I noticed that someone had the  badge, which states to be for answering a question with a score of 1 or more. I suppose that's really for "first (successful) answer".  Because the answer indicated had a score of &minus;3, and in fact no answers with non-negative scores.  So shouldn't this badge be cancelled?  (I know, it doesn't work that way now.) It is a one-shot determination, so if the answer happens to rise momentarily positive (the first vote was up, followed by two downs, for example, with the eventual average being 2 downs for each up) it is awarded, when that would seem to be "mistaken".  For many people, it might get re-awarded to a different answer. In general, I think you shouldn't be too quick to award it, but should wait for the average voting to show. Otherwise, why bother at all? Just say it's for answering, period.
What are badges? How do users get badges? How can users get tag badges? Can badges be lost/revoked/taken away after they are awarded? If so, how and when? How can I suggest a new badge?   For more information, see &quot;&quot; and the  in the .
Identify story: Overpopulation problem causes people to live one day a week This story is about a man that is able to live every day of the week, while everybody else has a single day of wakefulness assigned to them. Due to overpopulation, each human spends 6 out of 7 days of the week in stasis, and has jobs assigned to them in daily shifts. 1/7 of the population works on Monday. 1/7 works on Tuesday, &amp; so on. The main character has found away to circumvent the process and illegally live out his days consecutively by skipping the stasis program that is imposed on the general population and by working a different job everyday. Can somebody name this scifi story? It would be a very useful idea to be able to import PEAR projects that somebody maintains to their careers.stackoverflow.com profile. Most of what is required could be gleaned from the user specific rss profile URL; this takes the form of {username}.rss Removing overlapping part of polygons using QGIS? Here is what I'd like to achieve in QGIS.   Is there a processing tool to remove the overlapping part of polygons?    It is also very important that they do not get merged during the process and that they keep their original attributes since I need to work with them afterwards. Applied boolean modifier leaving shading artifacts with flat shading I tried to manipulate a cube with the Boolean modifier. I tried to cut out an edge of the cube with a cylinder. Also, the shading result looks awful in FLAT SHADING mode, so I turned on SMOOTH SHADING.   Has anyone an idea how to achieve a better result. (I also cut out a Sphere first, this made the result less awful).   I used Blender Internal and rendered with smooth shading for smoothness.  Using flat shading looks okay, except the "round" cutouts :/  Face normals are okay.  First image is smooth shading.    EDIT: User Pisurquatre suggested the EdgeSplit modifier, works perfect to me (Maybe a little tweaking here and there): Circle-Rectangle collision detection (intersection) How can I tell whether a circle and a rectangle intersect in 2D Euclidean space? (i.e. classic 2D geometry) I saw a commercial on televison for preparing spaghetti without boiling the noodles, just place them in water for about on hour.  Can this be done? How does the British-Irish Visa Scheme (BIVS) work? I am an Indian citizen and have applied for the Standard Visitor visa to the UK, and wish to travel to Ireland by using the BIVS scheme.    Do I need to fill up a separate application for BIVS endorsement or is it handed out automatically?  Also, If I enter the UK, and then travel to Ireland, can I re-enter the UK again? Dynamic LINQ OrderBy on IEnumerable / IQueryable I found an example in the  for Dynamic LINQ that allows you to use a sql-like string (e.g. OrderBy("Name, Age DESC")) for ordering. Unfortunately, the method included only works on IQueryable&lt;T&gt;. Is there any way to get this functionality on IEnumerable&lt;T&gt;? Using cd command in Windows command line, can't navigate to D:\ This may be a silly question, and I think I have looked elsewhere to find the answer...  Might be a path issue, but when I open the command line and type from the C:\&gt;: cd D:\  I cannot get to the D drive.  Even if I type: cd D:\&lt;folder name&gt;  The command.exe will auto-complete the line with the tab key, so it knows where I'm at.  It just doesn't print to screen the result or actually get me there.  This problem exists for the network drives as well. Now, if I use the chdir (cd) command like this: chdir D: or cd d: I get the print out of the D:\ below the command but it still says I'm in the C:\. I feel like I'm missing something simple. How do I change "Show Applications" button on GNOME 3 dash? I'm using Ubuntu with GNOME desktop.   I was wondering how to change the "Show Applications" button icon on the dash? Why does the base-emitter voltage of a BJT decrease with temperature? According to Sedra/Smith Microelectronic Circuits, \$v_{BE}\$ changes by \$-2\text{mV}/\text{°C}\$. I cannot understand how this could possibly be the case given the equations I am familiar with.  With all currents kept constant, we have:  \$\large{i_E = \frac{I_s}{\alpha}e^{v_{BE}/V_T}}\$  To keep \$i_E\$ constant, any change in \$V_T\$ would have to be accompanied by a change of the same factor in \$v_{BE}\$, otherwise \$\alpha\$ or \$I_s\$ would have to change, which as far as I understand is not possible.  So, how can \$v_{BE}\$ be inversely proportional to \$V_T\$? Bevel specific edges How do I use the latest GCC on Ubuntu? I want to compile my program with the latest version of gcc.  Ubuntu 14.04 comes with gcc 4.8.2, however there's 4.9.0 available, moreover, I see that it is available as a package: . I tried to install it  sudo apt-get install gcc-4.9   but it says  Reading package lists... Done Building dependency tree        Reading state information... Done Note, selecting 'gcc-4.9-base' for regex 'gcc-4.9' gcc-4.9-base is already the newest version. 0 upgraded, 0 newly installed, 0 to remove and 0 not upgraded.   Looks like it is already installed, just not as the default one? How do I utilize it to build my program? Run TTY script every time I open a GUI terminal Custom Cursor Image CSS I have a number of heroshot images, that have a modal popup when clicked.  I'm trying to get my cursor to turn into  magnifying glass whenever it is moved over the image. The following CSS does not appear to work even though my magnify.cur is present in the right location.  a.heroshot img {   cursor:url(/img/magnify.cur), pointer; }   Has anyone ever done anything similar? I don't mind a JavaScript solution if one exists.  EDIT: It works in Safari, but not in Firefox. First player rolls a 6-sided die 100 times and the second player 101 times. What are the odds of the second player getting more odd numbers than the first one?  I tried solving this via listing the options which led to a nasty sum that I wasn't able to solve neither by hand nor by using wolfram alpha. Clearly there has to be a more clever solution using the properties of a binomial distribution, but so far I haven't come across anything useful. Determine if $.ajax error is a timeout By viewing the polynomials as a difference of two squares, factorise the following polynomial: $$x^4+x^2+1.$$ I searched but couldn't find a way to solve this  Edit: By using Hans Lundmark hint, I get:  $$(x^2+1)^2-x^2$$   Is it fully factorized? What is the expected minimum euclidean distance between $N$ points uniformly and independently chosen on the shell(sphere) of 3-D ball of radius $R$?   Note that the expected minimum distance might be difficult to compute, so a good lower bound is also fine.   My approach  I take the approach similar to this   .   Suppose we put $N$ circles of radius $r$ uniformly on the surface of the ball. Therefore we could compute the expected minimum distance by the following expression \begin{align} E[D]=\int_0^\infty P(D&gt;t) dt. \end{align}  So, it remains to compute  $P(D&gt;t)$. Actually, since we are after a lower bound on  $E[D]$ it is enought to give a lower bound on $P(D&gt;t)$.  Let $S_i$ be the event that the pair of circles does not intersect which should be given by \begin{align} P(S_i) =  \frac{ Surf(R)-Surf(r)}{Surf(R)}= 1-\frac{  \pi r^2}{ 4\pi R^2}=1-\frac{ r^2}{ 4 R^2} \end{align}  Now the probability that euclidean minimum distance $D$ between $N$ balls is bigger than $t$ should be similar to  \begin{align} P[ D \ge t] \stackrel{?}{=}  P[ \cap S_i] \end{align} However, I think we somehow need to converte from the geodesic distance to euclidean distance but I am not very sure how to do that? "R cannot be resolved to a variable"?