align a div at the bottom I have a div which I want to align at the bottom of the page. For that I have used the following --  &lt;div style="position:fixed;bottom:0;overflow:auto;word-wrap:break-word;display:block" ng-show="noData || failedStatus"&gt;     &lt;div class="alert alert-danger" ng-show="noData" style="overflow:auto;word-wrap: break-word;"&gt;         &lt;!-- something --&gt;     &lt;/div&gt;     &lt;div class="alert alert-danger" ng-show="failedStatus" style="overflow:auto;word-wrap: break-word;"&gt;         &lt;!-- something --&gt;     &lt;/div&gt; &lt;/div&gt;   This code is aligning the div at the bottom but it is not wrapping the words inside the div.
How to align content of a div to the bottom Say I have the following CSS and HTML code:    #header {   height: 150px; } &lt;div id="header"&gt;   &lt;h1&gt;Header title&lt;/h1&gt;   Header content (one or multiple lines) &lt;/div&gt;     The header section is fixed height, but the header content may change.  I would like the content of the header to be vertically aligned to the bottom of the header section, so the last line of text "sticks" to the bottom of the header section.  So if there is only one line of text, it would be like:   ----------------------------- | Header title | | | | header content (resulting in one line) -----------------------------   And if there were three lines:   ----------------------------- | Header title | | header content (which is so | much stuff that it perfectly | spans over three lines) -----------------------------   How can this be done in CSS?
What is a substitute for red or white wine in a recipe? Is there a way to avoid SSH typing delay? Is it possible to perform a complete reinstall on an Android smartphone? I occasionally perform such a reinstall on my Windows PC. I repartition my hard drive, format the partitions, and then install Windows on a blank partition. Can this be done on an Android smartphone? (I have an unlocked Sony Xperia S.) I get this message while trying to install Ubuntu 12.04 as a guest in VirutalBox 4.1.18, on an Ubuntu 10.04 host.         This kernel requires the following feature not present on the CPU: pae   Some host specs:     The host's kernel is: Linux 2.6.32-41-generic-pae GNU/Linux   lscpu (host): Architecture: i686, CPU op-mode(s): 32-bit, 64-bit   grep --color=always -i PAE /proc/cpuinfo   does show pae in its output.   The 12.04 iso used is: ubuntu-12.04.0-desktop-i386.iso      As a comparison/check, I downloaded and installed Linux Mint 13 Cinnamon to the same host on the exact same VM (I just changed the .iso image). It worked fine. Its iso is: linuxmint-13-cinnamon-dvd-32bit.iso   It seems (to me) that I have pae.. what is going on here?   Update: I had assumed that Linux Mint also required pae (being Ubuntu based), but I've just run;   grep --color=always -i PAE /proc/cpuinfo   in the Mint VM.   It showed no output.   So it seems the issue may lie with VirtualBox.   If that is the case, how can I get Virtualbox into pae mode? How to assign symbols to functions If x and y are uncorrelated normal variates, x*exp(y) will have a symmetric unimodal distribution with positive excess kurtosis. Has this distribution been named and studied? Current measurement maximum time On my multimeter I have two current measurement connection. One for the mA and one for the A.  On the ampere one there's a time limitation but not on the mA one. Also they mention in the manual a kind of "cooldown time" after the measurement for the A one. So, basicaly, probing with this input can really slow down my project. (max 15 sec every 5 minutes)  My question is, I have to switch to that A one for a 600-700mA current (mA input is 500mA), as it's not a "big" current, can I bypass this limitation ? Is it design for higher current only or any measurement done on that input ? It's a middle-end multimeter but even the online datasheet is not complete about what. (Was excepting a kind of measured current/cooldown time graph) I have a JavaScript object. Is there a built-in or accepted best practice way to get the length of this object? const myObject = new Object(); myObject[&quot;firstname&quot;] = &quot;Gareth&quot;; myObject[&quot;lastname&quot;] = &quot;Simpson&quot;; myObject[&quot;age&quot;] = 21; Input impedance of Arduino Uno analog pins? I'm using the  on some small robotic arm motors, and reading in the analog voltage with an Arduino Uno.  I've had reasonably good results, but only after putting in an RC filter on the output.  However, in the application note on the datasheet, it says not to do that:  "the addition of an RC filter to the output of a sensor IC can result in undesirable device output attenuation — even for DC signals."  Then it gives a formula to calculate attenuation, but it depends on knowing the input impedance of whatever is reading the signal, so that's what I'm after here. Where does the phrase "Scare the Dickens out of..." originate from? Where does the phrase "Scare the Dickens out of..." originate from? And does it refer to Charles Dickens? Cool Text Highlighting in LaTeX LaTeX has soul. ConTeXt, of course, , where you can put sweet hightligher under your text for free!    In the pictured example, you can see highlights that not only highlight, but that do it with cool edges that look so good, you can smell fumes! The fearless author says he'll show how that was done in another post; this one shows only how to lay down a straight, yellow box. But, it is possible to do word-wrapped highlighting with cool-shapped boxes within ConTeXt, presumably.  The trouble for me is that getting into the ConTeXt world is steep climb right now, given the state of documentation and the flux of changes, so I need to continue with LaTeX. But the real problem is that now I can't live without the Organic-Looking Highlights of Awesomeness in that picture!  As I said, Non-Organic-Looking Highlights of the Ordinary are easy with the soul package:  \hl{Some awesome text that word wraps fine.}   It just taunts me on the page with its straight lines. So, if I sacrifice word wrapping, I can use PGF/TiKz and create the following macro:  \newcommand\hl[1]{%     \tikz[baseline,%       decoration={random steps,amplitude=1pt,segment length=15pt},%       outer sep=-15pt, inner sep = 0pt%     ]%    \node[decorate,rectangle,fill=navcolor,anchor=text]{#1\xspace};% }%   Which creates something like this:    Now, my color is too dark and my randomness needs adjusting, but you see that a similar effect could be achieved.  My trouble is that I don't understand TeX enough to dig into the Soul package and add the TiKz/PGF goodness that it so sorely needs. So, I leave it to you guys. Anyone want to tackle this?   Or is it yet another piece of fruit, just a touch out of reach, that shall have to wait until ConTeXt settles down? HTML escapes appearing in node titles on bespoke content type — substitutions? "going to do" vs. "going to be doing" Note: this post is different to , which just cites part from somewhere else, without any further analysis and thinking. Furthermore, none of the answers to that post answers my question - the situations where &quot;will (going to) be doing&quot; is more appropriate.   I will be participating in IELTS test next Tuesday afternoon.  In an online class, the teacher says the following at the beginning of the lesson.  I will be using NotePod (pointed out by red rectangle) to write relevant points we talk about today.   I guess &quot;will be using&quot; is more appropriate than &quot;will use&quot; there, right?  A  explains  This &quot;tense&quot; is used for reporting a scheduled activity with special focus on its timing— as if one is standing in the future and experiencing the flow of the time. The focus is on the temporary, ongoing, or repetitive nature of the activity.   I understand the grammar rules of those verb tenses and the difference between this group  I am going to drive home tomorrow.   I am going to be driving home tomorrow.  The former is just convey my activity in the near future while the latter probably makes an implication that &quot;don't disturb me as I am driving in that time&quot; However, in the following situations, I don't see the difference between 2 verb tenses.  Welcome back to this class. Today's lesson is for the IELTS students who are going to be taking the IELTS test. it's for everybody to be honest with you. this lesson is useful for anybody who needs to do some writing.   who are going to take the IELTS test ... Fastest Square Root Algorithm What is the fastest algorithm for finding the square root of a number?  I created one that can find the square root of "$987654321$" to $16$ decimal places in just $20$ iterations   I've now tried Newton's method as well as my own method (Newtons code as seen below)  What is the fastest known algorithm for taking the second root of a number?  My code for Newton's Method (*Edit: there was an error in my code, it is fixed in the comments below):        a=2   //2nd root     b=97654321   //base     n=1   //initial guess     c=0   //current iteration (this is a changing variable)     r=500000   //total number of iterations to run     while (c&lt;r)      {         m = n-(((n^a)-b)/(a*b))  //Newton's algorithm         n=m         c++;         trace(m + "  &lt;--guess   ...   iteration--&gt;  " + c)     } For which $n≥2$ does there exist $\sigma\in A_n$ with $|\sigma| &gt; n$? I have a proof that I'd like to be verified. My proof that there exists $\sigma\in A_n$ with $|\sigma| &gt; n$ if and only if $n\geq 8$: For $n&lt;8$, one can manually verify for each $n$ that there is no $\sigma\in A_n$ with order greater than $n$. Suppose $n\geq 8$. Case 1: $n$ is even and $3\not\mid n$. Then the permutation $(1,2,3)(4,\dots,n)$ is even and has order $\operatorname{lcm}(3,n-3)=3n-9\geq n+16-9\geq n + 7$. Case 2: $n$ is even and $3\mid n$. Then $n\geq 12$, so $n-2\geq 8$ is even and is not divisible by 3. Applying Case 1, we see that $(1,2,3)(4,\dots,n-2)\in A_n$ has order at least $n+5$. Case 3: $n$ is odd. Then apply one of the previous two cases with $n-1$ to get an even permutation with order at least $n+4$. $\blacksquare$ Is my proof correct? It doesn't seem very elegant. Is there some slicker proof? Reducibility of polynomials modulo p I've seen in every introductory material on LaTeX and in a lot of code of other people (event experts) that the \newcommand command is used by wrapping the first argument in braces, even if it seems unnecessary to me.   The braces are consistent with the usual way arguments are passed to LaTeX commands (since usually the braces are used even if the argument is made of a single token, for consistency and clarity), but in this single case they appear very ugly to me and in my opinion they lower readability.  But these are opinions, while the question is technical: is there any technical reason why one should write this:  \newcommand{\mycmd}[1]{Hello #1}   instead of this:  \newcommand\mycmd[1]{Hello #1} Issue with particles in blender donut tutorial I have been recently doing the blender donut tutorial and ran into a problem regarding the sprinkle particles. The particles (hair) keep appearing on the underside of the icing. As seen in the image below, most of the hair particles only come out from the bottom of the icing. I tried to readjust the normals but I realised the normals were already correct. (Blue colour donut in image) Someone suggested following through the weight painting part of the tutorial. However, the sprinkles still remain on the underside of the icing. What is the actual problem and how do I solve this?  Particle settings below  Edit: @moonboots Thanks the hair particle problem has been solved. How do I prevent the submersion of the sprinkle particles? For each positive even integer $n$, set   $$P_n = \displaystyle {{n} \choose {\frac{n}{2}}}\frac{1}{2^n}.$$  Show that $\displaystyle \lim_{n \to \infty} P_n$ exists and determine its value.   Here's what I have so far: each $P_n$ can be thought of as the probability of tossing a fair coin $n$ times, and obtaining exactly $\frac{n}{2}$ heads. My expectation is that $P_n \to 1$, since in any large trial, one would expect to record just about as many heads as tails. But I'm not sure how to mathematically justify this hunch.  Hints or solutions are greatly appreciated. Let $\{E_i\}_{i=1}^n$ be finitely many disjoint sets of real numbers (not necessarily Lebesgue measurable) and $E$ be the union of all these sets. Is it always true that  $$ m^\star (E)=\sum_{i=1}^N m^\star(E_n) $$ where $m^\star$ denotes the Lebesgue outer measure? If not, please give a counterexample. The Vitali set is a counterexample in the countable case, but I am not sure whether it is false in finite case.