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+\documentclass{article}
+\usepackage[fleqn]{amsmath}
+\usepackage[pdf,forpaper,cfg=hw,nopoints,useforms,
+% Try compiling this file under one of these options:
+% nosolutions,
+% answerkey,
+ vspacewithsolns,
+]{eqexam}
+
+\subject[CA]{College Algebra}
+\title[HW2]{HW \#2}
+\author{Dr.\ D. P. Story}
+\date{Spring 2011}
+\duedate{03/05/11}
+\keywords{Homework due \theduedate}
+
+% Make the end of solution label blank
+\def\exrtnlabelformat{}
+\def\exrtnlabelformatwp{}
+\solAtEndFormatting{\eqequesitemsep{3pt}}
+
+
+\begin{document}
+
+\maketitle
+
+\begin{exam}{HW}
+
+\begin{instructions}[]
+In preparation for the quiz on Thursday, solve each of these short
+problems in the space provided before looking at their solutions at the
+end of the document.
+\[
+ \text{\url{http://faculty.nwfsc.edu/web/math/storyd}}
+\]
+All class assignments and other announcements will be posted on
+this web site.
+\end{instructions}
+
+\begin{problem*}[4ea]
+Let $P(-4,2)$ and $Q(2,-3)$ be two points in the plane.
+\begin{parts}
+\item Find the distance $d(P,Q)$ between $P$ and $Q$.
+\begin{solution}[1in]
+We use the distance formula
+\[
+ d(P,Q)=\sqrt{(2+4)^2+(-3-2)^2}=\sqrt{61}
+\]
+to obtained the required answer.
+\end{solution}
+
+\item Find the midpoint $M$ between $P$ and $Q$.
+\begin{solution}[\sameVspace]
+We use the midpoint formula
+\[
+ M=\left(\frac{-4+2}{2},\frac{2+(-3)}{2}\right)=\left(-1,-\dfrac{1}{2}\right)
+\]
+to obtained the required answer.
+\end{solution}
+\end{parts}
+\end{problem*}
+
+\begin{problem*}[3ea]
+Complete each of the two sentences below with correct entries.
+\begin{parts}
+ \item The function $ g(x) = | x+2 | $ can be graphed from the library
+ function $ f(x) = |x| $ by shifting it \fillin[u]{.25in}{2} units
+ \fillin[u]{1.25in}{horizontally} (horizontally/vertically) \fillin[u]{.75in}{left} (left/right/up/down).
+% Make the solution label blank.
+\ifkeyalt
+\begin{solution}[]
+% We want this to be shown at the end of the file, but not if the author changes the option
+% to answerkey
+ The function $ g(x) = | x+2 | $ can be graphed from the library
+ function $ f(x) = |x| $ by shifting it \fillin[u]{.25in}{2} units
+ \fillin[u]{1.25in}{horizontally} (horizontally/vertically) \fillin[u]{.75in}{left}
+ (left/\penalty0right/\penalty0up/\penalty0down). % Note: \penalty0 is inserted to help TeX break the line after the forward slash /
+\end{solution}
+\fi
+ \item The function $ g(x) = 5 - x^2 $ can be graphed from the library
+ function $ f(x) = x^2 $ by first reflecting it with respect to the
+ \fillin[u]{.25in}{$x$} axis, then shifting it \fillin[u]{.25in}{5} units
+ \fillin[u]{1.25in}{vertically} (horizontally/vertically) \fillin[u]{.75in}{upward} (left/right/up/down).
+\ifkeyalt
+\begin{solution}[]
+% We want this to be shown at the end of the file, but not if the author changes the option
+% to answerkey.
+ The function $ g(x) = 5 - x^2 $ can be graphed from the library function
+ $ f(x) = x^2 $ by first reflecting it with respect to the
+ \fillin[u]{.25in}{$x$} axis, then shifting it \fillin[u]{.25in}{5} units
+ \fillin[u]{1.25in}{vertically} (horizontally/vertically)
+ \fillin[u]{.75in}{upward} (left/right/up/down).
+\end{solution}
+\fi
+\end{parts}
+\end{problem*}
+
+\begin{problem}[5]
+The circle $ x^2 + y^2 = 25 $ passes through
+the point $P(3,4)$. Let $\ell$ be the line passing though the origin and
+the point $P$. Find the equation of the line perpendicular to line $\ell$ and passing
+through point $P$.
+\begin{solution}[1in]
+The slope of the line perpendicular to $\ell$ is $ m = -\tfrac{3}{4} $, the line
+must pass through $(3,4)$; thus, the line is $ y-4 = -\tfrac{3}{4}(x-3) \implies
+y = -\tfrac{3}{4} x + \tfrac{25}{4} $. Thus,
+\[
+ \text{Ans:}\quad\boxed{ y = -\tfrac{3}{4} x + \tfrac{25}{4} }
+\]
+This is the equation of the line tangent to the circle at $P(3,4)$.
+\end{solution}
+\end{problem}
+
+\begin{problem}[3]
+If the slope the a line is negative, then the line is
+ \begin{answers}{4}
+ \bChoices
+ \Ans0 increasing\eAns
+ \Ans1 decreasing\eAns
+ \Ans0 constant\eAns
+ \Ans0 none of these\eAns
+ \eChoices
+ \end{answers}
+
+% The \texttt{answers} and \texttt{manswers} environments can now
+% be copied and pasted into the solutions environment as well.
+%
+\ifkeyalt
+\begin{solution}[.25in]
+If the slope the a line is negative, then the line is
+ \begin{answers}{4}
+ \bChoices
+ \Ans0 increasing\eAns
+ \Ans1 decreasing\eAns
+ \Ans0 constant\eAns
+ \Ans0 none of these\eAns
+ \eChoices
+ \end{answers}
+\end{solution}
+\fi
+\end{problem}
+\end{exam}
+\end{document}