Facebook is showing others my latest new added friends, saying something like "Aiwnas is now friends with X." Is it possible to stop Facebook from doing this? Or make it private for a customized list of my friends?
Is there a privacy setting on Facebook that will stop the posting that is made on your friend's wall when you make a new friend?  I do not want this to appear on my friend's wall when I make a new friend
Because the default 'Profile Graph' tool was not sufficient, I wrote a custom VBA script that creates scaled and exaggerated profile cross-sections in ArcGIS but I need more features:   Visualize more than 1 surface Visualize vertical and directional boreholes Ability to symbolize / edit the output Linked with underlying data - changes in terrain (or profile 3D line) reflected in output   A few years ago I tested an extension by A-Prime software called 'CrossView For ArcGIS' but I am not sure if it is compatible with ArcGIS 10 (Contacted A-Prime but they have not yet responded)   Area there any alternatives? Other 3rd party software? A functioning script? that can do this? With Ubuntu 16.04 laptop doesn't suspend when lid is closed I have a Dell 7737 laptop, and with Ubuntu 16.04 it no longer suspends when the laptop lid is closed. With 14.04 closing the lid put the laptop to suspend as expected. However, I reloaded the system with Ubuntu 16.04 and now closing the laptop does nothing. The laptop continues to run as if nothing happened.  When I manually select "Suspend" from the menu it works as expected. (Though it does prompt me to double login on resume. No, not encryption login followed by OS login. This is a double OS login.)  NOTE: Altering the System Settings|Power doesn't seem to affect anything.  Any ideas on what changed from 14.04 to 16.04 in this area? Prove that for real number $p\geq  0$ we have   $$\lim_{n\to\infty}\frac{(1^{1^p} \cdot2^{2^p}\cdots n^{n^p})^\frac{1}{n^{p+1}}}{n^\frac{1}{(p+1)}} = \exp\left(\dfrac{-1}{(p+1)^2}\right).$$ I'm using an iPhone, but I would imagine it's the same for other mobile devices as well.  If you browse to any stack exchange site, you get a very well made mobile site.  The only gripe that I have is that I can't find any way to view the number of visitors to my questions?  Am I missing something or could this possibly be integrated into the mobile renderer? Fell for fake sugar daddy. How do I get out? Suppose that $(s_n)$ converges to s. Prove that $(s_n^2)$ converges to $s^2$ I am really struggling with my proofs class, I don't really understand how to prove a statement like this, or what the epsilon is standing for..  Suppose that $(s_n)$ converges to s.  Prove that $(s_n^2)$ converges to $s^2$ directly without using the fact that $lim(s_nt_n)=st$  Suppose that $(s_n)$ converges to s. Then, since $(s_n)$ is convergent, there exists an $M_1$ such that $|S_n|&lt;M_1$, for all $n\in \!\,\mathbb{N} \!\,$. Let $M=M_1+|S|$ (I don't completely understand this step...). Then given $ε&gt;0$, there exists N such that n>N implies that $|s_n-s|&lt;ε/M$. Thus for n>N, we have $|s_n^2-s^2|=|s_n-s|*|s_n+s|&lt;(ε/M)(M)=ε$. Hence, $s_n^2$ converges to $s^2$. Windows Active Directory naming best practices? This is a  about Active Directory domain naming.   After experimenting with Windows domains and domain controllers in a virtual environment, I've realized that having an active directory domain named identically to a DNS domain is bad idea (Meaning that having example.com as an Active Directory name is no good when we have the example.com domain name registered for use as our website).   , but I'm still not sure about what other rules there are around naming Active Directory domains.  Are there any best practices on what an Active Directory name should or shouldn't be? Is there any way to hide taskbar labels and not combine taskbar buttons? There are three options for customizing taskbar buttons in Windows 7:   Always combine, hide labels Combine when taskbar is full Never combine   Is there any way to hide labels and not combine taskbar buttons? Or even better, to exclude only certain applications from combining?  My main problem is that I switch a lot between two instances of a single application, and having to hover over combined icon and then choosing the correct one. Alt-Tabbing is not a good alternative when I'm switching between more than three applications; clicking on taskbar only once is what I'm looking for. This is all in iTerm2 on OS X.  I used to have  export TERM='xterm-256color'   in my .bashrc. This meant Vim in tmux did use 256 colors.  And once I added  set -g xterm-keys on   then keyboard shortcuts with modifiers worked fine in Vim. Namely: shift+left/right that I mapped to switch Vim tabs, ctrl+up/down that I mapped to move ("bubble") lines, and shift+left/right which worked out of the box to jump by word in the command-line mode (e.g. when typing something like :e foo bar baz).  However, this setup had the problem that the Vim background color only shows behind text, as mentioned .  So I removed  export TERM='xterm-256color'   from my .bashrc and instead put this in my .tmux.conf:  set -g default-terminal "screen-256color"   That fixed the Vim background color, but broke the keyboard shortcuts - they do unexpected things (move the cursor, delete text) instead.  By using ctrl+v to insert the verbatim output from the key combinations (as described ), I was able to work around it:  map [1;5A &lt;C-Up&gt; map [1;5B &lt;C-Down&gt; map [1;2D &lt;S-Left&gt; map [1;2C &lt;S-Right&gt; cmap [1;2D &lt;S-Left&gt; cmap [1;2C &lt;S-Right&gt;   This makes the shortcuts work, but it doesn't feel like the right solution. Could anyone tell me what's happening here and how to fix it? Delete own query on data explorer Let $C$ be the Banach space of all complex continuous functions on $[0,1]$ endowed with the supremum norm. Let $B$ be the closed unit ball of $C$. Does there exist a $\Lambda\in C^*$ for which $\Lambda (B)$ is open? Note that $C$ is a locally convex space and $B$ is convex, balanced and closed set, by the separation theorem of convex sets,we know there exists $\Lambda\in C^*$ such that $|\Lambda f|\leq 1$ for all $f\in B$ and $|\Lambda f|&gt;1$ for $f\in B^c$. But how about this when we consider some concrete space above? How can I find $\gcd(a^m+1,a^n+1)$ with $a,m,n$ positive integers? How can I find $\gcd(a^m+1,a^n+1)$ with $a,m,n$ positive integers?    I have this idea:  Let $d=\gcd(m,n)$. Then there exist positive integers $x,y$ such that $mx-ny=d$ (WLOG). We shall find $G=\gcd(a^m+1,a^n+1)$.  If $m,n$ are odd, then $d$ is odd, therefore one and only one of $x,y$ is even. We have:  $$a^{ny}(a^d+1)=a^{mx}+a^{ny}=(a^{mx}-1)+(a^{ny}+1)=(a^{mx}+1)+(a^{ny}-1).$$  If $x$ is even and $y$ is odd, then $a^{m}+1\mid a^{mx}-1$ and $a^{n}+1\mid a^{ny}+1$, therefore $G\mid a^{ny}(a^d+1)$. If $x$ is odd and $y$ is even, then $a^{m}+1\mid a^{mx}+1$ and $a^{n}+1\mid a^{ny}-1$, thus $G\mid a^{ny}(a^d+1)$. However, since $\gcd(a^m+1,a^{ny})=\gcd(a^n+1,a^{ny})=1$, so $\gcd(G,a^{ny})=1$, hence $G\mid a^d+1$. We also have $a^d+1\mid a^{m}+1$ and $a^d+1\mid a^{n}+1$, so $a^d+1\mid G$. Thus $G=a^d+1$.  If $v_2(m)=v_2(n)=v_2(d)=k&gt;1$, then there exist some odd numbers $m_1,n_1,d_1$ such that $m=2^km_1,n=2^kn_1,d=2^kd_1$. We shall have $m_1x-n_1y=d_1$, so one and only one of $x,y$ is even, and we can use the same argument when $m,n$ are odd, so $G=a^d+1$.  However, if $v_2(m) \neq v_2(n)$, I cannot find any solutions for this. I think that $G \in \{1,2\}$, but I cannot prove or disprove it. How can I find $G=\gcd(a^m+1,a^n+1)$ if $v_2(m) \neq v_2(n)$ ? Moreover, is there anything from my arguments that can be improved ? How do I test for a mob, with a given name, in a certain location? I want to testfor a sheep, by the name of Lamb, in a certain location using coordinates.  I have already tried this:  testfor @e[x=999,y=80,z=-1034,type=Sheep,name=Lamb]   But it doesn't work. I have also tried:  testfor @e[x=999,y=80,z=-1034,type=Sheep]   But it says it found a sheep whether or not there is actually a sheep in that location. How would I prove that any binary tree that has n leaves has precisely $2n-1$ nodes ?  Given that a binary tree is either a single node "o" or a node with the left and right subtrees contains a binary tree        o  /       \ binary  binary tree    tree   Any help would be appreciated ! Block specific URL on web browsers I want to block certain URLs on browsers. I don't want to block whole site. For example, I want to block , but not whole  domain. Is there any way I can do it? Where does the term "fun fact" originate?-- namely, not with the compositional meaning but rather with the idiomatic usage to introduce some sort of unusual, esoteric, absurd or otherwise "controversial" fact? What are the shortcomings of the Mean Absolute Percentage Error (MAPE)? The  () is a common accuracy or error measure for time series or other predictions,  $$ \text{MAPE} = \frac{100}{n}\sum_{t=1}^n\frac{|A_t-F_t|}{A_t}\%,$$  where $A_t$ are actuals and $F_t$ corresponding forecasts or predictions.  The MAPE is a percentage, so we can easily compare it between series, and people can easily understand and interpret percentages.  However, I hear that the MAPE has drawbacks. I'd like to understand these drawbacks better so I can make an informed decision about whether to use the MAPE or some alternative like the MSE (), the MAE () or the MASE (). Tag score calculation wrong in Android App Get files in a directory sorted by last modified? Let $R$ a ring, and let $$\displaystyle R[[x]]=\left\{\sum_{k=0}^{\infty}a_k x^k\;\middle\vert\; a_k\in R\right\}$$ with addition and multiplication as defined for polynomials.  We have that $R[[x]]$ is a ring containing $R[x]$ as a subring.  How to prove that if $a_0\in R$ is a unit, then  $\displaystyle\sum_{k=0}^{\infty}a_k x^k$ is a unit in $R[[x]]$?