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Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | So those are the factors of six. And then 15, you could have said, well, one times 15, or three times five. And so you say the greatest common factor, well, three is the largest number that I've listed here that is common to all three of these factors. So once again, the greatest common factor of 12, six, and 15 is three. So when we're looking at the greatest common monomial factor, the coefficient is going to be three. And then we look at these powers of x. We have x to the fourth, we have x to the, let me do this in a different color. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | So once again, the greatest common factor of 12, six, and 15 is three. So when we're looking at the greatest common monomial factor, the coefficient is going to be three. And then we look at these powers of x. We have x to the fourth, we have x to the, let me do this in a different color. We have x to the fourth, x to the third, and x squared. Well, what's the largest power of x that's divisible into all of those? Well, it's going to be x squared. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | We have x to the fourth, we have x to the, let me do this in a different color. We have x to the fourth, x to the third, and x squared. Well, what's the largest power of x that's divisible into all of those? Well, it's going to be x squared. X squared is divisible into x to the fourth and x to the third, and of course, x squared itself. So the greatest common monomial factor is three x squared. This length right over here, this is three x squared. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | Well, it's going to be x squared. X squared is divisible into x to the fourth and x to the third, and of course, x squared itself. So the greatest common monomial factor is three x squared. This length right over here, this is three x squared. So if this is three x squared, we can then figure out what the width is. So what's, if we were to divide 12 x to the fourth by three x squared, what do we get? Well, 12 divided by three is four, and x to the fourth divided by x squared is x squared. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | This length right over here, this is three x squared. So if this is three x squared, we can then figure out what the width is. So what's, if we were to divide 12 x to the fourth by three x squared, what do we get? Well, 12 divided by three is four, and x to the fourth divided by x squared is x squared. Notice, three x squared times four x squared is 12 x to the fourth. And then we move over to this purple section. If we take six x to the third divided by three x squared, six divided by three is two, and then x to the third divided by x squared is just going to be x. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | Well, 12 divided by three is four, and x to the fourth divided by x squared is x squared. Notice, three x squared times four x squared is 12 x to the fourth. And then we move over to this purple section. If we take six x to the third divided by three x squared, six divided by three is two, and then x to the third divided by x squared is just going to be x. And then last but not least, we have 15 divided by three is going to be five. X squared divided by x squared is just one, so it's just going to be five. So the width is going to be four x squared plus two x plus five. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | If we take six x to the third divided by three x squared, six divided by three is two, and then x to the third divided by x squared is just going to be x. And then last but not least, we have 15 divided by three is going to be five. X squared divided by x squared is just one, so it's just going to be five. So the width is going to be four x squared plus two x plus five. So once again, the length, we figured that out, it was the greatest common monomial factor of these terms. It's three x squared, and the width is four x squared plus two x plus five. And one way to think about it is we just factored, we just factored this expression over here. |
Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3 | So the width is going to be four x squared plus two x plus five. So once again, the length, we figured that out, it was the greatest common monomial factor of these terms. It's three x squared, and the width is four x squared plus two x plus five. And one way to think about it is we just factored, we just factored this expression over here. We could write, we could write that, excuse me, I wanna see the original thing. We could write that three x squared times four x squared plus two x plus five, which is the entire width, well that's going to be equal to the area. That's going to be equal to our original expression, 12x to the fourth power plus six x to the third plus 15x squared. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | Maybe we can establish similarity between some of the triangles. There's actually three different triangles that I can see here. This triangle, this triangle, and this larger triangle. If we can establish some similarity here, maybe we can use ratios between sides somehow to figure out what BC is. When you look at it, you have a right angle right over here. In triangle BDC, you have one right angle. In triangle ABC, you have another right angle. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | If we can establish some similarity here, maybe we can use ratios between sides somehow to figure out what BC is. When you look at it, you have a right angle right over here. In triangle BDC, you have one right angle. In triangle ABC, you have another right angle. If we can show that they have another angle or another corresponding set of angles that are congruent to each other, then we can show that they're similar. Actually, both of those triangles, both BDC and ABC, both share this angle right over here. If they share that angle, then they definitely share two angles. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | In triangle ABC, you have another right angle. If we can show that they have another angle or another corresponding set of angles that are congruent to each other, then we can show that they're similar. Actually, both of those triangles, both BDC and ABC, both share this angle right over here. If they share that angle, then they definitely share two angles. They both share that angle right over there. Let me do that in a different color just to make it different than those right angles. They both share that angle there. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | If they share that angle, then they definitely share two angles. They both share that angle right over there. Let me do that in a different color just to make it different than those right angles. They both share that angle there. We know that two triangles that have at least two of their angles, that have at least two congruent angles, they are going to be similar triangles. We know that triangle, I'll write this, triangle ABC, we went from the unlabeled angle to the yellow right angle to the orange angle. Let me write it this way. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | They both share that angle there. We know that two triangles that have at least two of their angles, that have at least two congruent angles, they are going to be similar triangles. We know that triangle, I'll write this, triangle ABC, we went from the unlabeled angle to the yellow right angle to the orange angle. Let me write it this way. We went from the unlabeled angle right over here to the orange angle, or sorry, to the yellow angle, I'm having trouble with colors, to the orange angle, ABC. We want to do this very carefully here because the same points or the same vertices might not play the same role in both triangles. We want to make sure we're getting the similarity right. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | Let me write it this way. We went from the unlabeled angle right over here to the orange angle, or sorry, to the yellow angle, I'm having trouble with colors, to the orange angle, ABC. We want to do this very carefully here because the same points or the same vertices might not play the same role in both triangles. We want to make sure we're getting the similarity right. White vertex to the 90 degree angle vertex to the orange vertex, that is going to be similar to triangle. Which is the one that is neither right angle, so we're looking at the smaller triangle right over here, which is the one that is neither a right angle or the orange angle? It's going to be vertex B. Vertex B had the right angle when you think about the larger triangle, but we haven't thought about just that little angle right over there. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | We want to make sure we're getting the similarity right. White vertex to the 90 degree angle vertex to the orange vertex, that is going to be similar to triangle. Which is the one that is neither right angle, so we're looking at the smaller triangle right over here, which is the one that is neither a right angle or the orange angle? It's going to be vertex B. Vertex B had the right angle when you think about the larger triangle, but we haven't thought about just that little angle right over there. We're going to start at vertex B, then we're going to go to the right angle. The right angle is vertex D, and then we go to vertex C, which is in orange. We have shown that they are similar. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | It's going to be vertex B. Vertex B had the right angle when you think about the larger triangle, but we haven't thought about just that little angle right over there. We're going to start at vertex B, then we're going to go to the right angle. The right angle is vertex D, and then we go to vertex C, which is in orange. We have shown that they are similar. Now that we know that they are similar, we can attempt to take ratios between the sides. Let's think about it. We know what the length of AC is. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | We have shown that they are similar. Now that we know that they are similar, we can attempt to take ratios between the sides. Let's think about it. We know what the length of AC is. AC is going to be equal to 8, 6 plus 2. We know that AC, what's the corresponding side on this triangle right over here? You can literally look at the letters. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | We know what the length of AC is. AC is going to be equal to 8, 6 plus 2. We know that AC, what's the corresponding side on this triangle right over here? You can literally look at the letters. A and C is going to correspond to BC. The first and the third, first and the third. AC is going to correspond to BC. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | You can literally look at the letters. A and C is going to correspond to BC. The first and the third, first and the third. AC is going to correspond to BC. This is interesting because we're already involving BC. What is going to correspond to, and then if we look at BC on the larger triangle, so if we look at BC on the larger triangle, BC is going to correspond to what on the smaller triangle? It's going to correspond to DC. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | AC is going to correspond to BC. This is interesting because we're already involving BC. What is going to correspond to, and then if we look at BC on the larger triangle, so if we look at BC on the larger triangle, BC is going to correspond to what on the smaller triangle? It's going to correspond to DC. It's good because we know what AC is, and we know what DC is, and so we can solve for BC. I want to take one more step to show you what we just did here because BC is playing two different roles. On this first statement right over here, we're thinking of BC. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | It's going to correspond to DC. It's good because we know what AC is, and we know what DC is, and so we can solve for BC. I want to take one more step to show you what we just did here because BC is playing two different roles. On this first statement right over here, we're thinking of BC. BC on our smaller triangle corresponds to AC on our larger triangle. Then in the second statement, BC on our larger triangle corresponds to DC on our smaller triangle. In both of these cases, these are our larger triangles, and then this is from the smaller triangle right over here, corresponding sides. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | On this first statement right over here, we're thinking of BC. BC on our smaller triangle corresponds to AC on our larger triangle. Then in the second statement, BC on our larger triangle corresponds to DC on our smaller triangle. In both of these cases, these are our larger triangles, and then this is from the smaller triangle right over here, corresponding sides. This is a cool problem because BC plays two different roles in both triangles. Now we have enough information to solve for BC. We know that AC is equal to 9. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | In both of these cases, these are our larger triangles, and then this is from the smaller triangle right over here, corresponding sides. This is a cool problem because BC plays two different roles in both triangles. Now we have enough information to solve for BC. We know that AC is equal to 9. We know that AC is equal to 8. 6 plus 2 is 8. We know that DC is equal to 2. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | We know that AC is equal to 9. We know that AC is equal to 8. 6 plus 2 is 8. We know that DC is equal to 2. That's given. Now we can cross multiply. 8 times 2 is 16, is equal to BC times BC, is equal to BC squared. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | We know that DC is equal to 2. That's given. Now we can cross multiply. 8 times 2 is 16, is equal to BC times BC, is equal to BC squared. BC is going to be equal to the principal root of 16, which is 4. BC is equal to 4. BC is equal to 4, and we're done. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | 8 times 2 is 16, is equal to BC times BC, is equal to BC squared. BC is going to be equal to the principal root of 16, which is 4. BC is equal to 4. BC is equal to 4, and we're done. The hardest part about this problem is just realizing that BC plays two different roles and just keeping your head straight on those two different roles. Just to make it clear, let me actually draw these two triangles separately. If I drew ABC separately, it would look like this. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | BC is equal to 4, and we're done. The hardest part about this problem is just realizing that BC plays two different roles and just keeping your head straight on those two different roles. Just to make it clear, let me actually draw these two triangles separately. If I drew ABC separately, it would look like this. It would look like this. This is my triangle ABC. Then this is a right angle. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | If I drew ABC separately, it would look like this. It would look like this. This is my triangle ABC. Then this is a right angle. This is our orange angle. We know that the length of this side right over here is 8, and we know that the length of this side when we figure it out through this problem is 4. Then if we wanted to draw BDC, we would draw it like this. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | Then this is a right angle. This is our orange angle. We know that the length of this side right over here is 8, and we know that the length of this side when we figure it out through this problem is 4. Then if we wanted to draw BDC, we would draw it like this. BDC looks like this. This is BDC. That's a little bit easier to visualize because this is our right angle. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | Then if we wanted to draw BDC, we would draw it like this. BDC looks like this. This is BDC. That's a little bit easier to visualize because this is our right angle. This is our orange angle. This is 4, and this right over here is 2. I did it this way to show you that you kind of have to flip this triangle over and rotate it just to have kind of a similar orientation. |
Similarity example where same side plays different roles Similarity Geometry Khan Academy.mp3 | That's a little bit easier to visualize because this is our right angle. This is our orange angle. This is 4, and this right over here is 2. I did it this way to show you that you kind of have to flip this triangle over and rotate it just to have kind of a similar orientation. Then it might make it look a little bit clearer. If you found this part confusing, I encourage you to try to flip and rotate BDC in such a way that it seems to look a lot like ABC. Then this ratio should hopefully make a lot more sense. |
How to evaluate expressions with two variables 6th grade Khan Academy.mp3 | And I want to evaluate this expression when a is equal to 7 and b is equal to 2. And I encourage you to pause this and try this on your own. Well, wherever we see the a, we would just replace it with the 7. And wherever we see the b, we'd replace it with the 2. So when a equals 7 and b equals 2, this expression will be 7 plus 2, which, of course, is equal to 9. So this expression would be equal to 9 in this circumstance. Let's do a slightly more complicated one. |
How to evaluate expressions with two variables 6th grade Khan Academy.mp3 | And wherever we see the b, we'd replace it with the 2. So when a equals 7 and b equals 2, this expression will be 7 plus 2, which, of course, is equal to 9. So this expression would be equal to 9 in this circumstance. Let's do a slightly more complicated one. Let's say we have the expression x times y minus y plus x. Actually, let's make it plus 3x, or another way of saying it, plus 3 times x. So let's evaluate this when x is equal to 3 and y is equal to 2. |
How to evaluate expressions with two variables 6th grade Khan Academy.mp3 | Let's do a slightly more complicated one. Let's say we have the expression x times y minus y plus x. Actually, let's make it plus 3x, or another way of saying it, plus 3 times x. So let's evaluate this when x is equal to 3 and y is equal to 2. And once again, I encourage you to pause this video and try this on your own. Well, everywhere we see an x, let's replace it with a 3. Every place we see a y, let's replace it with a 2. |
How to evaluate expressions with two variables 6th grade Khan Academy.mp3 | So let's evaluate this when x is equal to 3 and y is equal to 2. And once again, I encourage you to pause this video and try this on your own. Well, everywhere we see an x, let's replace it with a 3. Every place we see a y, let's replace it with a 2. So this is going to be equal to 3 times y. And y is 2 in this case. 3 times 2 minus 2 plus this 3 times x. |
How to evaluate expressions with two variables 6th grade Khan Academy.mp3 | Every place we see a y, let's replace it with a 2. So this is going to be equal to 3 times y. And y is 2 in this case. 3 times 2 minus 2 plus this 3 times x. But x is also now equal to 3. So what is this going to be equal to? Well, this is going to be equal to 3 times 2 is 6. |
How to evaluate expressions with two variables 6th grade Khan Academy.mp3 | 3 times 2 minus 2 plus this 3 times x. But x is also now equal to 3. So what is this going to be equal to? Well, this is going to be equal to 3 times 2 is 6. This 3 times 3 is 9. So it simplifies to 6 minus 2, which is 4, plus 9, which is equal to 13. So in this case, it is equal to 13. |
Constructing and solving a two-step inequality Linear inequalities Algebra I Khan Academy.mp3 | However, both farms made up for some of this shortfall by purchasing equal quantities of apples from farms in neighboring states. What can you say about the number of apples available at each farm? Does one farm have more than the other, or do they have the same amount? How do I know? Let's define some variables here. Let's let m be equal to number of apples at Maple Farms. And who's the other guy? |
Constructing and solving a two-step inequality Linear inequalities Algebra I Khan Academy.mp3 | How do I know? Let's define some variables here. Let's let m be equal to number of apples at Maple Farms. And who's the other guy? River Orchards. So let's r be equal to the number of apples at River Orchards. So this first sentence, they say, let me do this in a different color. |
Constructing and solving a two-step inequality Linear inequalities Algebra I Khan Academy.mp3 | And who's the other guy? River Orchards. So let's r be equal to the number of apples at River Orchards. So this first sentence, they say, let me do this in a different color. They say, for the past few years, Old Maple Farms has grown about one thousand more apples than their chief rival in the region, River Orchards. So we could say, m is approximately river plus one thousand. Or, since we don't know the exact amount, it says it's about a thousand more. |
Constructing and solving a two-step inequality Linear inequalities Algebra I Khan Academy.mp3 | So this first sentence, they say, let me do this in a different color. They say, for the past few years, Old Maple Farms has grown about one thousand more apples than their chief rival in the region, River Orchards. So we could say, m is approximately river plus one thousand. Or, since we don't know the exact amount, it says it's about a thousand more. So we don't know it's exactly a thousand more. We can just say that in a normal year, Old Maple Farms, which we denote by m, has a larger amount of apples than River Orchard. So in a normal year, m is greater than r. It has about a thousand more apples than Old Maple Farms. |
Constructing and solving a two-step inequality Linear inequalities Algebra I Khan Academy.mp3 | Or, since we don't know the exact amount, it says it's about a thousand more. So we don't know it's exactly a thousand more. We can just say that in a normal year, Old Maple Farms, which we denote by m, has a larger amount of apples than River Orchard. So in a normal year, m is greater than r. It has about a thousand more apples than Old Maple Farms. Now, they say, due to cold weather this year, so let's talk about this year now, the harvest at both farms were down about a third. So this isn't a normal year. Let's talk about what's going to happen this year. |
Constructing and solving a two-step inequality Linear inequalities Algebra I Khan Academy.mp3 | So in a normal year, m is greater than r. It has about a thousand more apples than Old Maple Farms. Now, they say, due to cold weather this year, so let's talk about this year now, the harvest at both farms were down about a third. So this isn't a normal year. Let's talk about what's going to happen this year. In this year, each of these characters are going to be down by a third. Now, if I go down by a third, that's the same thing as being two thirds of what I was before. Let me do an example. |
Constructing and solving a two-step inequality Linear inequalities Algebra I Khan Academy.mp3 | Let's talk about what's going to happen this year. In this year, each of these characters are going to be down by a third. Now, if I go down by a third, that's the same thing as being two thirds of what I was before. Let me do an example. If I'm at x and I take away one third x, I'm left with two thirds x. So going down by a third is the same thing as multiplying the quantity by two thirds. So if we multiply each of these quantities by two thirds, we can still hold this inequality, we're doing the same thing to both sides of this inequality, and we're multiplying by a positive number. |
Constructing and solving a two-step inequality Linear inequalities Algebra I Khan Academy.mp3 | Let me do an example. If I'm at x and I take away one third x, I'm left with two thirds x. So going down by a third is the same thing as multiplying the quantity by two thirds. So if we multiply each of these quantities by two thirds, we can still hold this inequality, we're doing the same thing to both sides of this inequality, and we're multiplying by a positive number. If we were multiplying by a negative number, we would have to swap the inequality. So we can multiply both sides of this by two thirds. So two thirds of m is still going to be greater than two thirds of r. And you could even draw that in a number line if you like. |
Constructing and solving a two-step inequality Linear inequalities Algebra I Khan Academy.mp3 | So if we multiply each of these quantities by two thirds, we can still hold this inequality, we're doing the same thing to both sides of this inequality, and we're multiplying by a positive number. If we were multiplying by a negative number, we would have to swap the inequality. So we can multiply both sides of this by two thirds. So two thirds of m is still going to be greater than two thirds of r. And you could even draw that in a number line if you like. Let's do this in a number line. This all might be a little intuitive for you, and if it is, I apologize, but if it's not, it never hurts. So that's zero on our number line. |
Constructing and solving a two-step inequality Linear inequalities Algebra I Khan Academy.mp3 | So two thirds of m is still going to be greater than two thirds of r. And you could even draw that in a number line if you like. Let's do this in a number line. This all might be a little intuitive for you, and if it is, I apologize, but if it's not, it never hurts. So that's zero on our number line. So in a normal year, m has a thousand more than r. So in a normal year, m might be over here, and maybe r is over here. Let's say r is over there. Now, if we take two thirds of m, that's going to stick us someplace around, oh, I don't know, two thirds is right about there. |
Constructing and solving a two-step inequality Linear inequalities Algebra I Khan Academy.mp3 | So that's zero on our number line. So in a normal year, m has a thousand more than r. So in a normal year, m might be over here, and maybe r is over here. Let's say r is over there. Now, if we take two thirds of m, that's going to stick us someplace around, oh, I don't know, two thirds is right about there. So that is m, this is, let me write this, this is two thirds m. And what's two thirds of r going to be? Well, if you take two thirds of this, you get to right about there. That is two thirds r. So you can see two thirds r is still less than two thirds m. Two thirds r is still less than two thirds m, or two thirds m is greater than two thirds r. Now, they say both farms made up for some of the shortfall by purchasing equal quantities of apples from farms in neighboring states. |
Constructing and solving a two-step inequality Linear inequalities Algebra I Khan Academy.mp3 | Now, if we take two thirds of m, that's going to stick us someplace around, oh, I don't know, two thirds is right about there. So that is m, this is, let me write this, this is two thirds m. And what's two thirds of r going to be? Well, if you take two thirds of this, you get to right about there. That is two thirds r. So you can see two thirds r is still less than two thirds m. Two thirds r is still less than two thirds m, or two thirds m is greater than two thirds r. Now, they say both farms made up for some of the shortfall by purchasing equal quantities of apples from farms in neighboring states. So let's let a be equal to the quantity of apples both purchased. So they're telling us that they both purchased the same amount. So we could add a to both sides of this equation, it will not change the inequality. |
Constructing and solving a two-step inequality Linear inequalities Algebra I Khan Academy.mp3 | That is two thirds r. So you can see two thirds r is still less than two thirds m. Two thirds r is still less than two thirds m, or two thirds m is greater than two thirds r. Now, they say both farms made up for some of the shortfall by purchasing equal quantities of apples from farms in neighboring states. So let's let a be equal to the quantity of apples both purchased. So they're telling us that they both purchased the same amount. So we could add a to both sides of this equation, it will not change the inequality. As long as you add or subtract the same value to both sides, it will not change the inequality. So if you add a to both sides, you have a plus two thirds m is greater than two thirds r plus a. This is the amount that old maple farms has after purchasing the apples, and this is the amount that, what's it called, river orchards has. |
Constructing and solving a two-step inequality Linear inequalities Algebra I Khan Academy.mp3 | So we could add a to both sides of this equation, it will not change the inequality. As long as you add or subtract the same value to both sides, it will not change the inequality. So if you add a to both sides, you have a plus two thirds m is greater than two thirds r plus a. This is the amount that old maple farms has after purchasing the apples, and this is the amount that, what's it called, river orchards has. So after everything is said and done, old maple farms still has more apples. And you can see that here. Maple farms, normal year, this year they only had two thirds of the production, but then they purchased a apple. |
Constructing and solving a two-step inequality Linear inequalities Algebra I Khan Academy.mp3 | This is the amount that old maple farms has after purchasing the apples, and this is the amount that, what's it called, river orchards has. So after everything is said and done, old maple farms still has more apples. And you can see that here. Maple farms, normal year, this year they only had two thirds of the production, but then they purchased a apple. So let's say a is about, let's say that a is, I don't know, that many apples. So they got back to their normal amount. So let's say they got back to their normal amount. |
Constructing and solving a two-step inequality Linear inequalities Algebra I Khan Academy.mp3 | Maple farms, normal year, this year they only had two thirds of the production, but then they purchased a apple. So let's say a is about, let's say that a is, I don't know, that many apples. So they got back to their normal amount. So let's say they got back to their normal amount. So that's how many apples they purchased, so he got back to m. Now if r, if river orchards also purchased a apples, that same distance, a, if you go along here, gets you to right about over there. So once again, this is, let me do it a little bit different because I don't want it overlapping. So let me do it like this. |
Constructing and solving a two-step inequality Linear inequalities Algebra I Khan Academy.mp3 | So let's say they got back to their normal amount. So that's how many apples they purchased, so he got back to m. Now if r, if river orchards also purchased a apples, that same distance, a, if you go along here, gets you to right about over there. So once again, this is, let me do it a little bit different because I don't want it overlapping. So let me do it like this. So let's say this guy, m, I keep forgetting their names, old maple farms purchases a apples. Gets them that far, so that's a apples. But river orchards also purchases a apples. |
Constructing and solving a two-step inequality Linear inequalities Algebra I Khan Academy.mp3 | So let me do it like this. So let's say this guy, m, I keep forgetting their names, old maple farms purchases a apples. Gets them that far, so that's a apples. But river orchards also purchases a apples. So let's add that same amount. I'm just going to copy and paste it so it's the exact same amount. Copy and paste. |
Constructing and solving a two-step inequality Linear inequalities Algebra I Khan Academy.mp3 | But river orchards also purchases a apples. So let's add that same amount. I'm just going to copy and paste it so it's the exact same amount. Copy and paste. So old river orchards also purchases a. So it also purchases that same amount. So when all is said and done, river orchards is going to have this many apples in the year that they had less production but they went and purchased it. |
Constructing and solving a two-step inequality Linear inequalities Algebra I Khan Academy.mp3 | Copy and paste. So old river orchards also purchases a. So it also purchases that same amount. So when all is said and done, river orchards is going to have this many apples in the year that they had less production but they went and purchased it. So this right here is, this value right here is 2 thirds r plus a. That's what river orchards has and then old maple farms has this value right here which is 2 thirds m plus a. Everything said and done, old maple farms still has more apples. |
How to solve one-step equations Linear equations Algebra I Khan Academy.mp3 | All it's saying is something plus 7 is equal to 10, and you might be able to figure it out in your head, but if you want to do it a little bit more systematically, you're like, well, all I want on the left-hand side is an x. Well, if all I want on the left-hand side is an x, I'd want to get rid of the 7. I want to subtract 7 from the left-hand side, but if I want to maintain an equality here, whatever I do to the left-hand side, I also have to do to the right-hand side, going back to our scales. That's so that we can keep our scale balanced. So we can say that the left is still equal to the right, and so what we're going to be left with is x, and then the 7s cancel out, is equal to 10 minus 7 is equal to 3. So the unknown is 3, and you can verify it. 3 plus 7 is indeed equal to 10. |
How to solve one-step equations Linear equations Algebra I Khan Academy.mp3 | That's so that we can keep our scale balanced. So we can say that the left is still equal to the right, and so what we're going to be left with is x, and then the 7s cancel out, is equal to 10 minus 7 is equal to 3. So the unknown is 3, and you can verify it. 3 plus 7 is indeed equal to 10. Let's try one more. Let's say we have a minus 5 is equal to negative 2. So this is a little bit more interesting since we have all of these negative numbers here, but we can use the exact same logic. |
How to solve one-step equations Linear equations Algebra I Khan Academy.mp3 | 3 plus 7 is indeed equal to 10. Let's try one more. Let's say we have a minus 5 is equal to negative 2. So this is a little bit more interesting since we have all of these negative numbers here, but we can use the exact same logic. We just want an a over here on the left-hand side, so we have to get rid of this negative 5 somehow. Well, the best way of getting rid of a negative 5 is to add 5 to it, so I'll do that. So I will add 5 to the left-hand side, but if I want the left-hand side to stay equal to the right-hand side, whatever I do to the left, I have to do to the right, so I'm going to have to add 5 on the right-hand side as well. |
How to solve one-step equations Linear equations Algebra I Khan Academy.mp3 | So this is a little bit more interesting since we have all of these negative numbers here, but we can use the exact same logic. We just want an a over here on the left-hand side, so we have to get rid of this negative 5 somehow. Well, the best way of getting rid of a negative 5 is to add 5 to it, so I'll do that. So I will add 5 to the left-hand side, but if I want the left-hand side to stay equal to the right-hand side, whatever I do to the left, I have to do to the right, so I'm going to have to add 5 on the right-hand side as well. So on the left-hand side, I'm left with a, and then the negative 5 and the positive 5 cancel out, and on the right-hand side, and they're going to stay equal because I did the same thing to both sides, we have negative 2 plus 5, which is equal to 3. So a is equal to 3. Once again, you can verify it. |
Linear equation using segment Geometry 8th grade Khan Academy.mp3 | And they tell us that jk is equal to 7x plus 9. So this distance right over here is equal to 7x plus 9. Then they tell us that jl is equal to 114. So jl is the entire length of the segment. So this entire thing is equal to 114. And then they tell us that kl is equal to 9x plus 9. So this right over here is equal to 9x plus 9. |
Linear equation using segment Geometry 8th grade Khan Academy.mp3 | So jl is the entire length of the segment. So this entire thing is equal to 114. And then they tell us that kl is equal to 9x plus 9. So this right over here is equal to 9x plus 9. And they say find kl. So they essentially want us to figure out, what does 9x plus 9 equal? And to figure that out, we have to figure out what x equals. |
Linear equation using segment Geometry 8th grade Khan Academy.mp3 | So this right over here is equal to 9x plus 9. And they say find kl. So they essentially want us to figure out, what does 9x plus 9 equal? And to figure that out, we have to figure out what x equals. And lucky for us, they've given us all the information we need. They tell us that the entire segment is 114 long. And they don't give us any units. |
Linear equation using segment Geometry 8th grade Khan Academy.mp3 | And to figure that out, we have to figure out what x equals. And lucky for us, they've given us all the information we need. They tell us that the entire segment is 114 long. And they don't give us any units. It's just 114. It has the length 114. And we know that this segment, jk plus kl, added together is going to be equal to the length of the entire thing. |
Linear equation using segment Geometry 8th grade Khan Academy.mp3 | And they don't give us any units. It's just 114. It has the length 114. And we know that this segment, jk plus kl, added together is going to be equal to the length of the entire thing. So we could say that 7x plus 9. Actually, let me write it this way. We could write that jk, so the length of segment jk, plus the length of segment kl is going to be equal to 114. |
Linear equation using segment Geometry 8th grade Khan Academy.mp3 | And we know that this segment, jk plus kl, added together is going to be equal to the length of the entire thing. So we could say that 7x plus 9. Actually, let me write it this way. We could write that jk, so the length of segment jk, plus the length of segment kl is going to be equal to 114. 114. And we know that the length of segment jk is 7x plus 9. We know that the length of segment kl is 9x plus 9. |
Linear equation using segment Geometry 8th grade Khan Academy.mp3 | We could write that jk, so the length of segment jk, plus the length of segment kl is going to be equal to 114. 114. And we know that the length of segment jk is 7x plus 9. We know that the length of segment kl is 9x plus 9. So plus 9x plus 9. And this is going to be equal to 114. And now we just have to break out a little bit of our algebraic skills. |
Linear equation using segment Geometry 8th grade Khan Academy.mp3 | We know that the length of segment kl is 9x plus 9. So plus 9x plus 9. And this is going to be equal to 114. And now we just have to break out a little bit of our algebraic skills. So the first thing we might want to do is, let's see, we have two terms that have x's in them. If I have 7x's and I were to add that to another 9x's, that means I'm going to have 16x's. And then if I have, so this is just plain old 9, and I add it to another 9, that's going to give me 18. |
Linear equation using segment Geometry 8th grade Khan Academy.mp3 | And now we just have to break out a little bit of our algebraic skills. So the first thing we might want to do is, let's see, we have two terms that have x's in them. If I have 7x's and I were to add that to another 9x's, that means I'm going to have 16x's. And then if I have, so this is just plain old 9, and I add it to another 9, that's going to give me 18. And that's going to be equal to 114. Now I just subtract 18 from both sides. Let me do that explicitly. |
Linear equation using segment Geometry 8th grade Khan Academy.mp3 | And then if I have, so this is just plain old 9, and I add it to another 9, that's going to give me 18. And that's going to be equal to 114. Now I just subtract 18 from both sides. Let me do that explicitly. So I subtract 18 from both sides. On the left-hand side, I have 16x. And on the right-hand side, I'm going to have 114 minus 18. |
Linear equation using segment Geometry 8th grade Khan Academy.mp3 | Let me do that explicitly. So I subtract 18 from both sides. On the left-hand side, I have 16x. And on the right-hand side, I'm going to have 114 minus 18. Well, if I were to subtract 14, that would get me to 100. And I'm going to subtract 4 more than that. So that's going to get me to 96. |
Linear equation using segment Geometry 8th grade Khan Academy.mp3 | And on the right-hand side, I'm going to have 114 minus 18. Well, if I were to subtract 14, that would get me to 100. And I'm going to subtract 4 more than that. So that's going to get me to 96. And now we just divide both sides by 16. And let's see, this looks like 96 divided by 16. We could do it explicitly, but it looks like it's going to be 6. |
Linear equation using segment Geometry 8th grade Khan Academy.mp3 | So that's going to get me to 96. And now we just divide both sides by 16. And let's see, this looks like 96 divided by 16. We could do it explicitly, but it looks like it's going to be 6. 6 times 10 is 60. 6 times 6 is 36. 60 plus 36 is 96. |
Linear equation using segment Geometry 8th grade Khan Academy.mp3 | We could do it explicitly, but it looks like it's going to be 6. 6 times 10 is 60. 6 times 6 is 36. 60 plus 36 is 96. So we get x is equal to 6. Now we're not done yet. We're not looking just for x. |
Linear equation using segment Geometry 8th grade Khan Academy.mp3 | 60 plus 36 is 96. So we get x is equal to 6. Now we're not done yet. We're not looking just for x. We're looking for the length of KL. KL is 9x plus 9. Let me write that down. |
Linear equation using segment Geometry 8th grade Khan Academy.mp3 | We're not looking just for x. We're looking for the length of KL. KL is 9x plus 9. Let me write that down. So KL is equal to 9x plus 9. They told us that right over there is equal to 9x plus 9. We just figured out that x is equal to 6. |
Linear equation using segment Geometry 8th grade Khan Academy.mp3 | Let me write that down. So KL is equal to 9x plus 9. They told us that right over there is equal to 9x plus 9. We just figured out that x is equal to 6. So this is equal to 9 times 6 plus 9. And this is equal to 9 times 6 is 54 plus 9 is equal to 63. So KL is equal to 63. |
Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3 | So we're going to have to distribute this negative 4x squared over every term in the expression. So first we can start with negative 4x squared times 3x squared. So we can write that, we're going to have negative 4x squared times 3x squared. And to that we're going to add negative 4x squared times 25x. And to that we're going to add negative 4x squared times negative 7. So let's just simplify this a little bit. Now we can obviously swap the order. |
Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3 | And to that we're going to add negative 4x squared times 25x. And to that we're going to add negative 4x squared times negative 7. So let's just simplify this a little bit. Now we can obviously swap the order. We're just multiplying negative 4 times x squared times 3 times x squared. And actually I'll do out every step. Eventually you can do some of this in your head. |
Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3 | Now we can obviously swap the order. We're just multiplying negative 4 times x squared times 3 times x squared. And actually I'll do out every step. Eventually you can do some of this in your head. This is the exact same thing as negative 4 times 3 times x squared times x squared. And what is that equal to? Well negative 4 times 3 is negative 12. |
Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3 | Eventually you can do some of this in your head. This is the exact same thing as negative 4 times 3 times x squared times x squared. And what is that equal to? Well negative 4 times 3 is negative 12. And x squared times x squared, same base, we're taking the product. That's going to be x to the fourth. So this right here is negative 12x to the fourth. |
Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3 | Well negative 4 times 3 is negative 12. And x squared times x squared, same base, we're taking the product. That's going to be x to the fourth. So this right here is negative 12x to the fourth. Now let's think about this term over here. This is the same thing as, of course we have this plus out here. And this part right here is the exact same thing as 25 times negative 4 times x squared times x. |
Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3 | So this right here is negative 12x to the fourth. Now let's think about this term over here. This is the same thing as, of course we have this plus out here. And this part right here is the exact same thing as 25 times negative 4 times x squared times x. So let's just multiply the numbers out here. These were the coefficients. 25 times negative 4 is negative 100. |
Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3 | And this part right here is the exact same thing as 25 times negative 4 times x squared times x. So let's just multiply the numbers out here. These were the coefficients. 25 times negative 4 is negative 100. So it will be plus negative 100 or we can just say it's minus 100. And then we have x squared times x or x squared times x to the first power. Same base, we can add the exponents. |
Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3 | 25 times negative 4 is negative 100. So it will be plus negative 100 or we can just say it's minus 100. And then we have x squared times x or x squared times x to the first power. Same base, we can add the exponents. 2 plus 1 is 3. So this is negative 100x to the third power. And then let's look at this last term over here. |
Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3 | Same base, we can add the exponents. 2 plus 1 is 3. So this is negative 100x to the third power. And then let's look at this last term over here. We have negative 4x squared. So this is going to be plus. That's this plus right over here. |
Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3 | And then let's look at this last term over here. We have negative 4x squared. So this is going to be plus. That's this plus right over here. We have negative 4. We can multiply that times negative 7. And then multiply that times x squared. |
Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3 | That's this plus right over here. We have negative 4. We can multiply that times negative 7. And then multiply that times x squared. I'm just changing the order in which we multiply it. So negative 4 times negative 7 is positive 28. And then I'm going to multiply that times the x squared. |
Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3 | And then multiply that times x squared. I'm just changing the order in which we multiply it. So negative 4 times negative 7 is positive 28. And then I'm going to multiply that times the x squared. There's no simplification to do. No like terms. These are different powers of x. |
Dividing positive and negative numbers Pre-Algebra Khan Academy.mp3 | And what you'll see is that it's actually a very similar methodology, that if both are positive, you will get a positive answer. If one is negative, but not both, you're going to get a negative answer. And if both are negative, they will cancel out and you will get a positive answer. But let's apply it, and I encourage you to pause this video and try these out yourself and then see if you get the same answer that I'm going to get. So 8 divided by negative 2. So if I just had 8 divided by 2, that would be a positive 4. But since exactly one of these two numbers are negative, this one right over here, the answer is going to be negative. |
Dividing positive and negative numbers Pre-Algebra Khan Academy.mp3 | But let's apply it, and I encourage you to pause this video and try these out yourself and then see if you get the same answer that I'm going to get. So 8 divided by negative 2. So if I just had 8 divided by 2, that would be a positive 4. But since exactly one of these two numbers are negative, this one right over here, the answer is going to be negative. So 8 divided by negative 2 is negative 4. Now negative 16 divided by positive 4, we have to be very careful here. If I just had positive 16 divided by positive 4, that would just be 4. |
Dividing positive and negative numbers Pre-Algebra Khan Academy.mp3 | But since exactly one of these two numbers are negative, this one right over here, the answer is going to be negative. So 8 divided by negative 2 is negative 4. Now negative 16 divided by positive 4, we have to be very careful here. If I just had positive 16 divided by positive 4, that would just be 4. But because one of these two numbers is negative, and exactly one of these two numbers is negative, then I'm going to get a negative answer. Now I have negative 30 divided by negative 5. If I just had 30 divided by 5, that would be positive 6. |
Dividing positive and negative numbers Pre-Algebra Khan Academy.mp3 | If I just had positive 16 divided by positive 4, that would just be 4. But because one of these two numbers is negative, and exactly one of these two numbers is negative, then I'm going to get a negative answer. Now I have negative 30 divided by negative 5. If I just had 30 divided by 5, that would be positive 6. And because I have a negative divided by a negative, the negatives cancel out, and so my answer will still be positive 6. And I could even write a positive out here. I don't have to. |
Dividing positive and negative numbers Pre-Algebra Khan Academy.mp3 | If I just had 30 divided by 5, that would be positive 6. And because I have a negative divided by a negative, the negatives cancel out, and so my answer will still be positive 6. And I could even write a positive out here. I don't have to. But this is a positive 6. A negative divided by a negative, just like a negative times a negative, you're going to get a positive answer. 18 divided by 2. |
Dividing positive and negative numbers Pre-Algebra Khan Academy.mp3 | I don't have to. But this is a positive 6. A negative divided by a negative, just like a negative times a negative, you're going to get a positive answer. 18 divided by 2. And this is a little bit of a trick question. This is what you knew how to do before we even talked about negative numbers. This is a positive divided by a positive, which is going to be a positive. |
Dividing positive and negative numbers Pre-Algebra Khan Academy.mp3 | 18 divided by 2. And this is a little bit of a trick question. This is what you knew how to do before we even talked about negative numbers. This is a positive divided by a positive, which is going to be a positive. So that is going to be equal to positive 9. Now we start doing some interesting things. Here's this kind of a compound problem. |
Dividing positive and negative numbers Pre-Algebra Khan Academy.mp3 | This is a positive divided by a positive, which is going to be a positive. So that is going to be equal to positive 9. Now we start doing some interesting things. Here's this kind of a compound problem. You have some multiplication and some division going on. And so first, right over here, the way this is written, we're going to want to multiply the numerator out. And if you're not familiar with this little dot symbol, it's just another way of writing multiplication. |
Dividing positive and negative numbers Pre-Algebra Khan Academy.mp3 | Here's this kind of a compound problem. You have some multiplication and some division going on. And so first, right over here, the way this is written, we're going to want to multiply the numerator out. And if you're not familiar with this little dot symbol, it's just another way of writing multiplication. I could have written this little x thing over here. And what you're going to see is in algebra, the dot becomes much more common because the x gets used for other things. The times symbol, people don't want to confuse it with the letter x, which gets used a lot in algebra. |
Dividing positive and negative numbers Pre-Algebra Khan Academy.mp3 | And if you're not familiar with this little dot symbol, it's just another way of writing multiplication. I could have written this little x thing over here. And what you're going to see is in algebra, the dot becomes much more common because the x gets used for other things. The times symbol, people don't want to confuse it with the letter x, which gets used a lot in algebra. And so that's why they use the dot very often. So this just says negative 7 times 3 in the numerator. And then we're going to take that product and divide it by negative 1. |