How to disable auto-maximizing windows in Ubuntu 18.04 without using Unity Tweak Tool.
In Ubuntu 18.04, when moving the windows there is a resize effect that occurs when accidentally touching one of the screen edges.  This control was available in CCSM and Ubuntu Tweak in previous versions with Unity.  I'm not finding it in GNOME Tweaks.  I also searched the repository with:  apt search ubuntu.*tweak
OpenVPN restriction Ive set up an OpenVPN server at my place and handed out a certificate to a client who has an openvpn client at his place.  I need require that he should be able to connect to the vpn server only from one IP address (his current one).  How can I do this? htl5 voice recording with isualization XML - Data At Root Level is Invalid Would this look natural:     StackExchange is a helluva site! I found out about this site at Facebook.    Or should we use the preposition on:     StackExchange is a helluva site! I found out about this site on Facebook.    Or are both okay?   (Inspired by a ) Integral $\int\csc^3{x} \ dx$ I found these step which explain how to integrate $\csc^3{x} \ dx$. I understand everything, except the step I highlighted below.  How did we go from: $$\int\frac{\csc^2 x - \csc x \cot x}{\csc x - \cot x}\,dx%$$ to  $$\int \frac{d(-\cot x + \csc x)}{-\cot x + \csc x} \quad?$$  Thank you for your time! $$ \int \csc^3 x\,dx = \int\csc^2x \csc x\,dx$$ To integrate by parts, let $dv = \csc^2x$ and $u=\csc x$. Then $v=-\cot x$ and $du = -\cot x \csc x \,dx$. Integrating by parts, we have: $$\begin{align*} \int\csc^2 x \csc x \,dx &amp;= -\cot x \csc x - \int(-\cot x)(-\cot x\csc x\,dx)\\ &amp;= -\cot x \csc x - \int \cot^2 x \csc x\,dx\\ &amp;= -\cot x\csc x - \int(\csc^2x - 1)\csc x\,dx &amp;\text{(since }\cot^2 x = \csc^2-1\text{)}\\ &amp;= -\cot x \csc x - \int(\csc^3 x - \csc x)\,dx\\ &amp;= -\cot x\csc x - \int\csc^3 x\,dx + \int \csc x\,dx \end{align*}$$ From $$\int \csc^3 x\,dx = -\cot x\csc x - \int\csc^3 x\,dx + \int \csc x\,dx$$ we obtain $$\begin{align*} \int\csc^3x\,dx + \int\csc^3 x\,dx &amp;= -\cot x \csc x + \int\csc x\,dx\\ 2\int\csc^3 x\,dx &amp;= -\cot x\csc x + \int\csc x\,dx\\ \int\csc^3x\,dx &amp;= -\frac{1}{2}\cot x\csc x + \frac{1}{2}\int\csc x\,dx\\ &amp;=-\frac{1}{2}\cot x\csc x + \frac{1}{2}\int\frac{\csc x(\csc x - \cot x)}{\csc x - \cot x}\,dx\\ &amp;= -\frac{1}{2}\cot x \csc x + \frac{1}{2}\int\frac{\csc^2 x - \csc x\cot x}{\csc x - \cot x}\,dx\\ &amp;= -\frac{1}{2}\cot x \csc x + \frac{1}{2}\int\frac{d(-\cot x+\csc x)}{-\cot x +\csc x}\\ &amp;= -\frac{1}{2}\cot x\csc x + \frac{1}{2}\ln|\csc x - \cot x|+ C \end{align*}$$ Let $f\colon\mathbb R\to\mathbb R$ be twice differentiable with $f(x)\to 0$ as $x\to\infty$ and $f''$ bounded. Show that $f'(x)\to0$ as $x\to\infty$. (This is inspired by  to a different question) Whenever I start an application like Terminal or Preview, they are always opened as a small window on the top left corner of the screen. Is there any way to set them to open in the center of the screen? Preferably maximized too? How can we find premade adventures for a virtual tabletop? Me and a couple of friends are just getting into P&amp;P rpgs and are really excited to try something new. However, we have a problem that we live quite far apart from each other. Therefore we'd like to use one of excellent virtual tabletops that are available. But none of us have any real experience with creating an adventure and would thus like to use a premade and beginner friendly adventure. But we're having a hard time finding one with premade story, maps etc in an online format. Most seem to be printed books and the Roll20 marketplace hasn't really gotten started yet. So what we're looking for a is online way to play a premade adventure to get us started in the RPG world.  How can we find these adventures for a VTT? I noticed an unexpected different line spacing of the titlepage what drives me crazy. The left page is the MWE, the right page without the \\ in front of Name.  \documentclass{scrartcl}  \begin{document}  \centering \Huge \textbf{UUUUUUUU UUUUUUUUU UUUUUUUUUUUUU UUUUUUUUUUUUU} \vspace{\baselineskip}\\ \textbf{UUUUUUUU UUUUUUUUU UUUUUUUUUUUUU UUUUUUUUUUUUU} {\rmfamily\Large \vspace{5\baselineskip}\\Name}  \end{document}     Any idea what causes this? Thanks alot!  Cheers, Stephan An  is defined as $p = q\times d$ (for two point charges $\pm q$ separated by a distance $d$).    What is the physical meaning of this quantity?  Why does the direction of the electric dipole moment go from the negative charge to the positive charge? Another way of saying this, why electric dipole moment is a vector quantity? I'm currently using an entry-level DSLR with 18-55mm lens and having great trouble trying to creating a blurred background effect in my photos. Now matter what aperture or shutter speed I choose, I'm not able to get it.  There's a general question about , but what specific things can I do to get this effect with an entry-level lens? Hydrogen radial wave function infinity at $r=0$ When trying to solve the Schrödinger equation for hydrogen, one usually splits up the wave function into two parts: $$\psi(r,\phi,\theta)= R(r)Y_{\ell,m}(\phi,\theta).$$ I understand that the radial part usually has a singularity for the $1s$ state at $r=0$ and this is why you remove it by writing: $$R(r) = \frac{u(r)}{r}$$ But what is the physical meaning of $$R(r=0) = \infty~?$$ Wouldn't this mean that the electron cloud is only at the centre of the atomic nucleus? Displaying row/column of point in raster using QGIS? Is it possible in QGIS to display column and row index of pixel in raster?   I mean something similar to value tool. Help with proving that $\pi$ is irrational I was trying to prove that $\pi$ is irrational, just to see if I could do it. So far, I've tried to do this by using the fact that the sum   $$S=\sum\limits_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$ and arguing that $S$ is irrational instead, and thus implying that $\pi$ is also irrational.  To do this, I thought I could use partial sums $S_n$ of $S$:  $$S_n=\sum\limits_{k=1}^n \frac{1}{k^2}=\frac{A_n}{B_n}$$ where the fraction $A_n/B_n$ is written in lowest terms. We can note that the sequence $\lbrace S_n \rbrace$ is a strictly increasing sequence, with $\pi^2/6$ as its lowest upper bound, so I thought that maybe the sequences $\lbrace A_n \rbrace$ and $\lbrace B_n \rbrace$ both have no upper bound, and the sequences would tend to infinity, and if I were able to prove this, I thought I could use it to argue that $S$ is an irrational number, using the fact that every rational number can be written as the quotient between two finite integers.  We note that $A_n\geq B_n$ for all natural $n$, as $S_n\geq 1$, so all we would really need to prove in that case is that $\lbrace B_n \rbrace \rightarrow \infty$(if this assertion is even correct).  At first I thought that both $\lbrace A_n \rbrace$ and $\lbrace B_n \rbrace$ would be increasing sequences, but after checking with Maple I noticed that they weren't, sadly enough($S_9=\frac{9778141}{6350400}$ and $S_{10}=\frac{1968329}{1270080}$). However, they do indeed seem to get very large very quickly, so I'm thinking that my hypothesis about $\lbrace A_n \rbrace$ and $\lbrace B_n \rbrace$ is correct.  But I have trouble proving my hypothesis, and I'm kind of stuck, no knowing what to do. Is there a way to prove that $\lbrace B_n \rbrace \rightarrow \infty$? And of course, is this approach to prove that $\pi$ is irrational logically sound, or is it fundamentally flawed in some important aspect? In the latter case, what idea should I try next? Is there a rule for pronouncing “th” at the beginning of a word? Compute the (multiplicative) inverse of $4x+3$ in the field $\frac {\Bbb F_{11}[x]}{\langle x^2+1 \rangle}$? So I am finding a polynomial $px+q$ ($p,q \in \Bbb F_{11}$) which is multiplicative inverse of $4x+3$ in $\frac {\Bbb F_{11}[x]}{\langle x^2+1 \rangle}$.  i.e. $[(4x+3)+\langle x^2+1 \rangle][(px+q)+\langle x^2+1 \rangle]=1+\langle x^2+1 \rangle$   $\Rightarrow$  $(4x+3)(px+q)+\langle x^2+1 \rangle=1+\langle x^2+1 \rangle$  $\Rightarrow$  $4px^2+(4q+3p)x+3q+\langle x^2+1 \rangle=1+\langle x^2+1 \rangle$.  We see that the remainder,when $(4x+3)(px+q)$ is divided by $x^2+1$, is $1$. So by Division algorithm,  $$ \require{enclose} \begin{array}{r} 4p \\[-3pt] x^2+1 \enclose{longdiv}{4px^2+(4q+3p)x+3q} \\[-3pt] \underline{4px^2+4p} \\[-3pt] (4q+3p)x+(3q-4p) \\[-3pt] \end{array} $$    So I equate $(4q+3p)x+(3q-4p)=1$ and solve the simultaneous linear equations  $4q+3p=0, 3q-4p=1$. I get $p=6,q=1$  Hence $6x+1$ is the required inverse.  I am pretty sure that the answer is correct but is the method to achieve it right? Calculate text width with JavaScript Evaluation of $\sum_{n=0}^{\infty}\frac{1}{(n^4+n^2+1)n!}$ I want to trace a border in a box or minipage like this model. A border customizable in left angle on top and right angle in bottom. Somebody can help me? Thanks a lot JPierre Possible Duplicate:     Which one of these is correct?  It is I who am at fault. It is me who is at fault.  The word &quot;is&quot; is a conjugation of &quot;be&quot; which is a linking verb. I also want to know the same for 2nd person.  It is you who are at fault. It is you who is at fault.