I can't run sudo nautilus, but if I run only nautilus in terminal it runs.  And also I can't run native application via terminal using sudo.  Terminal shows me a warning if I run using sudo:    This pic shows, if I run nautilus in terminal it works:
I installed Ubuntu 17.10. Now I am having trouble with gksu:  $ gksu -dg synaptic No ask_pass set, using default! xauth: /tmp/libgksu-HgUjgQ/.Xauthority STARTUP_ID: gksu/synaptic/8760-0-alex-XPS-15-9530_TIME4974977 cmd[0]: /usr/bin/sudo cmd[1]: -H cmd[2]: -S cmd[3]: -p cmd[4]: GNOME_SUDO_PASS cmd[5]: -u cmd[6]: root cmd[7]: -- cmd[8]: synaptic buffer: -GNOME_SUDO_PASS- brute force GNOME_SUDO_PASS ended... Yeah, we're in... Unable to init server: Could not connect: Connection refused (synaptic:8767): Gtk-WARNING **: cannot open display: :1 xauth: /tmp/libgksu-HgUjgQ/.Xauthority xauth_env: (null) dir: /tmp/libgksu-HgUjgQ   If I don't use -g, the password dialog is disabled. So looks like a problem with creating a tty for root.  Any advice?
Given any $10$ consecutive positive  integers , does there exist one integer which is relatively prime to the product of the rest ? Evaluate the integral   $$\int\frac{1}{x^2\left(x^4+1\right)^{3/4}}\,dx$$   My Attempt:  Let $x = \frac{1}{t}$. Then $dx = -\frac{1}{t^2}\,dt$. Then the integral converts to   $$ -\int \frac{t^3}{(1+t^4)^{3/4}}\,dt $$  Now Let $(1+t^4) = u$. Then $t^3\,dt = \frac{1}{4}du$. This changes the integral to  $$ \begin{align} -\frac{1}{4}\int t^{-3/4}\,dt &amp;= -u^{1/4}+\mathcal{C}\\ &amp;= -\left(1+t^4\right)^{1/4}+\mathcal{C} \end{align} $$  So we arrive at the solution  $$\int\frac{1}{x^2\left(x^4+1\right)^{3/4}}\,dx = - \left(\frac{1+x^4}{x^4}\right)^{1/4}+\mathcal{C.}$$  Question: Is there any other method for solving this problem? Finding files by only knowing extension? How did Rita know she lost the power? Can I run a natural gas dryer from a portable propane tank? Broken screen while debug mode was disabled. How can I re-enable adb? The screen of my phone is physically broken, so it doesn't display anything, but the touch aspect does seem to be working, as I can unlock my phone, and get haptic feedback.  Before I go to get it replaced, I'd like to back up some personal data from the built-in memory to the SD card. I'm fairly certain that if I can get adb to recognize the device, then I can use  to see where I am touching the screen and do what I need to do. Problem is, when I plug it in, it's not recognized, while my emulator is.  I suspect that I might not have USB debugging enabled, which I think is necessary for adb to work. If this is the case, can someone give me screenshots of their Droid X (running 2.3.X) going from the home screen, through all the menus, to activating debug mode? Hopefully I can do it by dead reckoning and adb from there.  Edit: I haven't changed anything about the ROM or launcher; it's all stock. Ejecting a USB 3.0 drive connected to USB 3.0 port reconnects immediately I am currently learning electromagnetism and I am getting really annoyed using a physical hand to accomplish the right hand rule. I get it, i've seen diagrams, and i've solved problems with it but still somehow I mess up because I bend a finger or twist my wrist a strange way. I am tired of using my hand.  So my question is can I find an entirely mathematical basis for this?  Lets say for any given diagram I define the positive $x$ axis as going to the right, the positive $y$ axis as going up and the positive $z$ axis as going into the page. Then I just did unit vectors for everything to determine the missing component?  As I understand it $\vec{F} = q (\vec{V}\times\vec{B})$ and with that definition I am confident I could find out the direction of the electric field given the velocity of a particle of charge and the magnetic field but how can I re-arrange this so I can solve for other variables? For example what does $\vec{B}$ equal in terms of a cross product involving $\vec{V}$ and $\vec{F}$? Rotate (group) objects outwards on spherical particle system? I'm trying to create a mini-world with (among other things) trees on it. I have created an icosphere, modeled several trees and put them in a group. Next I added a particle system to the sphere, and chose hair. In the "render" section I selected Group and the name of the group I created. As expected the trees show up but they're all oriented in the same direction, while I would like them to pop up just like the default hairs do (along the normal of the individual faces). I tried fiddling with all kinds of settings, had them oriented in all kind of ways (e.g. towards the "north pole"). I also tried using the emitter instead of hair setting but without success as well. For $G$ a group and $H\unlhd G$, then $G$ is solvable iff $H$ and $G/H$ are solvable? I recently read the well known theorem that for a group $G$ and $H$ a normal subgroup of $G$, then $G$ is solvable if and only if $H$ and $G/H$ are solvable. In my book, only the fact that $G$ is solvable implies $H$ is solvable was proven. I was able to show that if $H$ and $G/H$ are solvable, then so is $G$, but I can't quite show that $G$ is solvable implies $G/H$ is solvable.   My idea was this. Since $G$ is solvable, there exists a normal abelian tower $$ G=G_0\supset G_1\supset\cdots\supset G_r=\{e\}. $$ I let $K_i=G_i/(H\cap G_i)$, in hopes of getting a sequence $$ G/H\supset K_1\supset\cdots\supset K_r=\{e\}. $$ My hunch is that the above is also a normal abelian tower. However, I'm having trouble verifying that $K_{i+1}\unlhd K_i$ and that $K_i/K_{i+1}$ is abelian.  Writing $H_i=H\cap G_i$, I take $gH_{i+1}\in K_{i+1}$ for some $g\in G_{i+1}$. If $g&#39;H_i\in K_i$, then I want to show $g&#39;H_igH_{i+1}g&#39;^{-1}H_i$ is still in $K_{i+1}$, but manipulating the cosets threw me off. I also tried to use either the second or third isomorphism theorems to show that $K_i/K_{i+1}$ is abelian, but I'm not clear on how to apply it exactly. I'd be grateful to see how this result comes through. Thank you. How does SO's (or other SE sites) versioning work of questions and answers? Do they use some standard piece of code or is it something homebrew?  With versioning I meant if for example a question is edited you'll see all the changes (a diff). Prove that $ \sum_{n=0}^\infty \frac{z^{2^n}}{1-z^{2^{n+1}}}=\frac{z}{1-z} $ Show that for $|z|&lt;1$, one have $$ \frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\cdots+\frac{z^{2^n}}{1-z^{2^{n+1}}}+\cdots = \frac{z}{1-z} $$ and $$ \frac{z}{1+z}+\frac{2z^2}{1+z^2}+\cdots+\frac{2^k z^{2^k}}{1+z^{2^k}}+\cdots = \frac{z}{1-z} $$ Use the dyadic expansion of an integer and the fact that $2^{k+1}-1=1+2+2^2+\cdots+2^k$. Unbiased estimator for geometric distribution parameter p How did Lucius Malfoy know Sirius's Animagus form? In Order of the Phoenix, Snape eggs on Sirius at Grimmauld Place:  ‘Tell me, how is Lucius Malfoy these days? I expect he’s delighted his lapdog’s working at Hogwarts, isn’t he?’ [Sirius] ‘Speaking of dogs,’ said Snape softly, ‘did you know that Lucius Malfoy recognised you last time you risked a little jaunt outside? Clever idea, Black, getting yourself seen on a safe station platform ... gave you a cast-iron excuse not to leave your hidey-hole in future, didn’t it?’ Order of the Phoenix - Chapter twenty-four, Occlumency - Page 460 - Bloomsbury  How did Lucius Malfoy know that Sirius was a dog Animagus? How did he know it was Sirius in the train station? Some facts to consider: According to the 's timeline, the characters who are considered to be of the Marauders era (born 1959) were only five years younger than Lucius Malfoy (born 1954), so Sirius and Snape, and the other Marauders-era students, and Lucius Malfoy, would have been at Hogwarts together for two years. Prisoner of Azkaban demonstrates it took Sirius, James, and Peter four to five years to master the Animagus transformation, which would have been at least two years after Lucius Malfoy would have left Hogwarts (Although we don't know when Lucius became a school governor.). A canon-based answer would be great; barring one, a speculative answer in the spirit of canon is welcome. Installed Ubuntu 13.04 on a Dell laptop I had laying around.  Install goes well but when I'm done, I only get a command line, no GUI at all.  Tried several versions, tried 32 vs 64 bit all give me the same result.  Any ideas on what to do/try? Why did Obi-Wan Kenobi choose that moment to appear to Luke? In The Empire Strikes Back, Luke Skywalker violently escapes from the lair of a wampa ice creature, finds himself in the middle of a life-threatening snowstorm on an inhospitable planet, collapses from cold and fatigue, and will likely be dead in a few minutes. As :     Luke lies face down in the snow, nearly unconscious. Slowly he looks   up and sees Ben Kenobi, barely visible through the blowing snow. It is   hard to tell if Kenobi is real or a hallucination.     Why did Obi-Wan Kenobi choose this moment to appear to Luke and deliver rather important instructions concerning next steps in his Jedi training?  It seems lucky that Luke remembered anything of the encounter at all. He probably thought he was hallucinating. In FHS-2.3, we have /media that holds mount points for removable media such as CD-ROMs and we have /mnt that holds temporarily mounted filesystems.  On the other hand, we have /run/media and /run/mount. For me, the CDs and USBs are mounted on /run/media.  I don't see any clear distinction between them(/media, /mnt, /run/mount) . What are their differences?     I have seen similar trend (mount on /run/media) in fedora 20 - GNOME 3.10.4 and ubuntu 14.04.1 (installed on virtual box) with GNOME 3.10.4. But when I plugged in a USB flash (with auto-mounter script) on a system with Centos 6 and GNOME 2.28.2 it was mounted on /media How to increase difficulty in Dungeon World I've been GMing for a couple of sessions now and I'm still struggling at making encounters or situations more difficult. I will describe dangerous situations and encounters in fiction but the mechanics make it too easy for my players to succeed.  For instance no matter how big I describe the Troll, if the fighter of the party charges him and swing his sword, roll+2 for Hack and slash, he'll easily hit the troll. Almost 40% chance to hit without consequences. I just spilt coffee on my laptop, what should I do? What is the value of $\sum\limits_{i=1}^\infty\frac{1}{p_{p_i}}$ where $p_{i}$ is the $i$th prime? What is the value of $\sum\limits_{i=1}^\infty\dfrac{1}{p_{p_i}}$ where $p_n$ is the nth prime (and so $p_{p_n}$ is the $k$th prime, where $k$ is the $n$th prime) ?  Thus $\frac{1}{3}+\frac{1}{5}+\frac{1}{11}+...=$ ??  How do I efficiently decide if $\sum_{i=1}^\infty\dfrac{1}{p_{p_i}}\gt1$ is true without using a computer ?  I assume this infinite sum is an irrational number. Has this already been proven ?