Rom

\(g(m)=5x+14\\ g(3) = 5(3)+14 = 29~gal\\ \text{it should be pretty obvious that the water in the pool increases at 5 gallons per minute}\\ g(0)=14~gal \text{ so there must have initially been 14 gallons in the pool}\)

This is a permutation.

There are 5! = 120 different permutations of 5 distinguishable objects taken 5 at a time.

deleted

\(\text{tracy had }n \text{ candies}\\ \text{she ate }\dfrac 1 3 \text{ of these leaving }\dfrac{2n}{3}\\ \text{she gave }\dfrac 1 4 \text{ of them to Rachel leaving }\dfrac 3 4 \dfrac{2n}{3}=\dfrac n 2\\ \text{Tracy and mom ate 30 candies from the }\dfrac{n}{2} \text{ that tracy had left}\\\text{Tracy's brother took from }1-5 \text{ candies, leaving }3 \text{ left}\)

\(\text{so condensing that down }\\ \dfrac n 2 - 30 -b = 3,~b\in 1,2,\dots 5\\ n=2(33+b)\\ \text{we note that }\dfrac n 3 \in \mathbb{N}\\ \text{so it must be that }b=3,~n=72\)

\(y = \dfrac 1 4 t^4 - t^2 + 5,~t\in [0,4]\)

\(\dfrac{dy}{dt} = t^3 -2t\\ \dfrac{dy}{dt}(1)= 1-2 = -1~m/s\)

\(\text{The fact that the sign of }\dfrac{dy}{dt} \text{ is negative shows that}\\ \text{the birds height is decreasing at }t=1\)

\(\text{Fido can reach a circle of radius }r\\ A_c = \pi r^2 \\ \text{the apothem of the hexagon is also }r \\ \text{a side will be such that }\tan(30^\circ) = \dfrac{\frac s 2}{r}\\ s = \dfrac{2r}{\sqrt{3}}\\ A_h = \dfrac 1 2 s a n = \dfrac 1 2 \cdot \dfrac{2r}{\sqrt{3}}\cdot r\cdot 6 = 2\sqrt{3}r^2\\ \dfrac{A_c}{A_h} = \dfrac{\pi r^2}{2\sqrt{3}r^2} = \dfrac{\pi}{2\sqrt{3}}= \dfrac{\sqrt{3}}{6} \pi\\ a = 3,~b=6,~ab=18\)

I'm showing 128 multiples of 7

618 total

Suppose we take this problem statement directly at it's word.

\(\text{The sum of the divisors of }n \text{ is }1 + \sqrt{n} + n \text{ i.e.}\\ \text{the divisors are in fact }1,~\sqrt{n},~n\)

\(\text{This implies 2 things}\\ n = p^2 \text{ for some integer }p\\ \text{all of the divisors of }n \text{ are }1, ~p,~p^2\\\)

\(\text{Thus we are looking at the squares of the prime numbers}\)

I'll let you assemble a list of the squares of the primes that are less than 2010.

You might also try to prove that the squares of the primes are the only numbers that satisfy the requirement.

\(n=4 \text{ is as far as you can go and has domain }x=\{16\} \\ f_4(16)=0\)

if you can figure the last one out you can figure this one out