I'm having problems with creating a DIN A 2 Layout in QGIS 3.0  Where can I define the page size? I defined it in the arrangement settings ("Page Setup") but when I create a rectangle the Layout is still in DIN A 4 ...
I am trying to change the page size in the QGIS 3.0 Composer but I am not able to see the Composition tab.  I have right clicked an empty space in the toolbar but did not see anything about composition.  I am using Windows 7.  I tried looking at the QGIS docs but they are still in testing.
How does a Windows 10 'digital license' activate the OS? $$ 3x+6y+5z=7 $$ The general solution to this linear Diophantine equation is as described    $$ x = 5k+2l+14 $$ $$ y = -l $$ $$ z = -7-k $$ $$ k,l \in \mathbb{Z} $$  If I plug the original equation into  the solution is: $$ y = 5n+2x+2  $$ $$ z =-6n-3x-1 $$ $$ n \in \mathbb{Z} $$  I can rewrite this as:  $$ x = l $$ $$ y = 5k+2l+2 $$ $$ z = -6k-3l-1 $$ $$ k,l \in \mathbb{Z} $$  However now two equations depend on two variables ($k,l$) and one on one variable $l$. In the first solution one equation depends on two variables and two on one variable.  Questions:  How can I come from a representation like the one from wolfram alpha for the general solution to one where all equations depend on one distinct variable except one equation.  Is there always such a representation? What should be done when no more Pokemon can be caught? (when full storage) Minimal and characteristic polynomial have same set of irreducible factors Let $V$ be a vector space over a field $F$ (not necessarily algebraically closed), and $T:V\rightarrow W$ be a $F$-linear map. If $c(T)$ and $m(T)$ denote the characteristic and minimal polynomial respectively, then I want to show that $c(T)$ and $m(T)$ have the same set of irreducible factors.  I know that the claim is true if $F$ is algebraically closed, because then the two polynomials have the same roots. The typical reasoning one gives when $F$ is not algebraically closed is: Take an algebraic closure of $F$, say $K$, and look at $m(T)$ and $c(T)$ in $K$ to conclude the proof. But as far as I know, the proof that any root of $c(T)$ is a root of $m(T)$ requires the fact that $V$ is a $K$-vector space. But we only know that $V$ is an $F$-vector space. I can't understand how to proceed after this. Statistical significance (p-value) for comparing two classifiers with respect to (mean) ROC AUC, sensitivity and specificity I have a test set of 100 cases and two classifiers.  I generated predictions and computed ROC AUC, sensitivity and specificity for both classifiers.  Question 1: How can I compute p-value to check if one is significantly better than the other with respect to all scores (ROC AUC, sensitivity, specificity)?    Now, for the same test set of 100 cases, I have different and independent feature assignments for each case. This is because my features are fixed but subjective and provided by multiple (5) subjects.  So, I evaluated my two classifiers again for 5 "versions" of my test set and obtained 5 ROC AUCs, 5 sensitivities and 5 specificities for both classifiers. Then, I computed the mean of each performance measure for 5 subjects (mean ROC AUC, mean sensitivity and mean specificity) for both classifiers.  Question 2: How can I compute p-value to check if one is significantly better than the other with respect to mean scores (mean ROC AUC, mean sensitivity, mean specificity)?    Answers with some example python (preferably) or MatLab code are more than welcome. What is the emergency phone number in different countries? Reading through my textbook I came across the following problem and I am looking for some help solving it.  Prove that,  $$\binom{3n}{r}=\binom{3n-1}{r} + \binom{3n-1}{r-1}$$  In a previous problem I solved the following but I wasn't sure how to use it to prove the question above. Prove,  $$\binom{n}{r}=\binom{n}{n-r}$$  Solution:  left Side,  $$\binom{n}{r}=\frac{n!}{(n-r)!r!}$$  right side,  $$\binom{n}{n-r}=\frac{n!}{(n-(n-r))!(n-r)!} = \frac{n!}{(n-r)!r!}$$  Therefore completing the proof,  but how do I solve my first question? thanks! Use the triangle inequality to show that $|a|+|b| \leq |a+b|+|a-b|$ How would I go about proving that  $$|a|+|b| \leq |a+b|+|a-b|$$  Using the triangle inequality?  I tried squaring both sides, yielding:  $$|a|^2 +|b|^2 +2|a||b| \leq 2|a|^2+2|b|^2+2||a|^2-|b|^2|$$  Is it correct to move the terms to the other side? I tried $$2|a||b| \leq |a|^2+|b|^2+2||a|^2-|b|^2|$$ Then it is obvious that  $$0\leq |a|^2+|b|^2-2|a||b| +2||a|^2-|b|^2|$$ $$0\leq (|a|-|b|)^2 +2||a|^2-|b|^2|$$ $$0\leq ||a|-|b||^2 +2||a|^2-|b|^2|$$  But I have neither used the triangle inequality, nor it looks mathematically rigorous for my real analysis class. Any tips and tricks for solving these problems? Adding a CTAN package to a MacPorts-maintained TeX installation I have installed TeXlive through MacPorts and now discover that I am missing some package. For clarity let's use a trivial example and say I am missing lipsum.sty.  I now have two options:  Option 1  I go to a CTAN site, look for lipsum.sty, download it, and install it in ~/Library/texmf/tex.  This is not bad, but it leaves it possible that I will end up with a second lipsum at some point. That's why the second option is desirable.  Option 2  2.1-- I find out which MacPorts package includes lipsum.sty  and  2.2-- I sudo port install the necessary package.  How do I do 2.1? WPF Get Element(s) under mouse How long can eggs be unrefrigerated before becoming unsafe to eat? Alternative definitions of wifebeater I was a li'l nonplussed to find out that the word  can mean other things besides "a man who beats up his wife". Its definition reads:        One who (usually as a repeated practice) beats one’s wife, or a husband prone to violence   A kind of sleeveless shirt, often but not exclusively worn as an undershirt.   (uncountable, UK, slang) Stella Artois, a brand of lager beer.       reckons that sense 2 is American and originates "apparently from the association of such a garment with men who commit domestic violence". It does not register the UK slang. I'm guessing that Stellas also have an unfortunate association with domestic violence.  Are these alternative senses popular in their respective countries or are they perhaps nonce words that have lingered on? Any idea how these disparaging associations came about? I expect that the brewers of Stella Artois are none too happy about it. Real Time LaTeX? Possible Duplicate:          I'm looking for an application in which I can write pure LaTeX in one half of the screen and on the other half to see the compiled document in real time. This seems really like an obvious way to do LaTeX authoring but it doesn't look like any software is able to do it. LyX is not what I'm looking for. I want to edit the actual LaTeX and see it rendered in real time. Why isn't this already being done somewhere? Implementing a sort of "tightened frenchspacing" By default, (La)TeX adds extra space between sentences. The exact rules are in the TeXbook, a summary by TH. can be found . What follows is is my "executive summary":   After any period (or quotation/exclamation mark) that doesn't follow a capital letter, TeX adds a font's "extra space" to the "normal (interword) space". For Computer Modern 10pt, the normal space is 3.33pt and the extra space 1.11pt, resulting in an intersentence space of 4.44pt (+33% compared to the nomal space, assuming no stretch and shrink). A sentence end after a capital letter is indicated by the \@ command: Ann is learning the ABC\@. To denote that a period after a lowercase letter doesn't end a sentence, one may use \@ as well as an explicit space: Prof.\@ Higgins or Prof.\ Higgins.   Intersentence spacing may be brought in line with normal spacing by issuing the \frenchspacing command. (This is the standard setting for certain languages, e.g. the  (n)german option of the babel package.) With \frenchspacing enabled, there's no need to use the \@ command. One may revert to (La)TeX's default behavour by issuing \nonfrenchspacing.  So, there's the choice between extra intersentence spacing and uniform spacing. What is missing in my opinion is a third option I'll call "tightened frenchspacing": normal (interword) spacing between sentences and decreased spacing after non-sentence ending periods. To be somewhat more precise (while still ignoring stretch and shrink), I envision the following:   Adhering to a font's ratio between "normal space" and "normal space plus extra space" (3:4 for Computer Modern), the intersentence space is set to the normal space. The space after non-sentence ending periods equals the normal space multiplied by the above ratio (resulting in a space of 2.5pt for Computer Modern). As with \nonfrenchspacing in effect, the \@ command is used to denote cases of special punctuation.   How should such a "tightened frenchspacing" option be implemented?  EDIT: Thanks to Mico for providing what may well be a stepping stone to a solution. Here's another attempt at clarification:   Simply reducing the spacing after any period is not what I had in mind. There should be a (tangible) difference between end-of sentence spaces and spaces after non-sentence-ending periods (only that the latter should be reduced instead the former augmented). If adhering to a font's normal space/extra space ratio is not possible, a solution that adds a fixed (reduced) space (say, a thin space) after non-sentence-ending periods will be accepted. For compatibility reasons, I'd prefer a solution that works with the traditional use of \@ (i.e allows "tightened frenchspacing" for existing documents without the need to change the document body). Should it turn out that a different set of "special punctuation" macros is needed, I'll accept this as a solution. Check tag classes with watir? What is the best way to install Guest Additions with all the resize and clipboard and file sharing features on Ubuntu 16.04? Add a downvote option in the VLQ review There are many answers that are well-formed and well-formatted but are simply plain wrong or misleading. When reviewing, I often have a hard time deciding which action to take, as:   Looks Good: is OK per se but I would never say "good" (would work if the wording was changed slightly), Edit: The question is formatted well and editing it to be right would go against the intent of the author, Recommend deletion: this looks like the most reasonable choice, but I wouldn't flag it if I were browsing the site in normal mode, Not sure: I am quite sure what is wrong but I end up clicking this because the other three options do not seem fitting.   My proposal is to add a fifth option - Downvote that would make it easier to choose in case of such answers. Prove or the disprove the existence of a limit of integrals Let $A \subset \mathbb{R}^2$ be the annulus defined, in polar coordinates, as  $$ A = \{ (r, \theta) \in [0, \infty) \times [0, 2\pi) : 2 \leq r \leq 3 \} \, .$$ Let $F: A \to A$, $F:(r, \theta) \mapsto (r, r \theta \text{ (mod } 2\pi))$. Let $g : A \to \mathbb{R}$ be a continuous function. Discuss the existence of the limit $$ \lim_{n \to \infty} \int_A g \circ F^{(n)}(x) \, \text{d}x \, ,$$ where $F^{(n)} = F \circ F \circ \dots \circ F$ denotes the composition of $F$ with itself $n$ times.  I cannot find a way to attack this problem. Here's what I tried: if $g \circ F^{(n)}$ converged, I think we could take the limit under the integral sign. So I tried to show the convergence of $F^{(n)}$. If it converged, then the limit function must send every point to a fixed point of $F$. So I determined the fixed points of $F$. But I don't think $F^{(n)}$ converges. Why are so many important verbs irregular? Freestyle Background color not white? I'm following  and I've set my compositor up in the exact same way as he did, except my background is like a slightly light grey and I don't know why.  In his video's he's getting a bright white as his background and I can't see where I'm going wrong.  Here's the render:   Here's the compositor window:    file.  Please help! I really want to know why it's not happening.