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Example 3 Subtracting polynomials Algebra I Khan Academy.mp3 | Simplify, 16x plus 14 minus the entire expression 3x squared plus x minus 9. So when you subtract an entire expression, this is the exact same thing as having 16x plus 14, and then you're adding the opposite of this whole thing, or you're adding negative 1, you're adding negative 1 times 3x squared plus x minus 9. Or another way to think about it is you can distribute this negative sign along all of those terms. That's essentially what we're about to do here. We're just adding the negative of this entire thing. We're adding the opposite of it. So this first part, I'm not going to change it. |
Example 3 Subtracting polynomials Algebra I Khan Academy.mp3 | That's essentially what we're about to do here. We're just adding the negative of this entire thing. We're adding the opposite of it. So this first part, I'm not going to change it. That's still just 16x plus 14. But now I'm going to distribute the negative sign here. So negative 1 times 3x squared is negative 3x squared. |
Example 3 Subtracting polynomials Algebra I Khan Academy.mp3 | So this first part, I'm not going to change it. That's still just 16x plus 14. But now I'm going to distribute the negative sign here. So negative 1 times 3x squared is negative 3x squared. Negative 1 times positive x is negative x, because that's positive 1x. Negative 1 times negative 9, remember you have to consider this negative right over there. That is part of the term. |
Example 3 Subtracting polynomials Algebra I Khan Academy.mp3 | So negative 1 times 3x squared is negative 3x squared. Negative 1 times positive x is negative x, because that's positive 1x. Negative 1 times negative 9, remember you have to consider this negative right over there. That is part of the term. Negative 1 times negative 9 is positive 9. Negative times a negative is a positive. So then we have positive 9. |
Example 3 Subtracting polynomials Algebra I Khan Academy.mp3 | That is part of the term. Negative 1 times negative 9 is positive 9. Negative times a negative is a positive. So then we have positive 9. And now we just have to combine like terms. So what's our highest degree term here? I like to write it in that order. |
Example 3 Subtracting polynomials Algebra I Khan Academy.mp3 | So then we have positive 9. And now we just have to combine like terms. So what's our highest degree term here? I like to write it in that order. We have only one x squared term, second degree term. We only have one of those, so let me write it over here. Negative 3x squared. |
Example 3 Subtracting polynomials Algebra I Khan Academy.mp3 | I like to write it in that order. We have only one x squared term, second degree term. We only have one of those, so let me write it over here. Negative 3x squared. And then what do we have in terms of first degree terms? Just an x, x to the first power. We have a 16x, and then from that we're going to subtract an x. Subtract 1x. |
Example 3 Subtracting polynomials Algebra I Khan Academy.mp3 | Negative 3x squared. And then what do we have in terms of first degree terms? Just an x, x to the first power. We have a 16x, and then from that we're going to subtract an x. Subtract 1x. So 16x minus 1x is 15x. If you have 16 of something and you subtract one of them away, you're going to have 15 of that something. And then finally, you have 14. |
Example 3 Subtracting polynomials Algebra I Khan Academy.mp3 | We have a 16x, and then from that we're going to subtract an x. Subtract 1x. So 16x minus 1x is 15x. If you have 16 of something and you subtract one of them away, you're going to have 15 of that something. And then finally, you have 14. You can view that as 14 times x to the 0, or just 14. 14 plus 9, they're both constant terms, or they're both being multiplied by x to the 0. 14 plus 9 is 23. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | We're told to study the growth of bacteria. A scientist measures the area in square millimeters occupied by a sample population. The growth of the population can be modeled by f of t is equal to 24 times e to the 0.4 times t power, where t is the number of hours since the experiment began. Here's the graph of f. So I guess f is going to be, the output of this function is going to be the number of square millimeters after t hours. All right, so here we have the graph. We see how as time goes on, the square millimeters of our little bacterial population keeps growing, and it clearly is growing, or it looks like it's growing exponentially. In fact, we know it's growing exponential because it's an exponential function right over here. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | Here's the graph of f. So I guess f is going to be, the output of this function is going to be the number of square millimeters after t hours. All right, so here we have the graph. We see how as time goes on, the square millimeters of our little bacterial population keeps growing, and it clearly is growing, or it looks like it's growing exponentially. In fact, we know it's growing exponential because it's an exponential function right over here. And they say, when does the population first occupy an area of 400 square millimeters? So pause this video and try to figure that out. All right, and this is a screenshot from the Khan Academy exercise. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | In fact, we know it's growing exponential because it's an exponential function right over here. And they say, when does the population first occupy an area of 400 square millimeters? So pause this video and try to figure that out. All right, and this is a screenshot from the Khan Academy exercise. So we wanna say, when does the population first occupy an area of 400 square millimeters? Let's see, 400 square millimeters is right over there. And so it looks like after seven hours that we are going to be 400 square millimeters or larger. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | All right, and this is a screenshot from the Khan Academy exercise. So we wanna say, when does the population first occupy an area of 400 square millimeters? Let's see, 400 square millimeters is right over there. And so it looks like after seven hours that we are going to be 400 square millimeters or larger. So it first hits it after seven hours. So seven hours, just like that. Now let's do the next few examples that build on this. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | And so it looks like after seven hours that we are going to be 400 square millimeters or larger. So it first hits it after seven hours. So seven hours, just like that. Now let's do the next few examples that build on this. So if I go back up to the top, and now we're told the same thing. We're using square millimeters to study the growth. This is the function. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | Now let's do the next few examples that build on this. So if I go back up to the top, and now we're told the same thing. We're using square millimeters to study the growth. This is the function. But then they add this next line. Here is the graph. Here is the graph of F, and the graph of the line Y equals 600. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | This is the function. But then they add this next line. Here is the graph. Here is the graph of F, and the graph of the line Y equals 600. So they added that graph there. And then they say, which statement represents the meaning of the intersection point of the graphs? All right, so let's look at the choices here. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | Here is the graph of F, and the graph of the line Y equals 600. So they added that graph there. And then they say, which statement represents the meaning of the intersection point of the graphs? All right, so let's look at the choices here. So, and it says choose all that apply. So pause this video and see if you can answer that. All right, so choice A says it describes the time when the population occupies 600 square millimeters. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | All right, so let's look at the choices here. So, and it says choose all that apply. So pause this video and see if you can answer that. All right, so choice A says it describes the time when the population occupies 600 square millimeters. So which statement represents the meaning of the intersection point of the graph? So they're talking about, they're talking about this point right over there. So does that describe the time when the population occupies 600 square millimeters? |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | All right, so choice A says it describes the time when the population occupies 600 square millimeters. So which statement represents the meaning of the intersection point of the graph? So they're talking about, they're talking about this point right over there. So does that describe the time when the population occupies 600 square millimeters? So that is the time when the population has indeed reached 600 square millimeters, because that's the line Y is equal to 600 square millimeters. So I like that choice. I will select it. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | So does that describe the time when the population occupies 600 square millimeters? So that is the time when the population has indeed reached 600 square millimeters, because that's the line Y is equal to 600 square millimeters. So I like that choice. I will select it. The next choice. It gives the solution to the equation 24 times E to the 0.4T is equal to 600. Well, if you think about it, this right over here in blue, we've already talked about it, that is Y is equal to 24 times E to the 0.4T power. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | I will select it. The next choice. It gives the solution to the equation 24 times E to the 0.4T is equal to 600. Well, if you think about it, this right over here in blue, we've already talked about it, that is Y is equal to 24 times E to the 0.4T power. This is Y is equal to 600. So the T value at which these two graphs equal, that means that they're both equal to the same Y value. Or another way to think about it, it means that that is equal to that, or that 24 times E to the 0.4T power is indeed equal to 600. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | Well, if you think about it, this right over here in blue, we've already talked about it, that is Y is equal to 24 times E to the 0.4T power. This is Y is equal to 600. So the T value at which these two graphs equal, that means that they're both equal to the same Y value. Or another way to think about it, it means that that is equal to that, or that 24 times E to the 0.4T power is indeed equal to 600. So I like this too. It gives a T value where this is true. So that's the solution to that equation. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | Or another way to think about it, it means that that is equal to that, or that 24 times E to the 0.4T power is indeed equal to 600. So I like this too. It gives a T value where this is true. So that's the solution to that equation. It describes the situation where the area the population occupies is equal to the number of hours. That's definitely not the case, because the area here is 600 square millimeters. The hours looks like it's a little bit after eight. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | So that's the solution to that equation. It describes the situation where the area the population occupies is equal to the number of hours. That's definitely not the case, because the area here is 600 square millimeters. The hours looks like it's a little bit after eight. So they're definitely not equal. It describes the area the population occupies after 600 hours. No, we don't have to look up there. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | The hours looks like it's a little bit after eight. So they're definitely not equal. It describes the area the population occupies after 600 hours. No, we don't have to look up there. This T axis doesn't even go to 600 hours. So we wouldn't select that as well. Now let's keep building and go to the next part of this. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | No, we don't have to look up there. This T axis doesn't even go to 600 hours. So we wouldn't select that as well. Now let's keep building and go to the next part of this. And it says, it says, so once again, we measure the area in square millimeters to figure out the growth of the population. The growth of, oh, so here, we have two populations here. It says the growth of population A can be modeled by F of T is equal to that. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | Now let's keep building and go to the next part of this. And it says, it says, so once again, we measure the area in square millimeters to figure out the growth of the population. The growth of, oh, so here, we have two populations here. It says the growth of population A can be modeled by F of T is equal to that. We've seen that already. But now they are introducing another population. The growth of population B can be modeled by G of T is equal to this, where T is the number of hours since the experiment began. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | It says the growth of population A can be modeled by F of T is equal to that. We've seen that already. But now they are introducing another population. The growth of population B can be modeled by G of T is equal to this, where T is the number of hours since the experiment began. Here are the graphs of F and G. So now we have two populations. They're both growing exponentially, but at different rates. And then it says, when do the populations occupy the same area? |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | The growth of population B can be modeled by G of T is equal to this, where T is the number of hours since the experiment began. Here are the graphs of F and G. So now we have two populations. They're both growing exponentially, but at different rates. And then it says, when do the populations occupy the same area? It says, round your answer to the nearest integer. And you could pause this video and try to think about that if you like. Well, you can see very clearly that it looks like they intersect right around there. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | And then it says, when do the populations occupy the same area? It says, round your answer to the nearest integer. And you could pause this video and try to think about that if you like. Well, you can see very clearly that it looks like they intersect right around there. So that's the point at which they're going to occupy the same area. It looks like it's about 175 square millimeters. But they're not asking about the area. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | Well, you can see very clearly that it looks like they intersect right around there. So that's the point at which they're going to occupy the same area. It looks like it's about 175 square millimeters. But they're not asking about the area. They're saying, when does it happen? And it looks like it happens after about five hours. So round to the nearest integer. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | But they're not asking about the area. They're saying, when does it happen? And it looks like it happens after about five hours. So round to the nearest integer. Let's say five hours. Now let's do the last part. So it's the same setup, but now they are asking us a different question. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | So round to the nearest integer. Let's say five hours. Now let's do the last part. So it's the same setup, but now they are asking us a different question. They are asking us, which statements represent the meaning of the intersection points of the graphs? All right, so choice A says, and then pause the video again and try to answer these on your own. All right, choice A says, it means that the populations both occupied about 180 square millimeters at the same time. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | So it's the same setup, but now they are asking us a different question. They are asking us, which statements represent the meaning of the intersection points of the graphs? All right, so choice A says, and then pause the video again and try to answer these on your own. All right, choice A says, it means that the populations both occupied about 180 square millimeters at the same time. So let's see this. That looks about right. I had estimated 175, but we could call that 180. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | All right, choice A says, it means that the populations both occupied about 180 square millimeters at the same time. So let's see this. That looks about right. I had estimated 175, but we could call that 180. And it looks like that does roughly happen at around the fifth hour. So it looks like they're occupying the same area at around the same time. So I like that choice. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | I had estimated 175, but we could call that 180. And it looks like that does roughly happen at around the fifth hour. So it looks like they're occupying the same area at around the same time. So I like that choice. It means that at the beginning, population A was larger than population B. Well, the point of intersection doesn't tell us what population was larger to begin with. We could try to answer it by looking over here. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | So I like that choice. It means that at the beginning, population A was larger than population B. Well, the point of intersection doesn't tell us what population was larger to begin with. We could try to answer it by looking over here. When time t equals zero, when time t equals zero, population A is the blue curve. It is F. And so it does look like population A was larger than population B at time t equals zero at the beginning. But that's not what the point of intersection tells us. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | We could try to answer it by looking over here. When time t equals zero, when time t equals zero, population A is the blue curve. It is F. And so it does look like population A was larger than population B at time t equals zero at the beginning. But that's not what the point of intersection tells us. So they're not just asking us for true statements. They're saying which statements represent the meaning, the meaning of the intersection point of the graphs. But that doesn't tell us about what the starting situation was. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | But that's not what the point of intersection tells us. So they're not just asking us for true statements. They're saying which statements represent the meaning, the meaning of the intersection point of the graphs. But that doesn't tell us about what the starting situation was. It gives a solution to the equation 24 times e to the 0.4t is equal to nine times e to the 0.6t. Well, we already talked about that in the last example where we only had one curve. And that actually is the case because y is equal to 24 times e to the 0.4t is the curve for population A. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | But that doesn't tell us about what the starting situation was. It gives a solution to the equation 24 times e to the 0.4t is equal to nine times e to the 0.6t. Well, we already talked about that in the last example where we only had one curve. And that actually is the case because y is equal to 24 times e to the 0.4t is the curve for population A. And then y is equal to nine times e to the 0.6t is the curve for population B. And so the point at which these two curves intersect, that's the point at which both this, we're at a t value that gives the same so that this expression will give you the same y value as this expression. Or another way to say it is we're at the t value where this is equal to this. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | And that actually is the case because y is equal to 24 times e to the 0.4t is the curve for population A. And then y is equal to nine times e to the 0.6t is the curve for population B. And so the point at which these two curves intersect, that's the point at which both this, we're at a t value that gives the same so that this expression will give you the same y value as this expression. Or another way to say it is we're at the t value where this is equal to this. So it does indeed give the solution to the equation. And then the last choice is it gives a solution to the equation 24 times e to the 0.4t is equal to zero? No, that would happen, if you wanna know when it's equal to zero, you would look at the curve y equals zero. |
Solving equations by graphing word problems Algebra 2 Khan Academy.mp3 | Or another way to say it is we're at the t value where this is equal to this. So it does indeed give the solution to the equation. And then the last choice is it gives a solution to the equation 24 times e to the 0.4t is equal to zero? No, that would happen, if you wanna know when it's equal to zero, you would look at the curve y equals zero. I'll do that in a different color, which is right over here. And see where it intersects the function f, which is equal to 24 times e to the 0.4t. But that's not what this point of intersection represents. |
Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3 | All right, let's tackle this one in purple first. And you might first notice that on both sides of the equation I have different bases, so it would be nice to have a common base. And when you look at it, you're like, well, 32 is not a power of eight, or at least it's not an integer power of eight, but they are both powers of two. 32 is the same thing as two to the fifth power, two to the fifth power, and eight is the same thing as two to the third power, two to the third power. So I can rewrite our original equation as instead of writing 32, I could write it as two to the fifth, and then that's going to be raised to the x over three power, x over three power, is equal to, instead of writing eight, I could write two to the third power, two to the third power, and I'm raising that to the x minus 12, x minus 12. Now if I raise something to a power and then raise that to a power, I could just multiply these exponents. So I could rewrite the left-hand side as two to the five x over three, five x over three power. |
Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3 | 32 is the same thing as two to the fifth power, two to the fifth power, and eight is the same thing as two to the third power, two to the third power. So I can rewrite our original equation as instead of writing 32, I could write it as two to the fifth, and then that's going to be raised to the x over three power, x over three power, is equal to, instead of writing eight, I could write two to the third power, two to the third power, and I'm raising that to the x minus 12, x minus 12. Now if I raise something to a power and then raise that to a power, I could just multiply these exponents. So I could rewrite the left-hand side as two to the five x over three, five x over three power. I just multiply these exponents, and that's going to be equal to two to the, and now I just multiply the three times x minus 12. So two to the three x minus 36, and now things have simplified nicely. I have two to this power is equal to two to that power, so these two exponents must be equal to each other. |
Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3 | So I could rewrite the left-hand side as two to the five x over three, five x over three power. I just multiply these exponents, and that's going to be equal to two to the, and now I just multiply the three times x minus 12. So two to the three x minus 36, and now things have simplified nicely. I have two to this power is equal to two to that power, so these two exponents must be equal to each other. Five x over three must be equal to three x minus 36. So let's set them equal to each other and solve for x. So five x over three is equal to three x minus 36. |
Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3 | I have two to this power is equal to two to that power, so these two exponents must be equal to each other. Five x over three must be equal to three x minus 36. So let's set them equal to each other and solve for x. So five x over three is equal to three x minus 36. Let's see, we could multiply, we could multiply everything by three. Let's do that. So if we multiply everything times three, you are going to get five x is equal to nine x minus, what is this, nine x minus 108, and now we can subtract nine x from both sides, and so we will get five x minus nine x is going to be negative four x is equal to negative 108. |
Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3 | So five x over three is equal to three x minus 36. Let's see, we could multiply, we could multiply everything by three. Let's do that. So if we multiply everything times three, you are going to get five x is equal to nine x minus, what is this, nine x minus 108, and now we can subtract nine x from both sides, and so we will get five x minus nine x is going to be negative four x is equal to negative 108. We're in the home stretch here. Divide, whoops, sorry about that. We could divide both sides by negative four, negative four, and we are left with x is equal to, what is this going to be, 27. |
Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3 | So if we multiply everything times three, you are going to get five x is equal to nine x minus, what is this, nine x minus 108, and now we can subtract nine x from both sides, and so we will get five x minus nine x is going to be negative four x is equal to negative 108. We're in the home stretch here. Divide, whoops, sorry about that. We could divide both sides by negative four, negative four, and we are left with x is equal to, what is this going to be, 27. X is equal to 27, and we are all done. We're all done, and if you substituted x back in there, you would get 32 to the 27 divided by three, so 32 to the ninth power is the same thing as eight to the 27 minus 12th power, so eight to the 15th, yeah, 27 minus 12, eight to the 15th power. So anyway, that was fun. |
Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3 | We could divide both sides by negative four, negative four, and we are left with x is equal to, what is this going to be, 27. X is equal to 27, and we are all done. We're all done, and if you substituted x back in there, you would get 32 to the 27 divided by three, so 32 to the ninth power is the same thing as eight to the 27 minus 12th power, so eight to the 15th, yeah, 27 minus 12, eight to the 15th power. So anyway, that was fun. Let's do the next one now. So this one looks interesting in other ways. We have rational expressions, we have an exponential up here, exponential down here, and the key realization here is, well, the first thing I'd like to do, let me write this 25 in terms of five. |
Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3 | So anyway, that was fun. Let's do the next one now. So this one looks interesting in other ways. We have rational expressions, we have an exponential up here, exponential down here, and the key realization here is, well, the first thing I'd like to do, let me write this 25 in terms of five. We know that 25 is the same thing as five squared, so we can rewrite this as five to the four x plus three over, instead of 25, I could rewrite that as five squared, and then I'm gonna raise that to the nine minus x, to the nine minus x, and that, of course, is going to be equal to five to the two x plus five. Now, five to the second, and then that to the nine minus x, I can just multiply these exponents, so this is going to be five to the four x plus three over five to the, two times nine is 18, two times negative x is negative two x, and that is going to be equal to, that is going to be equal to five to the two x plus five. And now, let's see, there's multiple ways that we could tackle it. |
Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3 | We have rational expressions, we have an exponential up here, exponential down here, and the key realization here is, well, the first thing I'd like to do, let me write this 25 in terms of five. We know that 25 is the same thing as five squared, so we can rewrite this as five to the four x plus three over, instead of 25, I could rewrite that as five squared, and then I'm gonna raise that to the nine minus x, to the nine minus x, and that, of course, is going to be equal to five to the two x plus five. Now, five to the second, and then that to the nine minus x, I can just multiply these exponents, so this is going to be five to the four x plus three over five to the, two times nine is 18, two times negative x is negative two x, and that is going to be equal to, that is going to be equal to five to the two x plus five. And now, let's see, there's multiple ways that we could tackle it. We could multiply both sides of this equation by five to the 18 minus two x, that's one way to do it, or we could say, hey, look, I have five to some exponent divided by five to some other exponent, so I could just subtract this blue exponent from this yellow one, so the left-hand side will simplify to five to the four x plus three minus, let me do this with a minus and a neutral color, minus 18 minus two x, 18 minus two x, and that, of course, is going to be equal to what we've had on the right-hand side, five to the two x plus five. Now we just have to simplify a little bit. Let's see, this is going to be, in fact, we could just say, look, I'm having trouble with my little pen tool. |
Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3 | And now, let's see, there's multiple ways that we could tackle it. We could multiply both sides of this equation by five to the 18 minus two x, that's one way to do it, or we could say, hey, look, I have five to some exponent divided by five to some other exponent, so I could just subtract this blue exponent from this yellow one, so the left-hand side will simplify to five to the four x plus three minus, let me do this with a minus and a neutral color, minus 18 minus two x, 18 minus two x, and that, of course, is going to be equal to what we've had on the right-hand side, five to the two x plus five. Now we just have to simplify a little bit. Let's see, this is going to be, in fact, we could just say, look, I'm having trouble with my little pen tool. Whoops. All right. So now we could say this exponent needs to be equal to that exponent because we have the same base, and so what we have here on the left-hand side, that I could rewrite as four x plus three minus 18 plus two x. I'm just multiplying the negative times both of these terms, so plus two x is going to be equal to two x plus five, two x plus five. |
Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3 | Let's see, this is going to be, in fact, we could just say, look, I'm having trouble with my little pen tool. Whoops. All right. So now we could say this exponent needs to be equal to that exponent because we have the same base, and so what we have here on the left-hand side, that I could rewrite as four x plus three minus 18 plus two x. I'm just multiplying the negative times both of these terms, so plus two x is going to be equal to two x plus five, two x plus five. So there's a bunch of different things we could do here. One, we could subtract two x from both sides. That'll clean it up a little bit. |
Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3 | So now we could say this exponent needs to be equal to that exponent because we have the same base, and so what we have here on the left-hand side, that I could rewrite as four x plus three minus 18 plus two x. I'm just multiplying the negative times both of these terms, so plus two x is going to be equal to two x plus five, two x plus five. So there's a bunch of different things we could do here. One, we could subtract two x from both sides. That'll clean it up a little bit. Two x from both sides. We could also subtract five from both sides. So let's just do that. |
Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3 | That'll clean it up a little bit. Two x from both sides. We could also subtract five from both sides. So let's just do that. So, well, let me just subtract it. Subtract five from both sides. I'm skipping some steps here, but I figure you're at this point reasonably comfortable with linear equations. |
Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3 | So let's just do that. So, well, let me just subtract it. Subtract five from both sides. I'm skipping some steps here, but I figure you're at this point reasonably comfortable with linear equations. So then, on the left-hand side, we are going to have four x, and then you have three minus 18 minus five. Three minus 18 is negative 15, minus five is negative 20, is going to be equal to zero. And then, because those cancel out, and so add 20 to both sides, you get four x is equal to 20. |
Geometric series introduction Algebra 2 Khan Academy.mp3 | So let's say this is the year, and we're gonna think about how much we have at the beginning of the year, and then this is the dollars in our account. And let's say that the bank is always willing to give us 5% per year, which is pretty good. It's very hard to find a bank account that will actually give you 5% growth per year. So that means if you put $100 in at the end of a year or exactly a year later, it'd be $105. If you put $1,000 in a year later, it'd be 1,050. It'd be 5% larger. And so let's say that we wanna put $1,000 in per year, and I wanna think about, well, what is going to be my balance at the beginning of year one, at the beginning of year two, at the beginning of year three, and then see if we can come up with a general expression for the beginning of year n. So year one, right at the beginning of the year, I put in $1,000 in the account. |
Geometric series introduction Algebra 2 Khan Academy.mp3 | So that means if you put $100 in at the end of a year or exactly a year later, it'd be $105. If you put $1,000 in a year later, it'd be 1,050. It'd be 5% larger. And so let's say that we wanna put $1,000 in per year, and I wanna think about, well, what is going to be my balance at the beginning of year one, at the beginning of year two, at the beginning of year three, and then see if we can come up with a general expression for the beginning of year n. So year one, right at the beginning of the year, I put in $1,000 in the account. That's pretty straightforward. But then what happens in year two? I'm going to deposit $1,000, but then that original $1,000 that I have would have grown. |
Geometric series introduction Algebra 2 Khan Academy.mp3 | And so let's say that we wanna put $1,000 in per year, and I wanna think about, well, what is going to be my balance at the beginning of year one, at the beginning of year two, at the beginning of year three, and then see if we can come up with a general expression for the beginning of year n. So year one, right at the beginning of the year, I put in $1,000 in the account. That's pretty straightforward. But then what happens in year two? I'm going to deposit $1,000, but then that original $1,000 that I have would have grown. So I'm going to deposit $1,000, and then the original $1,000 that I put at the beginning of year one, that has now grown by 5%. Growing by 5% is the same thing as multiplying by 1.05. So this is now going to be plus $1,000 times 1.05. |
Geometric series introduction Algebra 2 Khan Academy.mp3 | I'm going to deposit $1,000, but then that original $1,000 that I have would have grown. So I'm going to deposit $1,000, and then the original $1,000 that I put at the beginning of year one, that has now grown by 5%. Growing by 5% is the same thing as multiplying by 1.05. So this is now going to be plus $1,000 times 1.05. Fairly straightforward. Now what about the beginning of year three? How much would I have in the bank account right when I've made that first, that year three deposit? |
Geometric series introduction Algebra 2 Khan Academy.mp3 | So this is now going to be plus $1,000 times 1.05. Fairly straightforward. Now what about the beginning of year three? How much would I have in the bank account right when I've made that first, that year three deposit? Pause this video, see if you can figure that out. Well, just like at the beginning of year two and the beginning of year one, we're going to make a $1,000 deposit. But now the money from year two has grown by 5%. |
Geometric series introduction Algebra 2 Khan Academy.mp3 | How much would I have in the bank account right when I've made that first, that year three deposit? Pause this video, see if you can figure that out. Well, just like at the beginning of year two and the beginning of year one, we're going to make a $1,000 deposit. But now the money from year two has grown by 5%. So this is now going to be $1,000 times 1.05. And then that money that we originally deposited from year one, that was 1,000 times 1.05 in year two, that's going to grow by another 5%. And so this is going to be plus 1,000 times 1.05 times 1.05, we're growing by another 5%. |
Geometric series introduction Algebra 2 Khan Academy.mp3 | But now the money from year two has grown by 5%. So this is now going to be $1,000 times 1.05. And then that money that we originally deposited from year one, that was 1,000 times 1.05 in year two, that's going to grow by another 5%. And so this is going to be plus 1,000 times 1.05 times 1.05, we're growing by another 5%. Well, we could just rewrite this part right over here as 1.05 squared. So do you see a general pattern that's going to happen here? Well, as you go to year n, in fact, pause the video again and see if you could write a general expression. |
Geometric series introduction Algebra 2 Khan Academy.mp3 | And so this is going to be plus 1,000 times 1.05 times 1.05, we're growing by another 5%. Well, we could just rewrite this part right over here as 1.05 squared. So do you see a general pattern that's going to happen here? Well, as you go to year n, in fact, pause the video again and see if you could write a general expression. You're gonna have to do a little bit of this dot, dot, dot action in order to do it. But see if you could write a general expression for year n. Well, for year n, you're going to make that original $1,000 at the beginning of year n, and then you're going to have 1,000 times 1.05 for that $1,000 that you deposited at the beginning of year n minus one. And then this is just going to keep going, and it's going to go all the way to plus $1,000 to times 1.05 to the power of the number of years you've been compounding. |
Geometric series introduction Algebra 2 Khan Academy.mp3 | Well, as you go to year n, in fact, pause the video again and see if you could write a general expression. You're gonna have to do a little bit of this dot, dot, dot action in order to do it. But see if you could write a general expression for year n. Well, for year n, you're going to make that original $1,000 at the beginning of year n, and then you're going to have 1,000 times 1.05 for that $1,000 that you deposited at the beginning of year n minus one. And then this is just going to keep going, and it's going to go all the way to plus $1,000 to times 1.05 to the power of the number of years you've been compounding. So you could view this $1,000 as the one that you put in year one, and then how many years has it compounded? Well, when you go from one to two, you've compounded one year. When you go from one to three, you've compounded two years. |
Geometric series introduction Algebra 2 Khan Academy.mp3 | And then this is just going to keep going, and it's going to go all the way to plus $1,000 to times 1.05 to the power of the number of years you've been compounding. So you could view this $1,000 as the one that you put in year one, and then how many years has it compounded? Well, when you go from one to two, you've compounded one year. When you go from one to three, you've compounded two years. So when we're talking about the beginning of year n, you go up to the exponent that is one less than that. And so this is going to be to the n minus one power. So what we just did here is we've just constructed each one of these when we're saying, okay, how much money do we have in our bank account at the beginning of year three? |
Geometric series introduction Algebra 2 Khan Academy.mp3 | When you go from one to three, you've compounded two years. So when we're talking about the beginning of year n, you go up to the exponent that is one less than that. And so this is going to be to the n minus one power. So what we just did here is we've just constructed each one of these when we're saying, okay, how much money do we have in our bank account at the beginning of year three? Or how much do we have in our bank account at the beginning of year n? These are geometric series, and I'll write that word down, geometric series. Now, just as a little bit of a review, or it might not be a review, it might be a primer, series are related to sequences, and you can really view series as sums of sequences. |
Geometric series introduction Algebra 2 Khan Academy.mp3 | So what we just did here is we've just constructed each one of these when we're saying, okay, how much money do we have in our bank account at the beginning of year three? Or how much do we have in our bank account at the beginning of year n? These are geometric series, and I'll write that word down, geometric series. Now, just as a little bit of a review, or it might not be a review, it might be a primer, series are related to sequences, and you can really view series as sums of sequences. Sequences, and let me go down a little bit so that you can, so we have a little bit more space. A sequence is an ordered list of numbers. A sequence might be something like, well, let's say we have a geometric sequence. |
Geometric series introduction Algebra 2 Khan Academy.mp3 | Now, just as a little bit of a review, or it might not be a review, it might be a primer, series are related to sequences, and you can really view series as sums of sequences. Sequences, and let me go down a little bit so that you can, so we have a little bit more space. A sequence is an ordered list of numbers. A sequence might be something like, well, let's say we have a geometric sequence. In a geometric sequence, each successive term is the previous term times a fixed number. So let's say we start at two, and every time we multiply by three. So we'll go from two, two times three is six, six times three is 18, 18 times three is 54. |
Geometric series introduction Algebra 2 Khan Academy.mp3 | A sequence might be something like, well, let's say we have a geometric sequence. In a geometric sequence, each successive term is the previous term times a fixed number. So let's say we start at two, and every time we multiply by three. So we'll go from two, two times three is six, six times three is 18, 18 times three is 54. This is a geometric sequence, ordered list of numbers. Now, if we wanna think about the geometric series, or the one that's analogous to this, is that we would sum the terms here. So this would be two plus six plus 18 plus 54. |
Geometric series introduction Algebra 2 Khan Academy.mp3 | So we'll go from two, two times three is six, six times three is 18, 18 times three is 54. This is a geometric sequence, ordered list of numbers. Now, if we wanna think about the geometric series, or the one that's analogous to this, is that we would sum the terms here. So this would be two plus six plus 18 plus 54. Or we could even write it, and this will look similar to what we had just done with our little savings example, is this is two plus two times three plus two times three squared plus two times three to the third power. And so with a geometric series, you're going to have a sum where each successive term in the expression is equal to, if you put them all in order, is going to be equal to the term before it times a fixed amount. So the second term is equal to the first term times three, and we're summing them. |
Geometric series introduction Algebra 2 Khan Academy.mp3 | So this would be two plus six plus 18 plus 54. Or we could even write it, and this will look similar to what we had just done with our little savings example, is this is two plus two times three plus two times three squared plus two times three to the third power. And so with a geometric series, you're going to have a sum where each successive term in the expression is equal to, if you put them all in order, is going to be equal to the term before it times a fixed amount. So the second term is equal to the first term times three, and we're summing them. In a sequence, you're just looking at it. It's an ordered list, so to speak, but here you are actually adding up the ordered list. So what we just saw in this example is why a, one, what a geometric series is, but also a famous example of how it's useful, and this is just scratching the surface. |
Polynomial division introduction Algebra 2 Khan Academy.mp3 | So for example, if I had the polynomial, and this would be a quadratic polynomial, let's say x squared plus three x plus two, and I wanted to divide it by x plus one. Pause this video and think about what would that be? What would I have to multiply x plus one by to get x squared plus three x plus two? Well, one way to approach it is we could try to factor x squared plus three x plus two, and we've done that multiple times in our lives. We think about, well, what two numbers add up to three? You know, if I were to multiply them, I'd get two, and the ones that might jump out at you are two and one, and so we could express x squared plus three x plus two as x plus two times x plus one, and then all of that is going to be over x plus one. And so if you were to take x plus two times x plus one, and then divide that by x plus one, what is that going to be? |
Polynomial division introduction Algebra 2 Khan Academy.mp3 | Well, one way to approach it is we could try to factor x squared plus three x plus two, and we've done that multiple times in our lives. We think about, well, what two numbers add up to three? You know, if I were to multiply them, I'd get two, and the ones that might jump out at you are two and one, and so we could express x squared plus three x plus two as x plus two times x plus one, and then all of that is going to be over x plus one. And so if you were to take x plus two times x plus one, and then divide that by x plus one, what is that going to be? Well, you're just going to be left with an x plus two. This is going, I don't even have to put parentheses. This is going to be an x plus two. |
Polynomial division introduction Algebra 2 Khan Academy.mp3 | And so if you were to take x plus two times x plus one, and then divide that by x plus one, what is that going to be? Well, you're just going to be left with an x plus two. This is going, I don't even have to put parentheses. This is going to be an x plus two. And if we wanna be really mathematically precise, we would say, hey, this would be true as long as x does not equal negative one, because if x equals negative one in this expression or this expression, we're going to be dividing by zero, and we know that leads to all sorts of mathematical problems. But as we see, for any other x, as long as we're not dividing by zero here, this expression is going to be the same thing as x plus two, and that's because x plus two times x plus one is equal to what we have in this numerator here. Now, as we go deeper into polynomial division, we're going to approach things that aren't as easy to do just purely through factoring, and that's where we're going to have a technique called polynomial long division, polynomial long division, sometimes known as algebraic long division. |
Polynomial division introduction Algebra 2 Khan Academy.mp3 | This is going to be an x plus two. And if we wanna be really mathematically precise, we would say, hey, this would be true as long as x does not equal negative one, because if x equals negative one in this expression or this expression, we're going to be dividing by zero, and we know that leads to all sorts of mathematical problems. But as we see, for any other x, as long as we're not dividing by zero here, this expression is going to be the same thing as x plus two, and that's because x plus two times x plus one is equal to what we have in this numerator here. Now, as we go deeper into polynomial division, we're going to approach things that aren't as easy to do just purely through factoring, and that's where we're going to have a technique called polynomial long division, polynomial long division, sometimes known as algebraic long division. And if it sounds familiar, because you first learned about long division in fourth or fifth grade, it's because it's a very similar process, where you would take your x plus one, and you would try to divide it into your x squared plus three x plus two. And you do something very, and I'm gonna do a very quick example right over here, but we're gonna do much more detailed examples in future videos. But you look at the highest degree terms. |
Polynomial division introduction Algebra 2 Khan Academy.mp3 | Now, as we go deeper into polynomial division, we're going to approach things that aren't as easy to do just purely through factoring, and that's where we're going to have a technique called polynomial long division, polynomial long division, sometimes known as algebraic long division. And if it sounds familiar, because you first learned about long division in fourth or fifth grade, it's because it's a very similar process, where you would take your x plus one, and you would try to divide it into your x squared plus three x plus two. And you do something very, and I'm gonna do a very quick example right over here, but we're gonna do much more detailed examples in future videos. But you look at the highest degree terms. You say, okay, I have a first degree term and a second degree term here. How many times does x go into x squared? Well, it goes x times. |
Polynomial division introduction Algebra 2 Khan Academy.mp3 | But you look at the highest degree terms. You say, okay, I have a first degree term and a second degree term here. How many times does x go into x squared? Well, it goes x times. So you put the x in the first degree column, and then you multiply your x times x plus one. X times x is x squared. X times one is x. |
Polynomial division introduction Algebra 2 Khan Academy.mp3 | Well, it goes x times. So you put the x in the first degree column, and then you multiply your x times x plus one. X times x is x squared. X times one is x. And then you subtract this from that. So you might already start to see some parallels with the long division that you first learned in school many years ago. So when you do that, these cancel out. |
Polynomial division introduction Algebra 2 Khan Academy.mp3 | X times one is x. And then you subtract this from that. So you might already start to see some parallels with the long division that you first learned in school many years ago. So when you do that, these cancel out. Three x minus x, we are left with a two x. And then you bring down that two. So two x plus two. |
Polynomial division introduction Algebra 2 Khan Academy.mp3 | So when you do that, these cancel out. Three x minus x, we are left with a two x. And then you bring down that two. So two x plus two. And you say, how many times does x go into two x? Well, it goes two times. So you have a plus two here. |
Polynomial division introduction Algebra 2 Khan Academy.mp3 | So two x plus two. And you say, how many times does x go into two x? Well, it goes two times. So you have a plus two here. Two times x plus one, two times x is two x. Two times one is two. You can subtract these. |
Polynomial division introduction Algebra 2 Khan Academy.mp3 | So you have a plus two here. Two times x plus one, two times x is two x. Two times one is two. You can subtract these. And then you are going to be left with nothing. Two minus two is zero. Two x minus two x is zero. |
Polynomial division introduction Algebra 2 Khan Academy.mp3 | You can subtract these. And then you are going to be left with nothing. Two minus two is zero. Two x minus two x is zero. So in this situation, it divided cleanly into it, and we got x plus two, which is exactly what we had over there. Now, an interesting scenario that we're also going to approach in the next few videos is, what if things don't divide cleanly? For example, if I were to add one to x squared plus three x plus two, I would get x squared plus three x plus three. |
Polynomial division introduction Algebra 2 Khan Academy.mp3 | Two x minus two x is zero. So in this situation, it divided cleanly into it, and we got x plus two, which is exactly what we had over there. Now, an interesting scenario that we're also going to approach in the next few videos is, what if things don't divide cleanly? For example, if I were to add one to x squared plus three x plus two, I would get x squared plus three x plus three. And if I were to try to divide that by x plus one, well, it's not going to divide cleanly anymore. You could do it either approach. One way to think about it, if we know we can factor x squared plus three x plus two, is say, hey, this is the same thing as x squared plus three x plus two plus one. |
Polynomial division introduction Algebra 2 Khan Academy.mp3 | For example, if I were to add one to x squared plus three x plus two, I would get x squared plus three x plus three. And if I were to try to divide that by x plus one, well, it's not going to divide cleanly anymore. You could do it either approach. One way to think about it, if we know we can factor x squared plus three x plus two, is say, hey, this is the same thing as x squared plus three x plus two plus one. And then all of that's going to be over x plus one. And then you could say, hey, this is the same thing as x squared plus three x plus two over x plus one, x plus one over x plus one plus one over x plus one, plus one over x plus one. And we already figured out that this expression on the left, as long as x does not equal negative one, this is going to be equal to x plus two. |
Polynomial division introduction Algebra 2 Khan Academy.mp3 | One way to think about it, if we know we can factor x squared plus three x plus two, is say, hey, this is the same thing as x squared plus three x plus two plus one. And then all of that's going to be over x plus one. And then you could say, hey, this is the same thing as x squared plus three x plus two over x plus one, x plus one over x plus one plus one over x plus one, plus one over x plus one. And we already figured out that this expression on the left, as long as x does not equal negative one, this is going to be equal to x plus two. So this is going to be equal to x plus two, but then we have that one that we weren't able to divide x plus one into, so we're just left with a one over x plus one. And we'll study that in a lot more detail in other videos. What does this remainder mean? |
Polynomial division introduction Algebra 2 Khan Academy.mp3 | And we already figured out that this expression on the left, as long as x does not equal negative one, this is going to be equal to x plus two. So this is going to be equal to x plus two, but then we have that one that we weren't able to divide x plus one into, so we're just left with a one over x plus one. And we'll study that in a lot more detail in other videos. What does this remainder mean? And how do we calculate it if we can't factor a part of what we have in the numerator? And as we do our polynomial long division, we'll see that the remainder will show up at the end when we are done dividing. We'll see those examples in future videos. |
Periodicity of algebraic models Mathematics III High School Math Khan Academy.mp3 | So at time equals zero, she is, looks like about two, what is this, this would be one and a half, so it looks like she's about two meters off the ground. And then as time increases, she gets as high as, it looks like this is close to 30, maybe 34 meters, and then she comes back down, back to, looks like two meters, and up to 34 meters again. So let's read the question. So the question asks us, approximately how long does it take Divya to complete one revolution on the Ferris wheel? All right, so this is interesting. So this is when she's at the bottom of the Ferris wheel. So then she gets to the top of the Ferris wheel, and then she keeps rotating until she gets back to the bottom of the Ferris wheel again. |
Periodicity of algebraic models Mathematics III High School Math Khan Academy.mp3 | So the question asks us, approximately how long does it take Divya to complete one revolution on the Ferris wheel? All right, so this is interesting. So this is when she's at the bottom of the Ferris wheel. So then she gets to the top of the Ferris wheel, and then she keeps rotating until she gets back to the bottom of the Ferris wheel again. So it took her 60, and t is in terms of seconds, so it took her 60 seconds to go from the bottom to the bottom again. Another 60 seconds, she would have completed another revolution. And so, let me fill that in. |
Periodicity of algebraic models Mathematics III High School Math Khan Academy.mp3 | So then she gets to the top of the Ferris wheel, and then she keeps rotating until she gets back to the bottom of the Ferris wheel again. So it took her 60, and t is in terms of seconds, so it took her 60 seconds to go from the bottom to the bottom again. Another 60 seconds, she would have completed another revolution. And so, let me fill that in. It is going to take her 60 seconds. 60 seconds, and we of course can check our answer if we like. Let's do another one of these. |
Periodicity of algebraic models Mathematics III High School Math Khan Academy.mp3 | And so, let me fill that in. It is going to take her 60 seconds. 60 seconds, and we of course can check our answer if we like. Let's do another one of these. So here we have a doctor observes the electrical activity of Finn's heart over a period of time. The electrical activity of Finn's heart is cyclical, as we hope it would be, and peaks every 0.9 seconds. Which of the following graphs could model a situation if t stands for time in seconds, and e stands for the electrical activity of Finn's heart in volts? |
Periodicity of algebraic models Mathematics III High School Math Khan Academy.mp3 | Let's do another one of these. So here we have a doctor observes the electrical activity of Finn's heart over a period of time. The electrical activity of Finn's heart is cyclical, as we hope it would be, and peaks every 0.9 seconds. Which of the following graphs could model a situation if t stands for time in seconds, and e stands for the electrical activity of Finn's heart in volts? Well over here it looks like we peaked at zero seconds, and here we're peaking a little bit more than one. This looks like maybe at 1.1, and maybe at 2.2 and 3.3. This looks like it's peaking a little bit more than every one second, so like maybe every 1.1 seconds, not every 0.9 seconds, so I'd rule out a. |
Periodicity of algebraic models Mathematics III High School Math Khan Academy.mp3 | Which of the following graphs could model a situation if t stands for time in seconds, and e stands for the electrical activity of Finn's heart in volts? Well over here it looks like we peaked at zero seconds, and here we're peaking a little bit more than one. This looks like maybe at 1.1, and maybe at 2.2 and 3.3. This looks like it's peaking a little bit more than every one second, so like maybe every 1.1 seconds, not every 0.9 seconds, so I'd rule out a. This one is peaking, it looks like the interval between peaks is less than a second, but it looks like a good bit less than a second. It looks like maybe every 3 quarters of a second, or maybe every 4 fifths of a second. Not quite 9 tenths. |
Periodicity of algebraic models Mathematics III High School Math Khan Academy.mp3 | This looks like it's peaking a little bit more than every one second, so like maybe every 1.1 seconds, not every 0.9 seconds, so I'd rule out a. This one is peaking, it looks like the interval between peaks is less than a second, but it looks like a good bit less than a second. It looks like maybe every 3 quarters of a second, or maybe every 4 fifths of a second. Not quite 9 tenths. 9 tenths, this first peak would be a little bit closer to one, but this one is close. Choice C is looking good. The first, we're at zero, then the first peak, this looks pretty close to one, but it's less than one. |
Periodicity of algebraic models Mathematics III High School Math Khan Academy.mp3 | Not quite 9 tenths. 9 tenths, this first peak would be a little bit closer to one, but this one is close. Choice C is looking good. The first, we're at zero, then the first peak, this looks pretty close to one, but it's less than one. It looks like a tenth less than one. So I like choice C. Now choice D, it looks like we're peaking every half second, so it's definitely not that. So this looks like a peak of every 0.9 seconds. |
Periodicity of algebraic models Mathematics III High School Math Khan Academy.mp3 | The first, we're at zero, then the first peak, this looks pretty close to one, but it's less than one. It looks like a tenth less than one. So I like choice C. Now choice D, it looks like we're peaking every half second, so it's definitely not that. So this looks like a peak of every 0.9 seconds. This is the best representation that I, this is the best representation that I can think of. And you can actually verify that. If you have a peak every 0.9 seconds, you're gonna have 4 peaks in 3.6 seconds. |
Periodicity of algebraic models Mathematics III High School Math Khan Academy.mp3 | So this looks like a peak of every 0.9 seconds. This is the best representation that I, this is the best representation that I can think of. And you can actually verify that. If you have a peak every 0.9 seconds, you're gonna have 4 peaks in 3.6 seconds. So one, two, three, four. This looks like it's at 3.6. Over here, you have one, two, three, four. |
Periodicity of algebraic models Mathematics III High School Math Khan Academy.mp3 | If you have a peak every 0.9 seconds, you're gonna have 4 peaks in 3.6 seconds. So one, two, three, four. This looks like it's at 3.6. Over here, you have one, two, three, four. You've had 4 peaks in less than 3 seconds. So this definitely one, this one definitely isn't 0.9. So instead of just even forcing yourself to eyeball just between this peak and that peak, you can say, well, if we're every 0.9 seconds, how long would 3 peaks take, or 4 peaks? |
Zeros of polynomials (multiplicity) Polynomial graphs Algebra 2 Khan Academy.mp3 | All right, now let's work through this together. And we can see that all of the choices are expressed as a polynomial in factored form. And factored form is useful when we're thinking about the roots of a polynomial, the x values that make that polynomial equal to zero. The roots are also evident when we look at this graph here. We have a root at x equals negative four, a root at x equals negative 1 1 1 1 1, or negative 3 1 1 1 1 1, and a root at x is equal to one. So really what we have to do is say, which of these factors are consistent with the roots that we see? So let's go root by root. |
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