\documentclass[10pt,a4paper]{exam}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{tikz}
\usetikzlibrary{arrows}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{hyperref}
\usepackage{color}
\usepackage{tabularx,ragged2e}
\newcommand*\circled[1]{\tikz[baseline=(char.base)]{
            \node[shape=circle,draw,inner sep=2pt] (char) {#1};}}
\begin{document}

\begin{coverpages}
\begin{center}\huge MATH 1011 - EXAM 1 FALL 2016\\
\vspace*{2in}
\large Name: \makebox[3in]{\hrulefill}\\
\end{center}

\begin{flushleft}\vspace{0.5in}
Thursday 8 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} The \underline{associative law} says that for all algebraic expressions $a$ and $b$, 
$$a+b=b+a.$$ 
\textit{Explanation: the \underline{commutative law} says this. The \underline{associative law} says that 
$$a+(b+c) = (a+b)+c.$$}
		\part[1] \circled{\textbf{T}} \hspace{5pt} \textbf{F} \hspace{5pt} The \underline{distributive law} says that for all algebraic expressions $a,b,$ and $c$, 
		$$a(b+c) = ab+ac.$$
		\part[1] \circled{\textbf{T}} \hspace{5pt} \textbf{F} \hspace{5pt} The number $\dfrac{-5}{\phantom{-}3}$ is a \underline{rational number}.\\
		\textit{Explanation: Rational numbers are those numbers which are a quotient of integers. Since $-5$ and $3$ are integers, $\dfrac{-5}{3}$ is rational.}
		\part[1] \textbf{T}\hspace{5pt} \circled{\textbf{F}} \hspace{5pt} The number $\pi$ is an \underline{integer}. \\
		\textit{Explanation: we discussed in class that $\pi$ is, in fact, an irrational number. No integer is an irrational number because all integers are rational.}
		\part[1] \textbf{T}\hspace{5pt} \circled{\textbf{F}} \hspace{5pt} The number $3$ is a \underline{solution} of the equation
		$$2x+9=-10000.$$
		\textit{Explanation: a \underline{solution} of an equation is a value for the variable to take that makes the resulting equation true. If we substitute $x=3$ into the equation $2x+9=-10000$ we get
		$$2 \cdot 3 + 9 = -10000,$$
		but the left hand side simplifies to $6+9=15$, and so the resulting equation is
		$$15=-10000,$$
		which is not true. Therefore $3$ is not a solution of this equation.}
	\end{parts}
\question[10] Evaluate the given expressions at the specified values.
	\begin{parts}
		\part[5] $2x+7w$ where $x=1$ and $w=2$ \\
		\textit{Solution:} If $x=1$ and $w=2$ we may {\color{blue} \href{http://www.wolframalpha.com/input/?i=2*1+%2B+7*2}{calculate}}
		$$2\cdot 1 + 7 \cdot 2 = 2 + 14 = 16.$$
		\part[5] $|-2x+w|$ where $x=2$ and $w=-1$ \\
		\textit{Solution:} If $x=2$ and $w=1$ we may {\color{blue} \href{http://www.wolframalpha.com/input/?i=%7C-2*2%2B1%7C}{calculate}}
		$$|-2 \cdot 2 + 1| = |-4+1| = |-3| = 3.$$
	\end{parts}
\question[10] Find the prime factorization of the following numbers or state that they are prime.
	\begin{parts}
		\part[5] $36$  \\
		\textit{Solution:} Draw a "factor tree" or simply {\color{blue}\href{http://www.wolframalpha.com/input/?i=prime+factorization+of+36}{calculate}} follows:
		$$36=2 \cdot 18 = 2 \cdot 2 \cdot 9 = 2 \cdot 2 \cdot 3 \cdot 3.$$
		\part[5] $13$ \\
		\textit{Solution:} $13$ {\color{blue}\href{http://www.wolframalpha.com/input/?i=prime+factorization+of+13}{is}} a prime number.
		\vfill
	\end{parts}
\newpage
\question[15] Compute.
	\begin{parts}
		\part[5] $\dfrac{1}{2}+\dfrac{3}{5}$ \\
		\textit{Solution:} The common denominator here is $10$, meaning we will rewrite $\dfrac{1}{2}$ as $\dfrac{5}{10}$ and we will rewrite $\dfrac{3}{5}$ as $\dfrac{6}{10}$. Then we {\color{blue} \href{http://www.wolframalpha.com/input/?i=(1%2F2)%2B(3%2F5)}{calculate}}
		$$\dfrac{1}{2}+\dfrac{3}{5} = \dfrac{5}{10} + \dfrac{6}{10} = \dfrac{11}{10}.$$
		\part[5] $\dfrac{3}{4} \cdot \dfrac{5}{7}$ \\
		\textit{Solution:} We simply multiply across and {\color{blue} \href{http://www.wolframalpha.com/input/?i=(3%2F4)*(5%2F7)}{calculate}}
		$$\dfrac{3}{4} \cdot \dfrac{5}{7} = \dfrac{3 \cdot 5}{4 \cdot 7} = \dfrac{15}{28}.$$	
		\part[5] $\dfrac{\phantom{a}\dfrac{1}{2}\phantom{a}}{\dfrac{3}{2}}$
		\textit{Solution:} Recall to evaluate a fraction of fractions we flip the bottom and multiply, i.e. $\dfrac{\frac{a}{b}}{\frac{c}{d}}=\dfrac{a}{b} \cdot \dfrac{d}{c}$. Therefore we {\color{blue} \href{http://www.wolframalpha.com/input/?i=(1%2F2)+%2F+(3%2F2)}{calculate}}
		$$\dfrac{\frac{1}{2}}{\frac{3}{2}} = \dfrac{1}{2} \cdot \dfrac{2}{3} = \dfrac{1 \cdot 2}{2 \cdot 3} = \dfrac{2}{6} \stackrel{\mathrm{reduce}}{=} \dfrac{1}{3}.$$
	\end{parts}
\question[5] Calculate the expression and completely simplify:
$$2 \cdot 3 + [1-(2 \cdot 2 + 1)]^2.$$
\textit{Solution:} Using the order of operations and taking it exactly one step at a time, we {\color{blue} \href{http://www.wolframalpha.com/input/?i=2*3+%2B+%5B1-(2*2%2B1)%5D%5E2}{calculate}}
$$\begin{array}{ll}
2 \cdot 3 + [1-(2 \cdot 2+1)]^2 &= 2 \cdot 3 + [1 - (4+1)]^2 \\
&= 2 \cdot 3 + [1-5]^2 \\
&= 2 \cdot 3 + [-4]^2 \\
&= 2 \cdot 3 + (-4)(-4) \\
&= 2 \cdot 3 + 16 \\
&= 6 + 16 \\
&= 22.
\end{array}$$
\question[15] Solve the equation.
	\begin{parts}
		\part[5] $x-1=4$ \\
		\textit{Solution:} Add $1$ to both sides to {\color{blue} \href{http://www.wolframalpha.com/input/?i=solve+x-1%3D4}{get}}
		$$x=5.$$
		\part[5] $2x=26$ \\
		\textit{Solution:} Divide both sides by $2$ to {\color{blue} \href{http://www.wolframalpha.com/input/?i=solve+2*x%3D26}{get}}
		$$x=13.$$
		\part[5] $\dfrac{1}{2}x + \dfrac{1}{5} = \dfrac{3}{5}$ \\
		\textit{Solution:} First subtract $\dfrac{1}{5}$ from both sides to get
		$$\dfrac{1}{2} x = \dfrac{2}{5}.$$
		Now divide both sides by $\dfrac{1}{2}$ to {\color{blue} \href{http://www.wolframalpha.com/input/?i=solve+(1%2F2)*x+%2B+(1%2F5)+%3D+(3%2F5)}{get}}
		$$x = \dfrac{\frac{2}{5}}{\frac{1}{2}} = \dfrac{2}{5} \cdot \dfrac{2}{1} = \dfrac{4}{5}.$$
		\vfill
	\end{parts}
\question[10] Answer the following questions.
	\begin{parts}
		\part[5] What is $8 \%$ of $53$? Give your answer as a decimal. \\
		\textit{Solution:} This question translates to the equation
		$$x= 8 \cdot \dfrac{1}{100} \cdot 53.$$
		Simplify the right hand side to {\color{blue} \href{http://www.wolframalpha.com/input/?i=What+is+8%25+of+53%3F}{get}}
		$$x = \dfrac{8 \cdot 53}{100} = \dfrac{424}{100} = 4.24.$$
		\part[5] What $\%$ of $31$ is $7$? (decimal not necessary) \\
		\textit{Solution:} This question translates to the equation
		$$x \cdot \dfrac{1}{100} \cdot 31 = 7.$$
		Simplify this to get
		$$\dfrac{31}{100} x = 7.$$
		Now divide both sides by $\dfrac{31}{100}$ to get
		$$x = \dfrac{7}{\frac{31}{100}} = \dfrac{7}{1} \cdot \dfrac{100}{31} = \dfrac{700}{31},$$
		or in {\color{blue} \href{http://www.wolframalpha.com/input/?i=What+%25+of+31+is+7%3F}{other words}},
		$$x \% = \dfrac{7}{31}=22.58 \%.$$
	\end{parts}	
\question[10]  Do the following two problems.
	\begin{parts}
		\part[5] The relationship between the rate of travel (denoted $r$), the distance traveled (denoted $d$), and the time it took to travel (denoted $t$) is 
$$r = \dfrac{d}{t}.$$
		If you traveled at a rate of $50 \dfrac{\mathrm{miles}}{\mathrm{hour}}$ for $2$ hours, how far did you travel? \\
		\textit{Solution:} We are given the information $r=50 \dfrac{\mathrm{miles}}{\mathrm{hour}}$ and $t=2 \mathrm{\hspace{2pt} hours}$. Plugging this information into the equation yields
		$$50 \dfrac{\mathrm{miles}}{\mathrm{hour}} = \dfrac{d}{2 \mathrm{\hspace{2pt} hours}}.$$
		Therefore multiply both sides of the equation by $2 \mathrm{\hspace{2pt} hours}$ to get
		$$\left( 50 \dfrac{\mathrm{miles}}{\mathrm{hour}} \right) \cdot \left( 2 \mathrm{\hspace{2pt} hours} \right) = 100 \mathrm{\hspace{2pt} miles}.$$
		\part[5] The volume of a right circular cylinder with radius $r$ and height $h$ is 
		$$V = \pi r^2 h.$$
		If a right circular cylinder has radius $1$ inch and height $4$ inches, what is its volume? \\
		\textit{Solution:} We are told that $r=1 \mathrm{\hspace{2pt} inch}$ and $h=4 \mathrm{\hspace{2pt} inches}$. Substituting the values into the equation yields
		$$V = \pi \left( 1 \mathrm{\hspace{2pt} inch} \right)^2 \left( 4 \mathrm{\hspace{2pt} inches} \right)=4\pi \mathrm{\hspace{2pt} inches}^3.$$
	\end{parts}
\question[5] The sum of three consecutive numbers is 60. Find the numbers. \\
\textit{Solution:} Let $x$ be the first of the three numbers. Then $x+1$ is the second and $x+2$ is the third. Since we are told the sum of the three numbers is $60$ we get the following equation:
$$x+(x+1)+(x+2) = 60.$$
Collect like terms on the left hand side to get
$$3x+3=60.$$
Solve this equation by subtracting by $3$ and then dividing by $5$ to {\color{blue} \href{http://www.wolframalpha.com/input/?i=solve+3x%2B3%3D60}{get}} 
$$x = \dfrac{57}{3}=19.$$
\question[15] Solve \textbf{and} graph the following inequalities.
	\begin{parts}
		\part[5] $x-1>3$ \\
		\textit{Solution:} Add $1$ to both sides to {\color{blue} \href{http://www.wolframalpha.com/input/?i=solve+x-1%3E3}{get}}
		$$x>4.$$
		The graph is as follows: \\
		\begin{tikzpicture}
			\draw[<-, thick] (-5,0) -- (5,0);
			\draw (0,-0.05) node[below] {$4$};
			\draw[(->, ultra thick, blue] (0,0) -- (5,0);
		\end{tikzpicture}
		\part[5] $2x-3 \leq 7$ \\
		\textit{Solution:} Add $3$ and then divide by $2$ to {\color{blue} \href{http://www.wolframalpha.com/input/?i=solve+2x-3+%3C%3D+7}{get}}
		$$x \leq 5.$$
		The graph is as follows: \\
		\begin{tikzpicture}
			\draw[<-, thick] (-5,0) -- (5,0);
			\draw (0,-0.1) node[below] {$5$};
			\draw[<-, ultra thick, blue] (-5,0) -- (0,0);
			\draw[ultra thick, blue] (0,0) node {$\textbf{]}$};
		\end{tikzpicture}
		\part[5] $-2x+5<2$ \\
		\textit{Solution:} Subtract $5$ to get
		$$-2x < -3.$$
		Divide by $-2$ to {\color{blue} \href{http://www.wolframalpha.com/input/?i=solve+-2x%2B5%3C2}{get}}
		$$x > \dfrac{-3}{-2} = \dfrac{3}{2}.$$
		The graph is as follows: \\
		\begin{tikzpicture}
			\draw[<-, thick] (-5,0) -- (5,0);
			\draw (0,-0.05) node[below] {$\dfrac{3}{2}$};
			\draw[(->, ultra thick, blue] (0,0) -- (5,0);
		\end{tikzpicture}	
		\vfill
	\end{parts}
\end{questions}

\end{document}
