I would like to bring my girlfriend from the Phillipines to visit me in Trinidad. The easiest route seems to be manila intransit through Qatar and connect in Gatwick airport to Port of Spain Trinidad. From.my research it seems that no visa is required to transit Doha as the luggage will be checked all the way through to Gatwick. But as she will have to reclaim and recheck luggage she will require a visitor in transit visa. She will also require a visa on arrival into my country. She is currently unemployed in phillipines. I would really like to know if anyone has had experience in getting this transit visa. It seems as if you have to apply for it online but still also visit a visa application center in the Phillipines to have your biometric information collected. It all seems overwhelming. Grateful for any advice.
Is there a website or some other way to find out if I need a transit visa for a short stop in the UK on my way somewhere else? Ideally this should give me the answer for all nationalities, and take into account any other visas I may have.
Are there any disadvantages to using your companions for transporting Power Armor What are the complexity guarantees of the standard containers? We've heard it many times, nothing can escape the gravity of a black hole, even light once it's past the event horizon.  If this is true, how can a black hole emit anything?  are massive black holes that have consumed enormous amounts of matter and energy, and then blasted it back out into the universe.  How is this possible?  Time stands still in a black hole, that in itself should stop anything from moving. Is there a list of Ubuntu versions with default corresponding Linux kernel versions somewhere?  I would specifically like to know the most recent version of Ubuntu that still used Linux Kernel 2.x. Latex create a second keywordstyle using listing doesn't work What happens if you use a drone to build a building while he's carrying minerals? Kind of a basic question, but what happens? Do you lose it? Does it count as harvested?  In a similar fashion, what happens when a drone/scv/probe goes harvesting gas when he's holding a mineral? How to convert standard IP address format string to hex and long? Familiar form of address for a young, subordinate, woman that connotes respect (Female equivalent to 'Son') A(n often male) paternal figure could use the term 'son' in a fatherly way without referring to his biological Son.  Imagine a man has been verbally abused by a customer at work. His manger might say to him:     Come here son, I have something to tell you   about rude customers   Emphasis on the word I want gender-swapped  Is there a feminine equivalent for a subordinate that you respect?   From      son  (sŭn)   n.    1. One's male child.    2. A male descendant.    3. A man considered as if in a relationship of child to parent: a son of the soil.    4. One personified or regarded as a male descendant.    5. Used as a familiar form of address for a young man.    6. Son Christianity The second person of the Trinity.     Is there an equivalent for talking to a woman in a fatherly way?  The equivalent for TFD entry for daughter does not qualify it as 'familiar'        (often capital) a form of address for a girl or woman      Further more often such 'father-daughter' language patterns are negative, dismissive and lack respect. Do Chinese citizens require a visa to go to Hong Kong for less than 7 days? How to run Bash script on startup on Linux? I am running Linux 4.9.11-1.0.0 on my iMX7 dual development board. I have a test.sh file that I would like to run at startup. It currently only has a bunch of echo statements in it (just for testing purposes).   When I flick the switch on the board to turn it on, I see a bunch of initialization stuff come up. I would like the echo statements in my test.sh file to also be displayed in this initialization sequence on startup. Eventually once this is working, the plan is to do real stuff in place of the simple echo statements. I just want to make sure the script is running on startup at this point.  So far I have created the test.sh file and I am able to run it manually after the board has booted up. I have added the following to my /etc/rc.local file before the exit statement at the end:  sh '/home/root/test.sh'   I have verified that this rc.local file has #!/bin/sh -e at the top. I have also done:  $ chown root /etc/rc.local $ chmod 777 /etc/rc.local   Once this was all done, I verified that it should be working properly by typing:  $ /etc/init.d/rc.local start   Everything is working as expected up until this point and I am seeing the echo statements when I do that last step. However, when I go and try to reboot the board, I do not see the echo statements at all in the printed initialization stuff.  What am I missing? Maybe it is actually running on startup but I just don't see the output for some reason? @ variables in Ruby on Rails How to construct an example of a probability distribution for which $\mathbb{E}\left(\frac{1}{X}\right)=\frac{1}{\mathbb{E}(X)}$ holds, assuming $\mathbb{P}(X\ne0)=1$?    The inequality which follows from Jensen's inequality for a positive-valued RV $X$ is like $\mathbb{E}\left(\frac{1}{X}\right)\ge\frac{1}{\mathbb{E}(X)}$ (the reverse inequality if $X&lt;0$). This is because the mapping $x\mapsto\frac{1}{x}$ is convex for $x&gt;0$ and concave for $x&lt;0$. Following the equality condition in Jensen's inequality, I guess the distribution has to be degenerate for the required equality to hold. A trivial case where the equality holds is of course if $X=1$ a.e. Here is an example I found in a problem book: Consider a discrete random variable $X$ such that $\mathbb{P}(X=-1)=\frac{1}{9}, \mathbb{P}(X=\frac{1}{2})=\mathbb{P}(X=2)=\frac{4}{9}$. It is then easily verified that $\mathbb{E}\left(\frac{1}{X}\right)=\frac{1}{\mathbb{E}(X)}=1$.   This example shows that $X$ need not be positive (or negative) a.e. for the equality in the title to hold. The distribution here is not degenerate either.  How do I construct an example, possibly like the one I found in the book? Is there any motivation? If a trait would be advantageous to an organism then why hasn't it evolved yet?  Conversely, if a trait is not advantageous or mildly disadvantageous, why does it exist?  In other words why does evolution not make the organism more "perfect"?    This is a general question that would be applicable for any kind of trait. Please keep the answers precise and scientific.   Read this meta post for more information: Proving the measure of an increasing sequence of measurable sets is the limit of the measures Show that if $A_1\subseteq A_2\subseteq A_3\subseteq\cdots$ is an increasing sequence of measurable sets (so $A_j\subseteq A_{j+1}$ for every positive integer $j$), then we have $$m\left(\bigcup_{j=1}^\infty A_j\right)=\lim_{j\to\infty}m(A_j)$$  Here is my proof: According to the $\sigma$-algebra property, $\bigcup_{j=1}^{\infty}A_j$ is a measurable set, so it makes sense to talk about $m(\bigcup_{j=1}^{\infty}A_j)$. Firstly, I prove that $\lim_{j\to\infty}m(A_j)\leq m(\bigcup_{j=1}^{\infty}A_j)$. This is because for any given positive integer $N$, $A_N\subseteq \bigcup_{j=1}^{\infty}A_i$, according to monotonicity, we have $m(A_N)\leq m(\bigcup_{j=1}^{\infty}A_i)$. Take the limit,we will have $\lim_{j\to\infty}m(A_j)\leq m(\bigcup_{j=1}^{\infty}A_j)$. Secondly, I prove that $m(\bigcup_{j=1}^{\infty}A_j)\leq \lim_{j\to\infty}m(A_j)$. For any given positive integer $N$, $\bigcup_{j=1}^N A_j = A_N$. According to monotonicity,we have $m\left(\bigcup_{j=1}^N A_j\right)=m(A_N)\leq \lim_{j\to\infty}m(A_j)$. Take the limit, we will have $m\left(\bigcup_{j=1}^\infty A_j\right) \leq \lim_{j\to\infty} m(A_j)$. Combine the above two arguments, we will see that $$m\left(\bigcup_{j=1}^\infty A_j\right)=\lim_{j\to\infty} m(A_j)$$$\Box$  The above is my proof, unlike many books, my proof does not use the property of countable additivity. So I doubt my proof is correct. Who can point out where are my mistakes? Studying the equation $\ln(x)- 1+ x^2=0$ Why can't I press the Install button when installing applications from unknown sources? I'm trying to sideload an application, as seen in the picture below.    I have allowed the installation of applications from unknown applied actions in settings, but I can't seem to press the install button. The cancel button works correctly and installing applications via adb install works correctly. What's the problem?  EDIT: For those who have this problem, using ChainFire's  will correct it because the app uses hardware compositing, and Android doesn't notice a "layer" on top of the current screen. Compactness of $(M, d)$ in terms of continuous functions $f:M\to \mathbb R$ Suppose $(M,d)$ a metric space. I want to show that      If every continuous real-valued function on $M$ attains a maximum, then $(M, d)$ is compact.   I was trying to do this by assuming $M$ non-compact and showing explicitly a function which does not attain a maximum. If $M$ is not bounded it's enough to take the distance from a fixed point as we can always find a sequence which "exits" every compact and pushes the distance at infinity. But what if $M$ is bounded? In several cases I want to add a toolbar to the top of the iPhone keyboard (as in iPhone Safari when you're navigating form elements, for example).    Currently I am specifying the toolbar's rectangle with constants but because other elements of the interface are in flux - toolbars and nav bars at the top of the screen - every time we make a minor interface change, the toolbar goes out of alignment.  Is there a way to programmatically determine the position of the keyboard in relation to the current view? Prove $\forall n\geq 2,n\in\mathbb{Z}$, $(n+1)\mid(n^3+1)$ Question: Prove $\forall n\geq 2,n\in\mathbb{Z}$, $(n+1)\mid(n^3+1)$    I know that it is possible to solve by factoring $n^3+1$ and showing that $n+1$ is a multiple, but I would like to show this via induction.  Attempt: Using induction:  Let $n=2:3\mid 9$, so true for $n=2$  (1) Assume true for $n=k:(k+1)\mid(k^3+1)$, so $(k+1)r=k^3+1$ for some $r\in\mathbb{Z}$  (2) Let $n=k+1:(k+2)\mid(k+1)^3+1$, so $(k+2)s=(k+1)^3+1$ for some $s\in\mathbb{Z}$  From (2) we have $(k+1)s+s=(k+1)^3+1$, subbing (1) into (2), I get $k^3+1+s=(k+1)^3+1$  Therefore $k^3+s=(k+1)^3$.  Setting $s=3k^2+3k+1$, this holds.  Where am I going wrong, is the proof valid, any help would be greatly appreciated. If $f(x)\to 0$ as $x\to\infty$ and $f''$ is bounded, show that $f'(x)\to0$ as $x\to\infty$