html content added via jquery not getting methods (click,resizable) applied
Event binding on dynamically created elements?
Why do the major and minor groove exist in DNA? So I know that the major/minor groove arise from the offset pairing of the two strands in the helix.  All my textbook tells me is that "pitch causes the off-set pairing of DNA and this gives rise to the major and minor grooves."  I'm not too sure what this is saying.  What exactly is pitch and how does it give rise to the grooves? Let $V,W$ be nonzero normed spaces over $\mathbb{K}$ such that $V$ is finite-dimensional.  Let $E$ open in $\mathbb{K}$ and $p\in E$.  Let $f:E\rightarrow W$ be Gâteaux-differentiable at $p$.  Is $f$ necessarily Fréchet-differentiable at $p$ in this case?  I think this is not true in general, but cannot find a counterexample. What would be a counterexample? Moments of the truncated normal distribution (univariate) away from the mean Why can't we see the selected reasons when there are pending close votes for a question? When users vote to close a question, the original poster can't see the selected reason why until after the question is already closed.  If they could see this information, they might be able to edit the question to make it more acceptable and averting the need to close it.  For example, the owner of this  cannot see the reason why his question is being voted to close: Why is $R\cos{a} = mg$ in circular motion compared and not $R = mg\cos{a}$? Normally, if an object of mass $m$ is inclined to the horizontal at an angle $b$, we set the reaction force of the object on the inclined plane as $R = mg\cos{b}$ (if we resolve the force of gravity so the line of action coming out of the plane is perpendicular to it).  However in circular motion*. it's assumed that $R\cos{b} = mg$. In the example above, one would have to do this in order to arrive to the correct answer, instead of $R = mg\cos{b}$. Using $R = mg\cos{b}$ seems natural enough, as I am resolving vertically, however, both equations would produce two different values for $R$. Why is this?   To show what I mean: If we set the reaction force in this question as $mg\cos{a}$, then the centripetal force will be $mg\cos{b}\cos(\pi/2-b) = mg\cos{b}\sin{b} = \frac{1}{2}mg\sin(2b)$   Whereas If we use $R\cos{b} = mg$, $R = mg\sec{b}$ and the centripetal force will be $mg\sec{b}\sin{b} = mg\tan{b}$. This will end up with two different values for the radius of the circular motion, and hence two different final answers.  *In the circular motion questions I've seen in my mechanics module Misleading description of new Curious badge Suppress Chrome 'Failed to load resource' messages in console Does using an extender actually change the aperture of the lens? I just bought a Canon 2X extender and I always believed that it would only result in 2 stops' loss of light and not change the aperture.    However, when I tested this extender on a Canon 135mm f/2 lens, the maximum aperture I could set went down to F/4.  So it kind of confused me, because if I step down the aperture to f/5.6, is it actually setting the aperture to f/5.6 or f/2.8 (i.e., 1 stop below f/2)? How get eigenvectors without phase jump? Which rationals can be written as the sum of two rational squares? Which rational numbers can be written as the sum of two rational squares?  That is, for which rational numbers $a$, are there rational numbers $x$ and $y$ such that $a = x^2 + y^2$.  It is a famous theorem that if an integer can be written as the sum of two rational squares then it can be written as the sum of two integral squares, and then the solution is the famous one by Fermat, but I didn't find anything about the general case. Why was the previous Code of Conduct FAQ (and all answers) removed? I noticed that the  was deleted.   When the  came out, I tried to look for my old answer as well as others for comparison.  Why not just mark the old one superseded by [insert link]?  Note: I don't care about lost rep. There were a lot of useful links in answers and comments there (and some less-than-useful ones too).     Old CoC FAQ may be available  and . Source: (, ). I have the following element element(by.css(&quot;span[aria-label='Picker ID, TESTUSER']&quot;)  I want to assure the text on this filed is TESTUSER, so I tried:  expect(element(by.css(&quot;span[aria-label='Picker ID, TESTUSER']&quot;)).getAttribute('value')).toEqual('TESTUSER');  then I got the error: Expected null to equal 'TESTUSER' So I tried: element(by.css(&quot;span[aria-label='Picker ID, TESTUSER']&quot;)).getText().then(function (value) {         console.log(value);       })  And I got this error: Expected '' to be 'TESTUSER'.  How to get the value of this field and expect it is TESTUSER? There exists an injection from $X$ to $Y$ if and only if there exists a surjection from $Y$ to $X$. SharePoint Migration 2007 to 2013 Is zero odd or even? I came upon an interesting way to relatively quickly compute modular exponentiation with large numbers. However, I do not fully understand it and was hoping for a better explanation.  The method basically states if you have $x^y \bmod N$, it can be computed by repeatedly multiplying by $x$ modulo $N$:  $$x \bmod N \rightarrow x^2 \bmod N \rightarrow x^4 \bmod N \rightarrow x^8 \bmod N \rightarrow \ldots \rightarrow x^{2[\log y]} \bmod N$$  This is supposed to calculate in $\mathcal{O}(\log_2 N)$ time.  An example is given:  $$x^{25} = x^{11001_2} = x^{10000_2} * x^{1000_2} * x^{1_2} = x^{16} * x^8 * x^1$$  I don't really understand how this is faster. Aren't you just doing the same thing? You're just breaking down the exponents into representations of base $2$. Also where does $x^{2[\log y]}$ come from? If you're breaking it down into exponents of two where does a logarithm come from? (Aren't logs representations of numbers in exponents of $10$?)  I'm sorry if this appears to be a really stupid question - I have minimal knowledge in math.  Also, I wasn't sure if this belongs here or on stack exchange, so give me the word and I'll move it. How to get a cross-origin resource sharing (CORS) post request working How to save a copy of a multiplayer world in the new 0.7+ menus? How do you save a joined game after the 0.7.1 update? The menus have all changed since 0.6.x and I can't find the option to save a copy of a joined world anymore. (My friend has got a diamond house on survival and I want to save it but it doesn't let me!) How to prove $\left\{ \omega|X(\omega)=Y(\omega)\right\} \in\mathcal{F}$ is measurable, if $X$ and $Y$ are measurable? Given a probability space $(\Omega ,\mathcal{F} ,\mu)$. Let $X$ and $Y$ be $\mathcal{F}$-measurable real valued random variables.  How would one proove that $\left\{ \omega|X(\omega)=Y(\omega)\right\} \in\mathcal{F}$ is measurable.  My thoughts: Since $X$ and $Y$ are measurable, it is true, that for each $x\in\mathbb{R}:$ $\left\{ \omega|X(\omega)&lt;x\right\} \in\mathcal{F}$ and $\left\{ \omega|Y(\omega)&lt;x\right\} \in\mathcal{F}$.  It follows that $\left\{ \omega|X(\omega)-Y(\omega)\leq x\right\} \in\mathcal{F}$  Therefore $\left\{ -\frac{1}{n}\leq\omega|X(\omega)-Y(\omega)\leq \frac{1}{n} \right\} \in\mathcal{F}$, for $n\in\mathbb{N}$.  Therefore $0=\bigcap_{n\in\mathbb{N}}\left\{ -\frac{1}{n}\leq\omega|X(\omega)-Y(\omega)\leq \frac{1}{n} \right\} \in\mathcal{F}$.  Am working towards the correct direction? I appreciate any constructive answer! Proof a $2^n$ by $2^n$ board can be filled using L shaped trominoes and 1 monomino Suppose we have an $2^n\times 2^n$ board. Prove you can use any rotation of L shaped trominoes and a monomino to fill the board completely.  You can mix different rotations in the same tililng.