I am using Kubuntu 13.04. How to format a USB or external drive?
Using Ubuntu 11.10, how do I format an external or USB drive? I don't find an option to 'format' when I right click. In the new version, the drives don't come up on the desktop, but instead come on the Launcher.
Academic papers using Stack Exchange data Select node without namespace with XPath I have the following table:     \documentclass[a4paper,preprint,aps,                    unsortedaddress]{revtex4}     \usepackage[top=2.5cm,bottom=2.5cm,                 left=3.5cm,right=2.5cm]{geometry}     \usepackage{longtable,array,booktabs}      \ifdefined\mc        \renewcommand{\mc}{\multicolumn}     \else        \newcommand{\mc}{\multicolumn}     \fi      \begin{document}           \setlength\LTleft{-3.0cm}         \setlength\LTright{-2.0cm}          \scriptsize      \begin{longtable}{ l r @{--} p{1.3cm} &gt;{\centering}p{2.6cm}                         *{4}{p{1.25cm}} p{2.1cm}}                    %% headers and footers      \caption{Table} \\      \hline\hline     \endfirsthead      \mc{9}{c}{\tablename\ \thetable\ -- \emph{Continued from previous page}} \\      \hline     aaaaaaaaaaaaaaaaaaaaaaaa &amp; \mc{2}{c}{bbb--ccc} &amp; ddd &amp; eee &amp; fff &amp; ggg &amp; hhh &amp; iii \\     \hline     \endhead      \hline     \mc{9}{r}{\emph{Continued on next page}} \\     \endfoot      \hline\hline     \endlastfoot      %% body of table          aaaaaaaaaaaaaaaaaaaaaaaa &amp; \mc{2}{c}{bbb--ccc} &amp; ddd &amp; eee &amp; fff &amp; ggg &amp; hhh &amp; iii \\     aaaaaaaaaaaaaaaaaaaaaaaa &amp; bbb &amp; ccc &amp; ddd &amp; eee &amp; fff &amp; ggg &amp; hhh &amp; iii \\     aaaaaaaaaaaaaaaaaaaaaaaa &amp; bbb &amp; ccc &amp; ddd &amp; eee &amp; fff &amp; ggg &amp; hhh &amp; iii \\     aaaaaaaaaaaaaaaaaaaaaaaa &amp; bbb &amp; ccc &amp; ddd &amp; eee &amp; fff &amp; ggg &amp; hhh &amp; iii \\      \end{longtable}     \end{document}   which I would like to shift to the left by setting \setlength\LTleft{-3.0cm}  and \setlength\LTright{-2.0cm}. Furthermore, I have to apply \scriptsize for the table. However, none of these commands are working with revtex4. Could you please provide a solution to this problem? (This question is the same as in , but I cannot sign up with that username.)  If it is not possible, an alternative solution would also be useful for me. For this case, I attached the first page of my real table. What's the arc length of an implicit function? While an explicit function $y(x)$'s arc length $s$ is easily obtained as  $$s = \int \sqrt{1+|y'(x)|^2}\,dx,$$   is there any formula for implicit functions given by $f(x,y) = 0$? One can use the implicit differentiation $y'(x) = -\frac{\partial_y f}{\partial_x f}$ to obtain  $$s = \int\sqrt{1 + |\partial_y f / \partial_x f|^2}\,dx,$$  but that still requires (locally) solving for $y(x)$. Is there any formulation that does not require this, e.g. another implicit equation involving $s$?    Thoughts so far:  One could rewrite $s$ as  $$s = \int |\nabla f|\, |\partial_x f|dx,$$  or symmetrize to  $$s = \int |\nabla f|\, \underbrace{(|\partial_x f|dx + |\partial_y f|dy)}_{(*)}/2$$  where $(*)$ might be strongly related to $|df|$ I guess (though it's not identical due to the $|\cdot|$), but then? Check differentiability of multivariable functions I am trying to determine if the following functions are (totally)   differentiable at $(x,y)=(0,0)$.      $$f(x,y)=\sqrt{\vert xy \vert}$$      $$g(x,y)=e^{{{\vert x \vert}^3}y}$$   However I am not particularly sure where to start. The concept of the total derivative being the Jacobian matrix makes sense but I don't really understand how to work with functions that may not have a Jacobian matrix. Any help would be greatly appreciated! Cause of black artifacts in render? Does set difference distribute over set intersection? I am asked to prove that, if $A, B$ and $C$ are sets, then $$A-(B\cap C)=(A-B)\cap(A-C).$$  However, I think that either I have made an error in my working, or the wording of the problem contains a typographical error.    My working so far is as follows: $$\begin{align}x\in(A-(B\cap C)) &amp;\iff (x\in A)\wedge(x\not\in(B\cap C))\\ &amp;\iff(x\in A)\wedge((x\not\in B)\vee(x\not\in C))\\ &amp;\iff((x\in A)\wedge(x\not\in B))\vee((x\in A)\wedge(x\not\in C))\\ &amp;\iff(x\in(A-B))\vee(x\in(A-C))\\ &amp;\iff x\in((A-B)\cup(A-C)) \end{align}$$  First line to second line: by De Morgan's Law. Second line to third line: by distributivity of set intersection over set union. Third line to fourth line: by definitions of intersection and set difference.   If my working is correct, then I have shown that $$A-(B\cap C)=(A-B)\cup(A-C).$$ Am I correct? How would one write object-oriented code in C? What are some ways to write object-oriented code in C? Especially with regard to polymorphism.    See also this Stack Overflow question . Wrapfigure in an enumerate environment. I want to use wrapfigure in the enumerate environment. For example,   \documentclass[11pt, a4paper]{article} \usepackage{graphicx} \usepackage{float} \usepackage{wrapfig}  \begin{document} \normalsize {\bf Multiple Choice: } \begin{enumerate}     \item \begin{wrapfigure}{r}{4.5cm}           \includegraphics[scale=0.5]{Exam1_Fig1.jpg}           \end{wrapfigure}           As shown in the figure. This rock is phaneritic and contains quartz, K-feldspar, and plagioclase in nearly equal amounts. What is it? \\           (A) rhyolite \\           (B) basalt \\           (C) diorite \\           (D) ash-flow tuff \\           (E) granite \\ \end{enumerate} \end{document}   However, the figure inserted is not wrapped by the text, but put in a strange position. How to fix the problem?   Thank you! Why is python ordering my dictionary like so? Here is the dictionary I have  propertyList = {     "id":           "int",     "name":         "char(40)",      "team":         "int",     "realOwner":    "int",      "x":            "int",     "y":            "int",      "description":  "char(255)",      "port":         "bool",     "secret":       "bool",     "dead":         "bool",     "nomadic":      "bool",      "population":   "int",     "slaves":       "int", }   But when I print it out with "\n".join(myDict) I get this  name nomadic dead port realOwner secret slaves team y x population id description   I know that a dictionary is unordered but it comes out the same every time and I've no idea why. Reasons to use opamp I have a shapefile with a "id" field in which some non-unique ID numbers exist (don't ask).  When I try to import this to a PostGIS database using QGIS' DB Manager, it tries to re-use those IDs are unique primary key and fails.  The "Primary key" option is not ticked. "id" is shown in grey next to it.   How can I make DB Manager ignore an existing "id" column and just create a new SERIAL field or something like that?  QGIS 2.18. Why does demand for stock rise when a company appears to have high future value? So a company decides to raise money by selling shares of its stock. The stock is offered at an initial price of their determination. The stock is then traded in a market where forces of supply and demand, as well as speculation, affect its price.  When a company performs well (or makes decisions that will likely increase the company's earnings in the future), more people want to buy that stock than do people wanting to sell it. This creates demand for that stock which causes its sell/bid prices rise. When a company performs poorly, the opposite occurs; more people want to sell the stock than want to buy. This creates a surplus that causes sell/bid prices to fall.  Understanding this, people speculate over company performance to determine the future behavior of a stock, and attempt to "buy low, sell high" or "borrow high, return low" in order to make a profit.  But my question is regarding the first part of that process. What is it about company performance that causes the perceived value of its stock to rise? This is especially perplexing since I would guess that most people don't exercise their voting rights, and many stocks don't pay dividends.  It's not like a shareholder has any direct financial gains from increased company earnings (speculative stock pricing notwithstanding).  What is the intrinsic value of the stock?  It seems to me that all participants in a stock market tacitly agree that stock price will have a direct correlation with company performance. So another way of asking the question is, would an opposite agreement hold? What financially (this isn't a question about logistics, communication, or feasibility) stops all participants in a market from agreeing to value a particular company's stock inversely with respect to the company's performance?  Since I feel like I can't quite nail down my question, another form of it would be, why is stock market behavior driven by company performance? A locally constant sheaf on a locally connected space is a covering space; Proof? As part of my hobby i'm learning about sheaves from Mac Lane and Moerdijk. I have a problem with Ch 2 Q 5, to the extent that i don't believe the claim to be proven is actually true, currently. Here is the question repeated:     A sheaf on a locally connected space $X$ is locally constant if each point $x \in X$ has a basis of open neighborhoods  $\mathcal{N}_x$ such that whenever $U,V \in \mathcal{N}_x$ with $U \subset V$, the restriction $\mathcal{F}V \rightarrow \mathcal{F}U$ is a bijection. Prove that $\mathcal{F}$ is locally constant iff the associated etale space over $X$ is a covering.   I don't have a problem (i think) showing the "covering" $\rightarrow$ "locally constant" direction of the implication. My problem is in the reverse direction. In fact, in the process of trying to solve this problem i think i have come up with a very simple counterexample.  Here's my attempt at a counterexample:  Let $X = \{a,b,c\}$ be  finite space with open subsets $\{\{\},\{a\},\{a,b\},\{a,b,c\}\}$. I claim this is hyper-connected and locally connected.  Let $\mathcal{F}$ be the sheaf with $\mathcal{F}(\{a,b,c\}) = \{q\}$, $\mathcal{F}(\{a,b\}) = \{r,s\}$, and $\mathcal{F}(\{a\}) = \{t\}$, where $r = q|_{\{a,b\}}$, and $t = q|_{\{a\}} = r|_{\{a\}} = s|_{\{a\}}$. I claim this defines a locally constant sheaf. Indeed we can have $\mathcal{N}_{a} = \{\{a\}\}$, $\mathcal{N}_{b} = \{\{a,b\}\}$, and $\mathcal{N}_c = \{\{a,b,c\}\}$.  Now we can calculate the corresponding etale space $p : E \rightarrow X$ as follows: $E = \{\dot{q}(a), \dot{q}(b), \dot{q}(c), \dot{s}(b)\}$, and has these four distinct elements since $\dot{q}(a) = \dot{r}(a) = \dot{s}(a) = \dot{t}(a)$, and $\dot{q}(b) = \dot{r}(b)$. The action of $p$ follows by construction.  However, $p$ does not appear to me to be a covering space. In particular, the only neighborhoods of $b \in X$ are $\{a,b\}$and $\{a,b,c\}$ and we have $p^{-1}(\{a,b\}) = \{\dot{q}(a),\dot{q}(b),\dot{s}(b)\}$ and this cannot be partitioned into homeomorphic images of $\{a,b\}$ because $2 \nmid 3$. Similarly, $p^{-1}(\{a,b,c\}) = \{\dot{q}(a), \dot{q}(b), \dot{q}(c), \dot{s}(b)\}$ and this cannot be partitioned into homeomorphic images of $\{a,b,c\}$ since $3 \nmid 4$.  So i think i have a locally constant sheaf on a locally connected space whose corresponding etale space is not a covering. What have i done wrong? I have seen the claim mentioned elsewhere so i'm convinced i'm wrong somewhere. Perhaps someone knows where a correct proof is published otherwise? Problem:  If $\gcd(a, b) = 1$ and If $ab = x^2$ ,prove that $a$, $b$ must also be perfect squares; where $a$,$b$,$x$ are in the set of natural numbers  I've come to the conclusion that $a \ne b$ and $a \ne x$ and $b \ne x$ but I guess that won't really help me.. I understand that if the $\gcd$ between two numbers if $1$ then they obviously have no common divisors but where do I go from this point?  Any tips at tackling this would be great. It looks quite easy though I'm still trying to get my hand around these proofs! Any pointers in the right direction would be great.  Thank you in advance, Every word is marked as misspelled in TeXstudio Is propagation of uncertainties linear? I'm in doubt with one thing: let's imagine that we have $n+1$ quantities, $n$ of them being directly measured, and the other one being related to the first $n$ by a function $f : \mathbb{R}^n \to \mathbb{R}$. If the $x_i$ is one of the quantities being directly measured and if $\Delta x_i$ is it's uncertainty, then I've learned that the uncertainty of the quantity $y = f(x_1, \dots x_n)$ is given by:  $$\Delta y=\sum_{i=1}^{n}{\frac{\partial f}{\partial x_i}\Delta x_i}$$  My problem in understanding this is: well, if $f$ is such a function, it's derivative is the linear transformation $df : \mathbb{R}^n \to \mathbb{R}$ given by:  $$df=\sum_{i=1}^{n}\frac{\partial f}{\partial x_i}dx_i$$  Where $dx_i : \mathbb{R}^n \to \mathbb{R}$ is given by $dx_i(a_1 \dots a_n) = a_i$. Hence, what we are saying, is that $\Delta y = df ( \Delta x_1, \dots \Delta x_n)$, in other words, we are saying that the uncertainty of $y$ is given by a linear function of the vector of uncertainties of the quantities $x_i$. Why should that be true? I mean, what's the reasoning behind this? I really didn't get the point, my teacher just gave the equation without justifying or anything like that.  Any help and reference is appreciated. Why is my variable unaltered after I modify it inside of a function? - Asynchronous code reference Remove the shortcut of an application that was installed using Wine from the Unity Dash How do I remove an application shortcut from the Unity Dash (not from the Unity Launcher)?  My specific issue is I tried to install a game using Wine a while ago. I then uninstalled the game, then uninstalled Wine. Today I found that shortcuts to the game (and other related content that came with it) remain when I push Super and type the game name. (Wine is not visible though.) How to run Wireshark on Ubuntu 17.10 I used to run wireshark on ubuntu 17.04 with  ~/$ sudo wireshark   but now (just after update to 17.10) when I write the same I get an error  ~/$ sudo wireshark      QStandardPaths: XDG_RUNTIME_DIR not set, defaulting to '/tmp/runtime-root'     Invalid MIT-MAGIC-COOKIE-1 keyQXcbConnection: Could not connect to display :0