Rom

\(\text{For this problem they probably just want you to list out the 4 series elements and add them}\\ 100(1 + (-4) + 16 + (-64)) = 100(-51) = -5100\)

\(\text{Just eyeballing it you've got a regular hexagon with side length equal to}\\ \text{one third the side of the triangle. Let that triangle side length be }s\\ A_t = \dfrac{\sqrt{3}}{4}s^2\\ A_h = \dfrac{3\sqrt{3}}{2}\left(\dfrac{s}{3}\right)^2 = \dfrac{\sqrt{3}}{6}s^2\\ \dfrac{A_h}{A_t} = \dfrac{\frac{\sqrt{3}}{6}}{\frac{\sqrt{3}}{4}}=\dfrac 2 3\)

another way of doing (1)

\(h(x)=f(g(x))\\ h^{-1}(x) = g^{-1}(f^{-1}(x))\)

\(f^{-1}(x) = \dfrac{x-5}{7}\\ g^{-1}(x) = x+1\\ g^{-1}(f^{-1}(x))=\dfrac{x-5}{7}+1 = \dfrac{x+2}{7}\)

\(p(q(-1))=4(-1)+7 = 3\\ \text{linear functions are a bijection, i.e. }\\ p(x)=a \text{ and } p(y)=a \Rightarrow x = y\\ p(2)=3 \text{ and } p(q(-1))=3 \Rightarrow q(-1)=2\)

\((x+8)^2 = 44\\ (x+8)=\pm \sqrt{44} = \pm 2\sqrt{11}\\ x = -8 \pm 2\sqrt{11}\)

\(\text{Can always apply the quadratic formula}\\ x = \dfrac{9\pm \sqrt{81-4(5)(8)}}{10} = \\ \dfrac{9\pm\sqrt{81-160}}{10} = \\ \dfrac{9\pm i \sqrt{79}}{10}\)

\(\text{Couple ways to do this}\\ 1)~\text{convert both to proper fractions and multiply}\\ 19\dfrac 3 4 = \dfrac{79}{4}\\ 1\dfrac 2 5 = \dfrac 7 5\\ \dfrac{79}{4}\dfrac{7}{5} = \dfrac{553}{20}\)

\(2) \text{distribute }\\ \left(19+\dfrac 3 4\right)\left(1+\dfrac 2 5\right) = \\ 19 + \dfrac 3 4 + \dfrac{38}{5} + \dfrac{6}{20} = \dfrac{553}{20}\)

It's tough to tell what your answer is but

P[A AND B] = 0.15 = (0.75)(0.2) = P[A] P[B] so

P[A AND B] = P[A] P[B]

This means A and B are independent events

It's the quotient of some number, x, and 4

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