I used my Philippine passport to book an airline ticket, but I'm also an American citizen. Will this create problems with immigration?
Answers to a  mention that it is illegal for a US citizen to enter/leave their home country on a foreign passport. However this raises the following questions:   What is the maximum prescribed penalty for violating these rules? Are US immigration officials known for actually applying the penalty? E.g. the Cuba embargo banning visits to Cuba was almost never enforced, so it's possible this law is ignored as well.
Are those stars moving past the window when the ship is at warp? When the enterprise (ENT) is moving at warp we see numerous stars flying past in the windows:    However, according to , at warp 5 (the maximum speed of the 1st enterprise) it would take 7.4 days to get from the Sun to Alpha Centari, our nearest star(s) approximately 4 light years away. So to have stars whizzing past every few seconds would imply that the stars are very close together.  Is this an inaccurate depiction of star movement at warp, or is there some explanation like the warp bubble creates a lensing effect?  Now I realise the classic answer is "suspend your disbelief, it's a tv show", but the animators have at least thought about the star movement as they have added in a bit of red shift and blue shift which you can see in the image. Fix for Hibernate error "Use of the same entity name twice" How you fix the following Hibernate error:  What does "Use of the same entity name twice". Using the Pigeonhole Principle to show that $2$ of any $n+1$ numbers from $\{1,2,\ldots,2n\}$ sum to $2n+1$ Intuition for why the sum of Bernoulli trials is a sufficient statistic via cancellations on numerator and denominator? Suppose that $X_1, \ldots, X_n$ are iid Bernoulli trials with parameter $p$, and that $Y = \sum_{i=1}^nX_i$. Then, we have that $Y$ is a sufficient statistic by the following condensed proof:  $$ P(X_1 = x_1,  ... , X_n = x_n |Y = y) = \frac{p^y(1-p)^{n-y}}{\binom{n}{y}  p^y(1-p)^{n-y}} =\frac{1}{\binom{n}{y}}   \text{ if } \sum_{i=1}^{n}x_i = y $$  I am trying to see why this is the case. My reasoning is that there is some cancellation property on top and bottom that "kills" off the parameter $p$. In that whatever is on top through $p$ is perfectly matched by whatever is on the bottom minus scaling factors (in this case $\binom{n}{y}^{-1}$).   However, I am wondering what the deeper meaning behind this is, that the numerator of the joint distribution of the data and the sufficient statistic has its EXACT form on the bottom, except for constants which are only multiplicative (to facilitate cancellations). Am I onto something bigger and deeper here? What file sizes and formats are most appropriate for enlarging and projecting an image onto a large wallspace for painting? Running out of cards in Exploding Kittens In the Exploding Kittens game, the rules say     You won't ever run out of cards in the Draw Pile, so there is never any need for a reshuffle.   In every game I've played all the cards end up on the discard pile fairly quickly and there is nothing left in the draw pile, so we have to reuse them.  Is this meant to happen? Are we doing something wrong? Prove equation with binomial coefficient To show that $\sum a_nb_n$ is absolutely convergent Assume that $\sum a_n$ is convergent and $\sum b_n$ is absolutely convergent.To show that $\sum a_nb_n$ is  absolutely convergent  My try::Consider the sequence of partial sums of $\sum |a_nb_n|$  $S_n=|a_1b_1|+|a_2b_2|+......+|a_nb_n|$  Now $(a_1b_1+a_2b_2+......+a_nb_n)\leq ( (a_1^2+a_2^2+..+a_n^2))^\frac{1}{2}((b_1^2+b_2^2+..+b_n^2)^\frac{1}{2})$  How to proceed from here? I want to get a 360 and don't know wether to get the bigger (250GB), newer, more expensive Elite, or get one of the older, smaller (120GB), last gen ones. I know the arcade is going to be too small to save anything useful  How much hard disk space is recommend? What else is stored besides games? Is it possible to programmatically stop and start ArcGIS Server map services using Python? How do I install different (upgrade or downgrade) PHP version in still supported Ubuntu release? I am trying to update to install PHP 5.5, 5.6 or 7.0 in Ubuntu 12.04/14.04/16.04 LTS, and I can only get PHP 5.3.10 for Ubuntu 12.04, PHP 5.5.9 for Ubuntu 14.04 or PHP 7.0.4 for Ubuntu 16.04 using official repositories (using apt-get).  I'm not really sure how to do a manual update – as I need:   to play around with the new(est) PHP features to install older PHP version due support in the software Vieta Jumping: Related to IMO problem 6, 1988: If $ab + 1$ divides $a^2 + b^2$ then $ab + 1$ cannot be a perfect square. The famous IMO problem 6 states that if $a,b$ are positive integers, such that $ab + 1$ divides $a^2 + b^2$, then $\frac{ a^2 + b^2}{ab + 1 }$ is a perfect square, namely, $gcd(a,b)^2$.  How about a modification of this problem:    If $a,b$ are (strictly) positive integers, such that $ab + 1$ divides $a^2 + b^2$, then $ab + 1$ cannot be a perfect square.    I am looking for a proof of the claim above, or a counter-example.  One possible approach towards a proof is the following:   Suppose there is such a pair $(a,b)$ as above, then by the famous IMO problem 6 from 1988, $\frac{ a^2 + b^2}{ab + 1 } = g^2$ where $g = gcd(a,b)$. Since $ab + 1$ is a perfect square, $a^2 + b^2 = c^2$ for some integer $c$. So that $(a,b,c)$ is a Pythagorean tripple, therefore there exists positive integers $n,m,l$ with $n$ coprime to $m$, such that $a = l(n^2 - m^2)$ and $b = 2lnm$ - by Euclids formula.  Then by plugging in $a$ and $b$ in terms of $n,m,l$ into the original equation, and solving for $l$, it is possible to obtain the following:  There exists positive coprime integers $n,m$ such that   $\frac{(n^2 + m^2 + 1)(n^2 + m^2 - 1)}{2mn(n+m)(n-m)}$ is a perfect square.  If we put the quotient above into a program, as in this python code snipet:  N = 1000  for n in range(1,N):     for m in range(n+1, N):         A = (n*n + m*m + 1)*(n*n + m*m - 1)         B = 2*m*n*(n+m)*(m-n)         if A % B == 0:             print(A/B)   The quotient always is 2, regardless of whether or not $n$ and $m$ are co prime! So if the very strong implication   $2mn(n+m)(n-m) \mid (n^2 + m^2 + 1)(n^2 + m^2 - 1) \implies \frac{(n^2 + m^2 + 1)(n^2 + m^2 - 1)}{2mn(n+m)(n-m)} = 2$    holds, then the original problem will be solved. A problem about an $R$-module that is both injective and projective. Let $R$ be a domain that is not a field, and let $M$ be an $R$-module that is both injective and projective. Prove that $M= \left \{ 0 \right \}$.   This is exercise 7.52 of Rotman's Advanced Modern Algebra. Using theorems before exercises, because $M$ is injective and $R$ is a domain, I conclude that $$\forall m\in M ,\forall r\in R\ (r\neq 0) ,\exists {m}'\in M \Rightarrow m=r{m}'$$ and also because $M$ is projective there is a surjective $\psi$ from free $R$-module $F$ with basis $\left \{ e_{i} \right \}_{i\in I}$ to $M$ and thus we can conclude that for every $m\in M$ we have $$m=\sum r_{i}\Psi (e_{i})$$ now I don't know how should I use these together.  The idea of what is happening or a suggestion or a hint will be great. Which package to use for presentations? Beamer, Prosper, or Other Hi is it possible to exclude traffic from a certain country in google analytics?  If yes, how do I do it? If $A$ is a non-square matrix with orthonormal columns, what is $A^+$? If a matrix has orthonormal columns, they must be linearly independent, so $A^+ = (A^T A)^{−1} A^T$ . Also, the fact that its columns are orthonormal gives $A^T A = I$. Therefore,  $$A^+ = (A^T A)^{−1} A^T = (I)^{-1}A^T = A^T$$  Thus, $A^+ = A^T$. Am I correct? Thank you. network manager says "device not managed" Using maxTimeMS parameter with aggregation queries on Mongo 2.6 and Pymongo 2.7.1 Print  only? How do I print the indicated div (without manually disabling all other content on the page)?  I want to avoid a new preview dialog, so creating a new window with this content is not useful.  The page contains a couple of tables, one of them contains the div I want to print - the table is styled with visual styles for the web, that should not show in print. Some of the git commands have many options, and it would often be useful to search through them for the one I need - I was just looking for the option which controls the TAB width in git-gui, but there are about 200 completions for git config. An obvious workaround is to copy all the completions into an editor and search through them, but I'd rather do  [something] | grep tab   There are no man or info pages for compgen, help compgen doesn't even explain its own options, and there's no auto-complete for compgen (how's that for irony?).  PS:  doesn't work.  PPS: This is not a question about git-gui - The  was elsewhere.  PPPS: This is not about auto-completing commands, only command parameters.