diff --git "a/MathSc-Timestamp/test500.json" "b/MathSc-Timestamp/test500.json" deleted file mode 100644--- "a/MathSc-Timestamp/test500.json" +++ /dev/null @@ -1,5027 +0,0 @@ -[ - { - "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", - "qid": "x34OTtDE5q8_420_375" - }, - { - "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.", - "qid": "0h5Jd7sgQWY_1_1397" - }, - { - "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.", - "qid": "CQYELiTtUs8_190" - }, - { - "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.", - "qid": "CQYELiTtUs8_492" - }, - { - "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.", - "qid": "CQYELiTtUs8_305" - }, - { - "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.", - "qid": "rC165FcI4Yg_375" - }, - { - "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", - "qid": "TX7-Kdn6lJQ_630" - }, - { - "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,", - "qid": "6GB_kcdVMQo_201" - }, - { - "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", - "qid": "NnlAI4ZiUrQ_296" - }, - { - "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.", - "qid": "NnlAI4ZiUrQ_42_46" - }, - { - "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", - "qid": "eUF59jCFcyQ_43" - }, - { - "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.", - "qid": "eUF59jCFcyQ_630" - }, - { - "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.", - "qid": "eUF59jCFcyQ_614" - }, - { - "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", - "qid": "zTUnjPALX_U_237" - }, - { - "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.", - "qid": "GFiizJ-jGVw_303" - }, - { - "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.", - "qid": "Z4F88tTx9-8_377" - }, - { - "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.", - "qid": "qbCZbP6_j48_4" - }, - { - "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.", - "qid": "UPoXG1Z3sI8_192" - }, - { - "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.", - "qid": "6vy5CX6vK0I_604_604" - }, - { - "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.", - "qid": "XQTIKNXDAao_315" - }, - { - "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,", - "qid": "vYnreB1duro_180" - }, - { - "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.", - "qid": "JoGQYSTlOKo_120" - }, - { - "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.", - "qid": "JoGQYSTlOKo_80" - }, - { - "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", - "qid": "yAQSucmHrAk_285" - }, - { - "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.", - "qid": "uCW6154hPkc_267" - }, - { - "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,", - "qid": "uCW6154hPkc_592" - }, - { - "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.", - "qid": "1aJBToJrlvA_112" - }, - { - "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.", - "qid": "FmQoSenbtnU_377" - }, - { - "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.", - "qid": "FmQoSenbtnU_290" - }, - { - "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.", - "qid": "FmQoSenbtnU_620" - }, - { - "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", - "qid": "FmQoSenbtnU_690" - }, - { - "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.", - "qid": "ibdidr-bEvI_280" - }, - { - "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.", - "qid": "y55tzg_jW9I_179" - }, - { - "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,", - "qid": "j-iR7puLj6M_390" - }, - { - "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.", - "qid": "-QpkmwIoMaY_547" - }, - { - "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", - "qid": "-QpkmwIoMaY_510" - }, - { - "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", - "qid": "Kv-hRvEOjuA_98" - }, - { - "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", - "qid": "GR2GA7chA_c_930" - }, - { - "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.", - "qid": "NYtPw0WiUCo_568" - }, - { - "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", - "qid": "NYtPw0WiUCo_334" - }, - { - "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.", - "qid": "NYtPw0WiUCo_300" - }, - { - "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.", - "qid": "NYtPw0WiUCo_568" - }, - { - "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", - "qid": "Z-yv3Yq4Aw4_320" - }, - { - "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.", - "qid": "X9ypryY7hrQ_137" - }, - { - "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", - "qid": "nYFuxTXDj90_553" - }, - { - "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.", - "qid": "sXOIIEZh6qg_391" - }, - { - "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.", - "qid": "GwoX_BemwHs_710" - }, - { - "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", - "qid": "97POZGcfoY8_116" - }, - { - "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", - "qid": "BBIGR0RAMtY_405" - }, - { - "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.", - "qid": "dQCsA2cCdvA_100" - }, - { - "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", - "qid": "FNVvQ788wzk_680" - }, - { - "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.", - "qid": "LccmkSz-Y-w_248" - }, - { - "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.", - "qid": "GZx3U0dbASg_95" - }, - { - "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.", - "qid": "GZx3U0dbASg_375" - }, - { - "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.", - "qid": "GZx3U0dbASg_212" - }, - { - "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.", - "qid": "2ZgBJxT9pbU_553" - }, - { - "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.", - "qid": "cbkTTluHaTw_214" - }, - { - "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", - "qid": "D1NubiWCpQg_238" - }, - { - "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", - "qid": "-yCIMk1x0Pk_360" - }, - { - "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", - "qid": "zz4KbvF_X-0_800" - }, - { - "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.", - "qid": "ZrMw7P6P2Gw_595" - }, - { - "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.", - "qid": "Yiy84xYQ3es_247" - }, - { - "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": "", - "qid": "q3g3jsmCOEQ_319" - }, - { - "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": "", - "qid": "q3g3jsmCOEQ_487" - }, - { - "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.", - "qid": "d4bqNf37mBY_532" - }, - { - "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.", - "qid": "d4bqNf37mBY_268" - }, - { - "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.", - "qid": "d4bqNf37mBY_472" - }, - { - "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", - "qid": "UhIwMAhZpCo_223" - }, - { - "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", - "qid": "UhIwMAhZpCo_240_270" - }, - { - "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.", - "qid": "oIZV-ixRTcY_420" - }, - { - "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.", - "qid": "dstNU7It-Ro_63" - }, - { - "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,", - "qid": "MUQfl385Yug_150" - }, - { - "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.", - "qid": "-TpiL4J_yUA_149" - }, - { - "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.", - "qid": "EDlZAyhWxhk_67" - }, - { - "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.", - "qid": "JKvmAexeMgY_69" - }, - { - "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", - "qid": "8VgmBe3ulb8_300" - }, - { - "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", - "qid": "8VgmBe3ulb8_579_418" - }, - { - "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,", - "qid": "87_qIofPwhg_272" - }, - { - "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,", - "qid": "vWXMDIazHjA_445" - }, - { - "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.", - "qid": "vWXMDIazHjA_182" - }, - { - "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.", - "qid": "Qm_OS-8COwU_1080" - }, - { - "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": "", - "qid": "PupNgv49_WY_167" - }, - { - "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,", - "qid": "L5ErlC0COxI_281" - }, - { - "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.", - "qid": "L5ErlC0COxI_450" - }, - { - "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", - "qid": "1m9p9iubMLU_230" - }, - { - "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.", - "qid": "9uxvm-USEYE_137" - }, - { - "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.", - "qid": "BQMyeQOLvpg_334" - }, - { - "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.", - "qid": "BQMyeQOLvpg_816" - }, - { - "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.", - "qid": "BQMyeQOLvpg_762" - }, - { - "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?", - "qid": "6phoVfGKKec_150" - }, - { - "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.", - "qid": "BpBh8gvMifs_555" - }, - { - "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.", - "qid": "qfQv8GzyjB4_391" - }, - { - "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.", - "qid": "byjmR7JBXKc_235" - }, - { - "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,", - "qid": "qYjiVWwefto_124_186" - }, - { - "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.", - "qid": "6YRGEsQWZzY_229" - }, - { - "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.", - "qid": "PwDnpb_ZJvc_106" - }, - { - "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.", - "qid": "sXP7VhU1gYE_174" - }, - { - "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", - "qid": "H-E5rlpCVu4_125" - }, - { - "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.", - "qid": "H-E5rlpCVu4_300" - }, - { - "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.", - "qid": "H-E5rlpCVu4_387" - }, - { - "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.", - "qid": "V6Xynlqc_tc_107" - }, - { - "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", - "qid": "C5Lbjbyr1t4_361" - }, - { - "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", - "qid": "C5Lbjbyr1t4_300" - }, - { - "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.", - "qid": "yb7lVnY_VCY_42" - }, - { - "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.", - "qid": "yb7lVnY_VCY_75" - }, - { - "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.", - "qid": "Zn-GbH2S0Dk_269" - }, - { - "Q": "Why do you have to change the sign > to ?(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.", - "qid": "xdiBjypYFRQ_313" - }, - { - "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.", - "qid": "R948Tsyq4vA_250" - }, - { - "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", - "qid": "R948Tsyq4vA_55" - }, - { - "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.", - "qid": "MeU-KzdCBps_308" - }, - { - "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.", - "qid": "DkZnevdbf0A_104" - }, - { - "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.", - "qid": "-4bTgmmWI9k_328" - }, - { - "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.", - "qid": "u00I3MCrspU_20" - }, - { - "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.", - "qid": "uXjQ8yc9Pdg_623" - }, - { - "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", - "qid": "EGNlXtjYABw_1180" - }, - { - "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.", - "qid": "zC_dTaEY2AY_660" - }, - { - "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.", - "qid": "Y2-tz27nKoQ_221" - }, - { - "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.", - "qid": "cw3mp8oNASk_42" - }, - { - "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,", - "qid": "1uWZNW5PF-s_373" - }, - { - "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,", - "qid": "PPvd4X3Wv5I_70" - }, - { - "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", - "qid": "kQF9pOqmS0U_244" - }, - { - "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.", - "qid": "HKUJkMQsGkM_96" - }, - { - "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.", - "qid": "X6zD3SoN3iY_55" - }, - { - "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": "", - "qid": "WNuIhXo39_k_284" - }, - { - "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.", - "qid": "SP-YJe7Vldo_118" - }, - { - "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", - "qid": "tZKzaF28sOk_265" - }, - { - "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.", - "qid": "KibTbfkoPTs_185_211" - }, - { - "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.", - "qid": "XHHYA2Ug9lk_208" - }, - { - "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.", - "qid": "9iulv2QvKwo_14" - }, - { - "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.", - "qid": "Idxeo49szW0_152" - }, - { - "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.", - "qid": "XiAoUDfrar0_76_106" - }, - { - "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,", - "qid": "XiAoUDfrar0_309" - }, - { - "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", - "qid": "U-k5N1WPk4g_265" - }, - { - "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", - "qid": "U-k5N1WPk4g_223" - }, - { - "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.", - "qid": "3Md5KCCQX-0_465_636" - }, - { - "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.", - "qid": "s03qez-6JMA_215" - }, - { - "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.", - "qid": "s03qez-6JMA_124" - }, - { - "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.", - "qid": "X2jVap1YgwI_59" - }, - { - "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", - "qid": "X2jVap1YgwI_288" - }, - { - "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.", - "qid": "A_ESfuN1Pkg_261" - }, - { - "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,", - "qid": "MC0tq6fNRwU_460" - }, - { - "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": "", - "qid": "C7ducZoLKgw_145" - }, - { - "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", - "qid": "0671cRNjeKI_50" - }, - { - "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.", - "qid": "G2WybA4Hf7Y_308" - }, - { - "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.", - "qid": "McINBOFCGH8_235" - }, - { - "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", - "qid": "MC7l96tW8V8_299" - }, - { - "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", - "qid": "9wOalujeZf4_245" - }, - { - "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", - "qid": "9wOalujeZf4_245" - }, - { - "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", - "qid": "a70-dYvDJZY_240_600" - }, - { - "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.", - "qid": "riXcZT2ICjA_85" - }, - { - "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.", - "qid": "riXcZT2ICjA_560" - }, - { - "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,", - "qid": "riXcZT2ICjA_264" - }, - { - "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", - "qid": "riXcZT2ICjA_450" - }, - { - "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.", - "qid": "riXcZT2ICjA_77" - }, - { - "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", - "qid": "k9MEOgcc5KY_93" - }, - { - "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.", - "qid": "zltgXTlUVLw_28" - }, - { - "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": "", - "qid": "inlMrf2d-k4_62" - }, - { - "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": "", - "qid": "inlMrf2d-k4_120_131" - }, - { - "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.", - "qid": "UTs4uZhu5t8_39" - }, - { - "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.", - "qid": "_HJljJuVHLw_241" - }, - { - "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.", - "qid": "CecgFWTg9pQ_250" - }, - { - "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--", - "qid": "nOnd3SiYZqM_380" - }, - { - "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.", - "qid": "kbqO0YTUyAY_150" - }, - { - "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", - "qid": "kbqO0YTUyAY_180" - }, - { - "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.", - "qid": "l5VbdqRjTXc_160" - }, - { - "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.", - "qid": "l5VbdqRjTXc_354" - }, - { - "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.", - "qid": "9uwLgf84p5w_148" - }, - { - "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.", - "qid": "QxbJsg-Vdms_151" - }, - { - "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.", - "qid": "jb8mFpA1YI8_35" - }, - { - "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.", - "qid": "95logvV8nXY_407" - }, - { - "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.", - "qid": "fYQ3GRSu4JU_155" - }, - { - "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.", - "qid": "D2xYjiL8yyE_185" - }, - { - "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.", - "qid": "96uHMcHWD2E_228" - }, - { - "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", - "qid": "S-XKGBesRzk_440" - }, - { - "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.", - "qid": "wRBMmiNHQaE_280" - }, - { - "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.", - "qid": "q7eF5Ci944U_159" - }, - { - "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.", - "qid": "RLyXTj2j_c4_36" - }, - { - "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.", - "qid": "UFWAu8Ptth0_515" - }, - { - "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.", - "qid": "Za7G_eWKiF4_15" - }, - { - "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", - "qid": "eTwnt4G5xE4_97" - }, - { - "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.", - "qid": "D-EIh7NJvtQ_68_73" - }, - { - "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", - "qid": "Pb04ntcDJcQ_563" - }, - { - "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": "", - "qid": "gcnk8TnzsLc_190" - }, - { - "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.", - "qid": "uk7gS3cZVp4_71" - }, - { - "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", - "qid": "yUpDRpkUhf4_480" - }, - { - "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", - "qid": "vBlR2xNAGmo_201" - }, - { - "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.", - "qid": "jRrRqMJbHKc_78" - }, - { - "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", - "qid": "DL-ozRGDlkY_276" - }, - { - "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!", - "qid": "TINfzxSnnIE_462" - }, - { - "Q": "At 2:05, how did you get pi. When he said m 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", - "qid": "x5EJG_rAtkY_213" - }, - { - "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", - "qid": "tSHitjFIjd8_228" - }, - { - "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,", - "qid": "pzSyOTkAsY4_301" - }, - { - "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.", - "qid": "pzSyOTkAsY4_265" - }, - { - "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.", - "qid": "jAfJcgPGqgI_200_210" - }, - { - "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.", - "qid": "eVme7kuGyuo_138" - }, - { - "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", - "qid": "PiuhTj0zCf4_713" - }, - { - "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", - "qid": "hWhs2cIj7Cw_840" - }, - { - "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.", - "qid": "a_mzIWvHx_Y_42" - }, - { - "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", - "qid": "7Uos1ED3KHI_390" - }, - { - "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", - "qid": "MRAIgJmRmag_59" - }, - { - "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": "", - "qid": "i7idZfS8t8w_803" - }, - { - "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": "", - "qid": "i7idZfS8t8w_970" - }, - { - "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": "", - "qid": "i7idZfS8t8w_256" - }, - { - "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", - "qid": "JVDrlTdzxiI_431" - }, - { - "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", - "qid": "JVDrlTdzxiI_690" - }, - { - "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.", - "qid": "pi3WWQ0q6Lc_78" - }, - { - "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,", - "qid": "DPuK6ZgBGmE_305" - }, - { - "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.", - "qid": "rcLw4BlxaRs_113" - }, - { - "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.", - "qid": "Qwet4cIpnCM_496" - }, - { - "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", - "qid": "Qwet4cIpnCM_23" - }, - { - "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.", - "qid": "GH8-URjRQpQ_145" - }, - { - "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.", - "qid": "ios3QL9t9LQ_70" - }, - { - "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,", - "qid": "lEGS5ECgFxE_65" - }, - { - "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.", - "qid": "eTWCARmrzJ0_53" - }, - { - "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.", - "qid": "eTWCARmrzJ0_157" - }, - { - "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.", - "qid": "nGCW5teACC0_240" - }, - { - "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.", - "qid": "AmN0YyaTD60_222" - }, - { - "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.", - "qid": "BYTfCnR9Sl0_103" - }, - { - "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.", - "qid": "Tm98lnrlbMA_136" - }, - { - "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.", - "qid": "AOxMJRtoR2A_394_398_406" - }, - { - "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.", - "qid": "STyoP3rCmb0_118" - }, - { - "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.", - "qid": "STyoP3rCmb0_180" - }, - { - "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.", - "qid": "9VZsMY15xeU_294" - }, - { - "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", - "qid": "bNQY0z76M5A_645" - }, - { - "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.", - "qid": "bNQY0z76M5A_250" - }, - { - "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.", - "qid": "4h54s7BBPpA_22_56" - }, - { - "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.", - "qid": "sh-MP-dVhD4_105" - }, - { - "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.", - "qid": "sh-MP-dVhD4_44" - }, - { - "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.", - "qid": "RPhaidW0dmY_199" - }, - { - "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.", - "qid": "RPhaidW0dmY_111" - }, - { - "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", - "qid": "Dj1rbIP8PHM_83" - }, - { - "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", - "qid": "cqK3uRoPtk0_31" - }, - { - "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.", - "qid": "cqK3uRoPtk0_208" - }, - { - "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", - "qid": "QiwfF83NWNA_72" - }, - { - "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.", - "qid": "QiwfF83NWNA_320" - }, - { - "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.", - "qid": "iHnzLETGz2I_466" - }, - { - "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,", - "qid": "ZIqW_sXymrM_124" - }, - { - "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.", - "qid": "_y_Q3_B2Vh8_246" - }, - { - "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.", - "qid": "tuVd355R-OQ_210" - }, - { - "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.", - "qid": "tuVd355R-OQ_282" - }, - { - "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.", - "qid": "0pv0QtOi5l8_100" - }, - { - "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,", - "qid": "sxnX5_LbBDU_514" - }, - { - "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.", - "qid": "GdIkEngwGNU_463" - }, - { - "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!", - "qid": "Zp5z0wa0kgo_190" - }, - { - "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!", - "qid": "Zp5z0wa0kgo_176" - }, - { - "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!", - "qid": "Zp5z0wa0kgo_204" - }, - { - "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", - "qid": "E4HAYd0QnRc_645" - }, - { - "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": "", - "qid": "RI874OSJp1U_341" - }, - { - "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": "", - "qid": "2N62v_63SBo_455" - }, - { - "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.", - "qid": "WkspBxrzuZo_430" - }, - { - "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", - "qid": "6FrPLJY0rqM_84" - }, - { - "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", - "qid": "Yhlv5Aeuo_k_165" - }, - { - "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.", - "qid": "DMyhUb1pZT0_115" - }, - { - "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.", - "qid": "WAPZihVUmzE_456" - }, - { - "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.", - "qid": "oqCfqIcbE10_330" - }, - { - "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.", - "qid": "CDvPPsB3nEM_70" - }, - { - "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", - "qid": "Vc09LURoMQ8_504" - }, - { - "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?", - "qid": "boH4l1SgJbM_84" - }, - { - "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.", - "qid": "AR1uqNbjM5s_560" - }, - { - "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.", - "qid": "RhzXX5PbsuQ_20" - }, - { - "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.", - "qid": "55G8037gsKY_90" - }, - { - "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.", - "qid": "55G8037gsKY_288" - }, - { - "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.", - "qid": "1JT9oODsClE_190" - }, - { - "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.", - "qid": "xoXYirs2Mzw_110" - }, - { - "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": "", - "qid": "sjUhr0HkLUg_495" - }, - { - "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?", - "qid": "JGU74wbZMLg_362" - }, - { - "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.", - "qid": "JGU74wbZMLg_110" - }, - { - "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.", - "qid": "-YI6UC4qVEY_252_260" - }, - { - "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.", - "qid": "rYG1D5lUE4I_345" - }, - { - "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", - "qid": "rYG1D5lUE4I_311" - }, - { - "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.", - "qid": "rYG1D5lUE4I_401" - }, - { - "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?", - "qid": "u6zDpUL5RkU_161" - }, - { - "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.", - "qid": "NEaLgGi4Vh4_99" - }, - { - "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.", - "qid": "Alhcv5d_XOs_540" - }, - { - "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", - "qid": "viaPc8zDcRI_1200" - }, - { - "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.", - "qid": "viaPc8zDcRI_1125" - }, - { - "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", - "qid": "viaPc8zDcRI_843" - }, - { - "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.", - "qid": "gBxeju8dMho_21" - }, - { - "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": "", - "qid": "JnpqlXN9Whw_124" - }, - { - "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.", - "qid": "br7tS1t2SFE_240" - }, - { - "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.", - "qid": "_kbd6troMgA_320" - }, - { - "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.", - "qid": "K_OI9LA54AA_365" - }, - { - "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", - "qid": "K_OI9LA54AA_60" - }, - { - "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.", - "qid": "Bt60JVZRVCI_115" - }, - { - "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", - "qid": "muqyereWEh4_130" - }, - { - "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.", - "qid": "hmj3_zbz2eg_34" - }, - { - "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.", - "qid": "UCCNoXqCGZQ_180" - }, - { - "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.", - "qid": "AGFO-ROxH_I_361" - }, - { - "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": "", - "qid": "dZnc3PtNaN4_120" - }, - { - "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,", - "qid": "yV4Xa8Xtmrc_285" - }, - { - "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", - "qid": "7bO0TmJ6Ba4_473" - }, - { - "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.", - "qid": "TMmxKZaCqe0_540" - }, - { - "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.", - "qid": "TMmxKZaCqe0_588" - }, - { - "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?", - "qid": "TMmxKZaCqe0_505_518" - }, - { - "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.", - "qid": "5fkh01mClLU_620" - }, - { - "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?", - "qid": "FXgV9ySNusc_180" - }, - { - "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", - "qid": "FXgV9ySNusc_480" - }, - { - "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,", - "qid": "dXB3cUGnaxQ_240" - }, - { - "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.", - "qid": "jroamh6SIo0_125" - }, - { - "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.", - "qid": "b7JTVLc_aMk_255" - }, - { - "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,", - "qid": "b7JTVLc_aMk_120" - }, - { - "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": "", - "qid": "hIgnece9ins_1200" - }, - { - "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.", - "qid": "Xe6YlrCgkIo_320" - }, - { - "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.", - "qid": "bcCLKACsYJ0_200" - }, - { - "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.", - "qid": "Gn2pdkvdbGQ_562_306" - }, - { - "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.", - "qid": "0YErxSShF0A_100" - }, - { - "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.", - "qid": "CZdziIlYIfI_444" - }, - { - "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.", - "qid": "9tmtDBpqq9s_14" - }, - { - "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.", - "qid": "CDf_aE5yg3A_90" - }, - { - "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,", - "qid": "TugWqiUjOU4_210" - }, - { - "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.", - "qid": "8QihetGj3pg_59" - }, - { - "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,", - "qid": "ysVcAYo7UPI_56" - }, - { - "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", - "qid": "CJrVOf_3dN0_99" - }, - { - "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,", - "qid": "w56Vuf9tHfA_310" - }, - { - "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.", - "qid": "SeQeYVJZ2gE_661" - }, - { - "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.", - "qid": "6o7b9yyhH7k_413" - }, - { - "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", - "qid": "zM_p7tfWvLU_608" - }, - { - "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.", - "qid": "zM_p7tfWvLU_672" - }, - { - "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.", - "qid": "zWcfVC-oCNw_78" - }, - { - "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.", - "qid": "15s6B7K9paA_162" - }, - { - "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.", - "qid": "wJ37GJyViU8_102" - }, - { - "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", - "qid": "gl_-E6iVAg4_82" - }, - { - "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--", - "qid": "mLE-SlOZToc_496" - }, - { - "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.", - "qid": "c6-FJRda_Vc_88" - }, - { - "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.", - "qid": "dnjK4DPqh0k_350" - }, - { - "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,", - "qid": "dnjK4DPqh0k_329" - }, - { - "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.", - "qid": "-gB1y-PMWfs_152" - }, - { - "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", - "qid": "a3acutLstF8_77" - }, - { - "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.", - "qid": "F2mfEldxsPI_250" - }, - { - "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.", - "qid": "W-Nio466Ek4_166" - }, - { - "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.", - "qid": "zKotuhQWIRg_68" - }, - { - "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", - "qid": "xks4cETlN58_246" - }, - { - "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": "", - "qid": "mkyZ45KQYi4_37" - }, - { - "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.", - "qid": "he4kcTujy30_123" - }, - { - "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.", - "qid": "BItpeFXC4vA_49" - }, - { - "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.", - "qid": "IFU7Go6Qg6E_109" - }, - { - "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.", - "qid": "C2PC9185gIw_268" - }, - { - "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", - "qid": "Oc8sWN_jNF4_290" - }, - { - "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", - "qid": "Oc8sWN_jNF4_300" - }, - { - "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.", - "qid": "8CJ6Qdcoxsc_317" - }, - { - "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.", - "qid": "mVlCXkht6hg_369" - }, - { - "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.", - "qid": "711pdW8TbbY_148" - }, - { - "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.", - "qid": "711pdW8TbbY_94" - }, - { - "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.", - "qid": "LJqNdG6Y2cM_290" - }, - { - "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": "", - "qid": "XoEn1LfVoTo_510" - }, - { - "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": "", - "qid": "XoEn1LfVoTo_254" - }, - { - "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.", - "qid": "KV_XLL4K2Fw_86" - }, - { - "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", - "qid": "CrV1xCWdY-g_590" - }, - { - "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", - "qid": "CrV1xCWdY-g_465" - }, - { - "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.", - "qid": "WWv0RUxDfbs_372" - }, - { - "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.", - "qid": "Z7C69xP08d8_34" - }, - { - "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.", - "qid": "Z7C69xP08d8_150" - }, - { - "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--", - "qid": "VjLEoo3hIoM_152" - }, - { - "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.", - "qid": "KRFiAlo7t1E_170" - }, - { - "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", - "qid": "unSBFwK881s_94" - }, - { - "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.", - "qid": "unSBFwK881s_330" - }, - { - "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", - "qid": "PWiWkqHmum0_462" - }, - { - "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.", - "qid": "Ehd3cgRBvl0_131" - }, - { - "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", - "qid": "2ZzuZvz33X0_119" - }, - { - "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", - "qid": "2ZzuZvz33X0_101" - }, - { - "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", - "qid": "2ZzuZvz33X0_131" - }, - { - "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", - "qid": "2ZzuZvz33X0_136" - }, - { - "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.", - "qid": "2ZzuZvz33X0_225" - }, - { - "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.", - "qid": "2ZzuZvz33X0_171" - }, - { - "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.", - "qid": "tRHLEWSUjrQ_129" - }, - { - "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,", - "qid": "STcsaKuW-24_346" - }, - { - "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.", - "qid": "STcsaKuW-24_300" - }, - { - "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.", - "qid": "6agzj3A9IgA_113" - }, - { - "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", - "qid": "K4KDLWENXm0_315" - }, - { - "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.", - "qid": "qSPwUDmpnJ4_97" - }, - { - "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.", - "qid": "-6Fu2T_RSGM_421" - }, - { - "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.", - "qid": "rEtuPhl6930_180" - }, - { - "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.", - "qid": "rEtuPhl6930_179" - }, - { - "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": "", - "qid": "MyzGVbCHh5M_495_500" - }, - { - "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.", - "qid": "TgDk06Qayxw_317" - }, - { - "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.", - "qid": "lR_kUUPL8YY_192" - }, - { - "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,", - "qid": "zlq_jQYD1mA_160" - }, - { - "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.", - "qid": "_WOr9-_HbAM_50" - }, - { - "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", - "qid": "lvAYFUIEpFI_88" - }, - { - "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.", - "qid": "lvAYFUIEpFI_521" - }, - { - "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.", - "qid": "L0CmbneYETs_1040" - }, - { - "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", - "qid": "XyDMsotfJhE_94" - }, - { - "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", - "qid": "Ly86lwq_2gc_346" - }, - { - "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.", - "qid": "6m642-2D3V4_142" - }, - { - "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", - "qid": "YBYu5aZPLeg_41" - }, - { - "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.", - "qid": "0Dd-y_apbRw_745" - }, - { - "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.", - "qid": "-QHff5pRdM8_473" - }, - { - "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", - "qid": "-QHff5pRdM8_233_263_414_418" - }, - { - "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?", - "qid": "1wsF9GpGd00_145" - }, - { - "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.", - "qid": "qG3HnRccrQU_415" - }, - { - "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.", - "qid": "Wm0zq-NqEFs_16" - }, - { - "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,", - "qid": "tzQ7arA917E_236" - }, - { - "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,", - "qid": "CZ5ne_mX5_I_61" - }, - { - "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,", - "qid": "jOZ98FDyl2E_95" - }, - { - "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.", - "qid": "IbI-l7mbKO4_60" - }, - { - "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.", - "qid": "iDQ1foxYf0o_491" - }, - { - "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.", - "qid": "qcz1Cm_-l50_32" - }, - { - "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.", - "qid": "99pD1-6ZpuM_390" - }, - { - "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.", - "qid": "gEE6yIObbmg_25" - }, - { - "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.", - "qid": "y7QLay8wrW8_52" - }, - { - "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.", - "qid": "ik2CZqsAw28_315" - }, - { - "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.", - "qid": "8TIben0bJpU_244" - }, - { - "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.", - "qid": "jmbg-DKWuc4_85" - }, - { - "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.", - "qid": "h8EYEJ32oQ8_128" - }, - { - "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.", - "qid": "9IUEk9fn2Vs_458" - }, - { - "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.", - "qid": "yHo0CcDVHsk_173" - }, - { - "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.", - "qid": "6QZCj4O9sk0_24" - }, - { - "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": "", - "qid": "1LxhXqD3_CE_195" - }, - { - "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\").", - "qid": "Z8j5RDOibV4_78" - }, - { - "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.", - "qid": "tvnOWIoeeaU_69" - }, - { - "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.", - "qid": "gAV9kwvoD6s_89" - }, - { - "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.", - "qid": "Efoeqb6tC88_292" - }, - { - "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.", - "qid": "P6_sK8hRWCA_70" - }, - { - "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,", - "qid": "adPgapI-h3g_97" - }, - { - "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", - "qid": "adPgapI-h3g_246" - }, - { - "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", - "qid": "iX5UgArMyiI_90" - }, - { - "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", - "qid": "-cApVwKR1Ps_100" - }, - { - "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.", - "qid": "TvSKeTFsaj4_239" - }, - { - "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.", - "qid": "W3flX500w5g_216" - }, - { - "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.", - "qid": "d-2Lcp0QKfI_70" - }, - { - "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", - "qid": "S0Fd2Tg2v7M_801" - }, - { - "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.", - "qid": "lDY9JcFaRd4_392" - }, - { - "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.", - "qid": "lDY9JcFaRd4_432" - }, - { - "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", - "qid": "lDY9JcFaRd4_290" - }, - { - "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.", - "qid": "6EY0E3z-hsU_175" - }, - { - "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.", - "qid": "u9v_bakOIcU_339" - }, - { - "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.", - "qid": "u9v_bakOIcU_366" - }, - { - "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,", - "qid": "DaoJmvqU3FI_139" - }, - { - "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.", - "qid": "n0DLSIOYBsQ_355" - }, - { - "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,", - "qid": "okXVhDMuGFg_150" - }, - { - "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,", - "qid": "qFFhdLlX220_90" - }, - { - "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.", - "qid": "FtxmFlMLYRI_79" - }, - { - "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", - "qid": "IJySBMtFlnQ_210" - }, - { - "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.", - "qid": "VIVIegSt81k_205" - }, - { - "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.", - "qid": "VpuN8vCQ--M_387" - }, - { - "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,", - "qid": "dC1ojsMi1yU_293" - }, - { - "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,", - "qid": "WAoaBTWKLoI_226" - }, - { - "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.", - "qid": "QuZMXVJNLCo_450" - }, - { - "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,", - "qid": "i_0DXxNeaQ0_116" - }, - { - "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.", - "qid": "iqeGTtyzQ1I_190" - }, - { - "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.", - "qid": "0xrvRKHoO2g_6" - }, - { - "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", - "qid": "Mci8Cuik_Gw_175" - }, - { - "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", - "qid": "Mci8Cuik_Gw_137" - }, - { - "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.", - "qid": "ry81_iSHt6E_386" - }, - { - "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.", - "qid": "8C5kAIKLcZo_337" - }, - { - "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.", - "qid": "twMdew4Zs8Q_26" - }, - { - "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", - "qid": "IKsi-DQU2zo_170" - }, - { - "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.", - "qid": "IKsi-DQU2zo_114" - }, - { - "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.", - "qid": "d9pO2z2qvXU_121" - }, - { - "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.\"", - "qid": "VuJEidLhY1E_317" - }, - { - "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.", - "qid": "X3JqIZR1XcY_76" - }, - { - "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?", - "qid": "X7B_tH4O-_s_370" - }, - { - "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,", - "qid": "X7B_tH4O-_s_520" - }, - { - "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", - "qid": "FJ7AMaR9miI_240" - }, - { - "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.", - "qid": "pXo0bG4iAyg_133" - }, - { - "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.", - "qid": "pXo0bG4iAyg_536" - }, - { - "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.", - "qid": "d5lcGCbV5cM_272" - }, - { - "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.", - "qid": "AA6RfgP-AHU_575" - }, - { - "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.", - "qid": "AA6RfgP-AHU_440" - }, - { - "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,", - "qid": "hKkBlcnU9pw_76" - }, - { - "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", - "qid": "0zCcFSO8ouE_505" - }, - { - "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", - "qid": "EKvHQc3QEow_61" - }, - { - "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,", - "qid": "6CFE60iP2Ug_189" - }, - { - "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.", - "qid": "fGThIRpWEE4_580" - }, - { - "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", - "qid": "fGThIRpWEE4_170" - }, - { - "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.", - "qid": "k45QTFCHSVs_160" - }, - { - "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.", - "qid": "iUQR0enP7RQ_574" - }, - { - "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": "", - "qid": "xCIHAjsZCE0_300" - }, - { - "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": "", - "qid": "xCIHAjsZCE0_470" - }, - { - "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.", - "qid": "GiSpzFKI5_w_180" - }, - { - "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.", - "qid": "X1E7I7_r3Cw_249" - }, - { - "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,", - "qid": "iPwrDWQ7hPc_510" - }, - { - "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,", - "qid": "iPwrDWQ7hPc_279" - }, - { - "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,", - "qid": "iPwrDWQ7hPc_577" - }, - { - "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.", - "qid": "ZTbTYEMvo10_155" - }, - { - "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.", - "qid": "u8JFdwmBvvQ_100" - }, - { - "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.", - "qid": "u8JFdwmBvvQ_28" - }, - { - "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", - "qid": "-eAzhBZgq28_1000" - }, - { - "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.", - "qid": "Ihws0d-WLzU_221" - }, - { - "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.", - "qid": "Ihws0d-WLzU_244" - }, - { - "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.", - "qid": "R-6CAr_zEEk_254" - }, - { - "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.", - "qid": "7S1MLJOG-5A_35" - }, - { - "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.", - "qid": "yAH3722GrP8_220" - }, - { - "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.", - "qid": "yAH3722GrP8_201" - }, - { - "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", - "qid": "b-7kCymoUpg_130" - }, - { - "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": "", - "qid": "kZzoVCmUyKg_330" - }, - { - "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": "", - "qid": "kZzoVCmUyKg_33" - }, - { - "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.", - "qid": "lxjmR4pYIVU_85_110" - }, - { - "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.", - "qid": "xMz9WFvox9g_81" - }, - { - "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.", - "qid": "xMz9WFvox9g_156" - }, - { - "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", - "qid": "xMz9WFvox9g_371" - }, - { - "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.", - "qid": "TV5kDqiJ1Os_76" - }, - { - "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.", - "qid": "aASUZqJCHHA_60" - }, - { - "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.", - "qid": "FtxmFlMLYRI_50" - }, - { - "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.", - "qid": "wYrxKGt_bLg_626" - }, - { - "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.", - "qid": "wYrxKGt_bLg_390" - }, - { - "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.", - "qid": "wYrxKGt_bLg_632" - } -] \ No newline at end of file