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author | Karl Berry <karl@freefriends.org> | 2006-01-09 00:44:40 +0000 |
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committer | Karl Berry <karl@freefriends.org> | 2006-01-09 00:44:40 +0000 |
commit | b4fc5f639874db951177ec539299d20908adb654 (patch) | |
tree | 52f08823ca58fffe3db6a9b075635038c567626c /Master/texmf-dist/doc/latex/multenum/multienum.sample | |
parent | dec3d98ebe442d7ea93efbaa8dd2e2be8149a467 (diff) |
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git-svn-id: svn://tug.org/texlive/trunk@80 c570f23f-e606-0410-a88d-b1316a301751
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diff --git a/Master/texmf-dist/doc/latex/multenum/multienum.sample b/Master/texmf-dist/doc/latex/multenum/multienum.sample new file mode 100644 index 00000000000..b90f146386d --- /dev/null +++ b/Master/texmf-dist/doc/latex/multenum/multienum.sample @@ -0,0 +1,159 @@ +%This is a 2-page sample illustrating how to use the +%multienum package + +\documentclass{article} +\setlength{\textwidth}{6in} +\setlength{\textheight}{8.5in} +\setlength{\topmargin}{-0.5in} +\setlength{\oddsidemargin}{0.25in} +\usepackage{multicol,multienum} + + + +\begin{document} +\begin{center} +{\Large\bf Sample formating using {\tt multienumerate}} +\end{center} + +\bigskip +Sometimes we want to typeset the solutions to exercises. This +is easy to do using the {\tt multienumerate} environment. +\subsection*{Answers to All Exercises} +\begin{multienumerate} +\mitemxxxx{Not}{Linear}{Not}{Quadratic} +\mitemxxxo{Not}{Linear}{No; if $x=3$, then $y=-2$.} +\mitemxx{$(x_1,x_2)=(2+\frac{1}{3}t,t)$ or +$(s,3s-6)$}{$(x_1,x_2,x_3)=(2+\frac{5}{2}s-3t,s,t)$} +\mitemx{$(x_1,x_2,x_3,x_4)= (\frac{1}{4}+\frac{5}{4}s+\frac{3}{4}t-u,s,t,u)$ +or $(s,t,u,\frac{1}{4}-s+\frac{5}{4}t+\frac{3}{4}u)$} +\mitemxxxx{$(2,-1,3)$}{None}{$(2,1,0,1)$}{$(0,0,0,0)$} +\end{multienumerate} + + +\bigskip +\hrule + +\bigskip + +We can also enumerate the items using an even-only or odd only +counter. +\subsection*{Answers to Even-Numbered Exercises} +\begin{multienumerate}[evenlist] +\mitemxxxx{Not}{Linear}{Not}{Quadratic} +\mitemxxxo{Not}{Linear}{No; if $x=3$, then $y=-2$.} +\mitemxx{$(x_1,x_2)=(2+\frac{1}{3}t,t)$ or +$(s,3s-6)$}{$(x_1,x_2,x_3)=(2+\frac{5}{2}s-3t,s,t)$} +\mitemx{$(x_1,x_2,x_3,x_4)= (\frac{1}{4}+\frac{5}{4}s+\frac{3}{4}t-u,s,t,u)$ +or $(s,t,u,\frac{1}{4}-s+\frac{5}{4}t+\frac{3}{4}u)$} +\mitemxxxx{$(2,-1,3)$}{None}{$(2,1,0,1)$}{$(0,0,0,0)$} +\end{multienumerate} + +\hrule + +\subsection*{Answers to Odd-Numbered Exercises} +\begin{multienumerate}[oddlist] +\mitemxxxx{Not}{Linear}{Not}{Quadratic} +\mitemxxxo{Not}{Linear}{No; if $x=3$, then $y=-2$.} +\mitemxx{$(x_1,x_2)=(2+\frac{1}{3}t,t)$ or +$(s,3s-6)$}{$(x_1,x_2,x_3)=(2+\frac{5}{2}s-3t,s,t)$} +\mitemx{$(x_1,x_2,x_3,x_4)= (\frac{1}{4}+\frac{5}{4}s+\frac{3}{4}t-u,s,t,u)$ +or $(s,t,u,\frac{1}{4}-s+\frac{5}{4}t+\frac{3}{4}u)$} +\mitemxxxx{$(2,-1,3)$}{None}{$(2,1,0,1)$}{$(0,0,0,0)$} +\end{multienumerate} + +\bigskip +\hrule + +\bigskip + +Sometimes we want to create sublists which are +enumerated using an alpha counter. + +\begin{multienumerate} +\mitemx{Which of the following numbers is the solution of the +equation +$x+3=7$:} +\begin{multienumerate} +\mitemxxxx{1}{2}{3}{4} +\end{multienumerate} +\mitemx{The value of $\log_28$ is:} +\begin{multienumerate} +\mitemxxxx{1}{$-1$}{3}{$-3$} +\end{multienumerate} +\end{multienumerate} +\pagebreak + +\begin{multicols}{2} +\subsection*{Answers to All Exercises} +\begin{multienumerate} +\mitemxx{Not}{Linear} +\mitemxx{Not}{Quadratic} +\mitemxx{Not}{Linear} +\mitemx{$(x_1,x_2)=(2+\frac{1}{3}t,t)$ or +$(s,3s-6)$} +\mitemx{$(x_1,x_2,x_3)=(2+\frac{5}{2}s-3t,s,t)$} +\mitemx{$(x_1,x_2,x_3,x_4)= (\frac{1}{4}+\frac{5}{4}s+\frac{3}{4}t-u,s,t,u)$ +or $(s,t,u,\frac{1}{4}-s+\frac{5}{4}t+\frac{3}{4}u)$} +\mitemxx{$(2,-1,3)$}{None} +\mitemxx{$(2,1,0,1)$}{$(0,0,0,0)$} +\mitemxx{Not}{Linear} +\mitemxx{Not}{Quadratic} +\mitemxx{Not}{Linear} +\mitemx{$(x_1,x_2)=(2+\frac{1}{3}t,t)$ or +$(s,3s-6)$} +\mitemx{$(x_1,x_2,x_3)=(2+\frac{5}{2}s-3t,s,t)$} +\mitemx{$(x_1,x_2,x_3,x_4)= (\frac{1}{4}+\frac{5}{4}s+\frac{3}{4}t-u,s,t,u)$ +or $(s,t,u,\frac{1}{4}-s+\frac{5}{4}t+\frac{3}{4}u)$} +\mitemxx{$(2,-1,3)$}{None} +\mitemxx{$(2,1,0,1)$}{$(0,0,0,0)$} +\mitemxx{Not}{Linear} +\mitemxx{Not}{Quadratic} +\mitemxx{Not}{Linear} +\mitemx{$(x_1,x_2)=(2+\frac{1}{3}t,t)$ or +$(s,3s-6)$} +\mitemx{$(x_1,x_2,x_3)=(2+\frac{5}{2}s-3t,s,t)$} +\mitemx{$(x_1,x_2,x_3,x_4)= (\frac{1}{4}+\frac{5}{4}s+\frac{3}{4}t-u,s,t,u)$ +or $(s,t,u,\frac{1}{4}-s+\frac{5}{4}t+\frac{3}{4}u)$} +\mitemxx{$(2,-1,3)$}{None} +\mitemxx{$(2,1,0,1)$}{$(0,0,0,0)$} +\end{multienumerate} + +\subsection*{Multiple Choice} +\begin{multienumerate} +\mitemx{Which of the following numbers is the solution of the +equation +$x+3=7$:} +\begin{multienumerate} +\mitemxxxx{1}{2}{3}{4} +\end{multienumerate} +\mitemx{The value of $\log_28$ is:} +\begin{multienumerate} +\mitemxxxx{1}{$-1$}{3}{$-3$} +\end{multienumerate} +\mitemx{Which of the following numbers is the solution of the +equation +$x+3=7$:} +\begin{multienumerate} +\mitemxxxx{1}{2}{3}{4} +\end{multienumerate} +\mitemx{The value of $\log_28$ is:} +\begin{multienumerate} +\mitemxxxx{1}{$-1$}{3}{$-3$} +\end{multienumerate} +\mitemx{Which of the following numbers is the solution of the +equation +$x+3=7$:} +\begin{multienumerate} +\mitemxxxx{1}{2}{3}{4} +\end{multienumerate} +\mitemx{The value of $\log_28$ is:} +\begin{multienumerate} +\mitemxxxx{1}{$-1$}{3}{$-3$} +\end{multienumerate} +\end{multienumerate} +\end{multicols} + +\end{document} + + + |