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\begin{coverpages}
\begin{center}\huge MATH 1101 - EXAM 2 SPRING 2017\\
SOLUTION\\
\vspace*{2in}
\large Name: \makebox[3in]{\hrulefill}\\
\end{center}

\begin{flushleft}\vspace{0.5in}
Thursday 9 March 2017 \\
Instructor: Tom Cuchta
\end{flushleft}

\textbf{Instructions:}
\begin{itemize}
	\item  Show all work, clearly and in order, if you want to get full
	credit. If you claim something is true \textbf{you must show work backing up your claim}. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct).
	\item Justify your answers whenever possible to ensure full credit. 	
	\item  Circle or otherwise indicate your final answers.
	\item  Please keep your written answers brief; be clear and to the point.
	\item  Good luck!
\end{itemize}
\vspace*{10pt}

\end{coverpages}

\begin{questions}
\question[10] \begin{parts}
\part[3] Solve for $A$: $\dfrac{2A}{C}=B$.\\
\textit{Solution:}
\begin{align*}
	\dfrac{2A}{C}&=B\\
	C\cdot\dfrac{2A}{C}&=B\cdot C\\
	2A&=BC\\
	\dfrac{1}{2}\cdot 2A &= BC\cdot \dfrac{1}{2}\\
	A&=\dfrac{BC}{2}
\end{align*}
\vfill

\part[3] Solve for $T$: $S=\dfrac{5T}{3W-8X}$.\\
\textit{Solution:}
\begin{align*}
	S&=\dfrac{5T}{3W-8X}\\
	(3W-8X)\cdot S&=\dfrac{5T}{3W-8X}\cdot\left(3W-8X\right)\\
	S\left(3W-8X\right)&=5T\\
	\dfrac{1}{5}\cdot S\left(3W-8X\right)&=5T\cdot\dfrac{1}{5}\\
	\dfrac{S\left(3W-8X\right)}{5}&=T
\end{align*}
\vfill

\part[4] Solve for $X$: $S=\dfrac{5T}{3W-8X}$.\\
\textit{Solution:}
\begin{align*}
	S&=\dfrac{5T}{3W-8X}\\
	(3W-8X)\cdot S&=\dfrac{5T}{3W-8X}\cdot\left(3W-8X\right)\\
	S\left(3W-8X\right)&=5T\\
	3SW-8SX&=5T\\
	3SW-8SX+8SX&= 5T+8SX\\
	3SW&=5T+8SX\\
	3SW-5T&=5T+8SX-5T\\
	3SW-5T&=8SX\\
	\dfrac{1}{8S}\cdot\left(3SW-5T\right)&=8SX\cdot\dfrac{1}{8S}\\
	\dfrac{3SW-5T}{8S}&=X
\end{align*}
\vfill

\end{parts}

\question[8] \begin{parts}
\part[4] Tim is 3 years younger than Joe. Together the sum of their ages is $43$. How old are they?\\
\textit{Solution:} Let $t$ be Tim's age and $j$ be Joe's age. Since their ages sum to 43, we have that $t+j=43$. Further, since Tim is 3 years younger than Joe, we know that $t=j-3$. Combining, 
\begin{align*}
	t+j&=43\\
	(j-3)+j&=43\\
	2j-3&=43\\
	2j&=46\\
	j&=23
\end{align*}
So Joe is 23, making Tim 20.
\vfill

\part[4] The measure of the first angle in a triangle is 3 times the measure of the second angle. The measure of the third angle is $10^{\circ}$ less than the measure of the second angle. Use the fact that the sum of angles in a triangle is $180^{\circ}$ to find the measure of each angle.\\
\textit{Solution:}
Let $A$ be the first angle, $B$ the second, and $C$ the third. Then, we know that $A=3B$, $C=B-10$, and $A+B+C=180$. Combining,
\begin{align*}
	A+B+C&=180\\
	3B+B+(B-10)&=180\\
	5B-10&=180\\
	5B&=170\\
	B&=34
\end{align*}
Thus, $B=34^{\circ}$, making $A=102^{\circ}$ and $C=24^{\circ}$.
\vfill
\end{parts}

\question[8] Define $f(x)=3x^2+2x-1$. Compute...
\begin{parts}
\part[2] $f(-2)$\\
\textit{Solution:} $f(-2)=3(-2)^2+2(-2)-1=3(4)-4-1=12-5=7.$
\vfill
\part[2] $f(3)$ \\
\textit{Solution:} $f(3)=3(3)^3+2(3)-1=3(27)+6-1=81+5=86.$
\vfill
\part[4] $f(x+2)$\\
\textit{Solution:} 
\begin{align*}
	f(x+2)&=3(x+2)^2+2(x+2)-1\\
		&=3(x+2)(x+2)+2x+4-1\\
		&=3(x^2+4x+4)+2x+3\\
		&=3x^2+12x+12+2x+3\\
		&=3x^2+14x+15
\end{align*}
\vfill
\end{parts}

\question[6] Find the (largest) domain (in the real numbers) of
\begin{parts}
\part[3] $f(x)=\dfrac{1}{2x-3}$\\
\textit{Solution:} We must ensure there is no division by zero. Thus, we solve $2x-3=0$, which has solution $x=\dfrac{3}{2}$. Thus the domain is ``all real numbers except $\frac{3}{2}$", or $\left(-\infty,\dfrac{3}{2}\right)\cup\left(\dfrac{3}{2},\infty\right)$
\vfill

\part[3] $f(x)=\dfrac{\sqrt{x+1}}{x-5}$\\
\textit{Solution:} We need both the denominator to be non-zero and the argument of the square root to be positive ($\geq 0$). First, for the denominator, we solve $x-5=0$ or $x=5$. For the square root we solve $x+1\geq 0$ or $x\geq -1$. So the domain is ``all real numbers greater than or equal to -1 but not 5."
\vfill
\end{parts}
\newpage

\question[6] A company installs underground cables at a cost of $\$200$ for the first 40 feet and $\$7$ per foot after. 
\begin{parts}
\part[3] Express the cost $C$ in terms of the lengths of table L (for $L>40$ ft). \\
\textit{Solution:} We write $C(L)=200+75(L-40)$
\vfill
\part[3] Use the function in part (a) to determine the cost of $52$ feet of cable.\\
\textit{Solution:} Here, $L=52.$ So, $C(52)=200+75(52-40)=200+900=\$1100$
\vfill
\end{parts}


\question[6] 
\begin{parts}
\part[3] What is the $y$-coordinate of all points on the $x$-axis?\\
\textit{Solution:} All points on the $x$-axis have $y$-coordinate 0.
\vfill
\part[3] Sketch a picture of all points whose $y$-coordinate is greater than $3$. \\
\textit{Solution:} \\

\begin{tikzpicture}[scale=0.7]
	\draw[<->,very thick] (-2,0) -- (6,0);
	\draw[<->,very thick] (0,-2) -- (0,6);
	\draw[dotted] (-2,3) -- (6,3);
	\draw (0,3) node[below left] {$3$};
	\draw[fill=gray,nearly transparent,pattern=north east lines] (-2,3) rectangle (6,6);
\end{tikzpicture}
\vfill
\end{parts}

\question[12] 
\begin{parts}
\part[4] Is $(1,2)$ a solution of $x+y=3$? \\
\textit{Solution:} The point $(1,2)$ says that $x=1$ and $y=2$. Plugging these values into our equation we have $$1+2=3$$ which is \textbf{true} so $(1,2)$ is a solution of $x+y=3$.
\vfill
\part[4] Is $(3,-1)$ a solution of $2x-y=5$? \\
\textit{Solution:} The point $(3,-1)$ says that $x=3$ and $y=-1$. Plugging these values into our equation we have $$2(3)-(-1)=5$$ which is \textbf{not true} so $(3,-1)$ is not a solution of $2x-y=5$.
\vfill
\part[4] Find a solution of $x+4y=9$ with $x$-coordinate $1$.\\
\textit{Solution:} We set $x=1$ and solve for $y$.
\begin{align*}
	1+4y&=9\\
	4y&=8\\
	y&=2
\end{align*}
Thus, $(1,2)$ is a solution to $x+4y=9$.
\vfill
\end{parts}
\newpage
\question[10] 
\begin{parts}
\part[5] Solve the following system of equations by graphing:\\
$$\left\{ \begin{array}{ll}
-2x+y=-2 \\
x-y=2 \\
\end{array} \right.$$
\begin{tikzpicture} 
\draw[step=1.0,black,thin,color=gray,opacity=0.4] (-5,-5) grid (5,5);
\draw[line width=0.1cm] (-5,0) -- (5,0);
\draw[line width=0.1cm] (0,-5) -- (0,5);
\draw[color=white,fill=white] (1,-0.3) circle (1.61ex);
\draw (1,-0.3) node {$1$};
\draw[color=white,fill=white] (2,-0.3) circle (1.61ex);
\draw (2,-0.3) node {$2$};
\draw[color=white,fill=white] (3,-0.3) circle (1.61ex);
\draw (3,-0.3) node {$3$};
\draw[color=white,fill=white] (4,-0.3) circle (1.61ex);
\draw (4,-0.3) node {$4$};
\draw[color=white,fill=white] (-1,-0.3) circle (1.61ex);
\draw (-1,-0.3) node {$-1$};
\draw[color=white,fill=white] (-2,-0.3) circle (1.61ex);
\draw (-2,-0.3) node {$-2$};
\draw[color=white,fill=white] (-3,-0.3) circle (1.61ex);
\draw (-3,-0.3) node {$-3$};
\draw[color=white,fill=white] (-4,-0.3) circle (1.61ex);
\draw (-4,-0.3) node {$-4$};

\draw[color=white,fill=white] (0.3,2) circle (1.61ex);
\draw (0.3,2) node {$2$};
\draw[color=white,fill=white] (0.3,3) circle (1.61ex);
\draw (0.3,3) node {$3$};
\draw[color=white,fill=white] (0.3,4) circle (1.61ex);
\draw (0.3,4) node {$4$};
\draw[color=white,fill=white] (0.3,1) circle (1.61ex);
\draw (0.3,1) node {$1$};

\draw[color=white,fill=white] (0.3,-2) circle (1.61ex);
\draw (0.3,-2) node {$-2$};
\draw[color=white,fill=white] (0.3,-3) circle (1.61ex);
\draw (0.3,-3) node {$-3$};
\draw[color=white,fill=white] (0.3,-4) circle (1.61ex);
\draw (0.3,-4) node {$-4$};
\draw[color=white,fill=white] (0.3,-1) circle (1.61ex);
\draw (0.3,-1) node {$-1$};

\draw[<->, very thick,blue] (-1.5,-5) -- (3.5,5);
\draw[<->, very thick,blue] (5,3) -- (-3,-5);
\draw[blue,fill=blue] (0,-2) circle (1ex) node[left] {$(0,-2)$};
\end{tikzpicture}
\part[5] Solve the following system of equations algebraically: \\
$$\left\{ \begin{array}{ll}
2x+3y=6 \\
x-y=3
\end{array} \right.$$
\textit{Solution:} We solve the second equation for $y$ to find $y=x-3$. Substituting this into the first equation and solving for $x$ we have
\begin{align*}
	2x+3y&=6\\
	2x+3(x-3)&=6\\
	2x+3x-9&=6\\
	5x&=15\\
	x&=3
\end{align*}
Plugging this back into our second equation, we have that $y=3-3=0$. Thus the solution is $(3,0)$.
\vfill
\end{parts}

\question[6] 
\begin{parts}
\part[3] Find a positive angle coterminal with $82^{\circ}$.\\
\textit{Solution:} One such angle could be found by adding a revolution, i.e. the angle $442^{\circ}.$
\vfill
\part[3] Find a negative angle coterminal with $82^{\circ}$. \\
\textit{Solution:} One such angle could be found by making a clockwise revolution (``subtracting" a revolution), i.e. the angle $-278^{\circ}$.
\vfill
\end{parts}


\newpage

\question[12] 
\begin{parts}
\part[4] Convert to degrees: $5^{\circ} 18^{\prime} 59^{\prime \prime}$.\\
\textit{Solution:}
First, $$59''=59''\left(\dfrac{\frac{1}{3600}^{\circ}}{1''}\right)=\dfrac{59}{3600}^{\circ}=0.016388...$$
and $$18'=18'\left(\dfrac{\frac{1}{60}^{\circ}}{1'}\right)=\dfrac{18}{60}^{\circ}=0.3^{\circ}$$
Thus we see that $5^{\circ}18'59''=5^{\circ}+0.3^{\circ}+0.01639^{\circ}=5.31639^{\circ}$
\vfill
\part[4] Convert to radians: $45^{\circ}$.\\
\textit{Solution:}
$$45^{\circ}=45^{\circ}\dfrac{2\pi \text{ radians}}{360^{\circ}}=\dfrac{\pi}{4}\text{ radians}$$
\vfill
\part[4] Convert to degrees: $\dfrac{\pi}{10}$ radians. \\
\textit{Solution:}
$$\dfrac{\pi}{10}\text{ radians} = \dfrac{\pi}{10}\text{ radians} \left(\dfrac{360^{\circ}}{2\pi\text{ radians}}\right)=18^{\circ}.$$
\vfill
\end{parts}

\question[16] 
\begin{parts}
\part[8] Find the value of the six trigonometric functions of an angle $\theta$ that passes through the point $(2,3)$: \\
\textit{Solution:} First, we find $r$: $r=\sqrt{x^2+y^2}=\sqrt{4+9}=\sqrt{13}.$ Then, \\
\begin{tabular}{ll}
{} \\
$\sin(\theta)=\dfrac{3}{\sqrt{13}}$ & $\csc(\theta)=\dfrac{\sqrt{13}}{3}$ \\
{} \\
$\cos(\theta)=\dfrac{2}{\sqrt{13}}$ & $\sec(\theta)=\dfrac{\sqrt{13}}{2}$ \\
{} \\
$\tan(\theta)=\dfrac{3}{2}$ & $\cot(\theta)=\dfrac{2}{3}$
\end{tabular}
\vspace{0.5in}

\part[8] Given that $\sin(\theta)=\dfrac{2}{5}$, find $\cos(\theta)$ and $\tan(\theta)$. \\
\textit{Solution:} Since $\sin\left(\theta\right)=\dfrac{y}{r}=\dfrac{2}{5}$ we take $y=2$ and $r=5$ and solve $x^2+y^2=r^2$ for $x$, giving $x=\sqrt{r^2-y^2}$. Then $x=\sqrt{25-4}=\sqrt{21}.$

Then $\cos\left(\theta\right)=\dfrac{\sqrt{21}}{5}$ and $\tan\left(\theta\right)=\dfrac{2}{\sqrt{21}}.$
\vfill
\end{parts}

\end{questions}

\end{document}
