I read in a piece of news . " the amount of time that young people spend inactive." Inactive is an adjective . I consider it should be inactively . Is it right?
You can’t buy your kids clever. What’s more, if they’re merely above average, by sending them to some hideous Holland Park hothouse, you’re probably buying them miserable.  — Alex Proud,   ""   As far as I know, clever and miserable are both adjectives, not nouns. So why were they used with to buy? Is this an idiomatic way of saying You can't buy your kids intelligence and you're probably buying them (moral) misery? If so, what is the idiom used here? I'm not familiar with it.
I want to make a Launcher that will do privileged (requires root) commands, but I don't want to see the terminal or have to type my password into the terminal. So I found the command, gksudo that allows me to enter my password so that I can escalate my privilege level. Now I want to couple this, so that I can change the partition priority using this command cgpt add -i 6 -P 0 -S 0 /dev/sda (this command is privileged) and have it automatically reboot from this one launcher.  I setup a Luncher like this:  Name: ChromeOS  Description:  Command: gksudo gnome-terminal -e "cgpt add -i 6 -P 0 -S 0 /dev/sda;reboot"  Comment:  But it does not work. Any tips? How to get layer extent in OpenLayers or zoom to WMS layer extent? I have added WMS layers using follow:  basemap = new OpenLayers.Layer.WMS( "Layer Name1", "http://localhost:8080/geoserver/wms/workspacename", {layers: 'workspacename:layername'} );   utility = new OpenLayers.Layer.WMS( "Layer Name2", "http://localhost:8080/geoserver/wms/workspacename", {layers: 'workspacename:layername', transparent:"true"}, {isBaseLayer:false} );   Now I want to get zoom to my utility layer, How can I?  I have tried as following:  map.zoomToExtent(utility.getDataExtent()); What if an abbreviation is at the end of a declarative sentence? Should I duplicate a full stop sign? What if an abbreviation is at the end of a declarative sentence? Should I duplicate a full stop sign? If the following sentence begins with a proper noun, not putting two periods may be confusing, in theory. Edit: It used to be 'contraction' instead of 'abbreviation'. Is there a good reason to use "printf" instead of "print" in java? Why do Kurama and Gyuki appear as chakra while other bijuu don't? In Naruto, Kurama (the Nine-tailed bijuu) appears as chakra, as does Killer B's bijuu, Gyuki.      Other bijuu, however, don't appear that way. What is different between them? What is Skynet's ultimate goal after defeating the resistance? Showing that Y has a uniform distribution if Y=F(X) where F is the cdf of continuous X Let $X$ be a random variable with a continuous and strictly increasing c.d.f. function $F$ (so that the quantile function $F^{−1}$ is well-deﬁned). Deﬁne a new random variable $Y$ by $Y = F(X)$. Show that $Y$ has a uniform distribution on the interval $[0, 1]$.  My initial thought is that $Y$ is distributed on the interval $[0,1]$ because this is the range of $F$.  But how do you show that it is uniform? I've got a 32-bit Ubuntu installation running on 64-bit hardware. Now that , I would like to switch to 64-bit without having to reinstall the OS.  This is one of the :  Shawn installed his system using the 32-bit version of Ubuntu, but his hardware is 64-bit and he wants to switch over. He manually installs the amd64 versions of dpkg and apt, replacing the i386 versions and changing which architecture is used as the default; then he installs the amd64 ubuntu-minimal package; then he installs the amd64 ubuntu-desktop package. Over time the remaining i386 packages are replaced automatically on upgrade.  However, when trying to follow the instructions in there, I cannot find any 64-bit version of dpkg or apt.  Did this user story got implemented in a different way in the final spec, or do I need to do something differently?  In short, how can I switch my 32-bit installation to 64-bit? Did Dolores Umbridge Have Any Association with Voldemort (or Death Eaters) before His Return? I noticed that Dolores Umbridge was born during the first Wizarding War, so it's very likely she wasn't a Death Eater then (but she is pretty evil -- who knows?).  After that Voldemort was not around in a way that could affect many people, and most wouldn't know he was planning to rise again.  During that time, and up through Voldemort's return (in Goblet of Fire), did Umbridge have any connection with the Death Eaters or with Voldemort?  Was she doing what she did on her own, or was it because of an association with Voldemort or his allies? How to mount a VirtualBox shared folder at startup? How to use find command to search for multiple extensions I can get all jpg images by using:    find . -name "*.jpg"     But how can I add png files to the results as well? Getting "experimental" true displacement to work Installing MinionPro fonts on Ubuntu 10.04 running TexLive 2012 I followed the instructions to install MinionPro fonts from this website.   However, when I tried to run the tex file at the bottom, I have this error message:  ! LaTeX Error: File `MinionPro.sty' not found.   Running the command kpsewhich MinionPro.sty gives the following output:  /usr/local/share/texmf/tex/latex/MinionPro/MinionPro.sty   So the system knows where MinionPro.sty is, so why can't the file compile? Prove that $ m_{a} \geq m_{g} \geq m_{h} $ using strict inequalities unless $ a = b $. $ m_{a} = \frac{1}{2} (a + b) $   $ m_{g} =  \sqrt{ab} $   $ \frac{1}{m_{h}} = \frac{1}{2}(\frac{1}{a} + \frac{1}{b})$   Attempted Solution:  I believe I have shown the first step, which is $ m_{a} \geq m_{g} $ but i'm stuck on showing $ m_{g} \geq m_{h} $  $ \frac{a + b}{2} &gt; \sqrt{ab} $  $ (\frac{a + b}{2})^2 &gt; (\sqrt{ab})^2 $  $\frac{a^2 + 2ab + b^2}{4} &gt; ab$  $a^2 + 2ab + b^2 &gt; 4ab $  $a^2 -2ab + b^2 &gt; 0 $  $(a - b)^2 &gt; 0 $   so if $a = b$ we have $(a -b)^2 \geq 0$  Therefore, $m_{a} \geq m_{g} $  Now, i'm not sure if that is correct or not, if not can you please guide me to the complete solution or if I am correct on this part, provide a solution for $m_{g} \geq m_{h} $ How to avoid becoming stressed while playing endurance games? Although I am asking for Super Hexagon, I've noticed there is a multitude of games (rhythm games in particular) where the more stressed I am, the worse I play and/or are more likely to make a mistake. Often then the games are designed in such a way to induce stress, such as various ways of telling me how far I am, which also is telling me just how much it would suck if I were to fail right now.  This leads to a stress cycle, where I am noticing I am stressed, and I am aware of this being a bad thing, which makes me more stressed, over and around until I eventually fail.  This often means that I probably could've done &lt;some new level I just unlocked&gt; just fine, except I saw that it was new, got stressed about it and started the stress cycle.  So how can I:   Avoid getting stressed while playing Defuse the stress cycle once it has inevitably started   One thought I had for this is simply taking deep breaths, but particularly in Super Hexagon, there is no way to pause and taking deep breaths is distracting enough to make me fail. IE11 making background image text blurry I have a web application that's branded according to the user that's currently logged in. I'd like to change the favicon of the page to be the logo of the private label, but I'm unable to find any code or any examples of how to do this. Has anybody successfully done this before?  I'm picturing having a dozen icons in a folder, and the reference to which favicon.ico file to use is just generated dynamically along with the HTML page. Thoughts? What's the best way to invite a user to answer a question on Stack Overflow? On several occasions, I was reading a question thinking "I know a guy that knows the answer to this" and that guy (or gal) also happens to be a registered user. How can I bring this to the attention of this user? If $X$ is measurable space, $f : X\to [-\infty,+\infty]$, function such that $f^{-1}([-\infty,r))$ measurable for all rational numbers $r$,prove than $f$ is measurable  I know that $f$ is measurable iff set $\{x\in X : f(x)&lt;a\}$ is measurable for all $a\in \mathbb{R}$ but for $a\in\mathbb{Q}$ I don't know how to proceed. How many combinations can I make? let $n \gt 1$ be an integer, and consider $n$ people; $P_1, P_2,..., P_n$ let $A_n$ be the number of ways these $n$ people can be divided into groups, such that each group have either one or two people  determine $A_1, A_2, A_3$  So I have  $A_1 = 1$ way $A_2 = 4$ way $A_3 = 12$ way  But I am not sure if this is right... Can anyone confirm?  edit:  My way of getting this is  $A_1 = 1 $ because {P1} only has 1 way to sort  $A_2 = 4$ because {P1,P2} = {P1,P2},{P2,P1},{P1P2},{P2P1}  etc.