Error : The markup in the document following the root element must be well-formed
How to fix error: The markup in the document following the root element must be well-formed
Isometric Embedding of a separable Banach Space into $\ell^{\infty}$ How to tell Google that I have changed my website URLs? I have done major updates to my website, and renamed all my URLs.  How can I tell Google that I have renamed these URLs and let Google refresh its index?  My website URL: http://www.pndmasr.com/  My sitemap: http://www.pndmasr.com/sitemap.xml  I have uploaded my sitemap via Google Webmaster Tools many times.  However, every time I search Google for "pndmasr", I still get results with old pages. I have waited more than 3 days but the problem remains. Any suggestions?  Is there a problem with my sitemap? How can I get the HTML source of a given web address in C#? Recovering accidentally deleted files I accidentally deleted a file from my laptop. I'm using Fedora. Is it possible to recover the file? Custom sorting (non-alphabetical) Flash Player works on Firefox, but not on Chromium? I want to play online games, and other online stuff that requires Flash Player on the internet. Flash Player works just fine under Firefox, but in Chromium I just get a blank page. I don't know what is the exact problem or know to fix it. What's the difference between \usepackage{ngerman} and \usepackage[ngerman]{babel} I am quite proficient with the keyboard and try to only use the mouse when necessary. Given that context I was very familiar with the shortcut keys in MS Outlook 2003 and never needed to use the mouse to use that product. Now my company has recently upgraded to MS Outlook 2010 and I am finding it impossible to operate it without having to use the mouse. In particular replying to meeting requests with the old shortcut keys no longer seems possible.   I have held down the ALT key and sometimes it shows some shortcuts on the ribbon but this seems to be limited to ALT+H which tends to alternate between designating "Accept" and "Tentative" without any keys for the other options. Also as this appears to switch between "Accept" and "Tentative" one can't even use this consistently without first having to check what it means for each particular instance.  Any help with this would be greatly appreciated! How does ${!#} work in bash to get the last command-line argument? From the Advanced Bash-Scripting Guide, to get the last command-line argument:     Or: lastarg=${!#}    This is an   indirect reference to the $# variable.    Note that lastarg=${!$#} doesn't work. Every $k$ vertices in an $k$ - connected graph are contained in a cycle. Let $G$ be a $k$-connected graph. Meaning, $G$ has no fewer than $k$ vertices, and for every set of $k-1$ or fewer vertices, if we remove them from $G$, the graph stays connected (Of course, $G$ itself is also connected).  I want to prove that for any $k&gt;1$, if $G$ is $k$-connected, then every set of $k$ vertices is contained in a cycle.  I have tried some ways - mainly using induction by removing one of the vertices of the set from the graph, and/or using Menger's theorem to construct the cycle. But I always encounter problems with making sure that the cycle I'm building deosn't have repeating edges etc.  Help would be greatly appreciated :) Thanks! Why are addition and multiplication commutative, but not exponentiation? We know that the addition and multiplication operators are both commutative, and the exponentiation operator is not.  My question is why.  As background there are plenty of mathematical schemes that can be used to define these operators. One of these is   where $H_0(a,b) = b+1$   (successor op) $H_1(a,b) = a+b$   (addition op) $H_2(a,b) = ab $   (multiplication op) $H_3(a,b) = a^b$   (exponentiation op) $H_4(a,b) = a\uparrow \uparrow b$   (tetration op: $a^{(a^{(...a)})}$ nested $b$ times ) etc.    Here it is not obvious to me why $H_1(a,b)=H_1(b,a)$ and $H_2(a,b)=H_2(b,a)$ but not $H_3(a,b)=H_3(b,a)$  Can anyone explain why this symmetry breaks, in a reasonably intuitive fashion?  Thanks. Finding Stack Overflow/Exchange users based on geographic area? There is a very bad problem with the "Blank screen" function in Ubuntu 17.10. I refer to the setting accessed via Settings->Power->Blank Screen. This seems to actually perform a screen lock, since I need to enter my password after un-blanking the screen. (However, given the behavior described below it's not clear exactly what this function actually does - maybe it just logs you out? maybe it reboots the machine?)  The problem is that a simple screen lock, which one would expect to lock the screen without affecting the state of any applications, in fact closes all applications that are open when the screen lock activates. In other words, if you have the screen lock set to 5 minutes, and you walk away from your machine to get a sandwich or whatever, when you come back and unlock the machine all your apps have closed without saving their state. So far, the only work-around I have found is to entirely disable screen locking in Ubuntu 17.10. My question is, how can this be fixed so that screen locking works properly - that is: the screen locks on schedule after a period of inactivity, and when unlocked all applications that were running are still running?  I am asking a new question for the following reasons:  1) The existing answers to (and comments on)  do not address the issue, and in fact misrepresent the actual problem in ways that make me think the comment writers have not understood what is happening;  2) I cannot usefully address that question because I do not have sufficient reputation to either upvote the question or comment on it, and I cannot answer the question because I have no idea why this problem occurs;  3) I will probably never achieve a reputation of 15, much less 50, on StackExchange. Which is fine, that's not among my life goals.  4) But this question is important and deserves to be addressed. It is a drastically unacceptable bug. iOS app stalls phone when on weak wifi sign I've recently noticed a bug on the Stack Exchange IOS app. When I open the Stack Exchange app on weak wifi, it hangs on the blue load screen as it launches and either crashes the app or the entire phone lags and didn't respond for about a minute.  I'm running iOS 8.4.1, I have the most up to date version of the app, and by 'weak' wifi I mean two out of four bars of strength. Are these solutions of $2 = x^{x^{x^{\:\cdot^{\:\cdot^{\:\cdot}}}}}$ correct? Where is the icon used in this dialogue box? Where on my disk can I find the icon used in this dialogue box in OS X？    It may be Security.icns in OS X 10.7.3    I cannot find it in:   /System/Library/CoreServices/CoreTypes/Contents/Resources There are only LockedIcon.icns and UnlockedIcon.icns in this folder who looks like the icon in the dialogue box,the size of the icon in the dialogue box is 128x128px, but LockedIcon.icns and UnlockedIcon.icns is 32x32px.   lockedIcon.icns    So they are not the same icon.   /System/Library/CoreServices/Finder.app/Contents/Resources/ /System/Library/CoreServices/Problem Reporter.app/Contents/Resources /Applications/System Preferences.app/Contents/Resources /Applications/Utilities/Keychain Access.app/Contents/Resources /System/Library/CoreServices/Certificate Assistant.app/Contents/Resources/ Are $_SESSION variables stored on the client or the server? I am writing a bash script to look for a file if it doesn't exist then create it and append this to it:  Host localhost     ForwardAgent yes   So "line then new line 'tab' then text" I think its a sensitive format.  I know you can do this:  cat temp.txt &gt;&gt; data.txt   But it seems weird since its two lines. Is there a way to append that in this format:  echo "hello" &gt;&gt; greetings.txt no wifi in Ubuntu 15.04 I cannot see WiFi option in my Ubuntu 15.04. What should I do? Revert asap  lspci -nnk | grep 0280 -A2 output:-  03:00.0 Network controller [0280]: Qualcomm Atheros Device [168c:0042] (rev 30) Subsystem: Foxconn International, Inc. Device [105b:e09a]   rfkill list output:-  0: hci0: Bluetooth Soft blocked: no Hard blocked: no 1: acer-wireless: Wireless LAN Soft blocked: no Hard blocked: no      dmesg | grep ath [    9.545199] ath10k_pci 0000:03:00.0: pci irq msi-x interrupts 8 irq_mode 0 reset_mode 0 [   10.029684] ath10k_pci 0000:03:00.0: Direct firmware load for ath10k/cal-pci-0000:03:00.0.bin failed with error -2 [   10.029701] ath10k_pci 0000:03:00.0: Direct firmware load for ath10k/QCA9377/hw1.0/firmware-5.bin failed with error -2 [   10.029704] ath10k_pci 0000:03:00.0: could not fetch firmware file 'ath10k/QCA9377/hw1.0/firmware-5.bin': -2 [   10.029711] ath10k_pci 0000:03:00.0: Direct firmware load for ath10k/QCA9377/hw1.0/firmware-4.bin failed with error -2 [   10.029713] ath10k_pci 0000:03:00.0: could not fetch firmware file 'ath10k/QCA9377/hw1.0/firmware-4.bin': -2 [   10.029719] ath10k_pci 0000:03:00.0: Direct firmware load for ath10k/QCA9377/hw1.0/firmware-3.bin failed with error -2 [   10.029721] ath10k_pci 0000:03:00.0: could not fetch firmware file 'ath10k/QCA9377/hw1.0/firmware-3.bin': -2 [   10.029726] ath10k_pci 0000:03:00.0: Direct firmware load for ath10k/QCA9377/hw1.0/firmware-2.bin failed with error -2 [   10.029728] ath10k_pci 0000:03:00.0: could not fetch firmware file 'ath10k/QCA9377/hw1.0/firmware-2.bin': -2 [   10.029733] ath10k_pci 0000:03:00.0: Direct firmware load for ath10k/QCA9377/hw1.0/firmware.bin failed with error -2 [   10.029735] ath10k_pci 0000:03:00.0: could not fetch firmware (-2) [   10.029737] ath10k_pci 0000:03:00.0: could not fetch firmware files (-2) [   10.029739] ath10k_pci 0000:03:00.0: could not probe fw (-2) I was working on a problem that resulted in the calculation: $20 \equiv 10x \pmod{11}$.  I got the answer $x \equiv 2 \pmod{11}$ with the thought process: Since $10$ is a factor of $20$ I can rewrite $20 \pmod{11}$ as $10x \pmod{11}$ with $x=2$. But isn't this basically the same as using division, which we aren't supposed to do in modular spaces?  If I solve by multiplying $20$ by the multiplicative inverse of $10 \pmod{11}$ which is $10$, I get $200 \pmod{11}$ which simplifies to $2 \pmod{11}$ anyway. Did these numbers just happen to be the same or is my original method valid?