A word to describe a person who refuses to acknowledge illness in others
Word for the opposite of "hypochondriasis"
How do I install Komodo Edit? New to linux, help me install this program Komodo-Edit-7.1.0-10496-linux-x86_64  I'm running Ubuntu 12.04 64bit. Possessive of a word that is already possessive Is it true that $aH = bH$ iff $ab^{-1} \in H$ Let $H$ be a subgroup of a group $G$. The I know that for $a,b\in G$ we have $aH = bH$ if and only if $a^{-1}b \in H$.  My question is if it is also true that $aH = bH$ if and only if $ab^{-1} \in H$? Does it matter where the $-1$ goes?  I am guessing that this is not true since we keep putting the $-1$ on the left element, but I am also thinking that it might be true because they look like the same.  If $G$ is abelian, then it doesn't matter, so a possible counter example would have to involve a non-abelian group. If $H$ is a normal subgroup, then it should also be true because then $a^{-1}bH = Ha^{-1}b$. Show that the set of polynomials with rational coefficients is countable. Prove that to each $\epsilon >0$, there exists a $\delta >0$ so that the Lebesgue integral... Suppose $f$ is in $L^1$ space of $\mu$, where $\mu$ is the Lebesgue measure. Prove that to each $\epsilon &gt;0$, there exists a $\delta &gt;0$ so that the Lebesgue integral of the absolute value of $f$ is less than $\epsilon$ (over the set $E$) whenever the measure of $E$ is less than $\delta$.  I know that the integral of $f$ is finite. So for arbitrary $f$ we can find the integral of $f$ to be less than some $\epsilon$, but how would I connect that with the measure of $E$ less than $\delta$. Any help would be welcome. How can I recursively delete all files of a specific extension in the current directory? How do I safely delete all files with a specific extension (e.g. .bak) from current directory and all subfolders using one command-line? Simply, I'm afraid to use rm since I used it wrong once and now I need advice. is_active_sidebar() Always Returns False The abstract environment changes the \linewidth for the whole document In the article class, it seems that using the abstract environment has an effect to the rest of the document. More precisely, it seems that it redefines the \linewidth, which equals \textwidth without abstract, is smaller with the abstract. My feeling is that \linewidth is redefined for the abstract environment, but the value is not reseted at the end of the abstract.  Here is a minimal example:  \documentclass{article} \begin{document} \abstract{This is an abstract. This is an abstract.}\bigskip  This is text. This is text. This is text. This is text. This is text. This is text.\\  \rule{\textwidth}{3em} \rule{\linewidth}{3em} \end{document}   The output is not the same if you comment the \abstract line or not.  Do you have the same problem as I do when you compile (if not, I will upload some PDF files to show)? Is this a (known) bug in LaTeX2e? How can we circumvent the problem?  Note: I compile with the pdflatex command, and I use pdfTeX, Version 3.1415926-1.40.10 (TeX Live 2009/Debian). Does $k+\aleph_0=\mathfrak{c}$ imply $k=\mathfrak{c}$ without the Axiom of Choice? I'm currently reading a little deeper into the Axiom of Choice, and I'm pleasantly surprised to find it makes the arithmetic of infinite cardinals seem easy. With AC follows the Absorption Law of Cardinal Arithmetic, which states that for $\kappa$ and $\lambda$ cardinal numbers, the larger infinite and the smaller nonzero, then $\kappa+\lambda=\kappa\cdot\lambda=\max(\kappa,\lambda)$.  I was playing around with the equation $k+\aleph_0=\mathfrak{c}$ for some cardinal $k$. From the above, it follows that $\mathfrak{c}=k+\aleph_0=\max(k,\aleph_0)$, which implies $k=\mathfrak{c}$.  I'm curious, can we still show $k=\mathfrak{c}$ without the Axiom of Choice? Is it maybe possible to bound $\mathfrak{c}-\aleph_0$ above and below by $\mathfrak{c}$? But then I'm not quite sure such algebraic manipulations even mean anything, or work like that here. Certainly normal arithmetic does not! Thanks. How do I enable the "Universe" repository? How do I get to the Universe Repository in supported versions of Ubuntu? Finding distance between points in two different layers along route in ArcMap? I need to find the distance between points in separate layers along a stream network. I have source sites for macro-invertebrates and sites where they are lacking. I need to find the distance between the source sites and the sites where they are not through the stream network.   I tried the linear referencing toolbox but the locate features along route tool can only use one layer of points and has not way to differentiate between sources and not source sites.  Any ideas?  I am using ArcMap 10.3? What should I do when a hacker gives me hacked money? I was peacefully playing GTA Online (yeah, was killing someone coming out of ls customs) then a hacker came and killed me using explosive ammos, I called him a hacker and he teleported to me and gave me 790k. I know that people get banned for receiving this kind of money, what should I do to avoid the ban? How can I pause a render in progress and then resume it at a later time? I would like to pause a render that I have started, with the intention of resuming it at a later time (but during the same Blender session).  How can I do this? Is there a way to disable the hotkeys? Using PyQGIS to transform layer of polygons into layer of holes inside bbox? I'm currently using  but it sometimes falls over when the RSS or Atom feed isn't well formed. Are there any other options for parsing RSS and Atom feeds with PHP? Connectedness in the Quotient Space Windows Batch recursive delete old files That is,     why is $\varphi(AB)=\varphi(A)\varphi(B)$, if $A$ and $B$ are two coprime positive integers?   It's not just a technical trouble—I can't see why this should be, intuitively: I bellyfeel that its multiplicativity should be an approximation at most. And why the minimum value for $\varphi (n)$ should be $ \frac{4n}{15} $ completely passes me by. Photoshop CS6 Resize Rectangle with Rounded Corners This question is about Photoshop CS6. Hopefully the feature was added in this version. Before posting I looked up Google and  and  but didn't find an answer for CS6.  So the question is: Is there a feature within CS6 or a script that allows to resize a vector element with rounded corners without using the Direct Selection Tool (white arrow tool) and not changing the border radius? What would be great would be something like simply using the transform tool or something like a grid resize tool.