In &quot;take it easy&quot;, &quot;easy&quot; is an adjective. In &quot;take it seriously&quot;, &quot;seriously&quot; is an adverb. I'm baffled. Why not &quot;take it easily&quot;? Using an adverb to modify the verb makes more sense. For the same reason, I should say &quot;take it more slowly&quot;, instead of &quot;take it slower&quot;, right?
It's my understanding that "easily" is the regular adverb for "easy", e.g. "He makes friends easily".  But in some collocations one still uses "easy", e.g. "Take it easy", "Rest easy" or "Slow down easy". Is there any rule when "easy" would be used over "easily", or does one just have to know these collocations?
What happens to the Windows 7 key when upgrading to Windows 8? Microsoft is offering "Microsoft Windows 8 Professional Upgrade" at a discounted price compared to the OEM version. However it states:     If you currently have a personal computer running Windows 7, Windows XP or Windows Vista then you can upgrade to Windows 8 Pro (Professional).   My question is what happens to the Windows 7 key that I use to upgrade? Will it become the Windows 8 key or will it generate a new Windows 8 key?  Basically if I upgrade using my Windows 7 key from my laptop for my desktop will I be able to continue using the laptop on Windows 7 while concurrently using Windows 8 upgrade on my Desktop? If $a,b$ are the roots of the equation $2 x^2 -3 x +1 = 0$, find an equation whose roots are $a/(2b +3)$, $b/(2a +3)$ apex:inputField can only be used with sObjects I am trying to create a custom visualforce page that saves new opportunity contact roles to an opportunity. I am having trouble creating the first step of listing out all contacts that relate to that account. The error I am getting is...     Save error: Could not resolve the entity from  value binding '{!u.contact}'.   can only be used with SObjects, or objects that are Visualforce field component resolvable.   I do not understand why I am receiving this error. u.contact should resolve to a Contact object, which is an sObject. Am I misunderstanding something?  Here is my code...  Controller  public class ChooseContactController  {     public String accountId = ApexPages.currentPage().getParameters().get('accountId');     public Account account { get; set; }     public List&lt;UnfundedOpportunityContactRole&gt; unfundedOpportunityContactRoles { get; set; }      public ChooseContactController()      {          unfundedOpportunityContactRoles = new List&lt;UnfundedOpportunityContactRole&gt;();         List&lt;Contact&gt; contacts = [SELECT Id, Name FROM Contact WHERE AccountId =: accountId];         account = [SELECT Id, Name FROM Account WHERE Id =: accountId];          for(Contact c: contacts)          {             unfundedOpportunityContactRoles.add(new UnfundedOpportunityContactRole(c, 'The Man'));         }     }      public void Save() {}     public void Create() {}      public class UnfundedOpportunityContactRole     {         public Contact contact { get; set; }         public String role { get; set; }          public UnfundedOpportunityContactRole() { }          public UnfundedOpportunityContactRole(Contact contactParam, String roleParam)         {             this.contact = contactParam;             this.role = roleParam;         }     } }   VF page  &lt;apex:page controller="ChooseContactController"&gt;   &lt;apex:form &gt;     &lt;apex:pageBlock title="Choose a contact" mode="edit"&gt;       &lt;apex:pageBlockButtons &gt;         &lt;apex:commandButton action="{!create}" value="New"/&gt;         &lt;apex:commandButton action="{!save}" value="Save"/&gt;       &lt;/apex:pageBlockButtons&gt;         &lt;apex:pageBlockSection title="{!account.Name}"&gt;           &lt;apex:dataTable value="{!unfundedOpportunityContactRoles}" var="u"&gt;             &lt;apex:column headerValue="Contact"&gt;               &lt;apex:inputField value="{!u.contact}" /&gt;             &lt;/apex:column&gt;             &lt;apex:column headerValue="Role"&gt;               &lt;apex:inputField value="{!u.role}" /&gt;             &lt;/apex:column&gt;           &lt;/apex:dataTable&gt;         &lt;/apex:pageBlockSection&gt;     &lt;/apex:pageBlock&gt;   &lt;/apex:form&gt; &lt;/apex:page&gt; Someone from Mars is called a Martian. Someone from America is called an American. Someone from Europe is a European. Someone from Hyrule is a Hyrulean.  We are not far from colonizing other planets, let alone Earth's only moon. Obviously you wouldn't call someone from the moon a lunatic right? What if you are on Mars? What would I call a foreigner from Earth? Earthean doesn't sound right. Use property to show that if $f_n \rightarrow f$ in measure on finite measure set $E$ then $f_n^2 \rightarrow f^2$ in measure on $E$. Consider the following problem:     Let $E$ have finite measure, $\{f_n\} \rightarrow f$ in measure on $E$ and $g$ be a measurable function on $E$ that is finite almost everywhere. Use the fact that $\{f_n \cdot g\} \rightarrow f_n \cdot g$ to prove that $\{f_n^2 \} \rightarrow f^2$ in measure on $E$. Infer from this that if $\{g_n\} \rightarrow g$ in measure on $E$ then $\{f_n \cdot g_n \} \rightarrow f \cdot g$ in measure on $E$.   I'm struggling understand how I can apply the fact that $\{f_n \cdot g\} \rightarrow f_n \cdot g$ in measure to prove that $\{f_n^2 \} \rightarrow f^2$ in measure. I've been able to come up with a proof of the fact that if $\{g_n\} \rightarrow g$ in measure then $\{f_n \cdot g_n \} \rightarrow f \cdot g$ in measure, and this clearly implies the claim that $\{f_n^2 \} \rightarrow f^2$ in measure. Unfortunately my proof doesn't rely on the fact that $\{f_n \cdot g\} \rightarrow f_n \cdot g$ so it doesn't answer the problem since it explicitly asks us to prove it a certain way.  So far my best attempt to utilize the fact is as follows. Let $\eta &gt; 0$ then by triangle inequality and monotonicity of measure  \begin{align*}     \lim_{n\rightarrow \infty}\mu\left\{ \left|f_n^2 - f^2\right| &gt; \eta \right\} &amp;\leq \lim_{n\rightarrow \infty}\mu \left\{\left|f_n - f\right|\cdot \left|f_n\right| + \left|f_n - f\right| \cdot \left|f\right| &gt; \eta \right\} \\     &amp;\leq \lim_{n\rightarrow \infty}\mu\left\{\left|f_n - f\right|\cdot \left|f_n\right| &gt; \eta/2 \right\} + \lim_{n\rightarrow \infty}\mu\left\{\left|f_n - f\right| \cdot \left|f\right| &gt; \eta/2 \right\} \end{align*}  By the fact that $\{f_n \cdot f\} \rightarrow f \cdot f$ in measure \begin{equation*} \lim_{n\rightarrow \infty}\mu\left\{\left|f_n - f\right| \cdot \left|f\right| &gt; \eta/2 \right\} = \mu\left\{\left|f_n \cdot f - f\cdot f\right| &gt; \eta/2 \right\} = 0. \end{equation*} I don't see how to deal with the other limit without first proving the most general statement $\{f_n \cdot g_n \} \rightarrow f \cdot g$ in measure.   Any help or advice is greatly appreciated. What do you call a person who keeps talking about art; studies art, criticizes art, and thinks he'd be amazing with it. But he doesn't practice it A gold reviewer badge? For your continued efforts in reviewing posts, even though the system sometimes messes up and doesn't let you actually review things and keeps displaying them over and over and over. So also, for your patience in putting up with the glitches in the review panel too! Haha  Anyways, a gold reviewer badge: 20000 reviews, over 4000 actioned  I just multiplied both numbers by 20. Think those numbers seem challenging? Is anyone up near those numbers or should it be multiplied by 10 instead? I prefer the higher amount because it would be gold and something harder to work for. :) But we also have to consider the longevity of when all these posts are eventually reviewed and it's only new posts being created.  Anyways, my stats for those who are interested (I wouldn't get the badge to start because I only have ~2000 actioned):  Upvoted            189 Downvoted          11 Edited             926 Commented          127 Voted to Close     44 Flagged            447 Reviewed           21609   As far as the name of the badge, I really like "The Owl is Watching" but not sure if that would be too long. Sounds snazzy though. Ideas?    Personally, I was on a quest to try and clear out all the posts in a single tab before it started showing me the same things over and over (and over) and then it got frustrating because I wasn't even making progress. Quest denied. :(    Lately I've noticed a massive build-up of unreviewed posts. When the review panel was first implemented, there would only ever be maybe 50 unreviewed posts at a time (in a given tab) for my monthly view. Now I log in and there's a thousand or two thousand pending. I think a lot of people quit reviewing after the first badge, so having a gold badge with a high threshold would really improve activity in the review panel and keep people going to improve the community. Can you teleport farther away with misty step if you have a double from Invoke Duplicity? The Channel Divinity feature says:     you can cast spells as though you were in the illusion's space, but   you must use your own senses. (PHB 63)   If you were where your duplicate is, you could step to a space 30ft from its position. So if it is 60ft from you, you could teleport to a space 90ft from your actual position. But the spell says that:     you teleport up to 30 feet (PHB 260)   This would put a hard limit on the distance you can travel.  Which rule takes precedence here? Number theory proof mod p For a given prime $p$ and integer $k$,  $k∈&lt;1, p-1&gt;$ $S(k)$ is defined as a sum of remainders of $k, k^2, k^3, ..., k^{p-1}$ when divided by $p$. Prove that $S(k)=\frac{1}{2}p(p-1)$ is true for an odd number of different k. I managed to prove that $S(p-1)=\frac{1}{2}p(p-1)$ but can't see how to do it further. Shining a laser onto a mirror Theoretically, if I shined a laser at a mirror at an angle of 0 degrees so that the light was perfectly reflected back to the light source, then I should not be able to see the light because it is not reflected to my eyes.  | |  me | |============ laser | | |   However this is evidently not the case, as I am able to see the laser point on the mirror. Why is this the case? My initial assumption is that it is due to diffuse reflection on the imperfect mirror's surface, but I am not sure. How to run firefox and firefox ESR simultaneously Since multiple inheritance is bad (it makes the source more complicated) C# does not provide such a pattern directly. But sometimes it would be helpful to have this ability. For instance I'm able to implement the missing multiple inheritance pattern using interfaces and three classes like that: public interface IFirst { void FirstMethod(); } public interface ISecond { void SecondMethod(); }  public class First:IFirst  {      public void FirstMethod() { Console.WriteLine(&quot;First&quot;); }  }  public class Second:ISecond  {      public void SecondMethod() { Console.WriteLine(&quot;Second&quot;); }  }  public class FirstAndSecond: IFirst, ISecond {     First first = new First();     Second second = new Second();     public void FirstMethod() { first.FirstMethod(); }     public void SecondMethod() { second.SecondMethod(); } }  Every time I add a method to one of the interfaces I need to change the class FirstAndSecond as well. Is there a way to inject multiple existing classes into one new class like it is possible in C++? Maybe there is a solution using some kind of code generation? Or it may look like this (imaginary c# syntax): public class FirstAndSecond: IFirst from First, ISecond from Second { }  So that there won't be a need to update the class FirstAndSecond when I modify one of the interfaces.  EDIT Maybe it would be better to consider a practical example: You have an existing class (e.g. a text based TCP client based on ITextTcpClient) which you do already use at different locations inside your project.  Now you feel the need to create a component of your class to be easy accessible for windows forms developers. As far as I know you currently have two ways to do this:  Write a new class that is inherited from components and implements the interface of the TextTcpClient class using an instance of the class itself as shown with FirstAndSecond.  Write a new class that inherits from TextTcpClient and somehow implements IComponent (haven't actually tried this yet).   In both cases you need to do work per method and not per class. Since you know that we will need all the methods of TextTcpClient and Component it would be the easiest solution to just combine those two into one class. To avoid conflicts this may be done by code generation where the result could be altered afterwards but typing this by hand is a pure pain in the ass. Blender 2.80 AV Sync missing? I just installed Blender 2.8 and for the life of me I can not find AV Sync anywhere and video editing has become impossible. Is there some new setting I am unaware of? Find all polynomials $P:\mathbb{C}\rightarrow\mathbb{C}$ such that $$P(x^2)=P(x)^2 .$$   Here is what I tried:  First, it is easy to see the constant solutions, namely $P\equiv 0,P\equiv 1$.  Let $r$ be a root of $P$ (i.e. $P(r)=0$). It follows that all terms in the infinite sequence $r,r^2,r^4,\dots,r^{2n},\dots$ are roots of $P$. To avoid a polynomial with infinite roots, we have to have the sequence be periodic. So it turns out that either $r=0$, or the roots of $P$ are roots of unity of degree $2n$ for some $n\in\mathbb{N}$.  In the first case, we get $P(x)=xQ(x)$. Plugging in, we get $$x^2Q(x^2)=x^2Q(x)^2\implies Q(x^2)=Q(x)^2 ,$$ i.e. that $Q$ satisfies the same condition as $P$.  In the second case, we have $P(x)=(x^{2n}-1)Q(x)$. Plugging in, we get $(x^{2n}-1)Q(x^2)=(x^{2n}-2x^n+1)Q(x)^2$. I was not sure how to proceed from here. ICANN charges 18¢ per domain name registered regardless of its extension, but I pay $10 for a domain name per year with my registrar.   What do domain name registrars do for its customers to justify this extra $9.82?  Is there something that they do for our domain names other than registering it with ICANN?   I just read about the 7 people who hold the ICANN database key and got a little bit curious to know what the domain name servers do for us to justify this extra amount (12 times the actual fee)? MSO Review Queue Privileges are Misleading When I visit Stack Overflow, I see the review option in the bar at the top of the screen.    However when on MSO, I don't see it. I assumed this was because I didn't have enough reputation to access the queue. I was curious what the requirement was on meta, so I moseyed on over to the privileges page. To my surprise, I saw that the privilege is listed as only requiring 500 rep and informs me I already have it:    Confused, I manually typed in meta.stackoverflow.com/review to see what would happen. The page did in fact load, but as you can see meta has no "First Posts" or "Late Answers" queue.    Seeing on that page that a minimum of 2k reputation is required to access any queues on MSO, I report the following bug:  1) On the privileges list on MSO, the 500 rep privilege of "Access first posts and late answers review queues" should not appear.  I also have the following feature request:  2) On the privileges list on MSO, it would be useful if instead there was a 2k privilege that informed you that you have access to the "Suggested Edits" and "Low Quality Posts" queues. For f continuous on $[0,1]$, show that there exist points $\alpha_k$ such that $\sum \limits_{k=1}^n \frac{1}{f'(\alpha_k)} = n $ Suppose that $f$ is continous on $[0,1]$ , differentiable on $(0,1)$ , and $f(0)=0$ and  $f(1)=1$.For every integer $n$ show  that there must exist $n$ distinct points $\alpha_1,\alpha_2\cdots,\alpha_n$ in that interval so that    $\sum \limits_{k=1}^n \frac{1}{f'(\alpha_k)} = n $   Is it theorem or i can show it with basic differentiation knowledge .Please give me hint . How do I enable hibernation on Wubi? I installed Ubuntu on my Windows 7 Sager laptop using Wubi. Hibernate (i.e. suspend to disc) is not an option from the power icon, only suspend, shutdown, etc. Hibernate is also not an option from my battery/lid close options. I understand that hibernation is disabled by default in Ubuntu 12.04. I tried running pm-hibernate but I get the following message:  Looking for splash system... none  s2disk: Snapshotting system   and then the computer just hangs with a black screen. According to the documentation  if this fails then I can't enable hibernate but it offers no help in making pm-hibernate succeed.  Could swap be my problem? It looks like I have very small swap:  user@ubuntu:~$ cat /proc/swaps Filename                Type        Size    Used    Priority /host/ubuntu/disks/swap.disk            file        262140  0   -1   The advice on  is only for the author's set up (e.g. I don't have an Ubuntu install disk since I used Wubi) and he says that 'INFO: This will not work for 12.04, resume from hibernate work differently in 12.04.'  Any advice? I really need to get hibernate working to use my laptop as a, er, laptop.  Thanks P-adic expansion construction "ImportError: No module named _struct" configuring python-minimal When I run sudo apt-get upgrade, the dpkg error appears and interupts the process.  Error details:  Setting up python-minimal (2.7.3-0ubuntu7.1) ... Traceback (most recent call last):   File "/usr/local/lib/python2.7/runpy.py", line 162, in _run_module_as_main     "__main__", fname, loader, pkg_name)   File "/usr/local/lib/python2.7/runpy.py", line 72, in _run_code     exec code in run_globals   File "/usr/local/lib/python2.7/compileall.py", line 16, in &lt;module&gt;     import struct   File "/usr/local/lib/python2.7/struct.py", line 1, in &lt;module&gt;     from _struct import * ImportError: No module named _struct dpkg: error processing python-minimal (--configure):  subprocess installed post-installation script returned error exit status 255 dpkg: dependency problems prevent configuration of python:  python depends on python-minimal (= 2.7.3-0ubuntu7.1); however:   Package python-minimal is not configured yet.   And this problem leads to a series of dependency problems when processing the following packages:    python  gwibber-service  libgwibber3  libgwibber-gtk3  gwibber  gwibber-service-facebook  gwibber-service-identica  gwibber-service-twitter  python-all  python-dev  python-all-dev  python-apt  python-problem-report  python-apport  python-libxml2  unity-lens-gwibber  unity-scope-video-remote   Finally, nothing I can do for this problem. And several days later, the root directory is full and I even could not uninstall any software because of this problem.