Rom

\(x \cdot \dfrac{130}{100} = 69\\ x = 69 \dfrac{100}{130} = \dfrac{690}{13} \approx 53.077\)

\(\text{It can't be 70}\\ A_{board} =32^2 = 1024 ~sq~in\\ A_{photo}=3\cdot 5 = 15 ~sq ~in\\ \dfrac{1024}{15} = 68+\dfrac{4}{15}\\ \text{so with no overlapping 68 is the maximum possible, and is probably not attainable}\)

\(\sqrt{a}-\sqrt{b}=20\\ a = (20+\sqrt{b})^2\\ a-5b = (20+\sqrt{b})^2-5b\\ \dfrac{d}{db} (20+\sqrt{b})^2-5b = \\ 2(20+\sqrt{b})\cdot \dfrac{1}{2\sqrt{b}} - 5\\ \text{and we set this equal to 0}\)

\(\dfrac{20}{\sqrt{b}}+1 = 5\\ \sqrt{b}=5\\ b=25\\ \sqrt{a}-5=20\\ a=625\)

\(\text{Max value of $a-5b=625-5(25)=500$}\)

it means on average you win less than the 50c you paid to play.

\(\text{Most of this is red herrings}\\ \text{Given any event $B, ~P[B^c] = 1 - P[B]$}\\ P[B] = 0.35\\ P[B^c] = 1 - 0.35 = 0.65\)

\(-x^2 \leq 0\\ -x^2+5 \leq 5\\ y\leq 5\)

\(\text{The key thing is that the sum of the length of any two sides is greater than the length of the third side}\\ 7+10 > x^2\\ 17>x^2\\~\\ x^2+7>10\\ x^2 > 3\\ \text{integer values of $x$ are $2,3,4$}\)

only if you post it

\(N=12+15+25 = 52\\ P[70c]=\dfrac{12}{52}=\dfrac{3}{13}\\ P[40c]=\dfrac{15}{52}=\dfrac{5}{13}\\ P[30c]=\dfrac{25}{52}\\~\\ E=70 \dfrac{3}{13}+40 \dfrac{5}{13}+30\dfrac{25}{52}-50 = -\dfrac{205}{26}\approx -\$0.08\)