Vertical align div in a div I am designing a website using bootstrap with a 100% height and 100% width image at the first page. Now I added some text and two buttons above the image and want to horizontal and vertical align. I found several solutions for this here on Stackoverflow. But sadly none of them worked in my case (only the horizontal center worked). So here is my code, I hope you guys can find a solution...    #Startseite {  background: url(http://qnimate.com/wp-content/uploads/2014/03/images2.jpg) no-repeat center center fixed;   width: 100%;  position: relative;  height: 100%;  background-size: cover;  text-align: center;  display: table;  vertical-align: middle; } .CentermiddleStart {   display: inline-block; } #Startseite #button-right{     text-align: right; } #Startseite #button-left{     text-align: left; } &lt;head&gt;   &lt;meta name="viewport" content="width=device-width, initial-scale=1"&gt;     &lt;link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css"&gt; &lt;/head&gt;  &lt;body&gt; &lt;div id="Startseite"&gt;  &lt;div class="container CentermiddleStart"&gt;   &lt;div class="row"&gt;    &lt;div class="col-12"&gt;     &lt;h1&gt;Header&lt;/h1&gt;     &lt;h3&gt;Subheader&lt;/h3&gt;    &lt;/div&gt;   &lt;/div&gt;   &lt;div class="row"&gt;    &lt;div class="col-xs-6" id="button-right"&gt;     &lt;a href="#" class="btn btn-default btn-lg"&gt;Get started&lt;/a&gt;    &lt;/div&gt;    &lt;div class="col-xs-6" id="button-left"&gt;     &lt;a href="#" class="btn btn-default btn-lg"&gt;Get started&lt;/a&gt;    &lt;/div&gt;   &lt;/div&gt;  &lt;/div&gt; &lt;/div&gt; &lt;/body&gt;     So the image should be full size of the view and the buttons and text centered
How can I vertically align elements in a div? I have a div with two images and an h1. All of them need to be vertically aligned within the div, next to each other. One of the images needs to be absolute positioned within the div. What is the CSS needed for this to work on all common browsers? &lt;div id=&quot;header&quot;&gt;   &lt;img src=&quot;..&quot; &gt;&lt;/img&gt;   &lt;h1&gt;testing...&lt;/h1&gt;   &lt;img src=&quot;...&quot;&gt;&lt;/img&gt; &lt;/div&gt;
What's the nice 'trick' to showing that the following expression is irrational?  $2^{1/3} + 2^{2/3}$ What value to use to convert ADC value to voltage? While working on my project, I got to a point where I don't know the answer. I have a 10-bit ADC (the one incorporated in the ATMEGA328P) and it returns values up to 1023 (210 - 1).   The official documentation states that the equation to convert the values from the ADC to the corresponding voltage is:   $$ V = \frac{V_{in} \cdot ADC}{2^{10}} $$  But, if the values go up to 1023, then wouldn't it be better to replace the 210 by 210&nbsp;-&nbsp;1?  Thanks. Finding the derivative of $x\uparrow\uparrow n$ I am trying to find a general derivative for the function: $f(x)=x^{x^{x^{...^{x}}}}$however to do that I must find $f^{\prime }$ and $f^{\prime \prime}$...etc. I am now trying to write down a general expression for $f^{\prime \prime}$ and I have stumble opon the series in the picture below, I wonder, do anyone have an idea of what sort of expantion it migth be ? I have tried breaking it down into three different sums, but I wasn't very successful!  EDIT  So since this post has gotten some attention and the title has been change I will tell you what I am on about, although I already have a post just for the question in the title from before. I am as I said earlier trying to find a general derivative for the function $f(x)=x^{x^{x^{...^{x}}}}$ where $n$ is an arbitrary number, natural number and greater or equal to 2. I have already made success when I tried to find the first derivative, and I have also found some interesting pattern emerge when the derivatives as well as n incresses. But my question here was if anyone knew what this expation might be or look similar to be, since I have yet to be lernt series etc. I did not know that the question needed such background since it's straight a forward question. Run a batch file in a completely hidden way I'm looking for some way to run a batch file (.bat) without anything visible to the user (no window, no taskbar name, .etc..).  I don't want to use some program to do that, I'm looking for something cleaner. I've found  that uses VBScript, but I don't really like using VBS, either. Is the intersection of a closed set and a compact set always compact? In Ubuntu 18.04.1, I would like to use my left Super-key only to simply open the Show Applications menu (aka Dash) like a left click on the dock-icon (with the 9 dots) does and preferably close it again without showing me any open programs.   I can open it with Super + A and even close it again with Super R, but Super L appears to do nothing at all.   In Xfce, something similar was possible by adding a custom keyboard shortcut for Super L and setting the command to xfce4-popup-whiskermenu, which would open and close a menu with favourite applications. Unfortunately, I don't know the command that opens and closes the Show Applications menu.   Is there any way to add such a shortcut in the new Gnome Shell? Let $R$ be a ring such that $x^4=x$ for every $x\in R$. Is this ring commutative? How do I remove all specific characters at the end of a string in PHP? Expected prediction error - derivation Line-breaking widget layout for Android I'm trying to create an activity that presents some data to the user. The data is such that it can be divided into 'words', each being a widget, and sequence of 'words' would form the data ('sentence'?), the ViewGroup widget containing the words. As space required for all 'words' in a 'sentence' would exceed the available horizontal space on the display, I would like to wrap these 'sentences' as you would a normal piece of text.  The following code:  public class WrapTest extends Activity {     /** Called when the activity is first created. */     @Override     public void onCreate(Bundle savedInstanceState) {         super.onCreate(savedInstanceState);         LinearLayout l = new LinearLayout(this);         LinearLayout.LayoutParams lp = new LinearLayout.LayoutParams(                 LinearLayout.LayoutParams.FILL_PARENT,                 LinearLayout.LayoutParams.WRAP_CONTENT);         LinearLayout.LayoutParams mlp = new LinearLayout.LayoutParams(                 new ViewGroup.MarginLayoutParams(                         LinearLayout.LayoutParams.WRAP_CONTENT,                         LinearLayout.LayoutParams.WRAP_CONTENT));         mlp.setMargins(0, 0, 2, 0);          for (int i = 0; i &lt; 10; i++) {             TextView t = new TextView(this);             t.setText("Hello");             t.setBackgroundColor(Color.RED);             t.setSingleLine(true);             l.addView(t, mlp);         }          setContentView(l, lp);     } }   yields something like the left picture, but I would want a layout presenting the same widgets like in the right one.        Is there such a layout or combination of layouts and parameters, or do I have to implement my own ViewGroup for this? I need to show that for prime numbers of the form $p=4n+1$, $x=(2n)!$ solves the congruence $x^2 \equiv-1\pmod p$. I then need to show this implies p isn't a gaussian prime.   I have started to solve this using Wilson's theorem that a number $z$ is prime iff $(z-1)!\equiv-1\pmod z$. Therefore the endpoint of my proof should be that $(p-1)!\equiv-1\pmod p$.  As $p$ is of the form $p=4n+1$, I only need to prove that $4n!$ is congruent to $-1$ modulo $p$.   Here is my working so far: Starting with the congruence $x^2 \equiv -1\pmod p$:    $$\eqalign{x^2 = -1\pmod p&amp;\implies   x^2 + 1 = kp \implies (x-i)(x+i) = kp\cr &amp;\implies ((2n)!-i)((2n!)+i))=kp \implies 4n!- 1 =kp\cr}$$   This is where I start to run out of any ideas that seem to get me anywhere. Any tips would be greatly appreciated! text processing rows to columns for a block I have a file containing lists on Solaris:  List A hi hello hw r u  List B Hi Yes  List C Hello   I need to transpose the lists as shown below:  List A    List B    List C hi        Hi        Hello hello     Yes hw r u   How can I do this on Solaris? As mentioned on :     For something that can be done so   easily by accident, there should be   the ability to undo mistakes.   I know that I have clicked several times on a comment up-vote without actually wanting to.  I would suggest being able to cancel the up-vote (maybe only for a couple of minutes or even less than a minute, no problem, as long as I can undo a miss-click). I want to update ubuntu 11.04 throught apt-get. So I added the following lines in apt.conf  export http_proxy=http://username:password@proxyserver.net:port/ export ftp_proxy=http://username:password@proxyserver.netport/  export http_proxy=http://deepak:Deepak@123@12.1.1.1:3128   My question is: how do I insert special characters to a username or password?  For instance: my password for proxy is Deepak@123 and it is getting an error. Showing that $\Bbb R[x] / \langle x^2 + 1 \rangle$ is isomorphic to $\Bbb C$ question Is it correct to sometimes capitalize words mid-sentence in computer user interfaces? I was reading a blogpost here:  () One thing that threw me off was that after the first four large displayed equations, there is the statement &quot;it is sufficient to prove that if the theorem holds for $n=m+1$, then it holds for $n=m$.&quot; How is this type of induction valid? I browsed around for things like backward induction, reverse induction, and Cauchy induction, but couldn't find a justification for how this is valid. With the usual forward induction of verifying a base case and proving $P(n)\implies P(n+1)$, it's easy to intuitively understand how induction will show a property holds for all natural numbers (or at least starting at the base case). But with this reverse induction, it seems to me that if you prove $P(m+1)\implies P(m)$, then if you were able to verify a specific case like $P(15)$, then you would only know it's true for numbers up to $15$. How does it actually prove the property for all naturals? Help with proof of $|x+y| \geq |x|-|y|$ Strange black shading. Cannot fully light mesh. (Cycles) Difference between StringBuilder and StringBuffer