Accidentally drop table in database that has no backup
How do I get back some deleted records?
VLC will not install How do you prove that $\sqrt{2} + \sqrt{5}$ is irrational?    I tried to prove it by contradiction and got this equation:  $a^2/b^2 = \sqrt{40}$. New questions, Show Recommended, Sort Newest not working I found the following proof arguing for the irrationality of $\sqrt{2}$.     Suppose for the sake of contradiction that $\sqrt{2}$ is rational, and choose the least integer $q &gt; 0$ such that $(\sqrt{2} - 1)q$ is a nonnegative integer. Let $q'::= (\sqrt{2} - 1)q$. Clearly $0 &lt; q' &lt; q$. But an easy computation shows that $(\sqrt{2} - 1)q'$ is a nonnegative integer, contradicting the minimality of $q$.   Clearly, this is a vague proof in that it expects the reader to work out the math involved on his own. I took that as an exercise and worked out the following:  Assuming $\sqrt{2}$ to be rational, let $$\sqrt{2} = \frac{p}{q}$$ where $p$ and $q$ are least possible integers.  Clearly then, $\sqrt{2} - 1$ would also be rational, $$\sqrt{2} - 1 = \frac{p-q}{q}$$ where both $(p-q)$ and $q$ are also integers.  Multiplying both sides by $q$, we get $$(\sqrt{2} - 1)q = (p-q) = q'$$ where $q'$ is an integer. Using, $2&gt;\sqrt{2} &gt; 1$ , we get $$1&gt;\sqrt{2}-1&gt;0$$ proving that $q'$ is a non-negative integer.  As stated in the proof, computing $(\sqrt{2}-1)q'$ we get, $$(\sqrt{2}-1)q' = \frac{(p-q)^2}{q}$$which is not necessarily an integer (in fact necessarily not an integer). However, the proof claims it to be.  Am I missing something or something is wrong with how I proceeded?  Here is a link to the textbook where I found this proof: , (p 32, problem 1.14).  P.S. This is my first post here and my first time with TeX. I'd also like to know if there is anything wrong with the way my question is presented. I have seen similar posts but none seem to work. I simply need to remove the dots in the ToC between the chapter titles and the page numbers. I think the page numbers should line up (as if there were dots). This is using the memoir class. Here is my ToC info:  %%%% Do the ToC \let\oldtoc\tableofcontents \renewcommand{\tableofcontents}{\clearpage\pagestyle{toc}\oldtoc}  \renewcommand{\contentsname}{TABLE OF CONTENTS} \renewcommand{\listfigurename}{LIST OF FIGURES} \renewcommand{\listtablename}{LIST OF TABLES} \renewcommand*{\tocheadstart}{\vspace*{-\topfiddle}} \renewcommand*{\aftertoctitle}{\thispagestyle{plain}%   \par\nobreak \mbox{}\hfill{\normalfont Page}\par\nobreak} \renewcommand*{\cftchapterfont}{\normalfont} \renewcommand*{\cftchapterpagefont}{\normalfont} \renewcommand*{\cftchapterleader}{%   \cftchapterfont\cftdotfill{\cftchapterdotsep}} \renewcommand*{\cftchapterdotsep}{\cftdotsep} %\renewcommand*{\cftchaptername}{CHAPTER~} %%% no extra space before the entry \setlength{\cftbeforechapterskip}{0pt plus 0pt} %% no extra 'chapter' space in LoF/LoT \renewcommand*{\insertchapterspace}{}   I tried everything that was suggested. Nothing seemed to make any difference. I am stuck with this format. Any suggestions would be greately appreciated! Why should users never use normal sudo to start graphical applications? I'm trying to install , an open-source Microblogging application on my Windows development server using XAMPP as per the . The website cannot find PEAR, and throws the below errors:  Warning: require_once(PEAR.php) [function.require-once]: failed to open stream: No such file or directory in C:\xampplite\htdocs\laconica\lib\common.php on line 31 Fatal error: require_once() [function.require]: Failed opening required 'PEAR.php' (include_path='.;\xampplite\php\pear\PEAR') in C:\xampplite\htdocs\laconica\lib\common.php on line 31   PEAR is located in C:\xampplite\php\pear phpinfo() shows me that the include path is .;\xampplite\php\pear  What am I doing wrong? Why isn't the PEAR folder being included? Which application does the vi command open? When you type vi in the terminal, the following output is shown  ~                             VIM - Vi IMproved                                 ~                                                                              ~                               version 7.3.547                                  ~                           by Bram Moolenaar et al.                             ~           Modified by pkg-vim-maintainers@lists.alioth.debian.org              ~                 Vim is open source and freely distributable                    ~                                                                                ~                        Help poor children in Uganda!                                                                                So according to this, vi must be launching VIm. But when you type  vim it gives  aditya@aditya-desktop:~$ vim The program 'vim' can be found in the following packages:  * vim  * vim-gnome  * vim-tiny  * vim-athena  * vim-gtk  * vim-nox Try: sudo apt-get install &lt;selected package&gt;   This shows that vim is not installed. Notably man vi and man vim launch the same man pages.  So what does the vi command actually launch? What are the advantages of 10-bit monitors? To support 10-bit color the following are needed:   A monitor supporting it. A GPU supporting it (only AMD FirePro and NVIDIA Quadro support this?). Compatible software. Unless I am mistaken there are very few programs out there supporting 10-bit color. Photoshop is a notable example.   The questions are about how 10-bit monitors perform in comparison with 8-bit monitors:   In which situations would a 10-bit monitor give a noticeable advantage over an 8-bit monitor (say, for professional photography)? Have 10-bit monitors been compared against 8-bit monitors based on subjective or objective tests? What were the results? Human eyes can see only 10m colors, so would using a monitor with 1b colors make a difference? In a ring $(A,+, \cdot)$ if $aba = a$ then $bab = b$ and all non zero elements in $A$ are invertible. Let $\left(A,+, \cdot\right)$ be a ring with $1$  that satisfies the following condition:  For any nonzero $a\in A$, there exists a unique $b\in A$ such that $aba = a$.  Show that $b$ also satisfies $bab = b$, and that all nonzero elements of $A$ are invertible. How to convert Bedrock texture pack to Java Edition Minecraft How to revert to factory settings? I am trying to execute a web service call to our backend system when the opportunity is closed/won. I am attempting to do a  POST to our endpoint. All the api call is really carrying the account name and a few structural detail back (All listed below). My problem is that when the trigger executes I get the following error in the debug logs "System.CalloutException: You have uncommitted work pending. Please commit or rollback before calling out".  The Process Builder Passing in the company name:   APEX Class  public class Play2 { @TestVisible static HttpResponse res;  @InvocableMethod public static void sendNotification(List&lt;String&gt; name) {        HttpRequest req = new HttpRequest();     Http http = new Http();     String payload = JSON.serialize(         new Map&lt;String, String&gt; {             'companyName' =&gt; name.get(0),             'salesforceAccountType' =&gt; 'Company',             'salesforceSynchStatus' =&gt; 'Enabled'         }     );      req.setEndpoint('https://TheEndPoint');     req.setMethod('POST');     req.setHeader('Authorization', 'The Mighty API KEY');     req.setHeader('Content-Type', 'application/json');      req.setBody( payload );     req.setCompressed(true);       try {         System.debug('Point 1');         res = http.send(req);         System.debug('Point 2');     } catch(System.CalloutException e) {         System.debug('Callout error: '+ e);         System.debug(res.toString());     } } } I've posted a question or an answer, and I have received one or more downvotes. I am not really sure why I was down voted, or I think I know why, but I disagree.  Why did this happen, and what should I do about it? Why Bobbi and Hunter can no longer work for S.H.I.E.L.D.? Current code is:  \begin{enumerate}[label=LEVEL \arabic{*},leftmargin=*]     \item Description of Level 1 \quad L1 \\ \emph{more text}     \item Description of Level 2 \quad L2 \\ \emph{more text}  \end{enumerate}   but the \quad does not align L1 and L2 uniformly vertically. I want to replace \quad with a tab but the tabbing environment does not allow mixing with an enumerate! In Python, I can compile a regular expression to be case-insensitive using re.compile:  &gt;&gt;&gt; s = 'TeSt' &gt;&gt;&gt; casesensitive = re.compile('test') &gt;&gt;&gt; ignorecase = re.compile('test', re.IGNORECASE) &gt;&gt;&gt;  &gt;&gt;&gt; print casesensitive.match(s) None &gt;&gt;&gt; print ignorecase.match(s) &lt;_sre.SRE_Match object at 0x02F0B608&gt;   Is there a way to do the same, but without using re.compile. I can't find anything like Perl's i suffix (e.g. m/test/i) in the documentation. vectorized array creation from a list of start/end indices Novel about interstellar war: Both sides win, both sides think they lost I recall reading this novel in the &rsquo;90s; my recollection is that the author was better-known for bad fantasy than for SF (but the SF of his that I&rsquo;d read was better than his fantasy).  The story opened in media res just before a battle was joined; there is a dialogue between the captain of one of the ships and one of his crew. We never see the actual battle; we instead see the conquest of Earth by the aliens - and the conquest of the alien homeworld by Earth.  It is slowly revealed that the two sides contrived a way to both avoid the destruction of actual war, and solve their respective worlds&rsquo; respective problems of overpopulation, infrastructure decay, et multae ceterae. We also see that the solution was that the conquering aliens were secretly selecting Earthmen to rule their homeworld, while the humans were selecting aliens to rule Earth.  About halfway through, there is suddenly a third world that must be &lsquo;conquered&rsquo; and brought into the network. The plot gets complicated after that. How to get div height to auto-adjust to background size? How do I get a div to automatically adjust to the size of the background I set for it without setting a specific height (or min-height) for it?