A: Is Mr. Bob at home?      B: Sorry, he isn't at home. He___to Hong Kong for vacation?      A. went   B. has gone   C. is going   Which one is correct answer? and why it is not the other ones?
Non-na­tive speak­ers of­ten get con­fused about what the var­i­ous tens­es  and as­pects mean in English. With in­put from some of the folk here I've put to­geth­er a di­a­gram that I hope will pro­vide some clar­i­ty on the mat­ter.  I of­fer it as the first an­swer to this ques­tion. Con­sid­er it a liv­ing doc­u­ment. In­put is wel­come, and good sug­ges­tions will be  in­cor­po­rat­ed in­to the di­a­gram.    No­ta bene: What this is not is a dis­cus­sion of whether there are  more than two tens­es in English. , to which this ques­tion is not in­tend­ed to sup­ply ar­gu­ments one way or the oth­er.  Here, the aim is to pro­vide an overview of what con­struc­tions English-speak­ing peo­ple use for con­vey­ing in­for­ma­tion about ac­tions re­fer­ring to past, present, and fu­ture, and to pro­vide it first and fore­most to pre­cise­ly the peo­ple who are like­ly to use "tense" as a catch-all term in their search, rather than to lin­guists who know bet­ter.  Break­ing News There is now an ex­cel­lent ELU blog ar­ti­cle ti­tled  . It is high­ly rec­om­mend­ed read­ing.
Quotient of a local regular ring How can I prove this:     Let $A$ be a local regular ring with maximal ideal $\mathfrak m$ and $x \in \mathfrak m-\mathfrak m^2$. Then $A/(x)$ is a regular ring. Prove also that if $x\in\mathfrak m^2$, $x\ne 0$, the result does not hold anymore.   I don't know how to begin! Thanks to everyone who helps!  But if $d=\infty$? Note: This is not the same as , which is about the moderator tool. I am talking about the flag history page that is accessible through the user profile at .  I tend to have a pretty good hit rate for flags, well above 90% helpful on several sites. I also semi-regularly review the flag history page, both to see the status of any outstanding flags as well as see how flags I raised have been handled. Particularly, whenever a flag of mine is declined, I look over the post I raised the flag against with this new information in mind, to try to determine if I should change my flagging behavior in some way to better match the specific site's scope.  However, comment flags are difficult in this regard. Most often, there is only a binary status: helpful or declined. Helpful comment flags nearly always lead to the comment being deleted, so that obviously could present a bit of a problem, but that's not really relevant because a helpful flag means you should just keep doing what you've been doing. Declined comment flags, on the other hand, usually mean the comment is kept. It's the declined comment flags that I want to learn more about. But there's no way to!  All the flag says is which post it was on, how I flagged a comment, that I was the one to raise the flag (which frankly is rather useless information when you are already looking at your own flags summary), when the flag was raised and whether it's pending, helpful or declined. In a very few cases the moderator handling the flag may leave a helpful return comment.  This can get even more confusing if I raise multiple flags in quick succession against comments on the same question including any answers to it, because they basically all get bundled together.  Even in the case of a declined flag where the comment remains, there is no way to find your way back to the specific comment you flagged! Thus it is very difficult to recognize the error of your ways. Because comment flags (with few exceptions) are handled by diamond moderators only, the lack of feedback about which comment's flag was declined means it's easy to keep wasting the moderators' time without even intending to do so.  Hence my feature request: add a link to the specific comment that the specific flag was raised against, ideally only if that comment remains. Is TV harmful to my infant's development? I have a problem understanding the proof of Rencontres numbers (Derangements) I understand the whole concept of Rencontres numbers but I can't understand how to prove this equation  $$D_{n,0}=\left[\frac{n!}{e}\right]$$  where $[\cdot]$ denotes the rounding function (i.e., $[x]$ is the integer nearest to $x$). This equation that I wrote comes from solving the following recursion, but I don't understand how exactly the author calculated this recursion.  $$\begin {align*}  D_{n+2,0} &amp; =(n+1)(D_{n+1,0}+D_{n,0}) \\ D_{0,0} &amp; = 1 \\ D_{1,0} &amp; = 0  \end {align*} $$ I'd like to delete a file from my Time Machine backups, but there seems to be no option to do so in the Big Sur version of Time Machine. I go to the Time Machine menu and select &quot;Enter Time Machine&quot;. Then I see the view with the stack of windows. I browse to a file and select it. Now I want to delete it so it doesn't exist and take up space in backups. There is no option to delete it - either the action menu in the toolbar, or the control-click context menu. How do I delete a file from Time Machine backups? There used to be a way to do this. (I'd take a screenshot, but the key shortcuts for screenshots don't work in Time Machine. Not sure if this is just a Big Sur thing or if that's always the case.) Update: I should mention I'm using APFS for my Time Machine volume. That seems to be relevant. Also, I contacted Apple support about deleting entire snapshots (slightly different goal than the one here in this question). They said you can't do it anymore. It seems like there's some bugs here, because I have a &quot;Move to Trash&quot; option if I control-click one of my backups. It seems to work initially but then the backup comes back.  I don't like losing this much control over my files. I may look for another backup solution. In iOS 6 and below I could double tap the home button, tap and hold an app, and  click the red - to close an application(s).  In iOS 7 this does not work. How can I restart an app that's exploding without resetting the entire OS? What is the "N+1 selects problem" in ORM (Object-Relational Mapping)? The "N+1 selects problem" is generally stated as a problem in Object-Relational mapping (ORM) discussions, and I understand that it has something to do with having to make a lot of database queries for something that seems simple in the object world.  Does anybody have a more detailed explanation of the problem? Let $S$ and $T$ be finite non-empty sets such that $|S| = |T|$. Show that the function $f : S\to T$ is onto if and only if it is one-to-one. This is a recent homework bonus question assigned in my Proofs and Conjectures class. It (evidently) includes and evaluates our understanding of elementary-set theory and how to determine and prove that the function f is onto and one-to-one. I haven't been able to really grasp firmly how it is we do this. Help would be much appreciate, thank you. Troubleshoot High CPU usage by the "System" process Calculating polygon area within polygons in QGIS? I am new to QGIS - I want to estimate the area of the reef under a polygon grid. My shapefiles are already in meters, however I am having a hard time calculating the area under the grid layer. Attached are the photos. I want to calculate the total area of the bold black polygons (reefs) under each grid square (image 2). How much reef is inside each square, for example, grid1: 25 sq.km, grid 1 sq.km, etc. Each reef polygon has its corresponding area in the layer attribute table.    ADDITIONAL PHOTO: RED IS THE REEF  THE ATTRIBUTES OF THE REEF LAYER If a batsman hits the ball inside the ground and begins running between the wickets and now the ball is thrown back to the Fielder/Bowler/WK(FBWK)    If the FBWK catches the ball in one hand and hits the stumps with other hand before the batsman gets reached, is this a run out? If the FBWK catches the ball in hands and hits the stumps with leg or any other part of his body before the batsman gets reached, is this a run out?   This scenario can occur, when the ball was thrown far away from stumps. Can I leave the Schengen Area (and come back) while my non-eea spouse resident card application is being processed? My wife and I recently moved to Germany. She's a Swedish citizen, and I'm American. It'll be December before I can actually apply for my residence card, because my appointment to register our address is so far out - the local police station was all booked out and we didn't get an apartment right away.  I imagine that I won't be getting my residence card until well after Christmas, by which time my 90/180 days will be passed. What if I want to leave to, say, London for the holidays? Will I be allowed back in? I hit day 90 around the 21st of December. I could maybe leave on day 89 and come back on day 90 (with my wife and our marriage cert), or use this Polish solution: , since I can't be expelled from Germany for overstaying my visa while my application is processing.  Both of those seem a bit risky though. Am I reading too far into it, or is this a legitimate concern? Will I get some sort of documentation saying I have a residence card application processing? past simple or present perfect Please have a look at this sentence:     We haven't gotten a lot of Finnish groups that we liked here at Bananas but I'm pretty happy with this four tracker from Helsinki, apparently recorded in the same place where the Ramones played in the 70's.     The first verb is in present perfect (up to now) and the second is in simple past tense. I don't understand the simple past for the second verb.  Why "like" is not in its present perfect form as the first verb is? I have not done much proofs before this and need some guidance. I know that for a simple graph such as this :  node - node - node -node   Removing the first and last vertex will not disconnect the graph. So far, I can think of two cases: Where there are leaves, or nodes with degree of 1. Where there are no leaves, which means all nodes have degrees 2 or more.   In the first case, removing the leaf will not disconnect the graph. In the second case, removing any should not disconnect the graph. Problem is, I don't know how to formulate this into a good proof. Why does .NET use banker's rounding as default? According to the documentation, the  method uses a round-to-even algorithm which is not common for most applications. So I always end up writing a custom function to do the more natural round-half-up algorithm:  public static decimal RoundHalfUp(this decimal d, int decimals) {     if (decimals &lt; 0)     {         throw new ArgumentException("The decimals must be non-negative",              "decimals");     }      decimal multiplier = (decimal)Math.Pow(10, decimals);     decimal number = d * multiplier;      if (decimal.Truncate(number) &lt; number)     {         number += 0.5m;     }     return decimal.Round(number) / multiplier; }   Does anybody know the reason behind this framework design decision?  Is there any built-in implementation of the round-half-up algorithm into the framework? Or maybe some unmanaged Windows API?  It could be misleading for beginners that simply write decimal.Round(2.5m, 0) expecting 3 as a result but getting 2 instead. When I installed my SSD I just mounted with discard and didn't sweat it. However today I was reading about the pros and cons of using fstrim instead and decided to run the program to get an idea of how long it would actually take (still with my partitions mounted with discard). The command took several minutes on both my root and home partitions. For my home partition I used -v and got this:  $ sudo fstrim -v /home /home: 137494052864 bytes were trimmed   This is more than the amount of free space on the partition!  $ df -h /home Filesystem      Size  Used Avail Use% Mounted on /dev/sda2       206G   78G  118G  40% /home   Subsequent runs finish in less than a second, eg:  $ sudo fstrim -v /home /home: 0 bytes were trimmed   Surely if I have always had the partition mounted with discard, fstrim should not trim a large amount of data like that? The discard option is definitely enabled, here are the relevant fstab lines:  UUID=xxxxxxxx...    /          ext4   noatime,discard,errors=remount-ro  0      1 UUID=xxxxxxxx...    /home      ext4   noatime,discard,errors=remount-ro  0      2   And mount output lines:  /dev/disk/by-uuid/xxxxxxxx... on / type ext4 (rw,noatime,discard,errors=remount-ro,stripe=128,data=ordered) /dev/sda2 on /home type ext4 (rw,noatime,discard,errors=remount-ro,stripe=128,data=ordered)   The SSD is a TOSHIBA THNSNS256GMCP. Why does this happen? Fspy add-on not installing I have come across this .  Has anyone tried anything similar or does it work for you? Can you tell if an email customer / user enters is correct &amp; exists? Find this integral $$\operatorname I=\int\limits_{0}^{1}\dfrac{\arcsin{\sqrt{x}}}{x^4-2x^3+2x^2-x+1}\operatorname d\!x$$  My try: let $$f(x)=x^4-2x^3+2x^2-x+1$$ I found  $$f(1-x)=(1-x)^4-2(1-x)^3+2(1-x)^2-x+1=x^4-2x^3+2x^2-x+1=f(x)$$ so $$I=\int_{0}^{1}\dfrac{\arcsin{\sqrt{(1-x)}}}{x^4-2x^3+2x^2-x+1}dx$$ so $$2I=\int_{0}^{1}\dfrac{\arcsin{\sqrt{x}}+\arcsin{\sqrt{(1-x)}}}{x^4-2x^3+2x^2-x+1}dx$$ then I can't,Thank you very much I want to add a feature class within a Geodatabase to a map in ArcGIS Pro (1.4.1) using arcpy. I am converting some python code from ArcGIS Desktop to ArcGIS Pro.   This issue is very similar to:    and      However, instead of adding a feature class (or layer) to a map using the Python Window I am doing this from a  Script within a Toolbox. I have included a code snippet below.  lyrTest = r"C:\data\test.gdb\Layer1"  aprx = arcpy.mp.ArcGISProject("CURRENT") aprxMap = aprx.listMaps("MainMap")[0]  lyrFile = arcpy.mp.LayerFile(lyrTest) aprxMap.addLayer(lyrFile)   I get the following error message:  Traceback (most recent call last): File "C:\data\AddLayers.py", line 5, in &lt;module&gt;     aprxMap.addLayer(lyrTest) File "c:\users\user\appdata\local\programs\arcgis\pro\Resources\arcpy\arcpy\utils.py", line 191, in fn_ return fn(*args, **kw) File "c:\users\user\appdata\local\programs\arcgis\pro\Resources\arcpy\arcpy\_mp.py", line 1048, in addLayer return convertArcObjectToPythonObject(self._arc_object.addLayer(*gp_fixargs((add_layer_or_layerfile, add_position), True))) ValueError: C:\data\test.gdb\Layer1 Failed to execute (AddLayers).