ubuntu 16.04 included software
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How should I setup separate MX records for a subdomain? Let's say I have a domain that I run a web application on, for example cranketywidgets.com, and I'm using  for handling email for people working on that domain, for example, support@ cranketywidgets.com, jane@cranketywidgets.com, joe@cranketywidgets.com and so on.  Google's own mail services aren't always the best for sending automated reminder emails, comment notifications and so on, so the current solution I plan to pursue is to create a separate subdomain called mailer.cranketywidgets.com, run a mail server off it, and create a few accounts specifically for sending these kinds of emails.  What should the MX records and A records look like here for this?  I'm somewhat confused by the fact that MX records can be names, but that they must eventually resolve to an A record. What should the records look like here?  cranketywidgets.com - A record to actual server like 10.24.233.214  cranketywidgets.com - MX records for Google's email applications  mailer.cranketywidgets.com - MX name pointing to server's IP address  I would greatly appeciate some help on this - the answer seems like it'll be obvious, but email spam is a difficult problem to solve. Is there a way in blender or an addon that allows me to select a bunch of faces (or verts) and evenly distributes (or relaxes) them into quads clean quads.  Without screwing up the vert ID/order.  Edit: I would like to have the shapes of my quads to turn into perfect squares while maintaining the volume of the object and the vertex order. I try to download a file with wget and curl and it is rejected with a 403 error (forbidden).  I can view the file using the web browser on the same machine.  I try again with my browser's user agent, obtained by . I do this:  wget -U 'Mozilla/5.0 (X11; Linux x86_64; rv:30.0) Gecko/20100101 Firefox/30.0' http://...   and  curl -A 'Mozilla/5.0 (X11; Linux x86_64; rv:30.0) Gecko/20100101 Firefox/30.0' http://...   but it is still forbidden. What other reasons might there be for the 403, and what ways can I alter the wget and curl commands to overcome them?  (this is not about being able to get the file - I know I can just save it from my browser; it's about understanding why the command-line tools work differently)  update  Thanks to all the excellent answers given to this question. The specific problem I had encountered was that the server was checking the referrer. By adding this to the command-line I could get the file using curl and wget.  The server that checked the referrer bounced through a 302 to another location that performed no checks at all, so a curl or wget of that site worked cleanly.  If anyone is interested, this came about because I was reading  page to learn about embedded CSS and was trying to look at the site's css for an example. The actual URL I was getting trouble with was  and the curl I ended up with is  curl -L -H 'Referer: http://css-tricks.com/forums/topic/font-face-in-base64-is-cross-browser-compatible/' http://cloud.typography.com/610186/691184/css/fonts.css   and the wget is   wget --referer='http://css-tricks.com/forums/topic/font-face-in-base64-is-cross-browser-compatible/' http://cloud.typography.com/610186/691184/css/fonts.css   Very interesting. We had a rule check tonight that we couldn't answer. On the Pathfinder wiki I could only find info on natural 20s and 1s in regards to attack rolls and saves in combat.  Do they also apply to skill checks? Magento: Merge Product Description and Addition Information Tab We are working on a service that is quite new and unknown to people. It's is oriented on small business. The target audience is divided into 3 specific different groups. As for now, we have one landing page, which communicates to 3 groups.   We have very low conversion mainly because people don't understand how they can use the service and benefit from it.  So, we decided to change the logic:  There will be one entrance page, where user will have to choose his group (1 of 3) and then he will be forwarded to the page of the group he chose before, where all the explanations how to use the service are provided.  So instead of one page we will have four pages. Will it improve the communication with users and are there any suggestions how to deal with 3 different target audiences? Is it allowed/desirable to post code snippets? Instead of asking a question I would love to share the one or other code snippet I have created that is too small to make a full article on e.g. CodeProject.com or CodePlex.com.  Question: Would it be a good idea to post it as a Stack Overflow question? Or are there any other suggestions how to share it?  (Of course I could post it on my own weblog and then twitter about it, but I would love to put it in more of a programming context) Show by induction that for all natural numbers n>3  $(1+\frac{1}{n})^n&lt;n$  Let $(1+\frac{1}{n})^n&lt;n$ be true !   We show that $(1+\frac{1}{n+1})^{n+1})&lt;n+1$  $(1+\frac{1}{n})^n(1+\frac{1}{n})&lt;n(1+\frac{1}{n})$   $(1+\frac{1}{n})^{n+1}&lt;n+1$  So now we just need to show that   $(1+\frac{1}{n+1})^{n+1}&lt;(1+\frac{1}{n})^{n+1}$  So here , If I show that :   $(1+\frac{1}{n+1})&lt;(1+\frac{1}{n})$   Does that mean im done with exercise ? Facebook authorization fails on iOS6 when switching FB account on device Is there any way to get real-time compilation for LaTeX? I would really love to have a software or tool which provided real time compilation of my LaTeX document (to be able to see the final pdf or dvi document as I type the latex code), especially for when I am drawing pictures using tikZ. So far, I know of two things, but none of them work for me. They are:   : This is only for tikZ, which would be perfect for me, but unfortunately as far as I know it only works with Ubuntu and Debian, and I use Mac OS X (so I have never actually tried it, though I did dig a lot to see whether I would be able to install on my mac and from what I see it is really what I want - I am even considering installing Ubuntu just to use it!). : This is a minor mode for Emacs. It was extremely hard to get it to work on the mac (see my quest ), and it is not all that fast. The speed is not all that big of a deal for me, but the fact that it does not work with tikZ pictures is (the nodes all collapse in one point so that the words are all one on the top of the other).   So my question is: is there a software ideally like KtikZ or another type of editor which provides real-time compilation and that I can use on my mac? I guess if you know of a Windows/Linux option I would also like to know, and the same for a way to set up a script to get the document compiled every 5 seconds or so (I am sure I have seen this written somewhere, but I don't know whether I can do it with a mac or whether it is editor/pdf viewer dependent), but what I really want is a software/tool which would work with a mac.  To be clear, I am not after something like LyX, that is, I am not after a WYSIWYG-type thing, but rather something where I can type real LaTeX and see my code, but have at the same time another window showing me the pdf (or dvi) file compiled. Distance between two hyperplanes I have two parallel hyper planes $$a^Tx=b_1,a^Tx=b_2$$ where $a \in \mathbb{R}^n, x \in \mathbb{R}^n ,b \in \mathbb{R}$ and I want to find the distance between the two. I have read that the distance between the two hyperplanes is also the distance between the two points $x_1$ and $x_2$ where the hyperplane intersects the line through the origin and parallel to the normal vector $\vec a$. These points are given by $$x_1=\frac{b_1}{\|a\|^2_2}a$$ and $$x_2=\frac{b_2}{\|a\|^2_2}a$$ Then the distance is $|x_1-x_2|$ but I don't really understand how we got $x_1$ and $x_2$. I would like to put an address in my document as follows:  |                             | |                     Name    | |                     Street #| |                     City    | |                     ZIP     | |                             |   where the | mark the left and right page borders. I would like the items of the address (Name, Street etc.) to be left-flushed, but the box right-flushed so that the longest item touches the right page border.  Is there a simple way to accomplish this? Cost Distance Analyst In QGIS? Are there any tools, plugins or methods in QGIS which allows for an analysis of the best path between two points using least cost path method? As shown below ArcGIS has several distance tools. I have a sqlite (v3) table with this column definition:  "timestamp" DATETIME DEFAULT CURRENT_TIMESTAMP   The server that this database lives on is in the CST time zone.  When I insert into my table without including the timestamp column, sqlite automatically populates that field with the current timestamp in GMT, not CST.  Is there a way to modify my insert statement to force the stored timestamp to be in CST?  On the other hand, it is probably better to store it in GMT (in case the database gets moved to a different timezone, for example), so is there a way I can modify my select SQL to convert the stored timestamp to CST when I extract it from the table? What's a good beginner resource to learn group theory? How to simplify the summation kk! without using induction? How to put an acute on an i using Biber: issues with "\'{\i}" I use some packages with require utf8. Here is the main file:  \documentclass[a4paper,titlepage]{report}  \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage[american]{babel} \usepackage{csquotes} \usepackage[backend=biber,style=alphabetic]{biblatex} \addbibresource{refs.bib}  \begin{document} \nocite{*} \printbibliography[heading=bibintoc] \end{document}   And here is the reference file:  @MISC{Example,     author = {Author with spec\'{\i}al character},     title = {Title} }   I have no issues with composing special characters like \H{o} (ő) or \k{a} (ą). However, when I try to compose \'{\i} (í), I get the following inputenc error: "not set up for use with LaTeX". It seems like double compositions are not allowed. A solution is to use \'{i} instead, but not removing the dot is ugly. Surely there must be a better way?  Note: I have tried to add \DeclareUnicodeCharacter{00ED}{\'{\i}} to the preamble, but it doesn't help either. Russian SE has a slightly longer and more storied history than Japanese SO Proving $\sqrt2$ is irrational I used the method of contradiction by assuming that $\sqrt 2$ is a rational number. Then, by the definition of rational number, there exist two integers $p$ and $q$ whose ratio equals $\sqrt 2$. Thus, $$\frac pq = \sqrt2\tag{x}$$  Squaring both sides, $$p^2/q^2 = 2\tag{a}$$  or $$p^2 = 2q^2\tag{b}$$ This means that $p^2$ is an even number, implying that $p$ is even.  Now, any even integer can be written as $2^kf$ where $f$ is any odd integer and $k$ is some positive integer (the minimum value of $k$ and $f$ is $1$ since even numbers start from $2 = 2^1\cdot1$). For odd numbers, $k=0$.  For example, $8 = 2^3 \cdot 1$, $4= 2^2 \cdot 1$, $18= 2^1 \cdot 9$, $24 = 2^3 \cdot 3$, $-12= 2^2 \cdot(-3)$, etc.  Now, from $(b)$, $q^2$ can be even or odd.  Case 1: $q^2$ is even (thus meaning $q$ is even). Then $p= 2^{k_1} \cdot f_1$ (say) and $q = 2^{k_2} * f_2$ (say). Note here in this case, both conditions $k_1=k_2$ and $f_1=f_2$ can't hold simultaneously since that would mean $p/q =1$ and here $p/q =\sqrt2$ (from $(x)$)). Let's consider the condition when $k_1=k_2=k$ but $f_1\ne f_2$. Then $$\frac{p^2}{q^2}=\frac{(2^kf_1)^2}{(2^kf_2)^2} = \frac{f_1^2}{f_2^2} =2$$ (from $(a)$), i.e.: odd/odd ($f_1$ and $f_2$ are odd) can never equal $2$.  Now lets consider $f_1=f_2=f$ but $k_1\ne k_2$, thus $(2^{k_1}f)^2/(2^{k_2}f)^2= 2^{2(k_1-k_2)}= 2$ meaning $k_1 - k_2 = 0.5$ but $k_1,k_2$ are integers, so their difference can't be $0.5$. (Also here, $k_1-k_2$ must be greater or equal to $1$ since $2^{k_1-k_2}=\sqrt 2$, $k_1-k_2$ can't be negative since $\sqrt 2&gt;1$, but $k_1-k_2\ge1$ can't satisfy $(a)$ since the minimum value of $p/q$ in this case will be $2$ which is surely greater than $\sqrt2$.)  Now lets consider $f_1\ne f_2$ and $k_1\ne k_2$, thus $\frac{(2^{k_1} f_1)^2}{(2^{k_2} f_2)^2}= 2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$ can never equal $2$ since $\frac{f_1^2}{f_2^2}$ is either odd or "odd/odd" ie: it doesn't contain 2 as a factor and $2^{2(k_1-k_2)}$ is a power of $2$. So in $2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$, there's no chance of cancellation of the odd factor $\frac{f_1^2}{f_2^2}$.  Case 2: $q^2$ is odd (thus meaning $q$ is odd). $q=f'$ (say, here $k=0$ for $q$ but not for $p$), thus from $(a)$ $2^{2k}\frac{f^2}{f'^2} = 2$ but this isn't possible since a power of $2$ multiplied with odd factor can't equal $2$.  Thus, both case 1 and 2 suggest that for any possible combination of $k_1,k_2,f_1$ and $f_2$, $p/q\ne\sqrt2$, i.e.: for no value of integers $p$ and $q$, $p/q = \sqrt2$. Thus, this contradicts our assumption that $\sqrt2$ is rational. Therefore it must be irrational.  PLS NOTE that i would like to clarify that $f_1^2$/$f_2^2$ is either an odd integer or a fraction of odd/odd or 1/odd form, thus not containing  2 as a factor,for any $2^n$, if it has to be reduced to 2 , must be multiplied with $1/ 2^(n-1)$ but $f_1^2$/$f_2^2$ can't be of $1/2^[n-1]$ form since for odd nos, in $2^k$*f notation, k is 0 , so 2 vanishes as $2^k$ becomes 1 in this case thus f1/f2 can't be represented as  1/ $2^(n-1)$ in which n has to be existent.  therefore $2^{2(k_1-k_2)}\frac{f_1^2}{f_2^2}$ can never equal 2 as to further explain the 3rd condition of case 1  Difficulty: is my approach correct? This proof which I thought is different from proofs found on the internet or books since I have used $p$ and $q$ as any integers, which may have a common factor. So I am not sure if I am on right track. Will someone please check this out and make kind comments? Prove $f(S \cup T) = f(S) \cup f(T)$