I recently bought Dragon Age: Inquisition and I'm keen on playing it. I do have a German client, so the voice is set to German on standard.  In the ingame settings I'm able to change the text language to English, but not the voice.  I can read on the Interwebs that Bioware released a free DLC which enables English voice audio, but I really can't find where to activate that DLC. Several sites claim that there is a DLC section in the Origin-window, but I can't find it:    How can I activate the English voice language on another localized client?
I'd like to play Dragon Age Inquision in english instead of the language of my operating system, but it seems to default to that and I can't find a way to change it. I already tried setting the Origin client language to english, but that didn't change anything.  Where can I change the language for Dragon Age: Inquisition?
Why is the sum of two random variables a convolution? Change the display scaling on the fly I'm looking for a keyboard shortcut or a CLI instruction to change the display scale on the fly. Seems there's nothing in compiz-config manager to handle this? Why only light nuclei are able to undergo nuclear fusion not heavy nuclei? Is it because of the binding energy or the binding energy per nuclei .  I am having trouble with this whole binding energy and nuclear fusion concept. Dirac Delta or Dirac delta function? Non-blocking Synchronous AJAX Disable chat migration notification if one of the users has insufficient rep? Simple request, really.  Can we add a detection for the     Please avoid extended discussions in   comments. Would you like to   automatically move this discussion to   chat?   prompt, so that if one of the users in the "extended discussion" has under 20 rep it doesn't appear?  I think a lot of the users this apply to will have low rep and it may take some back and forth to get good info out of them.  New suggestion: how about if a sufficiently high-rep user is clicking on the link, the low-rep user gets a field promotion sufficient to chat? Low-rep users are particularly prone to post questions to lead in this direction of needing to move to chat. A better proof for $\det(P) = \pm1$ if $P$ is an orthogonal matrix Suppose $G$ is a group generated by elements $x$ and $y$ where $xy^2 = y^3x$ and $yx^3 = x^2y$ What can you prove about $G$? Suppose $G$ is a group generated by elements $x$ and $y$ where $xy^2 = y^3x$ and $yx^3 = x^2y$ What can you prove about $G$?  I've just been playing around with the relations but I can't seem to get anywhere. It seems as if I'm going in circles. I don't know what to look for in terms of how the equations should look in order to give me an insight into the structure of the group.  So given a group presentation, are there any general strategies of what to look for in the relations (as in how to rearrange them) in order to obtain insight about the group? Proving a sequence defined by a recurrence relation converges How can I prove that this iterative sequence converges to $2$? Can I use the convergence definition?  $$a_{n+1} = \frac{a_n}{2} + \frac{2}{a_n},\qquad  a_0 = 4  $$  Thanks for the help. How to make crs transformation of a QgsVectorLayer? I need to transform QgsVectorLayers in python from one crs into another crs. By googling I found the following solution (), which works for points only:  crsSrc = QgsCoordinateReferenceSystem(4326)    # WGS 84 crsDest = QgsCoordinateReferenceSystem(32633)  # WGS 84 / UTM zone 33N xform = QgsCoordinateTransform(crsSrc, crsDest) pt1 = xform.transform(QgsPoint(18,5))   Is there a way to make a crs transformation of an entire layer? How can I merge properties of two JavaScript objects dynamically? I need to be able to merge two (very simple) JavaScript objects at runtime.  For example I'd like to:  var obj1 = { food: 'pizza', car: 'ford' } var obj2 = { animal: 'dog' }  obj1.merge(obj2);  //obj1 now has three properties: food, car, and animal   Does anyone have a script for this or know of a built in way to do this?  I do not need recursion, and I do not need to merge functions, just methods on flat objects. This discussion informed our process for seven years, but eventually required updates and augmentation. For discussion regarding the successors to this process, please see:    Stack Exchange has some of the best Moderators around — seriously — and that is due in no small part to the communities who scrutinize and vote in our Moderator elections.  We have a rather formal process for electing moderators, but up until now  has been quite informal. On those rare, rare occasion when things just aren't working out, you can see why it becomes such a sticky business to remove someone from a democratically elected position. Our current process is one of &quot;we've got your back.&quot; The Community Team routinely monitors moderator activities, so if a moderator violates the  or fails to uphold the , we can and do intervene. If the disputes and problems persist without resolution, eventually we can ask someone to step down. That's not good enough. We need a much more transparent and prescribed process for when — and more importantly how — to intervene when on-going problems are going unresolved. In urgent situations, we'll step in to protect a site, but ideally the process should be impartial and not subject to accusations of Stack Exchange simply applying undue influence over a community when the whim strikes. The Question: Before we start bouncing around ideas internally, I'd like to hear what you think. What process should be in place to add some checks and balances to deal with higher issues like possible impeachment of a moderator? When and how would such a process come into play? How should such situations be resolved? How do we keep it from being abused? Have at it. This is a round table vetting of ideas. Why getting this error tlmgr Unknown directive? I haven't used tlmgr for a while but when I tried to install any package with it today, It returned following message:     (running on Debian, switching to user mode!) Unknown directive ...containerchecksum c59200574a316416a23695c258edf3a32531fbda43ccdc09360ee105c3f07f9fb77df17c4ba4c2ea4f3a5ea6667e064b51e3d8c2fe6c984ba3e71b4e32716955... , please fix it! at /usr/share/texlive/tlpkg/TeXLive/TLPOBJ.pm line 210,  line 5579.   I've tried to use fix specified on :     You have to download update-tlmgr-latest.sh from the tlpretest repository   and run it. Afterwards you will be able to work with tlmgr as usual   (announced by Norbert a few hours ago).   I've downloaded it from , but it returned another error:  Verifying archive integrity... All good. Uncompressing TeX Live Manager Updater............................................................................................................................................................................ ./runme.sh: Cannot find TeX Live root using kpsewhich --var-value=SELFAUTOPARENT. ./runme.sh: Please set your PATH as needed, otherwise it's hopeless.   Minimum example  Install cleanly vanilla TeXLive 2016 with tmlgr after having old TeXLive in the system successfully: there is no errors/warnings in /usr/local/texlive/2016/*.log, complete log example . Run the following but you get the same output as above when trying to update the system  root@masi:/home/masi/Downloads/t/install-tl-20161019# tlmgr update --self (running on Debian, switching to user mode!) Unknown directive ...containerchecksum c59200574a316416a23695c258edf3a32531fbda43ccdc09360ee105c3f07f9fb77df17c4ba4c2ea4f3a5ea6667e064b51e3d8c2fe6c984ba3e71b4e32716955... , please fix it! at /usr/share/texlive/tlpkg/TeXLive/TLPOBJ.pm line 210, &lt;$retfh&gt; line 5579. 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 to modify an invalid '/etc/sudoers' file? How do I edit an invalid sudoers file? It throws the below error and it's not allowing me to edit again to fix it.  Here is what happens:  $ sudo visudo &gt;&gt;&gt; /etc/sudoers: syntax error near line 28 &lt;&lt;&lt; sudo: parse error in /etc/sudoers near line 28 sudo: no valid sudoers sources found, quitting Similar matrices and field extensions Canonical copies of sets Sometimes one needs two disjoint copies of a given set $X$. This can be canonically done by taking $\{0\} × X$ and $\{1\} × X$. Sometimes we want to identify one of the copies with the original set $X$. So I wonder, is it possible to take a disjoint copy of $X$ straight ahead? And can this be done canonically? More precisely, are the following claims true? $ \def\dom{\operatorname{dom}} \def\rng{\operatorname{rng}} \def\On{\mathbf{On}} $   For every $X$ there is a one-to-one function $f$ such that $\dom(f) = X$ and $\rng(X) ∩ X = ∅$. There is an explicit formula $φ(X, f)$ such that for every $X$ there is exactly one $f$ satisfying $φ(X, f)$ and this $f$ satisfies 1. for $X$.   Since such construction of a disjoint copy is intuitively trivial, I'm interested in proofs of 1. and 2. without using regularity, choice, or infinity.   Some ideas:   We may simply put $f(x) := (x, X)$, but we need regularity to show that $(x, X) ∉ X$. We may consider $α = \sup(X ∩ \On)$ and pair members of $X$ with some ordinals above $α$, but we need AC to well-order $X$, and we need global choice to obtain 2. Let $&lt;$ be the transitive hull of $∈$. If the relation $&lt;$ is narrow, i.e. $\{y: y &lt; x\}$ is a set for every $x$, then we may choose $z ∉ ⋃\{(\leftarrow, x)_&lt;: x ∈ X\}$ and put $f(x) := (x, z)$ since $(x, z) = y ∈ X$ would imply $z &lt; y$. Note that we don't need the axiom of infinity to define $&lt;$, but I don't know how to prove its narrowness without infinity. what is the real use case of dependency injection? Difference between 2>&-, 2>/dev/null, |&, &>/dev/null and >/dev/null 2>&1 In Quantum Mechanics we usually consider operator-valued functions: these are functions that take in real numbers and gives back operators on the Hilbert space of the quantum system.  There are several examples of these. One of them is when we work with the Heisenberg picture, where we need to consider functions $\alpha : \mathbb{R}\to \mathcal{L}(\mathcal{H})$ such that $\alpha(t)$ is the operator at time $t$.  Another example is when we deal with exponentiation of operators, like when building the time evolution operator:  $$U(t,t_0)=\exp\left(-i\frac{H}{\hbar}(t-t_0)\right),$$  Here, the $\exp$ is usually understood as being defined via the eigenvalues of $H$.  The point is, the idea of a function $\alpha : \mathbb{R}\to \mathcal{L}(\mathcal{H})$ appears quite often in Quantum Mechanics, and sometimes one needs to differentiate these. In practice we do it formally, using all the properties we would expect, but I'm curious about how one would properly define this.  If we were dealing with bounded operators, then we could use the operator norm, which is available for this kind of operator, and define the derivative as we can usually we do when there's some norm around.  The point is that in Quantum Mechanics most of the time operators are unbounded.  So in the general case, how can one define the derivative of one operator-valued function?