any help/explanation is appreciated (i know this is such a newb question but nonetheless)  when I try to "sudo apt update" the command fails on one "link" (Failed to fetch   403  Forbidden [IP: 143.95.32.90 80] )  why is that? did the apt update command complete successfully? does this have anything to do with updating repositories in /etc/apt/sources.list?  Thank you in advance for any help folks, greatly appreciate any feedback.
PPA seems to be constantly offline. Whenever I use sudo apt-get update, this error is shown:  W: Failed to fetch http://ppa.launchpad.net/ 404  Not Found   How do I fix these errors?
Everything I've seen points to no (and I would love that to be true since applying for a visa online costs money) but I'm worried about getting refused due to the length of time I wish to stay, which is 6 months.  I went to university in the UK on a Tier 4 visa, but have never just been a tourist which is what I would like to do now. Since graduating, I have started working but am underwhelmed and would love to travel again. My original plan was Australia but now I would like to see the rest of the UK as well beforehand.  If I arrived with the following, would I be refused entry simply for not applying online?: an approved Australian work &amp; holiday visa, a mini itinerary, bank statements for the last 6 months. Script-r Symbol The near-ubiquitous undergraduate reference on electrodynamics, "Introduction to Electrodynamics" by David Griffiths make extensive use of a script-r symbol, defined to be \vector{\scriptr} \equiv \vector{r} - \vector{r}^{\prime} and looks like:    Although the question of how to produce such a symbol in LaTeX is quite a common one I am yet to find a satisfactory solution.  The closest I've found is from the physymb package and uses the Calligra fonts:    Sadly, there is not a bold variant of this character and it is far more cursive than I would like.  I am hence wondering if there are any better facsimiles. jQuery/JavaScript to replace broken images I've found  many queries to check edit counts, and I get different results for each one. Which one is the most accurate and up-to-date query? Or is there any way to see the edit count in stats? I have read that positional parameters start at $1 (for example: $1, $2, $3 and so on are positional parameters). But $0 is not a positional parameter.  But why $0 is not a positional parameter?  I think this could be a reason, but not sure:  Positional parameters only take their values when a script is executed. For example: if we do ./myScript Hello, then $1 will have the value Hello. But $0 can take its value on two occasions: when a script is executed (it would have the value of the script name), and when bash itself is executed without a script (it would have the value bash or -bash). Why does the Community user have votes cast? If Community is a background process and not a real person then why does its profile show that it has votes cast, and how does it cast them?  How can it decide which answers to upvote or downvote?  See the massive number of upvotes and downvotes cast in its profile: Following , I would like to know what exactly it means for a tag to be marked as blocked (formerly blacklisted) on a Stack Exchange site.  Does it mean I can't use that tag on my question? What are the reasons for blocking a tag? Who can block tags? Why am I being prompted to remove a tag from a question I'm editing? For example, if passed the following:  a = []   How do I check to see if a is empty? How can one change the timestamp of an old commit in Git? The answers to  describe a way to amend previous commit messages that haven't yet been pushed upstream.  The new messages inherit the timestamps of the original commits.  This seems logical, but is there a way to also re-set the times? How do you get vim to display wrapped lines without inserting newlines? I'm getting back in touch with my inner (g)vim due to an unscheduled MacBook mother(board) of a meltdown (my emergency backup Linux box won't run TextMate). All told I'm happy with vim's efficiency and power, but I'm mortified at how hard it is to get the kind of word wrap that even stupid HTML textareas achieve with no apparent effort.  Consider the text  Etiam ornare mollis tortor. Suspendisse sed neque. Nullam in elit. Cum sociis nullam.  By default, with an 80-character width vim displays this as  Etiam ornare mollis tortor. Suspendisse sed neque. Nullam in elit. Cum sociis nu llam.  This wrapping doesn't care about whitespace, so sometimes it just slices words into pieces (nullam in this case). Of course, you can turn on word wrap, and get this:  Etiam ornare mollis tortor. Suspendisse sed neque. Nullam in elit. Cum sociis nullam.  The problem is that vim inserts a newline at the linebreak, which I most emphatically do not want. In other words, I want the text to display exactly as vim displays it with word wrap turned on, but without inserting a newline. (This way it can be pasted into HTML textareas and email programs, among other places.)  Web searches have yielded nothing of use, despite diligent effort; I hope StackOverflow can succeed where my Google-fu has failed. Adding feature class to Map in CURRENT project using ArcPy in ArcGIS Pro? iPhone 4 passcode lock disabled and no access to iTunes When Jedi lose parts of their body, does their midi-chlorian count and strength in the Force diminish? From what I understand, midi-chlorian count is a strong indicator of ability in the Force (though there are other factors as well). So it seems like when Anakin loses his arm or Luke loses his hand, their midi-chlorian counts (and thus their Force ability) should slightly diminish.  Is this the case? Decrypt all RSA ciphertexts if you can break 1% of them I was given a cryptography question during an interview the other day.   Imagine you are given an efficient algorithm which for a given RSA public key (n,e) is able to decrypt 1% of the messages encrypted with that key.   1% means that the algorithm will deterministically decrypt 1% of all possible ciphertexts on the first call, and will never decrypt the other 99% of all possible ciphertexts with any number of invocations.   Figure out an algorithm that uses this one and can decrypt any message. Any ideas? I was reading a textbook. I found that it was mentioned the speed of sound increases with increase in temperature. But sound is a mechanical wave, and it travels faster when molecules are closer.   But an increase in temperature will draw molecules away from each other, and then accordingly the speed of sound should be slower. How is it possible that the speed of sound increases if temperature increases? What is the relation of speed of sound and temperature? Using Thunderbird and gmail, sent items appear as unread in "All Mail" I'm using a gmail account in Thunderbird. When I send a message, a new mail notification appears (sound and system try icon), an unread message appears in the gmail All Mail folder, and the account name turns blue.  I have to click on "Sent Mail" or "All Mail" to get rid of the notification. Unsubscribing from "All Mail" is not effective (it still turns the account name blue).  This seems to be popping up elsewhere (), but nobody has posted a helpful answer.  How do I keep Thunderbird from thinking that a sent message is new unread mail in the All Mail folder? How to use manual partitioning during installation? I'd like to see the full How-To on how to use manual partitioning during Ubuntu installation. The existing guides (at least those I found here) cover only automatic part and leave untouched the manual part (or extremely short and contain no pictures).  I'd like to cover such situations:     If your disk contains other systems: Effect on Google rankings if my site is down? Prove that $\sqrt{n} \le \sum_{k=1}^n \frac{1}{\sqrt{k}} \le 2 \sqrt{n} - 1$ is true for $n \in \mathbb{N}^{\ge 1}$ How many combinations can I make? let $n \gt 1$ be an integer, and consider $n$ people; $P_1, P_2,..., P_n$ let $A_n$ be the number of ways these $n$ people can be divided into groups, such that each group have either one or two people  determine $A_1, A_2, A_3$  So I have  $A_1 = 1$ way $A_2 = 4$ way $A_3 = 12$ way  But I am not sure if this is right... Can anyone confirm?  edit:  My way of getting this is  $A_1 = 1 $ because {P1} only has 1 way to sort  $A_2 = 4$ because {P1,P2} = {P1,P2},{P2,P1},{P1P2},{P2P1}  etc.