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diff --git a/texmf-dist/doc/latex/eqexam/examples/hw02.tex b/texmf-dist/doc/latex/eqexam/examples/hw02.tex new file mode 100644 index 00000000..0f03398b --- /dev/null +++ b/texmf-dist/doc/latex/eqexam/examples/hw02.tex @@ -0,0 +1,143 @@ +\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} |