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\noindent
\vfil \noindent
\large
\hfil Math 126 C - Autumn 2010 \hfil \pp
\hfil Mid-Term Exam Number One\hfil  \pp
\hfil October 26, 2010 \hfil \pp
\hfil Answers \hfil \pp
\normalsize

There were two versions of the exam in use.

\bigskip

Version A - Problem 1 asks for the angle between $\langle 3,4, -1 \rangle$ and
$\langle 5,2,8 \rangle$.

\begin{enumerate}

\item 1.2608 
\item $16 x -2 y +\frac{15}{2} z- 89 =0$
\item $\left( \frac{29}{9}, \frac{34}{9}, \frac{11}{9} \right)$
\item $\frac{2}{3} \left( 26^{3/2}-1 \right)$
\item (a) One point is $\frac{\pi^2 \sqrt{2}}{32}$
(b) At the point given in (a), the slope is $\frac{8+\pi}{8-\pi}$.
\end{enumerate}

Version B - Problem 1 asks for the angle between $\langle 5,-2,3 \rangle$
and $\langle 3,4,7 \rangle$.

\begin{enumerate}
\item 1.01453
\item $-10x-3y-z+60=0$
\item $\left( \frac{23}{2}, \frac{5}{4}, \frac{59}{4} \right)$
\item $\frac{4}{3} \left( 17^{3/2} -1 \right)$
\item  (a) One point is $\frac{\pi^2 \sqrt{2}}{32}$
(b) At the point given in (a), the slope is $\frac{8+\pi}{8-\pi}$.
\end{enumerate}



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