How do the Avengers know this fact?
In Age of Ultron, Thor seems to know something he possibly shouldn't?
Creating a responsive menu system for a site with a lot of content I am currently creating a website which has a lot of content - text. For this I need a serious menu structure with 5-10 menu items per menu and at least 15 menu-blocks per main menu item.   Further it has to be accessible via mobile.   Till now I have used panels and added menus inside the panels. But mobile isn't to happy about that and the list-names in the menus overlap the content.   How would you go about solving this issue? I trust you experts here! :-) Order of the product of two commuting elements with coprime orders in a group. I want to show that if $g,h\in G$ are group elements with finite coprime orders $m$ and $n$ and $gh=hg$ then the order of $gh$ is $mn$.    We have that  $$ (gh)^{mn}=g^{mn}h^{mn}=1 $$  using $gh=hg$, so $|gh|\leq mn$. On the other hand, $mn=\operatorname{gcd}(m,n)\operatorname{lcm}(m,n)=\operatorname{lcm}(m,n)$, since $m$ and $n$ are coprime. I am not sure how to proceed, I tried assuming that $|gh|&lt;mn$ and arriving at a contradiction, but it did not take me anywhere. Are two norms equivalent if they induce the same topology on a vector space? I recieve the below error when I am trying to call a Custom Lightning Component in a Visualforce page that is used in Communities.    Unrecognized Content-Security-Policy directive 'reflected-xss'.   logo180.png Failed to load resource: the server responded with a status of 404 (Not Found)   CommunityHomePage:1 Uncaught SyntaxError: Unexpected token w in JSON at position 0     at JSON.parse ()     at XMLHttpRequest.xhr.onreadystatechange (lightning/lightning.out.delegate.js?v=ZCuyRUYzK_kAf4cKCIeS2A:143)   My Component  &lt;aura:component controller="LeadController" implements="forceCommunity:availableForAllPageTypes"&gt;     &lt;aura:attribute name="theLead"          type="Lead"          default="{ 'sobjectType': 'Lead',                    'Rating__c': 2,                    'Company': '',                    'FirstName': '',                    'LastName': '',                    'Email': '' }"/&gt;     &lt;form&gt;          &lt;ui:inputText value="{!v.theLead.Company}" label="Company" class="input"/&gt;         &lt;ui:inputText value="{!v.theLead.FirstName}" label="First" class="input"/&gt;         &lt;ui:inputText value="{!v.theLead.LastName}" label="Last" class="input" /&gt;         &lt;ui:inputEmail value="{!v.theLead.Email}" label="Email" class="input" /&gt;         &lt;ui:button label="Save" press="{!c.save}"/&gt;     &lt;/form&gt; &lt;/aura:component&gt;   My Lightning App is   &lt;aura:application access="GLOBAL" extends="ltng:outApp" implements="forceCommunity:availableForAllPageTypes"&gt;     &lt;aura:dependency resource="c:CommunityChatter"/&gt; &lt;/aura:application&gt;   My VF Page is  &lt;apex:page standardController="Lead"&gt;     &lt;apex:includeScript value="{!$Site.BaseUrl}/lightning/lightning.out.js"/&gt;     &lt;script&gt;         $Lightning.use("c:MyLightApp", function() {             $Lightning.createComponent(                 "c:CommunityChatter",                 {},                 "theLead",                 function(cmp) {                     console.log("Component created!");                     console.log(cmp);                 });             });     &lt;/script&gt;      &lt;div id="theLead"&gt;&lt;/div&gt; &lt;/apex:page&gt; I don't have complex analysis at my beck and call, and I only have a low level of knowledge in topology, but I need to prove that this metric space (for any real $r$ and $R$ with $r &lt; R$)$$ X = \{ (x, y) \in \mathbb{R}^2 \ | \ r \leq x^2 + y^2 \leq R \}$$  with the Manhattan metric $d((x_1, y_1), (x_2, y_2)) = |x_1-x_2| + |y_1-y_2|$ is not simply connected.  I've already prooven that it's path connected, and now I need to show there are some points $P$ and $Q$ with two paths between them such that one cannot be continuously 'morphed' into the other.  $ \ $  What I have so far is as follows:  I take $P = (0, r)$ and $Q = (0, -r)$, with $f_0$ being a path from $P$ to $Q$ going clockwise around the circle radius $r$ and $f_1$ being much the same but going counterclockwise.  Now I assume there is a function $g : [0, 1]^2 \to X$ such that:   $g(s, 0) = f_0 (s)$ $g(s, 1) = f_1 (s)$ $g(0, t) = P$ $g(1, t) = Q$   To get the final result, I need to show that this function cannot be continuous, but for the life of me I cannot.  For some context, these are the topics which have been visited during the course, roughly in order of recentness.   Multiple connectedness Simple connectedness Pathwise connectedness Interior points Boundary points Open sets Compactness Complete metric spaces Bounded metric spaces Totally bounded metric spaces Closed sets Closure of a metric space Limit points Cauchy sequences Convergence Continuity Metric space quivalence Metric equivalence   Edit: I might have an argument that works, though it's far from rigourous. We can shrink $R$ to be as close to $r$ as we want, so we can essentially constrain the annulus down to a circle and thus force any path from the left side of the circle to the right side to go through $P$ or $Q$.  So holding $s \in (0, 1)$ constant and varying $t$ must produce a path through $P$ or $Q$ for any $s$. If for some $s$ it passes through $P$ and for some other $s$ it passes through $Q$, then there must be $s_0$ such that $\forall \epsilon &gt;0  \ \exists \delta \leq \epsilon$ st. $s_0$ produces a path through $P$ and $s_0 + \delta$ produces a path through $Q$.  Now we can consider the path $g(s, \frac{1}{2})$, and note that it must have a discontinuity at $s_0$.  Now consider the case where all $s$ produce paths through only one of $P$ or $Q$. WOLOG: $P$. Now, at $t = \frac{1}{2}$, $s$ arbitrarily close to $1$ are mapped away from $Q$, but $1$ is always mapped to $Q$ by definition, so $g(s, \frac{1}{2})$ has a discontinuity at $s=1$. Therefore $g(s, t)$ is not continuous.  This argument is definitely iffy to me, if no one has their own argument (using sufficiently low level concepts), then criticism on the above would be appreciated. Why are home appliances and other devices made to operate on dc? Can grep output only specified groupings that match? Say I have a file:  # file: 'test.txt' foobar bash 1 bash foobar happy foobar   I only want to know what words appear after "foobar", so I can use this regex:  "foobar \(\w\+\)"   The parenthesis indicate that I have a special interest in the word right after foobar.  But when I do a grep "foobar \(\w\+\)" test.txt, I get the entire lines that match the entire regex, rather than just "the word after foobar":  foobar bash 1 foobar happy   I would much prefer that the output of that command looked like this:  bash happy   Is there a way to tell grep to only output the items that match the grouping (or a specific grouping) in a regular expression? Android OS without Google Services? How to change the Play Store language? does not appear to actually define what a rigid shape is. Rather it defines 'same shape' and 'rigid transformations' without giving any definitions of what is necessary and sufficient for a shape to be considered rigid.  For instance, I've seen the following image:    I understand intuitively why the triangle is rigid and quadrilateral is non-rigid. It is also my understanding that inserting a single diagonal connection into the quadrilateral would make the shape rigid? However, given the following image (labeling the shapes $S_1$, $S_2$ and $S_3$ in sequence left to right):    If $S_1$ had a single diagonal connection AB, it would be possible to flip ADB over axis q to obtain a shape similar to $S_2$ with the addition of connection AB (my explanation may not be the greatest). This shape is clearly not the same as $S_1$, so how can we say $S_1$ with a single diagonal connection CD is rigid? I feel I am unclear on what the definition of rigidity actually is. Once upon a time,  DISPLAY=:0.0 totem /path/to/movie.avi   after ssh 'ing into my desktop from my laptop would cause totem to play movie.avi on my desktop.  Now it gives the error:   No protocol specified Cannot open display:    I reinstalled Debian squeeze when it went stable on both computers, and I guess I broke the config.   I've googled on this, and cannot for the life of me figure out what I'm supposed to be doing.   (VLC has an HTTP interface that works, but it isn't as convenient as ssh.)  The same problem arises when I try to run this from a cron job. Pytest is in PATH but not found Conditionals in English are usually formed by using if with normal word order; but for the three past (subjunctive) forms were, should, and had, it is also possible to express the conditional through subject–auxiliary inversion alone, with no if in the conditional clause.  Does forming conditionals in this manner differ semantically from if-conditionals? Is some aspect of the conditional statement or the conditionality emphasised more in one version than in the other? Or are there differences in how and when they are used? Or are they simply completely interchangeable?  For example:     If I were you, I would try it again   Were I you, I would try it again      If I had seen it, I would have told you   Had I seen it, I would have told you      If you should drink, don’t get behind the wheel   Should you drink, don’t get behind the wheel Activity has leaked window com.android.internal.policy.impl.PhoneWindow$DecorView@44f72ff0 that was originally added here How should you interview for QA *Automation* positions? Why use apparently meaningless do-while and if-else statements in macros? I am running Windows 8.1 N Pro x64 and there are 2 Internet Explorer processes that must be for 32 bit. It stays there even if Internet Explorer is closed. I tried ending the processes but they just come back. I run a x64 Internet Explorer. How can I stop those processes? How do I sort a std::vector by the values of a different std::vector? Algebric proof for the identity $n(n-1)2^{n-2}=\sum_{k=1}^n {k(k-1) {n \choose k}}$ Prove the identity:  $$n(n-1)2^{n-2}=\sum_{k=1}^n {k(k-1) {n \choose k}}$$  I tried using the binomial coefficients identity $2^n = \sum_{k=1}^n {n \choose k}$  but got stuck along the way. Can I connect to a specific access point by BSSID? Is there any way to get Mac OS X to connect to a specific access point by BSSID? I have a network with the same SSID for multiple access points, and I would sometimes like to specify which specific access point to connect to. Is this something Mac OS X can do natively, or is there a third party tool that can do this?