the domain I want is taken and according to ICANN it expires on 2021-01-05. The only status it has is &quot;clientTransferProhibited&quot;. I am not sure how to interpret this domain status... Does this status prohibit me from getting the domain once it has expired? And can I acquire the domain as soon as it expires according to the expiry date on ICANN?
I'm trying to purchase a domain name that was registered and apparently already expired. This is its status:  Status: clientHold http://www.icann.org/epp#clientHold Status: clientTransferProhibited http://www.icann.org/epp#clientTransferProhibited Status: redemptionPeriod http://www.icann.org/epp#redemptionPeriod Updated Date: 14-oct-2015 Creation Date: 01-sep-2014 Expiration Date: 01-sep-2015   From what I'm seeing from ICANN policies, it should already be available to be registered by third parties, but maybe I'm missing something. When will it become available for me to register it?
Earning a PhD help you learn tools and techniques in your field, and lets you become an expert in a very specific field of science. However, once you get your degree and try to enter the labor market, you're competing against people who are younger than you (i.e. recently graduated non-phd engineers) and may have actual work experience (i.e. non-phd engineers with a few years in the market).  There are, of course, some positions in which they cannot compete, including research and teaching position. Those are the minority, though... in most other fields it seems that holding a PhD is not as important as (for example) work experience.  So, I would like to know:   Is work experience more important than having a PhD in the many work environments? What skills can I acquire/demonstrate during my PhD to become more "valuable" than a non-PhD candidate for the same job? I have looked in several different Fate games' rulebooks, including Fate Core, DFRPG, Atomic Robo, FAE, and Jadepunk, which all describe Boosts as temporary Aspects you get to invoke for free, one time, and which disappear after being invoked.  In the other "invoke for free" situation I'm aware of (on Aspects newly-created via the Create Advantage action), they aren't singled out as "temporary."  I looked in many books but I'm interested in Fate Core.  So I haven't been able to figure out whether Boosts need to be invoked immediately or they disappear, without having been invoked at all? If $G$ is abelian, then the set of all $g \in G$ such that $g = g^{-1}$ is a subgroup of $G$ Prove that if $G$ is abelian then the set $H$ of all elements of $G$ that are their own inverses is a subgroup of $G$.   Naturally in an abelian group, $ab = ba$ for $a, b \in G$, however I'm not sure how to show the set elements that are their own inverses is a subgroup of $G$ using arbitrary elements. UV mapping not filling faces I have tried to map this simple object, but the uv mapping does not correspond to what i see in the textured object. It cuts the image a bit on top and a bit on the bottom of the faces. It does not fill the faces with the texture, even though i have mapped them correctly. Does anyone know what could be happening? Thanks in advance. $ f:\mathbb C\rightarrow \mathbb C$ defined by $f(z)=0$, if $Re(z)=0$ or $Im(z)=0$ and $f(z)=z$, otherwise. find() function in python2.7.5 Chance of selecting the last k pages in correct order from a set of n pages I have a set of documents numbered 1 to n. If I arrange the documents randomly, what is the chance of selecting the last k pages in the correct order if I select them at random from the documents?   Because order is important, I use the following formula:  $\displaystyle\frac{n!}{(n-k)!}$  From this I calculate how many ways there are to select k pages from n pages in the correct order. I need to have the last k pages. Which I can do in only one way. So is the chance   1/$\displaystyle\frac{n!}{(n-k)!}$?   It seems unlikely. It seems to me spontaneously that the chance should be smaller. How to show that $\sum_{k=1}^n k(n+1-k)=\binom{n+2}3$? How to get my 4yo daughter's fixation on my belly button under control? I don't know where to begin with this question. It's awkward and to be honest, it makes me feel rather uncomfortable. Simply put, I'm led to believe that my four year old daughter is obsessed with my belly button. It all started one day during the summer when she was three. I decided to take my family to the pool, and I guess I didn't see my daughter follow me into the changing room. Mind you, she's still at the age where it's appropriate to see her parents naked, so I wasn't surprised when, after I finished changing into my bikini, I turned around to see her staring at me.  I followed her gaze and saw her looking directly at my belly button. I asked her what she was looking at and she screamed "mommy, I can see your belly button!" which made me giggle. I picked her up and kissed her forehead, telling her that everyone has a belly button.  I think I might have accidentally given her a green light, because over the next year she constantly asks me if she can stick her finger in my belly button, and if I don't let her, she begins to scream.  For her fourth birthday, she requested that I wear nothing but a sports bra and underwear (in toddler speak. It sounded like "mommy, show your belly button for my birthday!" I could have simply worn a crop top but i didn't want to risk her throwing a temper tantrum)  My husband obviously doesn't mind, because he gets to look at my belly button all the time but it really upsets me that my daughter's finger is constantly in my belly button.  Tomorrow is her fifth birthday and I want to discuss with her that it's not appropriate to always be poking my belly button. Any ideas on how i can talk to her without her throwing a screaming fit? Why isn't the center of the galaxy "younger" than the outer parts? I understand that time is relative for all but as I understand it, time flows at a slower rate for objects that are either moving faster or objects that are near larger masses than for those that are slower or further from mass.  So, the illustrative example I always see is that if I were to leave earth and fly around at near light speed for a while or go into orbit around a black hole the time I experience would be substantially shorter than for those I left behind at home on earth and I'd come back to find that I've only aged however long I felt I was gone by my own clock but that on Earth substantially more time would have elapsed.  Following through on this model, the stars in orbit around the black hole at the center of the Milky Way are aging much much slower (relative to us), right?  So does it not follow that the center of the galaxy is some appreciable (I have no idea how to go about putting this in an equation so will avoid guessing at the difference) amount "younger" than the stuff further away from the center?   If this is not true, could someone please explain why not, and if it is true, can someone please point me to where I can calculate the age of the center of the galaxy :-)  And to be clear... what I'm asking is... if there was an atomic clock that appeared at the center of the galaxy when the center was first formed, and we brought it through a worm hole to earth today - how much time would have elapsed on that clock vs. the age we recon the galaxy currently is? (13.2 billion years) I'm trying to fix the following TikZ figure, where the circle device in the middle should be embraced by two bended arrows pointing counter-clockwise.    I have drawn these arrows with an arc path, but the tips of the arrows are unexpectedly both pointing upward.  Any idea how to get these arrow tips to follow the direction of the arc?  Here is my code for the above figure:  \documentclass[12pt]{article}  \usepackage{tikz} \usetikzlibrary{arrows} \usetikzlibrary{calc}  \begin{document} \begin{tikzpicture}   \draw (0,0) circle (1cm);   \draw[fill=gray!20]  (23:0.45cm)     -- (23:1cm) arc[radius=1, start angle=23, end angle=-203] (157:1cm)     -- (157:0.45cm) arc[radius=0.45, start angle=157, end angle=23] (23:0.45cm);   \draw[fill] (0,0) circle(2pt);   \draw[-&gt;,&gt;=stealth',semithick] (150:1.2cm) arc[radius=1.2, start angle=150, end angle=210] (210:1.2cm);  \draw[-&gt;,&gt;=stealth',semithick] (-30:1.2cm) arc[radius=1.2, start angle=-30, end angle=30] (30:1.2cm);   \foreach \x/\y in {1/0,2/0,3/0,4/1,5/1,6/1,7/0,0/1}     \node[rotate={\x*45-90}] at ($(0,0)!.75!\x*45:(1,0)$) {\y} ; \end{tikzpicture} \end{document} Projection of a point onto a simple convex polytope I have a convex polytope described by the intersection of a hypercube and two parallel lower and upper hyperplanes cutting through the hypercube. I want to find the projection of a point onto the polytope. I read the question  and see that the general problem of finding a projection onto a convex polytope can be solved with quadratic programming. However, taken alone, projection onto a hypercube or a hyperplane is trivial. For the hypercube, each dimension is defined by a lower+upper bounds and we just pull each dimension within those bounds, which take $O(n)$ time in $n$ dimensions. For the hyperplane just find the projection onto it. So I'm wondering whether there is simpler method for finding such a projection, maybe somehow combining the two independent projections. I am creating a model of a glass and I would like to put a drawing on it (a png picture).  I know how to do it using Blender Render, but not with Cycles Render. Help :) Subrings of integral domains have the same identity element. If $R$ is an integral domain and a subring $S$ has identity $1_S$, how would you show that $1_S=1_R$ (here $1_R$ is the identity of the ring $R$)? I am unsure about what an integral domain really is and how the subring comes into play here. I am currently looking at the various libraries that I can use to show a map in my custom Android App. I have found several like Mapsforge, OSMDroid, ArcGIS runtime for Android, Google Maps for Android and so on. I'm sure that I am missing a few.   Is there a comparison chart somewhere about the differences in functionalities and pros and cons of each one?   If there isn't one, we could make this question community wiki, to serve alongside  &amp; . Upgrading straight handlebars to drop handlebars I have a two-and-a-half year old  (with a road-bike-like frame and thin tires; it's somewhere in the loosely-defined "hybrid" range) that I use to commute to work, twenty-five miles round-trip, once or twice a week. The bike has straight handlebars, and having gotten used to the idea of commuting via bike, I'm interested in upgrading to drop handlebars. It looks like I'd need new break/shift levers to go with the handlebars; the whole upgrade looks like it'd run a couple hundred dollars. This is a sizable fraction of the original cost of the bike, which leads me to wonder if I'm better off saving my money for a new bike that's a little better adapted to my use.  Is upgrading my bike's handlebars a reasonable investment? I need a non trivial  examples of group isomorphism that maps $G\to H$ such that   $G/H \cong G$. During U-substitution, is the $du$ that we use as the infinitesimal in $\int f(u) \ du$ changing size? I am familiar and comfortable with the following definition of an integral (defined as the the Riemann Sum where n approaches infinity).    I have circled in red two terms to emphasize a concept that I was taught: the $\frac{b-a}{n}$ term effectively becomes the $dx$ term. This intuitively makes sense to me.   Additionally, I am familiar with the following statement (as expressed here: )    However, I have boxed in red something that I find rather confusing when trying to interpret it through the understanding that $\frac{b-a}{n}$ sorta kinda equals $dx$...  Specifically, the $du$ value appears to be changing throughout the interval $[ g(a), g(b)]$. To reframe this in the "Riemann Sum" version,we would have something like: $\text{ "value that varies depending on where you are"} * \frac{g(b)-g(a)}{n}$  This is very confusing to me...because, if this is the correct interpretation, I'm not quite certain I understand why this is allowed. Showing that a map is bijective and finding its inverse Let $S$ := {$(x, y) ∈ \mathbb R^2: x^2 + y^2 = 1$} be the unit circle. Let $p = (0, -1)$ ∈ $S$. Define a map $F$ from $\mathbb R$ × {$0$} to $S$ as follows: Given $(t, 0)$, let $F((t, 0)) = (x, y)$ ∈ $S$ be the intersection of $S$ with the line through $p$ and $(t, 0)$. Derive a formula for $F$. Show that $t$ → $F((t, 0))$ is a bijective map from $\mathbb R$ to $S$ \ {$p$}, and also derive a formula for its inverse.  If we express $S$ in polar coordinates: $$S:=\{(r,\varphi):r=1,\varphi∈(-\frac\pi{2},\frac{3\pi}{2})\}$$ then, from simple trigonometric relations, the inverse function $\Phi$ is: $$\Phi(\varphi)=\tan\frac{\frac\pi2-\varphi}2$$ and its inverse $F$ is: $$F(t)=\frac\pi{2}-2\arctan t$$ After changing coordinates back to cartesian, $F$ will look like: $$F(t)=(\cos(\frac\pi{2}-2\arctan t),\sin(\frac\pi{2}-2\arctan t))$$ which can be further simplified to: $$F(t)=(\frac{2 t}{t^2 + 1},\frac{1 - t^2}{t^2 + 1})$$ Is this the formula of $F$? Is it correct? And I know that to show that $\phi$ : $\mathbb R$ →  $S$ \ {$p$} is a bijective map, I have to show that $\phi$ is injective and surjective. But how do I do that? Can someone please help me show that $\phi$ is bijective? and how can I derive a formula for its inverse? I downloaded a file from the internet (I can link to it here, but in the interests of not annoying other users, I won't post the link), and I can't delete the file with the message:   You need permission to perform this action  You require permission from the computer's administrator to make changes to this file   Trouble is, I am the administrator and UAC is completely off.  If I go into the Properties/Security tab, I get the message   You do not have permission to view or edit this object's permission settings   Clicking "Advanced"/Owner, it says:   Current owner: Unable to display current owner.   If i try to change the owner, I get the same message:   Unable to set new owner on file  Access is denied.