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\noindent
\vfil \noindent
\large
\hfil Math 120 A, B - Winter 2009 \hfil \pp
\hfil Mid-Term Exam Number One\hfil  \pp
\hfil January 29, 2009 \hfil \pp
\hfil Answers \hfil \pp
\normalsize



There were two versions of the exam.

\bigskip

Version A - The left edge of the figure in problem 1 has length 5.

\begin{enumerate}

\item \[
\mbox{area} = \left\{
\begin{array}{ll}
-\frac{3}{10} x^2 + 5x & \mbox{if $0 \le x \le 5$,} \\
2x+\frac{15}{2} & \mbox{if $5 \le x \le 9$.} 
\end{array}
\right.
\]

\item (a) \$11.465 (b) \$7816.447.

\item 5.5163 hrs.

\item The minimum distance to the bear is 41.57597 feet.

\end{enumerate}

\bigskip
Version B - The left edge of the figure in problem 1 has length 10.

\begin{enumerate}

\item \[
\mbox{area} = \left\{
\begin{array}{ll}
-\frac{3}{10} x^2 + 10x & \mbox{if $0 \le x \le 10$,} \\
4x+30 & \mbox{if $10 \le x \le18$.} 
\end{array}
\right.
\]

\item (a) \$17.3841 (b) \$10914.09918.

\item 0.52249 hrs.

\item The minimum distance to the bear is 18.569533 feet.


\end{enumerate}




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