I have 3 partitions on my Ubuntu 18.04 desktop. It was good until yesterday. But don't know what happened. I Can't see 'Other Options' in the file manager when I open it. Looks weird. the file manager looks different than it usually used to look.. It doesn't have a left-side panel which it usually contains all the list which is in the middle. Please Help me.
When I open nautilus does not open the left bar where it says computer and down markers such as music, documents, videos, downloads. I reinstalled nautilus from the synaptic package manager but nothing happened. What I can do to make that bar visible again?
Was Stack Exchange named after Experts Exchange? Was Stack Exchange named after Experts Exchange, or is the former using the word "Exchange" a coincidence? Can the program counter on Intel CPUs can be read directly (that is without 'tricks') in kernel mode or some other mode? If $x_1,x_2,\ldots,x_n$ are the roots for $1+x+x^2+\ldots+x^n=0$, find the value of $\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}$ Let $x_1,x_2,\ldots,x_n$ be the roots for $1+x+x^2+\ldots+x^n=0$. Find the value of   $$P(1)=\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}$$   Source: IME entrance exam (Military Institute of Engineering, Brazil), date not provided (possibly from the 1950s)  My attempt: Developing expression $P(1)$, replacing the 1 by $x$, follows $$P(x)=\frac{(x_2-x)\cdots (x_n-x)+\ldots+(x_1-x)\cdots (x_{n-1}-x)}{(x_1-x)(x_2-x)\cdots (x_n-x)}$$    As $x_1,x_2,\ldots,x_n$ are the roots, it must be true that   $$Q(x)=(x-x_1)\cdots(x-x_n)=1+x+x^2+\ldots+x^n$$ and  $$Q(1)=(1-x_1)\cdots(1-x_n)=n+1$$ Therefore the denominator of $P(1)$ is  $$(-1)^{n} (n+1).$$ But I could not find a way to simplify the numerator.  Another fact that is probably useful is that  $$1+x^{n+1}=(1-x)(x^n+x^{n-1}+\ldots+x+1)$$ with roots that are 1 in addition of the given roots $x_1,x_2,\ldots,x_n$ for the original equation, that is  $$x_k=\text{cis}(\frac{2k\pi}{n+1}),\ \ k=1,\ldots,n.$$  This is as far as I could go...    Hints and full answers are welcomed. Why isn't Obi-Wan's ghost young? One of the changes made to the Special Editions of Star Wars: Episode VI: Return of the Jedi was at the end of the film.  Luke sees 3 ghosts during the celebration on the forest moon of Endor, Yoda, Anakin, and Obi-Wan.  In the Special Editions, the actor who played Anakin earlier in the film, Sebastian Shaw, is replaced with Hayden Christiansen who played young Anakin in Episodes 2 and 3.  My question is, why is Anakin the only one de-aged?  Shouldn't Obi-Wan also be his younger self, along with Yoda? When I open a window to air out the room, how does the smell disperse? Let $C$ be the coefficient of $x^2$ in the expansion of the product $(1 - x)(1 + 2x)(1 - 3x)\cdots(1 + 14x)(1 - 15x).$ Find $|C|.$   Just to begin,  $(1-x)(1+2x) = -2x^2 + x + 1$  $(1-x)(1+2x)(1-3x) = 6x^3 - 5x^2 - 2x + 1$   But expanding on like this take too long.  In the end the terms will be like:  $(1 - 15x)(a_0 + a_1x^1 + a_2x^2 + ....)$  Then just looking at the $x^2$, it will be:  $(a_2x^2 - 15a_1x^2) = x^2(a_2 - 15a_1)$  But it it still a very hard problem, any hints? Google Chrome: Is there a keyboard shortcut to open a link in a new tab? Is is possible to open a link in a new background tab in Google Chrome with a keyboard shortcut?  (I know it's possible with Ctrl+click and middle click, and I know that Shift+Enter opens the link in a new browser window.) Careers - Search Book JS error Fix for Hibernate error "Use of the same entity name twice" How you fix the following Hibernate error:  What does "Use of the same entity name twice". $C \otimes A \cong C \otimes B$ does not imply $A \cong B$ Let $R$ be a commutative unital ring and let $M$ be an $R$-module and let $S$ be a multiplicative subset of $R$.  Today I proved both of the following: $$ S^{-1} R\otimes_R S^{-1}M \cong S^{-1} M$$ and $$ S^{-1} R \otimes M \cong S^{-1} M$$  Now I'm slightly confused.      Either my proofs are wrong or $C \otimes A \cong C \otimes B$ does not imply $A \cong B$. But I can't come up with an example. Can someone give me an example? (Or tell me that my proofs are wrong.) Really stuck with this, \begin{cases}  2tx'(t)-1=x^2t \\  x(t=1)=1 \\  \end{cases} I don't understand the variables. What does it mean that $x$ differentiated have $t$ as a function value, like this: $x'(t)$? Don't get it :( Verbatim Environment in Figure Caption verbatim environment is not working in figure caption. Checked  and  is not working.  \documentclass[11pt, double space]{article} \usepackage{amsmath} \usepackage{graphicx} \usepackage{color} \usepackage[usenames,dvipsnames]{xcolor} \usepackage[english]{babel} \usepackage[autostyle, english = american]{csquotes} \MakeOuterQuote{"} \usepackage{cprotect}  \begin{document} \begin{figure}[!h] \caption{Sample 1 \cprotect\begin{verbatim} \tau(A) = e^{\chi_0A/(KD_p(K+A))} \end{verbatim} per hour.} \label{fig:singlecell_chemo} \end{figure}  \end{document}   I have tried cprotect package also. Any help. How does "failure" work in Dungeon World? How does it move the story forward? I am having trouble understanding the specifics of failure in Dungeon World. According to the rules, a GM should "make a move" against players when they fail, and that's alright, I get that.  What I don't get is how a GM is supposed to advance the plot by doing so. Let's take an example:  Rick the Fighter is climbing a massive chain that serves to hold down the wizards tower, preventing it from drifting off. He makes the move Defy Danger, to avoid slipping of, since it's raining hard, and the chain is slippery with algae and moss. He rolls and get a score of 5, meaning he fails.  In the above situation, he fails climbing the chain, so the GM should make a move. I have a hard time seeing what move would advance the story here, since I feel that the failure would imply that he doesn't get to the tower.  Is this one of those situations, where you just skip making a roll, and just let the players climb the chain, or does anyone have any idea how to handle the move, so that the story is advanced somehow?  I just want to be able to give the players some dangers that aren't traps and monsters, and I thought that a flying wizards tower was cool. I'm having a hard time seeing how I can advance the story on a failure in this situation.    Let me elaborate on the problem. I will not say that it's wrong to make the players succeed with some "cost", even on a failed roll, it just makes true "failure" impossible. That's my beef about it. The players will eventually succeed at anything that doesn't outright kill them. The question of whether P=NP is perhaps the most famous in all of Computer Science. What does it mean? And why is it so interesting?  Oh, and for extra credit, please post a proof of the statement's truth or falsehood. :) I'm planning to attend a Californian university as a visiting scholar (not paid by the university) on a J-1 visa, for a total of four weeks; a colleague of mine will do the same, but for 12 weeks. I've been told (personally, ie. no citable sources) that, if you're staying on a J-1, you're effectively a California resident for the time you're there, and, therefore, need a Californian driver's license.  I've read up on this and found  that states:     If you are a visitor in California over 18 and have a valid driver   license from your home state or country, you may drive in this state   without getting a California driver license as long as your home state   license remains valid.      If you become a California resident, you must get a California driver   license within 10 days. Residency is established by voting in a   California election, paying resident tuition, filing for a homeowner's   property tax exemption, or any other privilege or benefit not   ordinarily extended to nonresidents.   I'm not planning to vote in any elections, nor will I own a home there, but I'm not sure what would constitute "privileges or benefits not ordinarily extended to nonresidents".  Do we need to get driver's licenses in California, or can we use our foreign licenses? How about day trips to Arizona or Nevada?  Note: this seems to be related to , but my question is specifically about staying there on a J-1. Deleting an element from an array in PHP Is there an easy way to delete an element from an array using PHP, such that foreach ($array) no longer includes that element?  I thought that setting it to null would do it, but apparently it does not work. Version: Ubuntu 16.04  I have used it some weeks, but now it gives me this message when I'm trying to install software and update or upgrade:  Could not get lock /var/lib/apt/lists/lock - open (11: Resource temporarily unavailable) /var/lib/apt/lists/ locked   followed by this:    .  and another error:  ** (appstreamcli:4560): WARNING **: Exception: Unable to get write lock on /var/cache/app-info/xapian/default: already locked   What should I do? In a race there are n horses.In the race more than one horse may get the same position. For example, 2 horses can finish in 3 ways.   Both first horse1 first and horse2 second horse2 first and horse1 second   In how many ways,the race can finish so that a particular horse can never be first.If f(n) is the number of ways, is there any recurrence relation to get f(n). Why am I still getting a password prompt with ssh with public key authentication? I'm working from the URL I found here:    My ssh client is Ubuntu 64 bit 11.10 desktop and my server is Centos 6.2 64 bit. I have followed the directions. I still get a password prompt on ssh.  I'm not sure what to do next. Can't access pre populated SQLite database using PhoneGap/Cordova in Android