For example, there are 3 tags on Stack Overflow:        How can I merge all 3? It is not interesting to have different versions. I would like to keep xacml and have the others as synonyms. Is that possible?  EDIT - Version synonyms  I cannot follow the instructions in the answer to the duplicate question. See below...
What are tag synonyms? How can tag synonyms be distinguished from "real" tags? How are new tag synonyms created? Who can create tag synonyms? How can I delete/reverse/undo bad tag synonyms? What are merged tags?
How can I render an edge only mesh as solid lines? Can I partition a HDD for playing Xbox 360 games? I am planning to buy an xbox 360 console. my budget permits me acess to only the 4 GB one which needs extra space for many games' install. I have an external HDD of size 40GB. My question is can I divide it into partitions such thet I can use one for gaming on xbox and the other for data storage. Light fixture with a switch that has an outlet I hope this explains it better: in this room there is a switch with one set of wires and only the black one went through the hole in it, and another black one in the other hole, and is hooked to the outlet by being wrapped around the screw terminals. The ground is cut short and not grounded.  Now on to the outlet: the wire from the switch is connected to the top screw - but it's on the side that says "white".   There are two more sets of wires that connect to the outlet as follows:   the two black ones are on the the other screw terminal on the bottom (one on the "white" side)  the grounds are grounded to the screw terminal below that  the white ones are connected by the one from the switch to the bottom (black side), with one of the white ones from the other set on the screw terminal, and the one from the other cable is on the top screw terminal on the "white" side.   Now to the light: there are the black and white that come out of the box, and that is the set that has the ground cut short.  The original light fixture was a single 660w/250v rated socket.  The one my friend put up was a fixture with two 660w /250v sockets.  At the light only black and white connected, not ground... It worked then made noise and this is where I'm at.  When I wire test them, the black one from the switch and the white ones from either sets nothing...  but all the other ones light up.....! Some items in Terraria drop only very rarely from certain enemies. It's a pain having to farm so many enemies to try to get these rare items. Is there any way to make them drop more often? So i have this $3\times 3$ matrix  $$     \begin{pmatrix}     -2 &amp; 0 &amp; 0 \\     3 &amp; 1 &amp; -6 \\     0 &amp; 0 &amp; -2 \\     \end{pmatrix} $$  I want to check if the matrix is diagonizable. First thing i do is find the roots : namely $\lambda_1 = −2$ of multiplicity $2,$ and $\lambda_2 = 1$ of multiplicity $1.$  In order to ensure that the matrix is diagonalizable we need to find a basis of $\mathbb{R}^3$ whose elements are eigenvectors for $A.$ In particular, since we have the eigenvalue $\lambda_1=-2$ having multiplicity $2$ we need to check that the dimension of the eigenspace $E(\lambda_1)$ is equal to $2.$ We compute the eigenspaces of $A.$  $\DeclareMathOperator\Nul{Nul}$On the answer sheet is says that we need to find $\Nul(A+2Id)$... but why $2Id?$ Isn't is supposed to be $-2,$ so $\Nul(-2Id-A)$??   I do not understand this $\Nul(A+2Id).$ I always thought that we should compute $\Nul(-2Id-A),$ so what do i need to do ?   $\Nul(-2Id-A)$ gives me a different result from $\Nul(A+2Id).$  Does that mean that since $-2$ has a multiplicity of $2,$ we need to work with $2$ and not $-2$?  Somebody please help with a clear and simple explanation ! Thanks! Significance of the second focus in elliptical orbits How many distinct functions can be defined from set A to B? Maximize window shortcut What is the keyboard shortcut to maximize a window on OSX? I'd like to know the shortcut that is equivalent to double clicking on the window title. I do not want to make the window full screen. I have a web page that includes a bunch of images. Sometimes the image isn't available, so a broken image is displayed in the client's browser.  How do I use jQuery to get the set of images, filter it to broken images then replace the src?    --I thought it would be easier to do this with jQuery, but it turned out much easier to just use a pure JavaScript solution, that is, the one provided by Prestaul. For which $n$ does $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ imply $\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{a^n+b^n+c^n}$ I'm having trouble finishing a problem on an old national competition.  As the title states, the question says asks:  Given $a,b,c \neq 0,a+b=c$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$,  Find all integers $n$ such that $\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{a^n+b^n+c^n}$.  I know that if it is true for $n$, it is also true for $-n$ since $(a^n+b^n+c^n)(\frac{1}{a^n+b^n+c^n})=1$ and $(a^n+b^n+c^n)(\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n})=1$ and the statment follows for $-n$ given that we can make the substitution for $n$. Thus I found out that $n=-1$ satisfies the equality.  I also know that it fails for all even $n$. If $n$ were even, $(a^n,b^n,c^n)$ would all be positive, and Titu's Lemma asserts that $\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}\geq \frac{9}{a^n+b^n+c^n}$. Thus $n$ must be odd.  I also observed that $(2,-1,1)$ was a solution, and from here noted that this particular solution satisfies all odd $n$.  I have been trying to prove that all other solutions satisfy all odd $n$ with no success.  Any help would be greatly appreciated. Thanks! I  just got these 10 e-mails 5 hours ago from do-not-reply@stackexchange.com:      Recently, you asked a question on [X] Stack Exchange, but you haven't verified your email address yet.      Click on the following link to verify your email and set a password so that you can log in.      Here, [X] refers to all of the following:    GameDevelopment.SE SharePoint.SE Unix.SE  Skeptics.SE Japaneese.SE HomeImprovement.SE SciFi.SE PersonalFinanceMoney.SE  DatabaseAdmins.SE  TeX.SE    I haven't posted on any of these except for TeX.SE, but that was not recently, and through my real account.   =   I asked about this , and was told that someone seems to be impersonating me by using my email. Is it possible for staff to stop this?   Also, I'm pretty surprised there is no measure to stop people from using emails to existing SE accounts (see for example, on wordpress.com, if you use an e-mail which has an existing account on wp.com, you are asked to log in.).    EDIT:  I just got 2 more emails an hour ago for SharePoint.SE, and for AmateurRadio.SE. Iridescence In Cycles? Is it possible to create an  shader in Cycles? Specifically for something like , , , etc.. How to burn a Windows .iso to a USB device? I came across a reference to it recently on  and (as of now) it is not explained.  I suspect  might be it, but I don't know for sure. Get lists of dependent picklist options in Apex I have a pair of dependent picklists.  In Apex code, how can I determine what options are valid in the dependent field for each option in the controlling field?  I've tried using getPicklistValues(), but there doesn't seem to be any way of getting dependency information.  Example  Controlling_Field__c  Fruit, Vegetable, Dairy  Dependent_Field__c  (Fruit): Apple, Banana, Pear  (Vegetable): Tomato, Egglant, Lettuce  (Dairy): Milk, Cheese, Yogurt  What I'm Looking For  (In pseudocode, something like:)  controllingOptions = dependentfield.getPicklistValues(); for (option : controllingOptions) {     dependentOptions = dependentfield.getPicklistValuesFor(option);     do_something_with_it(); } There's a theorem of Kaplansky that states that if an element $u$ of a ring has more than one right inverse, then it in fact has infinitely many. I could prove this by assuming $v$ is a right inverse, and then showing that the elements $v+(1-vu)u^n$ are right inverses for all $n$ and distinct.  To see they're distinct, I suppose $v+(1-vu)u^n=v+(1-vu)u^m$ for distinct $n$ and $m$. I suppose $n&gt;m$. Since $u$ is cancellable on the right, this implies $(1-vu)u^{n-m}=1-vu$. Then $(1-vu)u^{n-m-1}u+vu=((1-vu)u^{n-m-1}+v)u=1$, so $u$ has a left inverse, but then $u$ would be a unit, and hence have only one right inverse.  Does the same theorem hold in monoids, or is there some counterexample? I want to synchronise two folders in real time under Windows 7.  Basically, I want to monitor a folder and synchronise each change (new files, changed files, deleted files) to another drive. It has to be in real time, so it gets synchronised instantly when a change happens. A one-direction synchronisation is enough.  I tried Microsoft's , but it does only syncing by hand or scheduled (thus not satisfying the "real-time" requirement).  Can this be achieved with Windows 7 itself? If not, are there other methods of doing this? Uxie and Azelf Are Not Appearing in Nameless Cavern How might a corrupt partition in TempDB result in DBCC CHECKDB reporting no issue? What's the recommend way of changing the font size in a particular table? Is there a better way than enclosing all values with, for example, the \tiny function.