As the title says. What's the point of the LTS if I have to update anyway at least every 2 years because the packages aren't updated to new versions and hence quickly contain non-supported versions?
Why are packages in the official Ubuntu repositories older than the latest (upstream) versions from Debian Sid, PPAs, the authors, etc.?
I have recently noticed that in chat when I type enough and the profile picture expands with my message, the hat stays the same size and it stays in the same place (I have the  app): Capturing SOAP requests to an ASP.NET ASMX web service Preview on the right please, optionally? This is a  about securing public DNS resolvers   Open DNS servers seem pretty neat and convenient, as they provide IP addresses that we can use consistently across our company regardless of where they are located. Google and OpenDNS provide this functionality, but I'm not sure that I want these companies to have access to our DNS queries.  I want to set up something like this for use by our company, but I hear a lot about this being dangerous practice (particularly in regards to ) and I want to make sure that we do this right. What things do I need to keep in mind when building this type of environment? I get “signal out of range” displayed on screen after boot up from fresh Ubuntu installation, if flashes grup in corner for half a sec then it goes to “No Input Signal”, anyone got a clue how to get the grub to load in diffrent resolution with out have access to the file system?   if not i guess my only option is to boot from the cd and edit som files to get grub to load in diffrent resolution?   as it is now i cant get access to grub menu, and i cant get access to login screen, all will give me “No Input Signal”, anyone worked around this? Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle   This question came up in a miscellaneous problem set I have been working on to refresh my memory on several topics I did earlier this year. I have tried changing $4\sin(A)\sin(B)\sin(C)$ to  $$4\sin(B+C)\sin(A+C)\sin(A+B)$$ by making substitutions by reorganizing $A+B+C=\pi$. I then did the same thing to the other side to get $$-2(\sin(B+C)\cos(B+C)+\sin(A+C)\cos(A+C)+\sin(A+B)\cos(A+B))$$ and then tried using the compound angle formula to see if i got an equality. However the whole thing became one huge mess and I didn't seem to get any closer to the solution. I am pretty sure there is some simpler way of proving the equality, but I can't seem to figure it out. Maybe there is a geometric interpretation or maybe it can be done using just algebra and trig. Any hint's would be appreciated (I would prefer an algebraic approach, but it would be nice to see some geometric proofs as well) My request is simple: Many times, there will be a question where some people feel like it should be closed. At the same time, I'll feel like it's a perfectly legit question, and it should therefore NOT be closed. Right now, I have to wait till it gets enough close votes (which it always does for some reason, as soon as 1 or 2 people vote to close, e/o else decides to vote as well) and then vote to re-open. How about, as soon as there's one vote to close, there should be an option to counter the close vote, something like "vote not to close" or whatever. This would then bring the "Vote to close" number down by one. Only when the vote to close outnumbers the vote not to close by 5, that's when a question is indeed closed. Any thoughts? JavaScript: getting ImageData without canvas I have been struggling to install VirtualBox Guest Additions in a Debian Virtual Machine (Debian 7, Debian 8 and Debian 9). To prove that $f_n(x) = \frac{nx}{1+n^2x^2}$ does not uniformly converge to $f(x) = 0$ on $[0,1]$ My Approach. To prove the given statement it is sufficient to show that $$\exists \epsilon &gt; 0, \exists x \in [0,1], \forall N \in \mathbb{N}, \exists n &gt; N $$ $$ \Rightarrow | f_n(x) - f(x)| &gt; \epsilon$$    Let $\epsilon = \frac{1}{10}, x = \frac{1}{n}, n =N+1$. Hence $| f_n(x) - f(x)| = \frac{1}{2} &gt; \epsilon = \frac{1}{10}$. Hence, we have shown for every $N \in \mathbb{N}$, how to choose $n$ such that $| f_n(x) - f(x)| &gt; \epsilon$. Thus, $f_n(x)$ does not converge uniformly on $[0,1]$.    Is there any logical flaw or wrong steps in my proof. Any help would be appreciated. This question follow's JeffE's comment :     In most industries, a PhD has negative value.   I am curious about what potential negatives there may be with having a PhD, particularly in industries.  For example, my PhD has applications for image sensors, and the potential of working in that field is appealing—would having a PhD be a potential barrier? Would it be a case of how explicit the practicalities of the PhD can be made?  Would this make any difference? Is Unicoins a April Fool’s Stunt? Is Unicoins a April Fool’s Stunt ? resizing a BUTTON through CSS Why is the d'Alembert's Principle formulated in terms of virtual displacements rather than real displacements in time? What is the difference between topological and metric spaces? What is the difference between a topological and a metric space? Make program first read from pipe, then from keyboard I think the following is true:  If $q$ is coprime to $a$ then $a\mid (nq-1)$ for some $n\in\mathbb Z$.  I have started to first reducing $n\in\mathbb Z$ to $n\in\{1,\ldots,a-1\}$ and then considering all the cases $q = ka + j$ for $j=1,2,3,\ldots,a-1$. I stopped at $j=3$ and found that this is kinda messy. Does anyone have an elegant proof for this (maybe using some results from number theory)?  [Note: question has been generalized from prime $q\nmid a$ to $\,q$ coprime to $a$] Is there a way to set my current location manually on Windows 8? Is there a way to set my current location manually on Windows 8? I don't have a GPS module, but the device is always located at the same place. The automatic detection via network sets a place 200 kilometers from here.  I searched for a setting that allows me to set the location by hand, but I didn't find one. Is GTA Online cross-platform? Let $\lambda(A)$ be the Lebesgue measure of $A$. There exists an interval $I$ such that $\lambda(E \cap I) / \lambda(I) < 1-\epsilon$ (Not mentioned in title but $\epsilon$ is a number greater than $0$ and $E$ a Lebesgue measurable subset of $\mathbb{R}^n$.) I know a question equivalent to this one has been asked (). But it was not answered. Also my thinking is a bit different. So I would love to get some help on this problem. So far I've got:  $$\frac{\lambda(E \cap I)}{\lambda(I)} &gt; 1- \epsilon \iff \frac{\lambda(I)-\lambda(E\cap I)}{\lambda(I)} &lt; \epsilon$$  And also, a theorem states that, for all $\epsilon$ there exists a closed set $F\subset E$ such that $\lambda(E\setminus F) &lt; \epsilon$ so it would sufice to prove  that there exists an interval $I$ such that $$\frac{\lambda(I)-\lambda(E\cap I)}{\lambda(I)}&lt;\lambda(E\setminus F).$$ I'm not sure if this helps me much but any help is appreciated. What would you do?