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Trolls, tolls, and systems of equations Algebra II Khan Academy.mp3 | So what's going to be the total value of my bills? What's going to be the value of the $5 bills plus the value of the $10 bills? And he tells me what that total value is. It's $5,500. So if I add these two things, they're going to add up to be $5,500. So the second statement we can represent mathematically with this second equation right over here. And what we essentially have right over here, we have two equations. |
Trolls, tolls, and systems of equations Algebra II Khan Academy.mp3 | It's $5,500. So if I add these two things, they're going to add up to be $5,500. So the second statement we can represent mathematically with this second equation right over here. And what we essentially have right over here, we have two equations. Each of them have two unknowns. And just using one of these equations, we can't really figure out what f and t are. You can pick a bunch of different combinations that add up to 900 here. |
Trolls, tolls, and systems of equations Algebra II Khan Academy.mp3 | And what we essentially have right over here, we have two equations. Each of them have two unknowns. And just using one of these equations, we can't really figure out what f and t are. You can pick a bunch of different combinations that add up to 900 here. You could pick a bunch of different combinations where if you work out all the math, you get 5,500 here. So independently, these equations, you don't know what f and t are. But what we will see over the next several videos is that if you use both of this information, if you say that there's an f and a t that has to satisfy both of these equations, then you can find a solution. |
Trolls, tolls, and systems of equations Algebra II Khan Academy.mp3 | You can pick a bunch of different combinations that add up to 900 here. You could pick a bunch of different combinations where if you work out all the math, you get 5,500 here. So independently, these equations, you don't know what f and t are. But what we will see over the next several videos is that if you use both of this information, if you say that there's an f and a t that has to satisfy both of these equations, then you can find a solution. And this is called a system of equations. This is a system of equations. Let me write that down. |
Area of a circle Perimeter, area, and volume Geometry Khan Academy.mp3 | The diameter of each wafer is 16 millimeters. What is the area of each candy? So the candy, they say it's the shape of circular disks. And they tell us that the diameter of each wafer is 16 millimeters. If I draw a line across the circle that goes through the center, the length of that line all the way across the circle through the center is 16 millimeters. So let me write that. So diameter here is 16 millimeters. |
Area of a circle Perimeter, area, and volume Geometry Khan Academy.mp3 | And they tell us that the diameter of each wafer is 16 millimeters. If I draw a line across the circle that goes through the center, the length of that line all the way across the circle through the center is 16 millimeters. So let me write that. So diameter here is 16 millimeters. And they want us to figure out the area of the surface of this candy, or essentially the area of this circle. And so when we think about area, we know that the area of a circle is equal to pi times the radius of the circle squared. And you say, well, they gave us a diameter. |
Area of a circle Perimeter, area, and volume Geometry Khan Academy.mp3 | So diameter here is 16 millimeters. And they want us to figure out the area of the surface of this candy, or essentially the area of this circle. And so when we think about area, we know that the area of a circle is equal to pi times the radius of the circle squared. And you say, well, they gave us a diameter. What is the radius? Well, you might remember the radius is half of diameter, so the distance from the center of the circle to the outside, to the boundary of the circle. So it would be this distance right over here, which is exactly half of the diameter. |
Area of a circle Perimeter, area, and volume Geometry Khan Academy.mp3 | And you say, well, they gave us a diameter. What is the radius? Well, you might remember the radius is half of diameter, so the distance from the center of the circle to the outside, to the boundary of the circle. So it would be this distance right over here, which is exactly half of the diameter. So it would be 8 millimeters. So where we see the radius, we could put 8 millimeters. So the area is going to be equal to pi times 8 millimeters squared, which would be 64 square millimeters. |
Area of a circle Perimeter, area, and volume Geometry Khan Academy.mp3 | So it would be this distance right over here, which is exactly half of the diameter. So it would be 8 millimeters. So where we see the radius, we could put 8 millimeters. So the area is going to be equal to pi times 8 millimeters squared, which would be 64 square millimeters. And typically, this is written with pi after the 64. So you might often see it as this is equal to 64 pi millimeters squared. Now, this is the answer, 64 pi millimeters squared. |
Area of a circle Perimeter, area, and volume Geometry Khan Academy.mp3 | So the area is going to be equal to pi times 8 millimeters squared, which would be 64 square millimeters. And typically, this is written with pi after the 64. So you might often see it as this is equal to 64 pi millimeters squared. Now, this is the answer, 64 pi millimeters squared. But sometimes it's not so satisfying to just leave it as pi. You might say, well, I want to get an estimate of what number this is close to. I want a decimal representation of this. |
Area of a circle Perimeter, area, and volume Geometry Khan Academy.mp3 | Now, this is the answer, 64 pi millimeters squared. But sometimes it's not so satisfying to just leave it as pi. You might say, well, I want to get an estimate of what number this is close to. I want a decimal representation of this. And so we could start to use approximate values of pi. So the most rough approximate value that tends to be used is saying that pi, a very rough approximation, is equal to 3.14. So in that case, we could say that this is going to be equal to 64 times 3.14 millimeters squared. |
Area of a circle Perimeter, area, and volume Geometry Khan Academy.mp3 | I want a decimal representation of this. And so we could start to use approximate values of pi. So the most rough approximate value that tends to be used is saying that pi, a very rough approximation, is equal to 3.14. So in that case, we could say that this is going to be equal to 64 times 3.14 millimeters squared. And we can get our calculator to figure out what this will be in decimal form. So we have 64 times 3.14 gives us 200.96. So we could say that the area is approximately equal to 200.96 square millimeters. |
Area of a circle Perimeter, area, and volume Geometry Khan Academy.mp3 | So in that case, we could say that this is going to be equal to 64 times 3.14 millimeters squared. And we can get our calculator to figure out what this will be in decimal form. So we have 64 times 3.14 gives us 200.96. So we could say that the area is approximately equal to 200.96 square millimeters. Now, if we want to get a more accurate representation of this, pi actually just keeps going on and on and on forever, we could use the calculator's internal representation of pi. In which case, we'll say 64 times, and then we have to look for the pi in the calculator. It's up here in this yellow, so I'll do this little second function, get the pi there. |
Area of a circle Perimeter, area, and volume Geometry Khan Academy.mp3 | So we could say that the area is approximately equal to 200.96 square millimeters. Now, if we want to get a more accurate representation of this, pi actually just keeps going on and on and on forever, we could use the calculator's internal representation of pi. In which case, we'll say 64 times, and then we have to look for the pi in the calculator. It's up here in this yellow, so I'll do this little second function, get the pi there. Every calculator will be a little different. But 64 times pi. Now we're going to use the calculator's internal approximation of pi, which is going to be more precise than what I had in the last one. |
Area of a circle Perimeter, area, and volume Geometry Khan Academy.mp3 | It's up here in this yellow, so I'll do this little second function, get the pi there. Every calculator will be a little different. But 64 times pi. Now we're going to use the calculator's internal approximation of pi, which is going to be more precise than what I had in the last one. And you get 201. So let me put it over here so I can write it down. So a more precise is 201. |
Area of a circle Perimeter, area, and volume Geometry Khan Academy.mp3 | Now we're going to use the calculator's internal approximation of pi, which is going to be more precise than what I had in the last one. And you get 201. So let me put it over here so I can write it down. So a more precise is 201. And I'll round to the nearest hundredth. So you get 201.06. So a more precise is 201.06 square millimeters. |
Converting decimals to fractions example 3 Decimals Pre-Algebra Khan Academy.mp3 | As a fraction, there are several ways of doing it. The way I like to do it is to say, well, 0.36, this is the same thing as 36 hundredths. Or one way to think about it is, this is in the hundredths place. Hundredths. Hundredths place. This is in the tens place. Or you could view this as 30 hundredths. |
Converting decimals to fractions example 3 Decimals Pre-Algebra Khan Academy.mp3 | Hundredths. Hundredths place. This is in the tens place. Or you could view this as 30 hundredths. You could view this as 3 tenths or 30 hundredths. So we could say that this is the same thing as 36 hundredths, or this is equal to 36 over 100. And we've already expressed it as a fraction, but now we can actually simplify it because both 36 and 100 have some common factors. |
Converting decimals to fractions example 3 Decimals Pre-Algebra Khan Academy.mp3 | Or you could view this as 30 hundredths. You could view this as 3 tenths or 30 hundredths. So we could say that this is the same thing as 36 hundredths, or this is equal to 36 over 100. And we've already expressed it as a fraction, but now we can actually simplify it because both 36 and 100 have some common factors. They are both divisible by, well, it looks like they're both divisible by four. So if we can divide the numerator by four and the denominator by four, we get we're doing the same thing to both, so we're not changing the value of the fraction. 36 divided by four is nine. |
Converting decimals to fractions example 3 Decimals Pre-Algebra Khan Academy.mp3 | And we've already expressed it as a fraction, but now we can actually simplify it because both 36 and 100 have some common factors. They are both divisible by, well, it looks like they're both divisible by four. So if we can divide the numerator by four and the denominator by four, we get we're doing the same thing to both, so we're not changing the value of the fraction. 36 divided by four is nine. And then 100 divided by four is 25. And now these two characters don't seem to share any other common factors, and so we've written it in simplified form. And we're done. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | So to answer this question, we need to know what it means to be consistent or inconsistent. So a consistent system of equations has at least one solution. And an inconsistent system of equations, as you can imagine, has no solutions. So if we think about it graphically, what would the graph of a consistent system look like? Well either, let me just draw a really rough graph, so that's my x axis, and that is my y axis. So if I have just two different lines that intersect, that would be consistent. So that's one line, and then that's another line. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | So if we think about it graphically, what would the graph of a consistent system look like? Well either, let me just draw a really rough graph, so that's my x axis, and that is my y axis. So if I have just two different lines that intersect, that would be consistent. So that's one line, and then that's another line. They clearly have that one solution where they both intersect, so that would be a consistent system. Another consistent system would be if they're the same line, because then they would intersect at a ton of points. Actually an infinite number of points. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | So that's one line, and then that's another line. They clearly have that one solution where they both intersect, so that would be a consistent system. Another consistent system would be if they're the same line, because then they would intersect at a ton of points. Actually an infinite number of points. So let's say one of the lines looks like that, and then the other line is actually the exact same line. So it's exactly right on top of it. So those two intersect at every point along those lines, so that also would be consistent. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | Actually an infinite number of points. So let's say one of the lines looks like that, and then the other line is actually the exact same line. So it's exactly right on top of it. So those two intersect at every point along those lines, so that also would be consistent. An inconsistent system would have no solutions. So let me again draw my axes. It will have no solutions. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | So those two intersect at every point along those lines, so that also would be consistent. An inconsistent system would have no solutions. So let me again draw my axes. It will have no solutions. The only way that you're going to have two lines in two dimensions have no solutions is if they don't intersect, or if they are parallel. So one line could look like this, and then the other line would have the same slope, but it would be shifted over. It would have a different y intercept. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | It will have no solutions. The only way that you're going to have two lines in two dimensions have no solutions is if they don't intersect, or if they are parallel. So one line could look like this, and then the other line would have the same slope, but it would be shifted over. It would have a different y intercept. So it would look like this. So that's what an inconsistent system would look like. You have parallel lines. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | It would have a different y intercept. So it would look like this. So that's what an inconsistent system would look like. You have parallel lines. This right here is inconsistent. So what we could do is just do a rough graph of both of these lines and see if they intersect. Another way to do it is you could look at the slope, and if they have the same slope and different y intercepts, then you'd also have an inconsistent system. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | You have parallel lines. This right here is inconsistent. So what we could do is just do a rough graph of both of these lines and see if they intersect. Another way to do it is you could look at the slope, and if they have the same slope and different y intercepts, then you'd also have an inconsistent system. But let's just graph them. So let me draw my x-axis, and let me draw my y-axis. So this is x, and then this is y. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | Another way to do it is you could look at the slope, and if they have the same slope and different y intercepts, then you'd also have an inconsistent system. But let's just graph them. So let me draw my x-axis, and let me draw my y-axis. So this is x, and then this is y. And then there's a couple of ways we could do it. The easiest way is really just find two points that satisfy each of these equations, and that's enough to define a line. So for this first one, let's just make a little table of x's and y's. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | So this is x, and then this is y. And then there's a couple of ways we could do it. The easiest way is really just find two points that satisfy each of these equations, and that's enough to define a line. So for this first one, let's just make a little table of x's and y's. When x is 0, you have 2y is equal to 13. So when x is 0, you have 2y is equal to 13, or y is equal to 13 over 2, which is the same thing as 6 1⁄2. So when x is 0, y is 6 1⁄2. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | So for this first one, let's just make a little table of x's and y's. When x is 0, you have 2y is equal to 13. So when x is 0, you have 2y is equal to 13, or y is equal to 13 over 2, which is the same thing as 6 1⁄2. So when x is 0, y is 6 1⁄2. I'll just put it right over here. So this is 0, 13 over 2. And then let's just see what happens when y is 0. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | So when x is 0, y is 6 1⁄2. I'll just put it right over here. So this is 0, 13 over 2. And then let's just see what happens when y is 0. When y is 0, then 2 times y is 0. You have x equaling 13. So we have the point 13, 0. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | And then let's just see what happens when y is 0. When y is 0, then 2 times y is 0. You have x equaling 13. So we have the point 13, 0. So this is 0, 6 1⁄2. So 13, 0 would be right about there. I'm just trying to approximate. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | So we have the point 13, 0. So this is 0, 6 1⁄2. So 13, 0 would be right about there. I'm just trying to approximate. 13, 0. And so this line right up here, this equation, can be represented by this line. Let me try my best to draw it. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | I'm just trying to approximate. 13, 0. And so this line right up here, this equation, can be represented by this line. Let me try my best to draw it. It would look something like that. Now let's worry about this one. Let's worry about that one. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | Let me try my best to draw it. It would look something like that. Now let's worry about this one. Let's worry about that one. So once again, let's make a little table, x's and y's. I'm really just looking for two points on this graph. So when x is equal to 0, you get this 3 times 0 is just 0. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | Let's worry about that one. So once again, let's make a little table, x's and y's. I'm really just looking for two points on this graph. So when x is equal to 0, you get this 3 times 0 is just 0. So you get negative y is equal to negative 11, or you get y is equal to 11. So you have the point 0, 11. So that's maybe right over there. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | So when x is equal to 0, you get this 3 times 0 is just 0. So you get negative y is equal to negative 11, or you get y is equal to 11. So you have the point 0, 11. So that's maybe right over there. 0, 11 is on that line. And then when y is 0, you have 3x minus 0 is equal to negative 11. Or 3x is equal to negative 11. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | So that's maybe right over there. 0, 11 is on that line. And then when y is 0, you have 3x minus 0 is equal to negative 11. Or 3x is equal to negative 11. Or if you divide both sides by 3, you get x is equal to negative 11 over 3. And this is the exact same thing as negative 3 and 2 thirds. So when y is 0, you have x being negative 3 and 2 thirds. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | Or 3x is equal to negative 11. Or if you divide both sides by 3, you get x is equal to negative 11 over 3. And this is the exact same thing as negative 3 and 2 thirds. So when y is 0, you have x being negative 3 and 2 thirds. So maybe this is about 6, so negative 3 and 2 thirds would be right about here. So this is the point negative 11 thirds, comma, 0. And so the second equation will look something like this. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | So when y is 0, you have x being negative 3 and 2 thirds. So maybe this is about 6, so negative 3 and 2 thirds would be right about here. So this is the point negative 11 thirds, comma, 0. And so the second equation will look something like this. It will look something like that. Now clearly, and I might have not been completely precise when I did this hand-drawn graph, clearly these two guys intersect. They intersect right over here. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | And so the second equation will look something like this. It will look something like that. Now clearly, and I might have not been completely precise when I did this hand-drawn graph, clearly these two guys intersect. They intersect right over here. And to answer their question, you don't even have to find the point that they intersect at. We just have to see very clearly that these two lines intersect. So this is a consistent system of equations. |
Consistent and inconsistent systems Algebra II Khan Academy.mp3 | They intersect right over here. And to answer their question, you don't even have to find the point that they intersect at. We just have to see very clearly that these two lines intersect. So this is a consistent system of equations. It has one solution. You just have to have at least one in order to be consistent. So once again, consistent system of equations. |
Solving proportions 2 exercise examples Algebra Basics Khan Academy.mp3 | Now there's a bunch of ways that you could do it. A lot of people, as soon as they see a proportion like this, they want to cross multiply. They want to say, hey, 3 times x minus 9 is going to be equal to 2 times 12. And that's completely legitimate. You would get, let me write that down. 3 times x minus 9 is equal to 2 times 12. So it would be equal to 2 times 12. |
Solving proportions 2 exercise examples Algebra Basics Khan Academy.mp3 | And that's completely legitimate. You would get, let me write that down. 3 times x minus 9 is equal to 2 times 12. So it would be equal to 2 times 12. And then you can distribute the 3. You'd get 3x minus 27 is equal to 24. And then you could add 27 to both sides. |
Solving proportions 2 exercise examples Algebra Basics Khan Academy.mp3 | So it would be equal to 2 times 12. And then you can distribute the 3. You'd get 3x minus 27 is equal to 24. And then you could add 27 to both sides. And you would get, let me actually do that. So let me add 27 to both sides. And we are left with 3x is equal to 51. |
Solving proportions 2 exercise examples Algebra Basics Khan Academy.mp3 | And then you could add 27 to both sides. And you would get, let me actually do that. So let me add 27 to both sides. And we are left with 3x is equal to 51. And then x would be equal to 17. And you can verify that this works. 17 minus 9 is 8. |
Solving proportions 2 exercise examples Algebra Basics Khan Academy.mp3 | And we are left with 3x is equal to 51. And then x would be equal to 17. And you can verify that this works. 17 minus 9 is 8. 8 twelfths is the same thing as 2 thirds. So this checks out. Another way you could do that, instead of just straight up doing the cross multiplication, you could say, look, I want to get rid of this 12 in the denominator right over here. |
Solving proportions 2 exercise examples Algebra Basics Khan Academy.mp3 | 17 minus 9 is 8. 8 twelfths is the same thing as 2 thirds. So this checks out. Another way you could do that, instead of just straight up doing the cross multiplication, you could say, look, I want to get rid of this 12 in the denominator right over here. Multiply both sides by 12. So if you multiply both sides by 12, on your left-hand side, you are just left with x minus 9. And on your right-hand side, 2 thirds times 12, well, 2 thirds of 12 is just 8. |
Solving proportions 2 exercise examples Algebra Basics Khan Academy.mp3 | Another way you could do that, instead of just straight up doing the cross multiplication, you could say, look, I want to get rid of this 12 in the denominator right over here. Multiply both sides by 12. So if you multiply both sides by 12, on your left-hand side, you are just left with x minus 9. And on your right-hand side, 2 thirds times 12, well, 2 thirds of 12 is just 8. And you could do the actual multiplication, 2 thirds times 12 over 1. 12 and 3. So 12 divided by 3 is 4. |
Solving proportions 2 exercise examples Algebra Basics Khan Academy.mp3 | And on your right-hand side, 2 thirds times 12, well, 2 thirds of 12 is just 8. And you could do the actual multiplication, 2 thirds times 12 over 1. 12 and 3. So 12 divided by 3 is 4. 3 divided by 3 is 1. So it becomes 2 times 4 over 1, which is just 8. And then you add 9 to both sides. |
Solving proportions 2 exercise examples Algebra Basics Khan Academy.mp3 | So 12 divided by 3 is 4. 3 divided by 3 is 1. So it becomes 2 times 4 over 1, which is just 8. And then you add 9 to both sides. So the fun of algebra is that as long as you do something that's logically consistent, you will get the right answer. There's no one way of doing it. So here you get x is equal to 17 again. |
Solving proportions 2 exercise examples Algebra Basics Khan Academy.mp3 | And then you add 9 to both sides. So the fun of algebra is that as long as you do something that's logically consistent, you will get the right answer. There's no one way of doing it. So here you get x is equal to 17 again. And you could also multiply both sides by 12 and both sides by 3. And then that would be functionally equivalent to cross multiplying. Let's do one more. |
Solving proportions 2 exercise examples Algebra Basics Khan Academy.mp3 | So here you get x is equal to 17 again. And you could also multiply both sides by 12 and both sides by 3. And then that would be functionally equivalent to cross multiplying. Let's do one more. So here, another proportion. And this time, the x is in the denominator. But just like before, if we want, we can cross multiply. |
Solving proportions 2 exercise examples Algebra Basics Khan Academy.mp3 | Let's do one more. So here, another proportion. And this time, the x is in the denominator. But just like before, if we want, we can cross multiply. And just to see where cross multiplying comes from, that it's not some voodoo, that you still are doing logical algebra, that you're doing the same thing to both sides of the equation, you just need to appreciate that we're just multiplying both sides by both denominators. So we have this 8 right over here on the left-hand side. If we want to get rid of this 8 on the left-hand side in the denominator, we can multiply the left-hand side by 8. |
Solving proportions 2 exercise examples Algebra Basics Khan Academy.mp3 | But just like before, if we want, we can cross multiply. And just to see where cross multiplying comes from, that it's not some voodoo, that you still are doing logical algebra, that you're doing the same thing to both sides of the equation, you just need to appreciate that we're just multiplying both sides by both denominators. So we have this 8 right over here on the left-hand side. If we want to get rid of this 8 on the left-hand side in the denominator, we can multiply the left-hand side by 8. But in order for the equality to hold true, I can't do something to just one side. I have to do it to both sides. Similarly, if I want to get this x plus 1 out of the denominator, I could multiply by x plus 1 right over here. |
Solving proportions 2 exercise examples Algebra Basics Khan Academy.mp3 | If we want to get rid of this 8 on the left-hand side in the denominator, we can multiply the left-hand side by 8. But in order for the equality to hold true, I can't do something to just one side. I have to do it to both sides. Similarly, if I want to get this x plus 1 out of the denominator, I could multiply by x plus 1 right over here. But I have to do that on both sides if I want my equality to hold true. And notice, when you do what we just did, this is going to be equivalent to cross multiplying. Because these 8's cancel out, and this x plus 1 cancels with that x plus 1 right over there. |
Solving proportions 2 exercise examples Algebra Basics Khan Academy.mp3 | Similarly, if I want to get this x plus 1 out of the denominator, I could multiply by x plus 1 right over here. But I have to do that on both sides if I want my equality to hold true. And notice, when you do what we just did, this is going to be equivalent to cross multiplying. Because these 8's cancel out, and this x plus 1 cancels with that x plus 1 right over there. And you are left with x plus 1 times 7. And I could write it as 7 times x plus 1 is equal to 5 times 8. Notice, this is exactly what you would have done if you were to cross multiply. |
Solving proportions 2 exercise examples Algebra Basics Khan Academy.mp3 | Because these 8's cancel out, and this x plus 1 cancels with that x plus 1 right over there. And you are left with x plus 1 times 7. And I could write it as 7 times x plus 1 is equal to 5 times 8. Notice, this is exactly what you would have done if you were to cross multiply. Cross multiplication is just a shortcut of multiplying both sides by both denominators. We have 7 times x plus 1 is equal to 5 times 8. And now we can go and solve the algebra. |
Solving proportions 2 exercise examples Algebra Basics Khan Academy.mp3 | Notice, this is exactly what you would have done if you were to cross multiply. Cross multiplication is just a shortcut of multiplying both sides by both denominators. We have 7 times x plus 1 is equal to 5 times 8. And now we can go and solve the algebra. So distributing the 7, we get 7x plus 7 is equal to 40. And then subtracting 7 from both sides, we are left with 7x is equal to 33. Dividing both sides by 7, we are left with x is equal to 33 over 7. |
Exponents of decimals.mp3 | What we're going to do in this video is get some practice evaluating exponents of decimals. So let's say that I have 0.2 to the third power. Pause this video, see if you can figure out what that is going to be. Well, this would just mean if I take something to the third power, that means I take three of that number and I multiply them together. So it's 0.2 times 0.2 times 0.2. Well, what is this going to be equal to? Well, if I take 0.2 times 0.2, that is going to be 0.04. |
Exponents of decimals.mp3 | Well, this would just mean if I take something to the third power, that means I take three of that number and I multiply them together. So it's 0.2 times 0.2 times 0.2. Well, what is this going to be equal to? Well, if I take 0.2 times 0.2, that is going to be 0.04. One way to think about it, two times two is four, and then I have one number behind the decimal to the right of the decimal here. I have another digit to the right of the decimal right over here, so my product is gonna have two digits to the right of the decimal, so it'd be 0.04. And then if I were to multiply that times 0.2, so if I were to multiply that together, what is that going to be equal to? |
Exponents of decimals.mp3 | Well, if I take 0.2 times 0.2, that is going to be 0.04. One way to think about it, two times two is four, and then I have one number behind the decimal to the right of the decimal here. I have another digit to the right of the decimal right over here, so my product is gonna have two digits to the right of the decimal, so it'd be 0.04. And then if I were to multiply that times 0.2, so if I were to multiply that together, what is that going to be equal to? Well, four times two is equal to eight, and now I have one, two, three numbers to the right of the decimal point, so my product is gonna have one, two, three numbers to the right of the decimal point. So now that we've had a little bit of practice with that, let's do another example. So let's say that I were to ask you, what is 0.9 squared? |
Exponents of decimals.mp3 | And then if I were to multiply that times 0.2, so if I were to multiply that together, what is that going to be equal to? Well, four times two is equal to eight, and now I have one, two, three numbers to the right of the decimal point, so my product is gonna have one, two, three numbers to the right of the decimal point. So now that we've had a little bit of practice with that, let's do another example. So let's say that I were to ask you, what is 0.9 squared? Pause this video and see if you can figure that out. All right, well, this is just going to be 0.9 times 0.9. And what's that going to be equal to? |
Exponents of decimals.mp3 | So let's say that I were to ask you, what is 0.9 squared? Pause this video and see if you can figure that out. All right, well, this is just going to be 0.9 times 0.9. And what's that going to be equal to? Well, you could just say nine times nine is going to be equal to 81. And so let's see, in the two numbers that I'm multiplying, I have a total of one, two numbers, or two digits, to the right of the decimal point, so my answer's going to have one, two digits to the right of the decimal point, so I'd put the decimal right over there, and I'll put the zero, so 0.81. Another way to think about it is 9 tenths of 9 tenths is 81 hundredths, but there you go. |
Order of operations PEMDAS Arithmetic properties Pre-Algebra Khan Academy (2).mp3 | So here we have all sorts of parentheses and numbers flying around, but in any of these order of operations problems, you really just have to take a deep breath. And remember, we're going to do parentheses first, parentheses, p for parentheses, then exponents. Don't worry if you don't know what exponents are, because this has no exponents in them. Then you're going to do multiplication and division. They're at the same level. Then you do addition and subtraction. So some people remember PEMDAS. |
Order of operations PEMDAS Arithmetic properties Pre-Algebra Khan Academy (2).mp3 | Then you're going to do multiplication and division. They're at the same level. Then you do addition and subtraction. So some people remember PEMDAS. But if you remember PEMDAS, remember multiplication, division, same level, addition and subtraction, also at the same level. So let's figure what order of operations say that this should evaluate to. So the first thing we're going to do is our parentheses, and we have a lot of parentheses here. |
Order of operations PEMDAS Arithmetic properties Pre-Algebra Khan Academy (2).mp3 | So some people remember PEMDAS. But if you remember PEMDAS, remember multiplication, division, same level, addition and subtraction, also at the same level. So let's figure what order of operations say that this should evaluate to. So the first thing we're going to do is our parentheses, and we have a lot of parentheses here. We have this expression in parentheses right there, and then even within that, we have these parentheses. So our order of operations say, look, do your parentheses first, but in order to evaluate this outer parentheses, this orange thing, we're going to have to evaluate this thing in yellow right there. So let's evaluate this whole thing. |
Order of operations PEMDAS Arithmetic properties Pre-Algebra Khan Academy (2).mp3 | So the first thing we're going to do is our parentheses, and we have a lot of parentheses here. We have this expression in parentheses right there, and then even within that, we have these parentheses. So our order of operations say, look, do your parentheses first, but in order to evaluate this outer parentheses, this orange thing, we're going to have to evaluate this thing in yellow right there. So let's evaluate this whole thing. So how can we simplify it? Well, if we look at just inside of it, the first thing we want to do is simplify the parentheses inside the parentheses. So you see this 5 minus 2 right there. |
Order of operations PEMDAS Arithmetic properties Pre-Algebra Khan Academy (2).mp3 | So let's evaluate this whole thing. So how can we simplify it? Well, if we look at just inside of it, the first thing we want to do is simplify the parentheses inside the parentheses. So you see this 5 minus 2 right there. We're going to do that first no matter what, and that's easy to evaluate. 5 minus 2 is 3. And so this simplifies to, I'll do it step by step. |
Order of operations PEMDAS Arithmetic properties Pre-Algebra Khan Academy (2).mp3 | So you see this 5 minus 2 right there. We're going to do that first no matter what, and that's easy to evaluate. 5 minus 2 is 3. And so this simplifies to, I'll do it step by step. Once you get the hang of it, you can do multiple steps at once. So this is going to be 7 plus 3 times the 5 minus 2, which is 3. And then you have, and all of those have parentheses around it. |
Order of operations PEMDAS Arithmetic properties Pre-Algebra Khan Academy (2).mp3 | And so this simplifies to, I'll do it step by step. Once you get the hang of it, you can do multiple steps at once. So this is going to be 7 plus 3 times the 5 minus 2, which is 3. And then you have, and all of those have parentheses around it. And of course, you have all this stuff on either side, the divide for, no, whoops, that's not what I want. I wanted to copy and paste that right there. So copy, and then, no, that's giving me the wrong thing. |
Order of operations PEMDAS Arithmetic properties Pre-Algebra Khan Academy (2).mp3 | And then you have, and all of those have parentheses around it. And of course, you have all this stuff on either side, the divide for, no, whoops, that's not what I want. I wanted to copy and paste that right there. So copy, and then, no, that's giving me the wrong thing. It would have been easier, let me just rewrite it. That's the easiest thing. I'm having technical difficulties. |
Order of operations PEMDAS Arithmetic properties Pre-Algebra Khan Academy (2).mp3 | So copy, and then, no, that's giving me the wrong thing. It would have been easier, let me just rewrite it. That's the easiest thing. I'm having technical difficulties. So divided by 4 times 2, and on this side, you had that 7 times 2 plus this thing in orange parentheses there. Now, at any step, you just look again. We always want to do parentheses first. |
Order of operations PEMDAS Arithmetic properties Pre-Algebra Khan Academy (2).mp3 | I'm having technical difficulties. So divided by 4 times 2, and on this side, you had that 7 times 2 plus this thing in orange parentheses there. Now, at any step, you just look again. We always want to do parentheses first. We keep wanting to do this until there's really no parentheses left. So we have to evaluate this parentheses in orange here. So we have to evaluate this thing first. |
Order of operations PEMDAS Arithmetic properties Pre-Algebra Khan Academy (2).mp3 | We always want to do parentheses first. We keep wanting to do this until there's really no parentheses left. So we have to evaluate this parentheses in orange here. So we have to evaluate this thing first. But in order to evaluate this thing, we have to look inside of it. And when you look inside of it, you have 7 plus 3 times 3. So if you just had 7 plus 3 times 3, how would you evaluate it? |
Order of operations PEMDAS Arithmetic properties Pre-Algebra Khan Academy (2).mp3 | So we have to evaluate this thing first. But in order to evaluate this thing, we have to look inside of it. And when you look inside of it, you have 7 plus 3 times 3. So if you just had 7 plus 3 times 3, how would you evaluate it? Well, look back to your order of operations. We're inside the parentheses here, so inside of it, there are no longer any parentheses. So the next thing we should do is, there are no exponents, there is multiplication. |
Order of operations PEMDAS Arithmetic properties Pre-Algebra Khan Academy (2).mp3 | So if you just had 7 plus 3 times 3, how would you evaluate it? Well, look back to your order of operations. We're inside the parentheses here, so inside of it, there are no longer any parentheses. So the next thing we should do is, there are no exponents, there is multiplication. So we do that before we do any addition or subtraction. So we want to do the 3 times 3 before we add the 7. So this is going to be 7 plus, and the 3 times 3 we want to do first. |
Order of operations PEMDAS Arithmetic properties Pre-Algebra Khan Academy (2).mp3 | So the next thing we should do is, there are no exponents, there is multiplication. So we do that before we do any addition or subtraction. So we want to do the 3 times 3 before we add the 7. So this is going to be 7 plus, and the 3 times 3 we want to do first. We want to do the multiplication first. 7 plus 9. That's going to be in the orange parentheses. |
Order of operations PEMDAS Arithmetic properties Pre-Algebra Khan Academy (2).mp3 | So this is going to be 7 plus, and the 3 times 3 we want to do first. We want to do the multiplication first. 7 plus 9. That's going to be in the orange parentheses. And then you have the 7 times 2 plus that on the left-hand side. You have the divided by 4 times 2 on the right-hand side. And now this, the thing in parentheses, because we still want to do the parentheses first, pretty easy to evaluate. |
Order of operations PEMDAS Arithmetic properties Pre-Algebra Khan Academy (2).mp3 | That's going to be in the orange parentheses. And then you have the 7 times 2 plus that on the left-hand side. You have the divided by 4 times 2 on the right-hand side. And now this, the thing in parentheses, because we still want to do the parentheses first, pretty easy to evaluate. What's 7 plus 9? 7 plus 9 is 16. And so everything we have simplifies to 7 times 2 plus 16 divided by 4 times 2. |
Order of operations PEMDAS Arithmetic properties Pre-Algebra Khan Academy (2).mp3 | And now this, the thing in parentheses, because we still want to do the parentheses first, pretty easy to evaluate. What's 7 plus 9? 7 plus 9 is 16. And so everything we have simplifies to 7 times 2 plus 16 divided by 4 times 2. Now, we don't have any parentheses left, so we don't have to worry about the p in PEMDAS. We have no e, no exponents in this, so then we go straight to multiplication and division. We have a multiplication, we have some multiplication going on there, we have some division going on here, and a multiplication there. |
Order of operations PEMDAS Arithmetic properties Pre-Algebra Khan Academy (2).mp3 | And so everything we have simplifies to 7 times 2 plus 16 divided by 4 times 2. Now, we don't have any parentheses left, so we don't have to worry about the p in PEMDAS. We have no e, no exponents in this, so then we go straight to multiplication and division. We have a multiplication, we have some multiplication going on there, we have some division going on here, and a multiplication there. So we should do these next, before we do this addition right there. So we could do this multiplication, we could do that multiplication. 7 times 2 is 14. |
Order of operations PEMDAS Arithmetic properties Pre-Algebra Khan Academy (2).mp3 | We have a multiplication, we have some multiplication going on there, we have some division going on here, and a multiplication there. So we should do these next, before we do this addition right there. So we could do this multiplication, we could do that multiplication. 7 times 2 is 14. We're going to wait to do that addition. And then here we have a 16 divided by 4 times 2. That gets priority of the addition, so we're going to do that before we do the addition. |
Order of operations PEMDAS Arithmetic properties Pre-Algebra Khan Academy (2).mp3 | 7 times 2 is 14. We're going to wait to do that addition. And then here we have a 16 divided by 4 times 2. That gets priority of the addition, so we're going to do that before we do the addition. But how do we evaluate that? Do we do the division first, do we do the multiplication first? And remember, I told you in the last video, when you have multiple operations of the same level, in this case division and multiplication, they're at the same level, you're safest going left to right, or you should go left to right. |
Order of operations PEMDAS Arithmetic properties Pre-Algebra Khan Academy (2).mp3 | That gets priority of the addition, so we're going to do that before we do the addition. But how do we evaluate that? Do we do the division first, do we do the multiplication first? And remember, I told you in the last video, when you have multiple operations of the same level, in this case division and multiplication, they're at the same level, you're safest going left to right, or you should go left to right. So you do 16 divided by 4 is 4, so this thing right here simplifies 16 divided by 4 times 2, it simplifies to 4 times 2, that's this thing in green right there. And then we're going to want to do the multiplication next. So this is going to simplify to, because multiplication takes priority of addition, this simplifies to 8, and so you get 14, this 14 right here, plus 8. |
Order of operations PEMDAS Arithmetic properties Pre-Algebra Khan Academy (2).mp3 | And remember, I told you in the last video, when you have multiple operations of the same level, in this case division and multiplication, they're at the same level, you're safest going left to right, or you should go left to right. So you do 16 divided by 4 is 4, so this thing right here simplifies 16 divided by 4 times 2, it simplifies to 4 times 2, that's this thing in green right there. And then we're going to want to do the multiplication next. So this is going to simplify to, because multiplication takes priority of addition, this simplifies to 8, and so you get 14, this 14 right here, plus 8. And what's 14 plus 8? That is 22. That is equal to 22. |
Circles radius, diameter, circumference and Pi Geometry Khan Academy.mp3 | So it's probably worthwhile for us to understand some of the properties of the circle. So the first thing when people kind of discovered the circle, and you just have to look at the moon to see a circle, but the first time they said, well, what are the properties of any circle? So the first one they might want to say is, well, a circle is all of the points that are equal distance from the center of the circle, right? All of these points along the edge are equal distance from that center right there. So one of the first things someone might want to ask is, what is that distance, that equal distance that everything is from the center right there? We call that the radius. The radius of the circle. |
Circles radius, diameter, circumference and Pi Geometry Khan Academy.mp3 | All of these points along the edge are equal distance from that center right there. So one of the first things someone might want to ask is, what is that distance, that equal distance that everything is from the center right there? We call that the radius. The radius of the circle. It's just the distance from the center out to the edge. If that radius is 3 centimeters, then this radius is going to be 3 centimeters. And this radius is going to be 3 centimeters. |
Circles radius, diameter, circumference and Pi Geometry Khan Academy.mp3 | The radius of the circle. It's just the distance from the center out to the edge. If that radius is 3 centimeters, then this radius is going to be 3 centimeters. And this radius is going to be 3 centimeters. It's never going to change. By definition, a circle is all of the points that are equidistant from the center point. And that distance is the radius. |
Circles radius, diameter, circumference and Pi Geometry Khan Academy.mp3 | And this radius is going to be 3 centimeters. It's never going to change. By definition, a circle is all of the points that are equidistant from the center point. And that distance is the radius. Now, the next most interesting thing about that, people might say, well, how fat is the circle? How wide is it along its widest point? Or if you just want to cut it along its widest point, what is that distance right there? |
Circles radius, diameter, circumference and Pi Geometry Khan Academy.mp3 | And that distance is the radius. Now, the next most interesting thing about that, people might say, well, how fat is the circle? How wide is it along its widest point? Or if you just want to cut it along its widest point, what is that distance right there? And it doesn't have to be just right there. I could have just as easily cut it along its widest point right there. I just wouldn't be cutting it someplace like that, because that wouldn't be along its widest point. |
Circles radius, diameter, circumference and Pi Geometry Khan Academy.mp3 | Or if you just want to cut it along its widest point, what is that distance right there? And it doesn't have to be just right there. I could have just as easily cut it along its widest point right there. I just wouldn't be cutting it someplace like that, because that wouldn't be along its widest point. There's multiple places where I could cut it along its widest point. Well, we just saw the radius. And we see that widest point goes through the center and just keeps going. |
Circles radius, diameter, circumference and Pi Geometry Khan Academy.mp3 | I just wouldn't be cutting it someplace like that, because that wouldn't be along its widest point. There's multiple places where I could cut it along its widest point. Well, we just saw the radius. And we see that widest point goes through the center and just keeps going. So it's essentially 2 radii. You get one radius there, and then you have another radius over there. And we call this distance along the widest point of the circle the diameter. |
Circles radius, diameter, circumference and Pi Geometry Khan Academy.mp3 | And we see that widest point goes through the center and just keeps going. So it's essentially 2 radii. You get one radius there, and then you have another radius over there. And we call this distance along the widest point of the circle the diameter. So that is the diameter of the circle. It has a very easy relationship with the radius. The diameter is equal to 2 times the radius. |
Circles radius, diameter, circumference and Pi Geometry Khan Academy.mp3 | And we call this distance along the widest point of the circle the diameter. So that is the diameter of the circle. It has a very easy relationship with the radius. The diameter is equal to 2 times the radius. Now, the next most interesting thing that you might be wondering about a circle is how far is it around the circle? So if you were to get your tape measure out, and you were to measure around the circle like that, what's that distance? We call that word the circumference of the circle. |
Circles radius, diameter, circumference and Pi Geometry Khan Academy.mp3 | The diameter is equal to 2 times the radius. Now, the next most interesting thing that you might be wondering about a circle is how far is it around the circle? So if you were to get your tape measure out, and you were to measure around the circle like that, what's that distance? We call that word the circumference of the circle. Now, we know how the diameter and the radius relates, but how does the circumference relate to, say, the diameter? If we know how it relates to the diameter, it's very easy to figure out how it relates to the radius. Well, many thousands of years ago, people took their tape measures out, and they keep measuring circumferences and radiuses. |
Circles radius, diameter, circumference and Pi Geometry Khan Academy.mp3 | We call that word the circumference of the circle. Now, we know how the diameter and the radius relates, but how does the circumference relate to, say, the diameter? If we know how it relates to the diameter, it's very easy to figure out how it relates to the radius. Well, many thousands of years ago, people took their tape measures out, and they keep measuring circumferences and radiuses. And let's say when their tape measures weren't so good, let's say they measured the circumference of the circle, and they would get, well, it looks like it's about 3. And then they measure the radius of this circle right here, or the diameter of that circle, and they'd say, well, the diameter looks like it's about 1. So they would say, let me write this down. |
Circles radius, diameter, circumference and Pi Geometry Khan Academy.mp3 | Well, many thousands of years ago, people took their tape measures out, and they keep measuring circumferences and radiuses. And let's say when their tape measures weren't so good, let's say they measured the circumference of the circle, and they would get, well, it looks like it's about 3. And then they measure the radius of this circle right here, or the diameter of that circle, and they'd say, well, the diameter looks like it's about 1. So they would say, let me write this down. So we're worried about the ratio of the circumference to the diameter. So let's say that somebody had some circle over here. Let's say they had this circle. |
Circles radius, diameter, circumference and Pi Geometry Khan Academy.mp3 | So they would say, let me write this down. So we're worried about the ratio of the circumference to the diameter. So let's say that somebody had some circle over here. Let's say they had this circle. And the first time, not that good of a tape measure, they measured around the circle. And they said, hey, it's roughly equal to 3 meters when I go around it. And when I measure the diameter of the circle, it's roughly equal to 1. |
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