Say I control , which states:     Whenever a creature dies, that creature's controller may draw a card.   If I control 2 creatures and my opponent controls one creature, and I cast  to wipe the board (including Fecundity), what happens?  Do I get to draw 2 cards?  Does my opponent get to draw 1 card?  Or do neither of us get to draw any?
has the following ability:     Whenever a creature you control with a +1/+1 counter on it dies, return that card to the battlefield under your control at the beginning of the next end step.   I had a number of creatures on the battlefield, including Marchesa herself, with one or more +1/+1 counters on them when  was played, destroying each creature on the board.  When Marchesa and several other creatures would be killed simultaneously,   Does Marchesa's ability trigger for herself? Does Marchesa's ability trigger for the other creatures under my control with +1/+1 counters?   In attempting to resolve the issue, we tried to look things up, leading to , which cites Rule 603.6d as proving that Marchesa's ability should trigger both for herself and for each other creature with +1/+1 counters.  Unfortunately, none of us could actually understand Rule 603.6d, aside from this example:     Two creatures are on the battlefield along with an artifact that has the ability Whenever a creature dies, you gain 1 life. Someone plays a spell that destroys all artifacts, creatures, and enchantments. The artifacts ability triggers twice, even though the artifact goes to its owners graveyard at the same time as the creatures.   The example certainly seems analogous to me, but without understanding the actual rule, rather than just the example, we weren't too sure.  A complete answer has to not only answer the question, but back up that answer with citation and explanation of the official rules on the matter.
How can I pause a render in progress and then resume it at a later time? Assume that $0 &lt; a_n \leq 1$ and that $\sum a_n=\infty$. Is it true that  $$ \sum_{n \geq 1} \frac{a_n}{(a_1+\ldots+a_n)^2} &lt; \infty $$ ?  I think it is but I can't prove it. Of course if $a_n \geq \varepsilon$ for some $\varepsilon &gt; 0$ this is obvious.  Any idea? Thanks. Surjectivity of Composition of Surjective Functions Suppose we have two functions, $f:X\rightarrow Y$ and $g:Y\rightarrow Z$. If both of these functions are onto, how can we show that $g\circ f:X\rightarrow Z$ is also onto? override parent theme configuration in child functions.php Determining coordinates of a SHP file i have a SHP file of a building converted from a DXF file. This SHP file is unprojected and have no specific coordinates (origin 0,0). I find WGS84 coordinates (lat, long) of this building from Google Maps. My aim is to replace these coordinates to the SHP file on QGIS.  Could you explain the process step by step?  (I read lots of similar questions, about affine transformation. But i don't know the paramaters.)    What i ask is how to make this transformation in QGIS (step by step)?   SHP file coordinates (not projected)  Scale: 1:142.670.713  X1,Y1: 16.70824, 253.74534  X2,Y2: 1.03521,-6.34918  X3,Y3: 244.00248,-7.49189  Target WGS84 coordinates   X1,Y1: 41.047773, 28.896241  X2,Y2: 41.045452, 28.895929   X3,Y3: 41.045436, 28.899039    I need a second vector layer with coordinates. I have just a SHP file of a building converted from a DXF file. That is not projected. I know where this building is, and can find the WGS84 coordinates from Google Maps. My aim is add WGS84 coordinates to this SHP file and export to replace the building on Google Maps.  My Questions are:  1) I think my SHP file has measurements in meters and unprojected. I have WGS84 lat long coordinates from Google Maps. How can i project this SHP file, for being ready to transformation? Right Click -> Set Layer CRS -> Choose WGS84 EPSG: 4326. Is this enough?  2) qgsAffine plugin just need transformation matrix paramaters. But the answer of @Jochen Schwarze is a bit complex for me. Is there a online calculator? (Just give the source XYs and target XYs and calculate the parameters.)    I think, I have a projection problem before affine transformation. I have a DXF file of a building from AutoCAD. I saved this DXF file as a SHP file in QGIS. But this SHP file is converted from a plain DXF file and is not referenced/projected, it is drawn from coordinates 0,0 in AutoCAD in meters. And the drawing has meters, WGS84 has decimal degrees.  When I set Layer CRS as WGS84 in QGIS, the coordinates don't change.  How can I project this SHP file as WGS84 in QGIS? You can't see the question owner's special color Get string inside parentheses, removing parentheses, with regex Why does my regular expression work in X but not in Y? I wrote a regular expression which works well in a certain program (grep, sed, awk, perl, python, ruby, ksh, bash, zsh, find, emacs, vi, vim, gedit, …). But when I use it in a different program (or on a different unix variant), it stops matching. Why? Let $f:\mathbb{R}_+ \to \mathbb{R}_+$ be a monotone decreasing function defined on the positive real numbers with $$\int_0^\infty f(x)dx &lt;\infty.$$ Show that $$\lim_{x\to\infty} xf(x)=0.$$   This is my proof: Suppose not. Then there is $\varepsilon$  such that for any $M&gt;0$ there exists $x\geq M$ such that $xf(x)\geq \varepsilon$. So we can construct a sequence $(x_n)$ such that $x_n \to \infty $ and $x_n f(x_n ) \geq \varepsilon$. So $$\frac{\varepsilon}{x_n}\leq f(x_n) \implies \sum_{n\in\mathbb{N}}\frac{\varepsilon}{x_n} \leq  \sum_{n\in\mathbb{N}} f(x_n) \leq \int_0^1 f(x)dx.$$ So we get a contradiction.  I feal like I have the correct idea but some details are wrong.  Any help would be appreciated. Accused of cheating on online exams Can I kill an indestructible creature with -X/-X and sufficient damage marked on it? Can I hit a 4/4 indestructible creature with a  (3 damage) and give it -1/-1 with  to kill it? (Does it matter which way around I cast those?)  I read  here and it says, damage is removed in cleanup step, so it should be on the creature still after both spells.     702.12. Indestructible   702.12a Indestructible is a static ability.   702.12b A permanent with indestructible can’t be destroyed. Such permanents aren’t destroyed by lethal damage, and they ignore the state-based action that checks for lethal damage (see rule 704.5g).   702.12c Multiple instances of indestructible on the same permanent are redundant. Why does my user number show up in the search box when clicking my profile information? Hey all-  I just noticed that if you click on your username on the top bar, when your profile shows up, in the search bar at the top you'll find an explicit search for your user account.  For my meta account, for example, I see user:157031 whenever I click on my profile.  Is this deliberate?  If so, can I suggest removing that text?  It means that I have to manually delete the contents of the search field if I want to do a search from my profile screen.  Thanks! Is there a word for someone who somewhat delights in their illnesses/medications? I'm trying to do something like the following:  enum E;  void Foo(E e);  enum E {A, B, C};   which the compiler rejects.  I've had a quick look on Google and the consensus seems to be "you can't do it", but I can't understand why.  Can anyone explain?  Clarification 2: I'm doing this as I have private methods in a class that take said enum, and I do not want the enum's values exposed - so, for example, I do not want anyone to know that E is defined as  enum E {     FUNCTIONALITY_NORMAL, FUNCTIONALITY_RESTRICTED, FUNCTIONALITY_FOR_PROJECT_X }   as project X is not something I want my users to know about.  So, I wanted to forward declare the enum so I could put the private methods in the header file, declare the enum internally in the cpp, and distribute the built library file and header to people.  As for the compiler - it's GCC. Why do I have two docks in Ubuntu desktop (with GNOME 3) after installing Dash to Dock? What's going on here? I've installed the  on Ubuntu 17.10. Everything was cool, and as I was trying to hone and tweak the look of my desktop, at some point the shortcuts bar (dock?) on the left is duplicated.  (Click image to enlarge) The default, which normally disappears with Dash to Dock, is there underneath the one that usually shows up when I install Dash to Dock. Why? I've tried to go through the options for Dash to Dock, the GNOME Tweak Tool, and the regular Ubuntu settings, but I can't figure it out. How many nonnegative integer solutions are there to the pair of equations  \begin{align}x_1+x_2+\dots +x_6&amp;=20  \\ x_1+x_2+x_3&amp;=7\end{align}  How do you find  non-negative integer solutions? I am trying to automate these steps so that I don't need to do this on every machines manually. I need to install latest app server software (abc.tar.gz) on all the unix boxes.  I need to do this in around 12 machines - "machine1" ... "machine12". I have a master machine "machine0" which has abc.tar.gz file so I was thinking to run my script from this machine only and install abc.tar.gz software by following below steps in all those 12 machines one by one. My unix account id is david and my process runs as golden user.  This is the steps I follow if I am installing abc.tar.gz in "machine1":  david@machine1:~$ sudo scp david@machine0:/home/david/abc.tar.gz . david@machine1:~$ sudo stop test_server david@machine1:~$ sudo su - golden golden@machine1:~$ cd /opt/process golden@machine1:/opt/process$ rm -rf * golden@machine1:/opt/process$ exit david@machine1:~$ sudo cp abc.tar.gz /opt/process david@machine1:~$ sudo chown -R golden /opt/process david@machine1:~$ sudo su - golden golden@machine1:~$ cd /opt/process golden@machine1:/opt/process$ tar -xvzf abc.tar.gz golden@machine1:/opt/process$ exit david@machine1:~$ sudo start test_server   How can I automate this so that it can do the same steps in all 12 machines one bye one? I mean install abc.tar.gz in machine1 first, then install on machine2 and go on..  I want to run this script from machine0 only.  Is there any way to automate this? Why is it difficult to differentiate between  and ? Is it because we don't clearly understand how both of these phenomenon takes place?   My thoughts: From an answer to one of my earlier question  I could understand that light waves behave differently when it strikes a slit which is of several times bigger than its wavelength and when it strikes a slit of size comparable to it its wavelength.In the former case the different points on the slit acts as wave-front sources producing circular wave fronts which interfere to form the diffraction pattern but in the later case plane wave-front striking a slit after passing through it forms circular wave-fronts and two such wave-fronts from two different slits interfere to form the interference pattern. But, I'm not satisfied by that because I don't get any intuition about points on the slit acting as wavefront sources. So, what is actually happening? Approximating $\arctan x$ for large $|x|$ I would like to know if there is reasonably fast converging method for computing large arguments of arctan.  Until now I've came across Taylor series that converges only on interval $(-1,1)$ and for increasing $|x|$ the rate decreases. Also I've found about continued fractions for arctan, which seems to converge more rapidly and can be used for any real number, but again with increasing $|x|$ I need to perform way too much iterations until I get at least 3 digits precision.  I'm therefore looking for a method that would allow me to compute arctan for large arguments with good precision and in reasonable amount of iterations. When is the new ball due for all forms of cricket? Is there a standard time fixed in Test, ODI and T20 formats of Cricket, when a fielding side captain can ask for a new ball?  If the captain of the fielding side doesn't ask for a new ball, will he be forced to have it mandatorily after a certain point by the match umpires?