So i have a Syrian passport and and Egyptian one, i work in Russia and my working visa is on my Syrian passport, i wanna travel to Indonesia, going there for Egyptians is visa free.  My question is, Can i leave Russia with my Syrian passport, and enter then leave Indonesia with my Egyptian one, and go back to Russia with my Syrian one (since the visa is on there)?  there are 4 countries involved, 1 visa, and 2 entries! i hope the description makes it clear ^
I am a citizen of two different countries, and have two passports.  How should I use my passports when traveling?
Example of a Borel set that is neither $F_\sigma$ nor $G_\delta$ I'm looking for subset $A$ of $\mathbb R$ such that $A$ is a Borel set but $A$ is neither $F_\sigma$ nor $G_\delta$. Is there an ingame way to change volume? I like the music, but I still prefer my own music.  Is there an ingame way to change the volume?   I can't find it. I have been studying the pronunciation of Greek-derived words in English, and I've found an odd anomaly. There are (at least) two possible pronunciation patterns for plural word-forms that end in -&#8288;mata (which correspond to singular word-forms that end in -&#8288;ma).   Antepenult stress (expected)  It appears that it is always possible to place the stress on the antepenult (third-to-last) syllable, such as:    /ˈlɛmə/, /ˈlɛmətə/  /ˈskiːmə/, /ˈskiːmətə/  /ˌtɛrəˈtoʊmə/, /ˌtɛrəˈtoʊmətə/~/ˌtɛrəˈtɒmətə/  /ˈstɪɡmə/, /ˈstɪɡmətə/   This is what would be expected from  (I believe most or all of these words came through Latin before entering English), since the penultimate "a" in the original Greek words is short.  Maybe Latin vowel quality and stress rules are not very relevant; certainly, few English speakers know about them nowadays. And even with antepenult stress, some of these pronunciations do not seem to follow all of the traditional rules for pronouncing Latin-derived terms in English. I was reminded by some comments in this  that  in antepenult stressed syllables, we would expect tense ("long") vowels to be shortened due to  (TSL), yielding pronunciations like /ˌtɛrəˈtɒmətə/ (which I just noticed is in fact listed in the Oxford English dictionary) and ?/ˈskɛmətə/ (which is not). (Compare to genus~genera /dʒiːnəs/~/dʒɛnərə/, and perhaps opus~opera /oʊpəs/~/ɒpərə/). However, there are other Latinate plural forms that typically don't have trisyllabic laxing (matrices, also I think helices), or where both the lax- and tense-voweled pronunciations coexist (codices, apices), so I guess the apparent lack of it in schemata should not be surprising. (There are even some exceptions to TSL in singular nouns like obesity.)  Penult stress (unexpected)  The other pronunciation seems to exist only for a couple of words, in which the penultimate "a" is commonly stressed and broadened: stigma ~ stigmáta  (listed by both the Merriam-Webster dictionary and the ), and also sometimes schéma ~ schemáta /skiːˈmɑːtə/ (listed only by the ). (Neither the Merriam-Webster dictionary nor the Oxford Learner's Dictionaries list *"lemmáta").  I'm wondering why this pronunciation exists. It's an irregular plural either way, and putting the stress on the penultimate syllable creates a stress change between the singular and the plural forms.  (It's true that all of these words are rare, and some people might pronounce some of their plurals differently. The pronunciations I give here are simply the ones listed in dictionaries.)  The plural form "stigmata" is probably most common in religious or theological contexts (where it refers to the wounds of Christ), so at first I wondered if the penultimate stress was supposed to be closer to the original Greek pronunciation, but according to , the stress is on the antepenult in Greek as well.  So, what's up with this? Did the stress perhaps change in Latin over time, is this a tradition of Greek accentuation that I'm not familiar with, is this a simple matter of ignorance of the original position of the stress that has now become entrenched with time and usage, or is this somehow explainable by established English stress patterns? (For example, do English speakers have some natural preference for a penultimate stress on polysyllabic words?)  If nobody knows of a reason, I suppose I can just put it down to people encountering the word mainly in text, not knowing which syllable to stress, and choosing the penult because it sounds better to them.  (I asked this question a while back on , but nobody knew of any reason.) Can ANOVA be significant when none of the pairwise t-tests is? Is it possible for one-way (with $N&gt;2$ groups, or "levels") ANOVA to report a significant difference when none of the $N(N-1)/2$ pairwise t-tests does?   In  @whuber wrote:      It is well known that a global ANOVA F test can detect a difference of means even in cases where no individual [unadjusted pairwise] t-test of any of the pairs of means will yield a significant result.   so apparently it is possible, but I do not understand how. When does it happen and what the intuition behind such a case would be? Maybe somebody can provide a simple toy example of such a situation?  Some further remarks:   The opposite is clearly possible: overall ANOVA can be non-significant while some of the pairwise t-tests erroneously report significant differences (i.e. those would be false positives). My question is about standard, non-adjusted for multiple comparisons t-tests. If adjusted tests are used (like e.g. Tukey's HSD procedure), then it is possible that none of them turns out to be significant even though the overall ANOVA is. This is covered here in several questions, e.g.  and . Update. My question originally referred to the usual two-sample pairwise t-tests. However, as @whuber pointed out in the comments, in the ANOVA context, t-tests are usually understood as post hoc contrasts using the ANOVA estimate of the within-group variance, pooled across all groups (which is not what happens in a two-sample t-test). So there are actually two different versions of my question, and the answer to both of them turns out to be positive. See below. Integral of the square of the normal distribution. Recommended visualization libraries for standalone applications Spontaneous electroweak symmetry breaking (i.e. $SU(2)\times U(1)\to U(1)_{em}$ ) is at scale about 100 Gev. So, for , gauge bosons $Z$ &amp; $W$ have masses about 100 GeV. But before this spontaneous symmetry breaking ( i.e. Energy > 100 GeV) the symmetry $SU(2)\times U(1)$ is not broken, and therefore gauge bosons are massless.  The same thing happens when we go around energy about $10^{16}$ GeV, where we have the Grand Unification between electroweak and strong interactions, in some bigger group ($SU(5)$, $SO(10)$ or others). So theoretically we should find gauge bosons $X$ and $Y$ with masses about $10^{16}$ GeV after GUT symmetry breaks into the Standard Model gauge group $SU(3)\times SU(2)\times U(1)$, and we should find massless X and Y bosons at bigger energies (where GUT isn't broken).  So this is what happened in the early universe: when temperature decreased, spontaneous symmetry breaking happened and firstly $X$ &amp; $Y$ gauge bosons obtained mass and finally $Z$ &amp; $W$ bosons obtained mass.   Now, I ask: have I understood this correctly? In other words, if we make experiments at energy above the electroweak scale (100 GeV) we are where $SU(2)\times U(1)$ isn't broken and then we should (experimentally) find $SU(2)$ and $U(1)$ massless gauge bosons, i.e. $W^1$, $W^2$, $W^3$ and $B$ with zero mass? But this is strange, because if I remember well in LHC we have just make experiments at energy about 1 TeV, but we haven't discovered any massless gauge bosons. I installed Ubuntu 18.04, and I want to customize the fonts. I have already change all the setting by tweak tool. But the system default font not be changed, is anything I need to do? Why doesn't Java allow to throw a checked exception from static initialization block? I lost my .keystore file? This has puzzled me for years: How does Blender decide the window where it will display the rendered image when I hit F12 to render? How to connect a Nokia Lumia 730 to Ubuntu 14.04? Cycles : shadows on smooth shaded mesh, strange behavior Find all integer roots of: $x^2(y-1)+y^2(x-1)=1$   Obviously $(2,1)$ and $(1,2)$ are two answers. But I was unable to manipulate the equation algebraically giving a useful form for finding all other possible answers! I also tried to view it as a quadratic equation in terms of $x$, but forming the Delta of that equation didn't help much other than if $y=1$ then certainly Delta would be a square and equation would have answer for $x$... For any prime $p > 3$, why is $p^2-1$ always divisible by 24? Let $p&gt;3$ be a prime. Prove that $24 \mid p^2-1$.   I know this is very basic and old hat to many, but I love this question and I am interested in seeing whether there are any proofs beyond the two I already know. Is there a way to crack the password on an Excel VBA Project? The infinite product of the sines of all positive integers is zero I tried to prove that $\prod\limits_{k = 1}^\infty  {\sin k = 0}$  Define $$P(m)=\prod\limits_{k = 1}^m  {\left|\sin k\right|}$$  then I take the logarithm of both sides $$\log P(m)=\log \prod\limits_{k = 1}^m  {\left|\sin k\right|}$$  which can be written as $$a_m=\log P(m)=\sum\limits_{k = 1}^m  {\log\,\left|\sin k\right|}$$ for any $k\in\mathbb{N}$ we have $0&lt;\left|\sin k\right|&lt;1$, so all terms of the sum above is strictly negative.  I say that the sequence diverges: $\left\{a_m\right\}_{m\in\mathbb{N}}\to\,-\infty$  This is a delicate point on which I have some doubts.  I know that  $$\int\limits_0^\pi  {\log \left|\sin x\right|{\text{ d}}x}  =  - \pi \log 2$$ and that the function $\log \left|\sin x\right|$ is periodic with period $\pi$,  so the integral from 1 to $\infty$ diverges to $- \infty$. Then the series diverges as well for the integral test for convergence.  As $\log P(m)\to\, -\infty$ we can say that $P(m)\to\,0$ as $m\to\,\infty$  Therefore $$\mathop {\lim }\limits_{m \to \infty } \prod\limits_{k = 1}^m {\left| {\sin k} \right| = \mathop {\lim }\limits_{m \to \infty } \prod\limits_{k = 1}^m {\sin k = \prod\limits_{k = 1}^\infty  {\sin k = } 0} }.\quad \blacksquare$$ Why does TeX prefer an overfull box to hyphenating here? What happens to waves when they hit smaller apertures than their wavelenghts? I was wondering this for quite a long time now. Let's say you have a water wave (like ripples, not the ones you see during tsunamis) with wavelength 10 m. Imagine you put a boundary with an opening of 1 m. Will diffraction be observed? According to my research, no. But, then, what would be seen? Different dependencies for different build profiles