NotThatSmart

2. The equation of a circle is $(x – h)^2 + (y – k)^2 = r^2$ where $(h,k)$ is the center of the circle and $r$ is the radius. 

Now, we can just plug in the information we were already given in the equestion. 

If the center is (11, -9), we have $h=11$ and $k=-9$ and radius of 12. 

We have $(x-11)^2+(y+9)^2 = 144$. 

Thanks! :)

I don't see a figure posted, but this problem probably involves the pythagorean theorem

Thank! :)

We can use casework to solve this problem. 

If 3 boxes are empty, then there are two possibilites. 

First, we could have 2 boxes with balls in them. 

Case 1 

Let's say one box and has 7 and the other has 1. 

There are 2 ways we can do this. 

Case 2

Now, let's say there is one boy with 6 and one box with 2, 

There are also 2 ways we can do this. 

Case 3

One box has 3 and the other has 5. 

Another 2 ways we can do this. 

Case 4

One box has 4 and the other has 4. 

Only one way to do this. 

$2 * 2 * 2 * 1 = 8$ ways we can do this in total

We choose 2 boxes out of the 5 we could have chosen from, which is just $5 \choose 2$ = 10. 

$10 * 8 = 80$ ways to do 2 boxes. 

Now, let's say only 1 box has all the balls. There are 5 ways to choose a box with all 8 balls in it. 

$80+5=85$. There are 85 ways we could do this!

Thanks! :)

We can use the information we already know to solve this problem!

First off, let's note that $(x+y)^3 = x^{3}+3x^{2}y+3xy^{2}+y^{3}$, which we can simplify to $8=5+6xy$. Solving for xy, we get $xy=1/2$. 

This may seem useless, but it will come into play later. 

Now, we know that $(x+y)^2 = x^2+2xy+y^2$, which contains x^2+y^2. Isolating this, we get $x^2+y^2 = (x+y)^2 - 2xy$. 

We already know all the terms! We can plug in 1/2 and 2 which we know from above! We get $x^2 + y^2 = 4 - 1 = 3$. 

So 3 is our answer!

Thanks! :)

First off, note that $(x+y)^3 = x^{3}+3x^{2}y+3xy^{2}+y^{3}$. 

We already know most of the terms, so we have $8 = 5+3xy(x+y)$, which is the same as $8=5+6xy$. Now, we have $xy = 1/2$. 

You may be wondering why in the world would we want xy, but it will make sense later. 

Now, when we see x^2+y^2, (x+y)^2 instantly comes to mind. 

We know that $(x+y)^2 = x^2+2xy+y^2$. This means that $x^2+y^2 = (x+y)^2 - 2xy$! We already know all these terms! We can easily plug in 2 and 1/2 to solve our problem!

$x^2 + y^2 = 4 - 1 = 3$. 

Our final answer is just 3. 

Thanks! :)

First, we can graph this problem. 

We can use the Pythagorean Theorem to complete this problem. 

First, we need similar triangles. We can extend AB towards the direction of A and meet it up with the line connecting the radiuses. We can call this point O. 

Let the center of the unit circle be N and the center of the larger circle M. Let's let x be the distance from the left edge of the smaller circle to O. 

Notice that triangles BMO and ANO are similar triangles. This means we can write

$BM / MO = AN / NO \\ 4 / (4 + 2 + x) = 1 / ( 1 + x) \\ 4 / ( 6 + x) = 1 /(1 +x) \\ 4(1 + x) = 1 (6 + x) \\ 4 + 4x = 6 + x \\ 3x = 2 \\ x = 2/3 \\ \\ OA = \sqrt {NO^2 - NA^2 } = \sqrt{(1 + 2/3)^2 - 1^2} = \sqrt{25/9 -1 } = \sqrt{ 16/9} = 4/3 \\ OB = \sqrt {MO^2 - MB^2 }= \sqrt{ (6 + 2/3)^2 - 4^2 } = \sqrt {400/9 -16} = \sqrt{ 256/9} = 16/3 \\ AB = OB - OA = 16/3 - 4/3 = 12/3 = 4$

You can also use translation for this problem, but my small brain can't figure it out!

Thanks! :)

I'm not exactly sure what the problem is asking, but I'll but the area in terms of the things we have. 

The volume of the pyramid is 1/3 of the base area mutliplied by the height. 

So we have the area of the pyramid as $\frac{1}{3}[ABCD]h$ where h is the height. We could probably figure out the height by using the paythagorean theorem through the slant height and the diagonal of ABCD!

I hope I answered your question!

Thanks! :)

First of all, let's move almost all the variables to one side of the equation and combine all the like terms. 

We get $x^2 - 21x + y^2 + 13y = 25$. 

Now, we complete the square for both x and y on the right side of the equation. 

$x^2 - 21x + 441/4 + y^2 + 13y + 169/4 = 25 + 441/4 + 169/4$

$(x - 21/2)^2 + (y + 13/2)^2 = 355 / 2$

Wait! This looks like the equation for a circle! 

Indeed, we have a circle at center $(21/2 , -13/2)$ with radius $\sqrt{355 / 2}$. 

The value of the largest possible x is just the radius added onto the x value of the center. 

$x = (21/2) + \sqrt{355 / 2}$

Thanks! :)