I have a question. How come that the flight from Toronto to Moscow takes the same time as the flight from Moscow to Toronto. Should it take much longer because of the earth rotation?  One direction we travel in direction or an orbit and traveling back we are going against the Earth orbit...  From Toronto 9hours From Moscow 9:45 hours.  If flying from Moscow we are traveling against the Earth rotation so we should add half of the travel to our current travel. It means that the travel from Moscow should take about 13 hours...  Thanks,
Say I'm flying from Sydney, to Los Angeles (S2LA), back to Sydney (LA2S).  During S2LA, travelling with the rotation of the earth, would the flight time be longer than LA2S on account of Los Angeles turning/moving away from our position?  Or, in the opposite direction, would the flight to Sydney be faster since the Earth turns underneath us and moves Sydney closer?  ===   Please ignore jet stream effects and all other variables; this is a control case in an ideal environment. By "dramatically" I suppose I mean a delay of 1 hour or more.
Vertical line for the the right margin in Notepad++ How to downgrade a package via apt-get? I'm trying to do some scenes which use a mixture of 3D backgrounds with 2D animated characters.The best way I've figured this out so far is to use a plane with an image sequence of pngs which hold the frames.  The problem with this method is Blender seams to only load image sequences based on image numbering. If I wanted to do a sequence which loops say, frames 0-5, then repeats 4-0 for the 'return' of the animation, I would have to load frames doubled up.  Is there a way, maybe with a script, to change the sequence order? I can see it as a useful plugin but my python skills are pretty much non existent. $P(\limsup A_n)=1 $ if $\forall A \in \mathfrak{F}$ s.t. $\sum_{n=1}^{\infty} P(A \cap A_n) = \infty$ Let $\mathfrak{F} = (A_n)_{n \in \mathbb{N}}$. Prove that $P(\limsup A_n)=1$ if $\forall A \in \mathfrak{F}$ s.t. $P(A) &gt; 0, \sum_{n=1}^{\infty} P(A \cap A_n) = \infty$. (Side question 1 Is second Borel-Cantelli out because we don't know if the $A_n$'s are independent?)  Suppose $\forall A \in \mathfrak{F}$ s.t. $P(A) &gt; 0, \sum_{n=1}^{\infty} P(A \cap A_n) = \infty$, but $P(\limsup A_n)&lt;1$. My prof gave us this convenient hint: Show $\exists M &gt; 0$ s.t. $P(\bigcap_{j=M}^{\infty} A_j^{c}) &gt; 0$.  $1 &gt; P(\limsup A_n)$  $1 - P(\limsup A_n) &gt; 0$  $P(\liminf A_n^{c}) &gt; 0$  By definition of $\liminf A_n^{c}$, $\exists M &gt; 0 $ s.t. $P(\bigcap_{j=M}^{\infty} A_j^{c}) &gt; 0$. (Side questions 2 and 3 Isn't this =1? ...Wait, is that what we're trying to prove? Haha)  $\to 1 - P(\bigcup_{j=M}^{\infty} A_j) &gt; 0$  $\to 1 &gt; P(\bigcup_{j=M}^{\infty} A_j)$  Main question: Now I guess we have to come up with some A s.t.  $\to 1 &gt; P(\bigcup_{j=M}^{\infty} A_j) \geq \sum_{n=1}^{\infty} P(A \cap A_n)$, or is there something else to do? PXE Boot Server Installation Steps in Ubuntu Server VM Today I came across an OBVIOUSLY good answer in /review. It had 7 upvotes, contained lists, paragraphs of explanation, and well-applied sample code. It was also moderately long. How does a question like this get flagged for review? I don't really know how the algorithm works.   If this is low-quality then I feel bad about all of my answers.  Here's a  if you care to see it in context. It's not really a huge problem (as Bart's comment emphasizes), but this seems like an extreme outlier in every measurable aspect. Harmonic Function with linear growth Is it possible to find the origin of a virus? Is there any method with which you can determine the origin for a given piece of malware in Windows?  One of my PCs was recently infected with the  password stealing trojan. Is there any way to tell from what source this virus was downloaded from? Is It Possible to NSLog C Structs (Like CGRect or CGPoint)? I want to be able to debug C structures without having to explicitly type every property that they consist of.  i.e. I want to be able to do something like this:  CGPoint cgPoint = CGPointMake(0,0); NSLog(@"%@",cgPoint);   Obviously the '%@' won't work, hence the question. How do I get a mesh data-block with modifiers and shape keys applied in Blender 2.8? I'm trying to update an exporter to work with Blender 2.8 but can't seem to get a mesh data-block modified by modifiers or shape keys. It always gives me the base mesh.  I'm currently using Object.to_mesh(). But reading Object.data or bmesh.from_mesh(obj) all returns only the base mesh. So what is the new function to call for getting a mesh data-block with the modifiers and shape keys applied? I'm sorry if this is a broad question but can someone explain to me how to interpret these regression diagnostic plots? I understand the Normal Q-Q show's how normal the spread of the data is, but the other three I am confused by. For the Residuals vs Leverage, what does it mean for a point to be outside of the bounds? What does it mean if the red lines are parabolic and curved like in the first two? Why do we care about residuals, and what do they tell us? I've searched the web multiple times for some guide to interpret these and I can't get any straight forward answers. Any help is appreciated, thank you. Photoshop CS6 Resize Rectangle with Rounded Corners This question is about Photoshop CS6. Hopefully the feature was added in this version. Before posting I looked up Google and  and  but didn't find an answer for CS6.  So the question is: Is there a feature within CS6 or a script that allows to resize a vector element with rounded corners without using the Direct Selection Tool (white arrow tool) and not changing the border radius? What would be great would be something like simply using the transform tool or something like a grid resize tool. How does the Girl-Genius web-comic relate to the Agatha H. books? After spending some time reading  I was convinced to buy the book.  I've ordered what I think is the  in my local bookstore (yay local) and it seems to relate to the first one or two years of of web content, but I can't be sure.  My question is: "how does the book series line up with the Web comic?"  Does it contain specific episodes/adventures, relate to strips from a specific time-span or is it simply an assortment of episodes?  UPDATE: They are amazing.  It is interplanetary war in the introduction!!!!  Yes! If we disbelieve $H_0$, why quote a p value calculated assuming $H_0$ was true? Hypothesis testing seeks to reject a null hypothesis ($H_0$) on the basis of an assumption made about the sample following a certain distribution. This assumption is conditional on $H_0$ being true. So what is the flaw with the following reasoning?     If I reject $H_0$ in favor of the alternative hypothesis ($H_1$), I   must then assume that (1) $H_1$ is true, and (2) that at the   same time the variable under examination follows the distribution   used to reject $H_0$. Since we also stated that the assumed   distribution is conditional on $H_0$ being true, it appears that $H_0$   has got to be true in order to justify the assumption made about the   distribution. Logically, it would therefore seem that both $H_0$ and   $H_1$ have to be true, which is a paradox, because they are mutually   exclusive! Do plugin files have to follow a specific convention to be "picked up" by WordPress? I need to solve $3x^2 + 6x +1 \equiv 0 \pmod {19}$  I saw the same problem here -     but didn't understand how he got to the conclusion   $x(x+2) \equiv 6 \pmod {19}$  and anyway im trying to solve it how we learned in class -   multiply both sides and the modulo by $4a$ then solve the two equations  $y^2 \equiv b^2 -4ac \pmod {4an}$  $2ax + b \equiv y \pmod {4an}$  so I tried multiplying the hole thing by $4a$ , that is $4 \times 3$  and got to $(2 \times 3x + 6)^2 \equiv 36 -4 \times 3 \pmod {19 \times 4 \times 3}$  now I am stuck . any help will be appreciated Is there any efficient algorithm for finding subgroups of a given finite group? I am implementing an algorithm which finds every subgroup of given group.  Here's my algorithm.  Let $G$ be a group of order $n$ with elements $g_1,\cdots,g_n$.  Then I consider each $\langle g_i\rangle $ which are all cyclic subgroups of $G$.  Of course, if $\langle g_i\rangle =\langle g_j\rangle $, we append only $\langle  g_i\rangle $ into our result set.  We call all distinct subgroups of the form $\langle g_i\rangle $ to be the first-generation. (Note that the number of the first-generation subgroups is at most $n$.)  To get $(i+1)$-th generation from $i$-th generation inductively, we construct the subgroup $S$ from each two elements of $i$-th generation $S_1, S_2$ to be  $$S=\langle S_1,S_2\rangle .$$  Obviously, if generated subgroup is already in previous generations, we do not call it $(i+1)$-th generation, hence every subgroups of $G$ has the unique generation number.  We do this construction until the number of the typical generation subgroups to be $0$, since we get all subgroups of $G$.  This works perfectly to any groups, but unfortunately, it is not fast although I implemented additional steps to avoid repeated computation.  So I searched some other algorithms but they are all vague to understand.  One of the algorithm is called the cyclic extension algorithm, but I can not get it easily unless I have studied some backgrounds for this method.  If you have enough knowledge about the cyclic extension algorithm or some other algorithms, please give me an easy explanation. I am in undergrad. course of math dept. Thanks. Why does getopts only work the first time? How to save the render result within Blender Is it possible to render an image and have blender remember it, so that next time I open the file the render result will not be blank?  I am not talking about saving to an external file.  Specifically I would like to render an image, close blender, and come back later and be able to do some compositing on the render without having to waste a half an hour rendering it again (my GPU drivers are messed up so I have to use the CPU).  Currently the render result is blank when ever I open a file. Do I need a visa to transit (or layover) in the Schengen area?