Is it grammatically correct to omit a comma after dialogue ending in an ellipsis? I know that Grammarly isn't always correct but before tonight, I didn't know punctuation directly after an ellipsis was grammatically correct, even in special circumstances such as dialogue.   Unless it's stylistic, Grammarly is insisting that what I currently have is incorrect (see photo below).  The sentence in question is:     “You want to marry me…” she led off in a breathy voice as she   grabbed my shirt collar. “Not out of love, but for mutual benefits.”   I would have never thought to include a comma there, even then, a comma would look so odd. From what I've read, it seems stylistic, but I'm asking is it grammatically correct to leave it as is?   At this point, I think I'd rather omit the ellipsis entirely and replace with a comma rather than add a comma to the end of the ellipsis.
Punctuation of a Direct Quote Is any particular punctuation necessary when a quotation is the subject of a sentence? For example, in the sentence below, should there be a comma after export? Or anything else?     "Laughter is America's most important export" is a well-known quote of Walt Disney's.
Do I need visa for UK visit if I have a work permit for Ireland? How to disable SSLv2 or SSLv3? I'm developing a site using Drupal 8 beta-14. I've created a view block of different terms and now I want to display it using code. How can I display it programatically?  I used to do it in Drupal 7 using this code but I'm confused about Drupal 8.  $block = module_invoke('block', 'block_view', '4'); $text_block = render($block['content']); How do I tell Finder to always show my Library folder on Mavericks? I thought there was a checkbox to suppress the hiding of my user Library folder on Mavericks, but I can't seem to find it. I appreciate hiding Library by default, but  have missed it being easily accessible.  How can I get the library folder to be visible in Finder, the Finder go menus / file save dialogs without needing to hold down shift? My Mac is getting really slow, what should I do? Is there any work-around if you are somewhere where images are blocked? I'm currently working in a bank, and while about 99.7% of the internet is blocked, Stack Exchange thankfully isn't (yet) however imgur is.  It means I can't view images posted alongside questions. The only workaround I have found is to search Google images for the URL, but you get a  which is a bit useless.  Does anybody know any clever tricks for viewing the images accompanying questions when you are somewhere where imgur is blocked?  Are there any plans to route the images stored on imgur through a Stack Exchange URL? Is it possible in Leaflet that popup opens on mouse-over, not on click?  This is working for just one marker at a time, but I need it for a bigger number of markers:  marker.on('mouseover', function(e){     marker.openPopup(); }); In Season 6, we find out the following about the magical properties of the wall:     BENJEN: This is where I leave you.   BRAN: You're not coming with us?   BENJEN: The Wall is not just ice and stone. Ancient spells were carved into its foundations. Strong magic to protect men from what lies beyond. And while it stands, the dead cannot pass. I cannot pass.   —"Winds of Winter" (Game of Thrones, S06E10)   So why would the men of Westeros need to worry about the Night King and his army? Couldn't they sit back and enjoy the wall's protection? Transform the NFSA with transition table  $$\begin{array}{ccc} X&amp;a&amp;b\\ A&amp;\{\}&amp;\{F\}\\ B&amp;\{\}&amp;\{\}\\ F&amp;\{\}&amp;\{A,S\}\\ S&amp;\{F\}&amp;\{A,F\} \end{array}$$  and starting state $S$ inta a DFSA and give the transition table of that DFSA. Place the letter $X$ (uppercase) in the top left corner of the matrix. Your states in the first column and inputs in the first row can be in the order of your choice.  The question then offers a $7\times 3$ matrix to be filled in.  Unsure how to transform the NFSA with transition table.  This is my answer but it is incorrect.  $$\begin{array}{ccc} X&amp;a&amp;b\\ \{S\}&amp;\{F\}&amp;\{A,F\}\\ \{F\}&amp;\{\}&amp;\{A,S\}\\ \{A\}&amp;\{\}&amp;\{F\}\\ \{A,F\}&amp;\{\}&amp;\{A,F,S\}\\ \{A,S\}&amp;\{F\}&amp;\{A,F\}\\ \{A,F,S\}&amp;\{F\}&amp;\{A,F,S\} \end{array}$$ Derangements with repetitive numbers I have a very simple problem.  Lets assume that I have a well shuffled deck of 52 cards. I start drawing the top card always and when the card matches its rank I lose. J=11 Q=12 K=13.  If there were only 13 cards I could easily use the $\ \frac{!n}{n!}$ for derangements in order to solve this. The problem is that there are 52 cards so when I pass 13 I start from 1 again so I don't know what is the probability to win. Example of the game     1st card: 4 - Continue   2nd Card: A - continue   3rd Card: K - Continue   4th Card: K - Continue   5th Card: 6 - Continue   6th Card: 9 - Continue   7th Card: 10 - Continue   8th Card: A - Continue   9th Card: J - Continue   10th Card: 3 - Continue   11th Card: 2 - Continue   12th Card: 8 - Continue   13th Card: A - Continue   1st card: 5 - Continue   2nd Card 2 LOSE   So actually I have to count from 1 to 13 4 times and if I draw all 52 cards then I win. What's the probability? What causes the texture of bread to be open, ie with lots of large holes, or close, ie a regular, uniform tender crumb with no large holes? How can I achieve either? A curious integral The following integral has been on my mind for a while $$\int_0^\infty \frac{\sin(x)}{e^x-1}\,\mathrm d x \tag{$\dagger$}$$    Let us indicate the integrand as $f(x)=\frac{\sin(x)}{e^x-1}$. The following are a couple of observations.   The integrand can be extended by continuity in $0$ since $$\lim_{x\to 0}f(x)=\lim_{x\to 0}\frac{\frac{\sin(x)}{x}}{\frac{e^x-1}{x}}=1$$ This is the original reason I started playing around with this integral. Mathematica yields the result $$\int_0^\infty \frac{\sin(x)}{e^x-1}\,\mathrm d x=\frac{\pi}{2}\textrm{Coth}(\pi)-\frac 12 \approx 1.076674047$$ which, following numerical evidence, seems correct. Complex analysis may be useful here, since the integrand is a holomorphic function on $\mathbb C$. I tried writing $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$, but I was not able to find an appropriate integration contour to solve the problem. The $-1$ in the denominator breaks the simmetry of the expression. This made most of my substitutions useless.   Can this integral be evaluated correctly, preferrably through complex analytic methods? How do I boot into recovery mode? Normal + Connected -> Uncountable Here's a solution (to the exercise "prove that if $X$ is a space with more than one element, and is normal and connected then $X$ is uncountable"):   by Urysohn's lemma, given $A$ and $B$, closed and disjoint in $X$, there exists a continuous function from $X$ into $[0,1]$ such that $f(A)={0}, f(B)={1}$.   If $X$ were countable, so it would be $f(X)\subset [0,1]$; choose $r\in (0,1)\setminus f(X)$, then $X=f^{-1}([0,r))\cup f^{-1}((r,1])$, so $X$ is disconnected.  But the full power of the lemma wasn't used, all that was used is that there exists a continuous function $X\longrightarrow [0,1]$, so the condition of being normal seems too much.  So my question is, what are the weakest conditions on $X$ for the existence of such continuous functions into $[0,1]$ (or $\mathbb{R}$, for that matter)? How do i make my ubuntu server public? I want to know how to make my ubuntu server ip public. I'm not asking for a domain, if people would be able to access my website by ip, it would be great! (just for now) Does someone have an idea how to do this? Suppose $f(x)$ is linear i.e. $f(x + y) = f(x) +f(y) $ and monotone on $[-\infty, +\infty]$, then $f(x) = ax$, a real.  We know $f(0)=0$, and if one assumes differentiability it is quite easy.      But how does monotone make up for differentiability? Linux: clnt_create: RPC: Program not registered NFS mounts recently automatically got un-mounted. When I checked, the NFS service status it was shown to be running.  [root@hsluasrepo]# service nfs status rpc.svcgssd is stopped rpc.mountd (pid 4083) is running... nfsd (pid 4148 4147 4146 4145 4144 4143 4142 4141) is running... rpc.rquotad (pid 4079) is running...  [root@hsluasrepo]# service rpcbind status rpcbind (pid  4203) is running...  [root@hsluasrepo common]# rpcinfo -p 10.80.3.154    program vers proto   port  service     100000    4   tcp    111  portmapper     100000    3   tcp    111  portmapper     100000    2   tcp    111  portmapper     100000    4   udp    111  portmapper     100000    3   udp    111  portmapper     100000    2   udp    111  portmapper   But showmount output was showing an error.  [root@hsluasrepo ]# showmount -e 10.80.3.154 clnt_create: RPC: Program not registered   After restarting the NFS service, showmount output displayed the NFS server's export list.  Can anyone tell me the root cause of this issue and How to avoid this problem in future?  /var/log/messages:  Jul 7 03:22:01 hsluasrepo rsyslogd: [origin software="rsyslogd" swVersion="5.8.10" x-pid="1188" x-info="rsyslog.com"] rsyslogd was HUPed Jul 7 03:22:02 hsluasrepo rhsmd: In order for Subscription Manager to provide your system with updates, your system must be registered with RHN. Please enter your Red Hat login to ensure your system is up-to-date. Jul 8 03:22:01 hsluasrepo rhsmd: In order for Subscription Manager to provide your system with updates, your system must be registered with RHN. Please enter your Red Hat login to ensure your system is up-to-date. Jul 8 16:36:55 hsluasrepo kernel: nfsd: last server has exited, flushing export cache Jul 8 16:36:55 hsluasrepo rpc.mountd[4083]: Caught signal 15, un-registering and exiting. Jul 8 16:36:55 hsluasrepo rpc.mountd[24463]: Version 1.2.3 starting Jul 8 16:36:55 hsluasrepo kernel: NFSD: Using /var/lib/nfs/v4recovery as the NFSv4 state recovery directory Jul 8 16:36:55 hsluasrepo kernel: NFSD: starting 90-second grace period Jul 8 16:37:32 hsluasrepo rpc.mountd[24463]: authenticated mount request from 10.60.5.208:1004 for /common/PROD (/common/PROD) Jul 8 16:38:09 hsluasrepo rpc.mountd[24463]: authenticated mount request from 10.60.5.181:869 for /common/PROD (/common/PROD)  Jul 8 16:38:43 hsluasrepo rpc.mountd[24463]: authenticated mount request from 10.60.5.180:825 for /common/PROD (/common/PROD) Jul 8 16:39:12 hsluasrepo rpc.mountd[24463]: authenticated mount request from 10.60.5.176:688 for /common/PROD (/common/PROD) Applying graduated renderer in PyQGIS? Just trying to run the following code already given in the pyqgis cookbook to create graduated colors symbology (as in example with 2 classes)  I define the source info for the shapefile but when I run execute the code within my test-plugin, it just adds the layer without classifiying. Just adds as usual, all polylines appear same.  Any solution or working code/method?  link to code in cookbook   from qgis.core import  (QgsVectorLayer, QgsMapLayerRegistry, QgsGraduatedSymbolRendererV2, QgsSymbolV2,QgsRendererRangeV2)  myVectorLayer = QgsVectorLayer('C:/my_test_shape_file.shp', 'test_shp_file', 'ogr')  myTargetField = 'target_field' myRangeList = [] myOpacity = 1  # Make our first symbol and range... myMin = 0.0 myMax = 50.0 myLabel = 'Group 1' myColour = QtGui.QColor('#ffee00') mySymbol1 = QgsSymbolV2.defaultSymbol(myVectorLayer.geometryType()) mySymbol1.setColor(myColour) mySymbol1.setAlpha(myOpacity) myRange1 = QgsRendererRangeV2(myMin, myMax, mySymbol1, myLabel) myRangeList.append(myRange1)  #now make another symbol and range... myMin = 50.1 myMax = 100 myLabel = 'Group 2' myColour = QtGui.QColor('#00eeff') mySymbol2 = QgsSymbolV2.defaultSymbol(myVectorLayer.geometryType()) mySymbol2.setColor(myColour) mySymbol2.setAlpha(myOpacity) myRange2 = QgsRendererRangeV2( myMin, myMax, mySymbol2, myLabel) myRangeList.append(myRange2)  myRenderer = QgsGraduatedSymbolRendererV2('', myRangeList) myRenderer.setMode(QgsGraduatedSymbolRendererV2.EqualInterval) myRenderer.setClassAttribute(myTargetField) myVectorLayer.setRendererV2(myRenderer) QgsMapLayerRegistry.instance().addMapLayer(myVectorLayer)      @Kelly's code works perfectly.   However I noticed that both the code (one I have typed in my first message and yours in the 2nd message) DO NOT work in QGIS v1.7.3 but WORKS in QGIS v1.8.0. I think this was a bug(?) that already resolved in v1.8.0  And one more question;  Do you have any sample code for (natural breaks, Jenks) classification of the "numeric_attribute_field" based on given number of classes (i.e. no of classes will be a variable in the code, say "n", and I will pass that from the plugin GUI i.e. n = spinBox.value() ) Resize Move Partitions on Dual Boot System Let $f \colon \mathbb{R} \rightarrow  \mathbb{R}$ be a smooth function such that $$f'(x) = f(1+x)$$ How can we find the general form of $f$? I thought of some differential equations, but not sure how to use them here. Thanks.