BuilderBoi

\({1 \over \sqrt{2} + \sqrt{8} + \sqrt{32} }\)

\({1 \over \sqrt{2} + 2\sqrt{2} + 4\sqrt{2} }\)

\({1 \over 7\sqrt{2} }\)

\({1 \over 7\sqrt 2} \times { \sqrt{2} \over \sqrt {2}} \)

\({\sqrt 2 \over 14} \)

\(2 + 14 = \color{brown}\boxed{16}\)

If 2 freshmen are at both ends, there are \(5 \times 4 \times 3 \times 2 = 60\) ways to order the freshman. 

Then, there are \(5!\) ways to order the rest (note that there will always be 2 sophomores next to each other). 

So, there are \(5! \times 60 = \color{brown}\boxed{7200}\) ways to order the students. 

Adding the two equations gives us \(2x = {184 \over 63}\), meaning \(x = {92 \over 63}\).

This means that \(y = {43 \over 63}\), so \(x^2 - y^2 = ({92 \over 63})^2 - ({43 \over 63})^2 = \color{brown}\boxed{5 \over 3}\)

We need to split this problem up into 4 cases: 

Case 1 - The last number is not a 1, 2, or 4: There are \({7! \over 3! 2!} = 420\) cases

Case 2 - The last number is a 1: There are \({7! \over 3!2!2!} = 210\) cases

Case 3 - The last number is a 2: There are \({7! \over 3!3!} = 140\) cases

Case 4 - The last number is a 4: There are \({7! \over 4!2!} = 105\) cases

So, there are \(420 + 210 + 140 + 105 = \color{brown}\boxed{875}\) cases.

Note that \(\triangle ABC \) is equilateral, so angle bisector \(AD\) is also an altitude of a 30-60-90 triangle. 

This means that the hypotenuse is \(24 \times {2 \sqrt3 \over 3} = 16 \sqrt 3\)

So the area of the triangle is \((16 \sqrt 3)^2 \times {\sqrt3 \over 4} = \color{brown}\boxed{192 \sqrt 3}\)

A picture would be nice...

Expected value of what? drawing one card, or two cards?

Split the hexagon into 6 equilateral triangles, and let the sides of the hexagon be s.

The longest diagonal would be the one between two opposite vertices, which is comprised of the sides of 2 equilateral triangles with side length s, so its length is 2s.

The perimeter of the hexagon is 6 times the side of the hexagon or 6s. 

So, the ratio is \({2s \over 6s} = {\color{brown}\boxed{1 \over 3}}\)

Note that \(9^4 = 9 \times 9^3\) and \(9^5 = 81 \times 9^3\)

We have the inequality \(9 \times 9^3 < x \times 9^3 <81 \times 9^3\), which is equal to \(9

The possible cases for x are 10, 11, 12, ... 80. 

Thus, there are a total of \(80 - 10 + 1 =\color{brown}\boxed{71}\) cases.

\(\color{brown}\boxed{\pm 10}\)