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How do file permissions work? Can a single PS Plus subscription be used on two different profiles on the same PS4? Both my girlfriend and I play on the PS4 at home, so there are 2 profiles on PS4. When we set our profile in PS4, we set the console as the primary console.  Now we want to buy PSN Plus subscription, and I want to know if the same subscription can be used in both profiles, or  only in one.   In case the same subscription can be used in both profiles, how can we set it up? Or is automatic? Proper use of the IDisposable interface I know from reading  that the "primary" use of the IDisposable interface is to clean up unmanaged resources.  To me, "unmanaged" means things like database connections, sockets, window handles, etc.  But, I've seen code where the Dispose() method is implemented to free managed resources, which seems redundant to me, since the garbage collector should take care of that for you.  For example:  public class MyCollection : IDisposable {     private List&lt;String&gt; _theList = new List&lt;String&gt;();     private Dictionary&lt;String, Point&gt; _theDict = new Dictionary&lt;String, Point&gt;();      // Die, clear it up! (free unmanaged resources)     public void Dispose()     {         _theList.clear();         _theDict.clear();         _theList = null;         _theDict = null;     }   My question is, does this make the garbage collector free memory used by MyCollection any faster than it normally would?  edit:  So far people have posted some good examples of using IDisposable to clean up unmanaged resources such as database connections and bitmaps.  But suppose that _theList in the above code contained a million strings, and you wanted to free that memory now, rather than waiting for the garbage collector.  Would the above code accomplish that? Firefox browser opens but URL is not launched Display Top X results for each unique value I am trying to figure out how I can output any machine that has had the latest 3 backups be larger than 10GB. My SQL Table looks similar to the below table:  ComputerID | BackupImageName | WindowsVolumeGUID | FileSize | TimeCreated 1234       | PCBackup.spi    | kjsdfai-dadad-ddd | 10400    | 05-22-20 1234       | PCBackup1.spi   | kjsdfai-dadad-ddd | 10500    | 05-23-20 1234       | PCBackup2.spi   | kjsdfai-dadad-ddd | 10300    | 05-24-20 1234       | PCBackup3.spi   | kjsdfai-dadad-ddd | 10900    | 05-25-20 1225       | PCBackup4.spi   | kjsdfa5-dadad-ddd | 10600    | 05-22-20 1225       | PCBackup5.spi   | kjsdfa5-dadad-ddd | 5000     | 05-23-20 1225       | PCBackup6.spi   | kjsdfa5-dadad-ddd | 10000    | 05-24-20 1225       | PCBackup7.spi   | kjsdfa5-dadad-ddd | 4588     | 05-25-20   I need to output all machines (computerID's) or Volumes (WindowsVolumeGUID) that have their latest 3 files in the list greater than 10000 file size. So from the above table it would output the computerID of 1234 because its latest 3 are greater than 10000 file size and even though the computerID 1225 has 2 files greater than 10000 it would not output the result because there is a break in-between the two large files. The search will need to take into account the WindowsVolumeGUID since some machines have more than one volume and I need to take into account the individual volume backups and count the latest entries based on that.  Alternatively, I can also accomplish the same thing if there is a way to output the top 3 results for each unique WindowsVolumeGUID and not just across the board.   Any help would be greatly appreciated! What does // mean in a path? How are Stormtroopers uniquely identified in the field? Multiple times in Star Wars: The Force Awakens, people refer to trooper FN-2187 by that specific correct name, while he is in armor.  But the armor is largely identical (some have little accessories like shoulder pads, or different weapon loadouts, but FN-2187 lacks these), which makes me wonder how do First Order (or even the old Imperials) identify who they're talking to?  How are Stormtroopers correctly identified while they're in armor? How can I make sure texts from friends with iPhones don't go to iMessage instead of my Android phone? I have always used an Android phone, but text messages from certain friends get hijacked by iMessage. How do I prevent iPhones and MacOS Messages from sending me iMessages? I need all messages to be SMS or MMS. Some friends with iPhones will text me, but I don't receive the message. Today I noticed the Message app on my Mac had notifications. I never open this app since I don't use iMessage, but I looked today, and found multiple &quot;lost&quot; messages from several weeks ago. Please note that this is NOT the same question as . My Android phone is not, and has never been, registered with iMessage, and I have already followed those instructions. Another poster () had the same question as me----but it was closed, being deemed as the same as the Removing phone number question. It's actually fairly different. I have tried:  Leaving my Messages app closed on my Mac. But somehow messages from these people get hijacked. Deregistering my mobile number using Apple's online Deregister iMessage tool. When I attempted this, I got the message &quot;Your phone number is not registered with iMessage.&quot; Finding a setting on my Android phone that addresses this. The problem is apparently (and logically seems to be) on the Apple end.  Things  says he tried:   Closing Messages.app. I still get macOS notifications of messages. In Messages &gt; Preferences… &gt; Accounts &gt; [my email address], untick &quot;Enable this account&quot;. Then my friends messages disappear and I never see them at all, at least until I reenable in Messages.app. Looking in System Preferences and on icloud.com and appleid.apple.com    to see if there is some setting I can disable—but I've not found    anything relevant. Using Apple's Deregister iMessage tool to unregister my mobile    number. When I attempted this it gave me the message &quot;Your phone number is not registered with iMessage.&quot; If $a,b$ are positive integers such that $\gcd(a,b)=1$, then show that $\gcd(a+b, a-b)=1$ or $2$. and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $ If  $a,b$ are positive integers such that $\gcd(a,b)=1$, then show that $\gcd(a+b, a-b)=1$ or $2$ and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $. Progress We have   $\gcd(a,b)=1\implies \exists u,v\in \mathbb Z$ such that $au+bv=1\implies a^2u'+bv'=1, u',v'\in \mathbb Z$. Let $\gcd(a+b, a-b)=d$. Then $\mid (a+b)x+(a-b)y, \forall x,y\in \mathbb Z$. How to show $\gcd(a+b, a-b)=1$ or $2$ and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $. On page 19 of the textbook  (by James, Witten, Hastie and Tibshirani--it is freely downloadable on the web, and very good), the following is stated:  Consider a given estimate $$\hat{Y} = \hat{f}(x)$$ Assume for a moment that both $$\hat{f}, X$$ are fixed. Then, it is easy to show that: $$\mathrm{E}(Y - \hat{Y})^2 = \mathrm{E}[f(X) + \epsilon - \hat{f}(X)]^2$$ $$ = [f(X) - \hat{f}(X)]^2 + \mathrm{Var}(\epsilon)$$  It is further explained that the first term represents the reducible error, and the second term represents the irreducible error. I am not fully understanding how the authors arrive at this answer. I worked through the calculations as follows: $$\mathrm{E}(Y - \hat{Y})^2 = \mathrm{E}[f(X) + \epsilon - \hat{f}(X)]^2$$ This simplifies to $[f(X) - \hat{f}(X) + \mathrm{E}[\epsilon]]^2 = [f(X) - \hat{f}(X)]^2$ assuming that $\mathrm{E}[\epsilon] = 0$. Where is the $\mathrm{Var}(x)$ indicated in the text coming from? Any suggestions would be greatly appreciated. In The Matrix Reloaded, the Architect reveals to Neo that Trinity is about to die. I can't help but wonder if Neo would still choose the left door if the Architect had withheld that information.  The Architect could have designed a better choice mechanism so as to trick Neo into making a choice in the machine's favor rather than a highly risky one.   Did I miss anything important in the movie? It is quite easy to show that (by using residue theorem) $$\int_0^{\infty} \frac{1}{1+x^n}dx = \frac{2 \pi i^{1+2/n}}{n(e^{2 \pi i / n} - 1)} $$ for $$n \ge 2$$  Is there any possibility to simplify $$\frac{2 \pi i^{1+2/n}}{n(e^{2 \pi i / n} - 1)}$$ or it is best result?  Thanks in advance! C++ guarantees that variables in a compilation unit (.cpp file) are initialised in order of declaration. For number of compilation units this rule works for each one separately (I mean static variables outside of classes).  But, the order of initialization of variables, is undefined across different compilation units.  Where can I see some explanations about this order for gcc and MSVC (I know that relying on that is a very bad idea - it is just to understand the problems that we may have with legacy code when moving to new GCC major and different OS)? Two monitors when only having one in 13.10 I just did a fresh install of Ubuntu 13.10 64bit, and it says that I have 2 monitors, when I actually only have one    I have a Dell XPS 15 with Nvidia graphics card if that helps in anyway How to open pdf file in Windows Phone 8? Lots of people would like to visit the EEA area to take holiday or to visit family or even to go for longer term visa arrangements in the Schengen zone.  In order to do this, they need a Schengen (or national) visa.  This appears to be a fairly straightforward process, but hints and tips can always be helpful.  Without focusing on any individual's circumstances, what are some helpful hints that can improve the likelihood of a successful application? $\mathbb R$ has the same cardinality of any interval Where should shell tools be installed? Are there Apple guidelines for where user shell tools in Mountain Lion should be installed? Some (e.g. TeX, or Apple's Git) end up in /usr, and others (e.g. X11) seem to end up in /opt (sometimes with a symlink from /usr), while others (e.g., Git) end up in /user/local, and I wonder what the difference is, and whether one is preferred over the other. How can I get a new browser session when opening a new tab or window on Firefox/Chrome? When opening new Internet Explorer window, a new browser session is issued. Therefore you can login with two different accounts on the same service (let's say ). The same does not apply when opening a new tab in the same window in Internet&nbsp;Explorer.  How can I get a new browser session when opening a new tab or window on Firefox?  How about Google Chrome? How can I disable the volume warning if you raise the volume too much?