The D key on my built-in keyboard (MacBook Pro) is stuck and it keeps looping into self diagnostic when I boot up. I would like to disable the keyboard during computer startup and go straight to the external keyboard but have not found a way to do this.  If I install Karabiner on my MacBook Pro and disable in Karabiner settings the built in keyboard, after I turn off the MacBook Pro and start it up again with the external keyboard plugged in, would the built in keyboard function during the start up or would it be disabled?   If not is there another way to disable the builtin keyboard during startup?
This is an attempt to write a canonical QA for this issue, as per the Meta post:   I expect it to be periodically edited with the goal of becoming a comprehensive information resource.  I have a MacBook Pro that has a key(Shift, Return,   Command, P, etc.) that doesn’t respond when pressed or is stuck in the down position.    It's possible that my favorite beverage was spilled (or some other liquid damage).  It's also possible that I have no idea what happened.  It was working yesterday and suddenly today it's no longer working.  Is there any way to bypass or remap the key?  Can I  have the system ignore it completely?  (This question can also apply to external USB keyboards or the Bluetooth keyboards)
I'm looking to build a collection of fantasy miniatures, preferably pewter - but I don't want to buy them one at a time from my local hobby store... Much too expensive, and they don't often carry anything I'm interested in buying, which makes it a long term venture as well. So I'm looking for ideas on how to obtain many miniatures at once, preferably for cheap. D&amp;D Miniatures aren't a good source for me as they're collectibles, and the prices on those are also prohibitive.  By the way - they don't have to be pewter, that's just my preference. =) Finding empty directories recursively Gimp won't allow me to copy and paste a transparent image On the latest version of GIMP on my PC, whenever I try to copy a transparent image from the internet, GIMP will automatically fill in the transparent areas with the background color I have selected.  This is a non-issue on my Mac, but I need to do this work on my PC, so how can I make it recognize the alpha of the image and not fill it with the background color?  The image canvas that I'm working with does have an alpha channel, but that does not seem to make a difference. For all $x,y∈\Bbb{R}$ define that $ x\equiv y$ if$ x^2=y^2$ For all $x,y\in\Bbb{R}$ define that $x\equiv y$ if $x^2=y^2$ . Then $\equiv$ is an equivalence relation on $\Bbb{R}$ , there are infinitely many equivalence classes, one of them consists of one element and the rest consist of two elements.  Solution  True. To show that $\equiv$ is reflexive we need to show that $\forall x\in\mathbb{R} :x\equiv x$. Let $x\in\mathbb{R} , x\equiv x \mbox{ if } x^{2}=x^{2}$, which is obvious.  $[x] ={y\in\mathbb{R} |x\equiv y} =[0]={y\in\mathbb{R}|0\equiv y}={0}$. Hence $y^{2} =0^{2} =0$ which implies that $y=0$.  Can anyone please give me correct answer to this question. Sometimes we assume that regressors are fixed, i.e. they are non-stochastic. I think that means all our predictors, parameter estimates etc. are unconditional then, right? Might I even go so far that they are no longer random variables?  If on the other hand we accept that most regressors in economics say are stochastic because no outside force determined them with some experiment in mind. Econometricians then condition on these stochastic regressors.  How is this different from treating them as fixed?  I understand what conditioning is. Mathematically, it means we make all observations and inference conditional on that particular set of regressors and have no ambitions to say that inferences, parameter estimates, variance estimates etc. would have been the same had we seen a different realization of our regressors (such is the crux in time series, where each time series is only ever seen once).  However, to really grasp the difference between fixed regressors vs. conditioning on stochastic regressors, I am wondering if anyone here knows of an example of an estimation or inference procedure that is valid for say fixed regressors but breaks down when they are stochastic (and will be conditioned on).  I am looking forward to seeing those examples! Find the center of the symmetry group $S_n$.   Attempt:  By definition, the center is $Z(S_n) = \{ a \in S_n : ag = ga \forall\  g \in S_n\}$. Then we know the identity $e$ is in $S_n$ since there is always the trivial permutation. Suppose $a$ is in $S_n$, but not equal to identity. Now we can imagine the permutation as bijective function that maps from $\{1,2,\dotsc,n\}$ to $\{1,2,\dotsc,n\}$.  So suppose $p$ is a permutation map. Then $p$ maps from a location $i$ to a location $j$. Take $p(i) = j$ where $i\neq j$. Let $k$ be in $\{1,2,\dotsc,n\}$, where $k$, $i$ and $j$ are all different elements. The cycle $r = (jk)$, then we will see if this commutes.  $rp(i) = rj$    Can someone please help me, I am stuck?  Thank you. How to extract code of .apk file which is not working? Control cannot fall through from one case label How do I transfer iPhone contacts to an Android phone? Let $R$ be a UFD. Then for any two elements $a,b \in R \setminus \{0\}$ $ab \sim \gcd (a,b){\operatorname{lcm} (a,b)}$.   My attempt :  Let $d = \gcd (a,b)$.Then $d|a$ and $d|b$.Then $\exists x,y \in R$ such that $a = dx$ and $b = dy$.Now it is to be shown that $c = dxy$ is the lcm of $a,b$.Which I find difficulty to show.It is clear that $a|c$ and $b|c$. Now if $u$ is any common multiple of $a$ and $b$.Then $dx|u$ and also since $y|b$ we have $y|u$ i.e. $dxy|u^2$ $\implies c|u^2$.But I have to show $c|u$ to complete the proof which I fail to prove.Please help me to complete the proof.  Thank you in advance. How can I do a full shutdown of windows 10 without logging in? How can I repair a ceiling fan that hums, but doesn't turn? What is the relation between normal extension and separable extension? What is the relation between normal extension and separable extension? Let F be the algebraic extension of K, if F is a separable extension of K,if and only if F is a normal extension of K? is this correct? Counterexamples to the avoidance lemma for arbitrary ideals Let $A$ be a commutative ring with $1$. Let $I$ and $J_k$, $k=1,\dots,n$ be ideals of $A$ with $I\subseteq \cup _{k=1}^n J_k$. Then I have obtained the following:  (1) If $J_k$, $k=1,\dots,n$, are prime ideals, then there exists some $j$ such that $I\subseteq J_j$.   (2) If $A$ is a principal ideal ring, then there exists some $j$ such that $I\subseteq J_j$.   I want some counterexamples that for all $k=1,\dots,n$, $I$ is not contained in $J_k$. How to get the counterexamples? Facing an error "*** glibc detected *** free(): invalid next size (fast)" Evaluating the limit of a sequence given by recurrence relation $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$. Is my solution correct? Editing/Creating files in C:\Program Files\WindowsApps What is in the space between a nucleus of an atom and its electrons? What is JSON and what is it used for? I've looked on Wikipedia and Googled it and read the official documentation, but I still haven't got to the point where I really understand what JSON is, and why I'd use it. I have been building applications using PHP, MySQL and JavaScript / HTML for a while, and if JSON can do something to make my life easier or my code better or my user interface better, then I'd like to know about it. Can someone give me a succinct explanation? Python metaheuristic packages