BuilderBoi

Now that I think of it, here's another way of reaching the same answer. 

We already know that \(CB = 12\sqrt2\) (look at the other solution to see how)

Now, let \(CE = a\) and \(AE = b\). We have \(a^2 + b ^2 = 81\), and \(a^2 + (b+9)^2 = ({12 \sqrt2})^2 = 288\). 

Subtracting the two equations gives us \(18b + 81 = 207\), meaning \(b = 7\). 

Plugging this into the first equation also gives us \(a = 4 \sqrt 2\)

First, we know that \(DB =DC = \sqrt{9^2 - 3^2} = 6\sqrt{2}\), which means \(CB = 12 \sqrt 2\)

Also, note that \(\triangle CEB \sim \triangle ADB\) by AAA. This is because both triangles have a right angle and share \(\angle B\), which means that all 3 traingles have the same angles, and thus, are similar. 

Now, by similiar triangles, we have \({AB \over CB} = {AD \over CE}\). Substituting what we know, we have \({9 \over 12 \sqrt 2} = {3 \over CE}\), meaning \(CE = \color{brown}\boxed{4 \sqrt2}\).

To solve for AE, we use the Pythagorean Theorem on \(\triangle AEC\), and find that \(AE = \sqrt{9^2 - {4\sqrt2}^2} = \color{brown}\boxed{7}\)

\(\sqrt[3]{250} = 5 \sqrt[3]{2}\)

\(\sqrt[3]{54} = 3\sqrt[3]2\)

\(5\sqrt[3]{2} + 3\sqrt[3]{2} = 8\sqrt[3]{2}\)

\(\sqrt[3]{8\sqrt[3]{2}} = \sqrt[3]{8} \times \sqrt[3]{ \sqrt[3]{2}} = 2\sqrt[9]{2}\)

\(2 + 9 + 2 = \color{brown}\boxed{13}\)

b) Start by giving each kid a sticker. Now, there are 15 stars and 4 bars for \({15 + 4 \choose 4} = \color{brown}\boxed{3,876}\) cases

a) There are 20 stars and 4 bars, so there are \({20 + 4 \choose 4} = \color{brown}\boxed{10,626}\) ways to give out the stickers

We have R _ R _ R

For each of the blank spaces, there are 2 choices, so there are \(2 \times 2 = \color{brown}\boxed{4}\) cases.

https://web2.0calc.com/questions/probability_46725#r2

Note that that \(\triangle XYD\) has a base of 12. This means that the height of the triangle can be at most 10/3. 

We want to find the shaded region's area over the triangle's area, which is 36. 

The area of the shaded region is the same as the area of the triangle minus the area of the non-shaded region. 

The nonshaded region is a right triangle similar to \(\triangle XYZ\) with a scale factor of 4/9, so its area is \(36 \times ({4 \over 9})^2 = {64 \over 9}\). 

This means that the shaded region has an area of \(36 - {64 \over 9} = {260 \over 9}\), so the probability is just \({{260 \over 9} \over {36}} = \color{brown}\boxed{65 \over 81}\)

The perimeter is \(2(5 + x)\). 

The area is \(5x\). 

Setting these equal gives us \(2(5 + x) = 5x\). 

Now, solve for x

There are \({12 \choose 3} = 220\) ways to choose the white balls, and there are 8 ways to choose the SuperBall.

There is only 1 way to match all 4 numbers, so the probability is \({1 \over 220 \times 8} = \color{brown}\boxed{1 \over 1760}\)