\documentclass[10pt,a4paper]{exam}
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            \node[shape=circle,draw,inner sep=2pt] (char) {#1};}}
\begin{document}

\begin{coverpages}
\begin{center}\huge MATH 1011 - EXAM 2 FALL 2016\\
\huge SOLUTION
\end{center}

\begin{flushleft}\vspace{0.5in}
Thursday 29 September 2016 \\
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 algebraically 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[5] Circle \textbf{T} for true and \textbf{F} for false.
	\begin{parts}
		\part[1] \textbf{T}\hspace{5pt} \circled{\textbf{F}} \hspace{5pt} $5x^2+3\sqrt{x}+1$ is a polynomial. \\
		\textit{Solution:} The $\sqrt{x}$ that appears makes this not a polynomial.
		\part[1] \circled{\textbf{T}} \hspace{5pt}\textbf{F} \hspace{5pt} The ordered pair $(0,1)$ is a solution of the equation $3y+2x=3$. \\
		\textit{Solution:} Plugging in $x=0$ and $y=1$ into the equation $3y+2x=3$ yields $3(1)+2(0)=3$, which is true. Therefore $(0,1)$ is a solution of $3y+2x=3$.
		\part[1] \textbf{T} \hspace{5pt} \circled{\textbf{F}} \hspace{5pt} The commutative law says that $a(b+c)=ab+ac$. \\
		\textit{Solution:} This is the distributive law, not the commutative law.
		\part[1] \circled{\textbf{T}} \hspace{5pt} \textbf{F} \hspace{5pt} A polynomial with three terms is called a trinomial. \\
		\textit{Solution:} This is true. A binomial has two terms, a trinomial has three terms, etc.
		\part[1] \textbf{T}\hspace{5pt} \circled{\textbf{F}} \hspace{5pt} A vertical line has zero slope. \\
		\textit{Solution:} Horizontal lines have zero slope. Vertical lines have ``undefined" or ``$\infty$" slope.
	\end{parts} 
\question[10] Write the slope and the $y$-intercept of the lines described by the following equations. \\
	\begin{parts}
		\part[3] $y=4x+1$ \\
		\textit{Solution:} This equation is already in ``slope-intercept" form $y=mx+b$ where $m$ denotes the slope and $b$ denotes the $y$-coordinate of the $y$-intercept (the $y$-intercept being a solution of the form $(0,b)$). Therefore \\
		\textbf{Slope:} $4$ \hfill  \textbf{$y$-intercept:} $(0,1)$ \hfill \hfill \\
		\part[3] $2y=6x+8$ \\
		\textit{Solution:} This equation is not in ``slope-intercept" form, so we will put it into this form. Divide by $2$ to get 
		$$y=3x+4.$$
		Now the equation is in slope-intercept form and we see \\
		\textbf{Slope:} $3$ \hfill  \textbf{$y$-intercept:} $(0,4)$ \hfill \hfill \\
		\part[4] $3y-6x=9$ \\
		\textit{Solution:} This equation is not in ``slope-intercept" form. To do this we solve for $y$ in the equation. First add $6x$ to both sides to get
		$$3y=6x+9.$$
		Now divide by $3$ to get
		$$y=2x+3.$$
		Now we have put the equation into ``slope-intercept" form, and so we may read off the slope and $y$-intercept. \\
		\textbf{Slope:} $2$ \hfill  \textbf{$y$-intercept:} $(0,3)$ \hfill \hfill \\
	\end{parts}
{\color{blue}\textit{NOTE:} you could also find slope by finding two solutions (plugging in one value and solving for the other) and then computing the slope from the two points; you could also find the $y$-intercept by plugging in $x=0$ and solving for $y$}
\newpage
\question[8] 
\begin{parts}
\part[4] Plot the points $(1,2)$, $(-3,4)$, $(-2,-3)$, and $(4,-1)$ on the following coordinate plane: \\
\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[color=black,fill=black] (1,2) circle (0.5ex);
\draw[color=black,fill=black] (-3,4) circle (0.5ex);
\draw[color=black,fill=black] (-2,-3) circle (0.5ex);
\draw[color=black,fill=black] (4,-1) circle (0.5ex);
\end{tikzpicture}

\part[4] Write the coordinates of the points that appear in the following graph: \\
\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[color=black,fill=black] (4,4) circle (0.5ex);
\draw (3.6,4.3) node {$(4,4)$};
\draw[color=black,fill=black] (-1,3) circle (0.5ex);
\draw (-1.5,3.3) node {$(-1,3)$};
\draw[color=black,fill=black] (-4,-1) circle (0.5ex);
\draw (-4.6,-0.7) node {$(-4,-1)$};
\draw[color=black,fill=black] (1,-3) circle (0.5ex);
\draw (1.5,-2.7) node {$(1,-3)$};
\end{tikzpicture}
\end{parts}
\newpage
\question[12] Graph the line in any way you choose.
	\begin{parts}
		\part[4] $y=2x+5$ \\
		\textit{Solution:} This equation is in ``slope-intercept" form and so it has slope $2$ and $y$-intercept $(0,5)$. 
		\begin{tikzpicture}
\draw[step=1.0,black,thin,color=gray,opacity=0.4] (-6,-6) grid (6,6);
\draw[line width=0.1cm] (-6,0) -- (6,0);
\draw[line width=0.1cm] (0,-6) -- (0,6);
\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] (5,-0.3) circle (1.61ex);
\draw (5,-0.3) node {$5$};
\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] (-5,-0.3) circle (1.61ex);
\draw (-5,-0.3) node {$-5$};

\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] (0.3,5) circle (1.61ex);
\draw (0.3,5) node {$5$};
\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,-5) circle (1.61ex);
\draw (0.3,-5) node {$-5$};
\draw[color=white,fill=white] (0.3,-1) circle (1.61ex);
\draw (0.3,-1) node {$-1$};

\draw[color=black,fill=black] (0,5) circle (0.75ex);
\draw[line width=1.5] (-5.5,-6) -- (0.5,6);
		\end{tikzpicture}
		\part[4] $3y+7x=4$ \\
		\textit{Solution:} We could do this problem by putting it into ``slope-intercept" form and reading off information. Instead we will find two solutions, plot them, and draw the line between them. First let $x=0$ which leaves us with the equation $3y=4$ which has solution $y=\dfrac{4}{3}$. This means $\left(0, \dfrac{4}{3} \right)$ is a solution. Now let $y=0$ yielding the equation $7x=4$, meaning $x=\dfrac{4}{7}$. Therefore $\left( \dfrac{4}{7}, 0 \right)$ is a solution. Now plot these two points and draw the line between them:
		\begin{tikzpicture}
\draw[step=1.0,black,thin,color=gray,opacity=0.4] (-3,-3) grid (3,3);
\draw[line width=0.1cm] (-3,0) -- (3,0);
\draw[line width=0.1cm] (0,-3) -- (0,3);
\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[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] (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] (0.3,1) 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,-1) circle (1.61ex);
\draw (0.3,-1) node {$-1$};

\draw[color=black,fill=black] (0,1.3333) circle (0.75ex);
\draw[color=black,fill=black] (0.57142857,0) circle (0.75ex);
\draw[line width=1.5] (-0.7142857,3) -- (1.85714,-3);
		\end{tikzpicture}
		\part[4] $x=11$ \\
		\textit{Solution:} Recall that any equation of the form $x=a$ denotes a vertical line at $x=a$. In this case we have a vertical line at $x=11$: \\
		\begin{tikzpicture}
%\draw[step=1.0,black,thin,color=gray,opacity=0.4] (-3,-3) grid (3,3);
\draw[line width=0.1cm] (-3,0) -- (3,0);
\draw[line width=0.1cm] (0,-3) -- (0,3);
\draw (1,-3) -- (1,3);
\draw[fill=white,color=white] (1,-0.3) circle (1.65ex);
\draw[line width=1.5] (1,-0.3) node {$11$};
		\end{tikzpicture}
	\end{parts}
\question[15] You rent a car with a full tank of gas on 3 September at noon and you return the car on 8 September at noon.  When the car was initially rented, the odometer read $5,120$ miles and when you returned the car, the odometer read $5,720$ miles. You returned the car with a full tank of gas and you had to put in a total of $20$ gallons of gas over the whole trip. The whole rental cost you a total amount of $\$200$ over the entire trip.
\begin{parts}
\part[5] What was your average rate of travel, measured in $\dfrac{\rm{miles}}{\rm{day}}$? \\
\textit{Solution:} Since you traveled for $5$ days and you traveled $5,720-5,120=600$ miles, the average rate of travel is 
$$\dfrac{600}{5} \dfrac{\mathrm{miles}}{\mathrm{day}}=120 \dfrac{\mathrm{miles}}{\mathrm{day}}.$$
\part[5] What was the average gas mileage, measured in $\dfrac{\mathrm{gallons}}{\mathrm{mile}}$?  \\
\textit{Solution:} As seen above, you traveled $600$ miles. Since you used $20$ gallons of gas, your average gas mileage was
$$\dfrac{20}{600} \dfrac{\mathrm{gallons}}{\mathrm{mile}} = \dfrac{1}{30} \dfrac{\mathrm{gallons}}{\mathrm{mile}}.$$
Note: if we measured the gas mileage as $\dfrac{\mathrm{miles}}{\mathrm{gallon}}$ we would have gotten $30 \dfrac{\mathrm{miles}}{\mathrm{gallon}}$. This would tell us how far each gallon of gas gets us, while the rate $\dfrac{\mathrm{gallons}}{\mathrm{mile}}$ tells us how much gas it takes go to one mile. 
\part[5] What was the average amount of money you spent per day, in appropriate units? \\
\textit{Solution:} The units here would be $\dfrac{\mathrm{dollars}}{\mathrm{day}}$. Since you spent $200$ dollars, you get
$$\dfrac{\mathrm{200}}{5} \dfrac{\mathrm{dollars}}{\mathrm{day}} = 40 \dfrac{\mathrm{dollars}}{\mathrm{day}}.$$
\end{parts}

\question[15] 
\begin{parts}
\part[5] Find an equation for the line with slope $2$ and $y$-intercept $(0,7)$. \\
\textit{Solution:} From the slope-intercept form of the line, such a line would have equation
$$y=2x+7.$$ 
\part[5] Find an equation of the line with slope $3$ that passes through the point $(11,13)$.   \\
\textit{Solution:} Using the point-slope form of the line we get
$$y-13 = 3(x-11).$$
\part[5] Find and equation of the line that contains the points $(3,-4)$ and $(2,1)$.  \\
\textit{Solution:} We need the slope. Compute it:
$$\mathrm{slope} = \dfrac{1-(-4)}{2-3} = \dfrac{1+4}{2-3} = \dfrac{5}{-1} = -5.$$
Therefore using the point-slope form of the line and the point $(2,1)$ we get
$$y-1 = -5(x-2).$$
\end{parts}
\vfill

\question[15] Simplify the following polynomial expressions completely.
\begin{parts}
\part[5] $(3x^2+5x-5)-(2x^2+4x-3)$ \\
\textit{Solution:} Calculate
$$\begin{array}{ll}
(3x^2+5x-5)-(2x^2+4x-3) &= 3x^2+5x-5 - 2x^2 - 4x + 3 \\
&= x^2 + x -2.
\end{array}$$
\vfill 
\part[5] $(x+3)(x+1)$\\
\textit{Solution:} Calculate
$$\begin{array}{ll}
(x+3)(x+1) &= (x+3)x + (x+3) (1) \\
&= x^2+3x + x+3 \\
&= x^2 + 4x + 3.
\end{array}$$
\vfill 
\part[5] $(x^2+5x+1)(2x^2-x+3)$ \\
\textit{Solution:} Calculate
$$\begin{array}{ll}
(x^2+5x+1)(2x^2-x+3) &= (x^2+5x+1)(2x^2) + (x^2+5x+1)(-x) + (x^2+5x+1)(3) \\
&= 2x^4+10x^3+2x^2-x^3-5x^2-x+3x^2+15x+3 \\
&=2x^4+9x^3+0x^2+14x+3 \\
&=2x^4+9x^3+14x+3 \\
\end{array}$$
\end{parts}
\question[15] Simplify completely.
\begin{parts}
\part[5] $x^2 x^3$ \\
\textit{Solution:} Using the rule that $a^n a^m = a^{n+m}$ we see
$$x^2x^3 = x^{2+3} = x^5.$$
\part[5] $(2x^2y^4)^2$ \\
\textit{Solution:} First we use the rule that powers distribute among a product to get
$$(2x^2y^4)^2 = 2^2 (x^2)^2 (y^4)^2.$$
Now we use the rule that $(a^n)^m=a^{nm}$ and so we continue the calculation:
$$(2x^2y^4)^2 = 4 x^{2\cdot 2} y^{4 \cdot 2} = 4x^4 y^8.$$
\part[5] $\left( \dfrac{xy}{z} \right)^3\left( \dfrac{3}{xy} \right)^2$ \\
\textit{Solution:} First we use the rule that powers distribute among a quotient to get
$$\left( \dfrac{xy}{z} \right)^3 \left( \dfrac{3}{xy} \right) = \left( \dfrac{(xy)^3}{z^3} \right) \left( \dfrac{3^2}{(xy)^2} \right).$$
Using the law that powers distribute among products yields
$$\left( \dfrac{xy}{z} \right)^3 \left( \dfrac{3}{xy} \right) = \left( \dfrac{x^3 y^3}{z^3} \right) \left( \dfrac{9}{x^2y^2} \right).$$
Now using the law $\dfrac{a^n}{a^m} = a^{n-m}$, we get
$$\left( \dfrac{xy}{z} \right)^3 \left( \dfrac{3}{xy} \right)=\dfrac{9}{z^3}x^{3-2}y^{3-2}=\dfrac{9xy}{z^3}.$$
\end{parts}
\question[5] Write a polynomial that describes the area of the big rectangle: \\
\begin{center}
\begin{tikzpicture}
\draw[thick] (0,0) -- (0,6);
\draw[thick] (2,0) -- (2,6);
\draw[thick] (0,2) -- (6,2);
\draw[thick] (0,0) -- (6,0);
\draw[thick] (4,0) -- (4,6);
\draw[thick] (0,4) -- (6,4);
\draw[thick] (6,0) -- (6,6);
\draw[thick] (0,6) -- (6,6);

\draw (-0.5,5) node {$x$};
\draw (-0.5,3) node {$2y$};
\draw (-0.5,1) node {$5z$};
\draw (1,-0.5) node {$x$};
\draw (3,-0.5) node {$3y$};
\draw (5,-0.5) node {$5z$};
\end{tikzpicture}
\end{center}
\textit{Solution:} Add up the areas of all the little squares:
$$x^2+3xy+5xz+2yx+6y^2+10yz+5zx+15zy+25z^2=x^2+6y^2+25z^2+5xy+10xz+25yz.$$

\end{questions}

\end{document}
