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Simplifying radicals examples.mp3 | So you undistribute it, do the distributive property in reverse. That would be the square root of 10 times three minus two, which is, of course, this is just one. So you're just left with the square root of 10. So all of this simplifies to square root of 10. Let's do a few more of these. So this says, simplify the expression by removing all factors that are perfect squares from inside the radicals and combining the terms. So essentially, the same idea. |
Simplifying radicals examples.mp3 | So all of this simplifies to square root of 10. Let's do a few more of these. So this says, simplify the expression by removing all factors that are perfect squares from inside the radicals and combining the terms. So essentially, the same idea. All right, let's see what we can do. So this is interesting. We have a square root of 1 1⁄2. |
Simplifying radicals examples.mp3 | So essentially, the same idea. All right, let's see what we can do. So this is interesting. We have a square root of 1 1⁄2. So can I, well, actually, what could be interesting is if I have a square root of something times the square root of something else, so the square root of 180 times the square root of 1 1⁄2. This is the same thing as the square root of 180 times 1 1⁄2. And this just comes straight out of our exponent properties. |
Simplifying radicals examples.mp3 | We have a square root of 1 1⁄2. So can I, well, actually, what could be interesting is if I have a square root of something times the square root of something else, so the square root of 180 times the square root of 1 1⁄2. This is the same thing as the square root of 180 times 1 1⁄2. And this just comes straight out of our exponent properties. It might look a little bit more familiar if I wrote it as 180 to the 1 1⁄2 power times 1 1⁄2 to the 1 1⁄2 power is going to be equal to 180 times 1 1⁄2 to the 1 1⁄2 power. Taking the square root, the principal root, is the same thing as raising something to the 1 1⁄2 power. And so this is the square root of 80 times 1 1⁄2, which is going to be the square root of 90, which is equal to the square root of 9 times 10. |
Simplifying radicals examples.mp3 | And this just comes straight out of our exponent properties. It might look a little bit more familiar if I wrote it as 180 to the 1 1⁄2 power times 1 1⁄2 to the 1 1⁄2 power is going to be equal to 180 times 1 1⁄2 to the 1 1⁄2 power. Taking the square root, the principal root, is the same thing as raising something to the 1 1⁄2 power. And so this is the square root of 80 times 1 1⁄2, which is going to be the square root of 90, which is equal to the square root of 9 times 10. And we just simplified square root of 90 in the last problem. That's equal to the square root of 9 times the square root or the principal root of 10, which is equal to 3 times the square root of 10. 3 times the square root of 10. |
Simplifying radicals examples.mp3 | And so this is the square root of 80 times 1 1⁄2, which is going to be the square root of 90, which is equal to the square root of 9 times 10. And we just simplified square root of 90 in the last problem. That's equal to the square root of 9 times the square root or the principal root of 10, which is equal to 3 times the square root of 10. 3 times the square root of 10. All right, let's keep going. So I have one more of these examples. And like always, pause the video and see if you can work through these on your own before I work it out with you. |
Simplifying radicals examples.mp3 | 3 times the square root of 10. All right, let's keep going. So I have one more of these examples. And like always, pause the video and see if you can work through these on your own before I work it out with you. Simplify the expression by removing all factors that are perfect squares. Okay, these are the same directions that we've seen the last few times. And so let's see. |
Simplifying radicals examples.mp3 | And like always, pause the video and see if you can work through these on your own before I work it out with you. Simplify the expression by removing all factors that are perfect squares. Okay, these are the same directions that we've seen the last few times. And so let's see. If I wanted to simplify this, this is equal to the square root of, well, 64 times 2 is 128, and 64 is a perfect square. So I'm gonna write it as 64 times 2 over, 27 is 9 times 3. 9 is a perfect square. |
Simplifying radicals examples.mp3 | And so let's see. If I wanted to simplify this, this is equal to the square root of, well, 64 times 2 is 128, and 64 is a perfect square. So I'm gonna write it as 64 times 2 over, 27 is 9 times 3. 9 is a perfect square. So this is going to be the same thing. And there's a couple of ways that we could think about it. We could say this is the same thing as the square root of 64 times 2 over the square root of 9 times 3, which is the same thing as the square root of 64 times the square root of 2 over the square root of 9 times the square root of 3, which is equal to, this is 8, this is 3. |
Simplifying radicals examples.mp3 | 9 is a perfect square. So this is going to be the same thing. And there's a couple of ways that we could think about it. We could say this is the same thing as the square root of 64 times 2 over the square root of 9 times 3, which is the same thing as the square root of 64 times the square root of 2 over the square root of 9 times the square root of 3, which is equal to, this is 8, this is 3. So it would be 8 times the square root of 2 over 3 times the square root of 3. That's one way to say it. Or we could even view the square root of 2 over the square root of 3 as the square root of 2 thirds. |
Simplifying radicals examples.mp3 | We could say this is the same thing as the square root of 64 times 2 over the square root of 9 times 3, which is the same thing as the square root of 64 times the square root of 2 over the square root of 9 times the square root of 3, which is equal to, this is 8, this is 3. So it would be 8 times the square root of 2 over 3 times the square root of 3. That's one way to say it. Or we could even view the square root of 2 over the square root of 3 as the square root of 2 thirds. So we could say this is 8 over 3 times the square root of 2 thirds. So these are all possible ways of trying to tackle this. So we could just write it, let's see, have we removed all factors that are perfect squares? |
Simplifying radicals examples.mp3 | Or we could even view the square root of 2 over the square root of 3 as the square root of 2 thirds. So we could say this is 8 over 3 times the square root of 2 thirds. So these are all possible ways of trying to tackle this. So we could just write it, let's see, have we removed all factors that are perfect squares? Yes, from inside the radicals. And we've combined terms. We weren't doing any adding or subtracting here. |
Simplifying radicals examples.mp3 | So we could just write it, let's see, have we removed all factors that are perfect squares? Yes, from inside the radicals. And we've combined terms. We weren't doing any adding or subtracting here. So it's really just removing the perfect squares from inside the radicals. And I think we've done that. So we could say this is going to be 8 thirds times the square root of 2 thirds. |
Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3 | I've drawn an arbitrary triangle right over here, and I've labeled the measures of the interior angles. The measure of this angle is x, this one's y, this one is z. And what I want to prove is that the sum of the measures of the interior angles of a triangle, that x plus y plus z, is equal to 180 degrees. And the way that I'm going to do it is using our knowledge of parallel lines or transversals of parallel lines and corresponding angles. And to do that, I'm going to extend each of these sides of the triangle, which right now are line segments, but extend them into lines. So this side down here, if I keep going on and on forever in the same directions, then now all of a sudden I have an orange line. And what I want to do is construct another line that is parallel to the orange line that goes through this vertex of the triangle right over here. |
Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3 | And the way that I'm going to do it is using our knowledge of parallel lines or transversals of parallel lines and corresponding angles. And to do that, I'm going to extend each of these sides of the triangle, which right now are line segments, but extend them into lines. So this side down here, if I keep going on and on forever in the same directions, then now all of a sudden I have an orange line. And what I want to do is construct another line that is parallel to the orange line that goes through this vertex of the triangle right over here. And I can always do that. I can just start from this point and go in the same direction as this line, and I will never intersect. I'm not getting any closer or further away from that line. |
Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3 | And what I want to do is construct another line that is parallel to the orange line that goes through this vertex of the triangle right over here. And I can always do that. I can just start from this point and go in the same direction as this line, and I will never intersect. I'm not getting any closer or further away from that line. So I'm never going to intersect that line. So these two lines right over here are parallel. This is parallel to that. |
Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3 | I'm not getting any closer or further away from that line. So I'm never going to intersect that line. So these two lines right over here are parallel. This is parallel to that. Now I'm going to go to the other two sides of my original triangle and extend them into lines. So I'm going to extend this one into a line. So I'll do that as neatly as I can. |
Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3 | This is parallel to that. Now I'm going to go to the other two sides of my original triangle and extend them into lines. So I'm going to extend this one into a line. So I'll do that as neatly as I can. So I'm going to extend that into a line. And you see that this is clearly a transversal of these two parallel lines. Now, if we have a transversal here of two parallel lines, then we must have some corresponding angles. |
Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3 | So I'll do that as neatly as I can. So I'm going to extend that into a line. And you see that this is clearly a transversal of these two parallel lines. Now, if we have a transversal here of two parallel lines, then we must have some corresponding angles. And we see that this angle is formed when the transversal intersects the bottom orange line. Well, what's the corresponding angle when the transversal intersects this top blue line? What's the angle on the top right of the intersection? |
Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3 | Now, if we have a transversal here of two parallel lines, then we must have some corresponding angles. And we see that this angle is formed when the transversal intersects the bottom orange line. Well, what's the corresponding angle when the transversal intersects this top blue line? What's the angle on the top right of the intersection? Angle on the top right of the intersection must also be x. The other thing that pops out at you is there's another vertical angle with x, another angle that must be equivalent. On the opposite side of this intersection, you have this angle right over here. |
Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3 | What's the angle on the top right of the intersection? Angle on the top right of the intersection must also be x. The other thing that pops out at you is there's another vertical angle with x, another angle that must be equivalent. On the opposite side of this intersection, you have this angle right over here. These two angles are vertical. So if this has measure x, then this one must have measure x as well. Let's do the same thing with the last side of the triangle that we have not extended into a line yet. |
Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3 | On the opposite side of this intersection, you have this angle right over here. These two angles are vertical. So if this has measure x, then this one must have measure x as well. Let's do the same thing with the last side of the triangle that we have not extended into a line yet. So let's do that. So if we take this one, so we just keep going. So it becomes a line. |
Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3 | Let's do the same thing with the last side of the triangle that we have not extended into a line yet. So let's do that. So if we take this one, so we just keep going. So it becomes a line. So now it becomes a transversal of the two parallel lines, just like the magenta line did. And we say, hey, look, this angle y right over here, this angle is formed from the intersection of the transversal and the bottom parallel line. What angle does it correspond to up here? |
Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3 | So it becomes a line. So now it becomes a transversal of the two parallel lines, just like the magenta line did. And we say, hey, look, this angle y right over here, this angle is formed from the intersection of the transversal and the bottom parallel line. What angle does it correspond to up here? Well, this is kind of on the left side of the intersection. It corresponds to this angle right over here, where the green line, the green transversal intersects the blue parallel line. Well, what angle is vertical to it? |
Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3 | What angle does it correspond to up here? Well, this is kind of on the left side of the intersection. It corresponds to this angle right over here, where the green line, the green transversal intersects the blue parallel line. Well, what angle is vertical to it? Well, this angle. So this is going to have measure y as well. So now we're really at the home stretch of our proof, because we will see that the measure, we have this angle and this angle. |
Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3 | Well, what angle is vertical to it? Well, this angle. So this is going to have measure y as well. So now we're really at the home stretch of our proof, because we will see that the measure, we have this angle and this angle. This has measure angle x. This has measure z. They're both adjacent angles. |
Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3 | So now we're really at the home stretch of our proof, because we will see that the measure, we have this angle and this angle. This has measure angle x. This has measure z. They're both adjacent angles. If we take the two outer rays that form the angle, and we think about this angle right over here, what's this measure of this wide angle right over there? Well, it's going to be x plus z. And that angle is supplementary to this angle right over here that has measure y. |
Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3 | They're both adjacent angles. If we take the two outer rays that form the angle, and we think about this angle right over here, what's this measure of this wide angle right over there? Well, it's going to be x plus z. And that angle is supplementary to this angle right over here that has measure y. So the measure of this wide angle, which is x plus z, plus the measure of this magenta angle, which is y, must be equal to 180 degrees, because these two angles are supplementary. So the measure of the wide angle, x plus z plus the measure of the magenta angle, which is supplementary to the wide angle, it must be equal to 180 degrees, because they are supplementary. Well, we could just reorder this if we want to put it in alphabetical order, but we've just completed our proof. |
Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3 | And that angle is supplementary to this angle right over here that has measure y. So the measure of this wide angle, which is x plus z, plus the measure of this magenta angle, which is y, must be equal to 180 degrees, because these two angles are supplementary. So the measure of the wide angle, x plus z plus the measure of the magenta angle, which is supplementary to the wide angle, it must be equal to 180 degrees, because they are supplementary. Well, we could just reorder this if we want to put it in alphabetical order, but we've just completed our proof. The measure of the interior angles of the triangle, x plus z plus y. We could write this as x plus y plus z, if the lack of alphabetical order is making you uncomfortable. We could just rewrite this as x plus y plus z is equal to 180 degrees. |
How to solve equations of the form x a = b Linear equations Algebra I Khan Academy.mp3 | We have x divided by 3 is equal to 14. So to solve for x, to figure out what the variable x must be equal to, we really just have to isolate it on the left-hand side of this equation. It's already sitting there. We have x divided by 3 is equal to 14. We could also write this as 1 3rd x is equal to 14. Obviously x times 1 3rd is going to be x over 3. These are equivalent. |
How to solve equations of the form x a = b Linear equations Algebra I Khan Academy.mp3 | We have x divided by 3 is equal to 14. We could also write this as 1 3rd x is equal to 14. Obviously x times 1 3rd is going to be x over 3. These are equivalent. So how can we just end up with an x on the left-hand side of either of these equations? These are really the same thing. Or another way, how can we just have a 1 in front of the x, a 1x, which is really just saying x over here? |
How to solve equations of the form x a = b Linear equations Algebra I Khan Academy.mp3 | These are equivalent. So how can we just end up with an x on the left-hand side of either of these equations? These are really the same thing. Or another way, how can we just have a 1 in front of the x, a 1x, which is really just saying x over here? Well, I'm dividing it by 3 right now. So if I were to multiply both sides of this equation by 3, that would isolate the x. And the reason that would work is if I multiply this by 3 over here, I'm multiplying by 3 and dividing by 3. |
How to solve equations of the form x a = b Linear equations Algebra I Khan Academy.mp3 | Or another way, how can we just have a 1 in front of the x, a 1x, which is really just saying x over here? Well, I'm dividing it by 3 right now. So if I were to multiply both sides of this equation by 3, that would isolate the x. And the reason that would work is if I multiply this by 3 over here, I'm multiplying by 3 and dividing by 3. That's equivalent to multiplying or dividing by 1. These guys cancel out. But remember, if you do it to the left-hand side, you also have to do it to the right-hand side. |
How to solve equations of the form x a = b Linear equations Algebra I Khan Academy.mp3 | And the reason that would work is if I multiply this by 3 over here, I'm multiplying by 3 and dividing by 3. That's equivalent to multiplying or dividing by 1. These guys cancel out. But remember, if you do it to the left-hand side, you also have to do it to the right-hand side. And actually, I'll do both of these equations at the same time because they're really the exact same equation. So what are we going to get over here on the left-hand side? 3 times anything divided by 3 is going to be that anything. |
How to solve equations of the form x a = b Linear equations Algebra I Khan Academy.mp3 | But remember, if you do it to the left-hand side, you also have to do it to the right-hand side. And actually, I'll do both of these equations at the same time because they're really the exact same equation. So what are we going to get over here on the left-hand side? 3 times anything divided by 3 is going to be that anything. We're just going to have an x left over on the left-hand side. And on the right-hand side, what's 14 times 3? 3 times 10 is 30, 3 times 4 is 12, so it's going to be 42. |
How to solve equations of the form x a = b Linear equations Algebra I Khan Academy.mp3 | 3 times anything divided by 3 is going to be that anything. We're just going to have an x left over on the left-hand side. And on the right-hand side, what's 14 times 3? 3 times 10 is 30, 3 times 4 is 12, so it's going to be 42. So we get x is equal to 42. And the same thing would happen here. 3 times 1 third is just 1. |
How to solve equations of the form x a = b Linear equations Algebra I Khan Academy.mp3 | 3 times 10 is 30, 3 times 4 is 12, so it's going to be 42. So we get x is equal to 42. And the same thing would happen here. 3 times 1 third is just 1. So you get 1x is equal to 14 times 3, which is 42. Now let's just check our answer. Let's substitute 42 into our original equation. |
How to solve equations of the form x a = b Linear equations Algebra I Khan Academy.mp3 | 3 times 1 third is just 1. So you get 1x is equal to 14 times 3, which is 42. Now let's just check our answer. Let's substitute 42 into our original equation. So we have 42 in place for x over 3 is equal to 14. So what's 42 divided by 3? And we could do a little bit of, I guess we'd call it medium-long division. |
How to solve equations of the form x a = b Linear equations Algebra I Khan Academy.mp3 | Let's substitute 42 into our original equation. So we have 42 in place for x over 3 is equal to 14. So what's 42 divided by 3? And we could do a little bit of, I guess we'd call it medium-long division. It's not really long division. 3 into 4, 3 goes into 4 one time. 1 times 3 is 3. |
How to solve equations of the form x a = b Linear equations Algebra I Khan Academy.mp3 | And we could do a little bit of, I guess we'd call it medium-long division. It's not really long division. 3 into 4, 3 goes into 4 one time. 1 times 3 is 3. You subtract. 4 minus 3 is 1. Bring down the 2. |
How to solve equations of the form x a = b Linear equations Algebra I Khan Academy.mp3 | 1 times 3 is 3. You subtract. 4 minus 3 is 1. Bring down the 2. 3 goes into 12 four times. So 3 goes into 42 14 times. So this right over here simplifies to 14. |
Adding numbers with different signs (example) Pre-Algebra Khan Academy.mp3 | Find the sum 15 plus negative 46 plus 29. So let's just think about this first part over here, where we have 15 plus negative 46, plus negative 46, and we'll worry about the plus 29 later on. So let's just do 15, 15, plus, plus negative, let me do that in a different color, plus negative 46, I'm doing that orange, plus negative 46. Let me draw a number line here, just so we can properly visualize what is going on. So that's my number line. That's my number line. We're starting at 15, so let me put, let's draw a zero over here. |
Adding numbers with different signs (example) Pre-Algebra Khan Academy.mp3 | Let me draw a number line here, just so we can properly visualize what is going on. So that's my number line. That's my number line. We're starting at 15, so let me put, let's draw a zero over here. And so we are starting at 15, 15 could be right over here, so this is 15. Maybe just draw, let me draw a big fat arrow to signify this is 15. 15 has an absolute value of 15, so the length of this arrow would be 15. |
Adding numbers with different signs (example) Pre-Algebra Khan Academy.mp3 | We're starting at 15, so let me put, let's draw a zero over here. And so we are starting at 15, 15 could be right over here, so this is 15. Maybe just draw, let me draw a big fat arrow to signify this is 15. 15 has an absolute value of 15, so the length of this arrow would be 15. Now, we're adding negative 46 to that 15. That is the same thing. This is equivalent to 15 minus 46. |
Adding numbers with different signs (example) Pre-Algebra Khan Academy.mp3 | 15 has an absolute value of 15, so the length of this arrow would be 15. Now, we're adding negative 46 to that 15. That is the same thing. This is equivalent to 15 minus 46. Minus 46, which means we are going to move 46 spots to the left of 15. Negative sign, or minus, means we're moving to the left on the number line. So we're going to move 46 to the left, so we're starting at 15, and we are going to move 46, we are going to move 46 to the left. |
Adding numbers with different signs (example) Pre-Algebra Khan Academy.mp3 | This is equivalent to 15 minus 46. Minus 46, which means we are going to move 46 spots to the left of 15. Negative sign, or minus, means we're moving to the left on the number line. So we're going to move 46 to the left, so we're starting at 15, and we are going to move 46, we are going to move 46 to the left. So the length of this arrow right here, the length of this arrow is going to be 46. And we're moving to the left. This is the negative 46 that we're adding to the 15. |
Adding numbers with different signs (example) Pre-Algebra Khan Academy.mp3 | So we're going to move 46 to the left, so we're starting at 15, and we are going to move 46, we are going to move 46 to the left. So the length of this arrow right here, the length of this arrow is going to be 46. And we're moving to the left. This is the negative 46 that we're adding to the 15. So we're going to end up, we're going to end up at some point over here. That point is clearly zero, because we were 15 to the right. Now we're going to move 46 to the left, so we're definitely going to be to the left of zero, so it's definitely going to be a negative number. |
Adding numbers with different signs (example) Pre-Algebra Khan Academy.mp3 | This is the negative 46 that we're adding to the 15. So we're going to end up, we're going to end up at some point over here. That point is clearly zero, because we were 15 to the right. Now we're going to move 46 to the left, so we're definitely going to be to the left of zero, so it's definitely going to be a negative number. And we can even think about, we can even think about the absolute value of that negative number. We can just visualize it. We moved this yellow arrow as a length of 15, this orange arrow has a length of 46, the blue arrow that I'm about to draw, which is the sum of these two, is going to be, is going to be right, is going to have this length right over here. |
Adding numbers with different signs (example) Pre-Algebra Khan Academy.mp3 | Now we're going to move 46 to the left, so we're definitely going to be to the left of zero, so it's definitely going to be a negative number. And we can even think about, we can even think about the absolute value of that negative number. We can just visualize it. We moved this yellow arrow as a length of 15, this orange arrow has a length of 46, the blue arrow that I'm about to draw, which is the sum of these two, is going to be, is going to be right, is going to have this length right over here. And just visually, how could we figure out the length of this blue part, if we know the length of this orange part and we know the length of this yellow part? What's this going to be the difference? It's just going to be the difference of these two. |
Adding numbers with different signs (example) Pre-Algebra Khan Academy.mp3 | We moved this yellow arrow as a length of 15, this orange arrow has a length of 46, the blue arrow that I'm about to draw, which is the sum of these two, is going to be, is going to be right, is going to have this length right over here. And just visually, how could we figure out the length of this blue part, if we know the length of this orange part and we know the length of this yellow part? What's this going to be the difference? It's just going to be the difference of these two. So the absolute value of the sum is going to be the difference between this length, 46 and 15. So this is 46, let me just figure that out, 46 minus 15, 6 minus 5 is 1, 4 minus 1 is 3. So the length of this is going to be 31, and it's going to be 31 to the left of 0. |
Adding numbers with different signs (example) Pre-Algebra Khan Academy.mp3 | It's just going to be the difference of these two. So the absolute value of the sum is going to be the difference between this length, 46 and 15. So this is 46, let me just figure that out, 46 minus 15, 6 minus 5 is 1, 4 minus 1 is 3. So the length of this is going to be 31, and it's going to be 31 to the left of 0. So this is going to be negative 31 right over here. So we know that this first part over here is negative 31, and then to that we are going to add 29. So we're going to add 29, let me do that in another color. |
Adding numbers with different signs (example) Pre-Algebra Khan Academy.mp3 | So the length of this is going to be 31, and it's going to be 31 to the left of 0. So this is going to be negative 31 right over here. So we know that this first part over here is negative 31, and then to that we are going to add 29. So we're going to add 29, let me do that in another color. So what does that mean? That means that we're going to start at negative 31, and we're going to move 29 to the right, we're adding 29. So we're going to move 29 to the right, so maybe that gets us right about there. |
Adding numbers with different signs (example) Pre-Algebra Khan Academy.mp3 | So we're going to add 29, let me do that in another color. So what does that mean? That means that we're going to start at negative 31, and we're going to move 29 to the right, we're adding 29. So we're going to move 29 to the right, so maybe that gets us right about there. Try and draw an arrow of length 29. I can draw a cleaner looking arrow than that. I'll do it right over here actually. |
Adding numbers with different signs (example) Pre-Algebra Khan Academy.mp3 | So we're going to move 29 to the right, so maybe that gets us right about there. Try and draw an arrow of length 29. I can draw a cleaner looking arrow than that. I'll do it right over here actually. So then we're going to move 29 over to the right. That's the 29 part. So now this is a positive 29. |
Adding numbers with different signs (example) Pre-Algebra Khan Academy.mp3 | I'll do it right over here actually. So then we're going to move 29 over to the right. That's the 29 part. So now this is a positive 29. So how do we figure out what this is? So this is going to land us, so this number, this is 29 right here that we're adding, this is going to land us right over here on the number line. So how do we figure out what number that is? |
Adding numbers with different signs (example) Pre-Algebra Khan Academy.mp3 | So now this is a positive 29. So how do we figure out what this is? So this is going to land us, so this number, this is 29 right here that we're adding, this is going to land us right over here on the number line. So how do we figure out what number that is? Well, once again we can just visualize it, and eventually you won't have to draw number lines and stuff, but I think it'll be useful here. The number that we're going to do, we're starting at negative 31, we're adding 29 to it, so it's going to make it less negative, but we're still adding less than 31, so we're not going to get all the way back to 0, but we're still going to have a negative number. So we're still going to have a negative number, but how can we figure out the absolute value of that negative number, its distance? |
Adding numbers with different signs (example) Pre-Algebra Khan Academy.mp3 | So how do we figure out what number that is? Well, once again we can just visualize it, and eventually you won't have to draw number lines and stuff, but I think it'll be useful here. The number that we're going to do, we're starting at negative 31, we're adding 29 to it, so it's going to make it less negative, but we're still adding less than 31, so we're not going to get all the way back to 0, but we're still going to have a negative number. So we're still going to have a negative number, but how can we figure out the absolute value of that negative number, its distance? Well, once again, that little white part right there, that white part plus this 29 is going to equal 31 if you just think of absolute value, if we don't think about the signs, if we just think about the length. Or another way to think about it, 31 minus 29 will give us the length of that white part, and of course it's going to be negative because the negative number here is larger than the positive number, and we're adding the two. So if we do 31 minus 29, you could borrow in all of that, but that's clearly just going to be equal to 2. |
Adding numbers with different signs (example) Pre-Algebra Khan Academy.mp3 | So we're still going to have a negative number, but how can we figure out the absolute value of that negative number, its distance? Well, once again, that little white part right there, that white part plus this 29 is going to equal 31 if you just think of absolute value, if we don't think about the signs, if we just think about the length. Or another way to think about it, 31 minus 29 will give us the length of that white part, and of course it's going to be negative because the negative number here is larger than the positive number, and we're adding the two. So if we do 31 minus 29, you could borrow in all of that, but that's clearly just going to be equal to 2. You could say that's 11, this is a 2, you subtract, this is equal to 2, but since it's negative 31 plus 29, it's going to be a negative 2. We're still going to stay negative. We haven't moved far to the right enough to pass 0, so this right here is going to be negative 2. |
Adding numbers with different signs (example) Pre-Algebra Khan Academy.mp3 | So if we do 31 minus 29, you could borrow in all of that, but that's clearly just going to be equal to 2. You could say that's 11, this is a 2, you subtract, this is equal to 2, but since it's negative 31 plus 29, it's going to be a negative 2. We're still going to stay negative. We haven't moved far to the right enough to pass 0, so this right here is going to be negative 2. Or another way to think about it, the length of this white bar, the absolute value is going to be 2. 2 plus 29 is 31, but we're operating to the left of 0, so it's negative 31. This is negative 2. |
Adding numbers with different signs (example) Pre-Algebra Khan Academy.mp3 | We haven't moved far to the right enough to pass 0, so this right here is going to be negative 2. Or another way to think about it, the length of this white bar, the absolute value is going to be 2. 2 plus 29 is 31, but we're operating to the left of 0, so it's negative 31. This is negative 2. Anyway, hopefully you found that useful. Well, anyway, let me make it clear. So our final answer is negative 2. |
Finding the inequality representing the graph example Algebra I Khan Academy.mp3 | So this is saying, you give me an x. So let's say we take x is equal to 1 right there. 3 times 1 plus 5, so 3 times x plus 5. So 3 times 1 is 3, plus 5 is 8. So 1, 2, 3, 4, 5, 6, 7, 8. This is saying that y will be less than 8. y will be less than 3 times 1 plus 5. So the y values that satisfy this constraint for that x are going to be all of these values down here. |
Finding the inequality representing the graph example Algebra I Khan Academy.mp3 | So 3 times 1 is 3, plus 5 is 8. So 1, 2, 3, 4, 5, 6, 7, 8. This is saying that y will be less than 8. y will be less than 3 times 1 plus 5. So the y values that satisfy this constraint for that x are going to be all of these values down here. Let me do it in a lighter color. It'll be all of these values. For x is equal to 1, it'll be all the values down here. |
Finding the inequality representing the graph example Algebra I Khan Academy.mp3 | So the y values that satisfy this constraint for that x are going to be all of these values down here. Let me do it in a lighter color. It'll be all of these values. For x is equal to 1, it'll be all the values down here. And it would not include y is equal to 8. y has to be less than 8. Now, if we kept doing that, we would essentially just graph the line of y is equal to 3x plus 5, but we wouldn't include it. We would just include everything below it, just like we did right here. |
Finding the inequality representing the graph example Algebra I Khan Academy.mp3 | For x is equal to 1, it'll be all the values down here. And it would not include y is equal to 8. y has to be less than 8. Now, if we kept doing that, we would essentially just graph the line of y is equal to 3x plus 5, but we wouldn't include it. We would just include everything below it, just like we did right here. So we know how to graph just y is equal to 3x plus 5. Let me write it over here. So if I were to write y is equal to 3x plus 5, we'd say, OK, 3 is the slope. |
Finding the inequality representing the graph example Algebra I Khan Academy.mp3 | We would just include everything below it, just like we did right here. So we know how to graph just y is equal to 3x plus 5. Let me write it over here. So if I were to write y is equal to 3x plus 5, we'd say, OK, 3 is the slope. Slope is equal to 3. And then 5 is the y-intercept. Now, I could just graph the line, but because that won't be included in the y's that satisfy this constraint, I'm going to graph it as a dotted line. |
Finding the inequality representing the graph example Algebra I Khan Academy.mp3 | So if I were to write y is equal to 3x plus 5, we'd say, OK, 3 is the slope. Slope is equal to 3. And then 5 is the y-intercept. Now, I could just graph the line, but because that won't be included in the y's that satisfy this constraint, I'm going to graph it as a dotted line. So we'll start with the y-intercept of 5. So 1, 2, 3, 4, 5. That's the y-intercept. |
Finding the inequality representing the graph example Algebra I Khan Academy.mp3 | Now, I could just graph the line, but because that won't be included in the y's that satisfy this constraint, I'm going to graph it as a dotted line. So we'll start with the y-intercept of 5. So 1, 2, 3, 4, 5. That's the y-intercept. And the slope is 3. So if you go over the 1, you go up 3. Let me do that in that darker purple color. |
Finding the inequality representing the graph example Algebra I Khan Academy.mp3 | That's the y-intercept. And the slope is 3. So if you go over the 1, you go up 3. Let me do that in that darker purple color. So it will look like this. It will look like that. And you see that point would be on it. |
Finding the inequality representing the graph example Algebra I Khan Academy.mp3 | Let me do that in that darker purple color. So it will look like this. It will look like that. And you see that point would be on it. That point would be on it. If you go back, you're going to go down by 3. So that point will be on it, that point and that point. |
Finding the inequality representing the graph example Algebra I Khan Academy.mp3 | And you see that point would be on it. That point would be on it. If you go back, you're going to go down by 3. So that point will be on it, that point and that point. And then I'll just connect the dots with a dotted line. So it will look. So that dotted line is the graph of y is equal to 3x plus 5. |
Finding the inequality representing the graph example Algebra I Khan Academy.mp3 | So that point will be on it, that point and that point. And then I'll just connect the dots with a dotted line. So it will look. So that dotted line is the graph of y is equal to 3x plus 5. But we're not going to include it. So that's why I made it a dotted line. So we're going to include the y's that are less than that. |
Finding the inequality representing the graph example Algebra I Khan Academy.mp3 | So that dotted line is the graph of y is equal to 3x plus 5. But we're not going to include it. So that's why I made it a dotted line. So we're going to include the y's that are less than that. So for any x, so you pick an x. Let's say x is equal to negative 1. If you evaluate 3x plus 5 for that x, you'd get here. |
Finding the inequality representing the graph example Algebra I Khan Academy.mp3 | So we're going to include the y's that are less than that. So for any x, so you pick an x. Let's say x is equal to negative 1. If you evaluate 3x plus 5 for that x, you'd get here. But we only care about the y's that are strictly less than that. So you don't include the line. It's everything below it. |
Finding the inequality representing the graph example Algebra I Khan Academy.mp3 | If you evaluate 3x plus 5 for that x, you'd get here. But we only care about the y's that are strictly less than that. So you don't include the line. It's everything below it. So for any x you pick, it's going to be below that line. You take the x, go up to that line, and everything below it. So for all of the x's, it's going to be this entire area. |
Finding the inequality representing the graph example Algebra I Khan Academy.mp3 | It's everything below it. So for any x you pick, it's going to be below that line. You take the x, go up to that line, and everything below it. So for all of the x's, it's going to be this entire area. It's going to be this entire area. It's going to be this entire area that's under the line. I'll do it in this orange so it's easier to see. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | Each red blood cell has a volume of approximately 90 times 10 to the negative 15 liters. How many red blood cells are there in a human body? Write your answer in scientific notation and round to two decimal places. So they tell us the total volume of blood in the human body. We have five liters. So we have five liters. And then they tell us that 40% of that is red blood cells. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | So they tell us the total volume of blood in the human body. We have five liters. So we have five liters. And then they tell us that 40% of that is red blood cells. So if we take the five liters and we multiply by 40%, this expression right here gives us the total volume of the red blood cells, 40% of our total volume of blood. Now, if this is a total volume of red blood cells and we divide by the volume of each red blood cell, then we're going to get the number of red blood cells. So let's do that. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | And then they tell us that 40% of that is red blood cells. So if we take the five liters and we multiply by 40%, this expression right here gives us the total volume of the red blood cells, 40% of our total volume of blood. Now, if this is a total volume of red blood cells and we divide by the volume of each red blood cell, then we're going to get the number of red blood cells. So let's do that. Let's divide by the volume of each red blood cell. So the volume of each red blood cell is 90 times 10 to the negative 15 liters. So let's see if we can simplify it. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | So let's do that. Let's divide by the volume of each red blood cell. So the volume of each red blood cell is 90 times 10 to the negative 15 liters. So let's see if we can simplify it. So one thing that we can feel good about is that the units actually do cancel out. We have liters in the numerator, liters in the denominator. So we're going to get just a pure number, which is what we want. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | So let's see if we can simplify it. So one thing that we can feel good about is that the units actually do cancel out. We have liters in the numerator, liters in the denominator. So we're going to get just a pure number, which is what we want. We just want how many red blood cells there are actually in the body. So let's just focus on the numbers here. So 5 times 40%. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | So we're going to get just a pure number, which is what we want. We just want how many red blood cells there are actually in the body. So let's just focus on the numbers here. So 5 times 40%. Well, 40% is the same thing as 0.4. So let me write that down. This is the same thing as 0.4. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | So 5 times 40%. Well, 40% is the same thing as 0.4. So let me write that down. This is the same thing as 0.4. 5 times 0.4 is 2. So our numerator simplifies to 2. And in the denominator, we have 90 times 10 to the negative 15, which isn't quite in scientific notation. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | This is the same thing as 0.4. 5 times 0.4 is 2. So our numerator simplifies to 2. And in the denominator, we have 90 times 10 to the negative 15, which isn't quite in scientific notation. Or actually, it definitely is not in scientific notation. It looks like it. But remember, in order to be in scientific notation, this number has to be greater than or equal to 1 and less than 10. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | And in the denominator, we have 90 times 10 to the negative 15, which isn't quite in scientific notation. Or actually, it definitely is not in scientific notation. It looks like it. But remember, in order to be in scientific notation, this number has to be greater than or equal to 1 and less than 10. It's clearly not less than 10. But we can convert this to scientific notation very easily. 90 is the same thing as 9 times 10. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | But remember, in order to be in scientific notation, this number has to be greater than or equal to 1 and less than 10. It's clearly not less than 10. But we can convert this to scientific notation very easily. 90 is the same thing as 9 times 10. Or you could even say 9 times 10 to the first. And then you multiply that times 10 to the negative 15. And then this simplifies to 9 times 10 to the, well, let's add these two exponents, 10 to the negative 14. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | 90 is the same thing as 9 times 10. Or you could even say 9 times 10 to the first. And then you multiply that times 10 to the negative 15. And then this simplifies to 9 times 10 to the, well, let's add these two exponents, 10 to the negative 14. And now we can actually divide. And let's simplify this division a little bit. This is going to be the same thing as 2 over 9 times 1 over 10 to the negative 14. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | And then this simplifies to 9 times 10 to the, well, let's add these two exponents, 10 to the negative 14. And now we can actually divide. And let's simplify this division a little bit. This is going to be the same thing as 2 over 9 times 1 over 10 to the negative 14. Well, what's 1 over 10 to the negative 14? Well, that's just 10 to the 14. So this right over here is just the same thing as 10 to the 14. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | This is going to be the same thing as 2 over 9 times 1 over 10 to the negative 14. Well, what's 1 over 10 to the negative 14? Well, that's just 10 to the 14. So this right over here is just the same thing as 10 to the 14. Now, you might say, OK, we just have to figure out what 2 ninths is. And we're done. We've written this in scientific notation. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | So this right over here is just the same thing as 10 to the 14. Now, you might say, OK, we just have to figure out what 2 ninths is. And we're done. We've written this in scientific notation. But you might already realize, look, 2 ninths is not greater than or equal to 1. How can we make this greater than or equal to 1? Well, we could multiply it by 10. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | We've written this in scientific notation. But you might already realize, look, 2 ninths is not greater than or equal to 1. How can we make this greater than or equal to 1? Well, we could multiply it by 10. If we multiply this by 10, then we've got to divide this by 10 to not change the value of this expression. But let's do that. So this is equal to, if we multiply, so I'm going to multiply this by 10. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | Well, we could multiply it by 10. If we multiply this by 10, then we've got to divide this by 10 to not change the value of this expression. But let's do that. So this is equal to, if we multiply, so I'm going to multiply this by 10. And I'm going to divide this by 10. So I haven't changed. I've multiplied and divided by 10. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | So this is equal to, if we multiply, so I'm going to multiply this by 10. And I'm going to divide this by 10. So I haven't changed. I've multiplied and divided by 10. So this is equal to 20 over 9 times 10 to the 14th divided by 10 is 10 to the 13th power. So what's 20 over 9? This is going to give us a number that is greater than or equal to 1 and less than 10. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | I've multiplied and divided by 10. So this is equal to 20 over 9 times 10 to the 14th divided by 10 is 10 to the 13th power. So what's 20 over 9? This is going to give us a number that is greater than or equal to 1 and less than 10. So let's figure it out. And they wanted us to round, I think they said, round our answer to two decimal places. So let's do that. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | This is going to give us a number that is greater than or equal to 1 and less than 10. So let's figure it out. And they wanted us to round, I think they said, round our answer to two decimal places. So let's do that. So 20 divided by 9. 9 doesn't go into 2. It does go into 20. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | So let's do that. So 20 divided by 9. 9 doesn't go into 2. It does go into 20. 2 times 9 is 18. Subtract. Get a remainder of 2. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | It does go into 20. 2 times 9 is 18. Subtract. Get a remainder of 2. I think you see where this show is going to go. Get a remainder of 2. 9 goes into 20 2 times. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | Get a remainder of 2. I think you see where this show is going to go. Get a remainder of 2. 9 goes into 20 2 times. 2 times 9 is 18. We're just going to keep getting 2's. So we get another 2. |
Red blood cells in human body (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | 9 goes into 20 2 times. 2 times 9 is 18. We're just going to keep getting 2's. So we get another 2. Bring down a 0. 9 goes into 20 2 times. So this thing right over here is really 2.2 repeating. |
Introduction to graphing inequalities Two-variable linear inequalities Algebra I Khan Academy.mp3 | Let's graph ourselves some inequalities. So let's say I had the inequality y is less than or equal to 4x plus 3. And on our xy-coordinate plane, we want to show all of 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. |
Introduction to graphing inequalities Two-variable linear inequalities Algebra I Khan Academy.mp3 | 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. I'll try to draw it a little bit neater than that. |