I am new to Ubuntu and trying to play around with Android development. While trying to install Android SDK, I can see the SDK being in .zip format. Does .zip file is a valid archive format for Ubuntu ?.
Just downloaded a .zip file from the internet. I want to use the terminal to unzip the file. What is the correct way to do this?
Where does matter go after reaching singularity in uncharged black holes? I am a layman in physics and just read about black holes on the internet. I read that matter encounters geodesic incompleteness in the singularity in an uncharged black hole.  I heard an analogy of geodesic incompleteness as a straight line on a paper reach a hole on the paper, so it cannot continue. But in this analogy, isn't the straight line possible to continue into 3D (continue down the paper)? So, if matter reaches the singularity, is it possible too (to reach another dimension)?   I also heard that the matter is annihilated when reaching the singularity, does it mean it disappear from this world, and violates conservation of energy? BAV99 forward voltage How do you spell the following date:  December 6 (AmE)  6 December (BrE)  Do you spell this date as 'December six' or 'December the sixth' in American English?  And in British English, do you spell it as 'Six December' or 'The sixth of December'? Can someone please suggest a query to directly see the number of answers I posted that were accepted. Right now I have to manually count from total given answers.  Hope there is an easy way. Why do I get the same result with rand() every time I compile and run? Determine the number of functions $f: \{1,2,3....,1999\}\to \{2000,2001,2002,2003\}$satisfying the condition that $f(1)+f(2)+...f(1999)$ is odd. Problems:Determine the number of functions $$f: \{1,2,3....,1999\}\to \{2000,2001,2002,2003\}$$ satisfying the condition that $f(1)+f(2)+...f(1999)$ is odd.   My Attempt: Each integer in domain has $4$ choices and therefore the total number of functions is $f$ is $4^{1999}.$ Since there are an equal number of functions that yield odd or even result, we can directly write that the number of functions satisfying the above condition to be $2*4^{1998}.$ I am unsure about this last claim and would like to prove it (if it is true.)Also is the answer to this problem correct?  Edit $1$: I tried to proceed further in the following manner:   For a given mapping $f$ let the number of integers assigned to $2000,2001,2002$ amd $2003$ be $p,q,r$ and $s$ respectively, where $p,q,r,s\in \mathbb{Z}^+$. Therefore the sum in question can be written as $f(1)+f(2)+...+f(1999)=2000p+2001q+2002r+2003s.$ Clearly the sum will be odd iff $q+s$  is odd. Also note that $p+q+r+s=1999\Rightarrow p+r$ is even. On the other hand if we let $q+s$ as even then $p+r$ must be even. Now the number of values of $p,q,r$ and $s$ for which the above two lemmas hold is the same and therefore if we let $f$ to be the function which maps elements form $\{1,2,3,...,1999\}$ to $\{2000,2001,2002,2003\}$ such that $q+s$ is even and $g$ to be the function  hich maps elements form $\{1,2,3,...,1999\}$ to $\{2000,2001,2002,2003\}$ such that $p+q$ is even we will get a bijection.(Maybe!) Limit of function of hyperbolic Baby Rudin claim: $1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}...$ converges This sequence is a rearrangement of the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}...$. Note that at this point in the text we do not have any theorem about the convergence of rearrangements.  Let $\{s_n\}$ be the sequence of partials sums of the series then for $n \ge 0$  $s_{3(n+1)} = \sum ^n _ {k=0} \frac{1}{4k+1} + \frac{1}{4k+3} - \frac{2}{4k+4}$  We can view it as the sequence(on $n$) of partials sums of  $\sum_0 a_n = \sum_0 \frac{1}{4n+1} + \frac{1}{4n+3} - \frac{2}{4n+4}$  Where $|a_n| = a_n = \frac{1}{4n+4}\{\frac{3}{4n+1}+\frac{1}{4n+3}\} \le \frac{1}{4n^2}$.  By the comparison test $s_{3(n+1)}$ converges to some real $\alpha$.  But $s_{3(n+1)+1} = s_{3(n+1)}+ \frac{1}{4n+5}$ and $s_{3(n+1)+2} = s_{3(n+1)}+ \frac{1}{4n+5}+\frac{1}{4n+7} $hence we have a partition of $\{s_n\}$ into subsequences which tend to $\alpha$ and this implies $s_n \rightarrow \alpha$.  Is my proof correct? Any alternative solutions are appreciated. How to check the tasks that are running at different run levels in debian? Path bar missing in file dialog after running software updater for Ubuntu 17.04 After running software updater, the path bar in the file dialog (for example the one you get when choosing Open File in in Google Chrome or Firefox) is missing. This is frustrating since it's no longer possible to copy a path from the terminal (for example) and paste it into the file dialog. Here is a screenshot of the dialog:     How can I get back the path bar? How do I turn on workspaces? (Why do I only have one workspace?) My server had been infected and the attacker spread a bunch of .5randomletters folders around my file system, with malicious code inside of them. I want to search for all the folders that begin with a dot, review them and then remove them.  How can I do that? Designer Added then removed by Visual Studio on load/unload Solving the matrix equation $AP=PB$, where $A$ and $B$ are similar Currently I'm stuck with this problem:     The following matrices are similar to each other.      $$A=\begin{bmatrix}-3&amp;0\\8&amp;4\end{bmatrix},\qquad B=\begin{bmatrix}-4&amp;4\\-2&amp;5\end{bmatrix},\qquad C=\begin{bmatrix}k&amp;1\\m&amp;-2\end{bmatrix}.$$      How do you find the matrix $P$ such that $AP = PB$?       And how can you calculate the values of $k$ and $m$?      Thank you in advance. Nested regression model Solve $x=y\frac{dy}{dx}-\left(\frac{dy}{dx}\right)^{2}$ Prove $g$ is a generator if $g^q=1 \pmod p$ Let $p$ be an odd prime and let $q=\frac{p-1}{2}$. If $g^q=-1\pmod p$ and $q$ is a prime, show that $g$ is a primitive root mod $p$.  I want to show $\{g,g^1,...,g^{p-1}\}=\{1,2,...p-1\}$. I argue by contradiction that if there're $x,y\in \{1,2,...p-1\},x&lt;y,g^x=g^y\pmod p$. Then $g^{y-x}=1\pmod p$. I don't know how to use the fact that $q$ is prime to lead to a contradiction. When I shoot a ball vertically upward, its velocity is decreasing since  there is a downward acceleration of about $9.8\,\mathrm{ms}^{-2}$.   I have read that at the top most point, when $v = 0$, the acceleration is still $9.8\,\mathrm{ms}^{-2}$ in the downward direction where $v=0$. That is, the acceleration is still the same.  But at the highest point, the ball is stationary, so it is not even moving. How can it accelerate? From time to time certain chat rooms (JavaScript and probably PHP, too) attract the attention of a fellow help vampire who thinks he can start sucking from the people in there.  Often their message end up being flagged as offensive (and those flags are likely to be validated, too, because people are annoyed by such behaviour) but it's kind of an inappropriate use of those flags.  I'm proposing a new chat flag type, which is especially meant for help vampires. Flagging it would present the flag to 10k users and owners of the room where the flag originated (no matter how much rep they have). When the flag is validated it would cause messages from this user to be shown with a low opacity for maybe three hours. It could also notify the user that help vampirism is a bad thing and that he needs to improve his behaviour.  Repeated flags of this type - obviously with some time in between in case multiple message which are really part of the same conversation are flagged - should then result in a temporary ban from the room where the messages got flagged. If that doesn't help the next step would probably be a temporary ban from all chat rooms.    For clarification, I'm not talking about the quick "this line doesn't work, why?" stuff but the kind of help vampires who get cocky even though they just got e.g. a link to a SO question which has a perfectly fine and fitting answer. Because hey, you didn't give them teh codez! How do I disable notifications in my personal profile for activity on pages I am an admin of? I'm an admin of several pages on Facebook (for technical stuff like RSS imports, etc.), but neither do I contribute content nor do I care about their content.  Still, as an admin, I get notifications (upper left hand, the globe icon) about activity (someone liked this, someone posted that) on those pages in my personal (!) profile .  Does anyone have the same issue? Is there any way to explicitly disable notifications for certain pages?  (This behavior seems to be new, by the way; I'm only seeing those notifications for a few weeks now.)