Getting my luggages at SFO port of entry and not boarding the next flight
Do you have to take the second leg of a domestic flight?
Oneboxing is a handy feature in chat for quickly previewing links, but the list of supported sites is very small, and occasionally boxes for sites such as Wikipedia can break and fetch any image/text.  Instead of having to request additional sites to be added to the onebox list (, , , , , and of course ), we could let the sites themselves handle the oneboxing with some &lt;meta&gt; tags, such as:  &lt;meta name="onebox-title" content="The Onebox Title"/&gt; &lt;meta name="onebox-image" content="/image.png"/&gt; &lt;meta name="onebox-summary" content="This is the summary for a site's onebox"/&gt;   Then when a link is placed on its own in chat, the oneboxing spider would look for these tags and, if they are found, show the appropriate data.  Support for these tags would certainly be bad (for lack of any real better term) at first, though most users of SO chat could easily add it to their personal / company's sites (should they be in such a position) and writing friendly suggestions to other sites to implement them isn't exactly a gargantuan effort.  This would also present an opportunity for currently oneboxed sites to set in stone what should be shown, such as each Wikipedia article setting the main image to be oneboxed instead of SO's spider picking one willy-nilly.  Finally, sites implementing these tags would lay the foundation for other sites and services to develop similar features to the onebox, without having to write per-site scrapers. What is the current URL to create a new Google Document please? Since Google last updated their Docs the old URL does not work. We were trying to come up with an easy way to generate a rational number in between two existing rational numbers with a fairly low numerator and denominator (the way we were doing this earlier was to find the average of the two rationals, but that results in a denominator of up to $c * d$. Does this inequality hold for all values of $a, b, c, d$? Let's get rid of the 10K flag queue are pretty cool... You get a birds-eye view of activity on the site, a &quot;dashboard&quot; view of what's happening. Some of the individual tools haven't scaled particularly well with Stack Overflow's growth, but the concept behind them is still sound: we trust you to enough to be a bee watcher now.  ...and then there's . What a let-down! Once upon a time, this queue contained spam and offensive flags, which one might reasonably assume were important enough to put in front of the more trusted members of the site. Re-flagging them brought them that much closer to deletion, while editing them offered salvation to some hapless author. Nowadays, it's a bunch of Not an Answer flags and a smattering of assorted cruft. 10K users can't even vote to delete these; only 20K users have that privilege. Re-flagging them does nothing but increase their priority in the moderator flag queue, where they frequently outrank more pressing issues;  has . Also, it's , and the behavior has diverged far enough from that of the moderator flag queue on which it is based that it has become an active hindrance to further development of the tooling there. Worst of all, we have  that's available to anyone with the editing privilege. It even has logic built in to prioritize answers likely to be deleted for users with . It's like we gave you a car for your 10th birthday, and then replaced it with a rusty bicycle when you turned 16. It was a nice idea, but it has outlived its usefulness. Proposed changes The /tools/flagged route goes away. Period. No replacement. The rest of the 10K tools stick around as informational pages. Not an Answer flags go into /review/low-quality, . Then we beef up the Low Quality review process to make better use of more experienced reviewers and solve this whole &quot;declined / helpful / disputed&quot; flag debate once and for all:  Effective # of reviews required == ReviewsRequired + # of applicable flags (where ReviewsRequired is 2 on Stack Overflow, 1 everywhere else). So 1 VLQ or NAA flag means EffectiveReviewsRequired=3. LQ tasks are not dequeued until one of the following conditions is met:  Post is edited from within review. Outcome: flags are marked &quot;helpful&quot; (current behavior). Post accumulates 3 Delete votes (can only happen when post scores &lt;= 0 and reviewers have &gt;= 20K rep). Outcome: post is deleted, flags are marked &quot;helpful&quot;. Task accumulates EffectiveReviewsRequired &quot;Looks Good&quot; reviews. Outcome: if the number of (Recommend)Delete reviews is &gt;= the number of Looks Good reviews, mark flags &quot;disputed&quot; and raise DisputedLowQuality mod flag. Otherwise, mark flags &quot;declined&quot;. Task accumulates 6* RecommendDelete + Delete reviews. Outcome: mark flags &quot;helpful&quot;. If the post scores &gt; 0 then raise DisputedLowQuality mod flag, else just delete post (current behavior).   As under the current system, flags on posts that've already completed one full review cycle without being deleted should skip /review and go directly into the mod queue.  End result  Flags get handled faster, more accurately, and with less wasted effort from 10K reviewers. Moderators are free to focus on situations that can't be handled by the community - the ! Developers are free to make enhancements to the moderator tooling without having to work around 10K user restrictions.  Questions Am I forgetting anything here? I haven't really spent much time in the 10K flag queue since back when it was filled with spam flags and therefore useful - is there a use-case that you depend on that would be lost with its removal? Post 'em below. *: on Stack Overflow, only . of a monic polynomial $f \in \mathbb F\left[x\right]$ in $1$ variable $x$ over a field $\mathbb F$ plays an important role in understanding the structure of finite dimensional $\mathbb F[x]$-modules.  It is an important fact that the characteristic polynomial and the minimal polynomial of $C(f)$ are both equal to $f$. This can be seen quite easily by induction on the degree of $f$.  Does anyone know a different proof of this fact? I would love to see a graph theoretic proof or a non inductive algebraic proof, but I would be happy with anything that makes it seem like more than a coincidence! I've seen them maneuver. They definitely can go backwards. How fast can they go in reverse? Is warp speed possible, or is reverse limited to impulse power? If it is limited to impulse power, can they travel backwards under impulse power as fast as they travel forwards under impulse power? In what order should I watch the various Stargate TV Series? I am unfamiliar with the (multiple) Stargate shows and films, other than the movie which I saw a while ago. I want to try watching the series to see if it is something that I like.   What are the non-subjective facts about the series (without spoilers) that I should be aware of when figuring out which series to start with, for example the original airing order of the series, chronological order of the series, frequency of references to a series other than the one that I'm watching, etc. How to select distinct field values using Solr? Ubuntu desktop suddenly points to home folder Summer in Saint Helena The Winter Bash site is located at http://winterba.sh/.  Only one problem:   . Saint Helena is in the Southern Hemisphere. It's Summer in Saint Helena.   So shouldn't it be the Summer Bash? Don't forget our Aussie/Kiwi/South African/Southern Brazilian/Argentine/Uruguayan/Zimbabwean/Swazi/Tokelauan/Tuvaluan/Pitcairn Islander/ friends! $(z_{2n})_{n\in \mathbb{N}}$ and $(z_{2n+1})_{n\in \mathbb{N}}$ converges to same limit. Show $(z_n)_{n\in \mathbb{N}}$ converges to the same limit. $(z_n)_{n\in \mathbb{N}}$ is a complex valued sequence such that subsequences $(z_{2n})_{n\in \mathbb{N}}$ and $(z_{2n+1})_{n\in \mathbb{N}}$ converges to same limit.  How could I show that the sequence $(z_n)_{n\in \mathbb{N}}$ converges to the same limit using $\epsilon$, $N$ definition?    For every $\epsilon &gt;0$,  $|z_{2n}-l|&lt;\epsilon$ whenever $n&gt;N_1$, and   $|z_{2n+1}-l|&lt;\epsilon$ whenever $n&gt;N_2$.  We want to show that for every $\epsilon &gt;0$, there exists $N$ such that $|z_{n}-l|&lt;\epsilon$ whenever $n&gt;N$. I tried something like  $|z_{n}-l|\le|z_n-z_{2n}|+|z_{2n}-l|$ but wasn't sure what to do next Prove that $T^2$ is the zero transformation if and only if $\mathrm{range}(T) \subseteq \ker(T)$. $\newcommand{\range}{\mathrm{range}}$The full question: $T^2$ means the composition, i.e. $T(T(x))$ Let $V$ be a vector space and let $T: V \rightarrow V$ be linear.  Prove that $T^2$ is the zero transformation if and only if $\range(T) \subseteq \ker(T)$. So far I know that for ($\implies$).  Assume $T^2$ is the zero transformation.  Then $\ker(T^2)=\{0 \in V: T^2(v)=0\}$.  Note that this is $\forall v \in V$. We also know that the $\range(T)=\{0\in V: \exists v \in V: T^2(v)=0\}$. Not sure where to go from here.  Also for the other way, I don't know where to go. How to export a individual layer with a fixed master layer in Photoshop? Let's say I have some layers and a single different one, or a group of layers I want to export merged together, in a way that every single exported layer is merged with a specific layer or group of layers.  In the example below I have some layers which need to be exported into JPG or PNG, merged with the very top layer called "Blurred". In my case, I want to export the layers (Jan 8 + Blurred) and (Layer 10 + Blurred) and so on. Acting as exporting layers merged with a master layer called "Blurred".  My question in a simple way: how to apply a Master layer(s)/Group(s) in Photoshop to be exported with every layer I have? Why should we typedef a struct so often in C? It's not clear to me how to choose Spells Known when creating a new character at a higher level.  Let's take the Bard as a working example first:  Bard level 2:   Spells Known: 5 Slots per Spell level: 1st &times; 3   Bard level 3:   Spells Known: 6 Slots per Spell level: 1st &times; 4, 2nd &times; 2   I'm interpreting this to mean that, if an existing 2nd-level Bard gains a new level (level 3), he's gaining 1 (one) new spell known. He can then choose the new spell known to be either 1st or 2nd, correct?  What I don't understand then, is the case of creating a new character starting at a higher level; say, creating a new Bard at level 3.  In that case you're forced to choose 6 spells, which must be from the list of 1st- and 2nd-level spells, correct?  When making this 3rd-level Bard from scratch, though, can I select 2nd level spells for all 6 spells known?  As another example, say you're creating a very high level Bard from scratch, starting at level 15. The book says your Spells Known is 19, and your spell slots are 1st × 4, 2nd × 3, 3rd × 3, 4th × 3, 5th × 2, 6th × 1, 7th × 1, and 8th × 1.  So, for those 19 Spells Known, you can choose any spells from 1st through 8th spell level. Am I free then to select as many spells as I can at 8th spell level (5 in number), working backwards, and in the end ignoring the 1st, 2nd, 3rd level spells entirely?  As another example, the Ranger doesn't know any spells at 1st level, nor have any spell slots, but the book says “You know two 1st-level spells of your choice from the ranger spell list.” Does it mean at 2nd level and above?  Long story short: When choosing Spells Known for a spellcasting character created above 1st level, am I allowed to choose my Spells Known at whatever spell level I like, just so long as I have at least one spell slot of that level?  I'm working on a D&amp;D 5e character generator and need to clarify this. Fibonacci Sequence and series limits 1) Let $F_1 = 1$, $F_2 = 1$, and $F_n = F_{n-1} + F_{n-2}$ for $n \geq 3$.   How do you prove that  $$\sum_{n=2}^\infty \frac1{F_{n-1} F_{n+1}} = 1$$   $$\sum_{n=2}^\infty \frac{F_n}{F_{n-1} F_{n+1}} = 1$$   Do you even use the formula $n(n+1)/2$?   I tried using $F_{n-1} = F_n - F_{n-2}$ and $F_{n+1} = F_n + F_{n-1}$ and then trying to figure out how to substitute those, but I've been going around in circles. How would you solve this? What tips can you give me to solve this? Is there any website that teaches this? I tried searching google and youtube but none were helpful.   2) Let $s_x = \tan(2/x)$. Where does this become nonincreasing? While using monotonic limits to establish there is a limit.   I was guessing you need to use upperbound and lowerbound but doesn't it go to infinity? How to disable SMS notification during phone call? I call somebody, or somebody calls me, we are talking. Now, during the call I got SMS and I cannot hear anything because the SMS notification is playing.  So how to disable this notification when call is active? Thank you in advance.  Samsung Galaxy Ace 2, Android 2.3.  Note: I am interested in setting this once and for good. Not something I have to remember to switch off each time I make/get a call. Imputing missing observation in multivariate time series I'm looking to view and archive all my old DMs in Twitter, since I would like to revive conversations that took place 1 year ago.  Acording to Twitter they don't delete old messages, they simply archive them in their database.  Is there a way to view and download all of these messages, just like you can view all your old public tweets with third party apps?  UPDATE: I found  but apparently it's not working for DMs now Use 'import module' or 'from module import'?