I am planning to travel to Vienna on 13th August from Vietnam with Qatar Airways and expect to arrive in Vienna at 20:25. However, my visa is only valid from 14th August. So will this prevent me from boarding and entering the country? I have tried contacting the airline regarding this issue but haven't been able to get in touch with them.  Thank you in advance!
I'll be traveling to Germany this May and my planes would arrive at 4 May 20.30 at Frankfurt Airport but the Schengen visa that I received is valid only at 5 May.  Is it possible to enter if it's only a few hours before the day change? Or do I have to wait until day change in the transit area?  Another thing, would the airline rejected me from boarding the plane?
How can I display just a portion of an image in HTML/CSS? How does transvoxel algorithm work and how can I implement It? How to change the "User License" type for customer portal users using a script? In the following sentence, I want to know if I've understand the correct meaning or not:     This version holds that the wave function does not so much describe the state of the physical system as the state of our knowledge of the system.    Can we say that it means the wave function describes the state of our knowledge  of the system more than it describes the state of the physical system  ?  It is worth noting that in the Cambridge Dictionary, we read:  not so much sth as sth ​ If you say that something is not so much one thing as something else, you mean it is more the second thing: They're not so much lovers as friends. I don't feel angry so much as sad. I want to communicate between several computers on my network (static Ethernet), through SSH. In order to do that I need to run ssh-add every time I log in on a specific machine, how can I do it so that it's set up once and it doesn't ask me for the passphrase every time I log in or reboot my machine?  I know that there is a way that you should add some lines to the bash_profile file, but I still need to type the password every time I reboot/log in to a specific machine.  if [ -z "$SSH_AUTH_SOCK" ] ; then     eval `ssh-agent -s`     ssh-add fi Change this behavior to allow for spelling corrections and the like: "Edits must be at least 6 characters" Now, it is easy to workaround this: add 10 characters, plus your one-character change, then remove these 10 characters in the next edit for the same post.  But this can hardly be what the designers intended. Instead, my suggestion is to change it to something like on Wikipedia, where you can mark an edit as "minor". Something like this:     Edits must be at least 6 characters, or select the checkbox "Minor, spelling or punctuational correction."   Edits like that will, of course, not count toward the edit-limit (before it becomes Community Wiki).  The suggestion often seen on Meta Stack Overflow is "just edit more in that post". But sometimes the post is just right, and there aren't any other spelling mistakes to correct. Why force a user to change more than necessary, or use an ugly hack? Can't get rid of header X-Powered-By:Express How do you spawn ONLY zombies in Singleplayer? When I ssh into my Ubuntu 18.04 box I get the motd  1 update could not be installed automatically   Which goes on to tell me to check the unattended-upgrades log.  The log shows no problems. How can I clear this bogus message? Is there a technical reason Android update story is so bad? Is there a technical reason why Android phones general lag a lot behind the stable released version of Android?    This applies to both new phones, for example a recent trip to the phone shop everything was 4.0 to 4.2.1, KitKat could not be seen. And also old phones, my girlfriend's iPhone 4 had an OTA update to the latest OS, this is a July 2010 phone, whilst a July 2010 Android is basically a paperweight.    Presumably Android itself doesn't concern itself with the actual hardware, and thus talks to the OS via abstractions? I guess then the handset maker just provides the drivers for the specific hardware, thus I can't see the issue with pushing OTA updates if the ABI remains stable (my phone doesn't grow new hardware features). How to change the name of an iOS app? I began an iPhone project the other day with a silly development code name, and now I want to change the name of the project since it's nearly finished.   But I'm not sure how to do this with Xcode, trying the obvious of changing the application's name in the info.plist file, causes the signing process to go wrong (I think...) and my app won't launch giving me a Launcher error.  I guess I could make a new project and copy paste everything over, but it seems so primitive that I'm hoping for a more civilized solution. Can you find all classes in a package using reflection? Is it possible to find all classes or interfaces in a given package? (Quickly looking at e.g. , it would seem like no.) What is the correct permission in /etc/fstab to mount NTFS? I have Windows and Arch Linux in dual with UEFI. I want to mount my another NTFS partition in Arch. I mounted the partition with mount /dev/sda5  /mnt/Apps command. Then added the output of genfstab -U /mnt/Apps in /etc/fstab file. The output is as follows:  UUID=01D158CC7C2A61A0 /mnt/Apps ntfs rw,nosuid,nodev,user_id=0,group_id=0,allow_other 0 0   But when I unmount and remount that partition all files and folders shows with root uid and gid and 0777 permission. I also tried with changing uid and gid to 1000 but same result. So, my question, what is the correct way to add NTFS partitions in fstab so that I can read &amp; write all file and folder both with normal user (1000:1000) and root (0:0)? Sending 2D arrays in Fortran with MPI_Gather Using conjugates to find a limit with a cubic root: $\lim_{h\to 0}\frac{\sqrt[3]{h+1}-1}{h}$ I currently have $$\lim_{h\to 0}\frac{\sqrt[3]{h+1}-1}{h}$$  Now, I know that when you have square root instead of a cubic root it's easy. You just multiply by the conjugate and simplify afterwards. If it were a sqrt I know that the limit is 1/2. I know that the result here is 1/3 but I can't seem to get there. I always end up with 1/4 due to getting something like: $$\frac{h+1-1}{h(\sqrt[3]{h+1}+1)(\sqrt[3]{h+1}+1)}$$  After simplifying and evaluating the limit you end up with $$\frac{1}{(\sqrt[3]{0+1}+1)(\sqrt[3]{0+1}+1)}$$  which turns out to be $1/4$.  What am I doing wrong? Let $p$ a cut point of a connected space $(X, \mathcal{T})$ and suppose $C$ and $D$ form a separation  of $X-\{p\}$, i.e $$X-\{p\}=C \cup D$$ Prove that $C \cup \{p\}$ is conected. I have this idea: Since $p$ is a cut point then $X-\{ p \}$ is not connected, and since $C$ and $D$ form a separation of $X- \{p \}$ then $X-\{p \} = C \cup D$, and also $C \cup D$ is not connected. And also $$C \cap \overline{D}= \overline{C} \cap D= \emptyset$$ I need an idea to continue please Let $G$ be abelian, $H$ and $K$ subgroups of orders $n$, $m$. Then G has subgroup of order $\operatorname{lcm}(n,m)$. Let $G$ be abelian, $H$ and $K$ subgroups of orders $n$, $m$. Then G   has subgroup of order $\operatorname{lcm}(n,m)$.   This is a statement that my lecturer mentioned in my (beginners') Abstract Algebra class. I'm not sure I understand why it's true.  What I have so far: Use the abelian group structure theorem on $\langle H, K\rangle$ (finite group generated by $H$ and $K$). Then $\langle H,K\rangle=C_{a_1}C_{a_2}\dotsm$ and $n, m |\langle H,K\rangle$. Which means it's also true that $\operatorname{lcm}(n,m)|\langle H,K\rangle$. Can I leverage this to say there's a subgroup of order $\operatorname{lcm}(n,m)$? Conjugacy classes in $A_n$. Suppose $n$ is a non negative integer $\geq 4$ and $\sigma\in S_n$ a permutation. Conjugacy classes in $S_n$ are completley determined by the cycle structure of $\sigma$. If we let the alternating group $A_n$.act on $S_n$ by conjugation, the orbits coincide with the conjugacy classes in $S_n$ except for those (even) permutations whose cycle structure is composed of even cycles (which corresponds to cycles of uneven length) of distinct order (fixed points are treated as cycles of length $1$.) See   For instance a $3$-cycle in $A_4$ has cycle structure $[3,1]$ which fits the bill, and a $5$-cycle in $A_5$ or $A_6$ works too, since the cycle structure is $[5]$ and $[5,1]$ respectively. On the other hand, a $3$-cycle in $A_5$ will does not satisfy the criteria as it has cycle structure $[3,1,1]$ with two fixed points.  In this case, the conjugacy class of $\sigma$ inside $S_n$ splits in $2$ orbits of the conjugation action of $A_n$, and if $\tau$ is any odd permutation, for instance $(12)$, then we have $$\mathrm{Conj}_{S_n}(\sigma)=\mathcal{O}(\sigma)\coprod\mathcal{O}(\tau\sigma\tau^{-1})$$  My question is How can we differentiate between the two orbits? Does it have geometric meaning?    I can see that this information can be useful when looking at representations of the alternating groups, since there are as many irreducible representations as conjugacy classes. As for some geometric interpretation, I ask because there is one for the three cycles in $A_4$. If $S_4$ is understood as the group of symmetries of a regular tetrahedron, then $A_4$ is the subgroup of direct symmetries, and a $3$-cycle corresponds to a rotation of angle $\frac{2\pi}{3}$ or $\frac{4\pi}{3}$ with axis passing through one of the vertices. Conjugacy classes inside $A_n$ preserve the angle, and that's how we can tell them apart. In this case the computations are so simple we don't really need this picture, but I think it is a nice way to understand the fact $3$-cycles in are not all conjugate inside $A_4$... Less than or equal sign If I know for two numbers a and b that    $${a &lt; b }$$  Then is it correct to say that    $$ a \leq b $$  I know that the second statement is true as long as the first one is. It seems OK as it is true but from the other side it seems kinda weird (to me), to say that, if you know, that a is strictly less that b.It's like you lose some extra information. Is it possible to build a single mysql query (without variables) to remove all records from the table, except latest N (sorted by id desc)?  Something like this, only it doesn't work :)  delete from table order by id ASC limit ((select count(*) from table ) - N)   Thanks.