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Let $f(x)=ax^2+bx+c$. Note that $f(1) = a \cdot 1^2 + b \cdot 1 + c = a + b + c$, so $a + b + c = f(1) = \boxed{-4}$.

Note that $\frac{1}{a^2}+\frac{1}{b^2}=\frac{a^2+b^2}{a^2b^2} =\frac{(a+b)^2 - 2ab} {(ab)^2}$. By Vieta's, $a+b = \frac{-(-11)}{5} = \frac{11}{5}$ and $ab = \frac{4}{5}$, so $$\frac{(a+b)^2 - 2ab}{(ab)^2} = \frac{\frac{121}{25} - 2 \cdot \frac{4}{5}}{\left(\frac{4}{5}\right)^2} = \frac{\frac{81}{25}}{\frac{16}{25}} = \boxed{\frac{81}{16}}.$$

By Chinese Remainder Theorem, there exists a unique residue $\pmod{315}$ for $N$. Hence, all we need to do is find a possible residue, and that will be the unique residue for $N$ when taken $\pmod{315}$. By $N \equiv 3 \equiv -4 \pmod{7}$ and $N \equiv 5 \equiv -4 \pmod{9}$, we get that $N \equiv -4 \pmod{63}$. Hence, $N = 63k + 59$ for some integer $k$.

Combining this with the last condition, we get $63k + 59 \equiv 3k + 4 \equiv 4 \pmod{5}$, so $3k \equiv 0 \pmod{5}$. Multiplying both sides of this congruence by $2$, we get that $6k \equiv k \equiv 0 \pmod{5}$, so hence $k \equiv 0 \pmod{5}$. Thus, $N \equiv 63 \cdot 0 + 59 \equiv 59 \pmod{315}$. 

Hence, we want to find the smallest four-digit $N$ where $N \equiv 59 \pmod{315}$, which gives us an answer of $315 \cdot 3 + 59 = \boxed{1004}$.

Let $f(x) = ax^3 + bx^2 + cx + d = (3x^2 - 5x + 4)(7 - x)$. Note that $8a + 4b + 2c + d = f(2)$, so hence we can just plug $x = 2$ into $(3x^2 - 5x + 4)(7 - x)$ for an answer of $$(3 \cdot 2^2 - 5 \cdot 2 + 4)(7 - 2) = (12 - 10 + 4)(7 - 2) = (6)(5) = \boxed{30}.$$

We get that he ran for $25 \cdot 7 = 175$ minutes, which is equal to $\boxed{\textbf{(C) }2\text{ hours and }55\text{ minutes}}$.

Let $x = \frac 1z$. Then $2x + 5 = 5x - 4 - 2x - 8 \implies 2x + 5 = 3x - 12 \implies 5 = x - 12 \implies x = 17$. Hence, $z = \boxed{\frac 1{17}}$.

By synthetic division, we get that $p(x) = (x - 2)(x^3 - x^2 - 20x + 50)$. By Rational Root Theorem, the possible roots of $x^3 - x^2 - 20x + 50$ are $\pm 1, \pm 2, \pm 5, \pm 10, \pm 25$, and $\pm 50$. Testing values, we get that $x = -5$ works, so by synthetic division, we get that $p(x) = (x - 2)(x + 5)(x^2 - 6x + 10)$.

By the quadratic formula, the roots of $x^2 - 6x + 10$ are $\frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 10}}{2 \cdot 1} = \frac{6 \pm 2i}{2} = 3 \pm i$. Thus, the roots of $p(x)$ are $\boxed{-5, 2, 3 + i, \text{and } 3 - i}$.

Note that $512 = \boxed{2^9} = 2^{3 \cdot 3} = (2^3)^3 = \boxed{8^3}$.

Squaring both sides, we get that $\left(m + \frac 1m \right)^2 = 8^2$. Expanding gives $m^2 + \frac{1}{m^2} + 2 = 8^2$, so hence $m^2 + \frac{1}{m^2} = 8^2 - 2 = 62$. Squaring both sides again, we get that $\left(m^2 + \frac{1}{m^2}\right)^2 = 62^2$. Expanding gives $m^4 + \frac{1}{m^4} + 2 = 62^2$, so hence $m^4 + \frac{1}{m^4} = 62^2 - 2 = \boxed{3842}$.

Let the measure of an exterior angle of the polygon be $x$. Then $x + 10x = 180^\circ$, so $x = \left(\frac{180}{11}\right)^\circ$. The measure of an exterior angle of a regular polygon with $\ell$ sides is equal to $\left(\frac{360}{\ell}\right)^\circ$. Thus, $\left(\frac{360}{\ell}\right)^\circ = \left(\frac{180}{11}\right)^\circ$, so the polygon has $\boxed{22}$ sides.

We can express all of the terms with exponent base $3$, giving $2y = 5 + 12 + 16 - 9 = 24 \implies y = \boxed{12}$.

Note that $\triangle ACE$ and $ABDE$ have the same base and height, and since the area of a rectangle is equal to $bh$ and the area of a triangle is equal to $\frac{bh}{2}$, we get that $\frac{[\triangle ACE]}{[ABDE]} = \boxed{\frac{1}{2}}$.

This is equivalent to the question "How many numbers can be expressed with 8 or less digits in binary?" 

Hence, every number from $1_2$ to $11{,}111{,}111_2$ works, for an answer of $\boxed{255}$