Once I plug headphones into my iMac I no longer have the option in the sound preferences panel to output audio to the built-in speakers. I have to unplug the headphones to get audio output through the built-in speakers. Since the plug is on the back of the display this is inconvenient. Does anyone know of an audio app that allows me to easily switch between built-in speakers and headphones for audio output?
I have a set of surround speakers, connected to my iMac through the headphone jack. I use a switch to control what goes to the speaker set (iMac, iPod, or MacBook/aux). So that I don't have to be constantly unplugging and replugging, I leave the audio cable in the iMac's audio port. This means that when I'm listening to something from another source, I can't hear anything from the iMac unless I unplug the cable.  Is there a way to force sound to come out of the internal speakers, even if there's something in the audio out port?
Lufthansa connecting flight at Munich, only 45 minutes in between. Will I be able to make it? I bought a single ticket from Lufthansa, of Mumbai to Paris. My first flight (LH765) lands at 5:55am at Munich, from where I have another Lufthansa flight (LH2226) to Paris at 6:40am. This is a single ticket so I guess its responsibility of the airline to get me on the second flight, but since it's Non-Schengen to Schengen, will customs etc. be a problem? I am assuming the checked-in luggage would transferred automatically by the airline. Will I be able to make it? Pretty much the problem asks if a function is continuous at the point $c$ and $f(c) &gt; 0$ then there exists a $d &gt; 0$ such that $\forall x$, $f(x) &gt; 0$ with $|x-c| &lt; d$.   I can understand what the problem means. That if a function is positive at a point then there is another point that's really close that'll also be positive. I can not prove this in a formal way though. I've tried using the intermediate value theorem and I do not know hot to implement it. Any ideas? Please require more reviewers on First Posts On the one hand, I see very few posts in the review queue when I visit.  On the other hand, I see a lot of pretty awful first posts sitting on the front page.  This leads me to suspect that robo-reviewers are blowing through the queue.  Given how empty the queue is, could you please increase the number of people who have to check in before a post leaves the queue?  Of course, some honey-pot solution to detect and disable the auto-clickers would be even better. Is there evidence of dark matter in our galaxy? Is there evidence of dark matter in our galaxy?  How can we measure this, say, how many percent of the center of our galaxy is dark matter?  I did not find the answer in the question I am dissecting some code and came across this,   $sql = 'SELECT page.*, author.name AS author, updator.name AS updator '      . 'FROM '.TABLE_PREFIX.'page AS page '      . 'LEFT JOIN '.TABLE_PREFIX.'user AS author ON author.id = page.created_by_id '      . 'LEFT JOIN '.TABLE_PREFIX.'user AS updator ON updator.id = page.updated_by_id '      . 'WHERE slug = ? AND parent_id = ? AND (status_id='.Page::STATUS_REVIEWED.' OR status_id='.Page::STATUS_PUBLISHED.' OR status_id='.Page::STATUS_HIDDEN.')';   I am wondering what the "?" does in the WHERE statement.  Is it some sort of parameter holder? Chance of selecting the last k pages in correct order from a set of n pages I have a set of documents numbered 1 to n. If I arrange the documents randomly, what is the chance of selecting the last k pages in the correct order if I select them at random from the documents?   Because order is important, I use the following formula:  $\displaystyle\frac{n!}{(n-k)!}$  From this I calculate how many ways there are to select k pages from n pages in the correct order. I need to have the last k pages. Which I can do in only one way. So is the chance   1/$\displaystyle\frac{n!}{(n-k)!}$?   It seems unlikely. It seems to me spontaneously that the chance should be smaller. I want to decrease the vertical space between the top of the page and the chapter heading. I have tried to follow the instructions for titlesec but I am not successful. I have tried the following two setups without any changes occuring.  First:  \documentclass[12pt]{report}   \usepackage[compact]{titlesec}   \begin{document}   \chapter{Characters}  The origin of the group determinant begins with Richard Dedekind in the late 1800's. He began to explore the idea after studying the discriminant in a normal field [A]. He made several observations about the group determinant, but was only able to prove some of them.   \end{document}   Second:  \documentclass[12pt]{report}   \usepackage{titlesec}   \titleformat{\chapter}[display]   {\normalfont\huge\bfseries}{\chaptertitlename\ \thechapter}{20pt}{\Huge}   \titlespacing{\chapter}{0pt}{0pt}{0pt}   \begin{document}   \chapter{Characters}    The origin of the group determinant begins with Richard Dedekind in the late 1800's. He began to explore the idea after studying the discriminant in a normal field [A]. He made several observations about the group determinant, but was only able to prove some of them.   \end{document} How and when is old-releases.ubuntu.com updated? The old-releases.ubuntu.com mirror hosts the repositories for EOL Ubuntu releases. The  currently lists up to 11.10 (or 12.04.3 if you count point upgrades), but the  lists up to 13.04. Why the discrepancy? When will the main page be updated? When will the newest EOL release (13.10) be shifted here? Combinatorial argument for $\sum\limits_{k=i}^{n}\binom{n}{k}\binom{k}{i} = \binom{n}{i}2^{n-i}$ I need to show that $$\sum\limits_{k=i}^{n}\binom{n}{k}\binom{k}{i} = \binom{n}{i}2^{n-i}$$  I know that $\displaystyle \binom{n}{k}\binom{k}{i}$ is counting the number of ways to pick $k$ elements from $n$ and then $i$ elements from those $k$, so $\displaystyle \sum\limits_{k=i}^{n}\binom{n}{k}\binom{k}{i}$ is counting the number of ways to do that $\forall k$.  I know that $\displaystyle \binom{n}{i} = \binom{n}{n-i}$ and that $2^{n-i}$ is the number of elements in the power set for $[n]\setminus[i]$.  How do I show that both sides are equivalent? I'm stumped on this one. Could anyone give me a pointer in the right direction? Unable to mount Windows (NTFS) filesystem due to hibernation Can someone explain __all__ in Python? I have been using Python more and more, and I keep seeing the variable __all__ set in different __init__.py files.  Can someone explain what this does? I've been having an argument with a colleague about this sentence, could you please let me know which one of us is correct:     There are no shortage of applications for our product in this   space.   She is convinced that are should be replaced by is, and I think it should stand as it is. Thanks for your help! $\gcd(a,b)\!=\!1\!=\!\gcd(a,c)\Rightarrow\gcd(a,bc)\!=\!1$ [coprimes to $\,a\,$ are product closed] How do I go about proving this?  If $\gcd(a,b)=1$ and $\gcd(a,c)=1$, then $\gcd(a,bc)=1$. I'm very confused with gcd proofs. find the value of limit Find the value of $$ \lim_{n\to\infty} \sum_{k=n+1}^{2n} \frac{1}{k} $$ I think it become a series with each denominator approach infinity. so the value of the limit is 0. am i true ? Find a real function $f:\mathbb{R}\to\mathbb{R}$ such that $f(f(x)) = -x$? I've been perusing the internet looking for interesting problems to solve. I found the following problem and have been going at it for the past 30 minutes with no success:     Find a function $f: \mathbb{R} \to \mathbb{R}$ satisfying $f(f(x)) = -x$ for all $x \in \mathbb{R}$.   I am also wondering, can we find $f$ so that is continuous?  I was thinking of letting $f$ be a periodic function, and adding half the period to x each time. I had no success with this, and am now thinking that such a function does not exist.  Source: Can i prevent/restrict standard user from installing/uninstalling software in Ubuntu 12.04 desktop? In  it says that flagged comments can be handled automatically if flagged repeatedly (3 + (score / 3) flags = deletion). Which is great; mods are busy people.  In , the venerable Bill the Lizard says that if all comments on a post are obsolete, you can flag the post (or one of the comments) with a custom flag saying that they're all obsolete, and the mod can remove them with one fell swoop. Problem is, that requires a mod.  What should I do when some of the obsolete comments are mine? I can't flag my own comments, and I don't like raising custom flags if I can avoid it.  You're wondering why I don't just delete them. There are two scenarios where I don't, at least not immediately:   Someone has commented with incorrect information. In that situation, I reply to the comment telling them how they've got it wrong. (Sometimes, of course, they quite correctly reply telling me how I got it wrong, which is neither here nor there&nbsp;&mdash; we still end up with obsolete comments. :-) ) Frequently, they reply saying "you're right" but (near universally) leave the comment lying around. I can (and do) then comment again saying "Let's clean these up" and flag their comments as obsolete. But I can't flag mine. Nor do I want to remove mine while the incorrect comment is still there; it leaves a confusing situation for anyone coming to the post later. Someone has commented with correct information. In that situation, if appropriate (e.g., the post author isn't busily editing at the same time and I know enough to do it properly), I'll edit the comment into the post and reply with "You're right, edited that in." and flag their comment as obsolete. These are rarely removed by the commenter. And again I can't flag mine.   The first scenario is the more problematic, since I can go back 10-20 minutes later and remove my "edited that in" comment and there's no harm if their comment is still there.  I currently bookmark and check back periodically, but that's labor-intensive and I'd like a better way.  Ideally, I'd want to tie my comment to one I've flagged so that it's deleted automatically if the other one is (by flagging, or by the commenter). That's a fair bit of development, though, I'm guessing.  What should I do? Go with a custom flag saying "The conversation between X and me is now obsolete," even though that means dumping it on a mod? In the Unity3D game engine, a common code sequence for getting remote data is this:  WWW www = new WWW("http://remote.com/data/location/with/texture.png"); yield return www;   What is the underlying mechanism here?  I know we use the yield mechanism in order to allow the next frame to be processed, while the download is being completed. But what is going on under the hood when we do the yield return www ?   What method is being called (if any, on the WWW class)? Is Unity using threads? Is the "upper" Unity layer getting hold of www instance and doing something?  EDIT:   This question is specifically about Unity3D internals. I'm not interested in explanations of how yield statement works in C#. Instead, I'm looking for an inside view of how Unity deals with these constructions, to allow, for example, to WWW to download a piece of data in a distributed manner across several frames. What is a good algorithm for estimating the median of a huge read-once data set? I need to do a lot of cleaning in my Virtual Machine whose Operating System is Ubuntu.  Normally in Windows, there is this kind of directory that contain temporary files and that should be cleaned :  C:\Documents and Settings\Administrator\Local Settings\Temp   Does it exist such a directory in Ubuntu? If yes, what is it?  Thank you a lot.