我刚刚完成了一份需要在午夜前提交的文件,但出于某种原因我无法导出为 PDF。它一直显示这张图片:
我猜是我缺少了某种包裹?
错误日志:
This is LuaTeX, Version beta-0.80.0 (MiKTeX 2.9) (rev 5238) (format=lualatex 2015.10.7) 7 OCT 2015 21:21
**Afleverings_Opgave_2.tex
(./Afleverings_Opgave_2.tex
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No file Afleverings_Opgave_2.aux.
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l.9 \begin{document}
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l.64 \item C
entrum -2,6 og radius = $|AB|=\sqrt{74}$
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l.64 \item C
entrum -2,6 og radius = $|AB|=\sqrt{74}$
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l.105 ... $r=\sqrt{56}\approx7,48331$\end{enumerate}
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l.105 ... $r=\sqrt{56}\approx7,48331$\end{enumerate}
Your command was ignored.
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l.107 \end{document}
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l.107 \end{document}
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[3] (./Afleverings_Opgave_2.aux))
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这是我使用的 TeX 格式的 LyX 文档的导出版本:
%% LyX 2.1.3 created this file. For more info, see http://www.lyx.org/.
%% Do not edit unless you really know what you are doing.
\documentclass[english]{article}
\usepackage[15]{fontenc}
\usepackage[latin9]{inputenc}
\usepackage{babel}
\begin{document}
\title{Afleveringsopgave 2}
\author{omitted}
\maketitle
1) Lad der været givet to vektorer i planen, $\vec{a}=({3\atop 8})$
og $\vec{b}=({1\atop 6})$
\begin{enumerate}
\item Beregn $\vec{a}*\vec{b}$
\begin{enumerate}
\item $(3*1)+(8*6)=51$
\end{enumerate}
\item Beregn vinklen mellem $\vec{a}$ og $\vec{b}$
\begin{enumerate}
\item $cos(v)=\frac{51}{\sqrt{3^{2}+8^{2}}*\sqrt{1^{2}+6^{2}}}=\frac{51}{\sqrt{2701}}$
\item $cos^{-1}(\frac{51}{\sqrt{2701}})\approx11,0937$
\item $180-11,0937\approx168,906$
\end{enumerate}
\item Beregn $\vec{a_{b}}$
\begin{enumerate}
\item $\vec{a_{b}}=\frac{({3\atop 8})*({4\atop 6})}{8^{2}+6^{2}}*({1\atop 6})=\frac{60}{100}*({1\atop 6})=({0,6\atop 3,6})$
\end{enumerate}
\end{enumerate}
2) Lad der i planen være givet punkterne $A(-2,6)$ og $B(5,1)$
\begin{enumerate}
\item Bestem en ligning (på formen $ax+by+c=0$) for linjen, som går gennem
punktet $P(1,1)$ og har $\vec{AB}$ som normalvektor.
\begin{enumerate}
\item $\vec{AB}=\vec{n}$
\item $\vec{n}=[{5-(-2)\atop 1-6}]=[{7\atop -5}]$
\item $7(x-1)-5(y-1)=0$
\item $7x-7-5y-5=0$
\item $7x-5y-12=0$
\end{enumerate}
\item Bestem en ligning (på formen $ax+by+c=0$) for linjen, som går gennem
punktet $Q(2,-3)$ og har$\vec{AB}$ som retningsvektor.
\begin{enumerate}
\item $\vec{r}=\vec{AB}=[{7\atop -5}]$
\item $\vec{n}=[{5\atop 7}]$
\item $5(x-2)+7(y-(-3))=0$
\item $5x-10+7y-(-21)=0$
\item $5x+7y+11=0$
\end{enumerate}
\item Bestem en ligning for den cirkel, som har centrum i punktet A og har
radius = $|AB|$
\begin{enumerate}
\item Centrum -2,6 og radius = $|AB|=\sqrt{74}$
\item $r=\sqrt{(5-(-2))^{2}+(1-6)^{2}}=\sqrt{74}$
\item $(x-(-2))^{2}+(y-6)^{2}\approx8,60233^{2}$
\item $(x+2)^{2}+(y-6)^{2}=74$
\end{enumerate}
\end{enumerate}
3) Lad der i planen være givet punkterne P(7,2) og Q(1,3)
\begin{enumerate}
\item Beregn vinklen mellem $\vec{PQ}$ og x-aksen
\begin{enumerate}
\item $\vec{x}=[{1\atop 0}]$
\item $\vec{PQ}=[{1-7\atop 3-2}]=[{-6\atop 1}]$
\item $cos(v)=\frac{[{-6*1\atop 1*0}]}{\sqrt{(-6)^{2}+1^{2}}*\sqrt{1^{2}+0^{2}}}=\frac{-5}{\sqrt{37}}$
\item $cos^{-1}(\frac{-5}{\sqrt{37}})\approx145,285$
\item $180-145,285=34,715$
\end{enumerate}
\item Beregn midtpunktet af linjestykket PQ
\begin{enumerate}
\item $M=(\frac{7+1}{2},\frac{2+3}{2})=(4,2.5)$
\end{enumerate}
\end{enumerate}
5) Bestem tallet t, således at vektorene $\vec{a}=({t+1\atop 4})$
og $\vec{b}=({t^{2}-2t\atop -t})$ er ortogonale.
\begin{enumerate}
\item $({t+1\atop 4})*({t^{2}-2t\atop -t})=0$
\item $t^{3}-2t+t^{2}-2t-4t=0$
\item $t^{3}-t^{2}-6t=0$
\item $t*(t^{2}-t-6)=0$
\item $t=0$
\item $t^{2}-t-6=0$
\item $1t^{2}-1t-6=0$
\item $d=(-1)^{2}-4*1*(-6)=25$
\item $t=\frac{-(-1)\pm\sqrt{25}}{2*1}=\{{3\atop -2}$
\item Løsnigen er enten 3 eller -2.
\end{enumerate}
6) Bestem centrum og radius for cirklen $x^{2}+y^{2}-16x+12y-44=0$
\begin{enumerate}
\item $(x-8)^{2}+(y+6)^{2}+44=0$
\item $(-44)+6^{2}+8^{2}=56$
\item Centrum (8,-6) og $r=\sqrt{56}\approx7,48331$\end{enumerate}
\end{document}
还附上了我的文件作为Dropbox 链接如果你想看看它。
希望你们能帮助我!:)