Is there a rule of thumb for removing spiciness from a dish, caused by things like jalepenos, cayenne pepper, black pepper, red pepper, szechuan peppercorn, etc?   I have read that honey and sugar may combat the spiciness in asian sauces, but have not heard of other ways to remove the kick found in many other dishes.
Sometimes when I'm making a sauce or soup that needs to be a bit spicy, I have no troubles spicing how I want to.  But if the dinner party is a bit late, or I made the sauce in advance to really draw out all the flavors of the spices, vegetables and meat, sometimes I end up with a way too spicy/hot dish.  You can always pour a bit water and try to dilute it a bit, but that also takes away all of the good flavor. So is there some trick that I'm missing I can use, when this happens to me (other than being more careful or using the spices later)?
Can I use the Unity launcher icon to minimize applications/windows? Add a note about downvoting on Meta What is difference between int.class and Integer.TYPE in java? Does "staff" take a plural verb? Which one of these two statements is correct?     Our staff do ...      Our staff does ...   And is staffs ever correct? Rejection of text containing SQL statements OK, I've been having weird problems for the last 1/2 hour trying to  on StackOverflow, and I keep getting this result (on Firefox 3.6)     The connection was reset      The connection to the server was reset while the page was loading.         The site could be temporarily unavailable or too busy. Try again in a few        moments.   If you are unable to load any pages, check your computer's network        connection.   If your computer or network is protected by a firewall or proxy, make sure        that Firefox is permitted to access the Web.      The problem seems to occur when I try to post some code with 'd--- table' in it, where 'd---' is replaced by the word drop.  I cannot post the following 3 lines, in either SO or this site, if they include the well-known SQL statements DROP.  stat.executeUpdate('d--- table if exists people;'); stat.executeUpdate('create table people (name, occupation);'); prep = conn.prepareStatement(...     'insert into people values (?, ?);');   test again:  'dro* table' 'drop ball' drop the table drop fable drop    t*ble droptable   drop a table  The evil words for me are "drop" and "table"... how come you guys can post? :(  edit: Could someone please suggest another online site that is not stackoverflow that is a quick way to try out a POST request of a text field?  UPDATE: Never mind. I don't know what the answer is, but the same problem occurs on Google's website. So it's not stackoverflow. Now to go to my IT Dept... Why does Captain America's Shield make noise? Proving any number of the form $2^n$ can't be written as the sum of k consecutive numbers. I think i have proved that any number of the form $2^n$ can't be written as the sum of k consecutive positive integers, but i would be grateful if i could have some clarification as to whether this is a correct and valid proof or not.  Let $g=x+(x+1)+(x+2)+(x+3)+...+(x+(k-1))$, hence g is the sum of k consecutive positive integers.  $g=kx+\sum_{d=1}^{k-1}d$  Hence, $g=kx+\frac{k(k-1)}{2}$  Rearranging for x we find, $x=\frac{g}{k}-\frac{(k-1)}{2}$  So, for certain values of $k$, we can be sure that a positive integer value of $x$ can be found, and hence $g$ can be written as the sum of $k$ consecutive positive integers.There are two cases for $k$ to considor:  Case 1)  $k$ is odd. If $x$ exists, it must be the case $k$ | $g$, as for odd values of $k$, $\frac{(k-1)}{2}$ will alwyas be an integer, and hence $x$ will.  Case 2)  $k$ is even. $k$ does not divide $g$, but $\frac{g}{k}$ must be a multiple of $\frac{1}{2}$, and hence equals $\frac{h}{2}$, for some odd $h$.  Letting $j=2^n$, assume that we can write $j$ as the sum of $k$ consecutive positive integers.  If $k$ is odd, then if a $k$ exists, it must divide $j$, by Case 1. But, $j$ has no odd factors as it is a power of two, hence $k$ could not be odd.  If $k$ is even, by Case 2 it mut be the case $k$ does not divide $j$. Also, $\frac{j}{k}=\frac{h}{2}$, for some odd $h$.  But, this implies $2j$ has an odd factor, $h$. but, as $2j$ is a power of $2$, it can't have odd factors. So, $k$ can't be odd.  Hence, $k$ is neither odd or even, and this is a contradiction to our assumption. Hence no $x$ exists, and hence and number of the form $2^n$ can't be written as the sum of $k$ consecutive positive integers.  Any feedback on this proof would be appreciated. Are files created in "/storage/emulated/0" supposed to appear in "/storage/sdcard0"? From what I guess, internal storage is available by every user. If emulated storage is a storage location, made for a specific user, are files created in /storage/emulated/0 supposed to appear in /storage/sdcard0? I assume no, because files that are user created should show up in emulated only, otherwise files created by the user would be available for all other users to see. myOpenID shuts down February 1, 2014; add an alternative login method to your account Janrain has , one of the many OpenID providers. MyOpenID will not be available starting February 1, 2014. (An   hinted at the same but did not set a date for end of service.)   This decision was made by Janrain, and Stack Exchange did not have any part in it. You can still use myOpenID to log in to Stack Exchange, but be aware that any outages in the myOpenID service will mean that you are unable to log in to Stack Exchange until Janrain fixes it – which may take several days, or weeks, or never. After February 1, you will no longer be able to use myOpenID to log in to Stack Exchange. Stack Exchange will not be able to answer any questions about why myOpenID is down or when it will be working again. If you do not have any other login credentials on your account by February 1, you will be locked out, and Stack Exchange will be unlikely to help you recover your account.  To avoid getting locked out of your Stack Exchange account, we strongly suggest that anyone using myOpenID as their primary (or even backup) login add a new login method to their Stack Exchange account.  We're also going to remove the "myOpenID" button from the sign-up/login page soon, in order to discourage new users from signing up using myOpenID.  Here's how to add a new login to your account. You must be logged into your account when you begin. (This information is also available .)  Open up your user profile page (by clicking on your login name at the top of the screen) and click the "my logins" link above the bio box on the right hand side.   You should see a list of all of your logins.  To add a new login, click "add more logins..." at the bottom of the box. Choose the new login from the list and follow the steps to confirm.    If you choose, you can remove your myOpenID login by clicking the same "my logins" link, then clicking "remove" next to myOpenID.  If you are adding a Stack Exchange OpenID, you will receive a confirmation email. You must click the link in the email to confirm your email address before the Stack Exchange OpenID will show up under "my logins". Please make sure that you are already logged into your account when you click the confirmation link; otherwise, you will create a new account that needs to be merged with your old one.    If you encounter any trouble adding logins, or accidentally create a separate account that you would like merged, please fill out our . Why many abstract nouns get an indefinite article while it's taught that abstract nouns shouldn't get it? Prove that the functions $f(z)$ and $\overline{f(\overline{z})}$ are simultaneously holomorphic.   I take this to mean that $f(z)$ is holomorphic if and only if $\overline{f(\overline{z})}$ is holomorphic.  Let $g(z)=\overline{f(\overline{z})}$. Note that $\overline{g(\overline{z})}=f(z)$. So it suffices to prove that if $f(z)$ is holomorphic, then $g(z)$ is holomorphic.  Write $f(z)=u(z)+iv(z)$. Since $f(z)$ is holomorphic, the real and imaginary parts satisfy the Cauchy-Riemann equations:  $$\frac{\partial{u(z)}}{\partial{x}} = \frac{\partial{v(z)}}{\partial{y}}, \frac{\partial{u(z)}}{\partial{y}} = -\frac{\partial{v(z)}}{\partial{x}}.$$  We have $g(z) = u(\overline{z})+i(-v(\overline{z}))$. To prove that $g(z)$ is holomorphic, we must prove that its real and imaginary parts satisfy the Cauchy-Riemann equations:  $$\frac{\partial{u(\overline{z})}}{\partial{x}} = \frac{\partial{(-v(\overline{z}))}}{\partial{y}}, \frac{\partial{u(\overline{z})}}{\partial{y}} = -\frac{\partial{(-v(\overline{z}))}}{\partial{x}}.$$  How can we obtain this from the above relations? Where can I find medicine? I know it is possible to be stung by bees in Animal Crossing: New Leaf.   I learned that the hard way when I was trying to find my furniture tree. Now, I have an eye swollen shut and all of my towns people are making fun of me. They're all being kind enough to inform me that I need Medicine, but they seem to be enjoying themselves at my expense too much to tell me where to actually find the Medicine.  Where can i get medicine? I'd really like it if my townspeople stopped making fun of me. How to find out when an Oracle table was updated the last time Can I find out when the last INSERT, UPDATE or DELETE statement was performed on a table in an Oracle database and if so, how?  A little background: The Oracle version is 10g. I have a batch application that runs regularly, reads data from a single Oracle table and writes it into a file. I would like to skip this if the data hasn't changed since the last time the job ran.  The application is written in C++ and communicates with Oracle via OCI. It logs into Oracle with a "normal" user, so I can't use any special admin stuff.  Edit: Okay, "Special Admin Stuff" wasn't exactly a good description. What I mean is: I can't do anything besides SELECTing from tables and calling stored procedures. Changing anything about the database itself (like adding triggers), is sadly not an option if want to get it done before 2010. Irreducible polynomial which is reducible modulo every prime How to show that $x^4+1$ is irreducible in $\mathbb Z[x]$ but it is reducible modulo every prime $p$?   For example I know that $x^4+1=(x+1)^4\bmod 2$. Also $\bmod 3$ we have that $0,1,2$ are not solutions of $x^4+1=0$ then if it is reducible the factors are of degree $2$. This gives that $x^4+1=(x^2+ax+b)(x^2+cx+d)$ and solving this system of equations $\bmod 3$  gives that $x^4+1=(x^2+x+2) (x^2+2x+2) \pmod 3$. But is there a simpler method to factor $x^4+1$ modulo a prime $p$? Convert from MySQL datetime to another format with PHP I have a datetime column in MySQL.   How can I convert it to the display as mm/dd/yy H:M (AM/PM) using PHP? How can I show that $ab \sim \gcd (a,b) {\operatorname{lcm} (a,b)}$ for any $a,b \in R \setminus \{0\}$? Let $R$ be a UFD. Then for any two elements $a,b \in R \setminus \{0\}$ $ab \sim \gcd (a,b){\operatorname{lcm} (a,b)}$.   My attempt :  Let $d = \gcd (a,b)$.Then $d|a$ and $d|b$.Then $\exists x,y \in R$ such that $a = dx$ and $b = dy$.Now it is to be shown that $c = dxy$ is the lcm of $a,b$.Which I find difficulty to show.It is clear that $a|c$ and $b|c$. Now if $u$ is any common multiple of $a$ and $b$.Then $dx|u$ and also since $y|b$ we have $y|u$ i.e. $dxy|u^2$ $\implies c|u^2$.But I have to show $c|u$ to complete the proof which I fail to prove.Please help me to complete the proof.  Thank you in advance. What do you do in Dungeon World when someone wants to perform an action they don't have a move for? I remember that a couple of years ago a friend showed me and some other people the following expression:  $$\lim_{n\to \infty}\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}.$$  As shown below, I can prove that this limit exists by the monotone convergence theorem. I also remember that my friend gave a very dubious "proof" that the value of the limit is $\frac{1}{e-1}$. I cannot remember the details of the proof, but I am fairly certain that it made the common error of treating $n$ as a variable in some places at some times and as a constant in other places at other times. Nevertheless, numerical analysis suggests that the value my friend gave was correct, even if his methods were flawed. My question is then:  What is the value of this limit and how do we prove it rigorously?  (Also, for bonus points, What might my friend's original proof have been and what exactly was his error, if any?)  I give my convergence proof below in two parts. In both parts, I define the sequence $a_n$ by $a_n=\frac{1^n+2^n+\cdots+(n-1)^n}{n^n}$ for all integers $n\ge 2$. First, I prove that $a_n$ is bounded above by $1$. Second, I prove that $a_n$ is increasing.   (1) The sequence $a_n$ satisfies $a_n&lt;1$ for all $n\ge 2$.  Note that $a_n&lt;1$ is equivalent to $1^n+2^n+\cdots+(n-1)^n&lt;n^n$. I prove this second statement by induction. Observe that $1^2=1&lt;4=2^2$. Now suppose that $1^n+2^n+\cdots+(n-1)^n&lt;n^n$ for some integer $n\ge 2$. Then  $$1^{n+1}+2^{n+1}+\cdots+(n-1)^{n+1}+n^{n+1}\le(n-1)(1^n+2^n+\cdots+(n-1)^n)+n^{n+1}&lt;(n-1)n^n+n^{n+1}&lt;(n+1)n^n+n^{n+1}\le n^{n+1}+(n+1)n^n+\binom{n+1}{2}n^{n-1}+\cdots+1=(n+1)^{n+1}.$$  (2) The sequence $a_n$ is increasing for all $n\ge 2$.  We must first prove the following preliminary proposition. (I'm not sure if "lemma" is appropriate for this.)  (2a) For all integers $n\ge 2$ and $2\le k\le n$, $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We observe that $k^2-1\le kn$, so upon division by $k(k^2-1)$, we get $\frac{1}{k}\le\frac{n}{k^2-1}$. By , we may find:  $$\frac{k+1}{k}\le 1+\frac{n}{k^2-1}\le\left(1+\frac{1}{k^2-1}\right)^n=\left(\frac{k^2}{k^2-1}\right)^n.$$  A little multiplication and we arrive at $\left(\frac{k-1}{k}\right)^n\le\left(\frac{k}{k+1}\right)^{n+1}$.  We may now first apply this to see that $\left(\frac{n-1}{n}\right)^n\le\left(\frac{n}{n+1}\right)^{n+1}$. Then we suppose that for some integer $2\le k\le n$, we have $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$. Then:  $$\left(\frac{k-1}{n}\right)^n=\left(\frac{k}{n}\right)^n\left(\frac{k-1}{k}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}\left(\frac{k}{k+1}\right)^{n+1}=\left(\frac{k}{n+1}\right)^{n+1}.$$  By backwards (finite) induction from $n$, we have that $\left(\frac{k}{n}\right)^n\le\left(\frac{k+1}{n+1}\right)^{n+1}$ for all integers $1\le k\le n$, so:  $$a_n=\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots+\left(\frac{n-1}{n}\right)^n\le\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}&lt;\left(\frac{1}{n+1}\right)^{n+1}+\left(\frac{2}{n+1}\right)^{n+1}+\left(\frac{3}{n+1}\right)^{n+1}+\cdots+\left(\frac{n}{n+1}\right)^{n+1}=a_{n+1}.$$  (In fact, this proves that $a_n$ is strictly increasing.) By the monotone convergence theorem, $a_n$ converges.  I should note that I am not especially well-practiced in proving these sorts of inequalities, so I may have given a significantly more complicated proof than necessary. If this is the case, feel free to explain in a comment or in your answer. I'd love to get a better grip on these inequalities in addition to finding out what the limit is.  Thanks! How do I clear the render border? In the 3d viewport I am using a border region to only render things inside the selected region. I did this with CtrlB. I cannot get rid of the region. How can I make the border disappear? Reactjs convert html string to jsx