ham and scalloped potatoes that had been cooked left out overnight- still safe to eat?
How do I know if food left at room temperature is still safe to eat?
I really tried hard to do this on my own, but I keep failing ... Using multi-column and multi-row commands confused me. I know some of you might frown upon such a request but I really need help getting this table implemented .. So I'm sorry if I upset anyone.  My preamble has :   \documentclass[12pt,a4paper]{report} \usepackage[lmargin=3.81cm,tmargin=2.54cm,rmargin=2.54cm,bmargin=2.52cm]{geometry} \linespread{1.5}     Is there a way I can convert this from excel to LaTeX form ? I tried an add-in before but it didn't work for me. Also online generators cannot combine cells the way I want them (unless I didn't find a better table generator ? ) What do I need to know about temperature and food safety? What general rules do I need to follow to keep my food safe? How do I know what temperature to cook something to, or whether my food is safe at room temperature? Generated Frattini Lemma Let $A$ be a normal subgroup of a group $G$. Let $B$ be a subgroup of   $A$ with the property that whenever $B_1 \leq A$ is conjugate to $B$   in $G$ then it is conjugate to $B$ in $A$. Prove that $G=AN_G(B)$   Help me some hints. Does a shortcode with a single attribute have to use an array? Why constructors will always have same name as of class and how they are invoked implicitly? Page range truncation with imakeidx This is a followup question to egreg's answer to  The answer runs into a problem if an index topic occurs in more than one page range. The following MWE, based on egreg's answer:  \begin{filecontents*}{subjind.ist} delim_0 ", \\checkrange{" delim_t "}" \end{filecontents*}  \documentclass{book} \usepackage{xparse} \usepackage{imakeidx}  \makeindex[title=Subject Index,options=-s subjind]  \ExplSyntaxOn \NewDocumentCommand{\checkrange}{&gt;{\SplitArgument{1}{--}}m}  {   \formatrange#1  } \NewDocumentCommand{\formatrange}{mm}  {   \IfNoValueTF { #2 }    { % no --, it's not a range     #1    }    { % it's a range     \egreg_range:nn { #1 } { #2 }    }  }  \bool_new:N \l__egreg_range_compare_bool  \cs_new_protected:Nn \egreg_range:nn  {   % print the start number   #1--   \int_compare:nNnTF { \str_count:n { #1 } } = { \str_count:n { #2 } }    {% same number of digits, remove equal ones at the start     \__egreg_range_compare:nn { #1 } { #2 }    }    {% different number of digits, print both     #2    }  } \cs_new_protected:Nn \__egreg_range_compare:nn  {   \bool_set_true:N \l__egreg_range_compare_bool   % do a loop on the digits in the first number   \int_step_inline:nn { \str_count:n { #1 } }    {     \bool_if:NTF \l__egreg_range_compare_bool      {       % if the digits coincide, print nothing       \str_if_eq:eeF { \str_item:nn { #1 } { ##1 } } { \str_item:nn { #2 } { ##1 } }        {         \str_item:nn { #2 } { ##1 }         % at the first differing digit, don't compare any longer         \bool_set_false:N \l__egreg_range_compare_bool        }      }      { \str_item:nn { #2 } { ##1 } }    }  } \ExplSyntaxOff  \begin{document} \setcounter{page}{95}\index{Topic0|(}Text\newpage \setcounter{page}{101}\index{Topic0|)}Text\newpage \setcounter{page}{124}\index{Topic1|(}\index{Topic2|(}Text\newpage \setcounter{page}{126}\index{Topic1|)}Text\newpage \setcounter{page}{134}\index{Topic2|)}Text\newpage \setcounter{page}{141}\index{Topic0}Text\newpage \index{Topic0}Text\newpage \index{Topic0}Text \printindex \end{document}   yields the following error message:  ! LaTeX3 Error: Too many '--' tokens when trying to split argument.  For immediate help type H &lt;return&gt;.  ...                                                l.3   \item Topic0, \checkrange{95--101, 141--143}   Entering non-stop yields the following output:    which omits "-3" from the second index entry of Topic0. The question is how to get around this.  Edit following up on egreg's ingenious solution.  I don't know if I should ask a new follow up question, so I'm posing it first here in this edit.  egreg's solution doesn't seem to handle index sub-topics.  If one adds the following 4 lines  \setcounter{page}{135}\index{Topic0!sub0|(}Text\newpage \setcounter{page}{136}\index{Topic0!sub0!subsub0|(}Text\newpage \setcounter{page}{139}\index{Topic0!sub0!subsub0|)}Text\newpage \setcounter{page}{140}\index{Topic0!sub0|)}Text\newpage   after  \setcounter{page}{134}\index{Topic2|)}Text\newpage   in the solution, one gets the following errors:  Output written in mwe.ind. Transcript written in mwe.ilg.  (mwe.ind ! Extra }, or forgotten \endgroup. l.4     \subitem sub0, 135--140}  ? r OK, entering \nonstopmode... ! Extra }, or forgotten \endgroup. l.5       \subsubitem subsub0, 136--139}  ) [145] (mwe.aux) )   and this output:    Alas I don't understand egreg's code well enough to make the required changes. Add warning when changing display name Merging columns from two separate files Looking for a post-apocalyptic trilogy where pilot is nursed back to health by natives of the land I read a series of books around 2005. I think it might have been a trilogy.  Basic story line is in a post apocalyptic world. The main character is a pilot for the Federation or something similar to that. He crash lands, is nursed back to health by the natives, comes to like and respect them, and does not want to fight against them.  Also in the later book I think it also involves another race who are very Japanese-ish. They drink sake and the "clan" names are things like Fu-Jitsu and Mitsu-Bishi. I am currently working on an update of tkz-euclide and new documentation but I just met the following problem. If I compile this code with pdflatex or lualatex and pgf 2.1 (The library babel does not exist with pgf 2.1) then the code compiles fine but with 3.0 pgf I must comment on the call \usetikzlibrary{babel} (pgf doc gives so I recommend to always load this library by saying \usetikzlibrary{babel}).  I'd like to know where the error occurred. Especially I would like to work around this problem.   With \usetikzlibrary{babel} I get the error Latex Error: ./untitled 2.tex:46 Undefined control sequence. l.1 \tkz@line@start  Bug in babel, bug in tikzlibrary{babel} bug in my code. I would like to keep  \usetikzlibrary{babel} but to avoid the error.  %!TEX TS-program = pdflatex \documentclass[a4paper]{article} \usepackage[french]{babel} \usepackage{tikz}  \usetikzlibrary{calc} %\usetikzlibrary{babel}  \makeatletter \pgfkeys{%     tkzdrawline/.cd,     start/.code         =  {\gdef\tkz@line@start{#1}},%     end/.code           =  {\gdef\tkz@line@end{#1}},%     start style/.code   =  {\tikzset{tkzstartstyle/.style={#1}}},%     end style/.code     =  {\tikzset{tkzendstyle/.style={#1}}},%    /tkzdrawline/.unknown/.code ={\let\searchname=\pgfkeyscurrentname\pgfkeysalso{%                          \searchname/.try=#1,/tikz/\searchname/.retry=#1}}}  %&lt;--------------------------------------------------------------–&gt; %                              Draw line %&lt;--------------------------------------------------------------–&gt; \pgfkeys{tkzdrawline/.cd, start/.code       =  {\def\tkz@line@start{#1}},% end/.code         =  {\def\tkz@line@end{#1}}, start style/.code =  {\tikzset{tkzstartstyle/.style={#1}}}, end style/.code   =  {\tikzset{tkzendstyle/.style={#1}}},   /tkzdrawline/.unknown/.code   = {\let\searchname=\pgfkeyscurrentname                                  \pgfkeysalso{\searchname/.try=#1,                                 /tikz/\searchname/.retry=#1}}}%         %&lt;--------------------------------------------------------------–&gt;  \def\tkzDrawLine{\pgfutil@ifnextchar[{\tkz@DrawLine}{%                                   \tkz@DrawLine[]}}%   \def\tkz@DrawLine[#1](#2,#3){%  \begingroup   \pgfkeys{tkzdrawline/.cd,start = {}, end = {}, start style={}, end style={},}   \pgfqkeys{/tkzdrawline}{#1}   \draw[ /tkzdrawline/.cd,#1] (#2) to%     node [at start,left,/tkzdrawline/.cd,tkzstartstyle] {\tkz@line@start}%     node [at end,right,/tkzdrawline/.cd,tkzendstyle] {\tkz@line@end} (#3);  \endgroup  }%   \makeatother   \begin{document}    \begin{tikzpicture}      \node (A) at (0,1) {};       \node (B) at (5,2) {};       \tkzDrawLine[end = {$\alpha$}](A,B)       \end{tikzpicture}  \end{document} In short, :     If you go before the Board of   Directors and ask for 1.5 billion   dollars to build a nuclear reactor, no   one will review or discuss the details   of the plant. They will assume that   experts have been over every inch of   the plans, and not want to look   foolish by asking a silly question.      However, if you ask the same group to   approve a 30 dollar expenditure for   lumber with which to build a bikeshed   (presumably a British term for the   smallest possible building) then be   prepared for a 45 minute discussion   about all aspects of the bikeshed,   including the color of the paint. The   explanation for this is that everyone   can grasp the scale of a bikeshed.   To transform this problem to Stack Overflow terms, questions that are trivial such as "How to convert a C string to a QString or "What is the difference between . and ->" get a lot more views, answers and votes than real questions of the form "How do I frob this widget" that bother real professionals. Fewer views mean less possible votes for good answers and hence less incentive for people to answer.  How can Stack Exchange solve this problem? Fix a wobbly Shimano freehub? What is the method to make files in a USB Flash Drive undeletable? Does these method work with USB 3.0 interface?  (I have surveyed some Flash Drive manufacturers, and I only see undeletable files for USB 2.0 drives, and not 3.0 drives. One manufacturer claimed that USB 3.0 cannot do undeletable file, so I am curious whether the claim is correct...) What's the Pythonic way to combine two sequences into a dictionary? Is there a more concise way of doing this in Python?:  def toDict(keys, values):   d = dict()   for k,v in zip(keys, values):     d[k] = v    return d Using a non-centralized structure for tex files - similar to \graphicspath The definite and zero article with the word "civilization" What's the negation of "I used to be"? Surely not "I didn't used to be"? Is there a formula for $(1+i)^n+(1-i)^n$? I'm wondering if there is a formula for the value of $(1+i)^n+(1-i)^n$?  I calculated the first terms starting with $n=1$ to be, in order, $2$, $0$, $-4$, $-8$, $-8$, $0$, $16$, $\dots$  So it seems to be some sequence of positive and negative powers of $2$ with $0$s thrown in. Is there a more explicit formulation of what $(1+i)^n+(1-i)^n$ is, based on $n$?   With the binomial theorem, I get it equal to $$ \sum_{k=0}^n\binom{n}{k}i^{n-k}(1+(-1)^{n-k}). $$ Can this be made nicer? Thanks. How to extrude and scale with an even offset? Problem: Prove that $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$.  My work: So I think I have to do a proof by induction and I just wanted some help editing my proof.  My attempt:  Let $P(n)=1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$.  Then $$P(1)=1^2=\frac{1(1+1)(2+1)}{6}$$ $$1=\frac{6}{6}.$$ So $P(1)$ is true.  Next suppose that $P(k)=1^2+2^2+\cdots+k^2=\frac{k(k+1)(2k+1)}{6}$ for $k \in \mathbb{N}$. Then adding $(k+1)^2$ to both sides of $P(k)$ we obtain the following: $$1^2+2^2+\cdots+k^2+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$$ $$=\frac{2k^3+3k^2+k+6(k^2+2k+1)}{6}$$ $$=\frac{2k^3+9k^2+13k+6}{6}$$ $$=\frac{(k^2+3k+2)(2k+3)}{6}$$ $$=\frac{(k+1)(k+2)(2k+3)}{6}$$ $$=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$$ $$=P(k+1).$$ Thus $P(k)$ is true for $k \in \mathbb{N}$. Hence by mathematical induction, $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ is true for $n \in \mathbb{N}$.