I want to travel to Bangladesh from Denmark via Heathrow Airport, so do I need a transit visa from the UK? My onward flight from Heathrow Airport within 2 hours and same as when I will return to Denmark. I am a Bangladeshi National with having danish permanent residence permit and my wife with a dependent visa from Denmark. Do I need to get UK transit visa? or I am ok to fly without a transit visa? Thanks!
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.
How do I disable this:    after taking screenshot? Are the "umbles" etymologically related? From      And we show telltale signs of early hypothermia, what the doctor and   the National Institutes of Health and other cold-weather cognescenti   call the "umbles."      First we lose some fine motor skills, McLaren says, "and we start to   fumble. And tumble. And stumble."      Then our cognitive abilities are compromised. "We mumble and we   grumble."   Are the "umbles", such as fumble, tumble, stumble, mumble and grumble, etymologically linked, either in terms of sharing a common ending, or entire words having a common ancestor?  Trying to do a search of the etymology of the umbles ended up giving the etymology of the word  instead. Are human bodies programmed to die? Following from this question: :  Is there any evidence that human bodies have systemic self-destruction built into their developmental program? I'm not talking about the cell death response, which I know is an important part of growth, development and cancer prevention.  I've read some things about telomere-shortening but don't know if this is a cause or an effect. This question has been asked before, at least twice by people who were trying to cheat on the PROMYS Europe 2020 Application, and once in a  because the OP showed no effort. I haven't been able to solve it, so I'm going to to ask it again.  PROMYS applications have closed, and I hope the effort shown below is enough to keep this question from being closed.     The Mathematical Forest is grown in a two-dimensional plane, where   trees can only grow on points with integer coordinates. To start with,   there are no trees at all. The foresters plant the first tree at   $(0,0)$. Each year, they carry out tree planting according to the   following rule. If there is a tree on the point (𝑚,𝑛) but there are   no trees on the points $(m+1,n)$ and $(m,n+1)$, then they can choose   to remove the tree on $(m,n)$ and plant new trees on the points   $(m,n+1)$ and $(m+1,n)$. For an integer $k\geq1$, the $k$th diagonal   consists of all points $(m,n)$ with $m+n=k−1$. Is it possible for the   foresters to arrange their planting so that eventually there are no   trees on the first $2$ diagonals? What about the first $3$ diagonals?    $4$ diagonals? Can you generalize?   It's easy to get to a position with no trees on the first two diagonals.  (It only takes $4$ plantings.)  I can prove it's impossible to reach a position with no trees on the first $4$ diagonals, and I believe it's impossible to reach a position with no trees on the first $3$ diagonals, but I can't prove it.  For the $k=4$ case, I used a pagoda function, as in a peg solitaire problem.  For $k=1,2,3,\dots$ define the potential of a tree on the $k$th diagonal as $\frac1{2^{k-1}}$, and the potential of the forest as the sum of the potentials of all the trees in it.  When a tree of potential $\frac1{2^{k-1}}$ is removed, it is replaced by $2$ trees of potential $\frac1{2^k}$ so the potential of the forest never changes.  Initially, the potential of the forest is $1$.  Since there are $k$ trees on diagonal $k$, the potential of all points beyond the fourth diagonal is $$\sum_{k=5}^\infty\frac k{2^{k-1}}=\frac34&lt;1,$$ so it is impossible that there are no trees on the first $4$ diagonals.  The potential of the fourth diagonal is $\frac12$, so this argument doesn't show that it's impossible to have no trees on the first $3$ diagonals.  I've done a lot of experimenting, though, and I believe that the statement is true.  (I even wrote a little computer game so I could experiment quickly.)  The best I've been able to do is to get down to one man on the third diagonal.   The black circles are trees eligible for removal, and the gray circles are other trees.  You can see that there is a "traffic jam" in front of the lone tree on diagonal $3$, and there seems to be no way to clear it.  I've been trying to come up with an argument that reflects this traffic jam, but I haven't been at all successful. We know that after $n$ removals there are $n+1$ trees, and I've been trying to prove somehow that they can't all be sufficiently far from the origin to allow all trees to get past diagonal $3$, but I haven't come close.   In the diagram above, if the tree at $(1,1)$ were moved to $(0,2)$, and the tree at $(1,2)$ were moved to $(2,0)$, then it would be possible to remove the tree at $(0,2)$, emptying the second diagonal.  In the fictitious position, we still have one tree on diagonal $2$ and two on diagonal $3$.   So, I think an argument along the lines I was trying cannot just be concerned with the distance from the origin; it must somehow take into account how the trees interfere with one another.  I haven't been able to do this.    I'd be grateful for solutions, hints, or counterexamples.  I wasn't really sure what tags to attach.  Please add any that seem appropriate. I am currently trying to develop my modeling skills, and following a tutorial found on YouTube. The tutorial teaches how to create an anvil. At 19:10 or so into the video, a subdivision surface modifier is used to create a smooth, soft look.  However, when I do the same, mine looks very sharp and generally bad.  Any ideas why? I will include the blend file below, I appreciate your time and suggestions.    This is his when he subdivides, which is very different from mine when I do the same at the same settings. (I cannot post more pictures because I am new, but if you check the blend file I attached, you should be able to see the difference.   This is what it looks like when I subsurf without the base face. In my experience, "finite" is pronounced (IPA) ˈfaɪnaɪt while "infinite" is ˈɪnfɪnɪt.  In general, the prefix "in" negates an adjective, but does not change the pronunciation. Based on this, I would expect ˈɪnfaɪnaɪt.  Is there a reason for this "deviation from the norm"? Are there dialects/regions that actually say ˈɪnfaɪnaɪt (or ˈfɪnɪt, for that matter)? In Age of Ultron, Thor seems to know something he possibly shouldn't? In a P.I.D., if $a^m = b^m$ and $a^n = b^n$ for $m, n \in \mathbb{N}$ with $\gcd(m,n) = 1$, then $a=b$ Let $R$ be a principal ideal domain, and $a, b \in R$ with $a^m = b^m$ and $a^n = b^n$ for $m, n \in \mathbb{N}$, and $\gcd(m, n) = 1$. I now want to show that we then already have $a = b$.   I think the statement is easily seen to be true if $a = 0$ (or $b = 0$), since if $a = 0$ and $a^m = b^m$, then $a = b$ since $R$ doesn't contain any nontrivial zero divisors, so $b ≠ 0$ already implicates $b^m = b \cdot b^{m-1} ≠ 0$.  I don't really know how to prove the statement for the general case, where I can't directly apply the fact that $R$ doesn't have any zero divisors except $0$. I most likely have to use the fact at some point that $gcd(a, b) = 1$, but I'm not sure how. Dropping one of the columns when using one-hot encoding My understanding is that in machine learning it can be a problem if your dataset has highly correlated features, as they effectively encode the same information.  Recently someone pointed out that when you do one-hot encoding on a categorical variable you end up with correlated features, so you should drop one of them as a "reference".  For example, encoding gender as two variables, is_male and is_female, produces two features which are perfectly negatively correlated, so they suggested just using one of them, effectively setting the baseline to say male, and then seeing if the is_female column is important in the predictive algorithm.  That made sense to me but I haven't found anything online to suggest this may be the case, so is this wrong or am I missing something?  Possible (unanswered) duplicate: I created a Compute Engine VM (Ubuntu 18.04), and when I log into it using the SSH button in the VM instances list I'm logged in as user username.  After a while, I realised I needed to be able to use another SSH client as well. As per  I enabled OS Login by adding a key-value pair in metadata, where key is enable-oslogin and value is TRUE.  I then tried to add my existing SSH key:  gcloud compute os-login ssh-keys add --key-file path/to/key.pub --ttl 0  which was successful except one major problem: it has added the key for user username_gmail_com and will only allow me to ssh to my VM as that user:  ssh username_gmail_com@myinstance  And it has also created that user on my VM. From the web interface, I'm logged in to the VM with that new user as well (/home/username_gmail_com). Of course, I can still access the original user's data (/home/username). When I disable OS Login the web interface logs me in as username again and SSH from another client is disabled.   Why is the username different when enabling OS Login? Can I change the username with OS Login enabled?   Edit  There has been an addition to the docs explaining exactly this question. Check here:  and on  all the way at the bottom under "Expected login behaviors". Why absence of electron is called hole? Best answer versus best solution Whenever I am given a system type with a certain transfer function, I am always asked to proceed to find out the 1. impulse response and 2. step response of the system. My question is, what is so special about these two response types that other responses, like for example parabolic response, is not what I am asked to find out? is it possible to raise the tip while raising the back and lowering the center of the tongue when making /R/ sound? Following , this is how we correctly make the r sound:   Let the throat vibrate since it is voiced. Round the lip a little. Raise the tip of the tongue towards the hard bump behind the upper front teeth,  but do not let the tip of tongue touch the bump. This also makes the center of the tongue go down. The back part of the tongue is raised because you pull it into a tight ball Fold and raise the sides of the tongue so they touch the upper side-teeth.   If we follow the above requirement, our tongue should have an s shape when we pronounce the r sound as depicted in the following sketch:    Is it possible to raise the tip while raising the back and lowering the center of the tongue when making an r sound?   How do native speakers make an r?  P.S. My tongue is very short anyway! I'm trying to find a simple way to create vertices where two objects intersect and delete the leftover internal planes + then join the meshes so that I have one hollow mesh to sculpt. The reason I need to know this is that on my first sculpt I just joined the meshes, but it didn't create vertices at the intersecting points so when I went to sculpt it just overlapped the meshes as you can see here in the neck.  Here is what I am working on now, it needs to be 1 hollow mesh My 12.04 x64 system won't boot into GUI, but I can still use virtual console with my own account, so apparently it's the Windows Management System that's causing it, how can I reinstall it from tty, what's the apt-get command? Don't automatically edit the post to insert "possible duplicate" links When a question is closed as a duplicate, its body is edited to auto-insert the link to that duplicate. I think there are a couple problems with that:   If the question is reopened, the auto-generated insert isn't automatically removed. It allows the poster (or anyone else) to roll back the post and protest the closure by reverting the automatic edit.    I propose that the links to the duplicate(s) are moved into a separate "header" field above the post instead of being edited into the body of the question. That way it could be removed or hidden if the question is reopened and it could not be edited or reverted without reopening the question.   There is some potential for loss of functionality there, since right now it is possible to edit the post to either include additional related links or remove obsolete/incorrect ones. That could be addressed by making the "duplicate field" editable at a certain reputation level (10k?) and enabling others to suggest edits for it that'd go through the usual approval/rejection process.  Thoughts? What commands (exactly) should replace the deprecated apt-key? Can I put 700c tires on rims made for 27 x 1.25"? How to open a web page from my application?