Is the genitive of  POTUS' or POTUS's
The school where I work has an S when it is an acronym (NEGS) it looks odd with an apostrophe to show ownership e.g. NEGS' tennis team...any thoughts?
Mac mini Mac OS 10.6 server migration to new hardware I have a Mac Mini with Mac OS X 10.6 Server installed. It has two external drives attached to it. One is for Time Machine and other contains data. I have bought the latest Mac Mini with Mac OS X 10.7 and two new hard drives. I want to migrate to the new hardware (the new Mac Mini and the new hard disks). Can Migration Assistant do that for me or do I have to transfer the data in the external drive manually? See , page 2: the internal logic diagram for the MM74HC138.    The diagram shows logic inverters with bubbles at the input or at the output. Is there an actual difference between them? jQuery not working when a new row is added to table HTML text input allow only numeric input Property of $\{0^n, 1^n, \ldots\}$ The differences between adjacent squares are the odd numbers:  0 1 4 9 16 25 …  1 3 5 7  9 …  The differences between adjacent odd numbers are 2 = 2!  I found that this truth is more general: If the differences between adjacent powers of $n$ are written out, and the differences of those differences etc, the $(n+1)$th sequence is {n!, n!, …}  Is this a well-known theorem? Does it have an easy proof? Non-trivial cyclic group with unique generator Let $G$ be a non-trivial cyclic group with an unique generator. Show that $G$ has exactly two elements.   Let $g$ be a generator of $G$. My first idea to approach the problem was to show that if $|G|=n &gt; 2$ then there is an integer $k$ between 1 and $n$ such that $\langle g^k \rangle = G$ or, equivalently, $g \in \langle g^k \rangle$. My second idea was to analyse the problem in 2 separate cases. For the first one, if $n$ is prime, let $k$ be a positive integer then, by Lagrange's Theorem, $|g^k| \mid n$ and, therefore, $\langle g^k \rangle = n$. Hence, we must conclude that, if $n &gt; 2$ the generator cannot be unique. For the second case, where $n$ is composite, I was hoping to get to a contradiction, but could not advance any further. Is there a way to develop either of these ideas or is it necessary a new approach? Non-Root Package Managers From my research, I seem to notice that all package managers insist on being used as a privileged user and must be installed into /.  Typically, what I like to do is create a throwaway account, compile some software, and install to $HOME for that account. I can try a variety of setups and then when I'm done, just destroy the account.  However, compiling software becomes tedious.  My experience is really just limited to yum, but I don't understand why I wouldn't be able to drop a repo file into ~/etc/yum.repos.d and have yum install everything into a home account.  Is there any reason why package managers must be used as a privleged user to install software? Sending email in .NET through Gmail I'd like to colorize a cell based on its contents. 0 = red, 100 = green, and linearly interpolated between.  function LinInt(x){       var ss = SpreadsheetApp.getActiveSheet();   var cell = ss.getActiveRange();   var hue;   hue = (x/100)*120;    var color = HSVtoRGB(hue, 40, 100);   cell.setBackground(color);    return x;     }     function HSVtoRGB(h, s, v) {     var r, g, b, i, f, p, q, t;     if (h &amp;&amp; s === undefined &amp;&amp; v === undefined) {         s = h.s, v = h.v, h = h.h;     }     i = Math.floor(h * 6);     f = h * 6 - i;     p = v * (1 - s);     q = v * (1 - f * s);     t = v * (1 - (1 - f) * s);     switch (i % 6) {         case 0: r = v, g = t, b = p; break;         case 1: r = q, g = v, b = p; break;         case 2: r = p, g = v, b = t; break;         case 3: r = p, g = q, b = v; break;         case 4: r = t, g = p, b = v; break;         case 5: r = v, g = p, b = q; break;     }   return '#'+((b | g &lt;&lt; 8 | r &lt;&lt; 16) / 0x1000000).toString(16).substring(2); }   When I do this I get an error saying that I am not allowed to call setBackground. I assume this is because I'm not allowed to change the color of one cell from another because I'm screwing up getting the current cell that the function is running in.  How do I get the cell that the function is running in so that I can call setBackground() on it?  For example, I'd like to type =LinInt(50) into cell A1 and have A1 be yellow and have the number 50 in it. Then when I type =LinInt(100) into cell A2 it is green and has the number 100 in it. How to find out when an Oracle table was updated the last time I have Ubuntu 14.04 and two monitors. When I open applications I need some of them to be on the primary monitor and others On the secondary. At the moment, apps always open in the primary monitor.   I have dash icons only on one (left) display, so I can't launch apps using the cursor or active window setting from Compiz. How to prevent Command-I/Command-Shift-I from opening Mac Mail when in browser? When in a browser (Chrome or Safari), the following keyboard shortcut will always bring up a window that says "Welcome To Mail":  Command+Shift+I  I would like to use this shortcut for another purpose.  Is there any way to disable this shortcut? Is there a way to reach a `protected` member of another object from a derived type? class MyBase {     protected object PropertyOfBase { get; set; } }  class MyType : MyBase {     void MyMethod(MyBase parameter)     {         // I am looking for:         object p = parameter.PropertyOfBase;  // error CS1540: Cannot access protected member 'MyBase.PropertyOfBase' via a qualifier of type 'MyBase'; the qualifier must be of type 'MyType' (or derived from it)     } }   Is there a way to get a protected property of a parameter of a type from an extending type without reflection? Since the extending class knows of the property through its base type, it would make sense if possible. Prove a function is constant under certain conditions Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be smooth and such that for every $x$    $$\int_{-\infty}^{\infty}\frac{|f(x)-f(y)|}{|x-y|^2}dy &lt; \infty$$   Prove that $f$ is constant. How do I change the location of macOS screen captures? As a developer, I take a lot of screen captures. I often take screen shots to share with Stack Exchange members in chat.  The problem is my Desktop is so cluttered with screen captures that they're now overlapping! I'd like to save all screenshots in a dedicated folder or directory.  In macOS, how do I change the default location where screen captures are saved?  These are the screen shots taken with CommandShift3 or CommandShift4 keyboard shortcuts.  I would prefer a solution which doesn't require installing any software. Calculate $\tan9^{\circ}-\tan27^{\circ}-\tan63^{\circ}+\tan81^{\circ}$ Calculate $\tan9^{\circ}-\tan27^{\circ}-\tan63^{\circ}+\tan81^{\circ}$?  The correct answer should be 4. Why are local variables not initialized in Java? Cannot dismiss scrollbar in code box, blocks view This question is a duplicate of  (but this question can't be marked as a duplicate because that question is on meta.stackoverflow.com, and this question is on meta.stackexchange.com).  I'm seeing a bug on stackoverflow.com. I'm not sure if the culprit is Stack Overflow, Google Chrome, or macOS.  (Below I refer to a "code box". By that I mean some text that begins with 4 spaces when editing a question or answer.)  When I hover over a code box that contains one long line of code, a horizontal scrollbar appears. When I stop hovering over the code box, very frequently (maybe 3 out of 4 times) the scrollbar does not disappear. Clicking elsewhere on the page also does not dismiss the scrollbar. The only way to dismiss it is to refresh the page.  ...After experimenting more, it appears to always happen if I enter the box from below, but not always when I enter from above. It seems to be triggered by hovering over the lower half of the box.  Also if I enter from the top and scroll to the right, then until I refresh the page again that scrollbox will behave normally (only show a scrollbar when scrolling, and disappear after). In that case the scrollbar also has a different appearance, being much less tall (maybe a quarter of the height of the code box rather than half as tall).  Further, if I get a box and scrollbar into that functional state and then hover over the lower part of the box for a bit, then sometimes eventually it will break and stay broken until refresh again.  And in any case hovering over the lower part of the box causes the scrollbar to get taller.  Here is a screenshot of the issue:        Google Chrome 71.0.3578.98 (Official Build) (64-bit)  macOS High Sierra 10.13.6  Here is a code box that repros the issue for me:  fooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo   Here is another:  foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo foo Inverse of this $3\times 3$ matrix using the Cayley–Hamilton theorem Find the inverse of the matrix $$\begin{pmatrix} -1 &amp; 2&amp; 0 \\ 1&amp; 1 &amp;0 \\ 2 &amp; -1&amp; 2 \end{pmatrix}$$ using the Cayley–Hamilton theorem.  Thanks! If $p,q,r$ are all primes,and $p|qr-1$,$q|pr-1$ and $r|pq-1$,find all possible values of $pqr$. If $p,q,r$ are all primes,and $p|qr-1$$~~~~~~~~~~~~~$$q|pr-1$$~~~~~~~~~~$ and $~r|pq-1$. Find all possible values of $pqr$.    My work: $qr-1=pk_1$ $pr-1=qk_2$ $pq-1=rk_3$ From the above equations, we can conclude that either $k_1,k_2,k_3$ are all even or one of the primes is $2$. I can also get that, $p^2q^2r^2=(pk_1+1)(qk_2+1)(rk_3+1)$ Now, I cannot proceed. Please help!