Rom

\(p = \dfrac{11}{50}\dfrac{10}{49}\dfrac{9}{48} = \dfrac{33}{3920}\)

\(\text{Rational coefficients means that if $\sqrt{2}+\sqrt{3}$ is a zero of the polynomial then}\\ \text{$-\sqrt{2}+\sqrt{3},~\sqrt{2}-\sqrt{3},$ and $-\sqrt{2}-\sqrt{3}$ must be as well}\\ p(x) = (x-(\sqrt{2}+\sqrt{3})) (x-(-\sqrt{2}+\sqrt{3})) (x-(\sqrt{2}-\sqrt{3})) (x-(-\sqrt{2}-\sqrt{3})) = \\x^4-10 x^2+1\)

\(p = \dfrac{\dbinom{7}{6}\dbinom{5}{4}\dbinom{6}{2}}{\dbinom{18}{12}} = \dfrac{25}{884}\)

\(\text{In order to win a player either has to roll 7 or 11, or the other player must roll 3, 4, or 18}\\ \text{Basically we are finding 3 element integer partitions of these numbers}\\ \text{In Kenji's case we have the element values 1-6 available 3 times each}\\ \text{In Manbu's case we have 1-4, 1-6, 1-8 available, once each}\\ P[\text{Kenji win}]=\dfrac{42}{216} + \dfrac{5}{192}-\dfrac{42}{216}\dfrac{5}{192} = \dfrac{1489}{6912} \approx 0.215422 \\~\\ P\text{Manbu win}] = \dfrac{37}{192}+\dfrac{5}{216}-\dfrac{37}{192}\dfrac{5}{216}=\dfrac{8767}{41472} \approx 0.211396\\ \text{Kenji has a slight advantage of winning on the first roll}\)

\(\text{Pick a square with probability 1}\\ \text{There are then 14 possible squares co-linear with this first square out of 63 possible squares}\\ \text{Having chosen the first two, there are 6 possible squares that will be co-linear with them out of 62 possible squares}\\ p = \dfrac{14}{63}\dfrac{6}{62} = \dfrac{2}{93}\)

\(ab-3a+4b=137\\ b(a+4)=137+3a\\ b = \dfrac{137+3a}{a+4},~a\neq -4\)

\(a - b = \\ a - \dfrac{137+3a}{a+4},~a \neq -4 = \\ \dfrac{a^2+4a-137-3a}{a+4}= \\ \dfrac{a^2 +a -137}{a+4}\)

\(\text{This will have a minimum absolute value of 0 when the numerator equals 0}\\ a^2 + a - 137 = 0\\ a = \dfrac{-1 \pm \sqrt{1+(4)(137)}}{2} = \dfrac{1}{2} \left(-1\pm 3 \sqrt{61}\right)\)

\(\text{Pick whether A is second or third. That's two choices. Pick the remaining letter. That's 6 choices}\\ \text{Thus a total of $2\cdot 6 = 12$ choices meeting the requirements}\\ p = \dfrac{1}{12}\)

\(\text{What they want you to recognize here is that if a polynomial with real coefficients has a complex zero}\\ \text{Then it's conjugate is also a zero}\\ \text{so we have }\\ p(x) = (x-5)(x+3)(x-(-1+3i))(x-(-1-3i)) = x^4-9 x^2-50 x-150\)

\(2x-5y=3\\ y = \dfrac{2x-3}{5} = \dfrac 2 5 x - \dfrac 3 5\\ m = \dfrac 2 5\\ m^\perp= -\dfrac{1}{m} = -\dfrac 5 2\\ n = \left(1, -\dfrac 5 2\right)\\ \hat{n} = \dfrac{2}{\sqrt{29}}\left(1,-\dfrac 5 2\right) = \left(\dfrac{2}{\sqrt{29}}, -\dfrac{5}{\sqrt{29}}\right)\)

\(p=\dfrac{\dbinom{8}{3}}{\dbinom{27}{3}} = \dfrac{56}{2925}\)