I'm trying to right doskey named that takes an argument. It should execute:  C:\User\Mercurius\Music\youtube-dl.exe -x --audio-format mp3 -f bestaudio "https://www.youtube.com/watch?v=ARGUMENT_HERE"   I tried this:  doskey youtube.com/watch?v=C\User\Mercurius\Music\youtube-dl.exe -x --audio-format mp3 -f bestaudio "https://www.youtube.com/watch?v=c2gSzYLJ8sY"   Then I tried using it but it failed:  C:\Users\Mercurius\Music&gt;youtube.com/watch?v= 'youtube.com' is not recognized as an internal or external command, operable program or batch file.   May you please help
I think the title is pretty self explanatory.   The question is how can I create an alias in a Windows cmd that will also work after I close and then reopen it. For example I would like an alias called ip to call the command ipconfig or an alias ls -l to call Dir.  My operating system is Windows 10, in case that makes a difference.
Why do I get "Required key not available" when install 3rd party kernel modules or after a kernel upgrade? This issue happens only on UEFI systems with enabled Secure Boot.  When I try to install DKMS modules like VirtualBox, Nvidia, or Broadcom drivers, they do not install and I get Required key not available when I try to modprobe them.  VirtualBox complains that vboxdrv is not loaded.  Broadcom wl driver is shown in lspci -k as a kernel module but is not in use. sudo modprobe wl throws Required key not available.  Also this issue may happen when I install some kernel modules from git sources.  This issue may appear after a kernel update as disabled wireless adapter, black screen after a reboot, etc.   How can I fix it? In setting up dual Boot with Windows and Ubuntu, which OS do I install first? I'd like to install both Ubuntu 12.04 and Windows XP on a Dell laptop, and I was thinking about using a dual boot structure, and using the bulk of my hard drive as empty hard drive space to share files between the two operating systems (so choice of file system type is very important in this set-up).     The kind of partitioning structure I would like to use is  Partition 1 - Ubuntu 12.04 (root) (20GB)  Partition 2 - Ubuntu /home        (20GB)  Partition 3 : Free Space          (560GB)  Partition 4 : Windows XP          (35GB)  Partition 5 : SWAP                (3GB)  (Total Hardrive Capacity is ~640GB)    My question is; what is the best way to go about setting up this kind this system?   Should I install Windows XP first and setup the partitions, and then install Ubuntu which I believe will install the GRUB bootloader for OS booting choice   or   Do I install Ubuntu first, setting up the available partitions and then perform a WIndows install?  Please let me know if there is anything in this setup that I have left out and should know about, including things related to setting particular partitions as logical or primary, and whether the boot partition and the filesystem partition should actually be two separate partitions. We have a fairly big MS SQL 2008R2 database that resides on a SSD drive. The drive itself only has ~110Gb of space, and the database files are the only files on the drive.  The database is in "Simple" recovery mode, and only has two files, .MDF and .LDF.   The disk is now nearly full: the MDF is currently 109Gb in size. However, SSMS tells me there's nearly 18Gb of "Space Available" (in the 'General' properties page), and if I go through the motions of Shrinking the file it also tells me there's 18Gb of space free. SSMS also tells me the database size is around 132Gb, which surprises me - that wouldn't fit on the drive!  From what I've read, shrink is a really bad idea. However, I'm starting to see replication errors (could not allocate space for object). We've previously tried shrinking the database, but within a few hours the file was back to it's original size.  How should we proceed - given there's apparently 18Gb of free space, should SQL automatically use that free space? Or is it as simple as: we really need more disk space? How to output HTML from JSP  block? I just started learning JSP technology, and came across a wall.  How do you output HTML from a method in &lt;%! ... %> JSP declaration block?  This doesn't work:  &lt;%!  void someOutput() {     out.println("Some Output"); } %&gt; ... &lt;% someOutput(); %&gt;   Server says there's no &ldquo;out&rdquo;.  U: I do know how to rewrite code with this method returning a string, but is there a way to do this inside &lt;%! void () {  } %> ? Though it may be non-optimal, it's still interesting. Inequality: $ab^2+bc^2+ca^2 \le 4$, when $a+b+c=3$. Let $a,b,c $ are non-negative real numbers, and $a+b+c=3$.  How to prove inequality $$ ab^2+bc^2+ca^2\le 4.\tag{*} $$    In other words, if $a,b,c$ are non-negative real numbers, then how to prove inequality $$ 27(ab^2+bc^2+ca^2)\le 4(a+b+c)^3.\tag {**} $$ $\color{gray}{\mbox{(Without using "universal" Lagrange multipliers method).}}$  Thanks! A Sequence converges to a limit $L \in \mathbb R$ if, and only if, an $N$-th tail of the sequences converges to $L$ for some $N \in \mathbb N$ Prove that $\sup \{-x \mid x \in A\} = -\inf\{x\mid x \in A\}$ I need to prove that $\sup \{-x \mid x \in A\} = -\inf\{x \mid x \in A\}$ and am having trouble moving the $-x$ out of the $\sup$ to $\inf$. Another thing is that I don't quite know how to prove $b = \inf$ (see below). Any help would be appreciated! Here my attempt:  Let $a = \sup\{-x \mid x \in A\}$  Then by definition $a \geq -x$ for $x \in A \implies -a\leq x$ for $x \in A$  Thus $-a$ is a lower bound of $A$. Assume $b$ is a lower bound of $A$  Then $b \leq -a \implies  a \geq -b$ and $a = \sup \{−x \mid x\in A\} = -\inf(A)$ What is a single word for a person who, as a habit, exaggerates a lot when relating information?   So for example, they relate something that happened but exaggerate so much that they are including clear falsehood and they are well known for doing that. What is the difference between "owing to" and "due to"? "Due to" seems more common than "owing to" in modern English. Is "owing to" simply an old-fashioned way of saying the same thing, or is there a rule to using it? Adding 3 electron spins I've learned how to add two 1/2-spins, which you can do with C-G-coefficients. There are 4 states (one singlet, three triplet states). States are symmetric or antisymmetric and the quantum numbers needed are total spin and total z-component.  But how do you add three 1/2-spins? It should yield 8 different eigenstates. Which quantum numbers do you need to characterise the 8 states?  It is not as easy as using C-G-coefficients and the usual quantum numbers as for the total momentum the doubly degenerate 1/2 state and quadruple degenerate 3/2 state can describe only 6 or the 8 states. You will need an additional quantum number for the degeneracy.  So how do you get the result?  (I actually tried out myself with a large 8x8 matrix. The total spin 1/2 is each doubly degenerate. For the additional quantum number I chose the cyclic permutation. Spin 1/2 states are neither symmetric nor antisymmetric. But what is the usual way to derive this?)  EDIT: For reference I'm adding my results for up to 4 spins from some time ago:   If you recall the basics of quantum mechanics with matrices it is actually a straightforward matrix diagonalization and requires no specialized knowledge. However, you still need to find an additional operator which breaks degeneracy. I chose the cyclic permutation, which seems to do the job. Please refer to the below answer, since I haven't checked all details. What screenshot tools are available? I have noticed that many people use images in their Ask Ubuntu answers. What software are people using? I remember watching this sci-fi movie where a huge metallic capsule gets dropped from a few hundred meters. The capsule had a chamber inside where only one person could be seated. During the drop, time dilated and the person would transcend to another world, lasting for days, even meeting dead people. Kind of a parallel reality.  It was an amazing movie, but I cannot recall its name. Had it not been vs If it were not for I've just seen the following sentence on a band's biography:     Had it not been for a studio theft, the band might never have made that song.   Then I was wondering if I can change this sentence to:     If it were not for a studio theft, the band might never have made that song.   Is the second sentence correct? Are they interchangeable?   Thanks in advance. What is the difference between "nothing but", "anything but", and "everything but"? Problem with patching babel package I am trying to use babel and biblatex package together but I am getting following error:    ! Package biblatex Error: Patching 'babel' package failed.    This is my code example:  \documentclass[12pt,a4paper,twoside,openright]{book} \usepackage{cmap} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage[slovak]{babel} \usepackage{xpatch} \usepackage{csquotes} \usepackage{biblatex} \begin{document} \enquote{quote} \end{document}   I would need the result look like this:    Well, it seems like I could get rid of the error when using xpatch package, but it works only when I delete 'slovak' option for babel package. That way the result look like this:    Please, is there any workaround for this problem? I am really stuck in this. I have sucessfully installed AniMove by activating experimental plugins (yay! thanks gis.stackexchange). But I can't find it in Qgis! Yes I have activated it. Where is it? Is this a common bug? I searched for it the toolbox, I looked up the names of the several single tools based on this  and only found "Kernel Density", which is filed under SAGA. But I don't have SAGA installed (and don't know if I need to?) Prove $x \equiv a \pmod{p}$ and $x \equiv a \pmod{q}$ then $x \equiv a\pmod{pq}$ Prove $x \equiv a \pmod{p}$ and $x \equiv a \pmod{q}$ then $x \equiv a\pmod{pq}$ for $p\neq q$ distinct primes.  Where can I start with this proof? It looks similar to the Chinese Remainder Theorem, but that deals with two different a values. I'm trying to make a page where a user provides an ID of a attachment and the site will set the headers and deliver that file instead of a 404 page. so /download/1754 would lookup the attachment with that Id and output it for viewing/download depending on the file type.  I just can't think of a wordpress friendly way to do this from inside a plugin. I considered using add_rewrite_rule but couldn't see a good way around it, also considered checking $_SERVER['REQUEST_URI'] to deliver the content before any HTML but then I'd have to stop the page from rendering a 404 and I was not sure of a Wordpress friendly way to do this.  My current idea was to have my plugin make a download page, assign a template and have that template run the logic instead of my plugin class. while I have used this before for another situation it was very messy as I had to write a lot of code to force users to be unable to update or edit the file and recreate/delete it on activation.deactivation. so ideally I'd like to avoid that. How do I do a SELECT * INTO [temp table] FROM [stored procedure]? Not FROM [Table] and without defining [temp table]?  Select all data from BusinessLine into tmpBusLine works fine.  select * into tmpBusLine from BusinessLine   I am trying the same, but using a stored procedure that returns data, is not quite the same.  select * into tmpBusLine from exec getBusinessLineHistory '16 Mar 2009'   Output message:     Msg 156, Level 15, State 1, Line 2   Incorrect syntax near the keyword   'exec'.   I have read several examples of creating a temporary table with the same structure as the output stored procedure, which works fine, but it would be nice to not supply any columns. Let $f$ be a bounded measurable function on a set of finite measure $E$.  For a measurable subset $A$ of $E$, show that $\int_A f = \int_E f \cdot \chi_A$     Lemma: Let $f$ be a bounded measurable function on a set of finite measure $E$.  Suppose $A$ and $B$ are disjoint measurable subsets of $E$.  Then $\int_{A \cup B} f = \int_A f + \int_B f$.   My first approach was to view the problem like this utilizing the above lemma:  $$\int_E  f \cdot \chi_A = \int_{E\setminus A} f \cdot X_A + \int_A f \cdot \chi_A $$  I'm not really sure how the result follows from the above.