I have problem with fully disable UAC (User Account Control) in Windows 8. I've never had such problems in Windows Vista, or Windows 7 - just have to disable UAC in Control Panel, reboot, and that's it.  Ok, step by step:  I've just make one account, during installation - account "Namek". In "Control Panel -> Accounts" I see only one account "Namek", and next to the avatar picture, it's:  Namek Local Account Administrator   Using:  cmd -&gt; net users   there are:  Administrator   Guest   Namek   using:  cmd -&gt; net users Administrator   and:  cmd -&gt; net users Namek   the only difference are: Comment - in Administrator it's built-in, in Namek it's empty Account active - Administrator no, Namek yes Password required - Administrator yes, Namek no (because during Windows 8 installation I left password empty).  So I started with:  Control Panel -&gt; Accounts -&gt; Family Safety section   and set slider to value:  Never notify   reboot, and now, I've problem when for example I try to save anything, using Paint, directly on C:\ drive. It says:  You don't have permission to save in this location. Contact the administrator to obtain permission. Would you like to save in the My Pictures folder instead?     Ok, I've read some tutorials on the Internet, and I see that the case isn't as easy as in Windows Vista, or Windows 7 -> just disable UAC in Control Panel... so next step:  In gpedit.msc, I have already set:  Computer Configuration -&gt; Windows Settings -&gt; Security Settings -&gt; Security Options ---&gt; User Account Control: Behavior of the elevation prompt for administrators in Admin Approval Mode   At value:  Elevate without prompting   So i didn't have to changed it.    In regedit:  HKEY_LOCAL_MACHINE\Software\Microsoft\Windows\CurrentVersion\Policies\System   At value:  EnableLUA   I had default 1, so I set it to 0. but again nothing has changed.    On another Windows 8 (on VirtualBox) I've try solution, to acti ve built-in Administrator account, typig: "cmd -> net user administrator /active:yes" and now after start Windows 8 I can login to Administrator account, and now I've got FULLY permissions, but when I'm logout, and login on normal account, with 'administrator' permissions (which I created during installation), I've again problem when for example I try to save anything, using Paint, directly on C:\ drive. It says: "You don't have permission...blabla"  It's annoying, that I always have to right click some program, and click Run as administrator. I do not want to activate in my normal system Administrator account (by net user administrator /active:yes), and work on it, because on Namek account, I've already set a lot of things... Any help, please :(?
I am using Windows 8 Enterprise on my development machine. Most of the time, I need full administrator for debugging, changing system files, etc.  In Windows 7, setting UAC to "never notify" would disable any administrator prompts. In Windows 8 this is no longer the case. Even with UAC disabled I get prompted to grant programs elevated privileges.  Is there a way disable this behaviour?  Note: I am fully aware of the repercussions. I have antivirus, firewall, etc and am generally quite careful about what I download or install on my machine.
How do I boot into recovery mode? The full problem is as stated in the title. I am here to check if this is a valid proof. I thought it would be easiest using Linear Algebra.     Recall that an infinite cyclic group is isomorphic to $\mathbb{Z}$. We wish to show that we do not have an isomorphism between $\mathbb{ZxZ\;and\;Z}$. Note that $\mathbb{ZxZ}$ is an infinite group (under addition of course). Now, in order for there to even be potential for an isomorphism, two spaces must have equal dimension. Since the $\mathbb{dim(\mathbb{ZxZ})}=2&gt;\mathbb{dim(\mathbb{Z})}=1$, we know that $\nexists$ an isomorphism between our spaces. Hence, $\mathbb{ZxZ}$ is not a cyclic group.   My question (besides a validity check): was there a better way to prove this? I just found this to be the easiest way.  EDIT Totally wrong with this one. Back to the cutting board. How can I apply a -p0 patch from any working directory? Why does an open interval NOT have measure zero? Illogical twoside margins ListPlot and filling between curves Storing C++ template function definitions in a .CPP file I have some template code that I would prefer to have stored in a CPP file instead of inline in the header.  I know this can be done as long as you know which template types will be used.  For example:  .h file  class foo { public:     template &lt;typename T&gt;     void do(const T&amp; t); };   .cpp file  template &lt;typename T&gt; void foo::do(const T&amp; t) {     // Do something with t }  template void foo::do&lt;int&gt;(const int&amp;); template void foo::do&lt;std::string&gt;(const std::string&amp;);   Note the last two lines - the foo::do template function is only used with ints and std::strings, so those definitions mean the app will link.  My question is - is this a nasty hack or will this work with other compilers/linkers? I am only using this code with VS2008 at the moment but will be wanting to port to other environments. Did the Council of Elrond know that destroying the Ring would destroy Sauron? Did they decide to destroy it simply because they were Wise and knew it was dangerous to wield? Did they create the Fellowship simply because they were heroic; and decided to face Sauron after the Ring was destroyed?  Or did they know that destroying the ring would defeat Sauron, and they wouldn't actually have to do anything once that is taken care of? How can I delete everything from my MacBook Pro before selling it? I have a Mac that runs Snow Leopard and I intend to upgrade it to Mountain Lion and sell it on ebay. However, I only have one hard-drive, and I'm sure I can't use something in my OS to delete my OS and all user data.  Is there something for this that's built into the firmware that I can do on boot, or otherwise?   I have source code on my Mac for multiple corporations' proprietary software and I can't risk somebody gaining access through the use of any sort of disk recovery tool. I plan to use a 7 pass erase rather than simply deleting the directory. Is binary logistic regression a special case of multinomial logistic regression when the outcome has 2 levels? I'm stuck in full screen in the node editor. Escape isn't working, and I have no menus and no idea how to leave except to quit Blender and reopen the program. Should I use #define, enum or const? In a C++ project I'm working on, I have a flag kind of value which can have four values. Those four flags can be combined. Flags describe the records in database and can be:   new record deleted record modified record existing record   Now, for each record I wish to keep this attribute, so I could use an enum:  enum { xNew, xDeleted, xModified, xExisting }   However, in other places in code, I need to select which records are to be visible to the user, so I'd like to be able to pass that as a single parameter, like:   showRecords(xNew | xDeleted);   So, it seems I have three possible appoaches:  #define X_NEW      0x01 #define X_DELETED  0x02 #define X_MODIFIED 0x04 #define X_EXISTING 0x08   or  typedef enum { xNew = 1, xDeleted, xModified = 4, xExisting = 8 } RecordType;   or   namespace RecordType {     static const uint8 xNew = 1;     static const uint8 xDeleted = 2;     static const uint8 xModified = 4;     static const uint8 xExisting = 8; }   Space requirements are important (byte vs int) but not crucial. With defines I lose type safety, and with enum I lose some space (integers) and probably have to cast when I want to do a bitwise operation. With const I think I also lose type safety since a random uint8 could get in by mistake.  Is there some other cleaner way?  If not, what would you use and why?  P.S. The rest of the code is rather clean modern C++ without #defines, and I have used namespaces and templates in few spaces, so those aren't out of question either. Are there any concrete rules that say which words (parts of speech) in a title should start with a capital letter? What would be a correct capitalization for the title of this question? Let $X_1$ and $X_2$ be independent and identically distributed exponential random variables with rate $\lambda$. Let $S_2 = X_1 + X_2$.  Q: Show that $S_2$ has PDF $f_{S_2}(x) = \lambda^2 x \text{e}^{-\lambda x},\, x\ge 0$.  Note that if events occurred according to a Poisson Process (PP) with rate $\lambda$, $S_2$ would represent the time of the 2nd event.   Alternate approaches are appreciated. The approaches provided are commonly used when learning queueing theory &amp; stochastic processes.     Recall the Exponential distribution is a special case of the Gamma distribution (with shape parameter $1$). I've learned there is a more general version of this  that can be applied. problem with surfaces on tkz-fct i am using this wonderful tool that tkz-fct is for creating nice documents but I have a weird result from recently. This is my document :  \documentclass[10pt,a4paper]{article} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage{calc} \usepackage{color} \usepackage{pifont}                     %pour utiliser la police Ding \usepackage[left=1.5cm,right=1.5cm,top=2cm,bottom=2cm]{geometry} %marges du document \usepackage[french]{babel} \usepackage[usenames,dvipsnames]{xcolor} \usepackage[L]{thmbox} \usepackage{eurosym} \usepackage{enumitem}               %pour modifer itemize et enumerate \usepackage{amsmath,amsfonts,amssymb} \usepackage{mathrsfs}               %pour utiliser mathscr \usepackage{array,multirow,makecell} \usepackage{fancyhdr}               %package pour modifier les en-têtes \usepackage{fancybox}               %pour boîtes spéciales \usepackage{tikz,tkz-tab,tkz-fct} \usepackage{tikz-dependency}        %permet de créer des schemas \usepackage{sectsty}                %permet de modifier les styles des sections (voir \sectionfont l.21) \usepackage{fourier-orns} \usepackage{shadethm}               %pour theoremes en gris \setcellgapes{1pt} \makegapedcells \newcolumntype{R}[1]{&gt;{\raggedleft\arraybackslash }b{#1}}   %pour redéfinir les colonnes dans tabular \newcolumntype{L}[1]{&gt;{\raggedright\arraybackslash }b{#1}} \newcolumntype{C}[1]{&gt;{\centering\arraybackslash }b{#1}} \renewcommand{\thesection}{\Roman{section}} %permet de changer la numérotation en chiffres romains des sections \newshadetheorem{defi}{Définition}  %crée un nouveau type de théorème \newshadetheorem{thm}{Théorème} \newshadetheorem{ppte}{Propriété} \sectionfont{\sectionrule{3ex}{3pt}{-1.5ex}{1pt}} %mettre un trait de séparation dans les sections (package sectsty) \pagestyle{fancy} %style en-têtes \lhead{} %en-tête gauche \chead{Lois à densité} %en-tête centrale \rhead{T\up{ale}} %en-tête droite \begin{document} \begin{tikzpicture}      \tkzInit[xmin=-4,xmax=4,ymax=1]      \tkzDrawX[label={},noticks]      %\tkzDrawY      \tkzFct[domain=-2:2]{exp(-x**2)}      \tkzDrawArea[color=blue, domain = -0.5:1]      \tkzText(-1,0.75){$\mathscr{C}_{f}$} \end{tikzpicture} \end{document}   But from now, it doesn't produce the result I would like as the surface start from the point (-0.5,f(-0.5)) and not as an integral from -0.5 to 1. This is a picture of the result :   Which is obviously not what I would like and this extends now to all my "old" docs which were "normal" before. I don't think I have done anything special except updating my Tex distribution ... If someone has an idea ?? The exchange of photons gives rise to the electromagnetic force Pardon me for my stubborn classical/semiclassical brain. But I bet I am not the only one finding such description confusing.  If EM force is caused by the exchange of photons, does that mean only when there are photons exchanged shall there be a force? To my knowledge, once charged particles are placed, the electromagnetic force is always there, uninterruptedly. According to such logic, there has to be a stream of infinite photons to build EM force, and there has to be no interval between one "exchange event" to another. A free light source from an EM field? The scenario is really hard to imagine.  For nuclei the scenario becomes even odder. The strong interaction between protons is caused by the exchange of massive pions. It sounds like the protons toss a stream of balls to one another to build an attractive force - and the balls should come from nothing.  Please correct me if I am wrong: the excitations of photons and pions all come from nothing. So there should be EM force and strong force everywhere, no matter what type of particles out there. Say, even electrical neutral, dipole-free particles can build EM force in-between. And I find no reason such exchanges of particles cannot happen in vacuum.   Hope there will be some decent firmware to refresh my classical brain with newer field language codes. Series $\frac{x^{3n}}{(3n)!} $ find sum using differentiation I recently flagged a question as "very low quality", just like I've flagged many questions in the past. However, my flag was found to be unhelpful this time. The same thing happened when I recently flagged a post as not an answer.  Are the moderators being more strict when it comes to handling quality related flags? If so, why? Should Luke's lightsaber have given away his plan? During Luke's confrontation of Kylo Ren toward the end of The Last Jedi we see that       the lightsaber Luke uses a blue lightsaber (which I thought was his original one but that really doesn't matter), not the green one of his own construction.   Given that Kylo Ren  and Rey has just     fought over and broken the blue lightsaber   Why did Kylo Ren not notice this "inconsistency" and realize that      Luke's physical body was not there? Microsoft Office alternative for Windows? Is there something similar to Office for Windows but free? I am student living on a budget, Office is not an option at the moment.