, 4, 5, 6, 7, 8 equal sections. So this right over here, this represents 1/8 of the whole. And so we see conceptually that 1/2 times 1/4, it completely makes sense, that 1/2 of 1/4 should be 1/8. And it hopefully makes sense that you get this 8 by multiplying the 2 times the 4. You started with 4 equal sections, but then you divided each of those 4 equal sections into 2 equal sections. So then you have 8 total equal sections that you split your whole into. Let's do another example, but now let's multiply two fractions that don't have 1's in the numerator. So let's multiply, let's multiply 2/3 times 4/5. And I encourage you now to pause the video and do something very similar to what I just did. 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."
    },
    {
        "Q": "Why didn't he just make the -6 a + 6 at 0:25? The way I am being taught right now is you take the first set of complex numbers leave them the same (2-3i stays the same) and then do the opposite of -(6-18) and this would be +6+18 when you remove the parentheses. The full equation should look something like this when the parentheses are removed: 2-3i+6+18. Why didn't he do it this way?",
        "A": "That would be true if it were (-6-18i) but look where the parentheses fall: -(6-18i). The basic rules of algebra apply and you have to distribute: -6+18i, as Sal had.",
        "video_name": "tvXRaZbIjO8",
        "timestamps": [
            25
        ],
        "3min_transcript": "We're asked to subtract. And we have the complex number 2 minus 3i. And from that, we are subtracting 6 minus 18i. So the first thing I'd like to do here is to just get rid of these parentheses. So we just have a bunch of real parts and imaginary parts that we can then add up together. So we have 2 minus 3i. And then we're subtracting this entire quantity. And to get rid of the parentheses, we can just distribute the negative sign. Or another way to think about it, we can say that this is negative 1 times all of this. So we can just distribute the negative sign. And negative 1 times 6 is negative 6. Let me do these in magenta. So this is negative 6. And then negative 1 times negative 18i-- well, that's just going to be positive 18i. Negative times a negative is a positive. And now we want to add the real parts, and we want to add the imaginary parts. So here's a real part here, 2. And then we have a minus 6. So we have 2 minus 6. And we want to add the imaginary parts. Let me do that in a different color. We have a negative 3i right over here. So negative 3i, or minus 3i right over there. And then we have a plus 18i or positive 18i. If you add the real parts, 2 minus 6 is negative 4. And you add the imaginary parts. If I have negative 3 of something and to that I add 18 of something, well, that's just going to leave me with 15 of that something. Or another way you could think about it, if I have 18 of something and I subtract 3 of that something, I'll have 15 of that something. And in this case, the something is i, is the imaginary unit. So this is going to be plus 15i. And we are done."
    },
    {
        "Q": "I love this song but what does Tau mean 0:50",
        "A": "Tau in this case means the ratio between the circumference of a circle and its radius.",
        "video_name": "FtxmFlMLYRI",
        "timestamps": [
            50
        ],
        "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": "1:14 - 1:20:\n\"Well, depending on the dimension, Because it's 1, 2, 3, There's 4 and even more.\"\nWhat is the fourth dimension?",
        "A": "The 4th dimension is time. As far as we know, it is the biggest dimension perceivable by humans. You can t see time, but you can measure it, and tell the difference between long and short periods of time.",
        "video_name": "FtxmFlMLYRI",
        "timestamps": [
            74,
            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 0:07 sal say to \"massage the equation\", but what does that mean?",
        "A": "It means manipulate the equation, converting it into an equivalent form that makes it easier to solve.",
        "video_name": "wYrxKGt_bLg",
        "timestamps": [
            7
        ],
        "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": "@ 10:26 in the video he says 5 is the same thing as 20/4 + 15/4. How is 5 the same as 20/4? Thanks.",
        "A": "Think of a fraction as literally the numerator divided by the denominator. When 20 is in the numerator, it s the same as saying 20 divided by 4. 20/4= 5.",
        "video_name": "wYrxKGt_bLg",
        "timestamps": [
            626
        ],
        "3min_transcript": "25, which is also equal to this expression. So let's add the left-hand sides and the right-hand sides. Because we're really adding the same thing to both sides of the equation. So the left-hand side, the x's cancel out. 35x minus 35x. That was the whole point. They cancel out, and on the y's, you get 49y plus 15y, that is 64y. 64y is equal to 105 minus 25 is equal to 80. Divide both sides by 64, and you get y is equal to 80/64. And let's see, if you divide the numerator and the denominator by 8-- actually you could probably do 16. 16 would be better. But let's do 8 first, just because we know our 8 times tables. So that becomes 10/8, and then you can divide this by 2, and 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."
    },
    {
        "Q": "At 6:30 Sal makes the bottom equation negative and the top one positive. Would it make a difference if I made the top one negative and the bottom one positive.",
        "A": "It doesn t matter. You want one equation to be negative and one equation to be positive so when you add them, one of the variables become 0 (eliminated). You could have both positives or both negatives and use subtraction. Subtraction is easier to mess up when subtracting negatives. So I recommend making one positive and one negative then use addition.",
        "video_name": "wYrxKGt_bLg",
        "timestamps": [
            390
        ],
        "3min_transcript": "Let me write that. Negative 10y is equal to 15. Divide both sides by negative 10. And we are left with y is equal to 15/10, is negative 3/2. So if you were to graph it, the point of intersection would be the point 0, negative 3/2. And you can verify that it also satisfies this equation. The original equation over here was 3x minus 2y is equal to 3. 3 times 0, which is 0, minus 2 times negative 3/2 is, this is 0, this is positive 3. Right? These cancel out, these become positive. Plus positive 3 is equal to 3. So this does indeed satisfy both equations. Let's do another one of these where we have to multiply, and the variables. Let's do another one. Let's say we have 5x plus 7y is equal to 15. And we have 7-- let me do another color-- 7x minus 3y is equal to 5. Now once again, if you just added or subtracted both the left-hand sides, you're not going to eliminate any variables. These aren't in any way kind of have the same coefficient or the negative of their coefficient. So let's pick a variable to eliminate. Let's say we want to eliminate the x's this time. And you could literally pick on one of the It doesn't matter. You can say let's eliminate the y's first. But I'm going to choose to eliminate the x's first. And so what I need to do is massage one or both of these equations in a way that these guys have the same coefficients, or their add the left-hand sides, they're going to eliminate each other. Now, there's nothing obvious-- I can multiply this by a fraction to make it equal to negative 5. Or I can multiply this by a fraction to make it equal to negative 7. But even a more fun thing to do is I can try to get both of them to be their least common multiple. I could get both of these to 35. And the way I can do it is by multiplying by each other. So I can multiply this top equation by 7. And I'm picking 7 so that this becomes a 35. And I can multiply this bottom equation by negative 5. And the reason why I'm doing that is so this becomes a negative 35. Remember, my point is I want to eliminate the x's. So if I make this a 35, and if I make this a negative 35, then I'm going to be all set."
    },
    {
        "Q": "At \"10:32\" how does he get 7x= 35/4 ? Wouldn't the it be 7x=20/4 because 5+15/4=20/4.",
        "A": "No 5 = 20/4. 15/4 + 20/4 = 35/4...",
        "video_name": "wYrxKGt_bLg",
        "timestamps": [
            632
        ],
        "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."
    }
]