Will the site association bonus come after I already have the account?
What is the association bonus, and how does it work?
R raster: masking a raster by polygon also remove cells partially covered? I want to mask out a raster based on the values of a polygon: i.e., set to NA all values of the raster that are not covered by the polygons. However, it seems that raster::mask() masks out cells (i.e. set them to NA) even if they are partially covered?  This is similar to the question: , but unlike crop(), mask() does not have the snap="out" option.   Check: here the right point of the triangle touches a cell, yet that cells is masked out.   library(raster) library(sp)  r &lt;- raster(xmn=1, xmx=5, ymn=1, ymx=5, nrows=4, ncols=4) r[] &lt;- 1:length(r)  Sr1 = Polygon(cbind(c(2,4,4,1,2),c(2,3,5,4,2))) Sr2 = Polygon(cbind(c(4.1,4,2),c(2,3,2)))  SpP = SpatialPolygons(list(Polygons(list(Sr1), "s1"), Polygons(list(Sr2), "s2")), 1:2)  plot(mask(r, SpP)) plot(SpP,  add=TRUE) Find the domain of composition of two functions Let $$f(x)= \frac{x}{x-9} \text{ and } g(x) = \frac{-6}{x+5} $$   Find the domain of $f \circ g.$    Please show the steps and substitution/algebra to the problem. Thank you. Im trying to run the following code as a powershell script,but I cant get it to work. 1st,the following works with one -Include item but I cant seem to get it to work with multiple ones. 2nd, the -Exclude perimeter doesn't seem to work. I still get files from the C:\Windows and C:\Program Files directorys .  $Include = "*.zip","*.rar","*.tar","*.7zip" $exclude = "C:\Windows","C:\Program Files" Get-ChildItem "C:\" -Include $Include -Exclude $Exclude -Recurse -Force -ErrorAction silentlycontinue | Select-Object -ExpandProperty FullName   Note: The purpose of this script is to find all compressed files on the system. I know this is probably really simple but I just cant seem to get it to work. I need something explained or corrected:  In my number theory class we proved that there are an infinite number of primes using Euler's Phi Totient. It went something like this:  Let $M = p_1 p_2 \dots p_n$ be the product of all primes. Consider $1 &lt; A \le M$:  Some prime must divide $A$, call it $q$. Since $q$ must be one of the primes, $q$ must divide $M.$  So the $gcd(A,M) &gt; 1.$ Thus $\phi(M) = 1$ ...?   Which is not even and contradicts the theorem that $\phi(N)$ is even for $N&gt;2.$ Therefore there exists an infinite number of primes.  I get confused by the statement "Thus $\phi(M) = 1$"....  Did i possibly copy this proof down wrong? or am I missing something? Thank you in advance.   Edit: by consider a such that...i meant consider an integer '$a$' I replaced it with $A$ to hopefully make it more clear.   Edit2: I'm sorry I am not familiar with using the equation editor. This is not a homework assignment, just studying for my exam. I just want to be able to understand this or clearify it. Random piece of plastic I have recently purchased a hybrid bike with suspension (not sure if that is relevant to this question). It came with two pieces of plastic. Any ideas what they may be for? Is true that $\sum_{k=1}^n\frac{[kx]}{k}\leq[nx]$, for every $x\in\mathbb{R}$ and for every $n\in\mathbb{Z}^+$? Is true that $\displaystyle\sum_{k=1}^n\dfrac{[kx]}{k}\leq[nx]$, for every $x\in\mathbb{R}$ and for every $n\in\mathbb{Z}^+$?  My work:  Let $x\in\mathbb{R}$ be given. For every $k=1,\dots,n$, define $i_k\in\{0,1,\dots,k-1\}$ such that  $$ [x]+\frac{i_k}{k}\leq x&lt;[x]+\frac{i_k+1}{k}\hspace{20mm} (*). $$  On one hand, we have that $[kx]=k[x]+i_k$ for every $k=1,\dots,n$.  So  $\displaystyle\sum_{k=1}^n\dfrac{[kx]}{k}=n[x]+\displaystyle\sum_{k=1}^n\dfrac{i_k}{k}=[nx]+\displaystyle\sum_{k=1}^n\dfrac{i_k}{k}-i_n$  Thus  $$\displaystyle\sum_{k=1}^n\dfrac{[kx]}{k}\leq[nx]\Leftrightarrow\displaystyle\sum_{k=1}^n\dfrac{i_k}{k}\leq i_n\,.$$  On the other hand, from $(*)$ is direct that $i_k=[k(x-[x])]$. Thus  $$\displaystyle\sum_{k=1}^n\dfrac{[kx]}{k}\leq[nx]\Leftrightarrow\displaystyle\sum_{k=1}^n\dfrac{[k(x-[x])]}{k}\leq [n(x-[x])]\,.$$  In other words, we can assume that $x\in[0,1)$.   Now the questions is:  Is true that $\displaystyle\sum_{k=1}^n\dfrac{[kx]}{k}\leq[nx]$, for every $x\in[0,1)$ and for every $n\in\mathbb{Z}^+$?  Any simpler approach? Any thoughts? Which Distribution Does the Data Point Belong to? $a,b,c$ are distinct real numbers that $$(a^2)(1-b+c)+(b^2)(1-c+a)+(c^2)(1-a+b)=ab+bc+ca.$$ Prove $$\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}=1.$$ I have tried two different approaches to this problem but am stuck.  First I started by expanding the bottom of each fraction. $\frac{1}{(a^2)-2ab+(b^2)}+\frac{1}{(b^2)-2bc+(c^2)}+\frac{1}{(c^2)-2ac+(a^2)}=1$  I also tried to expand $(a^2)(1-b+c)+(b^2)(1-c-a)+(c^2)(1-a+b)=ab+bc+ca$   $(a^2)-(a^2)b+(a^2)c+(b^2)-(b^2)c-(b^2)a+(c^2)-(c^2)a+(c^2)b=ab+bc+ca$ $(a^2)-(a^2)b+(a^2)c+(b^2)-(b^2)c-(b^2)a+(c^2)-(c^2)a+(c^2)b-ab-bc-ca=0$ Then I added one to both sides $(a^2)-(a^2)b+(a^2)c+(b^2)-(b^2)c-(b^2)a+(c^2)-(c^2)a+(c^2)b-ab-bc-ca+1=0+1$ Help will be appreciated. How do you determine the winner of a hand? In Texas Hold'em poker, there are some cases where a particular card, often called kicker, acts as a tiebreaker between players to determine who wins the pot, or if the pot has to be shared.  For some combinations such as three or four of a kind, there is no doubt: the kicker is the 5th card of the chosen combination, and determines the winner.  Example     Alice has 7 and 10   Bob has 7 and K   On the board are 7, 7, 2, 3, 4   In that case, both have a three of a kind. But because Bob has a king and Alice only a 10, he wins.  Most of poker websites explaining the rules clearly mention the role of the kicker applying to three of a kind and double pairs. But much fewer say something about colors, straights and full houses, and I found many contradicting answers. I also asked the question to some used online poker players and they also gave me contradicting answers.  I have tried to search for official tournament rules, but most of them only explain what happens with bad behaviors, bad deals, incorrect or confusing betting, showing cards when you shouldn't, acting when it's not your turn, etc. without mentioning subtleties about combinations at all.  To simplify my question, I will take three examples; I think it's better to start with examples before getting to the general answer if one exists. So, what's happening in the following 3 examples? Could you give a more generalized answer?  Example 1 - Flushes     Alice has 2&clubs; and 3&clubs;   Bob has 4&clubs; and 5&clubs;   On the board are 6&clubs;, 8&clubs;, 10&clubs;, 2&hearts;, 5&diams;   Contradicting answer 1: it's a tie, because the highest card included in the flush is the 10, which everybody chooses to include in their 5 showdown cards. Contradicting answer 2: Bob wins, because he has the greatest private card that is part of the flush  Example 2 - Straight     Alice has 6 and K   Bob has 6 and J   On the board are 4, 5, 7, 8, 10   Contradicting answer 1: it's a tie, because the greatest card in the straight is the 8 for both Alice and Bob Contradicting answer 2: Alice wins, because she owns an extra king, compared to the jack of Bob  Example 3 - Full house     Alice has 3 and 7   Bob has 3 and 6   On the board are 3, 3, 2, 2, 5   Contradicting answer 1: it's a tie, because one is supposed to choose only five cards to make a combination, and a full house is already five cards. There couldn't be any kicker, and thus their showdowns are strictly equals. Contradicting answer 2: Alice wins because of her extra 7, compared to the 6 of Bob  Thank you very much for your answers.  Note: I'm unable to post next to you, so I edit my own post; strange not be able to answer to an answer.  Ok, So if I summarize what you are saying :   In the flush case, Bob wins because at some point, their hand differs. Technically, I can continue comparing the highest private card to decide who wins. I had it correct. In the straight case, if the highest card of the straight is public, then it's technically always a tie, no matter what the players had as second private card (asuming that only one of the two cards was part of the straight). In a full house if both players have the same triplet and the same pair, it's always a tie, no matter what the players had as second private card (assuming again that only one of the two was part of the full house).   Please confirm me that what I summarized is correct.  Thank you for your answer. How do I get the last column text to be at the bottom of the row?  That is I want the XXX to be aligned with the line containing the arrow. I tried playing around with \multirow{3}{-3ex}{XXX}, but obviously I don't understand \multirow. Code: \documentclass{article} \usepackage{tabularx} \usepackage{booktabs} \usepackage{multirow}  \begin{document}     \begin{tabularx}{\linewidth}[t]{l X p{1.3cm}}         &amp; Header &amp; Title \\\cmidrule(lr){2-2}\cmidrule(lr){3-3}         a. &amp; Some text text text text text text text text text text text text text text              text text text text text text text text text text              Want XXX on this line $\rightarrow$%          &amp; XXX\\     \end{tabularx} \end{document} Which way does the scale tip? Why aren't search engines indexing my content? fake proof of $\forall a. \forall b. a = b \to 1 = 0$ I saw a less formal version of this fake proof that claimed to prove $2=1$ but because it assumed $a=b$ from the start I knew why it was wrong. It does seem however that the proof can be used to prove an certain implication I find very odd however. Clearly $\forall a. \forall b. a = b \to 0 = 1$ is false because I could pick $a$ and $b$ to be $1$ and then I would have derived 0 = 1 with no other conditions. So there must be something wrong with this proof that I am not seeing.  proof of $\forall a. \forall b. a = b \to 0 = 1$:  $a = b \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (assume)\\ a+a=a+b  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (x=y \to f(x) = f(y), f(x)=a+x)\\ 2a=a+b \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  (a+a=2a)\\ 2a-2b=a+b-2b \,\,\, (x=y \to f(x) = f(y), f(x)=x-2b)\\ 2(a-b)=a+b-2b \,\, (distributive)\\ 2(a-b)=a-b \,\,\,\,\,\,\,\,\,\,\,\,\,\, (-b=b-2b)\\ 2=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (x=y \to f(x) = f(y), f(x) =x/(a-b))\\ 1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (x=y \to f(x) = f(y), f(x) =x-1) $  now that I have "proven" $\forall a. \forall b. a = b \to 0 = 1$ I can now pick a and b to be 1.  $ 1 = 1 \to 1=0 \,\,\,\,\,\, (a=1, b=1, above)\\ 1 = 1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (\text{reflexivity})\\ 1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (\text{modus ponens}) \\ $  So now it seems that I have managed to derive $1=0$. Clearly I messed up given that no one has been able to show $ZFC$ inconsistent and ostensibly this fits well within the most studied aspects of $ZFC$. Either that or a random guy that makes meme's online found the most profound result in mathematics ever.  So where have I missed the mistake? strictly positive elements in $C^*$-algebra Let $A$ be a $C^*\text{-algebra}$ and $A_+$ denote the positive elements. An element $a\in A_+$ is called strictly positive if $\overline{aAa}=A$. Want to find the following:a)What are the strictly positive elements of $C_0(\Omega)$ where $\Omega$ is locally compact Hausdorff space, and $C_0(\Omega)$ is the space of continuous complex valued functions that vanishes at $\infty$.b)If $A$ is unital, then $a\in A_+$ is strictly positive iff $a$ is invertible.c)if $(e_n)$ is an approximate identity of $A$, then $a:=\sum_{n=}^{\infty}\frac{1}{2^n}e_n$is strictly positive.What I tried and know:a) I know that positive elements in this space are functions with positive images, and thus, strictly positive elements are functions with positive images and now real root. is this correct? and how can I show that the positive elements are these kind of functions.b) I figures out that if $a$ in invertible then $a$ is strictly positive, but no idea how to do the other direction.c)I did this but using another characterization of strictly positive elements but want to do it with this one, and I don't know how.Thank you for your help. This is by far the weirdest problem I've come across,  My mac can perfectly access internet through safari web browser, but when I tried to do   ping www.google.com   through the terminal, it's requests is being timed out.  None of the other apps are able to reach internet either. Dropbox, Skype, Spotify all showing offline, except the browser itself  I'm guessing something bad with the DNS, but spend 2 hours trying to fix this with no luck Received W-2 but was never paid I just now received a W-2 from a company that I worked for 3 days as a contractor. I had to leave country for 2 months due to a family emergency. Since the employment duration was very short, I was not paid. I was told that they don't want to charge the client because 3 days is not enough contribution.  Since I had to leave country the same day, I did not argue with them and told it's okay if they feel right.  Now I have a W-2 from them saying they have paid me about $1500. I contacted them but I have not received any reply yet.  I just want to be prepared. What can I do here? Will free-fall object into black hole exceed speed of light $c$ before hitting black hole surface? In Newtonian mechanics, if we throw an object in against direction of gravity with speed $v$ and it achieve max height of $h$. Now if we allow object to fall from that height $h$, it will eventually attain speed $v$ when it reach position where we launch it.   Now applying same idea to a black hole in general relativity. Speed require to escape black hole gravity is greater than $c$, so if we throw something into black hole with almost the speed of light, the object speed will exceed speed of light $c$ before hitting black hole surface! How relativity explain this? Can space-time curvature reduce speed of this freely falling object from attaining speed of light? In the midst of attempting to prove that for a commutative ring $A$ with identity, and an ideal $I$ of $A$, $I = \operatorname{rad}(I)$, where $\operatorname{rad}(I) = \{x: x^m \in I, m &gt;0\}$, implies that $I$ is an intersection of prime ideals, I've run into a little bit of confusion.   An answer to another question on this site, , indicates a strategy that I am attempting to use. I know that if $x$ is an element of $A\setminus I$, then so too is $x^m$ for any positive integer $m$. I use this fact to consider a subset $X = \{x^m : m &gt; 0\}$, which is closed under multiplication, and use Zorn's lemma to find an ideal of $A$ containing $I$ maximal with respect to the constraint that said ideal, $J$, is disjoint from X.  I want to prove $J$ is prime by contradiction. I suppose that $J$ were not prime; this also means that $J$ is not maximal in $A$. Then there would exist two elements, $f$ and $g$, of $A$ so that $fg$ is in $J$, but neither $f$ nor $g$ is in $J$. Following the prompting of the aforementioned answer, I know that $J + \langle a\rangle$ and $J + \langle b\rangle$ would be ideals strictly containing $J$, and since J is the maximal ideal disjoint from $X$, both $J +  \langle a\rangle$ and $J + \langle b\rangle$ contain at least some elements of $X$ (I am tempted to say all of them, but I'm not entirely sure that's true, in case for example $\langle a\rangle$ started containing powers of $x$ at $x^2$, in which case it isn't necessarily true that $\langle a\rangle$ contains $x$ itself), and so both $\langle a\rangle$ and $\langle b\rangle$ must themselves contain elements of $X$.  And this is where I am stuck. I feel like there is something I should know about multiplicative subsets of rings that would help me arrive at the desired contradiction, but I can't remember what it might be. I've been looking for a simple regex for URLs, does anybody have one handy that works well?  I didn't find one with the zend framework validation classes and have seen several implementations. Formatting floats differently based on placement