CentOS 7 DVD ISO vs Everything ISO different
What is the difference between CentOS "DVD" vs "Everything" ISOs
Less than or equal sign If I know for two numbers a and b that    $${a &lt; b }$$  Then is it correct to say that    $$ a \leq b $$  I know that the second statement is true as long as the first one is. It seems OK as it is true but from the other side it seems kinda weird (to me), to say that, if you know, that a is strictly less that b.It's like you lose some extra information. How do I correctly clone a JavaScript object? I have an object x. I'd like to copy it as object y, such that changes to y do not modify x. I realized that copying objects derived from built-in JavaScript objects will result in extra, unwanted properties. This isn't a problem, since I'm copying one of my own literal-constructed objects.  How do I correctly clone a JavaScript object? How much time do I have to wait before I can sell another car? I was wondering: How much time does it take until the cooldown resets and I can sell another car in the GTA Online LS Customs Garage? Algebraic closure of $\mathbb{C}(x)$ is isomorphic to $\mathbb{C}$ Let $x$ be transcendental over $\mathbb{C}$. Let $K$ be the algebraic closure of $\mathbb{C}(x)$. How to show that $K$ is isomorphic to $\mathbb{C}$? Subtracting (erasing) polygons from polygon using ArcGIS for Desktop? I have two polygon layers, one is green that represents all non-public right-of-way land. I have another polygon that is just a circle with a specified radius.  I would like my output to be a polygon layer that represents all the white space (the gaps) in the private land polygon layer that is also within the black circle.    How does one do this (In ArcGIS 10.2)? How to sort an array of objects containing null elements? How to play a YouTube clip in background/minimised? DOS commands for Epson dot matrix not working Our Epson dot matrix printer is not printing on the following DOS command.:  Type text.txt&gt;prn   But it does print perfectly on:  open TEXT.txt&gt;File&gt;Print    What is the issue here? Can we have discussions? Guys, I just wonder if people are allowed to have civilized discussion (on any IT topic) on this forum without actually being accused of starting a row. I'm asking this question for a reason that few days I've asked few questions referring to VM, JIT etc. and compared them to C++ way and despite the fact that I've explained myself that I'm not starting any agro with anyone, that I'm really just genuinely curious, my post was first closed and then removed. I'm asking: What's wrong with asking nonstandard questions which sometimes my be quite difficult to answer? Thanks Multiple returns from a function show elements of order 15 in $S_8$ satisfy a relation Let $G$ be a group.  Define a relation $\sim$ on $G$ by $a \sim b$ if there exists $g \in G$ such that $a = gbg^{-1}$.  Prove that all elements of order 15 in $S_8$ are related by $\sim$.  I noticed that in order to have order $15$ in $S_8$. We need a $5$ cycle and a $3$ cycle. Then, I am stuck. Can anyone give me some ideas about what to do? Thank you in advance. In FreeBSD and also in Linux, how can I get the numerical chmod value of a file? For example, 644 instead of -rw-r--r--? I need an automatic way for a Bash script. Prove that $G$ is abelian iff $\varphi(g) = g^2$ is a homomorphism I'm working on the following problem:     Let $G$ be a group. Prove that $G$ is abelian if and only if   $\varphi(g) = g^2$ is a homomorphism.   My solution: First assume that $G$ is an abelian group and let $g, h \in G$. Observe that $\varphi(gh) = (gh)^2 = (gh)(gh) = g^2h^2 = \varphi(g)\varphi(h)$. Thus, $\varphi$ is a homomorphism.  I'm having trouble completing the proof in the reverse direction. Assume that $\varphi$ is a homomorphism. We then know that $\varphi(gh) = \varphi(g)\varphi(h)$ and $\varphi(hg) = \varphi(h)\varphi(g)$. However, I don't see a way to use this to show that $gh = hg$.  Could anyone lend a helping hand? how to adapt the sorting of a custom type `@standard` in using biblatex/biber Relationship between complex number and vectors Sign of determinant when using $det A^\top A$ We have been given matrix: $$A =          \begin{pmatrix}         a&amp; b&amp; c &amp;d \\         b &amp;−a&amp; d&amp; −c\\         c&amp; −d &amp;−a&amp; b \\         d &amp;c&amp; −b&amp; −a\\         \end{pmatrix}     $$ ...and have been asked to calculate $\det(A)$ using $AA^T$.  We see that:  $$AA^T=\begin{pmatrix}         a^2+ b^2+ c^2+ d^2&amp; 0&amp; 0&amp;0\\         0 &amp;a^2+ b^2+ c^2+ d^2&amp; 0&amp; 0\\         0&amp; 0 &amp;a^2+ b^2+ c^2+ d^2&amp; b \\        0 &amp;0&amp; 0&amp; a^2+ b^2+ c^2+ d^2\\         \end{pmatrix}     $$   So, $\det(AA^T)= (a^2+ b^2+ c^2+ d^2)^4$  Now which should I choose: $(a^2+ b^2+ c^2+ d^2)^2 $ or $ -(a^2+ b^2+ c^2+ d^2)^2$ ?  Please explain me which one and why. Is it correct to use that/which in the relative clause when it is referring to the place as an object pronoun? What will I miss out on if I play Diablo 3 without Reaper of Souls? I own the base game of Diablo 3.  I didn't really like the "post-Inferno" content of the game, so I stopped playing shortly after I curb-stomped the final boss for the n-th time.    I understand now that a lot of things have changed in the base game.  There's also an expansion.  I don't plan to purchase the expansion, but some of these changes sound like they could make the game more fun for me.  I can guess that probably the new character and act are expansion-only content.  I know I'm probably , but if I go back and play now, what else will I be missing out on if I don't have the expansion? Why is lock(this) {...} bad? The  says that public class SomeObject {   public void SomeOperation()   {     lock(this)     {       //Access instance variables     }   } }  is &quot;a problem if the instance can be accessed publicly&quot;. I'm wondering why? Is it because the lock will be held longer than necessary? Or is there some more insidious reason? Identity for convolution of central binomial coefficients: $\sum\limits_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$ It's not difficult to show that  $$(1-z^2)^{-1/2}=\sum_{n=0}^\infty \binom{2n}{n}2^{-2n}z^{2n}$$  On the other hand, we have $(1-z^2)^{-1}=\sum z^{2n}$.  Squaring the first power series and comparing terms gives us  $$\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}2^{-2n}=1$$  that is,  $$\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}$$  My question: is there a more direct, combinatorial proof of this identity?  I've been racking my brains trying to come up with one but I'm not having much success.