Rom

\(\forall -\dfrac 2 3 \leq x, ~\min\left(2x+2,~\dfrac 1 2 x + 1, ~\dfrac 3 4 x + 7\right) = \dfrac x 2 + 1 \\ \forall x < -\dfrac 2 3, ~\min\left(2x+2,~\dfrac 1 2 x + 1, ~\dfrac 3 4 x + 7\right) = 2x+2 \\ x \to \infty \Rightarrow \dfrac x 2 + 1 \to \infty \\ \text{and thus the Min of these three functions is unbounded.}\\ \text{One might say it's maximum values is }\infty \\ \text{but it's probably better to just say it has no maximum value}\)

\(5^{23} \text{ is odd}\\ 7^{17} \text{ is odd} \\ 5^{23} + 7^{17} = \text{odd} + \text{odd} = \text{even} \\ \text{therefore 2 will the smallest prime divisor}\)

\(x^2 - 6x + 66 = \\ x^2 - 6x + 9 - 9 + 66 = \\ (x-3)^2 + 57 \\ b=-3,~c=57\)

Do you live in Norco?

\(\text{what they are probably after is } \tilde{\mu} = \dfrac{18}{50} \\ n_{pred}=n \cdot \tilde{\mu} = 125 \cdot \dfrac {18}{50} = 45\)

I helped get a horse cast in her stall to her feet today. 

Probably saved her from being euthed from a broken leg.

That was pretty favorite.

\(3b=8-2a\\ 2b+12 = \dfrac 23(8-2a)+12=\\ \dfrac{16-4a+36}{3} = \dfrac{52-4a}{3} \\ \text{If I understand the question were are looking for the values of }\\ a \in \{1,2,3,4,5,6\} \text{ such that } \dfrac{52-4a}{3} \in \mathbb{Z}\)

\(\dfrac{52-4a}{3} = 16+\dfrac{4-4a}{3} \text{ so we want values of }a \ni \dfrac{4-4a}{3} \in \mathbb{Z}\\ 4-4a = \{0, -4, -8, -12, -16, -20\}\\ \text{ and the two of those divisible by 3 are 0 and -12} \\ \text{corresponding to }a=1,~a=4\\ \text{ so there are 2 values of a that satisfy the original statement}\)

\(\text{the polynomial will be of the form}\\ p(x) = k(x+1)(x-1)(x-3) \\ p(2) = k(3)(1)(-1) = 5 \\ -3k = 5 \\ k=-\dfrac 5 3 \\ p(x) = \dfrac 5 3 (x+1)(x-1)(x-3) \\ p(x) = -\dfrac{5 x^3}{3}+5 x^2+\dfrac{5 x}{3}-5\)

\(total_{all} = (n_{boys}+n_{girls})m_{total} = (10+20)60 = 1800 \\ total_{girls} = n_{girls}m_{girls} = 20 \times 54 = 1080 \\ total_{boys} = total_{all}-total_{girls} = 1800 - 1080 = 720\\ m_{boys} = \dfrac{total_{boys}}{n_{boys}}= \dfrac{720}{10}=72\)

kind of a sexist problem :P

\(z = 5 e^{i \theta} \\ z^3 = 5^3 e^{i3\theta} \\ |z^3| = 5^3 = 125\)