I registered a domain example.com many years ago and had intentions of developing it into a social networking site, but then I got busy with other projects.  I recently realized that example.net is a very well established site &amp; organization (a sort of consortium of many businesses/groups). As a matter of fact, they even trademarked the keyword example, which makes developing example.com risky and pointless.  I have not received any offers/communication from the owner of example.net. Should I contact the owner of example.net with an offer for them to purchase it, or would that result in accusations of cyber-squatting?  Please note:   I registered the domain example.com a year before example.net I live in a different country than the company that owns example.net, hence I am unsure if the trademark law applies to me.
A search on google for "iphone developer" turns up loads of websites which have "iphone" in them, a trademarked name by Apple.  So I'm led to believe that having a domain such as iphonedeveloper.com is ok? Well, you're still using Apple's trademark, but it would be hard to brand yourself otherwise. You're an IPHONE DEVELOPER... right?  Well, what if I want to provide a website where users pay to get a list of the best offers from Ebay? I might have a domain like ebaydeals.com (I don't... i'm just speculating!).  Now I've heard that places like Ebay are really hot on the trigger and fire out emails to people who register domains like that straight away. But whats the difference?  In both cases I'm making money from the trademark, effectively, so is it just down to how lenient the company who owns the trademark is? Or are there rules? Is there a specific "line" you don't cross?  Thanks
Reverse String order Find the area where dog can roam A dog is tied to circular pillar by a rope. Radius of this pillar is $1m$ and length of rope is $\pi m$. What is an area where dog can roam?   I tried to find the area of all semicircles and then to find its sum. It is easy to find an area at front side of the pillar. It is $\displaystyle\frac12\pi^2\pi=\frac{\pi^3}{2}$. Problem is how to find remaining area. I tried to write this area using compass and straightedge, but I couldn't. Then I wrote this in AutoCAD and it looks like this: Is it possible to find the exact value of this area? How to get a cross-origin resource sharing (CORS) post request working How can I take a self-portrait without using a remote?  My camera is a Canon Powershot SX210 IS. I do not have a remote, but I do have a tripod. Slow math typing I have been using LaTeX for quite some time, but I find the math writing to be a bit frustrating. It seems to take me a relatively long time to code-write long math expressions, and sometimes it's hard to see where { start and end.  I am interested in writing as fast as possible.  Is code-typing each symbol really the fastest way? Is there any good program (in the spirit of Word's MathType) that enables producing math expressions by button clicking, that you recommend? I'm making a door with piston sliding sideways that reacts to pressure plates. As the pressure plates are sending the current to redstone at a very low level, I need to raise it somehow, and this leads to awkward stairs of redstone.    So I'm wondering whether there are better ways of sending the redstone current up, maybe even vertically. when I look at my Nexus 5 with android 5.1. battery screen, I can see something like this:    Screenshots (click images for larger variants)  Sum of all per-app percentages in list is 10. However as you can see, there is 60% battery consumed (40% remaining). So, what is the deal here? Who consumed 50% battery? Or what do the percentages next to apps mean?  As you can see, screen was on just for few short times, the same for awake, and yet battery life about 18 hours? I dont get it.  Also, I found this issue reported on AOSP, but noone from google seems to care: We often get questions on bicycles.stackexchange about identifying a bicycles manufacturer, model, make, and year.  Often these questions are downvoted or closed -- usually with a comment attached to them that it's not necessary to know the exact model/make/year of a bicycle if you want to fix, repair, or ride your bike. Why should owners not care about the model/make/year of their bicycle? Especially when it comes to less expensive bikes (aka BSOs) and BMX bicycles? This is intended as a canonical question that we can point closed questions to. Also see      (generally not helpful for identifying a brand/model) How to create a custom Windows 7 ISO file Table:   UserId, Value, Date.   I want to get the UserId, Value for the max(Date) for each UserId. That is, the Value for each UserId that has the latest date. Is there a way to do this simply in SQL? (Preferably Oracle)  Update: Apologies for any ambiguity: I need to get ALL the UserIds. But for each UserId, only that row where that user has the latest date. First Bash Script Problem Generate uniformly distributed weights that sum to unity? It is common to use weights in applications like mixture modeling and to linearly combine basis functions. Weights $w_i$ must often obey $w_i ≥$ 0 and $\sum_{i} w_i=1$. I'd like to randomly choose a weight vector $\mathbf{w} = (w_1, w_2, …)$ from a uniform distribution of such vectors.  It may be tempting to use $w_i = \frac{\omega_i}{\sum_{j} \omega_j}$ where $\omega_i \sim$ U(0, 1), however as discussed in the comments below, the distribution of $\mathbf{w}$ is not uniform.  However, given the constraint $\sum_{i} w_i=1$, it seems that the underlying dimensionality of the problem is $n-1$, and that it should be possible to choose a $\mathbf{w}$ by choosing $n-1$ parameters according to some distribution and then computing the corresponding $\mathbf{w}$ from those parameters (because once $n-1$ of the weights are specified, the remaining weight is fully determined).  The problem appears to be similar to the  (but, rather than picking 3-vectors whose $ℓ_2$ norm is unity, I want to pick $n$-vectors whose $ℓ_1$ norm is unity).  Thanks! If $A$ is an $m \times n$ and $B$ is an $n \times m$ matrix with $n &lt; m$ then prove that $C = AB$ is not invertible.  I am not getting any idea. May be I am missing vary basic point. $C = AB$ where both $A$ and $B$ are square matrices  will be invertible iff both $A$ and $B$ be invertible. I know it and its proof.  May be this question has been discussed earlier, but I am not seeing the link. Thank you for your help. prove $\frac {ab}{a+b} \geq \sum ^n_{i=1} \frac{a_ib_i}{a_i+b_i}$ "Delete this answered question?" when the only answer is deleted Looks like there's a small bug in the UI: I posted a question here on meta.SE which ended up being just something dumb I was doing and no use to anyone in the future. At first I posted an answer saying what the dumb thing was, but then I realized that the whole thing had no value, so I deleted the answer, and then deleted the question — whereupon it asked me:  Delete this answered question? We do not recommend deleting questions with answers because doing so deprives future readers of this knowledge. Repeated deletion of answered questions can result in your account being blocked from asking. Are you sure you wish to delete?  (Screenshot below) Probably shouldn't ask (or add to the tally) if the only answers are deleted ones... Claim: Let $X$ be a metric space. If $A,B\in X$ are disjoint, if A is compact, and if B is closed, then $\exists \delta&gt;0: |\alpha-\beta|\geq\delta\;\;\;\forall\alpha\in A,\beta\in B$.  Proof. Assume the contrary. Let $\alpha_n\in A,\beta_n\in B$ be chosen such that $|\alpha_n-\beta_n|\rightarrow0$ as $n\rightarrow \infty$.  Since A is compact, there exists a convergent subsequence of $\alpha_n\;(n\in\mathbb{N})$, $\alpha_{n_m}\;(m\in\mathbb{N})$, which converges to $\alpha\in A$.  We have   $$|\alpha-\beta_{n_m}|\leq|\alpha-\alpha_{n_m}|+|\alpha_{n_m}-\beta_{n_m}|\rightarrow0 \;\;\;as\;\;m\rightarrow\infty.$$ Hence $\alpha$ is a limit point of B and since B is closed $\alpha\in B$, contradiction.  Is my proof correct? I feel as though I am missing something simple which would trivialize the proof. Rewiring a switch controlled receptacle to a ceiling light I have run out of space on my WinXP virtual machine, which I only gave 10 GB space for when I created it.  Is there an easy way to increase it to, say, 20 GB?  I can't see any obvious option in VirtualBox settings.    The suggestion below gives this error  wim@wim-ubuntu:/media/data/winxp_vm$ VBoxManage modifyhd wim.vdi --resize 20000 VBoxManage: error: Cannot register the hard disk '/media/data/winxp_vm/wim.vdi' {46284957-2c09-4e70-8a49-bfbe0f7f681d} because a hard disk '/home/wim/VirtualBox VMs/winxp_vm/wim.vdi' with UUID {46284957-2c09-4e70-8a49-bfbe0f7f681d} already exists VBoxManage: error: Details: code NS_ERROR_INVALID_ARG (0x80070057), component VirtualBox, interface IVirtualBox, callee nsISupports Context: "OpenMedium(Bstr(pszFilenameOrUuid).raw(), enmDevType, AccessMode_ReadWrite, fForceNewUuidOnOpen, pMedium.asOutParam())" at line 210 of file VBoxManageDisk.cpp     Removing the .vdi from VirtualBox before calling VBoxManage command, then adding it back in, was successful.  But now I can't boot the virtual machine, I get this worrying screen:    By the way, it says FATAL: Could not read from the boot medium!  System halted.    The vdi must be reattached to the VM after VBoxManage command.  Further, the partition will need to be resized from WITHIN windows, because you will have this empty space:    I was able to resize the partition easily using a bit of freeware called . If a function $f$ is continuous $\implies$ $|f|$ is continuous.(Answered by Myself) $(i)$ Given a function $f:E \rightarrow \mathbb{R}$ define $|f|:E \rightarrow \mathbb{R}$ by $|f|(x)=|f(x)|$ for $x \in E$ show if $f$ is continuous at $c \in E$ then so is $|f|$.        $(ii)$ Now the converse is not true, find $f$ such that $|f|$ is continuous but $f$ is not. After update to OS X 10.11 I started having these bugs:   Finder sidebar text flashing  Very high CPU load:      I have Macbook Pro Retina 13" Mid 2014  I reset NVRAM.  I deleted com.apple.finder.plist, it didn't help.  Also I did this: , no longer flashes.