Assign different multimedia functions (like a notebook setup) to F1-F12 keys on desktop computers?
My keyboard has no "media" keys; can I control media without them?
Intersection Number of $B = Y^2 - X^3 + X$ and $F = (X^2 + Y^2)^3 - 4X^2Y^2$ using the fact $I(P,F \cap B) = ord_P^B(F) $. In "Settings" > "SD card &amp; phone storage settings" under "Internal Phone Storage" it lists "Available Space" as 13.78MB.  I'm trying to install an app that is 2.38MB.  Why do I get the "Insufficient storage available" error message?  I clearly have enough space.  If it's relevant, my phone is an LG Optimus V. How can I force a long string without any blank to be wrapped? Fibonacci sequence and Euclidean algorithm's connection. Let $f_1, f_2,\dots,f_i,\dots$ be the Fibonacci sequence with $f_1=f_2=1$. For any $a,b\in\mathbb{N}$ with $a&lt;b$, suppose the Euclidean algorithm takes $n$ long divisions to compute $\gcd(a,b)$. If $a\le f_j$, then $n\le j-1$, for $j\ge2$.   I want to prove this problem. But I can't see any connection between Fibonacci and Euclidean algorithm. If anyone can give a hint about first step, it will be very thankful.  My guess is to use induction. How to find an end point of an arc given another end point, radius, and arc direction? Given an arbitrary arc, where you know the following values: end point (x1,y1), radius (r) and arc direction (e.g. clockwise or counterclockwise from start to end), how can I calculate the other endpoint of the arc (and I say endpoint because I think of an arc as line segment that is curved, which happens to lie on the circumference of the circle and it has two end points).   Though that's just how I interpret the problem and maybe I'm not thinking about it correctly.  EDIT: Yes, I know the Origin point and the length of the arc. Sorry I forgot to mention. Also, the that circle would be on a 2D Cartesian plain (x and y axis.) Sorry, my math terminology is lacking.   Also, I'm not trying to cheat on homework or anything. This is a legitimate problem for an SVG graph I'm trying to create using JavaScript and I've tried my best but I need help.  Thanks in advance. I have absolutely no experience with Linux, and I desperately need to get my computer back up and running again with Windows.  How do I remove Ubuntu and reinstall Windows?    Editor's note: many of the answers are about removing Ubuntu from dual-boot but keeping Windows (which is a bit complicated), while other answers are about removing Ubuntu from single-boot (which is easy: basically just format the disk while installing Windows). The question as written is ambiguous between dual-boot or single-boot. I know how to successfully duplicate an object by using  bpy.ops.object.add_named(name = 'TemplateObjectName')   If the template object named 'TemplateObjectName' has animations, particle systems and modifiers attached to it, the previous call will create a duplicate object that also has those components. However, if I duplicate an object using   template_object = bpy.data.objects['TemplateObjectName'] new_object = bpy.data.objects.new('NewObjectName', template_object.data) collection.objects.link(new_object)   how do I attach the template object's animation, modifiers, particle systems...? Is it possible to recognise hardware (router or network card) just by its MAC address? Just wondering if that is possible. Are there any naming conventions that would allow one to deduce whether the MAC address belongs to a router (of a specific brand), or a network card, etc.. Is there a way to get new dwellers with higher than normal stats? Compute device list not showing card name when using OpenCL I have an AMD Radeon R7 M265, and when I force OPENCL with the batch file it only shows Iceland, Intel HD and the CPU.  I followed this instructions:   R, type CMD, press &#9166; Enter, go to your blender folder. ex:  cd C:\Program Files\Blender Foundation\Blender  Press &#9166; Enter and then type: set CYCLES_OPENCL_TEST=all Start Blender with blender.exe Inside blender go to File -> User preferences -> System Under compute device select OpenCL and then your card Close the preferences and go to the render panel and select device: GPU Compute.  I have the latest beta driver from AMD, but when I try to change the compute device it only shows this list:     Am I doing something wrong? Is my GPU is not supported? Or is it because it's in development and not every card is supported yet?  I've tried blender v2.74.5, v2.62, and v2.63. Every one gives the same list. How do I call a Scala Object method using reflection? After upgrading to Firefox 57 Quantum many website have rendering issues, such as empty frames, missing text or fonts, etc. I use Kubuntu 16.04 64 bit.  So far, I've tried:   Restarting Firefox in safe mode Disabling "hardware acceleration"   Wiping out my profile Purging Firefox and reinstalling Running Firefox from a clean user account   But nothing helped so far.  Any tips on how to solve this are much appreciated Why are Grothendieck's and Hartshorne's definitions of quasi-coherence equivalent? Hartshorne's Algebraic Geometry defines an $\mathcal O_X$-module $\mathscr F$ to be quasi-coherent if there is an open affine cover $(U_i=\operatorname{Spec} A_i)_{i\in\mathcal I}$ of $X$ such that each $\mathscr F\rvert_{U_i}$ is isomorphic to the sheaf of modules $\widetilde{M_i}$ associated to some $A_i$-module $M_i$. It is later proved that this is equivalent to the statement that for any open affine $U=\operatorname{Spec} A\subseteq X$, there exists an $A$-module $M$ with $\mathscr F\rvert_U\cong\widetilde M$.  On the other hand, Grothendieck's Éléments de Géométrie Algébrique defines $\mathscr F$ to be quasi-coherent if every point $x\in X$ admits an open neighborhood $U\subseteq X$ such that $\mathscr F\rvert_U$ is isomorphic to the cokernel of a morphism of free $\mathcal O_U$-modules.  Now I wonder how those definitions are equal. I tried proving it by myself, but both directions seem quite nontrivial to me... Any hints? Aside from the Force, is there magic in Star Wars? Why is this true: The only orthogonal projection that is also unitary from $\Bbb C^n$ to $\Bbb C^n$ is the identity Can anyone explain me please how to see this statement: the only orthogonal projection that is also unitary from $\Bbb C^n$ to $\Bbb C^n$ is the Identity.  how can I prove formally that? or how can I see that only the Identity satisfy it? Let's say it again, we all make mistakes, and I have just made one.  A brief history: I was doing some stuff on a VPS (Debian) I'm renting, when I noticed some strange behaviour. Using the netstat command I saw an non-authorized connection through SSH. I didn't know what to do, so I decided to close his connection using iptables:  iptables -A INPUT -p tcp --dport ssh -s IP -j DROP   But I am tired, and I wrote  iptables -A INPUT -p tcp --dport ssh -j DROP   and I kicked myself (and everyone else) out...  How do I fix this? There are five boxes consisting of several identical coins. There are five boxes consisting of several identical coins. The first box contains 4 coins, the second box contains 9 coins, the third box contains 7 coins, the fourth box contains 10 coins, and the fifth box contains 3 coins. Some coins are taken from the boxes such that at least a coin taken from each box. In how many ways in taking the coins, such that the numbers of the coins taken are 12?  My attempt was solving it using repetition combination of which, there are only seven coins remaining considering that at least a coin taken from each box. By using the repetition combination, i.e. taking 7 from 9 (five boxes and four borders), resulting in 36 ways. However, there is no 36 ways in the choices. Internet is working, but Firefox 67 browser cannot connect after update Why constructors will always have same name as of class and how they are invoked implicitly? Why is Facebook PHP SDK getUser always returning 0?