There seem to be tons of solved problems with this one, but neither of them seem to work for me...  I have created this little jsfiddle to show you:   And the CSS:  .footer {      width:798px;      border-top: 2px solid #2E181A;      clear: both;      padding: 5px 0 0 0;      background-color: inherit;      text-align: center;      bottom:0;      background-color: #E6D9BD;      position:relative;      height: 30px;      margin: -30px auto 2px auto;      z-index:30;  }   .container {      width: 788px;      margin: 0px auto 0px auto;      padding: 0px 0px 30px 0px;      border:5px solid #2E181A;      background-color: #E6D9BD;      min-height: 100%;      position:relative;      content: " "; /* 1 */      display: table; /* 2 */  }   .contentleft {      background-color: inherit;      margin:5px 10px 10px 10px;      padding:10px 5px 30px 5px;      float:left;      overflow: hidden;      width: 300px;      display:block;  }   .contentright {      background-color: inherit;      margin:5px 0px 10px 10px;      border:0px solid #2E181A;      padding:10px 5px 30px 5px;      float:left;      overflow: hidden;      width: 420px;      display:block;  }   I have set a top-border in the div.footer, and this should be visible and a little space between the border and the div.container.  Hope you will take a quick look at the code and see whatever I'm doing wrong!
I have a simple 2-column layout with a footer that clears both the right and left div in my markup.  My problem is that I can't get the footer to stay at the bottom of the page in all browsers.  It works if the content pushes the footer down, but that's not always the case.
Highlighting pattern strings in a large text Just like if we have any quadratic equation which has complex roots, then we are not able to factorize it easily. So we apply quadratic formula and get the roots.  Similarly if we have a cubic equation which has two complex roots (which we know conjugate of each other) and one fractional root, then we are not able to find its first root by hit &amp; trial.  So my question is like quadratic formula, is there exist any thing like cubic formula which help in solving cubic equations?  For example, I have an equation $$2x^3+9x^2+9x-7=0\tag{1}$$ and I have to find its solution which I am not able to find because it has no integral solution. Its solutions are $\dfrac {1}{2}$, $\dfrac{-5\pm \sqrt{3}i}{2} $, I know these solutions because this equation is generated by myself.  So how can I solve equations like these?  Also while typing this question, I thought about the derivation of quadratic formula, which is derived by completing the square method.  So I tried to apply ‘completing the cube’ method on the general equation $ax^3+bx^2+cx+d=0$ but it didn't help.  So please help me in finding a cubic formula or to solve the equations like given in example by an alternative method. Proof of the formula $1+x+x^2+x^3+ \cdots +x^n =\frac{x^{n+1}-1}{x-1}$ Reset the password in Ubuntu / Linux Bash in Windows I have installed Linux Bash on Windows 10 (available from Microsoft), but I've forgotten my password.  I've searched for a solution but I only find suggestions that involve doing something before the system boots, like accessing recovery mode, but that will not work in this case as this Ubuntu is just an application within a Windows system and doesn't have its own boot process.  How can I reset my password? How to change Blender's default startup settings? How can I change Blender from the default cube/lamp/camera default setup? Also, is it possible to change other features, such as the default view direction (Top, Side, etc) or the Editor type (3D View, Node Editor, Graph Editor, etc)? I was working on a problem in group theory, which asks about the automorphism group of a direct product of groups.  Okay, so I know that if $G,H$ are two groups whose orders are relatively prime, then $\operatorname{Aut}(G\times H)\cong\operatorname{Aut}(G)\times \operatorname{Aut}(H)$.  I also know that if $G,H$ are abelian and simple and isomorphic, then $\operatorname{Aut}(G\times H)\cong \operatorname{GL}_2(\mathbf{Z}_p)$ where $p$ is the order of $G$.  What I don't know is the following: what if $G,H$ are simple, yet not abelian? Let us for now focus on a more restricted case where $G\cong H$, in particular, $G=H$. What can we say about the group $\operatorname{Aut}(G\times H)$? (I heard something about this group being related to what is called "wreath product", which I do not know...)  Thanks bunch in advance! Shutdown button only logs out ubuntu 15.04 What are path and contour integrals? I am very confused on what path and contour integrals actually represent. I tried looking for an answer on Google but the only examples were related to Quantum Mechanics. For instance, real Riemann integrals can be thought as &quot;the area under the curve&quot;. What about path/contour integrals? Also, is there a difference between path and contour integrals, or are they the same thing? I've been thinking of contour integrals as path integrals where the path is a &quot;closed loop&quot;, but I'm not sure if that's the case. It's all so confusing, and not being able to visualise these concepts makes it feel like I'm just memorising formulas and theorems without any reason. I have been thinking about path integrals as integrating over a curve in three dimensions. For example, in an $xyz$ coordinate system (with $y$ pointing “up”), we have a path on the $x$-&quot;$z$&quot; plane and integrate over that path. That is, instead of integrating over and along &quot;$y=0$&quot;, we integrate over a certain path. But I don't know if this is a correct way to think of path integrals. This issue has been frustrating me a lot because I'm now studying theorems that involve path/contour integrals and I just can't understand what they really do or mean. By path integral I mean $$\int_{\gamma\vert_{[a,b]}}f(z)\hspace{0.2em}dz = \int_{a}^{b}f(\gamma(t))\cdot\gamma'(t)\hspace{0.2em}dt$$ where $f:U\subseteq\mathbb{C}\rightarrow\mathbb{C}$ is a function that satisfies all conditions for complex integration and $\gamma:[a,b]\rightarrow U\subseteq\mathbb{C}$ is at least a piece-wise regular path. Any help would be appreciated. I don't have any knowledge in hardware and would like to know which RAM should I buy for my laptop.  How do I discover what RAM I need &amp; how much my machine can take?  I have a Razer Blade 15 (RZ09-02705F76) 2018 but don't know if there is a specific type or format of RAM that suits it and don't know where to find this information. One-byte-off pointer still valid in C? How do I remove directories that begin with a dot after my server was infected? My server had been infected and the attacker spread a bunch of .5randomletters folders around my file system, with malicious code inside of them. I want to search for all the folders that begin with a dot, review them and then remove them.  How can I do that? I am having a strange problem in ArcGIS Toolbox.  I am trying to use the  tool from ArcToolbox. When I open the tool, there is no possibility to input Simplification Tolerance. Please see the attached image.  I can run the tool as a batch and from the Python window without any problem.  I am using ArcGIS 10(Editor) with Service Pack 5 on Windows 7 64Bit (English) and Windows Server 2008 64Bit (German). On both machines, I have the same problem. Any ideas?    I have reinstalled the software but the problem remains. Can't save my render as an image I seem to be having issues when I try and save the output file as a .png, or .jpeg for that matter. I've set up my user preferences, File, Render Output to my desktop as well as render output in the editor window. What am I missing here? I'm obviously very new to this program and your help would be appreciated. Thanks for your time Mean of a Cauchy Distribution A classic exponential inequality: $x^y+y^x>1$ This is a classic problem, but I haven't seen it on the site before.     Suppose $x$ and $y$ are real numbers that both lie in the interval   $(0,1)$. Prove  $$x^y+y^x&gt;1.$$   The following spoiler box contains a hint that indicates the solution I know. Does anyone know a more natural solution?       Use Bernoulli's inequality to show $$x^y\ge \frac{x}{x+y}.$$   Edit: I recently learned the source is the 1996 French math olympiad. I am planning on spending 90 days in Schengen zone countries.  I will be taking a cruise between Lisbon and Barcelona.  Do the days at sea count against the 90 days allowed for staying in the Schengen zone? What is inertia of a body? The definition of  is  "Inertia is the resistance offered by the body whenever its state of motion is changed."   What is inertia of a body? Is inertia actually a force exerted by the body?  If so, then why is there no application of inertia in numerical problems involving application of force on a body? What is the actual application of inertia in theory? How do I make the Unity Launcher background transparent? I found a way to make the icon backgrounds transparent for the Unity Launcher by changing the files under /usr/share/unity/4.  Does anyone know how to make the background transparent for the launcher itself? I already made it transparent under Compiz, but I want to remove the line that separates the launcher from the desktop. If anything, I just want to make it look like the icons are floating. How to format a USB drive? Let $M$ be a positive, continuous martingale that converges a.s. to zero as $t$ tends to infinity. I now want to prove that for every $x&gt;0$ $$ P\left( \sup_{t \geq 0 } M_t &gt; x \mid \mathcal{F}_0 \right) = 1 \wedge \frac{M_0}{x}. $$ My approach: I thought that rewriting the conditional probability to an expectation would help so we obtain that we must prove: $$ \mathbb{E} \left[ 1_{\sup_{t \geq 0 } M_t &gt; x}  \mid \mathcal{F}_0 \right] = 1 \wedge \frac{M_0}{x}.$$ A hint to me was given that I should consider stopping the process when it gets above $x$. Thus a stopping time that would do this is $\tau = \inf\{t\geq 0 : M_t&gt;x\}$. But now I'm stuck as I want to apply optional sampling results but we have an indicator which complicates things. How could I proceed from this? Any help is appreciated!