CubeyThePenguin

\(f(x) - g(x) + h(x) = (2x^2 + 3x - 9) - (5x + 11) + (-3x^2+1)\)

\(\boxed{-x^2-2x-19}\)

Let the side length of each triangle be 1. The total perimeter of the corrals is 12, so the side length of the square corral is 3.

Area of triangles: \(4(\frac{\sqrt{3} \cdot 1^2}{4}) = \sqrt{3}\)

Area of square: \(3^2 = 9\)

Ratio: \(\boxed{\frac{\sqrt{3}}{9}}\)

\(s(x) = (4x-7) + (x+1)^2 = x^2 +6x - 6\)

\(s(3) = 3^2 + 6(3) - 6 = \boxed{21}\)

\(5x \geq -85\)

\(\boxed{x \geq -17}\)

There are a total of \({7 \choose 2} = 21\) different pairs. We will count the pairs where the difference is 1 or 2. Note that a difference of 0 is not possible since the numbers have to be distinct.

Case 1: difference = 1

7-6, 6-5, 5-4, 4-3, 3-2, 2-1

6 pairs

Case 2: difference = 2

7-5, 6-4, 5-3, 4-2, 3-1

5 pairs

\(\text{probability} = 1 - \frac{11}{21} = \boxed{\frac{10}{21}}\)

If an integer is odd, then it is of the form \(2n+1.\)

Square it: \((2n+1)^2 = 4n^2 + 4n +1 = 4(n^2 + n) + 1 \equiv 1 \quad (\mod 4)\)

The equation simplifies to:

\((x-1)(x-2)(x-3) \dots (x-100) = 0\)

There are 100 solutions, namely x = 1 to x = 100.

\(f(x) = f(f(x))\)

\(x^2 - 2x = (x^2-2x)^2 - 2(x^2-2x)\)

Let \(x^2 - 2x = y.\)

\(y = y^2 - 2y\)

\(0 = y^2 - 3y\)

\(y = 0, 3\)

Now plug x back in.

\(x^2 - 2x = 0\)

\(\boxed{x = 0, 2}\)

\(x^2 - 2x = 3\)

\(\boxed{x = 3, -1}\)

\(f(g(x)) = f(x-3)\)

\(f(x-3) = (x-3)^2 + 1 = \boxed{x^2 - 6x + 10}\)