Fixed In-line Indentation
Signature/date line with fixed width
Rewriting Triple Integral in different order How can I rewrite, say: $$\int^1_0\int^{z^2}_0\int^y_0{f}dxdydz$$ In the orders $dzdxdy$ and $dydzdx$?  Would I have to solve this triple integral all the way or is there a simple way to do this? Creating all possible line segments between all points using QGIS How do you prove that $\sqrt{2} + \sqrt{5}$ is irrational?    I tried to prove it by contradiction and got this equation:  $a^2/b^2 = \sqrt{40}$. How does the two slit experiment work? Whilst going through 'A brief history of time', I faced difficulty in understanding the . How can an individual electron cause fringes on screen? I was unable to understand it? Please explain it simply. Character Limit in HTML Is it possible to use multiple here-docs in bash? Can one use multiple here-docs to provide input to a command in bash?  $ cat &lt;&lt;&lt;foo &lt;&lt;&lt;bar bar $ cat &lt;&lt;EOF1 &lt;&lt;EOF2 &gt; foo &gt; EOF1 &gt; bar &gt; EOF2 bar   Obviously, in both cases, the second here-doc is used as stdin, and replaces the first reference. Is the solution to use echos instead?  $ cat &lt;(echo -n foo) &lt;(echo bar) foobar   Also, for some reason, using a combination didn't work for me. Why would that be?  $ cat &lt;&lt;&lt;foo &lt;(echo bar) bar $ cat &lt;(echo -n foo) &lt;&lt;&lt;bar foo Swapping right shift and delete key using xmodmap I'm unsure how to swap these 2, this is the thought process I had but am unsure how to get it working:  remove sh = Shift_R remove dl = Delete  keysym Delete = Shift_R keysym Shift_R = Delete  add sh = Delete add dl = Shift_R   This is based on the code I use to swap caps lock and control.  Any thoughts?  xmodmap output:  dan@lantea:~$ xmodmap -pke | grep Shift_R keycode  91 = Shift_R NoSymbol Shift_R keycode 119 = Shift_R NoSymbol Shift_R dan@lantea:~$ xmodmap -pke | grep Delete keycode  62 = Delete NoSymbol Delete How to send emails and avoid them being classified as spam? How can I tell if a CVE has been fixed in Ubuntu's repositories? Every subspace of a vector space has a complement I want to see if my proof is true or I thought very trivially?    If $H$ is a subspace of a finite dimensional vector space $V$, show there is a subspace $K$ such that $H\cap K=0$ and $H+K=V$    So far I have tried :    $H\subseteq V$ is a subspace $\Rightarrow\,\exists\;K=(V-H)\subseteq V$ $K$ is a subspace because it's the sum of two subspace $V$ and $(-H)$ Now we should prove that $H\cap K=0$ We assume $H\cap K\neq0\Rightarrow\exists\,\vec u\neq0\,,\,\vec u\in H\cap K$ $\vec u\in H\cap K\Rightarrow \begin{cases}\vec u\in H\\\vec u\in K\,\Rightarrow\vec u\in V-H\Rightarrow\vec u\in V,\,\vec u\notin H\end{cases}$ That's a contradiction so our hypothesis was wrong and $\vec u=0$ Is there a way to insert Keyboard Glyphs in Google Docs? Here on the StackExchange framework we have a nice feature, which is apparently called "Keyboard Glyphs" (cf. ):     CTRL+Z   Is there a way (e.g. with an Add-On) to do the same in Google Docs? I browsed the Add-Ons but did not find anything, maybe it's hidden somewhere different (e.g. in a Code Formatting Plugin). How to customize my titlepage? How to remove duplicate entries in dash? What is a "DIE" package? In a list of ICs, along with the familiar package names such as QFN32, LQFP48, etc., I've seen a few ICs to be listed as DIE for the package size. I've never seen that description before as an IC package size, and  does not list it either.  What can it be?  I assume it's some kind of chip-scale package, but it does not reveal the silicon size or any other properties, like number of pins, etc. I read it in Reader's Digest in the seventies. Archaeologists interpreted everyday objects as the TV set and the toilet seat to be sacred and precious. Let $H$ be a subgroup of $G$ with $[G:H]=n$ and $G$ finite.  Let $G$ act on left cosets of $H$ by left multiplication.  Let $N$ be the kernel of the action.  Show that $[G:N]$ divides $n!$.  Ok so I have a theorem that says that if $G$ is simple then it embeds into $S_n$.  So if $G$ is simple than I'm done because $G$ divides $n!$ and thus $N$ divides $n!$ and hence $[G:N]$ divides $n!$.  Now I also know that $N$ is the largest normal subgroup of $G$ contained in $H$.  Thus if $G$ isn't simple then it contains some non-trivial normal subgroup $K$ and thus $K\cap H$ is normal in $H$ and by the 2nd Iso Theorem $H/K\cap H\cong KH/H$...    As you can see I'm not sure where to go from here. I was hoping to prove that $K\cap H$ wasn't trivial and thus that $N$ couldn't be trivial since it must contain $K\cap H$, but I'm not sure I can, and even if I could I'm not sure that would help me in showing that $[G:N]$ is small enough to divide $n!$.  Can anyone give some guidance?  Thanks. I was working on a worksheet for class and came across this problem:      Let $\varphi:R\to S$ be an onto ring homomorphism with the property that every ideal of $R$ is principal. Show that the same must be true for $S$.    I honestly do not know how to do this one! Is it possible to assign numbers generated in an Error-Message? I don't have administrator account on my mac On my mac I have 2 accounts and the both are normal account. So in this moment I don't have a Administrator account, only one that was in the past a administrator, but now is not.  How I can have again a Administrator account? Ignored users still appear in the 'other rooms you're in' message preview Well it might not be a huge issue to some (I suppose it depends on how you use the chat system) but when I "Ignore this user everywhere", their messages still appear on the sidebar for the 'other rooms you're in' if they say something. Albeit I can ignore this feature and the text within it, I just find it odd that I can still see it.   This is only a nuisance when you truly are annoyed by a user and ignoring them in the chat was the best possible alternative to leaving the room or starting an argument (no references here, but any long time chat user has been annoyed by some help vampire at one point in time or another and would just prefer to ignore them for good).   If you have no clue to the sidebar or preview I'm talking about: this is for you,    I have chosen to Ignore this user everywhere (Neal) and then switched to a new room, yet that users response appears in my 'preview' for the 'other rooms you're in' if he is also talking in them.  Note: My ignore of Neal was mostly due to jest, this in no way is to reflect on Neal but the chat feature (I hate to ever single anyone out here).