I try to conect to mysql with a webservice and AsyncTask
NullPointerException accessing views in onCreate()
SQL JOIN - WHERE clause vs. ON clause After reading it, this is not a duplicate of . The answer may be related (or even the same) but the question is different.    What is the difference and what should go in each?  If I understand the theory correctly, the query optimizer should be able to use both interchangeably. 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? Archimedes' principle for two liquid layers Problem: I have a cylindrical vessel of height $H$ and radius $R$. There are two liquid layers in the vessel. The first has density $D_1$ and height $h_1$, the second has density $D_2$ and height $h_2$. The second liquid is floating on the first liquid (thus $D_2 &lt; D_1$) and they are both entirely within the vessel ($h_1+h_2 \le H$). Now I have a cube of density $D$ and edge length $n$. The cube fits into the vessel ($n \le \sqrt2R$) but it might not be completely inside it ($n&gt;H$ is possible). I place it into the vessel such that one side is horizontal (the cube is not lopsided). Some liquid might overflow as a result. How do I use Archimedes' principle to calculate the topmost liquid level after placing the cube? I can ignore damping and other physical effects. EDIT: Note that the only values I know are $H, R, D_1, h_1, D_2, h_2, n$ and their abovementioned constraints.  My workings: I tried using the formula directly, where $\text{weight of cube}=\text{weight of displaced liquid}$. I first get $Dn^3=D_2n^2s_2$ where $s_2$ is the height submerged in the second liquid. If $s_2 \le h_2$ then the cube is only submerged in the second liquid. So the answer is $\min(H, h_1+h_2+{n^2s_2\over\pi R^2})$, accounting for overflows. This case is simple. But if $s_2&gt;h_2$ then the cube can be submerged in the first liquid too. I thought of subtracting the weight of the second liquid displaced to obtain the height submerged in first liquid, but realize that I don't know that. The weight displaced depends on the final liquid height, which can be affected by overflows, whether the cube fits totally within the two liquids, etc. In fact, it can be the case that after the liquid level rises, the cube now displaces more of the second liquid, causing it to not be submerged in the first (is that even possible?) It seems very messy and I have no idea how to start. :( Does anyone have a nice solution to this problem? How to find distance from the latitude and longitude of two locations? How to use \def to redefine \end{split}? When I use \def to redefine \end{split}, there is an error.  \documentclass{article} \usepackage{amsmath}  \begin{document} \def\be{\begin{equation}} \def\ee{\end{equation}} \def\bs{\begin{split}} \def\es{\end{split}}  \be     \bs         a\\         b     \es \ee  \end{document}   The error is:      Runaway argument?      a\ b \es \ee       ! Paragraph ended before \split was complete.   However, if I don't use \def\es{\end{split}} and just modify the code to be  \documentclass{article} \usepackage{amsmath}  \begin{document} \def\be{\begin{equation}} \def\ee{\end{equation}} \def\bs{\begin{split}} \def\es{\end{split}}  \be     \bs         a\\         b     \end{split} \ee  \end{document}   it's OK. What's the problem? Thank you! Is it possible to use "negative sound waves" to "cancel out" a sound to create silence? For every $n \in \mathbb{N}$, find a function $f$ which is differentiable $n$ times at $0$, but not $n+1$ times. For every $n \in \mathbb{N}$, (in this problem $\mathbb{N}$ starts at $n = 1$) find a function $f$ which is differentiable $n$ times at $0$, but not $n+1$ times.  The function I chose is the following:  $$ f_n(x) = \begin{cases} x^{n+1} &amp;\mbox{if } x \le 0\\ x^{n} &amp;\mbox{if } x &gt; 0 \end{cases} $$  Now I can't think of the best way to prove that $f$ is differentiable  $n$ times at $0$ but not $n+1$ times. ASP.NET MVC 3 localization with DisplayAttribute and custom resource provider I want to find out the number of possible combinations of $x$ numbers that sum to $y$. For example, I want to calculate all combination of 5 numbers, which their sum equals to 10.  An asymptotic approixmation is also useful. This question seems to be very close to number partitioning, with the difference that a number can be 0. See:    All possible partitions for sum 10 and 3 positions that can be zero, are 63 possiblities: (numbers shown as 3 digits)  019 028 037 046 055 064 073 082 091 109 118 127 136 145 154 163 172 181 190 208 217 226 235 244 253 262 271 280 307 316 325 334 343 352 361 370 406 415 424 433 442 451 460 505 514 523 532 541 550 604 613 622 631 640 703 712 721 730 802 811 820 901 910 What could happen if I use 2 cycle fuel in a 4 cycle engine? First off, this is NOT to be confused with the following question:    2 cycle oil is used for my weed eater. There is no way I could use a whole gallon in a year.  Sometimes I run out of gasoline for my 4 cycle lawn mower, and that 2 cycle mix fuel is sitting there getting really old.  If I run that in the lawn mower, am I simply going to create extra pollution or could the 2 cycle fuel cause any damage to the 4 cycle engine? I am a British citizen and US permanent resident. I rarely need visas. However, I am considering a Mediterranean cruise in the spring or summer of 2019.  Unlike British politicians, I like to plan ahead and be prepared. If I am going to need a Schengen visa, I want to assemble the documentation early next year, so I can apply to the appropriate country as soon as I have an itinerary and it is at most 90 days before travel. Given how things are going, I am preparing for a hard, no-agreement, Brexit.  For trade, that would mean WTO rules. For personal travel, is there a similar fall-back that would allow me visa-free visits to the Schengen countries? $\int f=0$ implies $f=0$ almost everywhere Let $f$ be nonnegative and measurable. Show that $\int f =0$ implies $f=0$ almost everywhere (except possibly on a set of measure zero).  My thoughts. Take a set E with $m(E) \neq 0$. Suppose $f \neq 0$ on E. That means $\int_E f \neq 0$. Since $f$ is nonnegative this implies $\int f &gt;0$. That contradicts the given information, so $f$ must be 0 almost everywhere. I recently installed Ubuntu 12.10 and it requires a passphrase to boot up (I installed it with an encrypted file system).  Do I have to reinstall to change to a standard unencrypted file system? Degree of minimum polynomial at most $n$ without Cayley-Hamilton? What books are recommended for an advanced undergraduate course in electrodynamics? What TV movie had lost space cadets who piloted yellow tripod robots, and a squid bad guy? I'm trying to find a movie I watched on TV, probably in the 90s. I think it was a made-for-TV movie.  It was about a ship with an AI. The ship was crewed by young adults. I think they were cadets?   There was an accident (like a warp drive malfunction) and they end up in another part of space where they meet a female alien whose ship had been shot down.  I remember they had manned robots they would pilot and they took them to a planet to get upgraded and they are attacked by more manned robots. They had three legs and they were yellow, I think.  I also remember the baddie had a holo-projector or something like that which allowed him to change his appearance. He commanded a massive fleet and he worked for someone he spoke to in a video chat that looked like a squid. What resolution would be needed for a billboard advertisement? Test -n gives unexpected result I understand test -n &lt;expression&gt; to return false if the expression evaluates to a string length of greater than 0.  Why then does the following happen?  Macbook:~ echo ${#undeclared_variable} 0 Macbook:~ test -n $undeclared_variable &amp;&amp; echo Foo Foo   Bash 4.x  My expectation is that since test -n $undeclared_variable should evaluate false, that Foo should not be echoed.  If I use [[  ]] instead of test, it works as expected. When incident light passes through a hydrogen gas, for example, does it have 50% chance (since it's a 1:1 ratio of protons to electrons) of getting absorbed by the proton? Any chance at all? If no, why not? Does a proton have a "bandgap"? If yes, what happens when a photon is absorbed by a proton? It is a known fact that if $R$ is a UFD, then $R[X_1,X_2,\dots]$ is also a UFD, but there is a subtlety that is making me uncomfortable.  The standard approach essentially goes something along the lines of...let $f$ be a polynomial, so it only involves finitely many variables, say up to $X_1,\dots,X_n$. Then since $R[X_1,\dots,X_n]$ is a UFD, $f$ has a factorization into irreducibles. Then $f$ cannot have a factorization involving indeterminates $X_N$ not in $R[X_1,\dots,X_n]$. If it did, then $f$ would have two factorizations in the UFD $R[X_1,\dots,X_n,\dots, X_N]$, a contradiction.  This seems to assume that the prime factorization in $R[X_1,\dots,X_n]$ is still a prime factorization in $R[X_1,\dots,X_n,\dots,X_N]$. But how can you see that irreducibles in $R[X_1,\dots,X_n]$ are still irreducibles in $R[X_1,\dots,X_m]$ for $m&gt;n$? This does not seem obvious.  Let $R_k=R[X_1,\dots,X_k]$. The explanantion I found is this: Suppose $p\in R[X_1,\dots,X_n]$ is irreducible in $R_n$. Suppose $p=ab$ for $a,b\in R_m$. Evaluating $X_1,\dots,X_n$ at $1$, you get  $$ \bar{p}=\overline{ab}\in R[X_{n+1},\dots,X_m] $$ But $\bar{p}\in R$, which implies that $a$ and $b$ do not involve any variables $X_{n+1},\dots,X_m$. So $a,b\in R_n$, and thus one is a unit, hence a unit in $R_m$.   The part I don't follow is how evaluation at $1$, implies $a$ and $b$ do not have indeterminates other than $X_1,\dots,X_n$. Isn't it possible that some of the higher indexed indeterminates disappear when you evaluate? Suppose for instance $p\in R_1$, and $a\in R_2$ is $a=-X_1X_2+X_2$. Then evaluting at $1$ gives $\bar{a}=0$ so we may not get the desired contradiction since $\overline{ab}=0\in R$?