Rom

You must not be asking this correctly.

The decimal representation for any integer terminates.  They have no non-zero digits to the right of the decimal point at all.

\(i \text{ is the value such that }i^2 = -1 \\ \text{you'll often see it written that }i=\sqrt{-1} \\ \text{but this is a bit of abuse of notation}\)

\((7-3i)(5i+4)-(3-i) = \\ 35i+28-15i^2-12i - (3-i) = \\ 28+15+(35-12)i - (3-i) = \\ 43+23i -3 + i = \\ 40+24i\)

\(P=b-1\\ (b-1)^2 = 31_b = 3b+1\\ b^2 -2b+1 = 3b+1\\ b^2 = 5b \\ b=5 \text{ (or 0, but 0 is not a valid base)}\\ b=5, ~P=4\\ (4_5)^2 = 16_{10} = 31_5\)

\(\text{Assuming }f(x) \text{ is invertible }\forall x \in \mathbb{R} \\ \text{then }\forall k\neq 0, \exists y = k x \ni f^{-1}(y) = k x \\ \text{i.e. }f(kx) \text{ is invertible.}\\ \text{Thus it must be the only value of k that makes the original statement is valid is }\\ k=0 \\ \text{and the sum of this is just 0} \\ \text{It's true that }\exists f(0) \ni f^{-1}(f(0))=0 \text{ but }f(0\cdot x) \text{ is not invertible}\)

If you were worried it had illegal material on it wth did you pick it up for?

Apply a hammer repeatedly to it or toss it into a fire.

\(t_1 = 1\\ t_{n+1} = t_n + n \\ t_2 = 1 + 2 = 3\\ t_3 = 3 + 3 = 6 \\ t_4 = 6 + 4 = 10 \\ t_5 = 10 + 5 = 15\)

This matches up with choice (c)

here's a little trick

\(9x = 10x - 1x \\ 9(2.333\bar{3}) = 10(2.333\bar{3}) - 2.333\bar{3} = \\ 23.333\bar{3} - 2.333\bar{3} = 21 \\ x = \dfrac{21}{9} = \dfrac 7 3\)

\(\text{Assume they both spent }N \text{ hours doing homework}\\ V_J = \dfrac{N}{5}\times 3 = \dfrac 3 5 N\\ V_L = \dfrac 1 2 N \\ \dfrac 3 5 > \dfrac 1 2 \\ \text{Jonathan gets to spend more time playing video games}\)

\(\text{Consider a win on the nth roll.}\\ \text{This roll sequence looks like }\\ \underset{n_T}{\underbrace{TTT\dots T}}\underset{n_H}{\underbrace{HHH\dots H}}T\\ 0\leq n_T \leq n-2,~1\leq n_H \leq n-1,~n_T+n_H +1 = n\)

\(\text{Thus there are }n-1 \text{ ways to accomplish a win on the nth roll}\\ \text{Each sequence has probability }p_n = 2^n \text{thus the probability of a win on the nth roll is given by}\\ P[n] = \dfrac{n-1}{2^n},~~n\in \mathbb{N},~2 \leq n\)

\(E[N] = \sum \limits_{n=2}^\infty~\dfrac{n(n-1)}{2^n}=6-2= 4\)

You can apply the same idea to solve 2 and 3

Below is the table of successive differences showing that the degree is 4

\(\left( \begin{array}{cc} 0 & \{63,8,-1,0,-1,8,63\} \\ 1 & \{55,9,-1,1,-9,-55\} \\ 2 & \{46,10,-2,10,46\} \\ 3 & \{36,12,-12,-36\} \\ 4 & \{24,24,24\} \\ \end{array} \right)\)