We want to setup a "small" server for on of our offices (8 Client Computers, local Domain not linked). We need to have an Exchange Server. Unfortunately we cannot use SBS since its support is running out.  We already do have a Windows Server 2012 R2 Essentials Licence and an Exchange Server 2013 Licence (+CALs).  As Exchange does not run on Essentials (thanks Microsoft for that!) we need to run a Standard Licence.  My Setup would be the following:   Host Machine Windows Server 2012 R2 Standard with Hyper-V on it 1 Virtual Machine Running Windows Server 2012 R2 Standard + Exchange Server 1 Virtual Machine Running Windows Server 2012 R2 Essentials (Domain Controller)   As Microsoft does not recommend running the domain controller on the same instance as the Exchange Server and we do have less than 25 Clients i would go for that setup. Seems a bit over complicated but as far as i understand it follows Microsoft's recommendations.  Do you have any recommendations concerning the hardware? Right now we have a Lenovo RD340 equipped with 8GB RAM, 1 SSD 128GB and 3*3TB HD in Raid 5. Maybe we should upgrade it to 16GB RAM, but as the Hyper-V can manage the required memory it could be enough right now. Where should we install the Guest VMs? On the SSD?
This is a  about  Related:      I have a question regarding capacity planning. Can the Server Fault community please help with the following:   What kind of server do I need to handle some number of users? How many users can a server with some specifications handle? Will some server configuration be fast enough for my use case? I'm building a social networking site: what kind of hardware do I need? How much bandwidth do I need for some project? How much bandwidth will some number of users use in some application?
Alternative definitions of wifebeater I was a li'l nonplussed to find out that the word  can mean other things besides "a man who beats up his wife". Its definition reads:        One who (usually as a repeated practice) beats one’s wife, or a husband prone to violence   A kind of sleeveless shirt, often but not exclusively worn as an undershirt.   (uncountable, UK, slang) Stella Artois, a brand of lager beer.       reckons that sense 2 is American and originates "apparently from the association of such a garment with men who commit domestic violence". It does not register the UK slang. I'm guessing that Stellas also have an unfortunate association with domestic violence.  Are these alternative senses popular in their respective countries or are they perhaps nonce words that have lingered on? Any idea how these disparaging associations came about? I expect that the brewers of Stella Artois are none too happy about it. Why does Microsoft Word 2010 show an error for the following sentence?     1.&nbsp;Where should this car be parked?   Word 2010 also suggests changing the sentence to     2.&nbsp;Where this car should be parked?   Is the sentence (1) wrong? If yes, why? What about sentence (2)? canonical-livepatch, Doesn't seem to patch anything How do I brew an X potion? What is the limit of In the Wikipedia article , it says     After Newtonian gravitation was found to be inconsistent with special relativity, . . .   I don't see how Newtonian gravitation itself is inconsistent with special relativity.  After all, Newton's Universal Law of Gravitation is completley analogous to Coulomb's Law, so it would seem that, if there were an analogous "gravitational magnetic field", one could formulate a theory of gravitation in exact analogy with Maxwell's Theory of Electromagnetism, and of course, this would automatically be consistent with Special Relativity.  What about this approach to gravitation does not work?  The only problem I could see with this is the lack of evidence for a "gravitational magnetic field".  That being said, gravity is "weak" as it is, and my guess is that it would be extremely difficult to set up an experiment in which a massive body were moving fast enough for this "magnetic field effect" to be obsevable.  EDIT:  As has been pointed out to me in the answers, Newton itself is inconsistent with SR.  Similarly, however, so is Coulomb's Law, yet, electromagnetism is still consistent with SR.  Thus, how do we know that it is not the case that Newton's Law is the special "static" case of a more general gravitomagnetic theory exactly analogous to Maxwell's Theory:  Let $\mathbf{G}$ be the gravitation field, let $\rho$ be the mass density, and $\mathbf{J}$ be the mass current density, let $\gamma _0$ be defined so that $\frac{1}{4\pi \gamma _0}=G$, the Universal Gravitation Constant, let $\nu _0$ be defined so that $\frac{1}{\sqrt{\gamma _0\nu _0}}=c$, and suppose there exists a field $\mathbf{M}$, the gravitomagnetic field, so that the following equations hold: $$ \vec{\nabla}\cdot \mathbf{G}=-\frac{\rho}{\gamma _0} $$ $$ \vec{\nabla}\cdot \mathbf{M}=0 $$ $$ \vec{\nabla}\times \mathbf{G}=-\frac{\partial \mathbf{M}}{\partial t} $$ $$ \vec{\nabla}\times \mathbf{M}=\nu _0\mathbf{J}+\gamma _0\nu _0\frac{\partial \mathbf{G}}{\partial t} $$ where theses fields would produce on a mass $m$ moving with velocity $\mathbf{v}$ in our inertial frame the Lorentz force $\mathbf{F}=m\left( \mathbf{G}+\mathbf{v}\times \mathbf{M}\right)$.  This theory would automatically be consistent with SR and would reduce to Newton's Law of Gravitation for the case of gravitostatics:  $\frac{\partial \mathbf{G}}{\partial t}=\frac{\partial \mathbf{M}}{\partial t}=\mathbf{J}=0$.  (To be clear, you can't just set the time derivatives equal to $\mathbf{0}$ as it seems I have done.  Upon doing so, you obtain the corresponding static theory, which is technically incorrect (as you can easily see because it won't be relativistically invariant), but is nevertheless often a useful approximation.)  My question can thus be phrased as:  without appealing directly to GR, what is wrong with this theory? How can I view all my unaccepted questions across SE? PHP PCRE (regex) doesn't support UTF-8? I installed kubuntu-desktop just to experiment with its interface.  Then I decided to remove and did the traditional apt-get purge and autoremove but it seems elements of kubuntu are still there. For example interface elements, scrollbars, bootscreen etc are still there.  Is there any method I can completely remove without installing a fresh copy of ubuntu? I understand that the alphabet for the English language is not strictly English as languages such as French, Dutch and many more use the same alphabet, with few additions in other languages. Is Roman alphabet the common term to refer to this set of symbols for writing texts? build folder usage issues I posted this question but for some reason I can't get back into my account:    Most things I see with 220V have both hot legs going through the switch or relay, current "returns" through the switch.  Is it against code to wire one leg directly to the equipment and the other leg is switched as opposed to having both legs switched.  It seems useless and a waste of wire to switch both legs.  Basically  vs  (minus the GFCI as I'm not allowed to have that)  As a bonus question.  I assume I need a "motor rated switch" as standard switches can weld due to the high startup current.  Is  one legal\safe? Ready to use leaderboard service. Is any? I'm looking for ready to use thirdparty leaderboard service where player's achivementes and scores can be placed. I'd intagrate it into my game to allow players push scores and locate themselves in the tables of tops. It would be nice whether anybody experienced share some points with me. Who invented "dooblidoo"? The word  is used by several different youtube channels as a different word for the youtube description bar. I've seen it used by the  and by .  Who was first to use and invent the term? Can you enter the Schengen area with a UK visa? I'm a UK passport holder; my wife isn't. We don't live in UK. If she gets a UK visit visa, does this entitle her to travel to the Schengen area or do I need to get her a separate Schengen visa? Protect .NET code from reverse engineering? Why is $\mathbb{E}[X] = 1 + \sum^\infty_{k=1}\mathbb{P}(X > k)$ true? I'm working through a problem regarding expected values in Markov chains, and at some point it says:     Recall from probability that if $X$ is a positive integer valued random variable, then $\mathbb{E}[X] = 1 + \sum^\infty_{k=1}\mathbb{P}(X &gt; k)$.   I know that by definition $\mathbb{E}[X] = \sum_a a\mathbb{P}(X = a)$, but I can't see how the above equality follows from this, nor am I sure if this is how to approach the problem. Consider the general linear regression model: $$y_i = \beta_0 + \beta_1x_{i1} + \beta_2x_{i2} + \cdots + \beta_px_{ip} + \epsilon_i = \mathbf{x}_i^t \beta + \epsilon_i$$  where $\textbf{x}_i = (1,x_{i1},x_{i2},\cdots,x_{ip})^T$, $\beta=(\beta_0,\beta_1,\cdots,\beta_p)^T$ and $\epsilon_i$ are iid N(0,$\sigma^2$).  I would like to see a complete proof of the following identity from first principles: $$\sum_{i=1}^n(y_i - \bar{y})^2 = \sum_{i=1}^n(\hat{y}_i - \bar{y})^2 + \sum_{i=1}^n(y_i - \hat{y}_i)^2$$ where $\hat{y}_i= \mathbf{x}_i^t \hat{\beta} $ ($\hat{\beta}$ is the least square estimator, $\bar{y}$ ia the sample mean of $y_i$).  I know that the two terms on the right can be obtained by subtracting and adding $\hat{y}_i$ on the left side. But this introduces a "cross term": $$\sum_{i=1}^n2(\hat{y}_i - \bar{y})(y_i - \hat{y}_i)$$  Many texts claim that this is zero, but I have not seen a general proof of this statement. How can this be shown? windows 7 desktop icons: new icon placement When moving/saving/etc a new file on the desktop, the icon appears second (just after "Computer"). I find it annoying, since I want my "special" icons to be first (such as the recycle bin, "Network", etc.).  Can this behavior be changed? A geometric reason why the square of the focal length of a hyperbola is equal to the sum of the squares of the axes? When I teach conics, I give a simple geometric argument for the Pythagorean relation for ellipses. (By "Pythagorean relation" I mean the fact that the square of the focal length $c$ is the difference of the squares of the semimajor and semiminor axes, i.e. $c^2=|a^2-b^2|$ for the ellipse defined by $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.) The picture argument goes as follows: just draw the obvious isosceles triangle from the two foci to one of the two points at the end of the minor axis. We see two right triangles; in each, one leg is the focal length $c$ and the other leg is the semiminor axis, while the hypotenuse is the semimajor axis. The result follows from the Pythagorean theorem applied to this right triangle.  But when it comes to the Pythagorean relation for hyperbolas, I can't think of an equally convincing picture to draw. Instead I have to give a rather inelegant algebraic argument for the relation $c^2=a^2+b^2$ (for the hyperbola defined by $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, with focal length $c$), directly from the definition of the hyperbola. The messy details are below.*  Now, obviously there is an asymmetry between the argument for the ellipse and the argument for the hyperbola: the ellipse argument is almost trivial and doesn't involve working straight from the locus definition, but the hyperbola argument is tedious and does require working straight from the definition. So, my question:     Is there an equally persuasive picture I could draw to convince high   school students of the Pythagorean relation for hyperbolas -- without having to get technical with points at infinity, say?   The natural triangle to draw is the right triangle with legs from the center to a vertex and from that vertex to an asymptote, which has legs of length $a$ and $b$; but then the problem becomes why the length along the asymptote should be equal to $c$. I doubt there is a purely (Euclidean) geometrical reason because this triangle lies on the asymptote, not on the hyperbola itself; I suspect some sort of limiting procedure will be required.    [*] The tedious algebraic details:     The foci lie at $(\pm c,0)$.   Say $a&gt;b$. Then hyperbola is the locus of points $(x,y)$ such that       $$\left|\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}\right|=2a$$      Assume for simplicity the expression between the absolute value bars   is positive. (The algebra is even more tedious if we don't make such a   simplifying assumption.) Then moving the right root to the other side   and squaring gives      $$(x+c)^2+y^2=4a^2+(x-c)^2+y^2+4a\sqrt{(x-c)^2+y^2}$$      Simplifying gives      $$xc-a^2=a\sqrt{(x-c)^2+y^2}$$      and upon squaring again we have      $$x^2c^2-2xca^2+a^4=a^2(x^2-2xc+c^2+y^2)=a^2x^2-2xca^2+c^2a^2+a^2y^2$$      which reduces to      $$x^2(c^2-a^2)-y^2a^2=a^2(c^2-a^2)$$      Defining $b^2:=c^2-a^2$ brings the equation into the form   $x^2/a^2+y^2/b^2=1$, and the Pythagorean relation follows from the   definition of $b$.