Sentence
stringlengths 120
4.63k
| video_title
stringlengths 28
104
|
|---|---|
Well, we define that as t, and then we got 12 minutes. Now, the next part of the question is how many kilometers east of the castle will they be at that time? So to figure out how many kilometers east of the castle, we just have to calculate this expression or this expression when t is equal to 12. So I'm going to use this first one. So you're going to have 225 over 60 times 12 is going to give us, let's see, 60 and 12 are both divisible by 12, so you get that one over five. And if you divide 225 by five, that is going to give us 45, and the units all work out to kilometers. So we answered the first two parts of the question. | Comparing linear rates example.mp3 |
So I'm going to use this first one. So you're going to have 225 over 60 times 12 is going to give us, let's see, 60 and 12 are both divisible by 12, so you get that one over five. And if you divide 225 by five, that is going to give us 45, and the units all work out to kilometers. So we answered the first two parts of the question. And the second part of this question, they tell us that Latanya and Jair both wrote correct inequalities for the times in minutes when the dragon is farther east of the castle than the griffin is. Latanya wrote t is greater than 12, and Jair wrote t is greater than 54. How did Latanya and Jair define their variables? | Comparing linear rates example.mp3 |
So we answered the first two parts of the question. And the second part of this question, they tell us that Latanya and Jair both wrote correct inequalities for the times in minutes when the dragon is farther east of the castle than the griffin is. Latanya wrote t is greater than 12, and Jair wrote t is greater than 54. How did Latanya and Jair define their variables? Well, Latanya defined her variables the exact same way that I define mine, because we got t is equal to 12 when the dragon passes up the griffin. So for t is greater than 12, the dragon is farther east of the castle than the griffin. So we did the exact same thing as Latanya. | Comparing linear rates example.mp3 |
How did Latanya and Jair define their variables? Well, Latanya defined her variables the exact same way that I define mine, because we got t is equal to 12 when the dragon passes up the griffin. So for t is greater than 12, the dragon is farther east of the castle than the griffin. So we did the exact same thing as Latanya. But what Jair did, if he got t is greater than 54, is he must have defined t as being equal to the number of minutes since the griffin, the slower, the first but slower creature, passed the castle. And so he would have gotten t is greater than 54. And then if you wanted to know how many minutes since the dragon passed the castle, because the dragon got there 42 minutes later, he would have subtracted 42 from that. | Comparing linear rates example.mp3 |
So we did the exact same thing as Latanya. But what Jair did, if he got t is greater than 54, is he must have defined t as being equal to the number of minutes since the griffin, the slower, the first but slower creature, passed the castle. And so he would have gotten t is greater than 54. And then if you wanted to know how many minutes since the dragon passed the castle, because the dragon got there 42 minutes later, he would have subtracted 42 from that. And that's how you connect these two numbers over here. But the important thing to realize is there's multiple ways to solve the same problem. What matters is to be very clear how you are defining that variable, and use it consistently throughout, and then interpret it correctly when you're answering the questions. | Comparing linear rates example.mp3 |
So let's find all of the x's that satisfy this. So the first thing I'd like to do is get rid of this 3. So let's subtract 3 from both sides of this equation. So the left-hand side is just going to end up being 4x. These 3's cancel out. That just ends up with a 0. No reason to change the inequality just yet. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
So the left-hand side is just going to end up being 4x. These 3's cancel out. That just ends up with a 0. No reason to change the inequality just yet. We're just adding and subtracting from both sides. In this case, subtracting. That doesn't change the inequality as long as we're subtracting the same value. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
No reason to change the inequality just yet. We're just adding and subtracting from both sides. In this case, subtracting. That doesn't change the inequality as long as we're subtracting the same value. We have negative 1 minus 3. That is negative 4. Negative 1 minus 3 is negative 4. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
That doesn't change the inequality as long as we're subtracting the same value. We have negative 1 minus 3. That is negative 4. Negative 1 minus 3 is negative 4. And then we'll want to divide both sides of this equation by 4. Let's divide both sides of this equation by 4. Once again, when you multiply or divide both sides of an inequality by a positive number, it doesn't change the inequality. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
Negative 1 minus 3 is negative 4. And then we'll want to divide both sides of this equation by 4. Let's divide both sides of this equation by 4. Once again, when you multiply or divide both sides of an inequality by a positive number, it doesn't change the inequality. So the left-hand side is just x. x is less than negative 4 divided by 4 is negative 1. x is less than negative 1. Or we could write this in interval notation. All of the x's from negative infinity to negative 1, but not including negative 1. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
Once again, when you multiply or divide both sides of an inequality by a positive number, it doesn't change the inequality. So the left-hand side is just x. x is less than negative 4 divided by 4 is negative 1. x is less than negative 1. Or we could write this in interval notation. All of the x's from negative infinity to negative 1, but not including negative 1. So we put a parenthesis right there. Let's do a slightly harder one. Let's say we have 5x is greater than 8x plus 27. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
All of the x's from negative infinity to negative 1, but not including negative 1. So we put a parenthesis right there. Let's do a slightly harder one. Let's say we have 5x is greater than 8x plus 27. Let's get all our x's on the left-hand side. The best way to do that is subtract 8x from both sides. You subtract 8x from both sides. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
Let's say we have 5x is greater than 8x plus 27. Let's get all our x's on the left-hand side. The best way to do that is subtract 8x from both sides. You subtract 8x from both sides. The left-hand side becomes 5x minus 8x. That's negative 3x. We still have a greater than sign. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
You subtract 8x from both sides. The left-hand side becomes 5x minus 8x. That's negative 3x. We still have a greater than sign. We're just adding or subtracting the same quantity to both sides. These 8x's cancel out and you're just left with a 27. So you have negative 3x is greater than 27. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
We still have a greater than sign. We're just adding or subtracting the same quantity to both sides. These 8x's cancel out and you're just left with a 27. So you have negative 3x is greater than 27. Now, to just turn this into an x, we want to divide both sides by negative 3. But remember, when you multiply or divide both sides of an inequality by a negative number, you swap the inequality. So if we divide both sides of this by negative 3, we have to swap this inequality. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
So you have negative 3x is greater than 27. Now, to just turn this into an x, we want to divide both sides by negative 3. But remember, when you multiply or divide both sides of an inequality by a negative number, you swap the inequality. So if we divide both sides of this by negative 3, we have to swap this inequality. It will go from being a greater than sign to a less than sign. And this is a bit of a way that I remember greater than. The left-hand side just looks bigger. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
So if we divide both sides of this by negative 3, we have to swap this inequality. It will go from being a greater than sign to a less than sign. And this is a bit of a way that I remember greater than. The left-hand side just looks bigger. This is greater than. If you just imagine this height, that height is greater than that height right there, which is just a point. I don't know if that confuses you or not. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
The left-hand side just looks bigger. This is greater than. If you just imagine this height, that height is greater than that height right there, which is just a point. I don't know if that confuses you or not. This is less than. This little point is less than the distance of that big opening. That's how I remember it. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
I don't know if that confuses you or not. This is less than. This little point is less than the distance of that big opening. That's how I remember it. But anyway, 3x over negative 3. So anyway, now that we divided both sides by a negative number, by negative 3, we swapped the inequality from greater than to less than. In the left-hand side, the negative 3's cancel out. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
That's how I remember it. But anyway, 3x over negative 3. So anyway, now that we divided both sides by a negative number, by negative 3, we swapped the inequality from greater than to less than. In the left-hand side, the negative 3's cancel out. You get x is less than 27 over negative 3, which is negative 9. Or in interval notation, it would be everything from negative infinity to negative 9, not including negative 9. If you wanted to do it as a number line, it would look like this. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
In the left-hand side, the negative 3's cancel out. You get x is less than 27 over negative 3, which is negative 9. Or in interval notation, it would be everything from negative infinity to negative 9, not including negative 9. If you wanted to do it as a number line, it would look like this. This would be negative 9. Maybe this would be negative 8. Maybe this would be negative 10. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
If you wanted to do it as a number line, it would look like this. This would be negative 9. Maybe this would be negative 8. Maybe this would be negative 10. You would start at negative 9, not include it because we don't have an equal sign here. You go all the way, everything less than that, all the way down as we see to negative infinity. Let's do a nice, hairy problem. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
Maybe this would be negative 10. You would start at negative 9, not include it because we don't have an equal sign here. You go all the way, everything less than that, all the way down as we see to negative infinity. Let's do a nice, hairy problem. Let's say we have 8x minus 5 times 4x plus 1 is greater than or equal to negative 1 plus 2 times 4x minus 3. This might seem very daunting, but if we just simplify it step by step, you'll see it's no harder than any of the other problems we've tackled. Let's just simplify this. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
Let's do a nice, hairy problem. Let's say we have 8x minus 5 times 4x plus 1 is greater than or equal to negative 1 plus 2 times 4x minus 3. This might seem very daunting, but if we just simplify it step by step, you'll see it's no harder than any of the other problems we've tackled. Let's just simplify this. You get 8x minus, let's distribute this negative 5. Let me say 8x and then distribute the negative 5. Negative 5 times 4x is negative 20x. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
Let's just simplify this. You get 8x minus, let's distribute this negative 5. Let me say 8x and then distribute the negative 5. Negative 5 times 4x is negative 20x. Negative 5, when I say negative 5, I'm talking about this whole thing. Negative 5 times 1 is negative 5. Then that's going to be greater than or equal to negative 1 plus 2 times 4x is 8x. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
Negative 5 times 4x is negative 20x. Negative 5, when I say negative 5, I'm talking about this whole thing. Negative 5 times 1 is negative 5. Then that's going to be greater than or equal to negative 1 plus 2 times 4x is 8x. 2 times negative 3 is negative 6. Now we can merge these two terms. 8x minus 20x is negative 12x minus 5 is greater than or equal to, we can merge these constant terms, negative 1 minus 6, that's negative 7. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
Then that's going to be greater than or equal to negative 1 plus 2 times 4x is 8x. 2 times negative 3 is negative 6. Now we can merge these two terms. 8x minus 20x is negative 12x minus 5 is greater than or equal to, we can merge these constant terms, negative 1 minus 6, that's negative 7. Then we have this plus 8x left over. I like to get all my x terms on the left-hand side, so let's subtract 8x from both sides of this equation. Let's subtract 8x from both sides of this equation. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
8x minus 20x is negative 12x minus 5 is greater than or equal to, we can merge these constant terms, negative 1 minus 6, that's negative 7. Then we have this plus 8x left over. I like to get all my x terms on the left-hand side, so let's subtract 8x from both sides of this equation. Let's subtract 8x from both sides of this equation. From both sides, I'm subtracting 8x. This left-hand side, negative 12 minus 8, that's negative 20. Negative 20x minus 5. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
Let's subtract 8x from both sides of this equation. From both sides, I'm subtracting 8x. This left-hand side, negative 12 minus 8, that's negative 20. Negative 20x minus 5. Once again, no reason to change the inequality just yet. All we're doing is simplifying the sides or adding and subtracting from them. The right-hand side becomes, this thing cancels out, 8x minus 8x, that's 0. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
Negative 20x minus 5. Once again, no reason to change the inequality just yet. All we're doing is simplifying the sides or adding and subtracting from them. The right-hand side becomes, this thing cancels out, 8x minus 8x, that's 0. You're just left with a negative 7. Now I want to get rid of this negative 5, so let's add 5 to both sides of this equation. The left-hand side, you're just left with a negative 20x. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
The right-hand side becomes, this thing cancels out, 8x minus 8x, that's 0. You're just left with a negative 7. Now I want to get rid of this negative 5, so let's add 5 to both sides of this equation. The left-hand side, you're just left with a negative 20x. The 5 and these 5s cancel out. No reason to change the inequality just yet. Negative 7 plus 5, that's negative 2. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
The left-hand side, you're just left with a negative 20x. The 5 and these 5s cancel out. No reason to change the inequality just yet. Negative 7 plus 5, that's negative 2. Now we're at an interesting point. We have negative 20x is greater than or equal to negative 2. If this was an equation or really any type of even inequality, we want to divide both sides by negative 20, but we have to remember, when you multiply or divide both sides of an inequality by a negative number, you have to swap the inequality, so let's remember that. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
Negative 7 plus 5, that's negative 2. Now we're at an interesting point. We have negative 20x is greater than or equal to negative 2. If this was an equation or really any type of even inequality, we want to divide both sides by negative 20, but we have to remember, when you multiply or divide both sides of an inequality by a negative number, you have to swap the inequality, so let's remember that. If we divide this side by negative 20 and we divide this side by negative 20, all I did is took both of these sides, divided by negative 20. We have to swap the inequality. The greater than or equal to has to become a less than or equal sign. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
If this was an equation or really any type of even inequality, we want to divide both sides by negative 20, but we have to remember, when you multiply or divide both sides of an inequality by a negative number, you have to swap the inequality, so let's remember that. If we divide this side by negative 20 and we divide this side by negative 20, all I did is took both of these sides, divided by negative 20. We have to swap the inequality. The greater than or equal to has to become a less than or equal sign. Of course, these cancel out. You get x is less than or equal to, the negatives cancel out. 2 over 20 is 1 over 10. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
The greater than or equal to has to become a less than or equal sign. Of course, these cancel out. You get x is less than or equal to, the negatives cancel out. 2 over 20 is 1 over 10. If we were writing it in interval notation, the upper bound would be 1 over 10. Notice we're including it because we have an equal sign, less than or equal, so we're including 1 over 10, and we're going to go all the way down to negative infinity, everything less than or equal to 1 over 10. This is just another way of writing that. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
2 over 20 is 1 over 10. If we were writing it in interval notation, the upper bound would be 1 over 10. Notice we're including it because we have an equal sign, less than or equal, so we're including 1 over 10, and we're going to go all the way down to negative infinity, everything less than or equal to 1 over 10. This is just another way of writing that. Just for fun, let's draw the number line right here. This is maybe 0. That is 1. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
This is just another way of writing that. Just for fun, let's draw the number line right here. This is maybe 0. That is 1. 1 over 10 might be over here. Everything less than or equal to 1 over 10. We're going to include the 1 tenth and everything less than that is included in the solutions. | Multi-step inequalities Linear inequalities Algebra I Khan Academy.mp3 |
Intercept form and I encourage you like always pause the video and see if you can do it So let's first think about point slope form point point slope Point slope form and point slope form is very easy to generate if you know a point on The line or if you know a point that satisfies where the x and y coordinates satisfy the linear equation and if you were to know the slope of The of the line that represents a solution set of that linear equation Now for sure we actually had we were given two points that are solutions that represent solutions to the linear equation To fully apply point slope or to apply point slope easily We just have to figure out the slope and what we could do is we could just evaluate What's the slope between the two points that we know and we just have to remind ourselves that slope slope is equal to Slope is equal to change in y over change in x Sometimes people say rise over run and what's that going to be? well if we say if we if we say that this the second point right over here if We say if we say this is kind of our if we starting at this point And we go to that point then our change in y going from this point to that point is going to be it's going to be equal to 1 minus 1 minus 9 1 minus 9 this point right over here is the point 6 comma 1 so we started at y equals 9 we finish at y equals 1 our Change in y is going to be 1 minus 9 we have a negative 8 change in y which makes sense We've gone down 8 so this is going to be equal to this is going to be equal to negative 8 That's our change in y and what's our change in x well we go from x equals 4 to x equals 6 So we end up at x equals 6 and we started at x equals 4 We started at x equals 4 so our change in x is 6 minus 4 which is equal to 2 Which is equal to 2 and you could have even done it visually to go from this point to this point your change in y your change in y is You went down 8 so your change in let me write this so your change in y is equal to negative 8 And what was your change in x to get and get to this point? Well your change in x is positive 2 So your change in x is equal to 2 And so what's your slope change in y over change in x negative 8 over 2 is equal to negative 4 So now that we have a now that we know the slope And we know a point and we know a and we know a point we actually know two points on the line We can express this in point slope form and so let's do that And the way I like to do is I always like to just take it straight from the definition of what slope is We know that the slope between any two points on this line is going to be negative 4 So if we take an arbitrary y that sits on this line And if we find the difference between that y and let's focus on this point up here so if we find the difference that y and this y and 9 and and It's over over the difference between some x on the line and this x and 4 and 4 this is going to be the slope between any x y on this line and This point right over here and the slope between any two points on a line are going to have to be constant So this is going to be equal to the slope of the line It's going to be equal to negative 4 and we're not in point slope form or classic point slope form just yet to do that We just multiply both sides times x minus 4 so we get y minus 9 We get y minus 9 is equal to our slope negative 4 times x minus 4 times X minus 4 and this right over here is our classic this right over here is our cat classic point slope form We have the point sometimes they even put parentheses like this But we could figure out the point from this point slope form the point that sits on this line Would things that make both sides of this equation equal to 0 so it would be x equals 4 y equals 9 Which we have right up there, and then the slope is right over here. It's negative 4 Now from this can we now express this this this linear equation in y-intercept form and Y intercept form just as a bit of a reminder. It's y is equal to mx plus B where this coefficient is our slope and this constant right over here is Allows us to figure out our y-intercept and to get this and to get this in this form We just have to simplify a little bit of this algebra, so you have y minus 9 y minus 9 is Equal to let's distribute this negative 4 and I'll just switch some colors Let's distribute this negative 4 negative 4 times x is negative 4x negative 4 times negative 4 is plus 16 And now if we just want to isolate the y on the left hand side we can add 9 to both sides Let's do that. Let's add 9 Let's add 9 to both sides Let's add 9 to both sides on the left hand side. We're just left with y and on the right hand side. | Point-slope and slope-intercept form from two points Algebra I Khan Academy.mp3 |
The function f is defined as follows. f of t is equal to negative two t plus five. So whatever we input into this function, we multiply it times negative two, and then we add five. So what is the input value for which f of t is equal to 13? So if f of t is equal to 13, that means that this thing over here is equal to 13 for some t, for some input. So we can just solve the equation and negative two t plus five is equal to 13. So let's do that. | How to match function input to output given the formula (example) Algebra I Khan Academy.mp3 |
So what is the input value for which f of t is equal to 13? So if f of t is equal to 13, that means that this thing over here is equal to 13 for some t, for some input. So we can just solve the equation and negative two t plus five is equal to 13. So let's do that. Negative two t plus five is equal to 13. Well, we can subtract five from both sides. I'm just trying to isolate the t on the left-hand side. | How to match function input to output given the formula (example) Algebra I Khan Academy.mp3 |
So let's do that. Negative two t plus five is equal to 13. Well, we can subtract five from both sides. I'm just trying to isolate the t on the left-hand side. So subtract negative five from the left. That's the whole reason why we did that, so those disappear. But we have to do it from the right as well. | How to match function input to output given the formula (example) Algebra I Khan Academy.mp3 |
I'm just trying to isolate the t on the left-hand side. So subtract negative five from the left. That's the whole reason why we did that, so those disappear. But we have to do it from the right as well. So you have 13 minus five is eight, and on the left-hand side, you still have your negative two t. So you have negative two t is equal to eight. Now to just have a t on the left-hand side, I want to divide both sides by negative two. And I'm left with t is equal to eight divided by negative two is equal to negative four. | How to match function input to output given the formula (example) Algebra I Khan Academy.mp3 |
But we have to do it from the right as well. So you have 13 minus five is eight, and on the left-hand side, you still have your negative two t. So you have negative two t is equal to eight. Now to just have a t on the left-hand side, I want to divide both sides by negative two. And I'm left with t is equal to eight divided by negative two is equal to negative four. So you input negative four, you input negative four into this function, into this function, and it will output 13. 13. Or we could write that f of negative four is equal to 13. | How to match function input to output given the formula (example) Algebra I Khan Academy.mp3 |
And I'm left with t is equal to eight divided by negative two is equal to negative four. So you input negative four, you input negative four into this function, into this function, and it will output 13. 13. Or we could write that f of negative four is equal to 13. But this is what they're looking for. This is the input value. Negative four is the input value for which f of t is equal to 13. | How to match function input to output given the formula (example) Algebra I Khan Academy.mp3 |
And we want to write that product in standard quadratic form, which is just a fancy way of saying a form where you have some coefficient on the second degree term, ax squared, plus some coefficient b on the first degree term, plus the constant term. So this right over here would be standard quadratic form. So that's the form that we want to express this product in. And I encourage you to pause the video and try to work through it on your own. All right, now let's work through this. And the key when we're multiplying two binomials like this, or actually when we're multiplying any polynomials, is just to remember the distributive property that we all by this point know quite well. So what we could view this as is we can distribute this x minus four, this entire expression, over the x and the seven. | Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3 |
And I encourage you to pause the video and try to work through it on your own. All right, now let's work through this. And the key when we're multiplying two binomials like this, or actually when we're multiplying any polynomials, is just to remember the distributive property that we all by this point know quite well. So what we could view this as is we can distribute this x minus four, this entire expression, over the x and the seven. So we could say that this is the same thing as x minus four times x, plus x minus four times seven. So let's write that. So x minus four times x, or we could write this as x times x minus four. | Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3 |
So what we could view this as is we can distribute this x minus four, this entire expression, over the x and the seven. So we could say that this is the same thing as x minus four times x, plus x minus four times seven. So let's write that. So x minus four times x, or we could write this as x times x minus four. That's distributing the, or multiplying the x minus four times x. That's right there. Plus seven times x minus four. | Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3 |
So x minus four times x, or we could write this as x times x minus four. That's distributing the, or multiplying the x minus four times x. That's right there. Plus seven times x minus four. Times x minus four. Notice, all we did is distribute the x minus four. We took this whole thing and we multiplied it by each term over here. | Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3 |
Plus seven times x minus four. Times x minus four. Notice, all we did is distribute the x minus four. We took this whole thing and we multiplied it by each term over here. We multiplied x by x minus four, and we multiplied seven by x minus four. Now we see that we have these, I guess you'd call them two separate terms, and to simplify each of them, or to multiply them out, we just have to distribute in this first, we have to distribute this blue x, and over here we have to distribute this blue seven. So let's do that. | Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3 |
We took this whole thing and we multiplied it by each term over here. We multiplied x by x minus four, and we multiplied seven by x minus four. Now we see that we have these, I guess you'd call them two separate terms, and to simplify each of them, or to multiply them out, we just have to distribute in this first, we have to distribute this blue x, and over here we have to distribute this blue seven. So let's do that. So here we could say x times x is going to be x squared. X times, we have a negative here, so we could say negative four is going to be negative four x. And just like that, we get x squared minus four x. | Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3 |
So let's do that. So here we could say x times x is going to be x squared. X times, we have a negative here, so we could say negative four is going to be negative four x. And just like that, we get x squared minus four x. And then over here, we have seven times x, so that's going to be plus seven x. And then we have seven times the negative four, which is negative 28. And we are almost done. | Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3 |
And just like that, we get x squared minus four x. And then over here, we have seven times x, so that's going to be plus seven x. And then we have seven times the negative four, which is negative 28. And we are almost done. We can simplify it a little bit more. We have two first degree terms here. If I have negative four x's, and to that I add seven x's, what is that going to be? | Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3 |
And we are almost done. We can simplify it a little bit more. We have two first degree terms here. If I have negative four x's, and to that I add seven x's, what is that going to be? Well, those two terms together, these two terms together are going to be negative four plus seven x's. Negative four plus seven. Negative four plus seven x's. | Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3 |
If I have negative four x's, and to that I add seven x's, what is that going to be? Well, those two terms together, these two terms together are going to be negative four plus seven x's. Negative four plus seven. Negative four plus seven x's. So all I'm doing here, I'm making it very clear that I'm adding these two coefficients, and then we have all the other terms. We have the x squared, x squared plus this, and then we have the minus 28. And we're at the home stretch. | Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3 |
Negative four plus seven x's. So all I'm doing here, I'm making it very clear that I'm adding these two coefficients, and then we have all the other terms. We have the x squared, x squared plus this, and then we have the minus 28. And we're at the home stretch. This would simplify to x squared. Now negative four plus seven is three, so this is going to be plus three x. That's what these two middle terms simplify to, to three x, and then we have minus 28. | Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3 |
And we're at the home stretch. This would simplify to x squared. Now negative four plus seven is three, so this is going to be plus three x. That's what these two middle terms simplify to, to three x, and then we have minus 28. Minus 28. And just like that, we are done. And a fun thing to think about, and notice it's in the same form. | Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3 |
That's what these two middle terms simplify to, to three x, and then we have minus 28. Minus 28. And just like that, we are done. And a fun thing to think about, and notice it's in the same form. If we were to compare, a is one, b is three, and c is negative 28. But it's interesting here to look at the pattern. When we multiply these two binomials, especially these two binomials, where the coefficient on the x term was a one. | Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3 |
And a fun thing to think about, and notice it's in the same form. If we were to compare, a is one, b is three, and c is negative 28. But it's interesting here to look at the pattern. When we multiply these two binomials, especially these two binomials, where the coefficient on the x term was a one. Notice, we have x times x, that's what actually forms the x squared term over here. We have negative four, let me do this in a new color. We have negative four times, that's not a new color. | Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3 |
When we multiply these two binomials, especially these two binomials, where the coefficient on the x term was a one. Notice, we have x times x, that's what actually forms the x squared term over here. We have negative four, let me do this in a new color. We have negative four times, that's not a new color. We have negative four times seven, which is going to be negative 28. And then how did we get this middle term? How did we get this three x? | Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3 |
We have negative four times, that's not a new color. We have negative four times seven, which is going to be negative 28. And then how did we get this middle term? How did we get this three x? Well, you had the negative four x plus the seven x, or you had the negative four plus the seven times x. You had the negative four plus the seven times x. So hopefully you see a little bit of a pattern here. | Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3 |
How did we get this three x? Well, you had the negative four x plus the seven x, or you had the negative four plus the seven times x. You had the negative four plus the seven times x. So hopefully you see a little bit of a pattern here. If you're multiplying two binomials where the coefficients on the x term are both one, it's going to be x squared, and then the last term, the constant term, is going to be the product of these two constants, negative four and seven. And then the first degree term, right over here, its coefficient is going to be the sum of these two constants, negative four and seven. Now this might, you could view this pattern, if you practice it, as just something that'll help you multiply binomials a little bit faster. | Multiplying binomials intro Mathematics II High School Math Khan Academy.mp3 |
So this table here, we are given a bunch of n's, n equals one, two, three, four, and we get the corresponding g of n. And one way to think about it is that this function g defines a sequence where n is the term of the sequence. So for example, we could say this is the same thing as the sequence where the first term is 168, second term is 84, third term is 42, and fourth term is 21, and we keep going on and on and on. Now let's think about what type of a sequence this is. If we think of it as starting at 168, and how do we go from 168 to 84? Well, one way you could say we subtracted 84, but another way to think about it is you multiply it by 1 1⁄2. So times 1 1⁄2. And then to go from 84 to 42, you multiply by 1 1⁄2 again. | Explicit & recursive formulas for geometric sequences High School Math Khan Academy.mp3 |
If we think of it as starting at 168, and how do we go from 168 to 84? Well, one way you could say we subtracted 84, but another way to think about it is you multiply it by 1 1⁄2. So times 1 1⁄2. And then to go from 84 to 42, you multiply by 1 1⁄2 again. Times 1 1⁄2. And to go from 42 to 21, you multiply by 1 1⁄2 again. So this right over here is a geometric series. | Explicit & recursive formulas for geometric sequences High School Math Khan Academy.mp3 |
And then to go from 84 to 42, you multiply by 1 1⁄2 again. Times 1 1⁄2. And to go from 42 to 21, you multiply by 1 1⁄2 again. So this right over here is a geometric series. We're starting at a term, and every successive term is the previous term times what's often called the common ratio, times 1 1⁄2. So how can we write g of n, how can we define this explicitly in terms of n? And I encourage you to pause the video and think about how to do that. | Explicit & recursive formulas for geometric sequences High School Math Khan Academy.mp3 |
So this right over here is a geometric series. We're starting at a term, and every successive term is the previous term times what's often called the common ratio, times 1 1⁄2. So how can we write g of n, how can we define this explicitly in terms of n? And I encourage you to pause the video and think about how to do that. So construct a, so if I say g of n equals, think of a function definition that describes what we've just seen here, starting at 168 and then multiplying by 1 1⁄2 every time you add a new term. Well, one way to think about it is we start at 168, and then we're gonna multiply by 1 1⁄2, we're gonna multiply by 1 1⁄2 a certain number of times. So we could view the exponent as the number of times we multiply by 1 1⁄2. | Explicit & recursive formulas for geometric sequences High School Math Khan Academy.mp3 |
And I encourage you to pause the video and think about how to do that. So construct a, so if I say g of n equals, think of a function definition that describes what we've just seen here, starting at 168 and then multiplying by 1 1⁄2 every time you add a new term. Well, one way to think about it is we start at 168, and then we're gonna multiply by 1 1⁄2, we're gonna multiply by 1 1⁄2 a certain number of times. So we could view the exponent as the number of times we multiply by 1 1⁄2. And how many times are we gonna multiply by 1 1⁄2? The first term, we multiply by 1 1⁄2 zero times. The second term, we multiply by 1 1⁄2 one time. | Explicit & recursive formulas for geometric sequences High School Math Khan Academy.mp3 |
So we could view the exponent as the number of times we multiply by 1 1⁄2. And how many times are we gonna multiply by 1 1⁄2? The first term, we multiply by 1 1⁄2 zero times. The second term, we multiply by 1 1⁄2 one time. Third term, we multiply by 1 1⁄2 two times. Fourth term, we multiply by 1 1⁄2 three times. So it seems like whatever term we're on, we're multiplying by 1 1⁄2 that term minus one times. | Explicit & recursive formulas for geometric sequences High School Math Khan Academy.mp3 |
The second term, we multiply by 1 1⁄2 one time. Third term, we multiply by 1 1⁄2 two times. Fourth term, we multiply by 1 1⁄2 three times. So it seems like whatever term we're on, we're multiplying by 1 1⁄2 that term minus one times. And you can see that this works. If n is equal to one, you're going to have one minus one, that's just gonna be zero, 1 1⁄2 to the zero is just one, so you're just gonna get a 168. If n is two, well, two minus one, you're gonna multiply by 1 1⁄2 one time, which you see right over here. | Explicit & recursive formulas for geometric sequences High School Math Khan Academy.mp3 |
So it seems like whatever term we're on, we're multiplying by 1 1⁄2 that term minus one times. And you can see that this works. If n is equal to one, you're going to have one minus one, that's just gonna be zero, 1 1⁄2 to the zero is just one, so you're just gonna get a 168. If n is two, well, two minus one, you're gonna multiply by 1 1⁄2 one time, which you see right over here. If n is three, you're gonna multiply by 1 1⁄2 twice. Three minus two is, or three minus one is two. You're gonna multiply by 1 1⁄2 twice, and you see that right over there. | Explicit & recursive formulas for geometric sequences High School Math Khan Academy.mp3 |
If n is two, well, two minus one, you're gonna multiply by 1 1⁄2 one time, which you see right over here. If n is three, you're gonna multiply by 1 1⁄2 twice. Three minus two is, or three minus one is two. You're gonna multiply by 1 1⁄2 twice, and you see that right over there. So this feels like a really nice explicit definition for this geometric series. And you could think of it in other ways. You could write this as g of n is equal to, let's see, one way you could write it as, you could write it as 168, and I'm just algebraically manipulating it, over two to the n minus one. | Explicit & recursive formulas for geometric sequences High School Math Khan Academy.mp3 |
You're gonna multiply by 1 1⁄2 twice, and you see that right over there. So this feels like a really nice explicit definition for this geometric series. And you could think of it in other ways. You could write this as g of n is equal to, let's see, one way you could write it as, you could write it as 168, and I'm just algebraically manipulating it, over two to the n minus one. Another way you could think about it is, well, let's use our exponent properties a little bit. We could say g of n is equal to, let's see, 1 1⁄2 to the n minus one, that's the same thing as 1 1⁄2, let me write this, it's equal to 168, let me do this in a different color. So this part right over here is the same thing as 1 1⁄2 to the n, so times 1 1⁄2 to the n times 1 1⁄2 to the negative one. | Explicit & recursive formulas for geometric sequences High School Math Khan Academy.mp3 |
You could write this as g of n is equal to, let's see, one way you could write it as, you could write it as 168, and I'm just algebraically manipulating it, over two to the n minus one. Another way you could think about it is, well, let's use our exponent properties a little bit. We could say g of n is equal to, let's see, 1 1⁄2 to the n minus one, that's the same thing as 1 1⁄2, let me write this, it's equal to 168, let me do this in a different color. So this part right over here is the same thing as 1 1⁄2 to the n, so times 1 1⁄2 to the n times 1 1⁄2 to the negative one. 1 1⁄2 to the negative one. Well, 1 1⁄2 to the negative one is just two, is just two, so this is times two. So we could rewrite this whole thing as, 168 times two is what, 336, 336, did I do that right? | Explicit & recursive formulas for geometric sequences High School Math Khan Academy.mp3 |
So this part right over here is the same thing as 1 1⁄2 to the n, so times 1 1⁄2 to the n times 1 1⁄2 to the negative one. 1 1⁄2 to the negative one. Well, 1 1⁄2 to the negative one is just two, is just two, so this is times two. So we could rewrite this whole thing as, 168 times two is what, 336, 336, did I do that right? 160 times two would be 320, plus 16, two times eight, so yeah, 336, and then times 1 1⁄2 to the n, times 1 1⁄2 to the n. So these are equivalent statements. This one makes a little bit more intuitive sense, it kind of jumps out at you, you're starting at 168 and you're multiplying by 1 1⁄2, whatever term you are, minus one times, but this is algebraically equivalent to this, to our original one. Now can we also define g of n recursively? | Explicit & recursive formulas for geometric sequences High School Math Khan Academy.mp3 |
So we could rewrite this whole thing as, 168 times two is what, 336, 336, did I do that right? 160 times two would be 320, plus 16, two times eight, so yeah, 336, and then times 1 1⁄2 to the n, times 1 1⁄2 to the n. So these are equivalent statements. This one makes a little bit more intuitive sense, it kind of jumps out at you, you're starting at 168 and you're multiplying by 1 1⁄2, whatever term you are, minus one times, but this is algebraically equivalent to this, to our original one. Now can we also define g of n recursively? And I encourage you to pause the video and try to do that. And in a lot of ways, the recursive definition is a little bit more straightforward, so let's do that. G, well, I'll make the recursive function a different, well, I'll still stick with g of n, since it's on this table right over here. | Explicit & recursive formulas for geometric sequences High School Math Khan Academy.mp3 |
Now can we also define g of n recursively? And I encourage you to pause the video and try to do that. And in a lot of ways, the recursive definition is a little bit more straightforward, so let's do that. G, well, I'll make the recursive function a different, well, I'll still stick with g of n, since it's on this table right over here. G of n is equal to, so let's see, if we're going to, when n equals one, if n is equal to one, we're starting at 168, 168, and if n is greater than one and a whole number, so if n, so this is going to be defined over all positive integers, and whole number, what are we going to do? Well, we're going to take 1⁄2 and multiply it times the previous term. So it's going to be 1⁄2 times g of n minus one. | Explicit & recursive formulas for geometric sequences High School Math Khan Academy.mp3 |
G, well, I'll make the recursive function a different, well, I'll still stick with g of n, since it's on this table right over here. G of n is equal to, so let's see, if we're going to, when n equals one, if n is equal to one, we're starting at 168, 168, and if n is greater than one and a whole number, so if n, so this is going to be defined over all positive integers, and whole number, what are we going to do? Well, we're going to take 1⁄2 and multiply it times the previous term. So it's going to be 1⁄2 times g of n minus one. And you can verify that this works. If n is equal to one, we just go right over here, it's going to be 168. G of two is going to be 1⁄2 times g of one, which is, of course, 168. | Explicit & recursive formulas for geometric sequences High School Math Khan Academy.mp3 |
So it's going to be 1⁄2 times g of n minus one. And you can verify that this works. If n is equal to one, we just go right over here, it's going to be 168. G of two is going to be 1⁄2 times g of one, which is, of course, 168. So 168 times 1⁄2 is 84. G of three is going to be 1⁄2 times g of two, which it is, g of three is 1⁄2 times g of two. So this is how we would define, this is the explicit definition of this sequence. | Explicit & recursive formulas for geometric sequences High School Math Khan Academy.mp3 |
And then we're told express the area of the entire rectangle and the expression should be expanded. So pause this video and see if you can work through this. All right, now let's work through this together. And what this diagram is showing us is exactly an indicative rectangle where its height is five and its width is three x squared minus x plus two. And what this shows us is that the area of the entire rectangle can be broken down into three smaller areas. You have this blue area right over here where the width is three x squared, the height is five. So what's that area going to be? | Multiply monomials by polynomials Area model Algebra 1 Khan Academy.mp3 |
And what this diagram is showing us is exactly an indicative rectangle where its height is five and its width is three x squared minus x plus two. And what this shows us is that the area of the entire rectangle can be broken down into three smaller areas. You have this blue area right over here where the width is three x squared, the height is five. So what's that area going to be? Well, it's going to be the height times width, five times three x squared. And that's the same thing as 15 x squared. Now what about this purple area right over here? | Multiply monomials by polynomials Area model Algebra 1 Khan Academy.mp3 |
So what's that area going to be? Well, it's going to be the height times width, five times three x squared. And that's the same thing as 15 x squared. Now what about this purple area right over here? Well, it's going to be height times width again, so it'll be five times negative x, which is the same thing as negative five x. And I know what some of you are thinking, how can you have a width of negative x? Well, we don't know what x is, so this is all a little bit abstract, but you could imagine having x be a negative value, in which case this actually would be a positive width. | Multiply monomials by polynomials Area model Algebra 1 Khan Academy.mp3 |
Now what about this purple area right over here? Well, it's going to be height times width again, so it'll be five times negative x, which is the same thing as negative five x. And I know what some of you are thinking, how can you have a width of negative x? Well, we don't know what x is, so this is all a little bit abstract, but you could imagine having x be a negative value, in which case this actually would be a positive width. And another thing you might be saying is, hey, this magenta area, when I just eyeball it, looks like it has a larger width than this blue area. How do we know that? We don't. | Multiply monomials by polynomials Area model Algebra 1 Khan Academy.mp3 |
Well, we don't know what x is, so this is all a little bit abstract, but you could imagine having x be a negative value, in which case this actually would be a positive width. And another thing you might be saying is, hey, this magenta area, when I just eyeball it, looks like it has a larger width than this blue area. How do we know that? We don't. They're just showing this as an indicative way. They might have actually the same width. They might have, one might be larger than the other, but this is just showing us that they're not necessarily the same. | Multiply monomials by polynomials Area model Algebra 1 Khan Academy.mp3 |
We don't. They're just showing this as an indicative way. They might have actually the same width. They might have, one might be larger than the other, but this is just showing us that they're not necessarily the same. It's very abstract. It's definitely not drawn to scale because we don't know what x is. All right, and then this last area is going to be the height, which is five, times the width, which is two. | Multiply monomials by polynomials Area model Algebra 1 Khan Academy.mp3 |
They might have, one might be larger than the other, but this is just showing us that they're not necessarily the same. It's very abstract. It's definitely not drawn to scale because we don't know what x is. All right, and then this last area is going to be the height, which is five, times the width, which is two. So that's just going to be equal to 10. So what we just saw is that the area of the whole thing is equal to the sum of these areas. And the sum of those areas is 15x squared plus negative five x, or we could just write that as minus five x, and then we have plus 10. | Multiply monomials by polynomials Area model Algebra 1 Khan Academy.mp3 |
All right, and then this last area is going to be the height, which is five, times the width, which is two. So that's just going to be equal to 10. So what we just saw is that the area of the whole thing is equal to the sum of these areas. And the sum of those areas is 15x squared plus negative five x, or we could just write that as minus five x, and then we have plus 10. Now I know what some of you are thinking. If I know that the height is five and the width is this value, well, couldn't I have just multiplied height times the entire width, three x squared minus x plus two, and then I would have just naturally distributed the five. And essentially, that's exactly what we did here. | Multiply monomials by polynomials Area model Algebra 1 Khan Academy.mp3 |
And the sum of those areas is 15x squared plus negative five x, or we could just write that as minus five x, and then we have plus 10. Now I know what some of you are thinking. If I know that the height is five and the width is this value, well, couldn't I have just multiplied height times the entire width, three x squared minus x plus two, and then I would have just naturally distributed the five. And essentially, that's exactly what we did here. Area models, you might have first seen them in elementary school, really to understand the distributive property. And so if you distribute the five, you get 15x squared minus five x plus 10. Same idea. | Multiply monomials by polynomials Area model Algebra 1 Khan Academy.mp3 |
And essentially, that's exactly what we did here. Area models, you might have first seen them in elementary school, really to understand the distributive property. And so if you distribute the five, you get 15x squared minus five x plus 10. Same idea. Let's do another example. So here we're given another rectangle. It has a height of two x. | Multiply monomials by polynomials Area model Algebra 1 Khan Academy.mp3 |
Same idea. Let's do another example. So here we're given another rectangle. It has a height of two x. We see that right there. A width of three x plus four. We see that over there. | Multiply monomials by polynomials Area model Algebra 1 Khan Academy.mp3 |
It has a height of two x. We see that right there. A width of three x plus four. We see that over there. Express the area of the entire rectangle. So same drill. See if you can pause this video and work on this on your own. | Multiply monomials by polynomials Area model Algebra 1 Khan Academy.mp3 |
We see that over there. Express the area of the entire rectangle. So same drill. See if you can pause this video and work on this on your own. All right, well, we can see that the height is two x. The width in total is three x plus four. But we can clearly see that the area of the entire thing can be broken up into this blue section and this magenta section. | Multiply monomials by polynomials Area model Algebra 1 Khan Academy.mp3 |
See if you can pause this video and work on this on your own. All right, well, we can see that the height is two x. The width in total is three x plus four. But we can clearly see that the area of the entire thing can be broken up into this blue section and this magenta section. The blue section's area is three x times two x. So three x times two x, which is equal to six x squared. And the magenta section is equal to width times height, its area. | Multiply monomials by polynomials Area model Algebra 1 Khan Academy.mp3 |
But we can clearly see that the area of the entire thing can be broken up into this blue section and this magenta section. The blue section's area is three x times two x. So three x times two x, which is equal to six x squared. And the magenta section is equal to width times height, its area. So that's going to be equal to eight x. And so the area of the whole thing is going to be the sum of these areas, which is going to be six x squared plus eight x. And we're done. | Multiply monomials by polynomials Area model Algebra 1 Khan Academy.mp3 |
And the magenta section is equal to width times height, its area. So that's going to be equal to eight x. And so the area of the whole thing is going to be the sum of these areas, which is going to be six x squared plus eight x. And we're done. And once again, we could have just thought about it as the height is two x. We're gonna multiply it times the width, times three x plus four. And then we just distribute the two x. | Multiply monomials by polynomials Area model Algebra 1 Khan Academy.mp3 |
And we're done. And once again, we could have just thought about it as the height is two x. We're gonna multiply it times the width, times three x plus four. And then we just distribute the two x. So two x times three x is six x squared. Two x times four is eight x. Same idea. | Multiply monomials by polynomials Area model Algebra 1 Khan Academy.mp3 |
This first equation, x minus four y is equal to negative 18. And the second equation, negative x plus three y is equal to 11. Now what we're gonna do is find an x and y pair that satisfies both of these equations. That's what solving the system actually means. As you might already have seen, there's a bunch of x and y pairs that satisfy this first equation. In fact, if you were to graph them, they would form a line. And there's a bunch of other x and y pairs that satisfy this other equation, the second equation. | Solving system with elimination Algebra Khan Academy.mp3 |
That's what solving the system actually means. As you might already have seen, there's a bunch of x and y pairs that satisfy this first equation. In fact, if you were to graph them, they would form a line. And there's a bunch of other x and y pairs that satisfy this other equation, the second equation. And if you were to graph them, it would form a line. And so if you find the x and y pair that satisfy both, that would be the intersection of the lines. So let's do that. | Solving system with elimination Algebra Khan Academy.mp3 |
And there's a bunch of other x and y pairs that satisfy this other equation, the second equation. And if you were to graph them, it would form a line. And so if you find the x and y pair that satisfy both, that would be the intersection of the lines. So let's do that. So actually, I'm just gonna rewrite the first equation over here. So I'm gonna write x minus four y is equal to negative 18. So we've already seen in algebra that as long as we do the same thing to both sides of the equation, we can maintain our equality. | Solving system with elimination Algebra Khan Academy.mp3 |
So let's do that. So actually, I'm just gonna rewrite the first equation over here. So I'm gonna write x minus four y is equal to negative 18. So we've already seen in algebra that as long as we do the same thing to both sides of the equation, we can maintain our equality. So what if we were to add, and our goal here is to eliminate one of the variables. So we have one equation with one unknown. So what if we were to add this negative x plus three y to the left-hand side here? | Solving system with elimination Algebra Khan Academy.mp3 |
So we've already seen in algebra that as long as we do the same thing to both sides of the equation, we can maintain our equality. So what if we were to add, and our goal here is to eliminate one of the variables. So we have one equation with one unknown. So what if we were to add this negative x plus three y to the left-hand side here? So negative x plus three y. Well, that looks pretty good because an x and a negative x are going to cancel out. And we are going to be left with negative four y plus three y. | Solving system with elimination Algebra Khan Academy.mp3 |
So what if we were to add this negative x plus three y to the left-hand side here? So negative x plus three y. Well, that looks pretty good because an x and a negative x are going to cancel out. And we are going to be left with negative four y plus three y. Well, that's just going to be negative y. So by adding the left-hand side of this bottom equation to the left-hand side of the top equation, we were able to cancel out the x's. We had x and we had a negative x. | Solving system with elimination Algebra Khan Academy.mp3 |