Rebel pronunciation
A term for words that change pronunciation with part of speech
What is this operator: &= I've had my Trek 3900 for about a year now, and I replaced the chain about 3 months ago, but otherwise it's running with the same parts it came with (not counting tubes and tires).  I had to replace my chain because it had gotten really rusty from riding in rain/mud/water on trails, and had started skipping.    My problem is, whenever I ride, if I get above or below a very light pressure when pedaling, the chain skips. I was wondering if this could be caused by my chain, which has gotten a little rusty but not as stretched as the other one was, or if I might need a new cassette?    I just readjusted my derailer, but it still skips.   Does anybody know what could be causing this? I would really love to play wizard/warlock multiclass with low charisma (just 13 for mutliclass). The handbook says that I can use Pact Magic slots to cast wizard spells, but doesn't specify which ability I use for spellcasting.     Pact Magic.   If you have both the Spellcasting class feature and the Pact Magic class feature from the warlock class, you can use the spell slots you gain from the Pact Magic feature to cast spells you know or have prepared from classes with the Spellcasting class feature, and [vice versa].   When I cast wizard spells using Pact Magic slots (or warlock spells using wizard's "Spellcasting" slots) do I use spellcasting ability of the slot, or do I get to choose? Given $m$ rows with $n$ variables where $m &lt; n$, and given that there exists a combination of some of the rows that gives a zero vector. Is it promised when computing the reduced-row echelon form, with Gaussian elimination to find a zero vector?  Update: I clearly don't know, how to pick up the right words to ask my question. So let me give you the full context and please help me say it right, so the question will make sense. I am solving step 3 in the quadratic sieve algorithm, here is a  description of it.  Suppose the composite number N is being factored. Bound $B$ is chosen, and the factor base is identified (which is called $P$), the set of all primes less than or equal to $B$. Next, positive integers $z$ are sought such that $z_2$ mod $N$ is B-smooth. Therefore it can be written, for suitable exponents $a_i$, $${\displaystyle z^{2}{\text{ mod }}N=\prod _{p_{i}\in &gt; P}p_{i}^{a_{i}}}$$ When enough of these relations have been generated (it is generally sufficient that the number of relations be a few more than the size of P), the methods of linear algebra can be used (for example, Gaussian elimination) to multiply together these various relations in such a way that the exponents of the primes on the right-hand side are all even: $${\displaystyle {z_{1}^{2}z_{2}^{2}\cdots z_{k}^{2}\equiv \prod &gt; _{p_{i}\in P}p_{i}^{a_{i,1}+a_{i,2}+\cdots +a_{i,k}}\ {\pmod {N}}\quad ({\text{where }}a_{i,1}+a_{i,2}+\cdots +a_{i,k}\equiv 0{\pmod &gt; {2}})}}$$  Now, let's say I found $m$ relations and the number of primes in $P$ is $n$, such that $m&lt;n$. It is given that there exists a combination of some of the $m$ relations that their product is a square number. My question is: When I perform the Gaussian elimination over F2 to find a null space combination, is it promised that I will find one? I successfully implemented the algorithm and the matrix part of my implementation works. But what I see is that if I found a null space(a combination of making a null space) and latter I add more relations(still $m &lt; n$), I lose the null space solution and not able to solve the matrix again. Each time I add a new relation I start solving it from the beginning, I don't remember any previous state. So what I want to know, could it be that there is a limitation with the Gaussian elimination algorithm, that it does not promise to find the null space if $m &lt; n$? How can I let a footnote continue on the next page? I have a rather long footnote which breaks a listing in two parts, unnecessarily.  Since the listing is more important then the footnote I'd rather lake to have the footnote split.   Can I customize (one or all) footnotes to have a maximal length before they are continued on the next page? Or can I manually split a footnote in two to get the effect? Subdividing a beveled mesh creates new vertices in wrong places Creating flow maps in QGIS I have several records from a survey and would like to show flow map from places mentioned in the survey. I have all places mentioned georeferenced and some records mentioned no places and others mentioned up to 16 places. I made a graphic so I can make myself clearer. Are there good reasons why it's a better practice to have only one return statement in a function?   Or is it okay to return from a function as soon as it is logically correct to do so, meaning there may be many return statements in the function? `du -sh` reports different used size than `df -h` Possible Duplicates:             df -h says that I have 494G used on /var:  Filesystem            Size  Used Avail Use% Mounted on /dev/sda3             564G  494G   41G  93% /var   But du -sh /var says it's 350G  What makes them different? Is there an easy way to get to www.stackexchange.com from any of the SE sites? One thing I've noticed about the relatively new sitewide layout is that I can't find a quick way to get back to www.stackexchange.com without finding StackExchange in the sites list. As a result, I am typing stackexchange in my toolbar a whole lot more often now. In particular, I tend to enjoy browsing the hot questions list and my own personal feed quite often so visit these pages on a daily basis. Is there an easy link to them that I'm missing?  (On the android app, there is a feed in the menu that I really appreciate, something like that on the desktop would be ideal!) Guests see the Protected notice , somewhat displaced:    Logged in, all is fine:    It actually belongs there, at the bottom, but then it has some design glitches, or it needs to be placed at the top.  Verified on Firefox and Chrome stable. How can I obfuscate (protect) JavaScript? How can editors and reviewers detect data manipulation? Efficiently replace all accented characters in a string? Calculate limit with L'Hospital's rule I want to calculate the limit $\displaystyle{\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}}$.   I have done the following:    It holds that $\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}=\frac{0}{0}$.   So, we can use L'Hospital's rule:  \begin{align*}\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}&amp;=\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x} \\ &amp;=\lim_{x\rightarrow 0}\frac{\left (x^2\cos \left (\frac{1}{x}\right )\right )'}{\left (\sin x\right )'} =\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )+x^2\cdot \left (-\sin \left (\frac{1}{x}\right )\right )\cdot \left (\frac{1}{x}\right )'}{\cos x} \\ &amp;=\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )-x^2\cdot \sin \left (\frac{1}{x}\right )\cdot \left (-\frac{1}{x^2}\right )}{\cos x} \\ &amp; =\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )+\sin \left (\frac{1}{x}\right )}{\cos x}=\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )+\sin \left (\frac{1}{x}\right )\right ) \\ &amp; =\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )\right )+\lim_{x\rightarrow 0}\left (\sin \left (\frac{1}{x}\right )\right )\end{align*}   We calculate the two limits separately   $\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )\right )$ :    \begin{equation*}\left |\cos \left (\frac{1}{x}\right )\right |\leq 1 \Rightarrow -1\leq \cos \left (\frac{1}{x}\right )\leq 1  \Rightarrow -2x\leq 2x\cdot \cos \left (\frac{1}{x}\right )\leq  2x\end{equation*} we consider the limit $x\rightarrow 0$ and we get \begin{equation*}\lim_{x\rightarrow 0} \left (2x\cdot \cos \left (\frac{1}{x}\right ) \right )=0\end{equation*}    How can we calculate the limit $\lim_{x\rightarrow 0}\left (\sin \left (\frac{1}{x}\right )\right )$ ? Participles to modify the subject of the main clause "Essos is an immense landmass located to the east of Westeros,   extending into the far east of the known world." (Here, "extending"   describes the Essos or does it describes the landmass? )   Having seen this sentence above, I question came to my mind.   Can I use participles to modify the subject of the main clause when there is no connection between main clause and participle clause. For example:      1- "He is a bookworm, living in Canada. (He is a bookworm + He lives in   Canada, there is no connection between being bookworm and living in   Canada)      2- He, living in Canada, is a book worm. (I suppose, this placement would be more correct) Show that $\frac{X}{X+Y}\sim Beta(\alpha,\beta)$ Let IG denote Inverse-Gamma distribution   .   If $X\sim IG(\alpha,1)$ and $Y\sim IG(\beta,1)$. Show that   $\frac{X}{X+Y}\sim Beta(\alpha,\beta)$   I tried with jacobian transformation taking $Z=\frac{X}{X+Y}$ and $W=Y$ then $$X=\frac{WZ}{1-Z};Y=W$$  $$\frac{\partial(z,w)}{\partial(x,y)}=\frac{1}{(1-z)^2}$$  $$f_{z,w}(z,w)=f_x(\frac{wz}{1-z})f_y(w)(1-z)^2$$ $$f_{z,w}(z,w)=\frac{1}{\Gamma{(a)}}(\frac{wz}{1-z})^{-\alpha-1}e^{-\frac{(1-z)}{wz}}\frac{1}{\Gamma{(\beta)}}w^{-\beta-1}e^{-\frac{1}{w}}(1-z)^2$$  but I'm stuck, in some place I read that $\frac{X}{X+Y}$ is a type 3 Beta distribution, but I can't show that. Is “all women who are mothers of daughters” right? Is the expression      All women who are mothers of daughters are here.   correct? I mean every woman who has at least one daughter is here. How do I backup my Stackexchange favorites for offline use? If $G/Z(G)$ is cyclic, then $G$ is abelian Continuing my work through Dummit &amp; Foote's "Abstract Algebra", 3.1.36 asks the following (which is exactly the same as exercise 5 in ):     Prove that if $G/Z(G)$ is cyclic, then $G$ is abelian. [If $G/Z(G)$ is cyclic with generator $xZ(G)$, show that every element of $G$ can be written in the form $x^az$ for some $a \in \mathbb{Z}$ and some element $z \in Z(G)$]      The hint is actually the hardest part for me, as the quotient groups are somewhat abstract. But once I have the hint, I can write:  $g, h \in G$ implies that $g = x^{a_1}z_1$ and $h = x^{a_2}z_2$, so \begin{align*}gh &amp;= (x^{a_1}z_1)(x^{a_2}z_2)\\\  &amp;= x^{a_1}x^{a_2}z_1z_2\\\ &amp; = x^{a_1 + a_2}z_2z_1\\\ &amp;= \ldots = (x^{a_2}z_2)(x^{a_1}z_1) = hg. \end{align*} Therefore, $G$ is abelian.  1) Is this right so far?  2) How can I prove the "hint"?