So, what I am trying to do is center my text in the header. And what I mean by this, is not the center of the page, but center it to where there is an equal gap on the top and bottom of the header. I don't want it to sit in the upper left corner with a large gap of blank space beneath it. I want it aligned to the left at the center of the header's left edge. If that makes sense. I just can't figure out how to do it correctly so any tips would be greatly appreciated. The header code: &lt;header&gt;     &lt;!-- &lt;img style=&quot;display: inline;&quot; src=&quot;images/someimage.jpg&quot; height=&quot;150&quot; width=&quot;184&quot; alt=&quot;sOME TEXT&quot; /&gt; --&gt;     &lt;h1 style=&quot;display: inline;&quot;&gt;SOME TEXT&lt;/h1&gt;     &lt;h1 style=&quot;display: inline;&quot;&gt;SOME TEXT&lt;/h1&gt;     &lt;h1 style=&quot;display: inline;&quot;&gt;SOME TEXT&lt;/h1&gt;     &lt;h1 style=&quot;display: inline;&quot;&gt;SOME TEXT&lt;/h1&gt; &lt;/header&gt;       The CSS associated with the header: header {     background-color: #af9483;     height: 140px;     background-image: url('');     background-repeat: no-repeat;     border-top: 5px #9f7e69 solid;     border-bottom: 5px #9f7e69 solid; }  h1 {     font-family: &quot;times new roman&quot;;     color: #523225;     text-align: center;     padding-bottom: 70px;     padding-right: 100px;     font-size: 2em; }
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;
Range of $(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$ is If $a_{1},a_{2},a_{3},a_{4}\in \mathbb{R}$ and $a_{1}+a_{2}+a_{3}+a_{4}=0$ and $a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4}=1.$ Then Range of $$E =(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$ is  Try: From $$(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$ $$=2(a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4})+2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{4}+a_{4}a_{1})$$ $$=2-2(a_{1}+a_{3})(a_{2}+a_{4})=2+2(a_{1}+a_{3})^2\geq 2$$ and equality hold when $\displaystyle a_{1}=-a_{3}$ and $a_{2}=-a_{4}$ Could some help me how to find its upper bound, Thanks I wish to show that $\mathbb{Q}(\sqrt2, \sqrt3)$ is not isomorphic to $\mathbb{Q}(\sqrt[4]2)$.    What I did so far was to show that $\mathbb{Q}(\sqrt2,\sqrt3) =\text{span}\{1,\sqrt2,\sqrt3,\sqrt6\}$.  Assuming they are isomorphic, denoting $f$ the isomorphism, then $f(1) = 1$, so $x^2-3$ has a root in $\mathbb{Q}(\sqrt2,\sqrt3)$ so $\sqrt 3\in\mathbb{Q}(\sqrt[4]2)$. ($\sqrt2 \in \mathbb{Q}(\sqrt[4]2)$, $\sqrt[4]2\cdot\sqrt[4]2=\sqrt2)$.   Thus under the assumption, $\mathbb{Q}(\sqrt2, \sqrt3)\subseteq \mathbb{Q}(\sqrt[4]2)$. As a vector spaces of the same dimension over $\mathbb{Q}$, it means that $\mathbb{Q}(\sqrt2, \sqrt3)= \mathbb{Q}(\sqrt[4]2)$.  Now all is left to do it to show a contradiction: $\sqrt[4]2\in\mathbb{Q}(\sqrt2,\sqrt3)$, then $\sqrt[4]2 = a+b\sqrt2+c\sqrt3+d\sqrt6$ so $2=(a+b\sqrt2+c\sqrt3+d\sqrt6)^4$, I believe that opening the expression, I would find out that it can't be an element of $\mathbb{Z}$. However this is a really inelegant way (if it really works), and I wish for a better way to do so.  An other approach: $\mathbb{Q}(\sqrt2,\sqrt3) = \mathbb{Q}(\sqrt2 + \sqrt3)$ by the irreducibility of $x^4-10x^2+1$ over $\mathbb{Q}$ (an some dimension theorem usage). Also, $x^4-10x^2+1$ splits in $\mathbb{Q(\sqrt2, \sqrt3)}$.   Is there a way to show that this polynomial doesn't split over $\mathbb{Q}(\sqrt[2]4)$? with different method then assuming that $\sqrt3\in\mathbb{Q}(\sqrt[4]2)$ and getting a contradiction? Maybe the fact that $\sqrt[4]2$ has complex roots might help? Extracting Windows 8 Start Screen Patterns Is there any way to extract the Windows 8 Start Screen patterns, in order to use them as standalone wallpapers on other systems?  For example see this screenshot:    I am interested in the dark blue background. I heard that this background is somehow adapted to your chosen color theme. So many different variations should exist.  Engadget has an article  briefly talking about these background patterns and the different color schemes. They call them "personalization tattoos".    Update:    I found the following page/guide:   This looks a very good way to achieve my goal. Unfortunately the guide seems to be dated.  I tried to find shsxs.dll on the RTM of Windows 8 without much luck. It looks like Microsoft changed the way they handle the background image. The file is nowhere to be found...  I also looked into uxtheme.dll, but it doesn't contain any interesting images.  Any new ideas? Any other *.dll with might be of interest here? This is, how I download various master branches from GitHub, and I aim to have a prettier script (and maybe more reliable?).  wget -P ~/ https://github.com/user/repository/archive/master.zip unzip ~/master.zip mv ~/*-master ~/dir-name   Can this be shorten to one line somehow, maybe with tar and pipe?  Please address issues of downloading directly to the home directory ~/ and having a certain name for the directory (mv really needed?). "This is considered..." vs. "This is considered as..." Does this "as" change the meaning or correctness of this sentence?     This is considered as socially desired.   This is considered socially desired.   Are both of them grammatically correct? Why or why not? How can we prove that $D(f)\backsimeq\operatorname{Spec} A_f$? Let $$D(f)=\operatorname{Spec}A - V(f)=\{p\in \operatorname{Spec} A:f\notin p\}$$ be the principal open subsets of $\operatorname{Spec}A$ with the Zariski topology. I'm trying to prove that $D(f)\backsimeq \operatorname{Spec} A_f$, where $A_f$ is the localization of $A$ at $f$.  If we consider the canonical homomorphism $\varphi:A\to A_f$ given by $\varphi (a) = \frac {a}{1}$, we can define $\varphi':\operatorname{Spec} A_f\to \operatorname{Spec} A$, given by $\varphi'(p)=\varphi^{-1}(p),\forall p\in \operatorname{Spec}A_f$.  I'm having trouble proving that this map is surjective.  Thanks. I want to clear the inputtext used to search, and clear all search results using a commandbutton. I tried the following on the 'Reset' button I added but it only clears the inputtext and the search results are not clearing.  &lt;apex:pageBlockSectionItem &gt;                     &lt;apex:outputPanel layout="block" Rendered="true"&gt;                     &lt;apex:inputText value="{!ReportSearchString}" /&gt;                     &lt;apex:commandButton value="Search" action="{!searchReports}" reRender="ReportSearchResults, "/&gt; &amp;nbsp;&amp;nbsp;                      &lt;apex:commandButton value="Back" action="{!BackToCreateFeedPage}"/&gt;&amp;nbsp;                     &lt;apex:commandButton value="Reset" onclick="this.form.reset();return false;"/&gt;                     &lt;/apex:outputPanel&gt;                 &lt;/apex:pageBlockSectionItem&gt;                             &lt;/apex:pageBlockSection&gt;                         &lt;apex:pageBlockSection id="ReportSearchResults"&gt;                 &lt;apex:outputText value="{!warningMessage}" rendered="{!!showTable}" /&gt;                 &lt;apex:pageBlockTable value="{!reportList}" var="item" rendered="{!showTable}"&gt;                     &lt;apex:column headerValue="Report Name" &gt;                         &lt;apex:commandLink value="{!item.Name}" action="{!BackToCreateFeedPage}"&gt;                             &lt;apex:param value="{!item.Id}" name="SelectedReport" assignTo="{!selectedReportId}" /&gt;                         &lt;/apex:commandLink&gt;   This is my search function, it works fine. I need to clear the results of this search on clicking 'Reset' button.   public void searchReports()     {         showTable = false;               reportList = new List&lt;Report&gt;();          reportSearchResult = new Map&lt;Id, Report&gt;();          selectedReport = new Report();         if((ReportSearchString != null &amp;&amp; ReportSearchString != '') &amp;&amp; ReportSearchString.length() &gt;= 3)         {             reportSearchResult = new Map&lt;Id, Report&gt;([Select Id, Name From Report Where Name Like :'%'+ReportSearchString+'%' limit 2000]);              warningMessage = 'No Result Found';         }         else         {             warningMessage = 'Search text should have atleast 3 characters';         }     } Covariant derivative ambiguity I'm studying general relativity and am running into an ambiguity with the covariant derivative. The covariant derivative acting on a scalar is, in a co-ordinate basis, simply $$\nabla_X f = X^a \nabla_a f = X^a \partial_a f $$ whilst when acting on vectors or tensors, the covariant derivative includes extra terms involving the connection coefficients $\Gamma$. My problem is that I'm not sure what happens when the covariant derivative acts on the components of a vector. So the $X^a$ in the above expression are each scalar functions, and so when we write $\nabla_Y X^a$, do we assume the covariant derivative acts on $X^a$ as scalars, or do we include connection coefficient terms?  To clarify, consider $$ \nabla_X Y = X^a \nabla_a(Y^b e_b) = X^a (\nabla_a Y^b) e_b + X^a Y^b (\nabla_a e_b) = X^a (\partial_a Y^b) e_b + X^a Y^b \Gamma^c{}_{ba} e_c $$ As you can see, when the covariant derivative acts on the $Y^b$, it acts as a simple partial derivative, since the $Y^b$ are here considered functions premultiplying the vectors $e_b$.   However, in this equation from  page: $$(\nabla_X \nabla_Y Z)^a = X^c \nabla_c Y^b \nabla_b Z^a = X^c Y^b \nabla_c \nabla_b Z^a + (X^c \nabla_c Y^b) \nabla_b Z^a = (\nabla^2_{X,Y} Z)^a + (\nabla_{\nabla_X Y} Z)^a$$ When the covariant derivative acts on the $Y^b$ term it treats it like a vector.  What's going on here? Thanks. Case-insensitive attribute-value selector with Jquery How do I get object names to appear in the render? In my scene I have multiple objects (30+). In the viewport I have the handy control of selecting (in the Properties Editor) Object> Display> Name and copying to selected.  This is beautiful for diagrams etc. as wherever my view is at whatever distance these labels have been placed instantly, always face me and are always the same size.  How can you get these to display in a rendered view?  Is the ONLY option to manually create a label for each object, get them to track to your camera (Probably have to track to an empty miles behind your camera rotation otherwise they offset if your camera isn't facing exactly front on to an object) and add some sort of scaling property to change depending on how far away you are? Here's the very dumb way:  def divisorGenerator(n):     for i in xrange(1,n/2+1):         if n%i == 0: yield i     yield n   The result I'd like to get is similar to this one, but I'd like a smarter algorithm (this one it's too much slow and dumb :-)  I can find prime factors and their multiplicity fast enough.  I've an generator that generates factor in this way:  (factor1, multiplicity1) (factor2, multiplicity2) (factor3, multiplicity3) and so on...  i.e. the output of   for i in factorGenerator(100):     print i   is:  (2, 2) (5, 2)   I don't know how much is this useful for what I want to do (I coded it for other problems), anyway I'd like a smarter way to make  for i in divisorGen(100):     print i   output this:  1 2 4 5 10 20 25 50 100     UPDATE: Many thanks to Greg Hewgill and his "smart way" :) Calculating all divisors of 100000000 took 0.01s with his way against the 39s that the dumb way took on my machine, very cool :D  UPDATE 2: Stop saying this is a duplicate of  post. Calculating the number of divisor of a given number doesn't need to calculate all the divisors. It's a different problem, if you think it's not then look for "Divisor function" on wikipedia. Read the questions and the answer before posting, if you do not understand what is the topic just don't add not useful and already given answers. If I play a spell with {X} in the mana cost, and I have a card out that reduces the casting cost by 2 generic mana, what happens to the X? Does it start at -2?  For example: I have  in play. If I play  for just 1 blue mana, do we draw 2 cards? Installing Reaver 1.4 I am trying to install reaver 1.4, but when I run ./configure ; make I get this message at the end  int ret_val = EXIT_FAILURE, r = 0;   I have searched a fair bit and cannot find an answer, any help would be appreciated. I apologize if this has been answered and I've just failed to find it. Please help me to find elements $a,b$ in a ring $R$ such that $a\mid b$ and $b\mid a$, but there does not exist any unit $u$ in $R$ such that $a=ub$. How to make a standalone document with one equation? Unable to install "": snap "" has "install-snap" change in progress I just finished the installation of the Ubuntu 18.04, but whenever I try to install any application from Ubuntu Software the same error occurs (for example "vlc"):     unable to install "vlc": snap "vlc" has "install-snap" change in progress   I hope somebody can tell me what I've done wrong. What are the options for a non-EU national who wants to extend his stay in Switzerland (totaling less than 90 days)? Calculating CR of monsters that do not deal calculable damage I have a database which contains 5 tables. Each table contains 24 rows and each row contains 4 columns.  I want to display these records in Excel sheet. The heading of each table is the name of the table, but I am unable to merge the columns for heading.  Please help me. returning a local variable from function in C