Can't get into my account can't remember password and don't have email account and phone # on account
I have an old Facebook account that I created, uploaded a couple of photos and simply ignored (i.e. no friends added). Now, I can't remember the email address associated with the account and I don't remember the password either. How can I find out which email address I had associated with that account?  P.S. I have created a new Facebook account and my picture can be used to identify me (if I can get someone at Facebook to take a look). Other than that, I'm willing to have that old account deleted altogether if recovery isn't possible.  Is there a way to resolve this issue?
How to delete a file from Time Machine backups in Big Sur? Can I disable full-disk encryption? Do neutrinos not couple to the Higgs field? Look at this if block:  #!/bin/bash  str="m.m" if [[ "${str}" =~ "m\.m" ]]; then     echo "matched" else     echo "not matched"     exit 1 fi  exit 0   This should print "matched", but it doesn't. Where am I going wrong? Batch renaming files I have a directory full of images:  image0001.png image0002.png image0003.png ...   And I would like a one-liner to rename them to (say).  0001.png 0002.png 0003.png ...   How do I do this? Close Vote Review Audits are too obvious The other day, I got an obviously good question in a Close Vote Review.  Even if I didn't read the question,  is a bit of a giveaway.  Today I got one where   Also, on sites other than SO (where I understand the queues are huge), the number of reviews needed is often low enough (single digits) that you usually notice when you get an extra one.  I don't mind - I'm quite happy to get easy-to-pass audits - but maybe they could be a little harder... Is it common for native speakers to put a question mark at the end of declarative sentence? Let $X\subset \mathbb{R}$ Lebesgue measurable, $|X|<|\mathbb{R}|$, is it true that $X$ is null? Let $X\subset \mathbb{R}$ Lebesgue measurable, $|X|&lt;2^{\aleph_0}$, is it true that $X$ is null?  Of course I am not assuming the Continuum Hypothesis.   EDIT: It might be helpful to know that all Borel measurable sets have cardinality either $\aleph_0$ or $2^{\aleph_0}$. Then a measurable set of cardinality strictly between those two must be Lebesgue but not Borel measurable. I'm being forced to use my Macbook more and more with my metered 4G connection. I really need to keep data usage limited. Windows has an  mode which really helps with this.   What's the best way to handle this issue in the OSX world? Perhaps some use of the firewall? Is there any sort of logging happening as to what is using the network?  Thanks for your help :-) Contour method to show that $\int_0^\infty\frac{x-\sin x}{x^3} \, dx=\frac\pi4$ Show that $$\int_{0}^{\infty} \frac{x - \sin(x)}{x^3} \, dx = \frac{\pi}{4}$$   My attempt is as follows: Let $$f(z) = \frac{z - i e^{iz}}{z^3}$$ and consider the contour on $[\epsilon, R]$ followed by a semicircular arc in the counter clockwise direction, then on $[-R, -\epsilon]$, then the semicircular clockwise contour avoiding the origin. We have, then, that  $$0 = \int_{\Gamma} f(z) dz = \int_{[\epsilon, R]} f(t) dt + \int_{C_R}f(Re^{it})Rie^{it}dt + \int_{[-R, -\epsilon]}f(t)dt + \int_{C_{\epsilon}}f(\epsilon e^{-it})\epsilon i e^{-it} dt$$  Then the first and third integrals ( $I_1$ and $I_3$) combine so that  $$I_1 + I_3 = 2\int_{\epsilon}^R \frac{t - \sin{t}}{t^3}\,dt$$  Further,   $$|I_{C_R}| \leq \int_0^\pi \left|\frac{Re^{it} - ie^{-R\sin{t}}e^{iRcos{t}}}{R^2 e^{2it}} \right|dt \rightarrow 0 \text{ as } R\rightarrow \infty$$  (I've omitted the details, it isn't too bad to bound)  However, I'm having trouble computing the limit  $$\lim_{\epsilon \rightarrow 0}\int_{C_{\epsilon}} f(\epsilon e^{-it})\epsilon i e^{-it} dt$$  No matter which way I look at it, it seems like this limit does not exist. Perhaps I'm seeing something wrong or have I chosen a bad $f(z)$? Structure of ideals in the product of two rings $R$ and $S$ are two rings. Let $J$ be an ideal in $R\times S$. Then there are $I_{1}$, ideal of $R$, and $I_{2}$, ideal of $S$ such that $J=I_{1}\times I_{2}$.  For me it's obvious why $\left\{ r\in R\mid \left(r,s\right)\in J\text{ for some } s\in S\right\}$ is an ideal of $R$ (and the same for $S$) so I can prove $J$ is a subset of the product of two ideals and also that the product of these two ideals is an ideal. But I can't see how to prove equality without assuming existence of unity or commutativity. Trying basically to show that if $\left(r,s\right)\in J$ then also $\left(r,0\right),\left(0,s\right)\in J$. Thank you for your help! I want to make a Launcher that will do privileged (requires root) commands, but I don't want to see the terminal or have to type my password into the terminal. So I found the command, gksudo that allows me to enter my password so that I can escalate my privilege level. Now I want to couple this, so that I can change the partition priority using this command cgpt add -i 6 -P 0 -S 0 /dev/sda (this command is privileged) and have it automatically reboot from this one launcher.  I setup a Luncher like this:  Name: ChromeOS  Description:  Command: gksudo gnome-terminal -e "cgpt add -i 6 -P 0 -S 0 /dev/sda;reboot"  Comment:  But it does not work. Any tips? I am able to successfully use wp.hooks.applyFilters() and wp.hooks.addFilter(), but I'm not able to render a React Component using wp.hooks.addAction(). However, the console.log() runs just fine within the callback for an action. This is my code:  import { createHooks } from '@wordpress/hooks';  let globalHooks = createHooks();  const Header = () =&gt; {     return (         &lt;h1&gt;Header Component&lt;/h1&gt;     ); }  const RevisedHeader = () =&gt; {     return (         &lt;h1&gt;Revised Header Component&lt;/h1&gt;     ) }  export default class App extends Component {      constructor( props ) {         super();          globalHooks.addFilter( 'replace_header', 'revisedHeaderCallback', this.revisedHeaderCallback );         globalHooks.addAction( 'after_replace_header', 'afterRevisedHeaderCallback', this.afterRevisedHeaderCallback );     };      revisedHeaderCallback( header ) {         return &lt;RevisedHeader /&gt;;     }      afterRevisedHeaderCallback() {         return &lt;RevisedHeader /&gt;;     }      render() {          return(             &lt;div className="App"&gt;                 { globalHooks.applyFilters( 'replace_header', &lt;Header /&gt; ) }                 { globalHooks.doAction( 'after_replace_header' ) }             &lt;/div&gt;         );     } } I really enjoyed that site. I've only posted on it once, but I enjoyed seeing other people's posts and appreciated reading about entrepreneurial dilemmas. I am trying to define an alternative \item command that would allow me to use a different color for some of the items. My code has the following form:  \documentclass{article} \usepackage{color}  \definecolor{gray}{rgb}{0.7,0.7,0.7} \newcommand{\grayitem}[1]{{\color{gray} \item #1}}  \begin{document}  \begin{enumerate} \item First item. \grayitem Second item. \item Third item. \end{enumerate}  \end{document}   I was hoping that  \grayitem Second item.   would produce the same result as  {\color{gray} \item Second item.}   but for some reason, these are not equivalent. They produce the following results, respectively:  &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;   Why are these two lines of code not equivalent? What am I doing wrong and how can I change the definition of \grayitem to produce the second result?  I have also tried removing a pair of braces, like this  \newcommand{\grayitem}[1]{\color{gray} \item #1}   but then the third item becomes gray as well.  The solution should also work for multi-paragraph items. How long would the conference be for? I want to know if this is grammatically correct " How long would the conference be for?". Java Strings: "String s = new String("silly");" I'm a C++ guy learning Java. I'm reading Effective Java and something confused me. It says never to write code like this:  String s = new String("silly");   Because it creates unnecessary String objects. But instead it should be written like this:  String s = "No longer silly";   Ok fine so far...However, given this class:  public final class CaseInsensitiveString {     private String s;     public CaseInsensitiveString(String s) {         if (s == null) {             throw new NullPointerException();         }         this.s = s;     }     :     : }  CaseInsensitiveString cis = new CaseInsensitiveString("Polish"); String s = "polish";    Why is the first statement ok? Shouldn't it be   CaseInsensitiveString cis = "Polish"; How do I make CaseInsensitiveString behave like String so the above statement is OK (with and without extending String)? What is it about String that makes it OK to just be able to pass it a literal like that? From my understanding there is no "copy constructor" concept in Java? Prove $|\arctan x - \arctan y|\leq |x-y|$ How many Naruto Shippuden dubbed episodes are there? I am watching Naruto and I am currently on episode 164. How many episodes of Shippuden do they have dubbed, and are they still dubbing it? Just recently, the right-click context menus in Windows 7 have been acting in a kind of left-handed way. That is, when possible, the initial menu appears to the left of the mouse pointer. When I hover over a submenu option, the next level of context menu appears to the left of the first and so-on.  Basically, what normally happens when you're right at the right-hand edge of the screen is happening all the time - except when I'm at the left-hand edge. Perhaps it's some kind of left-handed users option, but that doesn't seem likely - more like a right-to-left language option, except that I've never even tried to use a right-to-left language.  I haven't checked every app. for the issue - probably only Windows Explorer, definitely not in Firefox, for the moment that's as far as I've checked.  Scratch that - this affects menu bars as well as context menus. The menus dangle down to the left of the menu-bar item instead of the right, and submenus open to the left too. This seems to apply to most applications, but not too Firefox or Thunderbird.  I find this extremely annoying.  I have no idea how this occurred, though it isn't the only weird glitch I've had in Windows 7 in the last few days - see   I assume there's a setting somewhere that has got changed, but I can't find anything. There's also a chance that I've suffered some kind of registry corruption, so I might need to apply a fix in the registry - not something that scares me, but I need to know what to fix and how.  So - does anyone know what setting has been changed, and how I can fix it?  Also, beyond CCleaner (which I already have), are there any tips on making sure that the registry stays clean and sane?  Finally, both Microsoft Security Essentials and Malwarebytes Antimalware seem happy that I don't have a malware problem. Should my recent glitches make me doubt that?