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{\LARGE{
\hfil Math 120 A, B Autumn 2012 \hfil \pp
\hfil Mid-Term Exam Number One\hfil  \pp
\hfil October 18, 2012 \hfil \pp
\hfil Answers \hfil \pp
}}

There were two versions of the exam.

Version A - In problem 1, the function's domain is $0 \le x \le 10$.


\begin{enumerate}
\item 
\[
f(x) = \left \{ \begin{array}{cc}
6-\sqrt{4-(x-2)^2} & \mbox{if $0 \le x \le 4$,} \\
-2(x-6)+2 & \mbox{if $4 \le x \le 6$,}\\
2+\sqrt{16-(x-10)^2} & \mbox{if $6 \le x \le 10$}
\end{array}
\right.
\]

\item (a) $ \dsps x=\frac{15}{15}t, y=\frac{36}{13}t$ (b) $\dsps x=5-\frac{2}{3} t, y=-4+2t$

\item 2.068377 hours

\item 7.1922359 days from now

\end{enumerate}

Version B - In problem 1, the function's domain is $0 \le x \le 5$.

\begin{enumerate}

\item 
\[
f(x) = \left \{ \begin{array}{cc}
3-\sqrt{1-(x-1)^2} & \mbox{if $0 \le x \le 2$,} \\
-2(x-2)+3 & \mbox{if $2 \le x \le 3$,}\\
1+\sqrt{4-(x-5)^2} & \mbox{if $3 \le x \le 5$}
\end{array}
\right.
\]

\item (a) $\dsps x=\frac{8}{5}t, y=\frac{6}{5}t$ (b) $\dsps x=5-\frac{5}{3}t, y=6-\frac{10}{9}t$

\item 3.6131983 hours

\item 8.6946976 days from now.


\end{enumerate}





\end{document}


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