I have had this question since I learned about there to be (which was a long time ago), and I know there used to be means something that used to exist, i.e, it no longer exists, for instance: There used to be a house across my apartment = One house used to exist, but no longer does, it's something that existed in the past, but not anymore. So, based on the facts that were mentioned, every there to be has a negative form and interrogative, therefore, what's there used to be's ?  My question has been identified as a possible duplicate of another one, but the one suggered isn't the same as mine, I'm talking about the verb there to be, the other question is about the verb "used to", both questions are different from each other.
I just made a minor , and  the grammar correction was unnecessary.  The original text was     Another factor--sometimes life happens and you can't fly. In the old days you could simply sell your ticket to someone else, now you either have to eat a hefty change fee or lose it outright. That's money in their pockets that they didn't used to get.   It was corrected to:     Another factor--sometimes life happens and you can't fly. In the old days you could simply sell your ticket to someone else, now you either have to eat a hefty change fee or lose it outright. That's money in their pockets that they didn't use to get.    Was this grammatical correction an improvement? Was the original text grammatically correct? Is the edited text grammatically correct?
In response to  question, I started to think about what it means when a Wild Magic Surge results in a "You cast spell-X".  So, "specific beats general" gives:   You can cast a second spell when you would not normally be allowed to You can cast a spell you don't know If the surge forces you to cast a concentration spell you would lose any concentration spell that you currently had active Subject to what the specific entry says you can target it how you like Unless the specific entry prohibits it you could use metamagic on the spell   I am pretty confident in all of the above but if anyone disagrees please address in an answer.  Here comes the question.  How does this impact on your available spell slots?  "You cast ..." to me suggests that if you have an applicable slot available then you have to use it to cast the spell because that is the general rule for casting spells but if you don't then "specific beats general" and it is cast anyway.  Specific examples:     You cast fireball as a 3rd-level spell centered on   yourself   This must be a 3rd level spell:   if you have a 3rd level slot available you have to use it? if you don't, if you have 5 sorcery points and a bonus action available must you use these to get that slot? if you don't have any of that then the spell goes off anyway?      You cast confusion centered on yourself.   No level specified and confusion is a 4th level spell:   if you have a 4th level slot available you have to use it? if you have a 4th, a 6th and an 8th level slot you can choose which one to use? if you have only a 5th level slot but also 6 sorcery points and a bonus action, you could choose to use the 5th level slot or use the points to create a 4th level slot? if you don't have any 4th level or higher slots but you have 6 sorcery points and a bonus action available must you use these to get that slot? if you don't have any of that then the spell goes off anyway?   And so on ... Isekai manga where the mc is reincarnated through his own magic std::string formatting like sprintf Colored borders around windows after update After some gorgeous updates I witness this: How prove this two symmetric matrices $AB=0$ Let $A,B$ be real symmetric matrices, and for any $n\in \Bbb N^{+}$, and for all $x,y\in \Bbb R$, we have $$tr(xA+yB)^n=x^ntr(A^n)+y^ntr(B^n).$$  Show that $AB=0$.  My try:since $$(xA+yB)^n=x^nA^n+\binom{n}{1}x^{n-1}A^{n-1}yB+\cdots+\binom{n}{n-1}xA(yB)^{n-1}+y^nB^n$$ so $$tr\left(x^nA^n+\binom{n}{1}x^{n-1}A^{n-1}yB+\cdots+\binom{n}{n-1}xA(yB)^{n-1}+y^nB^n\right)=x^ntr(A^n)+y^ntr(B^n)$$  then  $$tr\left(\binom{n}{1}x^{n-1}A^{n-1}yB+\cdots+\binom{n}{n-1}xA(yB)^{n-1}\right)=0$$ so $$\binom{n}{1}x^{n-1}ytr(A^{n-1}B)+\cdots+\binom{n}{n-1}xy^{n-1}tr(AB^{n-1})=0$$ then I can't  I found this is a nice result, maybe this is an odd problem, and I can't solve it. Thank you. Finding $\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!} +\frac{5}{3!+4!+5!}+\cdots+\frac{2008}{2006!+2007!+2008!}$ How to find :  $$\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!} +\frac{5}{3!+4!+5!}+\cdots+\frac{2008}{2006!+2007!+2008!}$$ Deleting your own closed questions fails silently on the mobile site There is some sort of delay for deletion of your own closed questions (2 days, I think) for non-mods. Now, when you try deleting a closed question on the desktop site, it nicely tells you "please wait xyz to allow for possible reopening" (in a box similar to the "please consider adding a comment explaining why you downvoted" thingy).  On mobile, it just reloads the page—there is no indicator that deletion is not allowed.  Either completely disable the delete button on mobile for closed questions, or give it a popup as well. Not letting the user know what's wrong is confusing. Nesting block level elements inside the  tag... right or wrong? Going through a text file line by line in C I've been learning C#, and I'm trying to understand lambdas.  In this sample below, it prints out 10 ten times.  class Program {     delegate void Action();     static void Main(string[] args)     {         List&lt;Action&gt; actions = new List&lt;Action&gt;();          for (int i = 0; i &lt; 10; ++i )             actions.Add(()=&gt;Console.WriteLine(i));          foreach (Action a in actions)             a();     } }   Obviously, the generated class behind the lambda is storing a reference or pointer to the int i variable, and is assigning a new value to the same reference every time the loop iterates.  Is there a way to force the lamda to grab a copy instead, like the C++0x syntax  [&amp;](){ ... } // Capture by reference   vs.  [=](){ ... } // Capture copies Difference between the i386 download and the amd64? What is the difference between standard i386 download and the amd64 download of Ubuntu 11.04? I am currently running the i386 on an amd machine, would my system work better with the amd64 download instead? (my machine was running Windows 7 64bit, but I neglected to look at that prior to installing Ubuntu). There are two checkboxes for Run As Administrator in the Properties dialog for a shortcut in Windows 7. One is under the Shortcut tab in the Advanced Properties dialog, and the other is under the Compatibility tab.  I've seen this under Windows 7 and Windows Server 2008. It might be in Vista and other versions, but I'm not sure.  What is the difference between these two settings? How do I find the files installed by a particular package? Where is the .deb file to look at? I have installed pulseaudio-module-bluetooth using apt.  $ type pulseaudio-module-bluetooth bash: type: pulseaudio-module-bluetooth: not found $ which pulseaudio-module-bluetooth $ whereis pulseaudio-module-bluetooth pulseaudio-module-bluetooth:   Clearly, I'm looking for the wrong thing.  The package does not simply install a command that is the same name as the package.  Alright, then.  I want to find out what all commands (or executables) this package installed, and their locations.  The answers to "" tell me how to find the files installed by the package if I have a .deb file.  I'm not installing directly from a .deb file, though.  I'm using apt.  Where is the .deb file that that used?  Is there a copy on my system somewhere that I can query with the commands in the answers to that question?  Where?  If there isn't a local copy on my system, can I get one with apt?  How?  Is there some handy apt (or similar) command that wraps this up for me, so that I do not have to run dpkg-deb directly?  What is it?  Can I find the package's file list entirely on-line, without explicitly downloading any .deb files and before installing anything with apt?  How? I came across this problem and I believe Lagrange's theorem is the key to its solution.  The question is:     Let $p$ be an odd prime. Prove that there is some integer $x$ such that $x^2 \equiv −1 \pmod p$ if and only if $p \equiv 1 \pmod 4$.   I appreciate any help.  Thanks. Four tasks in parallel... how do I do that? I have a bunch of PNG images on a directory. I have an application called pngout that I run to compress these images. This application is called by a script I did. The problem is that this script does one at a time, something like this:  FILES=(./*.png) for f in  "${FILES[@]}" do         echo "Processing $f file..."         # take action on each file. $f store current file name         ./pngout -s0 $f R${f/\.\//} done   Processing just one file at a time, takes a lot of time. After running this app, I see that the CPU is just 10%. So I discovered that I can divide these files in 4 batches, put each batch in a directory and fire 4, from four terminal windows, four processes, so I have four instances of my script, at the same time, processing those images and the job takes 1/4 of the time.  The second problem is that I lost time dividing the images and batches and copying the script to four directories, open 4 terminal windows, bla bla...  How do that with one script, without having to divide anything?  I mean two things: first how do I from a bash script, fire a process to the background ? (just add &amp; to the end?) Second: how do I stop sending tasks to the background after sending the fourth tasks and put the script to wait until the tasks end? I mean, just sending a new task to the background as one tasks end, keeping always 4 tasks in parallel? if I do not do that the loop will fire zillions of tasks to the background and the CPU will clog. $(n^2 - 1)!$ is a multiple of $(n!)^n$ Question: Find all positive integers $n$ such that $(n^2 - 1)!$ is a   multiple of $(n!)^n$.   I can show that the required condition fails if $n$ is a prime, and by direct computation, it also fails for $n=4$. After doing a few small examples by hand, I believe that the required answer is 1 and all composites greater than 4. However, I cannot seem to prove this.  For each prime $p&lt;n$, I tried to count the number of appearances of $p$ on both sides, but I get that I am supposed to compare $n(\lfloor{\frac{n}{p}} \rfloor + \lfloor{\frac{n}{p^2}} \rfloor +\dots)$ against $\lfloor{\frac{n^2-1}{p}} \rfloor + \lfloor{\frac{n^2-1}{p^2}} \rfloor + \dots$, which is not obvious to compare.  Any help will be gretaly appreciated. Generating a range of numbers in MySQL How do I generate a range of consecutive numbers (one per line) from a MySQL query so that I can insert them into a table?  For example:  nr 1 2 3 4 5   I would like to use only MySQL for this (not PHP or other languages). Given an instance of System.Reflection.Assembly. How do you redirect to a page using the POST verb? When you call RedirectToAction within a controller, it automatically redirects using an HTTP GET.  How do I explicitly tell it to use an HTTP POST?  I have an action that accepts both GET and POST requests, and I want to be able to RedirectToAction using  POST and send it some values.  Like this:  this.RedirectToAction(     "actionname",     new RouteValueDictionary(new { someValue = 2, anotherValue = "text" }) );   I want the someValue and anotherValue values to be sent using an HTTP POST instead of a GET. Does anyone know how to do this? There was the following paragraph in the article that came under the title, “GOP and the rise of Anti-knowledge” written by  (October 29, 2015):     “English unfortunately doesn’t have a precise word for the German   “Fachidiot,” a narrowly specialized person accomplished in his own   field but a blithering idiot outside it. In any case, a surgeon is   basically a skilled auto mechanic who is not bothered by the sight of   blood and palpitating organs. We need the surgeon’s skills on pain of   agonizing death, and reward him commensurately but that does not make   him a Voltaire.”    In Japanese we have a word, 専門バカ, of which literal translation is “specialist idiot” meaning exactly same as defined the above - a narrowly specialized person accomplished in his own field but a blithering idiot outside it.  I’m really curious about the absence of short descriptions of the specialist who knows nothing more than his specialty area in English like German “Fachidiot,” and Japanese "専門バカ."  Isn’t there really any short word counterpart to “Fachidiot,” specialist’s ignorance, for instance, in English?  P.S.  I found "narrow-minded specialisit" as a German-to-English translation of "fachidiot" in the web site, Interglot translation dictionary.