One of my friend got job and I want to tell this news to my other friends. How should I say it? He is selected for a job. He was selected for a job. He has been selected for a job.
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.
patching SQL Server with Availability Group and Secondary Replica What should a main game loop do? Generally, what are the core things that one should do in the game loop, and what are some things that one shouldn't do in the game loop? How to determine if a type implements a specific generic interface type Singleton: How should it be used Zero divisor in $R[x]$ Let $R$ be commutative ring with no (nonzero) nilpotents. If $f(x) = a_0+a_1x+\cdots+a_nx^n$ is a zero divisor in $R[x]$, how do I show there's an element $b \ne 0$ in $R$ such that $ba_0=ba_1=\cdots=ba_n=0$? This time when I used certbot it created a certificate for "www.$website" instead of $website. This is not what I need.  Here it is:    $ website="my_website123.com"   $ sudo certbot certonly --standalone -d $website -d www.$website --email admin@$website              Saving debug log to /var/log/letsencrypt/letsencrypt.log   Plugins selected: Authenticator standalone, Installer None   Starting new HTTPS connection (1): acme-v01.api.letsencrypt.org   Cert not yet due for renewal    You have an existing certificate that has exactly the same domains or certificate name you requested and isn't close to expiry.   (ref: /etc/letsencrypt/renewal/www.my_website123.com.conf)    What would you like to do?   -------------------------------------------------------------------------------   1: Keep the existing certificate for now   2: Renew &amp; replace the cert (limit ~5 per 7 days)   -------------------------------------------------------------------------------   Select the appropriate number [1-2] then [enter] (press 'c' to cancel): 2   Renewing an existing certificate   Performing the following challenges:   http-01 challenge for my_website123.com   http-01 challenge for www.my_website123.com   Waiting for verification...   Cleaning up challenges    IMPORTANT NOTES:    - Congratulations! Your certificate and chain have been saved at:      /etc/letsencrypt/live/www.my_website123.com/fullchain.pem      Your key file has been saved at:      /etc/letsencrypt/live/www.my_website123.com/privkey.pem      Your cert will expire on 2018-10-18. To obtain a new or tweaked      version of this certificate in the future, simply run certbot      again. To non-interactively renew *all* of your certificates, run      "certbot renew"    - If you like Certbot, please consider supporting our work by:       Donating to ISRG / Let's Encrypt:   https://letsencrypt.org/donate      Donating to EFF:                    https://eff.org/donate-le   Why did it create one for "www" instead of the bare domain?   On all my own servers this exactly command has always created a cert. without "www" - this is what I need. I even tried yesterday on a different server -- without "wwww".  How can I make it create a cert. without "www"? Why do some Kree have white skin? This seems bizarre to me.  All the Kree we've seen thus far (MCU based) were blue skinned... Ronan the Accuser being the most famous  What do Kree people look like?  Is there anything concrete with their appearance?  It seems odd that Danvers (Captain Marvel) and Jude Law's character (Yon-Rog?) and the "Supreme Intelligence" are all white skinned. How many RSA keys before a collision? What ensures big holes in my bread? What visa does my wife need to travel to the UK with me, if I'm going there on a business visa? I am resident of India and just got UK Business visa ( 6 months, multiple entry) .. I am going for a week on business trip. I want to take my wife along with me and at the end of trip.. I want to extend my trip for 3 days and we want to spend some time on leisure.   What visa I should apply for my wife.. visitor visa? should she mention in the visa form that she is going along with me and staying with me while I am on work and later we are spending time on leisure.   While filling my business visa form, there was a field asking if anyone else is going with me.. I selected no.. as there were no plans.  Also is it ok to spend time on leisure on UK Business visa Could gravity accelerate light? Autoexpand UITextView height dynamically in ios Continuously invertible operators Is there an easy way to remove overlapping faces? In the example remove doubles doesn't work  You might be able to do the following: A to select all faces. X &gt; Only Faces. F to create faces, but after deleting the faces I'm left with something like this.  If I press A to select all and then press F to create faces then there are still overlapping. Blend file Is the set of all functions $f:Z_+\rightarrow Z_+$ that are eventually constant countable. I keep getting an error in line 38 saying "Extra alignment tab has been changed to \cr." \documentclass [12pt]{article} \oddsidemargin=0pt \topmargin=0pt \textwidth=6.5in \textheight=8.5in \begin{document} \begin{center} {\large\bf\LaTeX\ Example 6} \end{center} We can typeset array structures \begin{equation} \mathbf{A}\times \mathbf{B}=\left| \begin{array}{ccc} \mathbf{i} &amp; \mathbf{j}  &amp; \mathbf{k} \\      a_1  &amp;      a_2        &amp; a_3 \\       b_1  &amp;      b_2        &amp; b_3  \end{array}  \right| \end{equation} \begin{quote} \begin{verbatim} \begin{equation} \mathbf{A}\times \mathbf{B}=\left| \begin{array}{ccc} \mathbf{i} &amp; \mathbf{j}  &amp; \mathbf{k} \\  a_1  &amp;  a_2   &amp; a_3 \\  b_1  &amp;  b_2   &amp; b_3 \end{array} \right| \end{equation} \end{verbatim} \end{quote} Or \begin{equation} \mathbf{A}=\left( \begin{array}{ccc} a_{11} &amp; a_{12} &amp; \cdots &amp; a_{1n} \\ a_{21} &amp; a_{22} &amp; \cdots &amp;  a_{2n} \\ \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ a_{n1} &amp; a_{n2} &amp; \cdots &amp; a_{nn} \end{array} \right) \end{equation} \end{document} Solving $x - a \log(x)=b$ Let $a&gt;0$ and $b \in \mathbb{R}$: Assume there exists an $x &gt;0 $ s.t. $$x - a\log(x) = b$$ holds. How can it be determined in closed-form? If $f:E\rightarrow S$ is an elliptic curve over a scheme $S$ (so $f$ is proper and smooth of relative dimension one with geometrically connected fibers of genus one, equipped with a section $0:S\rightarrow E$), then is the sheaf $\underline{\omega}_{E/S}:=f_*\Omega_{E/S}^1$ actually free of rank one? According to the statement of Grothendieck-Serre duality in Hida's book Geometric Modular Forms and Elliptic Curves, there should be a canonical isomorphism $\underline{\omega}_{E/S}\cong\mathcal{Hom}_{\mathscr{O}_S}(R^1f_*\mathscr{O}_E,\mathscr{O}_S)$, and the first result on elliptic curves in this book is that $R^1f_*\mathscr{O}_E\cong\mathscr{O}_S$. It's also not clear to me whether $S$ is being assumed (locally) Noetherian, but if that's necessary for Grothendieck-Serre duality to hold, then I'm fine with assuming it. I can't find another reference with a statement of Grothendieck-Serre duality in this generality which does not use the language of derived categories (which I unfortunately don't understand).  The reason I'm kind of skeptical about this is that in Hida's book, as well as in Katz-Mazur, it is said that $\underline{\omega}_{E/S}$ is invertible, so that an $\mathscr{O}_S$-basis $\omega$ for $\underline{\omega}_{E/S}$ can be found locally on $S$. If the invertible sheaf in question were really trivial then one would be able to choose a global $\mathscr{O}_S$-basis for $\underline{\omega}_{E/S}$, and there would be no reason to talk about doing so locally. Hida goes on to say that, choosing an $\mathscr{O}_S$-basis $\omega$ locally on $S$ allows one to regard $(\Omega_{E/S}^1,\omega)$ as a relative effective Cartier divisor in $E/S$, which also doesn't make complete sense to me because if we can only find $\omega$ locally, how are we getting a global section $\omega\in H^0(E,\Omega_{E/S}^1)=H^0(S,\underline{\omega}_{E/S})$ (unless $\underline{\omega}_{E/S}$ really is trivial)? Proving this identity $\sum_k\frac{1}{k}\binom{2k-2}{k-1}\binom{2n-2k+1}{n-k}=\binom{2n}{n-1}$ using lattice paths Construction of a Borel set with positive but not full measure in each interval