Why is there no option to download 32 bit or 64 bit for Ubuntu?
Why is there no 32 bit ISO for the new Ubuntu 18.04?
Which of these two phrases has more negative undertone?      "The interview I had was not very good"    or      "The interview I had was not good" I have a client that has a Links page in SharePoint 2013. On the page is a column called URL that is set as a hyperlink column, but they want the column to accept pw://example.com/... instead of just ...  Is there a way to change the hyperlink column to allow for this behavior? I considered creating it as a single line of text and then making that text box a hyperlink-enabled box. Is there another native option? If not, which of these would be easiest? I am new to SharePoint so any help would be appreciated! Thanks in advance!   URL    Type the Web address: (Click here to test)  pw://test  Invalid URL: pw://test How can I match multiple occurrences with a regex in JavaScript similar to PHP's preg_match_all()? I am trying to parse url-encoded strings that are made up of key=value pairs separated by either &amp; or &amp;amp;.   The following will only match the first occurrence, breaking apart the keys and values into separate result elements:  var result = mystring.match(/(?:&amp;|&amp;amp;)?([^=]+)=([^&amp;]+)/)   The results for the string '1111342=Adam%20Franco&amp;348572=Bob%20Jones' would be:  ['1111342', 'Adam%20Franco']   Using the global flag, 'g', will match all occurrences, but only return the fully matched sub-strings, not the separated keys and values:  var result = mystring.match(/(?:&amp;|&amp;amp;)?([^=]+)=([^&amp;]+)/g)   The results for the string '1111342=Adam%20Franco&amp;348572=Bob%20Jones' would be:  ['1111342=Adam%20Franco', '&amp;348572=Bob%20Jones']   While I could split the string on &amp; and break apart each key/value pair individually, is there any way using JavaScript's regular expression support to match multiple occurrences of the pattern /(?:&amp;|&amp;amp;)?([^=]+)=([^&amp;]+)/ similar to PHP's preg_match_all() function?  I'm aiming for some way to get results with the sub-matches separated like:  [['1111342', '348572'], ['Adam%20Franco', 'Bob%20Jones']]   or   [['1111342', 'Adam%20Franco'], ['348572', 'Bob%20Jones']] Previous question:    That was when my buttons died.  On other days, my text dies.    And some times it's just italics.    Now, I like blaming @skiwi for this, and he's often nearby when it happens.  But today, he's not here, and it happened again.    I'm running Windows 8 and using Chrome 40.0.2214.94 m.  Refreshing and such does not fix the issue.  Inspect element -> console -> error log:  http://chat.stackexchange.com/chats/8595/messages/new Failed to load resource: the server responded with a status of 409 (Conflict) http://i.stack.imgur.com/2ZJdJ.png?s=128&amp;g=1&amp;g&amp;s=16 Failed to load resource: the server responded with a status of 502 (Bad Gateway) http://i.stack.imgur.com/boUZp.jpg?g&amp;s=16 Failed to load resource: the server responded with a status of 502 (Bad Gateway) http://i.stack.imgur.com/U8P3C.jpg?s=128&amp;g=1&amp;g&amp;s=32 Failed to load resource: the server responded with a status of 502 (Bad Gateway) http://i.stack.imgur.com/boUZp.jpg?g&amp;s=16 Failed to load resource: the server responded with a status of 504 (Gateway Time-out) http://chat.stackexchange.com/messages/19891816 Failed to load resource: the server responded with a status of 409 (Conflict)   What's going on here?  I should note, I can view italics by selecting, then alt-tabbing:    So it reeks of a client-side issue. What's hiding the text from me?  EDIT:  30 april 2015 it happened again =/ Zero visibility while editing a message in a chat room I'm gearing up to deploy a Drupal 7 site and I can't find any documentation about what the recommended security-conscious file and directory permissions should be set to.  Specifically default/files/ (also sub directories?), settings.php, .htaccess and anything else I should be aware of. As a non-native speaker, I feel that it is ridiculous that can and can't could sound so similar in American accent.  Just now, I was just listening to a video in which the speaker with an American accent says     Fat cells can't reproduce themselves.   The speaker dragged the word "can't" longer for emphasis, but perhaps as a result, what I heard at first was     Fat cells can reproduce themselves.   Of course from the context I later figured out what he was really saying, but I felt extremely surprised by how these two words which literally mean the opposite could sound so similar in American accent. How is that not affecting American people's daily conversations? As a non-native speaker, is there a way for me to further clarify them when my listeners are confused? (Like when you say "one-eight" to clarify that you were trying to say "eighteen" as opposed to "eighty".) Urgent Sydney transit visa question How do I make exponential equation with points?  Here are my points:  x = 0, y = 0 x = 1, y = 4.75000 x = 2, y = 9.26250 x = 3, y = 13.54937 x = 4, y = 17.62191   this is same as   x = 0, y = 0 x = 1, y = (0 + 5) * 0.95  x = 2, y = ((5 * 0.95) + 5) * 0.95 x = 3, y = (((5 * 0.95) + 5) * 0.95 + 5)) * 0.95 x = 4, y = ((((5 * 0.95) + 5) * 0.95 + 5)) * 0.95 + 5) ) 0.95   as x value increase by 1, y value is added by 5 and multiplied by 0.95 i tried to make an equation for this but i couldn't  also i tried to use natural exponential but the initial value is 0 so it didn't work  please help me If $|G/H|=4$ then $G$ is union of three proper subgroups Let $G$ be a finite group and $N$ a normal subgroup of $G$ of index $4$. Prove that $G$ has a subgroup of index $2$ (hence normal) and if $G/H$ is not cyclic then $G$ is equal to the union of three proper subgroups.   I understand that $G/H$ is isomorphic to $\mathbb{Z}_4$ or $\mathbb{Z}_2\times \mathbb{Z}_2$. Let $K^*$ be a subgroup of $G/H$ of order $2$. Then $K^*\cong K/H$ for some $K$ such that $H$ is normal in $K$ and hence $[G:K]=[G/H:K/H]=4/2=2$.  If $G/H$ is not cyclic then $G/H\cong \mathbb{Z}_2\times \mathbb{Z}_2$. I don't know how to prove $G$ is the union of three subgroups. Could it be true that as $G/H$ is the union of three proper subgroups $K_1^*,K_2^*,K_3^*$ then the corresponding groups $K_1,K_2,K_3$ have union equal to $G$? Would anyone give me a hint to prove this?  Thank you. I installed the  on my Ubuntu&nbsp;11.04 (Natty Narwhal) with   sudo apt-get install lamp-server^   Now I navigate to /var/www and use   wget http://ftp.drupal.org/files/projects/drupal-7.x-dev.tar.gz   I get the following error.   --2011-08-03 13:59:00--   http://ftp.drupal.org/files/projects/drupal-7.x-dev.tar.gz Resolving  ftp.drupal.org... 64.50.233.100, 64.50.236.52 Connecting to  ftp.drupal.org|64.50.233.100|:80... connected. HTTP request sent,  awaiting response... 200 OK Length: 2757101 (2.6M)  [application/x-gzip] drupal-7.x-dev.tar.gz: Permission denied   Cannot write to `drupal-7.x-dev.tar.gz' (Permission denied).   How do I resolve this error?  I tried to download the same file in the downloads folder, and it went on without an issue. So it seems to be some permission error, but I'm not sure about that or how to solve it. Are delphi variables initialized with a value by default? I'm new to Delphi, and I've been running some tests to see what object variables and stack variables are initialized to by default:  TInstanceVariables = class   fBoolean: boolean; // always starts off as false   fInteger: integer; // always starts off as zero   fObject: TObject; // always starts off as nil end;   This is the behaviour I'm used to from other languages, but I'm wondering if it's safe to rely on it in Delphi?  For example, I'm wondering if it might depend on a compiler setting, or perhaps work differently on different machines.  Is it normal to rely on default initialized values for objects, or do you explicitly set all instance variables in the constructor?  As for stack (procedure-level) variables, my tests are showing that unitialized booleans are true, unitialized integers are 2129993264, and uninialized objects are just invalid pointers (i.e. not nil).  I'm guessing the norm is to always set procedure-level variables before accessing them? Let $q: \mathbb{N} \to \mathbb{Q}$ be a bijection and denote the image of $k \in \mathbb{N}$ by $q_k$.  Let $f: \mathbb{R} \to (0,1)$, $$ f(x) = \underset{q_k \leq x}{\sum_{k \in \mathbb{N}}} 2^{-k} $$  Show that:  1) $f$ is strictly increasing  2) $f$ is continuous at $x$ iff $x \in \mathbb{R} \setminus \mathbb{Q}$.  My work:  Let $x, y \in \mathbb{R}$, $x &lt; y$. Then there exists $q_n \in \mathbb{Q}$ s.t. $x &lt; q_n &lt; y$. Now $$ f(x) = \underset{q_k \leq x}{\sum_{k \in \mathbb{N}}} 2^{-k} &lt; \underset{q_k \leq x}{\sum_{k \in \mathbb{N}}} 2^{-k} + 2^{-n} \leq \underset{q_k \leq y}{\sum_{k \in \mathbb{N}}} 2^{-n} = f(y) $$ so $f$ is strictly increasing.  I think my proof of 1) is ok, but how can I prove 2)? help with some Hilbert style proofs in a propositional logic axiom system. I'm very new to classical propositional logic, and my lecturer is using a system with the following axioms:  A1. X→(Y→X)  A2. (X→(Y→Z))→((X→Y)→(X→Z))  A3. (¬Y→¬X)→(X→Y)  Use uniform substitution and Modes Ponens (MP) as rule of inference. Deduction Theorem also allowed.  I'd like to deduce the following:  (a) X, Y ⊢ ¬(X→(¬Y)) (conjunction, but expressed in terms of NOT and IMPLIES)  (b) X→((X→Y)→Y)  (c) (X→Y)→(((¬Y)→X)→Y)  (d) (((¬Y)→X)→Y)→(X→Y)  Remark: The last two are to show that the axiom (A3) can be replaced with a different axiom:  B3. (¬Y→¬X)→(((¬Y)→X)→Y).  Solution: Bram28 proves (b) and (c) below. I demonstrate (a) and (d).  (d) is alright; first, I claim that X, ((¬Y→X)→Y) ⊢ Y.  Axiom (A1) gives ⊢X→(¬Y→X) so that ((¬Y→X)→Y)⊢X→Y by Hypothetical Syllogism (HS), which Bram28 shows to be valid.  Now that we have ((¬Y→X)→Y)⊢X→Y just apply the deduction theorem to get what we want: ⊢((¬Y→X)→Y)→(X→Y).  (a) is slightly trickier. Axiom (A1) gives ⊢ Y→(X→Y) so that X, Y ⊢ (X→Y).  Next, by part (b), we have as a theorem that ⊢X→((X→¬Y)→¬Y).  Therefore X, Y ⊢(X→¬Y)→¬Y.  But ⊢(¬¬(X→¬Y))→(X→¬Y) by DN Elim, which Bram28 shows to be valid.  By HS, X, Y ⊢(¬¬(X→¬Y))→¬Y.  By Axiom (A3) and MP, X, Y ⊢(Y→¬(X→¬Y)).  Also, rather trivially X, Y ⊢Y.  Finally by MP we conclude X, Y ⊢¬(X→¬Y) as required. Using notify-send with cron Suppose you have an object with zero for the value of all the derivatives of position. In order to get the object moving you would need to increase the value of the velocity from zero to some finite value. A change is velocity is acceleration, so the value of the acceleration would have to increase from zero to some value. You would also need to increase the jerk from zero to some value to have a change in acceleration.  Is there an infinite series of higher derivatives of position for this to work? Or am I missing something? Is there to make the login screen appear on the external display in 18.04? Let $L$ and $R$ be $n \times n$ matrices. I am trying to solve the following minimization problem  $$ L^{k+1} := \arg \min_{L} \lambda \|L\|_{\ast} + \frac{1}{2\mu}\|L-R^{k}\|_{F}^{2} $$ $$ \mbox{s.t.}\ L \succeq 0 $$  where $\|\cdot\|_{\ast}$ and $\|\cdot\|_{F}$ denote the nuclear and Frobenius norms, respectively. Also, $k$ expresses the $k$th iteration when solving an optimization problem successively.  I don't know how to solve this problem. Would you tell me the way? For which values of $\alpha$ and $\beta$ does the integral $\int\limits_2^{\infty}\frac{dx}{x^{\alpha}\ln^{\beta}x}$ converge? Values of $\sum_{n=0}^\infty x^n$ and $\sum_{n=0}^N x^n$ Why does the following hold:  \begin{equation*} \displaystyle \sum\limits_{n=0}^{\infty} 0.7^n=\frac{1}{1-0.7} = 10/3\quad ? \end{equation*}  Can we generalize the above to      $\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ ?   Are there some values of $x$ for which the above formula is invalid?  What about if we take only a finite number of terms? Is there a simpler formula?     $\displaystyle \sum_{n=0}^{N} x^n$   Is there a name for such a sequence?    This is being repurposed in an effort to cut down on duplicates, see here: Coping with  duplicate questions.  and here: .