Unable to use Robo3t
What's the easiest way to run GUI apps on Windows Subsystem for Linux as of 2018?
My only problem is making them line up three-across and have equal spacing. Apparently, spans can not have width and divs (and spans with display:block) don't appear horizontally next to each other. Suggestions?  &lt;div style='width:30%; text-align:center; float:left; clear:both;'&gt; Is what I have now. XML DML for 2008 R2..How to modify nested Elements What are the rules about using an underscore in a C++ identifier? It's common in C++ to name member variables with some kind of prefix to denote the fact that they're member variables, rather than local variables or parameters. If you've come from an MFC background, you'll probably use m_foo. I've also seen myFoo occasionally.  C# (or possibly just .NET) seems to recommend using just an underscore, as in _foo. Is this allowed by the C++ standard? I have a point $(X,Y)$ on an elliptical curve $E(a,b)$ where $a=-3$ and $B$ is a large number that is in hexadecimal from -51BD. To compress this point oficially in a program, we know that every $X$ on the curve has two $Y$'s, one even and one odd. Therefore, we only need to store whether the corresponding $Y$ point we are storing is even or odd. So to store the point $(X,Y)$ now, we shift $X$ ridding of its most significant bit and add $1$ if the $Y$ point is odd. Therefore, one can think of the compressed point being $2 \cdot X+B$ ($B$ is the bit we added to tell if $Y$ was odd or even) since the multiplication by $2$ is implied because of the shift. I understand this completely. It's recovering the original point that has me confused.  Since a point on the curve is given by the equation $y^2=x^3+ax+b$, we can find square roots officially in $\mathrm{GF}(p)$. The following from my spec is what confuses me.  If $p$ is prime and congruent to $3$ mod $4$, one of its square roots is $z^{(p+1)/4} \bmod p$ and the other is $p-z^{(p+1)/4} \bmod p$. How would I solve this? What is $z$?  For example, I know how to extract the $x$ point so let's say I simplify the right side of the equation to be $7$ and $p$ is $11$. Would I be trying to solve for the square root that's equal to $7$ mod $11$? \footnote breaks @-constructs in \index How many ordered pairs $(A,B)$ they are if $A$ and $B$ are subsets of $\{ 1,2,3,4,5 \}$ such that $A \cap B = \{1\}$? How many ordered pairs $(A,B)$ they are if $A$ and $B$ are subsets of $\{ 1,2,3,4,5 \}$ such that $A \cap B = \{1\}$?  If 1 is in both sets :  Size of $A$ is $&gt;1$ and size of $B$ is $&gt;1$.  $\Rightarrow$ Size of $A \cup B =4$  $$5 \cdot (3 \cdot 3 \cdot 3 \cdot 3) = 405$$ is not correct.  What am I missing here? Let's assume a very simplified situation. There is only one seller asking for 10$ and there is only one buyer bidding 20$. At what price would the transaction go?   Some answers say it won't, because the ask price has to match the bid price. () Other answers say, that the time ordering is important. () Which answer is correct? What is the final price in the presented situation? Let $R$ be the subring of $M_2(\mathbb{R})$ defined by $$R=\left\{\begin{pmatrix}a&amp;b\\0&amp;d\end{pmatrix}\mid a\in\mathbb{Q},b,d\in\mathbb{R}\right\}$$  Show that $R$ is right-Artinian but not left-Artinian.  The "obvious" approach to show it is not left-Artinian is to find an $A\in R$ such that $$RA\supset R(A^2)\supset R(A^3)\supset\dots$$  is an infinite decreasing chain of proper left submodules of $R$ (as a module over itself). However, after spending a lot of time in this endeavor I've come to the conclusion that no such element $A$ exists (I can post a proof if anybody is interested but it'd be long and not worth the time for me otherwise). Does anybody have an idea for another approach I can take to this? I have no idea what else submodules of $R$ would look like. Do I need an ESTA if entering the US from Canada by train As a UK citizen, I will be flying to Vancouver (for which I will need an eTA) and then taking an AMTRAK train to Seattle. I will stay in Seattle for several days before returning to Vancouver and flying back to the UK. Do I need an ESTA for entering the US during this trip? See these three pages:          All three of them say:     Earned at least X total score for at least 20 answers in the  tag.   This same text also appears on the tooltip for the tag badge:   Also note the double colon after 'gold badge'  As well as on these pages:          But the  clearly states:        Bronze Badge:  You must have a total score of 100 in at least 20 non-community wiki answers to achieve this badge.   Silver Badge: You must have a total score of 400 in at least 80 non-community wiki answers to achieve this badge.   Gold Badge: You must have a total score of 1000 in at least 200 non-community wiki answers to achieve this badge.        Edit:  It has been noted in the comments that other users see different numbers in these pages, but in all cases it appears they see the same number of answers are required for all tag badge description, regardless of the badge's type. What is this sand covered insect? I have found in a wooden house, in the mountains of central Spain this very curious insect:    I was wondering if that was its skin or whether it's decomposition or sand or saw-dust.  That particular region of Spain does have stink-bugs ... but this one doesn't look like anything I've seen before.  A google image search returns mostly arachnids ... but it seems to have only 6 legs. How can we differentiate between change and progress in the area of history and natural sciences? I am a beginner using QGIS (1.8.0 Lisboa) on a Mac (OS X 10.6.8 Snow Leopard). I work on a file with two layers in the following sequence (top to bottom):    a vector polygon layer with a few outlines of various buildings in France and Italy (CRS: EPSG:4326 - WGS 84 - "on the fly" re projection is activated ) a Google satellite layer uploaded via open layer plugin.   Sometimes when changing the scale (scrolling in and out) the Google layer updates a little wrong in size or position compared to the polygon layer. But as soon as I "pan" the map everything reconstructs right. Now the problem I have concerns the print composer.  The result of any kind of export - pdf, or image or svg - of my map (vector layer + Google layer in the background) differs from what is shown in the preview of the print composer: the vector layer and the Google layer positions are mismatching and in a different scale. I can send some pictures showing screenshots of the print composer preview and the final results if required.  I tried almost everything:    new files with different CRS settings "cache" update preview "render" update preview different paper sizes,  different scales,   The problem appears only using an open layer map (Google, OpenStreetMap...). If I try the same using two vector layer (shapefile) - one with the buildings, another one with the city in background - everything works out great.  ...exasperating! Since it is a real important research project I would be very grateful for some help and tips. I set a command block to /tp. It was supposed to activate only if you had 0 time left, and to activate the command block with a hopper clock to keep giving it power.   But it is not checking if you have 0 time left, so it keeps teleporting me. I can't get close enough to the command block to break it, so I'm stuck in a neverending teleport loop.  Any ideas? My question relates to Chapter 3, Exercise 8 in &quot;Baby Rudin&quot;. It states: If $\sum_n a_n$ converges, and if $\{b_n\}$ is monotonic and bounded, prove that $\sum_n a_n b_n$ converges. My attempt would have been: Since $\{b_n\}$ is monotonic and bounded, $\{b_n\}$ converges and it exists $\inf \{b_n\}$ as well as $\sup \{b_n\}$. But then we have $$ \left| \sum_n a_n \inf \{b_n\} \right| \leq \left| \sum_n a_n b_n \right| \leq  \left| \sum_n a_n \sup \{b_n\} \right| \leq \max \left( \left| \sup \{b_n\} \right|, \left| \inf \{b_n \} \right| \right)\varepsilon \leq \tilde{\varepsilon} $$ since $\sum_n a_n$ converges and $\max \left( \left| \sup \{b_n\} \right|, \left| \inf \{b_n \} \right| \right)$ is a finite number. This would imply, by the comparison test, that $\sum_n a_n b_n$ converges as well. $\quad \Box$ But as there was a way longer, more rigorous proof chosen in this , I'm a bit suspicious that my proof is not complete. Am I missing something?  EDIT: Thanks everyone! @hermes: The last part of your proof gave me the following idea. As $\lim_{n \to \infty} b_n = c$ and $\{b_n\}$ is monotonic, couldn't we just set $b_n = c - c_n$ with a monotonically decreasing sequence $\{c_n\}$ which has $\lim_{n \to \infty} c_n = 0$. Then we have $$\sum_n a_n b_n = \underbrace{c \sum_n a_n}_{\text{converges by assumption}} - \underbrace{\sum_n a_n c_n}_{\text{converges by Theorem 3.42}} \leq \varepsilon_1 - \varepsilon_2 = \varepsilon $$ since  Theorem 3.42 Suppose  the partial sums of $\sum_n a_n$ form a bounded space $\quad \checkmark$ $c_0 \geq c_1 \geq c_2 \geq \dots \quad \checkmark$ $\lim_{n \to \infty} c_n = 0 \quad \checkmark$  Then $\sum_n a_n c_n$ converges.  Thus $\sum_n a_n b_n$ converges as well. $\quad \Box$ Now that should hold, I think. So I wouldn't need to go through all the estimates. How to know the partition where grub is installed I am Dual Booting Maverick with Natty. I want to know which partition has Grub installed.   I want to make sure that grub is installed in the maverick partition not in natty, so that when I remove the natty partition, I don't have booting problem in the absence of grub.  On maverick there is no menu.list file in /boot/grub but natty has one.  Here is the output of fdisk -l  gaurav@gaurav-desktop:~/Desktop$ sudo fdisk -l  Disk /dev/sda: 320.1 GB, 320072933376 bytes 255 heads, 63 sectors/track, 38913 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x9cdb9cdb     Device Boot      Start         End      Blocks   Id  System /dev/sda1   *           1        7295    58592773+  83  Linux /dev/sda2            7295       25475   146033664   83  Linux Partition 2 does not end on cylinder boundary. /dev/sda3           25476       32194    53970367+  83  Linux /dev/sda4           32195       35483    26414081    5  Extended /dev/sda5           32195       32444     1999872   82  Linux swap / Solaris /dev/sda6           32444       35483    24413184   83  Linux gaurav@gaurav-desktop:~/Desktop$ How do I start DMing a new game after the Red Box? Entering Unicode characters in LaTeX How do I enter Unicode characters in LaTeX?  What packages do I need to install and what escape sequence do I type to specify Unicode characters in an ASCII source file? Suppose 27 cubes are stacked together, suspended in the air, to form a larger cube.  The cube is then painted on all the exposed surfaces and dried.  The smaller cubes are then randomly permuted in both spatial position and spatial orientation to form another large cube.  What is the probability that new larger cube is identical to the original?  This is not homework, it was passed on by a friend who found it somewhere on the internet, and they do not remember where. Moving a magnet close to a conductor induces a current. If it consists of a superconducting material with resistance $R=0$, then my textbook says:     Then the induced current will continue to flow even after the induced emf has disappeared.   This makes sense physically - there is no resistance to stop charge flow. But then the book draws this conclusion:     Thanks to this persistent current, it turns out that the flux through the loop is exactly the same as it was before the magnet started to move, so the flux through a loop of zero resistance never changes.   If the flux $\Phi$ never changes in a superconductor, from Faraday's law this means - from what I have learned - that no electromotive force $\mathcal{E}$ is induced:  $$\mathcal{E}=-\frac{\mathrm{d} \Phi}{\mathrm{d}t}=0 \:\:\:\:\text{ when }\frac{\mathrm{d} \Phi}{\mathrm{d}t}=0$$  My conclusion is therefor: There would never be induced any current at all. Current can never be induced in a superconductor loop. Is this the case or am I misunderstanding my book?