what is the default content of /etc/apt/sources.list in Ubuntu 18.04 LTS (Bionic Beaver)
What is the correct output of cat /etc/apt/sources.list?
Is there a word which describes terms that are spelled like a particular pronunciation? Some examples: &quot;never&quot; might be written as &quot;neva&quot; &quot;I don't know&quot; is sometimes written as &quot;iono&quot; or &quot;I dunno&quot; on various internet media. My question is: is there a word for this particular technique of spelling a term in the way it is (sometimes) pronounced? Can't install wine from winehq.org on Ubuntu (actually Lubuntu) 18.04 LTS I have followed all the instructions on .  In particular, I have run the following commands: sudo dpkg --add-architecture i386  wget -nc https://dl.winehq.org/wine-builds/winehq.key sudo apt-key add winehq.key sudo apt-add-repository 'deb https://dl.winehq.org/wine-builds/ubuntu/ bionic main'   The first three of the above seemed to work fine.  However, the last command gives me the following error: Some packages could not be installed. This may mean that you have requested an impossible situation or if you are using the unstable distribution that some required packages have not yet been created or been moved out of Incoming. The following information may help to resolve the situation:  The following packages have unmet dependencies:  winehq-stable : Depends: wine-stable (= 5.0.0~bionic) E: Unable to correct problems, you have held broken packages.  I'm running Lubuntu 18.04.3 on a USB live stick.  (However, if I remember correctly, a couple days ago I got the exact same error on a normal hard disk Lubuntu 18.04.3 installation.) I've read that a lot of people get similar errors.  Unfortunately I haven't found any solutions online that worked. Any help would be much appreciated.  Thanks! EDIT:  Thanks for the comments and suggestions everyone.  I'll try them out later today. In the mean time, here's something I don't understand.  Ubuntu 18.04 has been around for almost two years now.  In all that time, surely somebody has successfully installed wine, right?  What did they do to get it working? Should my flair be including sites with less than 200 reputation? Let $\Omega$ be a simply-connected domain. Let $z_1,z_2\in\Omega$. Prove that exists an univalent function such that $f(\Omega)=\Omega$ and $f(z_1)=z_2$.   Since $\Omega$ is simply connected, one can apply Riemann mapping theorem to get $g:\Omega\to\mathbb{D}$ ($\mathbb{D}$ is the unit disk) with $g(z_1)=0,g(\Omega)=\mathbb{D}$. I thought taking $$f(z)=g^{-1}(g(z)+g(z_2))$$ thus $f(z_1)=g^{-1}(g(z_1)+g(z_2))=z_2$. On the other hand, $f$ does not map $\Omega$ to itself.  How can  I build a map which will satisfy the requirements above? If $F_n$ is the $n$-th fibonacci number, then prove that, $$F_n={n-1 \choose 0 }+{n-2 \choose 1 }+{n-3 \choose 2 }+\ldots$$  I tried the idea of using Pascal's triangle, but it seems to need some adjustment, so I am stuck, please help.    Any combinatorial argument would be of greater help than rigorous proof.  Thank you. In Windows, you can run android application using BlueStack.   I was wondering, is there any way these apps can be run on Ubuntu 12.04? "Looking forward to hearing from you" - May I omit "I am" at the beginning? How to change Function (Fn) Key behavior on wireless Logitech keyboard Use parameter expansion to modify output of another expansion I'm trying to use the Bash parameter expansions to modify the output of a command substitution or another parameter expansion. The following nested expansions work quite well in Zsh; but result in a &quot;bad substitution&quot; error in Bash: ${${PWD##*/}//trunk/latest}  or ${$(basename $PWD)//trunk/latest}  the output should be the last folder of the $PWD, replaced by latest when my current directory is trunk so /home/user/trunk should become latest Is there a Bash equivalent allowing to chain expansions without relying on variables or pipes? Or do Bash expansions only allows the input to be a string or a plain variable? The number $2^{3^n}+1$ is divisible by $3^{n+1}$ and not divisible by $3^{n+2}$. Prove that for every $n\in \mathbb N$, $2^{3^n}+1$ is divisible by $3^{n+1}$ and not divisible by $3^{n+2}$.   I was able to prove that $2^{3^n}+1$ is divisible by $3^{n+1}$ using induction. First, I check whether the statement is true for $n=1$. The statement holds since: $$2^{3^1}+1=9$$ $$3^{1+1}=9$$ Second, I assume that the statement is true for every $n\in \mathbb N$. Finally, I take that $n=n+1$. From that, the original expression becomes $(2^{3^n})^3+1$, which is a sum of cubes, and the same as: $$(2^{3^n}+1)((2^{3^n})^2-2^{3^n}+1)$$ As I assumed that $2^{3^n}+1$ is divisible by $3^{n+1}$, and one of the factors is $2^{3^n}+1$, the entire expression must be divisible by $3^{n+1}$, and so the statement is true for every $n\in \mathbb N$.  Now, I don't know how to prove that $2^{3^n}+1$ isn't divisible by $3^{n+2}$. Please explain how I can do so step-by-step, without using complicated terms. I love the subset of collective nouns known as the terms of venery. These are collective nouns specific to a particular group of animals. Some of the more inventive examples are: a murder of crows, a crash of rhinos, a mischief of mice, and a puddling of ducks (specifically swimming ducks).  Is there a standard for these group names and is there a central reference for them as with other zoological taxonomy?  I can see in the Wikipedia articles that the practice has its origins in Medieval Hunting tradition, but the Kangaroo (Mob), for instance would have been unknown at that time, so how do new collections come into being? Get the subdomain from a URL Getting the subdomain from a URL sounds easy at first.  http://www.domain.example   Scan for the first period then return whatever came after the "http://" ...  Then you remember  http://super.duper.domain.example   Oh. So then you think, okay, find the last period, go back a word and get everything before!   Then you remember  http://super.duper.domain.co.uk   And you're back to square one. Anyone have any great ideas besides storing a list of all TLDs? MySQL Hashing Function Implementation I know that php has md5(), sha1(), and the hash() functions, but I want to create a hash using the MySQL PASSWORD() function.  So far, the only way I can think of is to just query the server, but I want a function (preferably in php or Perl) that will do the same thing without querying MySQL at all.  For example:  MySQL hash -> 464bb2cb3cf18b66  MySQL5 hash -> *01D01F5CA7CA8BA771E03F4AC55EC73C11EFA229  Thanks! What size images do I need for common print sizes, such as wallet size, 8x10, and 16x20?  Is there a general formula?  How could that size change if, for example, a 16x20 was printed on canvas rather than photopaper? What are the advantages of using the optical viewfinder over live preview to take photos on a DSLR? Forgive this novice question, I am basically a complete newbie to photography.  Really, all I've ever used for most my of short life have been point-and-shoots; I've only recently started to pick up some curiosity.  Anyway, from what I've gathered, reading through many of the  questions, most serious photographers use the optical viewfinder as their primary tool to determine what makes a good shot.  Live-preview isn't used much at all; it seems its help is to primarily set up the shot.  Reading a Wikipedia article seems to confirm this view, adding that      [l]ive preview in DSLRs does not typically serve as their principal means of framing and previewing before taking a photograph, with this function still being mainly performed with optical viewfinder. While initially largely a novelty feature, live-preview functionality has become more common on DSLR cameras...    My questions are:   What are the myriad advantages of using the viewfinder over live preview, what does it help one accomplish in achieving the goals of photography? What are some interesting ways serious photographers use live preview to take better shots they might have otherwise missed? What is the size of an enum in C? English equivalent of the expression “a longer arm than résumé” There is by now a captcha-problem What are the badges I can earn on each site, and what are the exact criteria for earning each badge? Formerly List of all badges with full descriptions.  What are badge name's requirements? Why didn't I get badge name ? Which badges can I earn multiple times?    Jump to:                    Note: Some badges are awarded based on score. The term score means the total number of upvotes minus the total number of downvotes.  Any badge with a *** next to it is one of 20 badges that count towards the  displayed on a moderator candidate in an election. The candidate score is a total of 40 points: the first 20 are awarded based on the user's rep divided by 1000 and rounded down; the second 20 represents the total number of unique badges earned of the 20 that count. Note that if the same badge is earned multiple times, it will only count once towards the candidate score.  Any badge with a &quot;(retired)&quot; next to it is no longer awarded, but is retained by users who previously earned them. See  for more info on what it means for a badge to be retired.       Visit the  of the  on any site to see a complete list of badges that you can filter by earned, unearned, or type. Prove by mathematical induction that $2n ≤ 2^n$, for all integer $n≥1$? I need to prove  $2n \leq 2^n$, for all integer $n≥1$ by mathematical induction?  This is how I prove this:  Prove:$2n ≤ 2^n$, for all integer $n≥1$  Proof: $2+4+6+...+2n=2^n$  $i.)$ Let $P(n)=1         P(1): 2(1)=2^1\implies 2=2$.        Hence, $P(1)$ is true.  $ii.)$  Assume that $P(n)$ is true for $n=k$, i.e, $2+4+6+...+2k=2^k$, and prove that $P(n)$ is also true for $n=k+1$, i.e, $2+4+6...+2(k+1)=2^{(k+1)}$  from the assumption add $2(k+1)$ on both sides so we have   $2+4+6...2k+2(k+1)=2^k+2(k+1)$  I'm confused with $2^k+2(k+1)$, I don't know how to make $2^k$ be equivalent to $2^{k+1}$.  I feel i'm doing something wrong.   Any help would be appreciated!