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! :)