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Setting up proportions to solve word problems 7th grade Khan Academy.mp3 | I have three word problems here. What I want to do in this video is not solve the word problems, but just set up the equation that we could solve to get the answer to the word problems. And essentially, we're going to be setting up proportions in either case. So in this first problem, we have nine markers cost $11.50. And then they ask us, how much would seven markers cost? Now let's just set x to be equal to our answer. So x is equal to the cost of seven markers. |
Setting up proportions to solve word problems 7th grade Khan Academy.mp3 | So in this first problem, we have nine markers cost $11.50. And then they ask us, how much would seven markers cost? Now let's just set x to be equal to our answer. So x is equal to the cost of seven markers. So the way to solve a problem like this is to set up two ratios and then set them equal to each other. So you could say that the ratio of nine markers to the cost of nine markers, so the ratio of the number of markers, so 9 to the cost of the nine markers to $11.50, this should be equal to the ratio of our new number of markers, 7, equal to the new number of markers, so that's 7, to whatever the new cost, or whatever the cost of seven markers are, to x. So let me do x in green. |
Setting up proportions to solve word problems 7th grade Khan Academy.mp3 | So x is equal to the cost of seven markers. So the way to solve a problem like this is to set up two ratios and then set them equal to each other. So you could say that the ratio of nine markers to the cost of nine markers, so the ratio of the number of markers, so 9 to the cost of the nine markers to $11.50, this should be equal to the ratio of our new number of markers, 7, equal to the new number of markers, so that's 7, to whatever the new cost, or whatever the cost of seven markers are, to x. So let me do x in green. 2 to x. So this is a completely valid proportion here. The ratio of nine markers to the cost of nine markers is equal to seven markers to the cost of seven markers. |
Setting up proportions to solve word problems 7th grade Khan Academy.mp3 | So let me do x in green. 2 to x. So this is a completely valid proportion here. The ratio of nine markers to the cost of nine markers is equal to seven markers to the cost of seven markers. So then you could solve this to figure out how much those seven markers would cost. And you could flip both sides of this and it would still be a completely valid ratio. You could have $11.50 to 9. |
Setting up proportions to solve word problems 7th grade Khan Academy.mp3 | The ratio of nine markers to the cost of nine markers is equal to seven markers to the cost of seven markers. So then you could solve this to figure out how much those seven markers would cost. And you could flip both sides of this and it would still be a completely valid ratio. You could have $11.50 to 9. So the ratio between the cost of the markers to the number of markers you're buying is equal to, so $11.50 to 9 is equal to the cost of seven markers is equal to the ratio of the cost of seven markers to the number of markers, which is obviously 7. So all I did is flip both sides of this equation right here to get this one over here. You could also think about the ratios in other ways. |
Setting up proportions to solve word problems 7th grade Khan Academy.mp3 | You could have $11.50 to 9. So the ratio between the cost of the markers to the number of markers you're buying is equal to, so $11.50 to 9 is equal to the cost of seven markers is equal to the ratio of the cost of seven markers to the number of markers, which is obviously 7. So all I did is flip both sides of this equation right here to get this one over here. You could also think about the ratios in other ways. You could say that the ratio of nine markers to seven markers is going to be the same as the ratio of their cost, is going to be equal to the ratio of the cost of nine markers to the cost of seven markers. And then obviously, you could flip both of these sides. So you could say that the ratio of seven markers, same magenta color, the ratio of seven markers to nine markers is the same thing as the ratio of the cost of seven markers to the cost of nine markers. |
Setting up proportions to solve word problems 7th grade Khan Academy.mp3 | You could also think about the ratios in other ways. You could say that the ratio of nine markers to seven markers is going to be the same as the ratio of their cost, is going to be equal to the ratio of the cost of nine markers to the cost of seven markers. And then obviously, you could flip both of these sides. So you could say that the ratio of seven markers, same magenta color, the ratio of seven markers to nine markers is the same thing as the ratio of the cost of seven markers to the cost of nine markers. So that is $11.50. So all of these would be valid proportions, valid equations that describe what's going on here. And then you would just have to essentially solve for x. |
Setting up proportions to solve word problems 7th grade Khan Academy.mp3 | So you could say that the ratio of seven markers, same magenta color, the ratio of seven markers to nine markers is the same thing as the ratio of the cost of seven markers to the cost of nine markers. So that is $11.50. So all of these would be valid proportions, valid equations that describe what's going on here. And then you would just have to essentially solve for x. So let's do this one right over here. Seven apples cost $5. How many apples can I buy with $8? |
Setting up proportions to solve word problems 7th grade Khan Academy.mp3 | And then you would just have to essentially solve for x. So let's do this one right over here. Seven apples cost $5. How many apples can I buy with $8? So once again, we can say, so we're going to assume that what they're asking is how many apples, let's call that x. x is what we want to solve for. So seven apples cost $5. So we have the ratio between the number of apples, 7, and the cost of the apples, 5, is going to be equal to the ratio between another number of apples, which is now x, and the cost of that other number of apples. |
Setting up proportions to solve word problems 7th grade Khan Academy.mp3 | How many apples can I buy with $8? So once again, we can say, so we're going to assume that what they're asking is how many apples, let's call that x. x is what we want to solve for. So seven apples cost $5. So we have the ratio between the number of apples, 7, and the cost of the apples, 5, is going to be equal to the ratio between another number of apples, which is now x, and the cost of that other number of apples. And it's going to be $8. And so notice here, in this first situation, what was unknown was the cost. So we kind of had the number of apples to cost, number of apples to cost. |
Setting up proportions to solve word problems 7th grade Khan Academy.mp3 | So we have the ratio between the number of apples, 7, and the cost of the apples, 5, is going to be equal to the ratio between another number of apples, which is now x, and the cost of that other number of apples. And it's going to be $8. And so notice here, in this first situation, what was unknown was the cost. So we kind of had the number of apples to cost, number of apples to cost. Now in this example, the unknown is the number of apples. So number of apples to cost, number of apples to cost. Now we could do all of the different scenarios like this. |
Setting up proportions to solve word problems 7th grade Khan Academy.mp3 | So we kind of had the number of apples to cost, number of apples to cost. Now in this example, the unknown is the number of apples. So number of apples to cost, number of apples to cost. Now we could do all of the different scenarios like this. You could also say the ratio between 7 apples and x apples is going to be the same as the ratio between the cost of 7 apples and the cost of 8 apples. And obviously, you could flip both sides of either of these equations to get two more equations. And any of these would be valid equations. |
Setting up proportions to solve word problems 7th grade Khan Academy.mp3 | Now we could do all of the different scenarios like this. You could also say the ratio between 7 apples and x apples is going to be the same as the ratio between the cost of 7 apples and the cost of 8 apples. And obviously, you could flip both sides of either of these equations to get two more equations. And any of these would be valid equations. Now let's do this last one. So we have a cake recipe for five people. I'll use new colors here. |
Setting up proportions to solve word problems 7th grade Khan Academy.mp3 | And any of these would be valid equations. Now let's do this last one. So we have a cake recipe for five people. I'll use new colors here. A cake recipe for five people requires two eggs. So we want to know how many eggs. So this we'll call x. |
Setting up proportions to solve word problems 7th grade Khan Academy.mp3 | I'll use new colors here. A cake recipe for five people requires two eggs. So we want to know how many eggs. So this we'll call x. And you don't always have to call it x. You could call it e for x. Well, e isn't a good idea, because e represents another number once you get to higher mathematics. |
Setting up proportions to solve word problems 7th grade Khan Academy.mp3 | So this we'll call x. And you don't always have to call it x. You could call it e for x. Well, e isn't a good idea, because e represents another number once you get to higher mathematics. But you could call them y or z or any variable, a, b, or c, anything. How many eggs do we need for a 15-person cake? So you could say the ratio of people to eggs is constant. |
Setting up proportions to solve word problems 7th grade Khan Academy.mp3 | Well, e isn't a good idea, because e represents another number once you get to higher mathematics. But you could call them y or z or any variable, a, b, or c, anything. How many eggs do we need for a 15-person cake? So you could say the ratio of people to eggs is constant. So if we have five people for two eggs, then for 15 people, we are going to need x eggs. This ratio is going to be constant. 5 over 2 is equal to 15 over x. |
Setting up proportions to solve word problems 7th grade Khan Academy.mp3 | So you could say the ratio of people to eggs is constant. So if we have five people for two eggs, then for 15 people, we are going to need x eggs. This ratio is going to be constant. 5 over 2 is equal to 15 over x. Or you could flip both sides of this. Or you could say the ratio between 5 and 15 is going to be equal to the ratio between the number of eggs for five people. Let me do that in that blue color. |
Setting up proportions to solve word problems 7th grade Khan Academy.mp3 | 5 over 2 is equal to 15 over x. Or you could flip both sides of this. Or you could say the ratio between 5 and 15 is going to be equal to the ratio between the number of eggs for five people. Let me do that in that blue color. The ratio between the number of eggs for five people and the number of eggs for 15 people. And obviously, you could flip both sides of this equation. So all of these, we've essentially set up the proportions that describe each of these problems. |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | So as you can tell, light is very fast. 3 times 10 to the 8th meters per second. If it takes 5 times 10 to the 2nd power seconds for light to travel from the sun to the earth. So just let's think about that a little bit. So 5 times 10 to the 2nd, that's 500. 500 seconds, you have 60 seconds in a minute. So 8 minutes would be 480 seconds. |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | So just let's think about that a little bit. So 5 times 10 to the 2nd, that's 500. 500 seconds, you have 60 seconds in a minute. So 8 minutes would be 480 seconds. So 500 seconds would be about 8 minutes 20 seconds. So it takes 8 minutes 20 seconds for light to travel from the sun to the earth. What is the distance in meters between the sun and the earth? |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | So 8 minutes would be 480 seconds. So 500 seconds would be about 8 minutes 20 seconds. So it takes 8 minutes 20 seconds for light to travel from the sun to the earth. What is the distance in meters between the sun and the earth? So they're giving us a rate. They're giving us a speed. They're giving us a time. |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | What is the distance in meters between the sun and the earth? So they're giving us a rate. They're giving us a speed. They're giving us a time. And they want to find a distance. So this goes straight back to the standard. Distance is equal to rate times time. |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | They're giving us a time. And they want to find a distance. So this goes straight back to the standard. Distance is equal to rate times time. So they give us the rate. The rate is 3 times 10 to the 8th meters per second. So it's 3 times 10 to the 8th meters per second. |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | Distance is equal to rate times time. So they give us the rate. The rate is 3 times 10 to the 8th meters per second. So it's 3 times 10 to the 8th meters per second. That right there is the rate. They give us the time. The time is 5 times 10 to the 2nd seconds. |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | So it's 3 times 10 to the 8th meters per second. That right there is the rate. They give us the time. The time is 5 times 10 to the 2nd seconds. So times 5 times 10 to the 2nd seconds. I'll just use that with an s. So how many meters? So what is the distance? |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | The time is 5 times 10 to the 2nd seconds. So times 5 times 10 to the 2nd seconds. I'll just use that with an s. So how many meters? So what is the distance? And so we can just reassociate these, or actually move these around from the commutative and the associative properties of multiplication. So this right here is the same thing. And actually, you can multiply the units. |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | So what is the distance? And so we can just reassociate these, or actually move these around from the commutative and the associative properties of multiplication. So this right here is the same thing. And actually, you can multiply the units. That's called dimensional analysis. And when you multiply the units, you kind of treat them like variables. You should get the right dimensions for distance. |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | And actually, you can multiply the units. That's called dimensional analysis. And when you multiply the units, you kind of treat them like variables. You should get the right dimensions for distance. So let's just rearrange these numbers. This is equal to 3 times 5. I'm just recommuting and reassociating these numbers. |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | You should get the right dimensions for distance. So let's just rearrange these numbers. This is equal to 3 times 5. I'm just recommuting and reassociating these numbers. So 3 times in this product, because we're just multiplying everything. 3 times 5 times 10 to the 8th times 10 to the 2nd. And then we're going to have meters per second. |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | I'm just recommuting and reassociating these numbers. So 3 times in this product, because we're just multiplying everything. 3 times 5 times 10 to the 8th times 10 to the 2nd. And then we're going to have meters per second. So we could write meters per second times seconds. And if you treated these like variables, this seconds would cancel out with that seconds right there. And you'd just be left with the unit meters, which is good. |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | And then we're going to have meters per second. So we could write meters per second times seconds. And if you treated these like variables, this seconds would cancel out with that seconds right there. And you'd just be left with the unit meters, which is good. Because we want a distance in meters, in just meters. So how does this simplify? This gives us 3 times 5 is 15. |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | And you'd just be left with the unit meters, which is good. Because we want a distance in meters, in just meters. So how does this simplify? This gives us 3 times 5 is 15. 15 times 10 to the 8th times 10 squared. We have the same base. We're taking the product, so we can add the exponents. |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | This gives us 3 times 5 is 15. 15 times 10 to the 8th times 10 squared. We have the same base. We're taking the product, so we can add the exponents. So this is going to be 10 to the 8 plus 2 power, or 10 to the 10th power. Now you might be tempted to say that we're done, that we have this in scientific notation. But remember, in scientific notation, this number here has to be greater than or equal to 1 and less than 10. |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | We're taking the product, so we can add the exponents. So this is going to be 10 to the 8 plus 2 power, or 10 to the 10th power. Now you might be tempted to say that we're done, that we have this in scientific notation. But remember, in scientific notation, this number here has to be greater than or equal to 1 and less than 10. This clearly is not less than 10. So how do we rewrite this? We can write 15 as 1.5. |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | But remember, in scientific notation, this number here has to be greater than or equal to 1 and less than 10. This clearly is not less than 10. So how do we rewrite this? We can write 15 as 1.5. This clearly is greater than 1 and less than 10. And to get from 1.5 to 15, you have to multiply by 10. One way to think about it is 15 is 15.0. |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | We can write 15 as 1.5. This clearly is greater than 1 and less than 10. And to get from 1.5 to 15, you have to multiply by 10. One way to think about it is 15 is 15.0. And so you have a decimal here. If we're going to move the decimal 1 to the left to get us 1.5, we're essentially dividing by 5. If we're moving the decimal 1 to the left to make it 1.5, that's essentially dividing by 10. |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | One way to think about it is 15 is 15.0. And so you have a decimal here. If we're going to move the decimal 1 to the left to get us 1.5, we're essentially dividing by 5. If we're moving the decimal 1 to the left to make it 1.5, that's essentially dividing by 10. Moving the decimal to the left means you're dividing by 10. If we don't want to change the value of the number, we need to divide by 10 and then multiply by 10. So this and that are the same number. |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | If we're moving the decimal 1 to the left to make it 1.5, that's essentially dividing by 10. Moving the decimal to the left means you're dividing by 10. If we don't want to change the value of the number, we need to divide by 10 and then multiply by 10. So this and that are the same number. Now 15 is 1.5 times 10. And then we have to multiply that times 10 to the 10th. Not x to the 10th. |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | So this and that are the same number. Now 15 is 1.5 times 10. And then we have to multiply that times 10 to the 10th. Not x to the 10th. Times 10 to the 10th power. This right over here. 10 is really just 10 to the first power. |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | Not x to the 10th. Times 10 to the 10th power. This right over here. 10 is really just 10 to the first power. So we can just add the exponents. Same base, taking the product. So this is equal to 1.5 times 10 to the 1 plus 10 power. |
Speed of light and distance from sun (scientific notation word problem) Pre-Algebra Khan Academy.mp3 | 10 is really just 10 to the first power. So we can just add the exponents. Same base, taking the product. So this is equal to 1.5 times 10 to the 1 plus 10 power. Or 10 to the 11th power. And we are done. This is a huge, huge distance. |
Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3 | So we're told to factor the polynomial below by its greatest common monomial factor. So what does that mean? So we have these two terms, and I want to figure out their greatest common monomial factor, and then I want to express this with that greatest common monomial factor factored out. So how can we tackle it? Well, one way to start is I can look at just the constant terms. I can look at, or not the constants, the coefficients, I should say. So I have the eight and the 12, and I can say, well, what is just the greatest common factor of eight and 12? |
Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3 | So how can we tackle it? Well, one way to start is I can look at just the constant terms. I can look at, or not the constants, the coefficients, I should say. So I have the eight and the 12, and I can say, well, what is just the greatest common factor of eight and 12? The GCF of eight and 12. And there are a lot of common factors of eight and 12. They're both divisible by one, they're both divisible by two, they're both divisible by four, but the greatest of their common factors is going to be four. |
Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3 | So I have the eight and the 12, and I can say, well, what is just the greatest common factor of eight and 12? The GCF of eight and 12. And there are a lot of common factors of eight and 12. They're both divisible by one, they're both divisible by two, they're both divisible by four, but the greatest of their common factors is going to be four. So that is equal to four. So let me just leave that there. And then we can think about what is, well, let me actually write it right over here. |
Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3 | They're both divisible by one, they're both divisible by two, they're both divisible by four, but the greatest of their common factors is going to be four. So that is equal to four. So let me just leave that there. And then we can think about what is, well, let me actually write it right over here. I'll put a four here. And now we can move on to the powers of x. We have an x squared and we have an x. |
Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3 | And then we can think about what is, well, let me actually write it right over here. I'll put a four here. And now we can move on to the powers of x. We have an x squared and we have an x. And we can say, what is the largest power of x that is divisible into both x squared and x? Well, that's just going to be x. X squared is clearly divisible by x, and x is clearly divisible by x, but x isn't going to be, isn't going to have a larger power of x as a factor. So this is the greatest, you could view this as the greatest common monomial factor of x squared and x. |
Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3 | We have an x squared and we have an x. And we can say, what is the largest power of x that is divisible into both x squared and x? Well, that's just going to be x. X squared is clearly divisible by x, and x is clearly divisible by x, but x isn't going to be, isn't going to have a larger power of x as a factor. So this is the greatest, you could view this as the greatest common monomial factor of x squared and x. Now we do the same thing for the y's. So we have a y and a y squared. If we think in the same terms, the largest power of y that's divisible into both of these is going to be just y to the first power, or y. |
Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3 | So this is the greatest, you could view this as the greatest common monomial factor of x squared and x. Now we do the same thing for the y's. So we have a y and a y squared. If we think in the same terms, the largest power of y that's divisible into both of these is going to be just y to the first power, or y. And so four xy is the greatest common monomial factor. And to see that, we can express each of these terms as a product of four xy and something else. So this first term right over here, so let me pick a color. |
Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3 | If we think in the same terms, the largest power of y that's divisible into both of these is going to be just y to the first power, or y. And so four xy is the greatest common monomial factor. And to see that, we can express each of these terms as a product of four xy and something else. So this first term right over here, so let me pick a color. So this term right over here, we could write as four xy, that one's actually, that color's hard to see, let me pick a darker color. We could write this right over here as four xy times what? And I encourage you to pause the video and think about that. |
Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3 | So this first term right over here, so let me pick a color. So this term right over here, we could write as four xy, that one's actually, that color's hard to see, let me pick a darker color. We could write this right over here as four xy times what? And I encourage you to pause the video and think about that. Let's see, four times what is equal, is going to get us to eight? Well, four times two is going to get us to eight. X times what is going to get us to x squared? |
Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3 | And I encourage you to pause the video and think about that. Let's see, four times what is equal, is going to get us to eight? Well, four times two is going to get us to eight. X times what is going to get us to x squared? Well, x times x is going to get us to x squared. And then y times what is going to get us to y? Well, it's just going to be y. |
Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3 | X times what is going to get us to x squared? Well, x times x is going to get us to x squared. And then y times what is going to get us to y? Well, it's just going to be y. So four xy times two x is actually going to give us this first term. So actually, let me just rewrite it a little bit differently. So it's four xy times two x is this first term, and you can verify that. |
Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3 | Well, it's just going to be y. So four xy times two x is actually going to give us this first term. So actually, let me just rewrite it a little bit differently. So it's four xy times two x is this first term, and you can verify that. Four times two is going to be equal to eight. X times x is equal to x squared. And then you just have the y. |
Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3 | So it's four xy times two x is this first term, and you can verify that. Four times two is going to be equal to eight. X times x is equal to x squared. And then you just have the y. Now let's do the same thing with the second term. And I just want to do this to show you that this is their largest common monomial factor. So the second term, and I'll do this in a slightly different color, do it in blue. |
Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3 | And then you just have the y. Now let's do the same thing with the second term. And I just want to do this to show you that this is their largest common monomial factor. So the second term, and I'll do this in a slightly different color, do it in blue. I want to write this as the product of four xy and another monomial. So four times what is 12? Well, four times three is 12. |
Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3 | So the second term, and I'll do this in a slightly different color, do it in blue. I want to write this as the product of four xy and another monomial. So four times what is 12? Well, four times three is 12. X times what is x? Well, it's just going to be one, so we don't have to write up anything here. And then y times what is y squared? |
Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3 | Well, four times three is 12. X times what is x? Well, it's just going to be one, so we don't have to write up anything here. And then y times what is y squared? It's going to be y times y is y squared. And you can verify. If you multiply these two, you're going to get 12xy squared. |
Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3 | And then y times what is y squared? It's going to be y times y is y squared. And you can verify. If you multiply these two, you're going to get 12xy squared. Four times three is 12. You get your x. And then y times y is y squared. |
Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3 | If you multiply these two, you're going to get 12xy squared. Four times three is 12. You get your x. And then y times y is y squared. So so far, I've written this exact same expression, but I've taken each of those terms and I factored them into their greatest common monomial factor and then whatever is left over. And now I can factor the four xy out. I can actually factor it out. |
Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3 | And then y times y is y squared. So so far, I've written this exact same expression, but I've taken each of those terms and I factored them into their greatest common monomial factor and then whatever is left over. And now I can factor the four xy out. I can actually factor it out. So this is going to be equal to, if I factor the four xys out, you could kind of say I undistribute the four xy. I factor it out. This is going to be equal to four xy times 2x plus, when I factor four xy from here, I get the three y left over. |
Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3 | I can actually factor it out. So this is going to be equal to, if I factor the four xys out, you could kind of say I undistribute the four xy. I factor it out. This is going to be equal to four xy times 2x plus, when I factor four xy from here, I get the three y left over. So that's three y. And we're done. And you can verify it. |
Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3 | This is going to be equal to four xy times 2x plus, when I factor four xy from here, I get the three y left over. So that's three y. And we're done. And you can verify it. If you were to go the other way, if you were to distribute this four xy and multiply it times 2x, you'd get 8x squared y. And then when you distribute the four xy onto the three y, you get the 12xy squared. And so we're done. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | We need to divide 0.25 into 1.03075. Now the first thing you want to do when your divisor, the number that you're dividing into the other number, is a decimal, is to multiply it by 10 enough times so that it becomes a whole number, so you can shift the decimal to the right. So every time you multiply something by 10, you're shifting the decimal over to the right once. So in this case, we want to shift it over to the right once and twice. So 0.25 times 10 twice is the same thing as 0.25 times 100, and we'll turn the 0.25 into 25. Now if you do that with the divisor, you also have to do that with the dividend, the number that you're dividing into. So we also have to multiply this by 10 twice. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | So in this case, we want to shift it over to the right once and twice. So 0.25 times 10 twice is the same thing as 0.25 times 100, and we'll turn the 0.25 into 25. Now if you do that with the divisor, you also have to do that with the dividend, the number that you're dividing into. So we also have to multiply this by 10 twice. Or another way of doing it is shift the decimal over to the right twice. So we shift it over once, twice, it will sit right over here. And to see why that makes sense, you just have to realize that this expression right here, this division problem, is the exact same thing as having 0.25 as having 1.03075 divided by 0.25. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | So we also have to multiply this by 10 twice. Or another way of doing it is shift the decimal over to the right twice. So we shift it over once, twice, it will sit right over here. And to see why that makes sense, you just have to realize that this expression right here, this division problem, is the exact same thing as having 0.25 as having 1.03075 divided by 0.25. And so we're multiplying the 0.25 by 10 twice. We're essentially multiplying it by 100. Let me do that in a different color. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | And to see why that makes sense, you just have to realize that this expression right here, this division problem, is the exact same thing as having 0.25 as having 1.03075 divided by 0.25. And so we're multiplying the 0.25 by 10 twice. We're essentially multiplying it by 100. Let me do that in a different color. We're multiplying it by 100 in the denominator. This is the divisor. We're multiplying it by 100. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | Let me do that in a different color. We're multiplying it by 100 in the denominator. This is the divisor. We're multiplying it by 100. So we also have to do the same thing to the numerator if we don't want to change this expression. We don't want to change the number. So we also have to multiply that by 100. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | We're multiplying it by 100. So we also have to do the same thing to the numerator if we don't want to change this expression. We don't want to change the number. So we also have to multiply that by 100. And when you do that, this becomes 25, and this becomes 103.075. Now let me just rewrite this. Sometimes if you're doing this in a workbook or something, you don't have to rewrite it as long as you remember where the decimal is. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | So we also have to multiply that by 100. And when you do that, this becomes 25, and this becomes 103.075. Now let me just rewrite this. Sometimes if you're doing this in a workbook or something, you don't have to rewrite it as long as you remember where the decimal is. But I'm going to rewrite it just so it's a little bit neater. So when we multiplied both the divisor and the dividend by 100, this problem becomes 25 divided into 103.075. These are going to result in the exact same quotient. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | Sometimes if you're doing this in a workbook or something, you don't have to rewrite it as long as you remember where the decimal is. But I'm going to rewrite it just so it's a little bit neater. So when we multiplied both the divisor and the dividend by 100, this problem becomes 25 divided into 103.075. These are going to result in the exact same quotient. They're the exact same fraction if you want to view it that way. We just multiplied both the numerator and the denominator by 100 to shift the decimal over to the right twice. Now that we've done that, we're ready to divide. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | These are going to result in the exact same quotient. They're the exact same fraction if you want to view it that way. We just multiplied both the numerator and the denominator by 100 to shift the decimal over to the right twice. Now that we've done that, we're ready to divide. So the first thing, we have 25 here. And there's always a little bit of an art to dividing something by a multiple digit number. So we'll see how well we can do. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | Now that we've done that, we're ready to divide. So the first thing, we have 25 here. And there's always a little bit of an art to dividing something by a multiple digit number. So we'll see how well we can do. So 25 does not go into 1. 25 does not go into 10. 25 does go into 103. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | So we'll see how well we can do. So 25 does not go into 1. 25 does not go into 10. 25 does go into 103. We know that 4 times 25 is 100. So 25 goes into 103 4 times. 4 times 5 is 20. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | 25 does go into 103. We know that 4 times 25 is 100. So 25 goes into 103 4 times. 4 times 5 is 20. 4 times 2 is 8, plus 2 is 100. We knew that. 4 times 4 quarters is $1. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | 4 times 5 is 20. 4 times 2 is 8, plus 2 is 100. We knew that. 4 times 4 quarters is $1. It's $0.10. And now we subtract. 103 minus 100 is going to be 3. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | 4 times 4 quarters is $1. It's $0.10. And now we subtract. 103 minus 100 is going to be 3. And now we can bring down this 0. So we bring down that 0 there. 25 goes into 30 one time. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | 103 minus 100 is going to be 3. And now we can bring down this 0. So we bring down that 0 there. 25 goes into 30 one time. And if we want, we can immediately put this decimal here. We don't have to wait until the end of the problem. This decimal sits right in that place. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | 25 goes into 30 one time. And if we want, we can immediately put this decimal here. We don't have to wait until the end of the problem. This decimal sits right in that place. So we could always have that decimal sitting right there in our quotient or in our answer or our quotient. So anyway, we were at 25 goes into 30 one time. 1 times 25 is 25. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | This decimal sits right in that place. So we could always have that decimal sitting right there in our quotient or in our answer or our quotient. So anyway, we were at 25 goes into 30 one time. 1 times 25 is 25. And then we can subtract. 30 minus 25, well, that's just 5. We could do all of this borrowing business or regrouping. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | 1 times 25 is 25. And then we can subtract. 30 minus 25, well, that's just 5. We could do all of this borrowing business or regrouping. This can become a 10. This becomes a 2. 10 minus 5 is 5. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | We could do all of this borrowing business or regrouping. This can become a 10. This becomes a 2. 10 minus 5 is 5. 2 minus 2 is nothing. But anyway, 30 minus 25 is 5. Now we can bring down this 7. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | 10 minus 5 is 5. 2 minus 2 is nothing. But anyway, 30 minus 25 is 5. Now we can bring down this 7. 25 goes into 57 two times. 25 times 2 is 50. 25 goes into 57 two times. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | Now we can bring down this 7. 25 goes into 57 two times. 25 times 2 is 50. 25 goes into 57 two times. 2 times 25 is 50. And now we subtract again. 57 minus 50 is 7. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | 25 goes into 57 two times. 2 times 25 is 50. And now we subtract again. 57 minus 50 is 7. And now we're almost done. We can bring down this 5. We bring down that 5 right over there. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | 57 minus 50 is 7. And now we're almost done. We can bring down this 5. We bring down that 5 right over there. 25 goes into 75 three times. 3 times 25 is 75. We say 3 times 5 is 15. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | We bring down that 5 right over there. 25 goes into 75 three times. 3 times 25 is 75. We say 3 times 5 is 15. Regroup the 1. We can ignore that. That was from before. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | We say 3 times 5 is 15. Regroup the 1. We can ignore that. That was from before. 3 times 2 is 6 plus 1 is 7. So you can see that. And then we subtract. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | That was from before. 3 times 2 is 6 plus 1 is 7. So you can see that. And then we subtract. And then we have no remainder. So 25 goes into 103.075 exactly 4.123 times, which makes sense because 25 goes into 100 about 4 times. This is a little bit larger than 100, so it's going to be a little bit more than 4 times. |
Dividing by a multi-digit decimal Decimals Pre-Algebra Khan Academy.mp3 | And then we subtract. And then we have no remainder. So 25 goes into 103.075 exactly 4.123 times, which makes sense because 25 goes into 100 about 4 times. This is a little bit larger than 100, so it's going to be a little bit more than 4 times. And that's going to be the exact same answer as the number of times that 0.25 goes into 1.03075. This will also be 4.123. So this fraction or this expression is the exact same thing as 4.123. |
Simplifying radicals examples.mp3 | So this first exercise, and these are all from Khan Academy, it says simplify the expression by removing all factors that are perfect squares from inside the radicals and combine the terms. If the expression cannot be simplified, enter it as given. All right, let's see what we can do here. So we have negative 40, the negative square root of 40, I should say. Let me write it a little bit bigger so you can see that. So the negative square root of 40 plus the square root of 90. So let's see, what perfect squares are in 40? |
Simplifying radicals examples.mp3 | So we have negative 40, the negative square root of 40, I should say. Let me write it a little bit bigger so you can see that. So the negative square root of 40 plus the square root of 90. So let's see, what perfect squares are in 40? So what immediately jumps out at me is that it's divisible by four, and four is a perfect square. So this is the negative square root of four times 10 plus the square root of, well, what jumps out at me is that this is divisible by nine. Nine is a perfect square, so nine times 10. |
Simplifying radicals examples.mp3 | So let's see, what perfect squares are in 40? So what immediately jumps out at me is that it's divisible by four, and four is a perfect square. So this is the negative square root of four times 10 plus the square root of, well, what jumps out at me is that this is divisible by nine. Nine is a perfect square, so nine times 10. And if we look at the 10s here, 10 does not have any perfect squares in it anymore. If you wanted to do a full factorization of 10, a full prime factorization, it would be two times five. So there's no perfect squares in 10. |
Simplifying radicals examples.mp3 | Nine is a perfect square, so nine times 10. And if we look at the 10s here, 10 does not have any perfect squares in it anymore. If you wanted to do a full factorization of 10, a full prime factorization, it would be two times five. So there's no perfect squares in 10. And so we can work it out from here. This is the same thing as the negative of the square root of four times the square root of 10 plus the square root of nine times the square root of 10. And when I say square root, I'm really saying principal root, the positive square root. |
Simplifying radicals examples.mp3 | So there's no perfect squares in 10. And so we can work it out from here. This is the same thing as the negative of the square root of four times the square root of 10 plus the square root of nine times the square root of 10. And when I say square root, I'm really saying principal root, the positive square root. So it's the negative of the positive square root of four. So that is, so let me do this in another color so it can be clear. So this right here is two. |
Simplifying radicals examples.mp3 | And when I say square root, I'm really saying principal root, the positive square root. So it's the negative of the positive square root of four. So that is, so let me do this in another color so it can be clear. So this right here is two. This right here is three. So it's going to be equal to negative two square roots of 10 plus three square roots of 10. So if I have negative two of something and I add three of that same something to it, that's going to be what? |
Simplifying radicals examples.mp3 | So this right here is two. This right here is three. So it's going to be equal to negative two square roots of 10 plus three square roots of 10. So if I have negative two of something and I add three of that same something to it, that's going to be what? Well, that's going to be one square root of 10. Now, if this last step doesn't make full sense, actually, let me slow it down a little bit. I could rewrite it this way. |
Simplifying radicals examples.mp3 | So if I have negative two of something and I add three of that same something to it, that's going to be what? Well, that's going to be one square root of 10. Now, if this last step doesn't make full sense, actually, let me slow it down a little bit. I could rewrite it this way. I could write it as three square roots of 10 minus two square roots of 10. That might jump out at you a little bit clear. If I have three of something and I were to take away two of that something, in that case, the square roots of 10s, well, I'm gonna be left with just one of that something. |
Simplifying radicals examples.mp3 | I could rewrite it this way. I could write it as three square roots of 10 minus two square roots of 10. That might jump out at you a little bit clear. If I have three of something and I were to take away two of that something, in that case, the square roots of 10s, well, I'm gonna be left with just one of that something. I'm just gonna be left with one square root of 10, which we could just write as the square root of 10. Another way to think about it is we could factor out a square root of 10 here. So you undistribute it, do the distributive property in reverse. |
Simplifying radicals examples.mp3 | If I have three of something and I were to take away two of that something, in that case, the square roots of 10s, well, I'm gonna be left with just one of that something. I'm just gonna be left with one square root of 10, which we could just write as the square root of 10. Another way to think about it is we could factor out a square root of 10 here. So you undistribute it, do the distributive property in reverse. That would be the square root of 10 times three minus two, which is, of course, this is just one. So you're just left with the square root of 10. So all of this simplifies to square root of 10. |