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So I've got the shout in my shouts list and I've unlocked all three parts of it, but I can't seem to be able to use it all, just the first part.  Am I missing something? I still have a quest for 'Learn the first word of Durnehviir's shout' but it's showing as me having already got it.  What gives? What is the preferred way to find focused control in WinForms app? I made the jump from Jumsoft's Money app to GnuCash a while back. It appealed to all my OCD tendencies. I don't know how to use Perl, or stuff of the sort, but with detailed instructions I manage to get things done... when I can find them.  I set up Finance::Quote to pull quotes after I set up GnuCash. I even managed to get it to pull quotes for Korean funds. And every thing had been working well, until tonight. I only get back a pop-up that states "There was an unknown error while retrieving the price quotes." That's where I need help. What can I do?  I dug around and found that it seems that it has something to do with Yahoo Finance ending it's API support. I also learned that Finance::Quote can pull data from many sources, so I don't really understand why one missing API would break it. I also saw that someone out there is trying to use AEX to pull the quotes instead for GnuCash specifically. I found Finance::Quote on cpan (I also first heard of cpan), but I don't understand what I'm looking at. But that's all beside the point. I don't know how to fix this.  Any help would be appreciated. I'm on Windows 10, GnuCash version 2.6.17, and have Strawberry Perl 5.18.2.2-32bit installed.  edit Thanks everyone who responded. It seems to be working more or less. The currency and Korean funds still haven't updated, but you've shown this horse to water. I'll do my best to drink it now. Plurals of acronyms, letters, numbers — use an apostrophe or not? When I was in high school back in the 1970s, I was taught that to make a plural of an acronym, a letter, or a number, one should add an apostrophe and "s". Like I would have written this sentence, "... back in the 1970's ..." I would write "one CD, two CD's". Etc. I followed this rule faithfully for years until a co-worker told me it was wrong. Now I can't find any source that agrees with what I was taught.  Is this a rule that has changed over time? Was the convention in the 70s that one should use an apostrophe but this has changed and now we do not? Or were my high school English teachers just confused? Order of "not" with infinitive This is one thing that keeps bugging me, and maybe there's a direct answer. Grammatically, which one is more correct of these two?  Does it make a difference?  I tried not to do that. I tried to not do that. If  $\mathbb R^*$ and $\mathbb C^*$ be respectively multiplicative groups of non-zero real and complex number then whether $\operatorname{Hom}(\mathbb Z_6,R^*\oplus C^*)$  is  isomorphic to $\mathbb Z_{12}$? or $\mathbb Z_6$? or $\mathbb Z_6\oplus \mathbb Z_2$? Given two positive integers $a$ and $b$, prove that  $\sum_{k=0}^{n} {k \choose a} {n-k \choose b} = {n+1\choose a+b+1}$  I think there should be a good combinatorial proof for this, given the simplicity of the right-hand side... my hunch is to form all possible subsets of size $a+b+1$ by dividing the set into $k$ and $n-k$ elements, then choosing $a$ from the first, $b$ from the second, iterating over all values of $k$.  I can't quite make this work, so I've also been trying to rewrite ${n+1 \choose a+b+1}$ to make it clear.  Open to any proof method, all help appreciated. Why does factoring eliminate a hole in the limit? How can one prove the Bohr-Sommerfeld quantization condition $$ \oint p~dq ~=~2\pi n \hbar $$ from the WKB ansatz solution $$\Psi(x)~=~e^{iS(x)/ \hbar}$$ for the Schroedinger equation? With $S$ the action of the particle defined by Hamilton-Jacobi equation $$ \frac{\partial S}{\partial t}+ \frac{(\nabla S)^{2} }{2m}+V(x)~=~0 .$$ Kinetic energy conservation in a collision Kinetic energy conservation in collision.  In any type of collision type of collision if we take the system to comprised of both bodies then the net external force is zero. So the work done by external force is also zero. By the relation $W=K_f-K_i$ as $W$ is zero hence kinetic energy is conserved. But this not the case.   Can you tell me what is the flaw in my argument? Some people say that the argument has neglected the fact that some kinetic energy is transformed into heat. But heat is created due to the interactions between bodies during collision which is thus an internal force and thus shouldn't affect kinetic energy. I have got myself a dxf which is built with units in the wrong scale. How do I upscale that in QGIS?   Note it does open in QGIS, just in the wrong map units and location. How would one denote the set of all subsets of $A$ which have size $2$?  Would  $$\binom{A}{2}$$  be a good idea? Prove equality between binomial coefficients. Time complexity of nested for-loop I need to calculate the time complexity of the following code:  for (i = 1; i &lt;= n; i++) {   for(j = 1; j &lt;= i; j++)   {    // Some code   } }   Is it O(n^2)? I've already done a lot of searching (in particular: ) to try to prove this statement:      $f: A \to B$ is injective if and only if it has a left inverse. [I'm going to also assume $A$ and $B$ are nonempty.]   My attempt so far:     $\Leftarrow$ Suppose $f$ has a left inverse and $f(a_1) = f(a_2)$ for    $a_1, a_2 \in A$. Then there is a function $g: B \to A$ such that   $g\left(f(a_1)\right) = a_1$ and $g\left(f(a_2)\right) = a_2$. Since   $g$ is well-defined, it follows that $g\left(f(a_1)\right) =  g\left(f(a_2)\right)$, and hence $a_1 = a_2$.         $\Rightarrow$Now   suppose $f$ is injective. Then $f(a_1) = f(a_2) \implies a_1 = a_2$.   Furthermore, since $A$ is nonempty, there is an element $x \in A$.   Define $g: B \to A$ by  \begin{equation*} g(b) = \begin{cases}  f^{-1}(b), &amp; b \in \text{Im}(f) \\ x, &amp; b \in A \setminus  \text{Im}(f)\text{.} \end{cases} \end{equation*} This mapping is   well-defined, since if $f$ is injective, $f^{-1}(b)$ is unique.   My main question: does this imply that if $f: A \to B$ is injective that any mapping $g: \text{Im}(f) \to A$ is bijective? Otherwise, I don't understand why we can even use $f^{-1}$ on an element. [Please read the link above for more details - in proof 1.] I was just reading Harry Potter and the Deathly Hallows, and I was wondering, when Hermione Granger changed into Bellatrix Lestrange, she should've obtained the Dark Mark, temporarily.  Now, my question is: would the Dark Mark work if pressed? Can you recommend a Java library for reading (and possibly writing) CSV files? $a\mid b,\ c\mid d\,\Rightarrow\ ac\mid bd $ $\ \, \bf\small [Divisibility\ Product\ Rule]$ I just need to check the reasoning in my proof is correct, I think it is valid although I'm not totally convinced because I can't follow the logic; does proving that $x$ is an integer prove that $ac|bd$?  Theorem: Let $a$, $b$, $c$, $d$ be integers. If $a|b$ and $c|d$, then $ac|bd$.  Proof: $aj=b$ and $ck=d$ for integers $j$, $k$. Then, $ac|bd$ implies $acx=ajck$ and thus $x=jk$ for some $x$. Since the product $jk$ is an integer, $x$ is an integer and thus $acx=ajck$ and thus $ac|bd$. Sometimes when adding an object to the scene I see that all options (on toolbar or in window after pressing F6) are disabled. Can anyone tell me when can this happen? What is causing it and how to revert it - make it enabled again. It worked for me when I restarted Blender and reloaded the scene. However I would really like to know what can cause it? Are there any rules in baseball which prohibit teams from "borrowing" players via trade? Every season near the trade deadline, we see teams that are out of it trading expensive players and/or free agents-to-be, typically for younger players or prospects.  For example, Matt Garza, set to become a free agent after the season, was traded to the Rangers for a few prospects.  It made sense for the Cubs to trade Garza, because he wasn't under contract any further than this season, and thus doesn't have much value to them.  Compare Garza to someone like Mike Trout, who is likely the most valuable property in baseball, given his skill and inexpensive cost over the next several years.  There's no way the Angels would trade him, because he'll be the cost-controlled centerpiece of the team for years to come.  However, with the Angels out of it this year, what if they wanted to trade Trout to a contender (say the Pirates), but only for the remainder of this season?  That is, they'd get several top prospects now, and would have an agreement with the Pirates that they'd trade him back in the offseason (for something pretty worthless, say for a journeyman minor leaguer).  The Pirates get Trout for the stretch run and are happy to give up a few prospects.  Trout doesn't have much value to the Angels this year, but he has a ton of value for next year.  So they get a few prospects for him, in exchange for the part of his career which has very little value to them.  So in effect, the Pirates are renting or borrowing him for the remainder of 2013.  Why don't trades like this happen?  Are there any rules prohibiting trades like this?