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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 \(\boxed{255}\)