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{\LARGE{
\hfil Math 120 A - Spring 2013 \hfil \pp
\hfil Mid-Term Exam Number One\hfil  \pp
\hfil April 25, 2013 \hfil \pp
\hfil Answers \hfil \pp
}}

There were two versions of the exam.

Version A: In problem 1, Veeree's house is 4 km north of Tom's.

\begin{enumerate}
\item $\frac{17}{\sqrt{41}}+\frac{11}{\sqrt{41}} = 4.37286533...$ km
\item \[
\mbox{area} = \left \{ \begin{array}{ll}
\frac{1}{2}x(4+\frac{4}{3}x+4) & \mbox{if $0 \le x \le 3$} \vspace{0.1in}\\ 
\frac{1}{2}(3)(8+4)+\frac{1}{2}(x-3)(8-\frac{8}{7}(x-10)) & \mbox{if $3 \le x \le 10$} 
\end{array}
\right.
\]

\item 5.903575 km

\item (a) $x=2.5t, y=5+0.25t$ (b) $x=-3+6t, y=4-2t$ (c) 0.476534 seconds

\end{enumerate}

Version B: In problem 1, Veeree's house is 6 km east of Tom's house.

\begin{enumerate}

\item 8.2219219 km

\item \[
\mbox{area} = \left \{ \begin{array}{ll}
\frac{1}{2}x(2+\frac{3}{2}x+2) & \mbox{if $0 \le x \le 4$} \vspace{0.1in} \\
\frac{1}{2}(4)(2+8)+\frac{1}{2}(x-4)(8-\frac{8}{5}(x-9)) & \mbox{if $4 \le x \le 9$} 
\end{array}
\right.
\]

\item 15.11223 km

\item (a) $x=-2+\frac{5}{4}t, y=5-\frac{9}{4}t$ (b) $x=-5+\frac{5}{2}t, y=\frac{1}{2}t$ (c) 1.9178082 seconds

\end{enumerate}

\end{document}


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