While watching Doctor Who, I found something that troubled me. The Daleks attempted to blow up the TARDIS; but the last time we saw the TARDIS explode it took the entire universe with it. There was another moment in which the Daleks attempted to destroy the TARDIS at the Crucible. Why did the Daleks attempt to destroy the TARDIS knowing it would blow up the universe and them along with it.
From the Cracks in time and space story arc it's known that if a TARDIS is blown up (possibly while in the Time Vortex) it rips creation itself apart. Also, from the Stolen Earth / Journey's End we learn the Daleks' masterplan which is subjugating creation itself via destroying it with the Reality Bomb (unfading Z-neutrino energy radiation).  In the Last Great Time War all sorts of temporal weapons were used and it's known that Daleks have the ability of time travel as well. Why didn't they use the Time Vortex back then to win the war once and for all via detonating a TARDIS while inside?
Quad tree vs Grid based collision detection Cases environment compresses fractions? what should I do to stop the cases environment compressing the first line of the following? \begin{equation} \alpha_{mn} = \begin{cases} \frac{16\mu^2}{mn\pi^2\left[\pi^2(m^2/a^2 + n^2/b^2) - \mu^2\right]} &amp;\mbox{for } m \mbox{ and } n \mbox{ odd;}  \\ 0 &amp;\mbox{for } m \mbox{ or } n \mbox{ even;} \end{cases} \\[1ex] \end{equation}  I'm not sure to do tex on this site, but if anyone tries it they see the first case is compressed because it's a fraction. proving that $f(x) = x^s$ is holder continuous with holder exponent s Boolean b1 = true; Boolean b2 = true; String b3 = 'b1 &amp;&amp; b2'; Boolean b4 = Boolean.valueOf(b3); system.debug('test Bool '+ b4);   Given a string how can you convert into expression without breaking the sting. How to select the architecture to download? Let $A$ and $B$ be two sets of real numbers. Define the distance from $A$ to $B$ by $$\rho (A,B) = \inf \{ |a-b| : a \in A, b \in B\} \;.$$ Give an example to show that the distance between two closed sets can be $0$ even if the two sets are disjoint. Prove that $A = B = C = D$ $A, B, C, D$ are sets, where $A \not = \emptyset$ and $B \not = \emptyset$. Equation $(A \times B) \cup (B \times A) = C \times D$ is true.  Prove that $A = B = C = D$.  I tried to find $A, B, C, D$ separately, but that didn't get me anywhere. Any ideas how to solve it? JavaScript post request like a form submit how implement the jqueryUI in visualforce Ill implement the some UI but its not working in the visualpage , can u know how implement jquery ui in viusalpage ,  Kindly tell the procedure ,     Im using this url its not working ,,,,, $$\lim_{n\to \infty}\frac1{2n}\log\left({2n \choose n}\right)$$  Now, $\log(n!) = \Theta (n\log(n))$ so I think we could write,   $$\lim_{n\to\infty}\frac{1}{2n}\left(\log\left(2n!\right) - \log\left(n!^2\right)\right) = \frac{1}{2n}\left(\log\left(2n!\right) - 2\log\left(n!\right)\right) $$  $$\lim_{n\to\infty}\frac{1}{2n}\left(2n\log\left(2n\right) - 2n\log\left(n\right)\right) $$  $$\log\left(2n\right) - \log\left(n\right) = \log(2)$$  Is this legitimate? I feel this might be wrong since, the $\Theta$ notation conceals a constant factor. Show that $\mu (E \setminus A)>0$ if $A$ is a nonmeasurable subset of $X$ and $A \subseteq E$, $E$ is measurable. This question was asked , but the question wasn't asking for an answer, just a clarification on the proof by contradiction they already had, so I'm reposting it to ask for help arriving at the answer.  Because the asker above said they proved it by contradiction, that is what I'm attempting. But I'm having issues wrapping my brain around the concept of set measurability. So let's give it a good college try with what I can understand.  For clarity, $\mu$ is an outer measure.  A is nonmeasurable, so $\exists B \subseteq X$ such that $\mu (B) \ne \mu (B \bigcap A) + (B \bigcap A^c)$.  E is measurable, so $\mu (B) = \mu (B \bigcap E) + (B \bigcap E^c)$.  This essentially means $\mu (B) \ge \mu (B \bigcap E) + (B \bigcap E^c)$ (Since monotonicity is alredy implied by the definition of outer measure).  So we can attempt a proof (or the start of one) by contradiction.  Assume $\mu (E \setminus A)=0$. So $E \setminus A$ is a null set.  So $\mu (B) \ge \mu (B \bigcap E) + (B \bigcap E^c)$. I assume I'm supposed to show a contradiction using that A is nonmeasurable. I'm unsure how.  Another way I think could work would be to use: $\mu (E \setminus A)= \mu (E \bigcap A^c) = 0$ and arrive at a contradiction given that A is nonmeasurable, but again, I'm unsure how.  I just need some direction from here. Thank you in advance for any advice. I am trying to import an Excel table with around 10,000 XY points into ArcMap 10.2. I am using State Plane NAD 1983 Alabama West (US Feet) as the Coordinate System in my map data frame. I then add the excel table to my Table of Contents in the map then select Display XY Data. Next, I choose the same coordinate system as my data frame then select OK. Once it creates the Events layer for my points they are not in the correct location.   Should I switch everything in my map to WGS84 or is there a solution for my points to be added onto my current map with the state plane coordinate system? In the PGF/TikZ manual, sometimes I see the option right of=somenode instead of right=of somenode. They look very similar, but the effects are different.  The distance between nodes positioned with the latter option is boundary-to-boundary, as stated in the manual. However, with the first option, it seems that the distance is shorter. I couldn't find any explanation of the first option in the manual.  My question is: what is the difference between the two options? Is there any explanation in the manual that I missed? Are "beginner questions" allowed on Stack Overflow? Prove that $z^2 \equiv ab\pmod p$ is solvable if and only if both or neither of $x^2 \equiv a \pmod p$, $x^2 \equiv b \pmod p$ are solvable. Suppose the $p$ is an odd prime not dividing $ab$.  Prove that $z^2 \equiv ab \pmod p$ is solvable if and only if both or neither of $x^2\equiv a\pmod{p}$ and $x^2 \equiv b \pmod p$ are solvable. Counting unique records across 2 columns and computing a score based on boolean values I have a table that looks like this in Google Big Query:  custID  InteractionDate Purchased?  Sales 1       20150312           F         0 1       20150312           T         200 1       20150314           T         150 2       20150221           T         400 2       20150221           F         0 2       20150221           T         120   ..that I need to transform to the following structure:   custID InteractionDate Success Sales  1     20150312          0.5    200  1     20150314           1     150  2     20150221         0.66    520   ..where in this second table the success column is defined as  count(purchase=Yes)/(count(purchase=Yes) + count(purchase=No)),   and sales is just the sum.  I'm quite new to BQ and no-SQL languages. I simply don't know how to do this, I've searched and seen examples that suggest using partition over but I simply haven't seen any example that does just this. Thanks a lot in advance. Does a monk's Unarmored Movement speed increase apply to the fly speed gained from a Fly spell? I installed Dev C++ using Wine on Ubuntu 18.04 and the program is compiling but it not showing any output screen. Why do employers require you to spread your 401(k) contributions throughout the year to get the maximum match? I think the title says it all. I can't see any reason employers have to incentivize employees to contribute to their 401(k) evenly throughout the year. Instead couldn't they use a relatively simple formula that is based on how much of their income the employee has contributed to their 401(k) thus far for the year? If that or their current contribution is over the threshold, they get the match this pay period, even if they didn't contribute anything.  EDIT: To clarify, my employer matches 100% of the first 3% of my contribution and 50% of the next 2%. The match is added every pay period along with my contribution. I am pretty sure there is no "true-up" feature, so if I were to hit the $17.5K limit early, and not be able to contribute at least 5% of my salary, I would not get the full match. So I have to be careful to make my contributions evenly throughout the year to get the maximum match. Just trying to figure out my employer's motivation for requiring this. I know that this sounds like Strange Days, which I haven't seen, but from the Wikipedia description I don't think that's it.  I saw this back in the early-to-mid-90's on TV, pretty sure it was a premium channel. It follows a research group who develops a way to record experiences. Not thoughts, but at least the sights, sounds, smells, touch, etc. They record it onto this reel-to-reel tape, it looked striated and golden in color. The protagonist is the head of the research team.  I don't remember a great deal of the plot, but there were conspiracies surrounding the invention. The lab clown decides to give it a whirl by putting the bike-helmet-sized apparatus on his head and going out to have a little fun, in the classic wine, women, and song fashion. Once he returns, an older member of the team finds the tape and finds a section where the clown is having sex. Later on, the protagonist, who is the lead researcher, finds the older member shut up in his house seemingly for days and nearly comatose. The older researcher had the playback device over his head and is non-responsive. Worried, the protagonist removes the helmet and puts it on himself, only to find that the older researcher had snipped a few seconds of tape and had them running on a non-stop loop; the moment of orgasm for the clownish researcher. The older researcher eventually comes to and awkwardly apologizes.  I think there were also experiments in recording the sensorum of a lab monkey, and that played into the plot. I remember the plot seemed fairly thin and dull for what seemed to be a pretty neat idea. I don't desperately feel the need to watch the movie, but I'm mildly curious if my impressions would differ 20 years later now.