Rom

\(\text{To find the remainder when dividing by $(x+1)$ we evaluate the polynomial at $x=-1$}\\ (-12)(-1)^{17} + 3(-1)^5 - 9(-1)^2 -1 = \\ 12-3-9-1=\\ -1\\~\\ \text{D is the correct choice. Well done!}\)

\(\dfrac{x^2+5}{2} = 7\\ x^2 + 5 = 14\\ x^2 = 9\\ x = \pm 3\)

\(\sqrt{\dfrac{20-5x}{x}}=5\\ \dfrac{20-5x}{x} = 25\\ 20-5x=25x\\ 20=30x\\ x=\dfrac 2 3\)

\(\text{Just as an alternative way to calculate this}\\ z = 1+i = \sqrt{2} e^{i \pi/4}\\ z^4 = (\sqrt{2})^4 e^{4(i\pi/4)} = 4e^{i\pi} = -4\)

\(\text{$v$ projected onto $(0,1,0)$ is $(v \cdot (0,1,0))(0,1,0)$}\\ v \cdot (0,1,0) = v_y\\ v \perp (0,1,0) = (0,v_y,0)\\ \text{The matrix that will perform this is}\\ P=\begin{pmatrix}0&0&0\\0&1&0\\0&0&0\end{pmatrix}\)

D

(x-4) as a factor corresponds to the zero being at x=4

\(\dfrac{25}{75} = \dfrac{1}{3} \times 100\% \approx 33.3%\)

\(S = \sum \limits_{k=0}^\infty ~x^k= \dfrac{1}{1-x}\\ \dfrac{dS}{dx} = \sum \limits_{k=1}^\infty k x^{k-1} = \dfrac{1}{(1-x)^2}\\ \sum \limits_{k=1}^\infty k x^k = \dfrac{x}{(1-x)^2}\\~\\ \sum \limits_{k=1}^\infty ~\dfrac{k}{3^k} = \dfrac{\frac 1 3}{\left(1-\frac 1 3\right)^2}= \dfrac 3 4\)

\(\text{just plug the points in and solve the resulting equations}\\ 2^2 + b(2) + c = 3\\ 4^2 + b(4) + c = 3\\~\\ 4 + 2b + c = 3\\ 16+4b+c = 3\\ \text{subtract twice (1) from (2)}\\ 8-c=-3\\ c=11\)

\(\text{The best way to proceed here is to row reduce each matrix to reveal it's rank}\\ \text{You'll need that info for the second part of the question anyway}\\~\\ \text{$\textbf{A}$ is the identify matrix. You should know immediately that it's rank 3 and thus invertible}\\~\\ \text{$\textbf{B}$ has to be row reduced}\\ \text{$\textbf{B} \sim \left( \begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ \end{array} \right)$}\\ \text{We see that $\textbf{B}$ is rank 2 and thus not invertible}\)

\(\text{$\textbf{C} \sim \left( \begin{array}{ccc} 1 & -2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right)$}\\ \text{$\textbf{C}$ is seen to be rank 1 and thus not invertible}\)

\(\text{$\textbf{D}$ row reduces to the identity matrix and is thus rank 3 and invertible}\)

\(\text{As far as the output goes rank 3 outputs to all 3D space, rank 2 to a plane, rank 1 to a line}\\ \text{$\textbf{A}$ and $\textbf{D}$ output to all 3D space}\\ \text{$\textbf{B}$ outputs to a plane}\\ \text{$\textbf{C}$ outputs to a line}\)