I'm running Ubuntu 15.04 with "Cinnamon" installed, and I wish to install  this exstension :   I tried to install it this way:  1) Copied the folder "CoverflowAltTab@dmo60.de" to "~/.local/share/cinnamon/extensions/".  2) Enabled the extension in the cinnamon settings. (Eventually you have to restart cinnamon: Press Alt+F2, type 'r' and enter)  The problem is that I can`t locate this path : ~/.local/share/cinnamon/extensions/
How do you show a hidden folder (whose name starts with .) in File Manager a.k.a. Nautilus on Ubuntu?  I am somewhat new to Ubuntu and can't seem to find the "view" menu.
Can helium disappear from Earth? Helium is lighter than air, so it should fly off from Earth. Is it possible that in the future we will run out helium? Should 'drive by' downvoting be more effectively caught? I have an answer that I'm certain is technically correct. However, it continues to be downvoted. The answer has already been noted in comments on other answers to be better. It goes more in-depth than the others, yet it continues to be downvoted.  The already upvoted answers were nowhere close initially, and I had already commented on them to say why it is wrong...  Isn't there a limit for that? Shouldn't something be catching this? after upgrading to 19.04 the Grub menu have become hidden. It still works and I can boot to both Windows 10 and Ubuntu by blindly finding the entry and pressing enter. I have tried to edit the /etc/default/grub and running sudo update-grub afterward but no change. Here is how it looks now:  # If you change this file, run 'update-grub' afterwards to update # /boot/grub/grub.cfg. # For full documentation of the options in this file, see: #   info -f grub -n 'Simple configuration'  GRUB_DEFAULT=0 GRUB_TIMEOUT=10 GRUB_DISTRIBUTOR=`lsb_release -i -s 2&gt; /dev/null || echo Debian` GRUB_CMDLINE_LINUX_DEFAULT="quiet splash" GRUB_CMDLINE_LINUX="" GRUB_TIMEOUT_STYLE=menu GRUB_HIDDEN_TIMEOUT_QUIET=false  # Uncomment to enable BadRAM filtering, modify to suit your needs # This works with Linux (no patch required) and with any kernel that obtains # the memory map information from GRUB (GNU Mach, kernel of FreeBSD ...) #GRUB_BADRAM="0x01234567,0xfefefefe,0x89abcdef,0xefefefef"  # Uncomment to disable graphical terminal (grub-pc only) #GRUB_TERMINAL=console  # The resolution used on graphical terminal # note that you can use only modes which your graphic card supports via VBE # you can see them in real GRUB with the command `vbeinfo' #GRUB_GFXMODE=640x480  # Uncomment if you don't want GRUB to pass "root=UUID=xxx" parameter to Linux #GRUB_DISABLE_LINUX_UUID=true  # Uncomment to disable generation of recovery mode menu entries #GRUB_DISABLE_RECOVERY="true"  # Uncomment to get a beep at grub start #GRUB_INIT_TUNE="480 440 1"   I can't use either Shift or Esc to make it visible too. After pressing either of these, Grub doesn't work anymore and I need to turn my laptop off and on in order to boot.  Edit: Not sure if it helps or not, but I just realized that if I go to the BIOS when I exit BIOS the grub menu shows up with no problem. The next time I boot, the grub menu is invisible again. How do you test that a Python function throws an exception? How does one write a unittest that fails only if a function doesn't throw an expected exception? Services Actions Missing in Central Admin Quick way to subtract white outline from knotwork pattern in Illustrator? I am working with a pretty complex traced AI that I received, and I want to basically make it so that all the visually black stuff is shapes, and all the white outlines (invisible here) are removed. Most of the spaces when a line crosses over another is created by a white outline.   I know about expanding appearance and the pathfinder tool, but because there are so many lines that are crisscrossing, I haven't had an easy way of minusing the fronts away. I could expand appearance and sort every object carefully, but I feel like there may be a quicker way of doing this.  If I put it in Photoshop as a smart object in 'screen' mode, it looks right.. I could put that back in Illy and trace it, but I don't want to complicate the paths. Run a shell script as another user that has no password I would like to run a script from the main ubuntu shell as a different user that has no password.  I have full sudo privileges, so I tried this:  sudo su -c "Your command right here" -s /bin/sh otheruser   Then I have to enter my password, but I am not sure if that script is now really running under that user.  How can I confirm that the script is really running under that user now? Show that $a \star b=a \cdot b+a+b$ is binary operation for the group $\Bbb{ Q} \setminus \{-1\}$ The group $\left(\Bbb{ Q}\setminus\{-1\},\star\right)$ has as its underlying set the rational numbers different from $-1$ and the operation $\star$ is defined as $a \star b=a \cdot b+a+b$ where multiplication/addition are the usual operations with rational numbers.Show that this is a binary operation.  What I did (Please help verify):  identity element has to be 0 since $a\star 0 = a+0+a(0)$ so for $a\star b=0$ we have  $ab+a+b=0$ $$ab+a=-b \\ a(b+1)=-b \\ a=\frac{-b}{b+1}$$ so $b$ cant be $-1$ since that solution is not in $\Bbb{Q}$. Similar question might have been answered here. Is this operation true for all groups? I think so.  Thanks for the tips and help. Can I use the Stack Exchange concept in my own sites? I am thinking of making a site which will have a forum. I already have a lot of users (in an old forum, which I want to renew).  I like the Stack Exchange concept (voting, reputation, etc).  Can I build one of my own on my site? (I don't intend to copy any source code, just the concept itself.)  Is the concept patented or otherwise legally protected, or is it free to use? BibTeX loses capitals when creating .bbl file Titles of articles I'm about to cite contain upper case letters and when using BibTeX it converts them to lower case ones. This happens only in the title and only the first letter conserves its case. For example, when I cite an article about HF, the reader won't know if it is about Hafnium (Hf) or fluorine acid (HF).  I know that I can fix it manually in the .bbl file but I would like to avoid it, or fix it automatically. How to prevent "Delhi Belly"? How to simplify the summation kk! without using induction? $$\sum_{k=1}^nk(k!)$$  I know the answer is (n+1)!-1..I can solve this question using principle of mathematical induction...but I would like to know if there is any other alternative approach Proving $\sum\limits_{k=0}^{n}\cos(kx)=\frac{1}{2}+\frac{\sin(\frac{2n+1}{2}x)}{2\sin(x/2)}$ I am being asked to prove that $$\sum\limits_{k=0}^{n}\cos(kx)=\frac{1}{2}+\frac{\sin(\frac{2n+1}{2}x)}{2\sin(x/2)}$$ I have some progress made, but I am stuck and could use some help. What I did: It holds that $$\sum\limits_{k=0}^{n}\cos(kx)=\sum\limits_{k=0}^{n}Re(\cos(kx))=\sum\limits_{k=0}^{n}Re(\cos(x)^{k})=Re(\sum\limits_{k=0}^{n}\cos(x)^{k})=Re\left(\cos(0)\cdot\frac{\cos(x)^{n}-1}{\cos(x)-1}\right)=Re\left(\frac{\cos(x)^{n}-1}{\cos(x)-1}\right) $$ For any $z_{1},z_{2}\in\mathbb{C}$ we have it that if $z_{1}=a+bi,z_{2}=c+di$ then $$\frac{z_{1}}{z_{2}}=\frac{z_{1}\overline{z2}}{|z_{2}|^{2}}=\frac{(a+bi)(c-di)}{|z_{2}|^{2}}=\frac{ac-bd+i(bc-ad)}{|z_{2}|^{2}}$$ hence $$Re\left(\frac{z_{1}}{z_{2}}\right)=\frac{Re(z_{1})Re(z_{2})-Im(z_{1})Im(z_{2})}{|z_{2}|^{2}}$$ Thus, $$Re\left(\frac{\cos(x)^{n}-1}{\cos(x)-1}\right)=\frac{(\cos(nx)-1)(\cos(x)-1)-\sin(nx)\sin(x)}{(\cos(x)-1)^{2}+\sin^{2}(x)}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{\cos^{2}(x)-2\cos(x)+1+\sin^{2}(x)}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{-2\cos(x)+2}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{-2(\cos(x)-1)}= \frac{=\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{-2(-2\cdot\sin^{2}(x/2))}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{4\sin^{2}(x/2)}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{4\sin^{2}(x/2)}=\frac{\cos(x(n+1))-\cos(nx)-\cos(x)+1}{4\sin^{2}(x/2)} $$ This is the part where I am stuck, I would appriciate any help or hint on how to continue. Edit: Given the corrections by André I get: $$(\cos(nx+x)-1)(\cos(x)-1)+\sin(nx+x)\sin(x)=\cos(nx+x)\cos(x)-\cos(nx)-\cos(x)+1+\sin(nx+x)\sin(x)$$ so $$\cos(nx+x)\cos(x)+\sin(nx+x)\sin(x)=\cos(xn+x-x)-\cos(nx)=0$$ Edit 2: I found anoter mistake in the above, I will try to correct Edit 3: When multiplying correctly the above it works out :-) In the inline math mode ($...$), if the formula is too long, LaTeX will try to break it on operators, e.g.  very long text followed by a very long equation like $a+b+c+d+e+f+g+h+i+j+k+l$ etc   may be rendered as  very long text followed by a very long equation like a+b+c+d+e+f+g+h+i+ j+k+l etc   However, the break won't happen if they are separated by commas, e.g.  very long text followed by a very long equation like $a,b,c,d,e,f,g,h,i,j,k,l$ etc   will overflow the page like  very long text followed by a very long equation like a,b,c,d,e,f,g,h,i,j,k,l etc   How to make LaTeX able to insert line breaks after a comma too? Finding Maximum value of consecutive positive integers with constraints Let, $x,y,z$ be consecutive positive integers such that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} &gt; \frac{1}{10}$ I need to find the maximum value of $x+y+z$ ?  My attempt:  I tried guessing first, To maximise $x+y+z$ it is necessary to have high value for $x$, But not too high which will contradict the constrain of the problem. As we can roughly split the $0.1$ into three parts so I started with numbers closer to $30$ with trial an error I got that for $x=29$ we get the optimum and final result would be 90 (=max. value of $x+y+z$).   Is there any way to arrive at this conclusion with using inequalities? Any hints will be appreciated... How to get back sudo on Ubuntu? Word meaning a job someone is installed in where they don’t do anything If you click on the Stack Exchange menu or the inbox menu or the achievements/reputation menu, you can then move the mouse between those three menus without any additional clicks.  This behaviour does not appear to be applying to the Winterbash menu (or the "help" menu for that matter, but that's on the other side of the bar).  Could we get this sorted, since otherwise the behaviour is a little inconsistent, with extra clicking on a menu bar where in all other circumstances clicking would close the menu! When selecting a new site in the app, the user name remains displayed in the bar above.       How to reproduce:   Look at a user's profile. Click "All (number) sites" Click one of the icons that are displayed in the left.   The user name at the top remains displayed.     App Version: 1.0.89 Device Manufacturer: asus Device Model: K011   OS Version: 4.4.2 (JP_K011-JP_user_3.1.23.189_20141124-user-20141124) I get this error whenever I try to install programs using the terminal:  home@ubuntu:~$ apt-get install myunity E: Could not open lock file /var/lib/dpkg/lock - open (13: Permission denied) E: Unable to lock the administration directory (/var/lib/dpkg/), are you root?   Also I'm unable to install updates using the terminal.