Rom

\(3x+1 = 3(x-2) + 7\\ \dfrac{3x+1}{x-2}=\dfrac{3(x-2)+7}{x-2} = 3 + \dfrac{7}{x-2}\\ A=3\\ B=7\)

Let p be the probability of an even roll.

For option A

E[moves]=4p - 3(1-p) = 7p-3

For option B

E[moves]=5p-4(1-p) = 9p-4

Playing with the numbers it turns out that p=P[even roll]=3/4

I suggest you confirm this for yourself.

EA=21/4 - 12/4 = 9/4

EB = 27/4 - 16/4 = 11/4

EB offers the greater likelihood of moving closer to the finish line.

\(\text{In general an arithmetic sequence has the explicit form }\\ s_n = n d + s_0\\ \text{It's recursive form is }\\ s_n = s_{n-1}+d\\ \text{In both cases }s_0 \text{ is specified}\)

\(fine_5=2.25=5d+fine_0\\ fine_{21}=6.25=21d+fine_0\\ fine_{21}-fine_5 = 4= 16d\\ d=\dfrac 1 4\\ 2.25=\dfrac 9 4 = \dfrac 5 4 + fine_0\\ fine_0=1\)

\(a) ~fine_n = fine_{n-1}+0.25,~fine_0=1\\ b)~fine_n = (0.25)n + 1\)

\(c) ~\text{Hopefully clearly he'll choose the explicit formula. Think about it.}\)

\(d)~fine_{60}=60(0.25)+1 = 16\)

\(\text{Given the removable singularity at }x=1 \text{, and the vertical asymptote at }x=-1\\ \text{and the fact that }q(x) \text{ is quadratic}\\ q(x) =q_0 (x-1)(x+1)\\ \text{because of the removable singularity we also know that }p(x) \text{ has }(x-1) \text{ as a factor}\)

\(q(2)=3\\ q_0 (1)(3)=3\\ q_0=1\\ q(x) = (x+1)(x-1)=x^2-1\)

\(\text{As }\dfrac{p(x)}{q(x)} \text{ has a horizontal asymptote of }0\\ \text{the degree of }p(x) \text{ must be less than that of }q(x)\\ \text{so }p(x)=p_0(x-1)\\ p(2)=1\\ p_0=1\\ p(x)=x-1\)

\(p(x)+q(x) = (x-1)+(x^2-1) = x^2 + x - 2\)

The first two primes greater than 20 are 23 and 29.

23 x 29 = 667

\(f(x) \text{ is even, so }f(-x)=f(x)\)

\(g(2)=f(2+3)-5 = f(5)-5=-2\\ f(-5)=f(5)\\ f(-5)-5=-2\\ f(-8+3)-5 = -2\\ g(-8)=-2\\ (a,b)=(-8,-2)\)

a quadriateral with two sets of adjacent sides having equal length.

Think of the usual example of a kite.

All the choices are off by a factor of 5 or more just based on the volume of the warehouse and the volume of the cylinders

\(\text{A line has the same slope at all points so }\\ m = \lim \limits_{t\to \infty}\dfrac{y(t)}{x(t)} = -\dfrac 3 2\\ \dfrac{b}{a}=-\dfrac 3 2\\ b=-\dfrac{3a}{2}\\ a+b=3\\ a+\left(\dfrac{-3a}{2}\right)=3\\ -\dfrac{a}{2}=3\\ a = -6\\ b = 9\\ <-6,9>\)

\(x \in \mathbb{Z} \Rightarrow \lfloor x \rfloor + \lfloor -x \rfloor = 0\\ x \not \in \mathbb{Z} \Rightarrow \lfloor x \rfloor + \lfloor -x \rfloor = -1\)