A simple derivation of the Centripetal Acceleration Formula?

Question: Could someone show me a simple and intuitive derivation of the Centripetal Acceleration Formula $a=v^2/r$, preferably one that does not involve calculus or advanced trigonometry?
Answer: Imagine an object steadily traversing a circle of radius $r$ centered on the origin. Its position can be represented by a vector of constant length that changes angle. The total distance covered in one cycle is $2\pi r$. This is also the accumulated amount by which the position has changed...
Now consider the velocity vector of this object: it can also be represented by a vector of constant length that steadily changes direction. This vector has magnitude $v$, so the accumulated change in velocity is $2 \pi v$.
The magnitude of acceleration is then $\frac{\text{change in velocity}}{\text{elapsed time}}$, which we can write as:
$$a = \frac{2 \pi v}{\left(\frac{2\pi r}{v} \right)} = \frac{v^2}{r} \,.$$
Q.E.D.

Aside: that derivation is used in many algebra/trig-based textbooks.
Comment: This is a great explanation but could someone explain why the accumulated change in velocity is 2*pi*v?
Comment: @Conceptuality So, by assumption the speed of the particle is constant, right? But as it goes around the circle its direction of motion changes. That means the velocity vector keeps the same length but turns through a whole circle. The tip of the velocity vector describes a circle of radius $v$, so the distance that the tip moved was $2 \pi v$.
Comment: @theenigma017 Angles are formally dimensionless (this is why angular velocity and angular frequency are the same measure in SI). Which is why $2 \pi r$ is a distance and not some other quantity, Which means that $2 \pi v$ is a speed. The first three comments may help, or you can just think of this as comparing distance and speed.
Comment: Notice that after one full turn the change in position is also zero. What we are interested in here really the average value of the instantaneous acceleration, but to get it requires calculus (or at least the machinery of limits), which the OP doesn't want. So instead we consider not displacement but distance and also the equivalent of distance for velocity (which doesn't have a conventional name) rather than $\vec{v}_f - \vec{v}_i$. This gives the centripetal acceleration for all curves (knowing $r$ and $v$), but we must add the "transverse" acceleration by hand.
Comment: @dmckee, in the similar triangle proof, the change of velocity is simply the distance between the tips of two velocity vectors draw tail-to-tail. But would "delta v" simply be zero since magnitude is constant?---------how do you really mean sure the change in direction anyway in degree?
Comment: @Doeser I'm not entirely sure that I understand the question. I should note that the argument here is a little round-about. We're letting the "position distance" and "velocity distance" on a whole circle stand in for correct quantities which are the infinitesimal displacement and infinitesimal change in velocity, and we only get away with it because of the uniformity of the motion. In short this is a cheat that we can only trust because we know how to get the answer another way; but it has the advantage of being very accessible to students who don't have calculus.
Comment: @dmckee, I was talking about the other popular proof online. You said "infinitesimal change in velocity", but isn't there no difference in the magnitude of velocity in uniform circular motion?
Comment: @Doeser Sure, the magnitude stays the same, but velocity is a *vector* and its direction changes. See the figure in [diracpaul's answer](http://physics.stackexchange.com/a/193628/520), but imagine that the angle $\phi$ is taken to be very small.
Comment: 2 pi R is the distance aka circumference that's all good ...
But 2 pi V is an angle * velocity ? what is the intuition ?

also used in below answer
Comment: The version I learnt, which is nicer, presents it as $$\frac{2\pi r}v=T=\frac{2\pi v}a\qquad\implies\qquad a=\frac{v^2}r$$where there is nothing to have to draw---what we are doing is saying that in one period $T$, the position vector executes a full circle, and so did the velocity vector, each satisfying its own circumference traversal time equality, and we are merely equating them. No need to argue so much.
Answer:

In right-angled-triangle ABC
\begin{equation}
\dfrac{|\dfrac{\Delta \mathbf{\vec v}}{2}|}{|\vec V|}=Sin\dfrac{\Delta \vec \theta}{2}
\tag{01}
\end{equation}
If
\begin{equation}
\theta
\tag{02}
\end{equation}
is small
\begin{equation}
v(t) \approx v(t + \delta t ) = \mathbf{\vec v}
\tag{03}
\end{equation}
\begin{equation}
\dfrac{|\dfrac{\delta \mathbf{\vec v}}{2}|}{|\vec V|}=Sin\dfrac{\delta \vec \theta}{2}
\tag{03}
\end{equation}
For small angle
\begin{equation}
\delta \theta \approx Sin {\delta \theta}
\tag{04}
\end{equation}
So on rearranging
\begin{equation}
\dfrac{\delta  \mathbf{\vec v}}{2} =\dfrac{\delta \vec \theta}{2} \times \vec v
\tag{05}
\end{equation}
\begin{equation}
\delta \mathbf{\vec v} =\delta \vec \theta \times \vec v
\tag{06}
\end{equation}
\begin{equation}
\dfrac{\delta \mathbf{\vec v}}{\delta \mathbf{t}} =\dfrac{\delta \vec \theta}{\delta \mathbf{t}} \times \vec v
\tag{07}
\end{equation}
\begin{equation}
\mathbf{\vec a}=\dfrac{\delta \mathbf{\vec v}}{\delta \mathbf{t}}
\tag{08}
\end{equation}
\begin{equation}
\vec \omega =\dfrac{\delta \vec \theta}{\delta \mathbf{t}}
\tag{09}
\end{equation}
\begin{equation}
\mathbf{\vec a}=\vec \omega \times \vec v
\tag{10}
\end{equation}
\begin{equation}
\mathbf{a}=\omega \times v
\tag{11}
\end{equation}
And since
\begin{equation}
\mathbf{v}=\omega \times r
\tag{12}
\end{equation}
So
\begin{equation}
\mathbf{a}=\dfrac {v^{2}}{\mathbf{r}}
\tag{13}
\end{equation}
Comment: Why is $|\Delta\mathbf v|=v|\Delta\phi|$?
Comment: With a small angle limit this is, of course, the correct derivation. But you might want to be explicit that  $|\Delta \mathbf{v}| = v |\Delta \phi|$ is correct in that limit. I have mixed success in the classroom with this derivation (and with the one I used, too).
Comment: @Allawonder did you work it out? I am stuck at the same question.
Comment: V(vector) = an angle * velocity ? what is the intuition ?

also used in above answer
Answer: You can do this derivation by breaking the position of the orbiting particle down into components. It isn't short, but I think it's useful because it supplements the algebra with concrete physical analogies. I'll organize it into four parts: decomposition, oscillation, energy, and symmetry.
Decomposition
The position of a particle moving along a circular path can be described by two half-out-of-phase sine waves -- or equivalently, a sine wave and a cosine wave:

(via)
This is easy to derive: suppose the particle is moving at a constant angular velocity $\omega$ around a circle of radius $r$. Then $\theta = \omega t$, and basic trigonometry tells us that the position of the particle given $\theta$ is given by $x = r \cos \theta$ and $y = r \sin \theta$. We can substitute to get $x = r \cos (\omega t)$ and $y = r \sin (\omega t)$.
Oscillation
It turns out that there's another kind of movement that's described by sine waves: the oscillation of a weight on a spring. In such a system,
$$x(t) =A \cos \left(\sqrt{k \over m}t \right)$$
where $A$ is the amplitude (i.e. the difference between the spring's maximum length and its resting length), $k$ is the spring's force constant, from Hooke's law, and $m$ is, of course, the mass. Deriving this equation in the usual way requires not only calculus, but also differential equations -- so I'm going to ask you to take my word for it until a bit later.
For our purposes, this means that the mechanics of the orbiting particle can be simulated by two oscillating springs: one for the $x$ component, and one for the $y$ component that is identical to the first, but that is half-out-of-phase with it. To picture what this looks like, look back at the animation above, and pretend the blue dot and the red dot are attached to springs that are at $0$ when resting.
Now that we have this spring-based model, we can use it to determine the force being applied to the particle along the $x$ axis! By Hooke's law, the force a spring exerts on an attached weight is $F = -kx$. Let's try to use that formula to determine the acceleration of the particle when the $x$ spring is at its maximum length. We know its maximum length will be $r$ in this case -- this corresponds to the moment when the particle is at $x = r, y = 0$. And we know that $F = ma$. So by substitution, $ma = -kr$; divide by $m$, and
$$ a = -k \frac{r}{m}$$
Our work is half done. But now we have a new problem: we don't know what $k$ is -- after all, there's no actual spring, so we can't measure anything. We need to know what value of $k$ a spring will have if it moves the same way as the particle does along the $x$ axis. To solve this problem, we have to start thinking about conservation laws.
Energy
Let's think about what happens when the $x$ spring oscillates. When the spring is at $x = r$, it exerts the largest force on the particle, but the particle isn't moving at all. This is right at the peak of the wave. That means $v = 0$. From that point, the force that the spring applies to the particle will accelerate it from $0$ to its largest velocity $-v_\text{max}$.1 And by the time the particle reaches $x = 0$, the spring will be applying no force; that's the spring's resting state. Since the spring is applying no force, we've reached $-v_\text{max}$ -- the direction of future acceleration will be in the opposite direction, slowing the particle down until it reaches $x = -r$.
So this means that at $x = r$, we have $F = -F_\text{max}$, $a = -a_\text{max}$, and $v = 0$. And at $x = 0$, we have $F = 0$, $a = 0$, and $v = -v_\text{max}$.
Here's what's happening: the energy in the system is moving back and forth between maximum kinetic energy (at $x = 0$ when the spring is not stretched or compressed at all) and maximum potential energy (at $x = r$, when the spring is fully stretched). And because of conservation of energy, these two maxima must be equal; in other words, $E_\text{max}$ = $P_\text{max}$.
The formula for kinetic energy is $E = \frac{1}{2} mv^2$ -- that's basic Newtonian mechanics. What's the formula for $P$ in this case? This is the hardest part of the derivation. The potential energy stored in the spring is equal to the amount of negative work done to stretch it to $x = r$. So we have to remember the formula for work: $W = Fd$, where $d$ stands for distance traveled -- i.e. $x$, assuming that we start at $x = 0$. But then we have a problem. $F = -kx$ is not constant -- it's a function of $x$.
In general, this would mean we have to do calculus. But luckily enough, $F = -kx$ is a linear function, and so the value we want is equal to the area of the triangle formed by the $x$ axis and the line $F = -kx$:

(via)
In the above chart, $k = 1$, the distance traveled $d = x_\text{max} = 1$, and the area of the given triangle represents the value you get when you multiply $F(x)$ by distance traveled, adjusting for changes in the value of $F$ as the distance increases. But since the height of the triangle is $-kx_\text{max}$ and the base of the triangle is $x_\text{max}$, we can just use plain old geometry. The area of a triangle is $\rm \frac{1}{2} base \times height$ -- or, here, because $x_\text{max} = r$
$$W = -\frac{1}{2} kr^2$$
And since potential energy is equal to negative work:
$$P = \frac{1}{2} kr^2$$
Isn't it interesting how similar that is to $E = \frac{1}{2} mv^2$? If you think about that for a while, and consider that $E + P$ has to be a constant value, you might start to realize why an oscillating spring follows a sinusoidal path! (Hint: write the formula for a circle, but use $\sqrt{E + P}$ instead of $r$, and $v$ instead of $y$.)
But you can come back to that later. We're very close! Setting $v = v_\text{max}$:
$$E + P = E_\text{max} = P_\text{max} = \frac{1}{2} kr^2 = \frac{1}{2} mv^2$$
What happens when we solve for $k$?
\begin{align}\frac{1}{2} kr^2 &= \frac{1}{2} mv^2\\ \implies kr^2 &= mv^2\\ \implies k &= \frac{mv^2}{r^2}\end{align}
Now we can substitute that into our formula for acceleration above:
\begin{align}a &= -k \frac{r}{m}\\ &= -\frac{mv^2}{r^2} \frac{r}{m}\\ &= -\frac{v^2}{r}\end{align}
You might be wondering why the negative sign appears in this version. But recall that the acceleration is technically in the opposite direction of the displacement. So when $x = r, y = 0$, the acceleration is in the direction of $-r$. If it were otherwise, the particle would be accelerating outwards!2
Symmetry
The last step of this derivation requires a trick. We began by breaking movement in two dimensions into movement along two one-dimensional components. We then used imaginary springs to describe the particle's movement along those two components. And now, we face a final question: how do we pick our $x$ and $y$ components?
They have to be at right angles to one another, but that's only half the battle -- we have to find the "right place" to start, the "real" $x$ coordinate. The trouble is that we can't. The circular path along which the particle moves is rotationally symmetrical. There's nothing about the circle to tell us where it "starts" or "ends."
This means that the above line of reasoning holds no matter where we start. We can pick any point on the circle as the $x=1, y=0$ point, and the above will be valid. So wherever the particle is, we simply set that point as our $x = r, y = 0$ point, and everything else falls into place.
If we wanted to make more work for ourselves, we could work out the details trigonometrically, using the above formulas, adjusting them for the $y$ axis, and then recombining the $x$ and $y$ values using vector algebra. But we don't need to -- the symmetry argument is more powerful in this case.
1. Here "largest" really means "most negative," because the particle is moving in the negative $x$ direction. These values will really be $-F_\text{max}$ and $-v_\text{max}$. Later, when the particle is moving in the opposite direction, these values will be positive.
2. Making sense of the signs takes a lot of subtle detail work; in particular, you have to understand why negative work becomes positive potential energy. It also helps to think about $r$ as a vector (which has a direction) instead of a magnitude (which doesn't). Fortunately, physical intuition provides a reliable guide in this case; if something turns out obviously wrong, double-check your thinking.
Answer: In order to move through a concave path, an agent has to impart force to otherwise a linearly-moving object. The object , by virtue of its motion, under the absence of any external force, always travels or tends to travel in the direction of the velocity vector at the concerned instant.
So, when the object has to transverse a curve trajectory, the main requisite is the introduction of a force that manipulates the direction of velocity such that the resultant locus is the required curvilinear path, otherwise the object would travel straight.
The direction of the force is evidently the direction of the acceleration or the limit of change of velocity with respect to time. So, to find the direction, let's think of an infinitesimal situation.
Let for a short amount of time $\Delta t$, the distance traveled is $v(t)\Delta t$ along a circular arc of radius $r$. The angle transversed is then $$\Delta\theta = \dfrac{v(t) \Delta t}{r}$$.
Imagine the bisector of the angle. Now, consider the changes in velocity parallel & perpendicular to this bisector. Initially the velocity has a component $v\sin(\frac{\Delta\theta}{2})$ away from the center & $v\cos(\frac{\Delta\theta}{2})$ transversely.Afterwards, it has a component $v\sin(\frac{\Delta\theta}{2})$ toward the center & $v\cos(\frac{\Delta\theta}{2})$ transversely as before. Thus the change of velocity is of magnitude $2v\sin(\frac{\Delta\theta}{2})$ toward the center of the arc.
As $\Delta\theta$ is vanishingly small, $\sin(\frac{\Delta\theta}{2})$ becomes indistinguishable as $\dfrac{\Delta\theta}{2}$. Thus, we can put $$|\Delta v(t)| = v^2 \dfrac{\Delta\theta}{r}$$. And the direction is towards the center. Thus the force is rotating the position vector along the curved trajectory and the change is radially inward regardless of whether it is traced clockwise or counter-clockwise.

The picture becomes more vivid if we calculate using polar co-ordinate.
First, we write the position vector as $\mathbf{r} = r\cdot \mathbf e_r$ . Now consider the change of $\mathbf{r}$ with time.  Its change during $\Delta t$ is $r\Delta\theta \cdot \mathbf e_{\theta}$. $e_r \, \textrm{and} \, e_{\theta}$ are mutually perpendicular, the first being outward radially from the center . Therefore velocity is $$\mathbf v = \dfrac{\mathrm d\mathbf{r}}{\mathrm dt} = r\frac{\mathrm d\theta}{\mathrm dt}\cdot \mathbf e_{\theta} = \omega r \cdot \mathbf e_{\theta}$$.
By putting $r = 1$, we get $$\frac{\mathrm d}{\mathrm dt} (\mathbf e_r) = \omega \mathbf e_{\theta}\;.$$
Similarly, a change of $\theta$ implies a change of $\mathbf e_{\theta}$.It can be seen that $$\frac{\mathrm d}{\mathrm dt} (\mathbf e_{\theta}) = -\omega \cdot \mathbf e_r\;.$$ Now we differentiate the velocity, $$\mathbf{a} = \omega r \dfrac{\mathrm d}{\mathrm dt} (\mathbf e_{\theta}) = -{\omega}^2 r\cdot \mathbf e_r \;.$$ This result falls down automatically with the correct direction which is opposite of $\mathbf e_r$ i.e. toward the center radially.
Answer:
$$\vec{a}_{avg}=\frac{∆\vec{v}}{∆t}$$
So,
$$ a_{avg}=\frac{|∆\vec{v}|}{∆t}$$
From (a) and (b) in the figure, it is clear that the velocities triangle is similar to the triangle $OP_{1}P_{2}$
So $$\frac{|∆\vec{v}|}{v_{1}}=\frac{|∆\vec{s}|}{R}$$
$$\frac{a_{avg}∆t}{v_{1}}=|∆\vec{s}|/R$$ then $$ a_{avg}=\frac{|∆\vec{s}|v_{1}}{R∆t}$$
However, as t goes to zero, we have $|∆\vec{s}|=∆s$ and $\frac{∆s}{∆t}=v_{1}$. Therefore, $$ a=\frac{v_{1}^{2}}{R}$$ However, since speed is constant we can simply drop the subscript 1 to get
$$ a=\frac{v^{2}}{R}$$
Answer: You said to not include calculus, but I will allow myself to take simple derivatives since velocity and acceleration are calculus quantities anyway.
For this proof I will assume two things

The radius is constant, i.e. $\vec r\cdot \vec r=r^2=c$
The acceleration is in opposite direction to the position vector, i.e. $\vec a\cdot \vec r=-ar$. It is centripetal acceleration after all. I think this should be provable from 1. with some more work.

Now I will use the identity $\frac{d}{dt}(\vec a\cdot \vec b)=\frac{d\vec a}{dt}\cdot\vec b+\vec a\cdot\frac{d\vec b}{dt}$ on 1. to get
\begin{align}
\frac{d}{dt}(\vec r\cdot \vec r)&=0\\
\vec v\cdot \vec r+\vec r\cdot \vec v&=0\\
\vec v\cdot\vec r&=0.
\end{align}
This is a well-known proof to show the velocity is orthogonal to the position vector. Applying the derivative once more we get
\begin{align}
\frac{d}{dt}(\vec v\cdot\vec r)&=0\\
\vec a\cdot\vec r+v^2&=0
\end{align}
Now we can use 2.:
\begin{align}
-ar+v^2&=0\\
a=\frac{v^2}{r}
\end{align}
Note that I didn't assume that $a$ or $v$ are constant, so this equality should hold whenever 1. and 2. are true.
Comment: Perhaps the shortest derivation uses the cross product: $$\vec v=\vec {\omega} \times \vec r$$ $$\text {so}\ \ \ \ \ \ \vec a=\vec {\omega} \times \vec v +\dot{\vec {\omega}} \times \vec r.$$But $\dot{\vec {\omega}}=0$, so $$\vec a=\vec {\omega} \times \vec v =\omega v(-\hat r)=-\frac{v^2}r\ \hat r.$$
Comment: Though if we let the rotation occur in the complex plane: $$z=|z|e^{i\omega t}$$ So $$\frac{dz}{dt}=i \omega |z|e^{i\omega t}\ \ \ \ \ \text{and} \ \ \ \ \ \frac {d_2 z}{dt^2}=-\omega^2 |z| e^{i\omega t}=-\omega^2 z.$$
Answer: Here is a rather quirky derivation that doesn't use calculus notation. I believe that the general idea dates back to Huygens (c 1659).
A small arc of a circle is almost indistinguishable from part of a parabola. You can show using Pythagoras's theorem that if $x$ is the semi-length of the chord that runs between the ends of an arc of a circle of radius $r$, and $h$ is the sagitta of the arc (the distance between the centre of the arc and the centre of the chord) then
$$x^2=2rh\ \ \ \ \ \ \ \ \text {provided $h$ is small enough to neglect $h^2$.}$$
In order to follow the path, a body at the midpoint of the arc at $t=0$ and moving parallel to the chord at speed $u$ must have an acceleration $a=\frac {u^2}r$ at right angles to the chord. We know this from ordinary projectile theory... At time $t$ from passing the centre of the arc,
$$x=ut\ \ \ \ \ \text{and}\ \ \ \ \ h=\tfrac 12 at^2\ \ \ \ \ \text{so}\ \ \ \ \ \ x^2=2\left(\frac{u^2}a\right) h$$
Comparing with our first equation, $r=\frac{u^2}a$, that is $a=\frac {u^2}r$.
Answer: A most elementary way to look at what is  changing in the radial direction only.
The time it takes from $v$ at rim to stationary state $v=0$ at center traversing radius length $r$ along radius is
$$ T=\frac{r}{v-0} =\frac{r}{v}$$
Note that this is not a uniform motion along a straight line in the tangential direction.
Acceleration along the radius (centripetal) then is
$$ T=\frac{v-0}{T} =\frac{v}{r/v} = \frac{v^2}{r}. $$


How do I change the ":" symbol in the minibuffer?

Question: When you switch to command-line-mode, with :, you see a ":" symbol identifying that you're indeed in command line mode. Is there any way, other than modifying source code, I can change this symbol?
Answer: The answer is short and might be unpleasing for you but: no.
You can't change that character without modifying the sources.
Answer: I am really not too sure, but by digging through the plugin vim-easymotion which displays a prompt "Search for characters: ", I found that it uses:
https://github.com/osyo-manga/vital-over
Which, based on the example, seems to be able to customise the command or search prompt.
I did not have time to experiment deeply with it, but it might be a lead.
Comment: My guess would be that they use a terminal window like fzf-vim for example, where it is easy to change the prompt. However I don't use easymotion and I don't have time to dig the code right now so I could be completely wrong. It's too bad that vital-over doesn't have a complete english documentation by the way.


ISP connections Vs IX Connections

Question: I would like to know the difference between ISP connections and IX connections? Basically we are evaluating some of the ISPs to get our edge circuits. While doing some research on it, came across to peering connections to IX locations. How do they help us? When I search in peering database, companies are having ISP links as well as IX links. How do they use it? Any pointers will be helpful? Sorry if it is a beginner question, could not find the correct difference and usage of these two models.
Comment: Did any answer help you? If so, you should accept the answer so that the question doesn't keep popping up forever, looking for an answer. Alternatively, you can provide your own answer and accept it.
Answer: The Internet is really just a bunch of networks connected together. Consumers and businesses connect to ISPs as customers. ISPs connect to other ISPs, either as customers of larger ISPs, or as peers at Internet Exchange Points (IXPs). At IXPs, ISPs can often peer with multiple other ISPs and CDNs, reducing costs by having a single location for connections, instead of running connections to multiple places in order to connect to multiple peers.
Comment: This is a different question, and you should really ask it in a different question, although it's actually off-topic since questions about networks over which you have no control are off topic, and you would really need to ask the provider(s), but if you were a cloud provider, and there was an ISP which, itself, was your customer, and/or the ISP had many or important or high-paying customers of yours, what would you do? It makes sense to have as direct a connection as possible to your customers, or your competition may win their business.
Comment: But, for example if you check any enterprise or cloud service provider in peeringdb, in addition to multiple ISP connections, they also peer publically in IXs. In this case, how they use this peering line? They send their traffic through this line, if so why they have multiple ISP connections also?

ISP connections Vs IX Connections


Mac keybindings not working in OneNote

Question: I've remapped my home and end keys to behave like Windows using https://damieng.com/blog/2015/04/24/make-home-end-keys-behave-like-windows-on-mac-os-x/
Shift+end highlights text from the cursor to the end of the line.
Shift+home highlights to the beginning of the line.
This works in all applications I've used since remapping - EXEPT MS OneNote. Anyone know why OneNote might behave differently? MS Outlook, MS Word, Mac Text Edit, Mac Notes - all work as I'd like.
What would cause OneNote to respond with an error beep as compared to the other apps? I don't have any custom key mappings in system/keyboard setup for OneNote.
MacOS: 12.3.1
MacBook Pro M1, 14"
Answer: Because, as the article you link says:

This remapping does the following in most Mac apps - including Chrome
(some apps manage their key handling directly):

OneNote apparently does.

Mac keybindings not working in OneNote


Is there a way to search a specific subject across state law?

Question: I'm doing research comparing a specific type of law, for example regarding marijuana possession, across states. What is the best way to search state law? Do I need to go state-by-state to find a state-specific database, or is there a better way? I'm finding this not to be straightforward at all - very unlike searching federal law and legislation.
Comment: Try Googling the question. There are many 50-state summaries out there.
Comment: E.g. https://www.google.com/search?client=firefox-b-1-d&q=50+state+summary+marijuana+law; https://www.google.com/search?client=firefox-b-1-d&sxsrf=ACYBGNRcsAUgNilwZVnJPFHrssFutnb_ig%3A1574470743990&ei=V4TYXenkO5TZ9AP99pDQCg&q=50+state+summary+tenant+rights&oq=50+state+summary+tenant+rights&gs_l=psy-ab.3..33i299.26910.30239..30621...1.5..0.129.1327.8j6......0....1..gws-wiz.......0i71j0i22i30j33i160.5EeWK4zo-Vs&ved=0ahUKEwip7J3LkP_lAhWULH0KHX07BKoQ4dUDCAo&uact=5
Comment: The National Conference of State Legislatures has lists of state laws for some topics, and advocacy groups also have been known to do this legwork.
Answer: It is inherently and intrinsically harder to search the laws of 50 states and compare them than it is to review a single body of federal law.
Also, case law and statute law, are researched quite differently, even though they can cover the same topics. Word searches work well for some kinds of case law issues, but poorly for statutes and regulations. Some issues, however, are hard to research with word searches, even if they involve case law.
Westlaw codes the important legal holdings of every published case in categories based upon their "key" system which has a subject-matter based digest of all case law. If you find one case with a key code that is on point, you can search all 50 states for that.
The "Restatements of Law" (e.g. Restatement of Contracts, Restatement of Torts) purports to be a summary of the majority common law rules in the named subject area, although it often reflects the lead author and supporting drafting committee's view on the "better rule" rather than the majority rule. These are very influential and often cited by courts in states where there is no case law on point. Some have been updated several times to reflect changes in the common law over time in different subject areas (e.g. the original Restatement of Contracts favored form over substances in a way that the later version do not). The American Bar Association signs off on these before they are published.
The National Conference of State Legislatures promulgates "Uniform" and "Model" acts that states may adopt with or without amendment and tracks which states have adopted them and with what modifications.
Most areas of law are the subject of legal treatises that make comprehensive 50 state summaries of relevant case law, one or two of which are typically the "leading treaties" and widely used and relied upon.
Many law review articles will do summaries of as many states' laws as possible on a given issue.
Many lobbying and special interest groups and also nolo.com prepare 50 state summaries. A good, reasonably up to date and accurate source on many topics is Wikipedia.
Ultimately, all secondary sources rely on somebody digging through state statutes in every state, analyzing them and coding them accurately, usually with citations to the source material. In fast changing areas of law this can be tricky, since laws of the same subject could be codified in different kinds of places (e.g. one might put something in a criminal code, while another might put it in a public health code or a regulation rather than a statute).
Every legal search engine (there are about half a dozen main ones, one or two of which are free) allows for 50 state searches in some form of primary legal materials like statutes and published appellate cases.
Answer: It looks like Westlaw is the most efficient way to do this.
Pending legislation may be more difficult though.


From diagram, assuming that the reaction is 40% complete in the vessel: a. The weight of the methanol added per day b. The amount of Hl recycled

Question:
Here is my solution on the problem and the diagram below is the actual given of the problem. I hope you could help me correct it.
Comment: What's your question?
Comment: (1) I assume that this is a check my work question? (2) Were you given any information other than what is in the diagram?
Comment: Note the posting a picture of written text instead of typing that text is discouraged for several reasons. 1/It cannot be searched for. 2/It cannot be reused as reference or quoting, putting extra effort on shoulders of responders. 3/It can be challenging to read and interpret. 4/ Low effort of question preparation leads often to low effort of question answering, even to question closure.
Answer: Edit 5/16 3pm Another equation just occurred to me...
Edit 5/16 6pm fixed a math mistake in equation (2)

This isn't an answer yet, I'm trying to clarify the question.
(1) I assume that this is a check my work question?
(2) Were you given any information other than what is in the diagram below?

So using the variables indicated in the diagram:
R = recycled HI, units = lb/day
X = feed of $\ce{CH3OH}$, units = lb/day
W = waste stream, units = lb/day
Y = product stream, units = lb/day
Additional points:

The waste and product outputs give % values. Weight % or Mole fraction %?
Weight % would seem to be the right choice to me.
The diagram would seem to indicate that the total input to the reactor is $\pu{2000 + R lbs/day}$ of $\ce{HI}$.

(A) - 2000 lb/day of HI feed
Assume 2000 lb/day of $\ce{HI}$ is input into the reactor plus whatever is recycled. We can only assume that the recycling has reached a steady state, so 2000 lbs/day of HI, or its equivalent, must be in the input stream to the separator. So summing the two steams from the separator:
\begin{align}
\pu{1 lb \ce{HI}} &= \pu{1 lb \ce{HI}} \times \pu{2200 g/lb} \times \dfrac{\pu{141.939 g \ce{CH3I}/mol}}{\pu{127.911 g \ce{HI}/mol}}\times  \dfrac{1}{\pu{2200 g/lb}} \\
&= \pu{1.1097 lb \ce{CH3I}} \\
\end{align}
$\therefore 2000 = 0.8265\times W  + \dfrac{0.816}{1.1097}\times Y = 0.8265\times W + 0.73533\times Y\tag{1} $
(B) Now for $\ce{CH3OH}$ feed
$$\pu{1 lb \ce{CH3OH}} = \dfrac{141.939}{32.04} = \pu{4.4301 lb \ce{CH3I}}$$
$\therefore X = \dfrac{0.816}{4.4301}\times Y + 0.184 \times Y = 0.36819 \times Y\tag{2}$
(C) Balance water and $\ce{CH3I}$
The moles of water in the waste stream has to equal the moles of $\ce{CH3I}$ in the product stream.
$$\dfrac{0.1735\times W}{18.015} = \dfrac{0.816\times Y}{141.94}$$
$$\therefore W = 0.59693 \times Y\tag{3}$$
Substituting (3) into (1)
$$2000 = 0.8265\times (0.59693 \times Y) + 0.73533\times Y\tag{1a} $$
or $Y = 1,627.75 \approx \pu{1628 lbs/day}$
and $W = 0.59693\times 1627.75 = 971.65 \approx \pu{972 lbs/day}$
and $X = 0.36819 \times 1,627.75 = 599.32 \approx \pu{599 lbs/day}$
(D) For R -- "40% Complete" of what?

40% of HI is reacted?
40% of $\ce{CH3OH}$ is converted to $\ce{CH3I}$?

If the 40% percent is for the fraction of $\ce{CH3OH}$ which reacted, then  81.6:18.4 is not 60:40 in weight % or mole per cent for $\ce{CH3I:CH3OH}$.
\begin{array}{|c|c|c|c|c|}\hline
& \pu{M.W.} &\pu{mol \%} & \pu{\dfrac{\pu{g}}{\pu{mol}}} & \pu{wt \%} \\ \hline
\ce{CH3I} & 141.939 & 82.65 & 117.31 & 95.47 \\ \hline
\ce{CH3OH} & 32.04 & 17.35 & 5.56 & 4.53 \\ \hline
\end{array}
\begin{array}{|c|c|c|c|c|}\hline
& \pu{M.W.} &\pu{wt \%} & \pu{\dfrac{\pu{mol}}{\pu{100 g}}} & \pu{mol \%} \\ \hline
\ce{CH3I} & 141.939 & 82.65 & 0.58229 & 51.814 \\ \hline
\ce{CH3OH} & 32.04 & 17.35 & 0.54151 & 48.186 \\ \hline
\end{array}
The only way to use the information seems to be to assume that 40% of the total $\ce{HI}$ feed, 2000 + R, is converted to $\ce{CH3I}$.
$\therefore (2000 +R)\times 0.4 = \dfrac{0.816}{1.1097}\times Y = 0.73533\times Y\tag{4}$
Using Y from above, $R = 992.33 \approx \pu{992 lbs/day}$

Note - In comments below user Chet Miller makes a convincing argument about two points.

The 40% reacted would refer to the % of the limiting reactant which reacted. This would mean that the 40% referred to $\ce{CH3OH}$.
The OP's diagram is incorrectly drawn. Being a chemical engineer he proposes a diagram which makes more sense. There has to be a second separator of some sort.

I doubt that you could get all the HI out of water stream, but concentrated HI is about 57% HI by weight. So the reactor output and the waste stream, W, would have to be above atmospheric pressure.

Given Chet's proposed diagram below $R = 0.8265\times W = 803.06 \approx \pu{803 lbs/day}$
Comment: Shouldn't the recycle stream be coming from the separator, and have the same concentration as in the waste stream?  For 40% complete, they mean that the limiting reactant is 40% consumed.
Comment: @ChetMiller - RE: "*recycle stream be coming from the separator*" ---I don't know. I took the diagram literately, without making any assumptions about the actual process. Obviously some kind of separation would have to be happening if just HI was recycled from the reactor output. That being said I don't think  you'd recycle waste from the separator shown in the diagram. You wouldn't want to put the water back into the reactor.
Comment: RE: *40% complete, they mean that the limiting reactant is 40% consumed.* --- Could be, I'm not extremely familiar with how chem engineering problems are setup. I also wonder how accurate the OP's drawing is.
Answer: It's much easier to work in terms of lb-mole flow rates than in terms of lb or in terms of metric units.  So I will express X,Y, W, and R in terms of lb-moles/day.  2000 lb/day of HI is 15.6 lb-moles/day.  Since the stoichiometry is such the number of moles of product from the reaction are equal to the number of moles of reactant, the molar flow must be constant.  Therefore we can use molar balances on the system.  The overall molar balance on the system is $$15.6+X=Y+W$$ In addition, a carbon balance on the system is $$X=Y$$ and thus $$W = 15.6\ lb-moles/day$$.
The mole fractions in stream Y are:$$CH_3I=0.4998$$ and $$CH_3OH=0.5002$$
The mole fractions in stream W are:  $$HI=0.4012$$and$$H_2O=0.5998$$
So an overall balance on iodine yields: $$15.6=0.4012W+0.4988Y=(0.4012)(15.6)+0.4998W$$From this, it follows that $$Y=18.76\ lb-moles/day=X$$
These values agree with the mass flow rates determined by @MaxW.


If you open a black hole on Earth, is it escapable?

Question: What if the Relativistic Heavy Ion Collider at Brookhaven created an unstable blackhole; could you survive this and how?
Comment: Welcome to Worldbuilding. If you have a minute, please take the [tour] and have a look at the [help]. Please observe our [Code of Conduct]. Writing in capital letters is the equivalent of screaming and considered rude.
Comment: A: I'm not entirely sure that 'unstable black hole' is a real concept. B: What mass black hole are we talking here? C: The 'and how' part of the question is pretty broad, and very much depends on a host of secondary properties of the black hole, the planet, how the black hole forms and the preparedness of the person doing the escaping.
Comment: (1) That's the *Relativistic* HIC, not "Realistic". (2) What's an "unstable" black hole? (3) Since $m = E / c^2$, one can easily compute the maximum possible mass for whatever kind of hole of whatever color the Brookhaven machine could have produced. Hint: it's not large. RHIC [draws about 77 MW](https://science.energy.gov/~/media/np/nsac/pdf/Reports%20Archives/Operating_Costs_BNL_RHIC.pdf); one full day of operation consumes 6.7 TJ, corresponding to a mass of 74 milligrams (0.04 avoirdupois drams in Ancient American). Such a tiny black hole will have a very very very short life...
Comment: If the energy and speed of the collision are high enough (I suspect that the RHIC at BNL would have to be substantially upgraded for this to happen) it is theoretically possible for tiny black holes to be produced, Hower their existence would be fleeting. In the sense that they are fleeting (their Hawking Radiation is sufficient to 'deplete' them before they can gather additional mass) perhaps that could classify them as 'unstable'.
Comment: If you do some googling, you will find some quite well thought out arguments regarding what would happen if a particle accelerator created a black hole.  The general consensus is that there is no threat at all, unless some "new" physics is discovered at such high energies, but we can barely even speculate on what would happen if new physics came around beacuse it's...well.. not known.
Answer: The government website says it can't, but we say it can! Unless scientists can somehow bring an incredible amount of mass to earth, the answer is yes, easily. Configuring particles to form an event horizon does not somehow increase their mass or gravity. You could conceivably pack lots of mass close enough to form an event horizon, but it would be incredibly small and incredibly brief. At any appreciable distance, that black hole's gravity would be exactly the same as what those particles had in their previous configuration.
Answer:
What if the Realistic Heavy Ion Collider at Brookhaven created an unstable black hole; could you survive this and how?

That is a bit strange, not sure what is an unstable black hole, so I will rephrase it to:

What if suddenly a black hole appeared on Earth; could you survive this (gravity)?

Also, I will only take into account gravity and not other effects as time dilatation nor Hawking radiation.
It depends on the mass of the black hole. But for a particle collider, of course.
A black hole doesn't differ in other objects about its gravity.
You may need to know about the Surface Gravity. On Earth, that is around 9.8 m/s 2, obviously easy to escape.
Given the following formula:
$$g = \frac{GM}{r^2}$$
Where $M$ is the mass of the black hole in kilograms, $r$ the radius (or distance between you and the body mass) in meters, and $G$ is the gravitational constant, (i.e: $6.67408(31)\times10^{-11}\text{m}^3\text{kg}^{-1}\text{s}^{-2}$) you can calculate the gravity.
With this value, you can calculate the gravitational force of an object (even a black hole) of any given mass to any given distance from it. It is to your choice to determine if you are able to escape or not with your current technology level (remember that if $g >= c$, where $c$ is the speed of light ($299,792,458 \text{ m/s}$) you can never escape, no matter how much you try).
Remember, a black hole of a few kilograms won't produce any relevant gravitation well after a few millimeters from it (instead, you should be careful about its Hawking radiation, not gravity).
A black hole produced in a particle collider would only have the mass of a few atoms, it would be harmless since its Schwarzschild radius (and gravity) will be incredibly small, it will literally die of hunger since particles won't fit on its "mouth" nor on its gravity well!
Another formula that may be useful for you is the Schwarzschild radius mentioned above, also know as gravitational radius, a distance from a black hole where nothing can escape. That is:
$$r_s = \frac{2GM}{c^2}$$
If you want to know the mass of a black hole enough strong to pull anything from a given range use:
$$M = \frac{c^2}{2G}R_s$$
If we resolve it with the radius of Earth which is $6.371\text{ km}$ do:
$$M = \frac{c^2}{2G}6.371.000\text{ m}$$
$$M = 4.28 \times 10^{33} \text{ kg} = 4,289,704,786.36 \text{ Yg}$$
Or $2,157.28 \text{ M}_\odot$, in solar masses.
Other Facts
As said in a comment, I ignored the Hawking radiation. Well, not anymore.
Hawking radiation is a process where the black hole emits energy, thus lowering its mass and shrinking. Due to this radiation, black holes has a limited lifespan and tend to evaporate. However, the evaporation is inversely proportional to the mass of the black hole, so a normal one will die after the thermal death of the universe.
Lifetime
The lifetime of a black hole is calculated given the following formula:
$$t = M^3\frac{5120 \pi G^2}{\hbar c^4}$$
Where $\hbar$ is the reduced Planck constant which is $\hbar = \frac{h}{2\pi}$, where $h$ is the Plack constant equal to $6.62607015 \times 10^{-34} \text{Js}$.
For a black hole of the mass of two hydrogen atoms (since you want to make it in the collider) would be very small. An hydrogen atoms has an atomic weight of 1.008 u. Where $u = 1.66053904020 \times 10^{-27} \text{ kg}$, very small. So its mass will have $3.3476467050432 \times 10^{-27} \text{ kg}$.
$$t = (3.3476467050432 \times 10^{-27})^3\frac{5120 \pi G^2}{\hbar c^4}$$
$$t = 3,155\times 10^{-95} \text{ seconds}$$
As you can see, very little. At that time the black hole will disappear in less time than the photons will take to reach the nearest atom of air in their surroundings.
But, let's do the opposite, we will try with the big black hole from above:
$$t = (4.28 \times 10^{33})^3\frac{5120 \pi G^2}{\hbar c^4}$$
$$t = 6.639 \times 10^{84} \text{ s}$$
Or, $2.105 \times 10^{77} \text{ years}$
Heat
The emission of Hawking radiation produces heat. As everything related to Hawking radiation, it is inversely proportional to the mass of the black hole. The formula is:
$$T = \frac{1}{M}\frac{\hbar c^3}{8 \pi k_b G}$$
Where $T$ is the temperature in degrees kelvin and $k_b$ is the Boltzmann constant equal to $k = 1.3806485279 \times 10^{-23} \text{ Jk}^{-1}$
$$T = \frac{1}{3.3476467050432 \times 10^{-27}}\frac{\hbar c^3}{8 \pi k_b G}$$
$$T = 3.365 \times 10^{49} \text{ k}$$
That is very hot! More that the hottest stars!
And with the big black hole:
$$T = \frac{1}{4.28 \times 10^{33}}\frac{\hbar c^3}{8 \pi k_b G}$$
$$T = 2.86 \text{ k}$$
... very little. Almost nothing. That is why big black holes last longer, they emit very little energy.
Luminosity
The emission of Hawking radiation produces light. As everything related to Hawking radiation, it is inversely proportional to the mass of the black hole. The formula is:
$$L = \frac{1}{M^2}\frac{\hbar c^6}{15360 \pi G^2}$$
Small black hole:
$$L = \frac{1}{(3.3476467050432 \times 10^{-27})^2}\frac{\hbar c^6}{15360 \pi G^2}$$
$$L = 1.063 \times 10^{59} \text{ W}$$
Like you can see, both the temperature and luminosity are incredible big, but since the duration of the black hole is very small, it won't cause any damage. For been exactly, the energy emitted during it lifetime is based on the mass and energy equivalence:
$$E = mc^2$$
So:
$$E = 3.3476467050432 \times 10^{-27}c^2$$
$$E = 3.008 \text{ J}$$
Like I said, harmless.
Or $1.063 \times 10^{35} \text{ YW}$
More that the brightest stars or quasars, but just for an instant.
Big black hole:
$$L = \frac{1}{(4.28 \times 10^{33})^2}\frac{\hbar c^6}{15360 \pi G^2}$$
$$L = 0.082 \text{ W}$$
P.D: Other facts math take from Hawking Radiation Calculator.
Comment: A reasonable answer, but you're ignoring black hole evaporation, which becomes entirely applicable for holes this small.
Comment: @WhatRoughBeast done


Issues creating a glint or shine on a logo in After Effects that is exportable to HTML

Question: I am trying to replicate a moving logo with a glint that is exportable to HTML.
I followed this video which uses light sweep, the animation is working, but no light effects came through.
I read that Bodymovin does not support effects, so I used alpha matte instead to achieve the same effect as per this video but its still having problems. The light is showing up, but there is no matting, please see the image attached.
In summary, I am trying to make a logo with a glint/lightsweep effect that can work with Bodymovin. What am I doing wrong? Is there a way to get either of these projects to work? Here is the after effects project showing the look I'm trying to achieve.
Answer: As per Bodymovin documentation, these are the supported features:

Shapes

paths, ellipses, stars, rectangles
fills, stroke, gradient fills and gradient strokes
dashes
trim paths (partial)
rounded corners
transforms (position, anchor point, rotate, scale, opacity, skew, skew axis)

Solids
Images
Comps
Nulls
Texts

Most fonts are supported, they can be exported as shapes or as text provided you load the font on the browser
A single text declination is supported per textbox.
Not supported: text decorations like Small caps, or Vertical Scale.

Cameras (partial)
Expressions

Expressions. Partially supported. More and more expressions are supported on each release.
If you have any trouble with one, let us know.
Expression sliders are supported.

General supported properties

transform (rotate,scale,position,anchor point, opacity)
time remap
masks (additive, and in some cases difference, inverted and subtractive)
luma and alpha masks (with the svg renderer)

Supported effects (with the svg/html renderer)

fill
pro levels
drop shadow
tint
tritone

So as long as you use vector shapes and vector effects / transformations you should be good. Try using a vector mask and your glint could be another vector shape moving from side to side.


Possible values of $x+y+z$

Question: Given that $xyz=11(x+y+z)$, find the possible values of $x+y+z$ such that $x\leq y$ satisifies $x^2+y^2=2018$
I know that $x,y$, or $z$ should be divisible by $11$. I am having trouble going to the next step
Comment: What Kind of numbers are $x,y,z$?
Comment: Any condition on $z$?
Comment: I suppose, $x,y,z\in \mathbb{Z}$.
Comment: No conditions were given for z
Answer: Since $2018=2\cdot 1009$, we have only the divisors $1,2,1009,2018$, so we have
$$
r_2(2018)=4(d_1(n)-d_3(n))=4,
$$
where $$d_i(n)=\sum_{d\mid n,d\equiv i \bmod 4}1.$$
So there are only $4$ possibilities, up to sign and permutation, namely
essentially only
$$
13^2+43^2=2018.
$$
Then we solve $xyz=11(x+y+z)$ for all $4$ possibilities $(x,y)=(\pm 13,\pm 43)$. This is not solvable for integer $z$.
Answer: First $x^2+y^2=2018$ has as only solutions $(x,y)=(\pm13,\pm43),(\pm43,\pm13)$ so by the restriction $x\le y$ we have two solutions $(x,y)=(\pm13,43)$. This is not compatible with the equation $xyz=11(x+y+z)$ because in both cases we would have
$$-z=\frac{330}{570}\\z=\frac{616}{548}$$ Thus there are no solutions.
Answer: For real numbers, we have
$$z=\frac{11x+11y}{xy-11}=\frac{22(x+y)}{(x+y)^2-2040}$$
subject to $-2\sqrt{1009}\leq x+y\leq 2\sqrt{1009}$
Answer: To use as little machinery as possible but at the cost of a bit of computation, note that we must have $\frac 12\cdot 2018 \lt y^2 \lt 2018$, which gives $33 \le y \le 44$, only $12$ possibilities.  Trying each to check if $x$ is an integer gets us that the only choice for $x,y$ is $x=13, y=43$.  That would require $13\cdot 43z=11(13+43+z)$ which clearly has no solution with $z \ge 1$ because the left side is much too big.
Answer: I think the problem also states that $x,y,z$ must be integers.
Assume that $x$ is divisible by $11$, then $x=11k$ ($k$ is an integer) and $(11k)^2+y^2=2018$.
We will have $y^2=2018-121k^2$. Because $y^2\ge 0$, we can prove that $k^2$ must be in this set of numbers: $(1;4;9;16)$.
We have four cases:

$k^2=1\Rightarrow y=43.55$, result eliminated
$k^2=4\Rightarrow y=39.17$, result eliminated
$k^2=9\Rightarrow y=30.48$, result eliminated
$k^2=16\Rightarrow y=9.05$, result eliminated

So there are no cases available for "$x$ is divisible by $11$" that satisfy $x^2+y^2=2018$, this means "$y$ is divisible by $11$" cannot be true as well.
Because of this, someone should attempt it by solving the equation $x^2+y^2=2018$ for integers $x,y$ instead, you can make a table that consists of $45$ cases from $1$ to $45$ for $x$ to find $y$ (or fewer cases if you use the condition $x\le y$ to eliminate some cases). This method is really tedious, but you should get $x=13$ and $y=43$, then you can solve for $z$.
However, $z=1.12$ which means $z$ is not an integer, this means there are no three integers $x,y,z$ satisfy all the conditions.
Comment: Why can't $z$ be the one divisible by $11$?

Possible values of $x+y+z$


My Cubot X12´s screen is busted and im still recieving messages which I cannot read or listen to.

Question: So my phone is busted and I get messages from my friends which i can´t read. Is there a way I can read the messages from an another device or even control the phone exteranaly?
Answer: Put the sim into another phone


How to list all the files changed by a perforce change list

Question: I have a specific change list, and from the command line I want to list all files that were a part of that change list.  How do I do this?
Comment: Probably the easiest way: `p4 files @=`
Comment: @user114245 That's much more precise, you should add an answer
Answer: That's the describe command. To describe a particular changelist, you want p4 describe <changelist number>.
Update:
If you only want the file names, you can use the files command with the -F option to override the output format: p4 -Ztag -F "%depotFile%" files @=<changelist>
See http://www.perforce.com/blog/130826/fun-formatting for more information about the -F option.
Comment: I had to put @= in quotes, eg. "@=1234" if 1234 is CL number
Comment: @coffee If you're comfortable with Python, you can run `p4 -G describe -s ` and it will cause the output to be formatted as marshalled Python dictionary objects, which you can then parse and retrieve only the files.
Comment: This is the command I need!  However, it's pretty verbose.  I can exclude the diff'd code with -s, but is there a way ONLY list the files changed?
Comment: If you'd like to set up a convenient p4 describe tool in your visual client, select Tools | Manage Custom Tools, New | Tool, and in the resulting dialog set the following:
* Name: Describe Changelist
* Add to applicable context menus: True
* Application: p4
* Arguments: describe -s %c
* Run tool in terminal window: True.

This provides a convenient context menu item usable on changelists other than the default and outputs to a terminal window for convenient copying.
Answer: user114245 gave the best answer, but only in a comment. I'm adding as an answer to give it more visibility, and to improve a little.
For change 12345, this is the closest you can get with just a p4 command
p4 files @=12345

which gives only this output
//depot/file1#3 - delete change 3 (text)
//depot/file2#3 - edit change 3 (text)
//depot/file5#1 - add change 3 (text)

If you want to remove the extraneous info about each file, you'll need to process that output through more tools on the command line. Assuming a standard unixy environment, you can use a single sed command like so
p4 files @=12345 | sed s/#.*//

to get the desired result
//depot/file1
//depot/file2
//depot/file5

The currently accepted answer by Mike is this
p4 describe 12345

which gives all of this extra detail in the output
Change 12345 by day@client1 on 2013/06/21 00:25:28

    Some example changes

Affected files ...

... //depot/file1#3 delete
... //depot/file2#3 edit
... //depot/file5#1 add

Differences ...

==== //depot/file2#3 (text) ====

1c1
< This is file 2
---
> This is file 2 - edited

That was improved on by Doug's answer which uses grep and awk to filter out the noise and just leave the files changed, but the command is quite long
p4 describe -s 12345 | grep '^\.\.\.' | awk '{print $2}'

I think the solution given here is neater.
Comment: `p4 fstat -T "depotFile" @=12345` is pretty good too. But you have to trim off the `... depotFile ` leading text.
Answer: Mike O'Connor's answer helped me too.
As for just getting a list of files changed, couldn't it just be as easy as this?
p4 describe -s {change number} | grep '^\.\.\.' | awk '{print $2}'

that's what I'm going to use.
Comment: You can move the regular expression pattern into your awk script: `awk '/^\.\.\./{print $2}'`
Answer: If your changelist is still pending, you can run p4 opened or p4 opened -c <changeListNumber> for a specific (unsubmitted) changelist number.
From the official documentation:
"List files that are open in pending changelists."
Answer: p4 change -o
It has the list of files at the end.
Comment: For **submitted** changelists, no it doesn't. The Files section is only shown for pending changelists. The output for a submitted changelist actually has a comment describing the Files section which says `What opened files from the default changelist are to be added to this changelist.  You may delete files from this list. (New changelists only.)`
Comment: With version 2013.1, the "Files" section isn't even shown for pending changelists, from what I can see.


Spearman correlation between two rasters in R

Question: I have two rasters of the same area. The only difference is that they were created using two different methods. I would like to compare them and see how correlated they are pixel by pixel.
Any ideas how to do this in R or how to do a spearman correlation.
I wanted to do something similar to this (http://www.hakimabdi.com/20150101/an-r-function-to-test-pixelwise-correlation-between-two-time-series-of-raster-data/) with the result presented as a third raster.
Comment: Well, let's stop and think about this for a second. The example you provide is for a time-series n=t(1..i). With two rasters, for any given pixel, you have an n=2. How can you base a correlation on two observations? Now, do you want a correlation based on a moving window or for the entire raster? A moving window correlation would allow for creating a new raster of the correlation coefficients but would be highly dependent on the size or shape of the focal window.
Comment: Hi Jeffrey, thanks for the answer. The example I gave in my previous message was more to show the kind of output I " hope " I could get. I'm fully aware that this is for time-series. What I basically want to do is to compare two rasters for the same geographic areas done in two different ways. Looking if there regions that are similar in both maps or completely different, and check how significant is the difference for example. What I have done so far is doing one minus the other but I cannot say that it's statistically significant.
Comment: Computing the Spearman correlation will tell you little. To determine statistical significance it would help to have an explicit model of the sources of variation between the two rasters.  To compare the rasters, it is far better to *look* at a map of their differences and move on from there. But, as @JeffreyEvans notes, a "pixel by pixel" correlation makes no sense and is not possible to compute.
Comment: The link appears to be broken. If a code snippet that you tried and were stuck on had been included as formatted text within your question body then this question would have retained much more value.
Answer: Here are two approaches to compute correlation coefficients with Raster objects (notwithstanding the comments on your question about utility; that I concur with).
<code># generate some data
library(raster)
set.seed(89)
b <- brick(system.file("external/rlogo.grd", package="raster"))
d <- b
d[] <- runif(ncell(b)*nlayers(b))
s <- stack(b[[1:2]], d[[1:2]], b[[3]], d[[3]])

# let's say we have two Raster objects with three layers
x <- s[[1:3]]
y <- s[[4:6]]

# stack these
z <- stack(x,y)

r <- calc(z, fun=function(x) cor(x[1:3], x[4:6], method='spearman'))

## Now, to do local correlation with two layers
r1 <- b[[1]]
r2 <- b[[2]]
rc <- corLocal(r1, r2, method='spearman')
</code>
Comment: @JaberBelkhiria if this solved your problem, you should mark it as the solution.
Comment: Fails because `d` does not exist when you do `d[] = runif(....)`. Did `d` come from somewhere else?


Lists with Parent Child Relationshiips - add records

Question: I am creating a CRM system in SharePoint 2010, and I would like to be able to have the main list have the customers, and then sub lists with contact information, notes information, task information and libraries.  I would like to have a page that shows the profile of the customer with webparts of each of the associated items, and when I add an item, I want it to automatically know who the customer is and associate it to the correct customer.
Comment: So what is your exact problem here, you haven't really asked a question.
Answer: PirateEric has actually answered this kind of question, and if I'm not wrong you are looking for the same:
How can I pass a Parameter via Query String to populate a lookup field on a New Item Form/DVWP?
So you have to pass the CustomerId in Query String to this page and use JavaScript/JQuery to set the value of the lookup, if you don't want the user to change the value.. you just make it read-only or even hide the lookup control..
Answer: You can create webpart page which displaying your list in webpart zone. Then you can link the webpart. here is the sample of parent-child relationship:
http://www.sharepointhillbilly.com/Lists/Posts/Post.aspx?ID=4


Russian books on Fourier Series

Question: Russians are known on having very strong literature in Mathematics and Sciences overall. So as I am willing to improve my knowledge and problem solving in real-analysis, especially in Fourier series, so I wonder what are the books written and used by the Russians( I have a somewhat good level in Russian)?
Comment: see here https://books.google.de/books/about/Fourier_Series.html?id=XqqNDQeLfAkC&redir_esc=y i hope this will help you
Comment: Ok, I just saw that Dr Graubner has already posted Tolstov's book which comes as no surprise given its status. Sorry for that.. At least I hope the other two will be of some interest.
Answer: At the top of my head I would recommend Georgi Tolstov's "Fourier Series" (1960 or 1962 version) which is also available in English but rather easy to find in Russian as well.
It provides for a very smooth and elegant introduction to the topic as far as my experience allows me to say. Descriptive and very well balanced. In any case it is considered a classic.
Another choice would be Anatolyi Aksionov's "Mathematical Analysis. Fourier Series" which is rather short (~90 pages) and part of a series on textbooks on analysis by the same author. I have not read it though.
A book on the topic written by a Russian, but in English (and perhaps the most recent one covering both of these "conditions") is Alexander Koldobsky's "Fourier Analysis In Convex Geometry" which is certainly more specialised and focuses on the application of Fourier analytic methods in solving the Busemann-Petty problem (for more on this check here).
Answer: I hope it can help you :
Season $6-7$ of volume $3$ of  The fundamentals of mathematical analysis by Grigorii Fichtenholz
фихтенгольц - курс дифференциального и интегрального исчисления том 3:
original language book
author: Фихтенгольц, Григорий Михайлович

Russian books on Fourier Series


Is the sentence "Je me sentais de la même façon que tu te sens maintenant" correct?

Question: Is "Je me sentais de la même façon que tu te sens maintenant" correct? I want it to mean "I felt the same way that you're feeling now" Can I use "en ce moment" at the end instead of "Maintenant", and if there's anything wrong can you explain?
Comment: This sounds ok to me, as a native speaker, and this sounds better that what the current answers propose. Also "en ce moment" is perfectly idiomatic.
Comment: @AlexandreC. Même que je suis américain, je préfère la phrase dans la question.
Answer: It is correct since there is no mistake, but is not the good way to express it. If you are tied to the litterate translation, you can use it. If you just want to express it, you could use :

J'avais le même sentiment que tu as maintenant

"Que tu te" is heavy in french and won't be widely used in familiar language except in the subjonctif tense which is not very used itself.
You can even remove the second verb and write :

J'avais le même sentiment que toi à présent

And yes, en ce moment can replace maintenant like à présent did in my second translation.
Answer: You could also say:
  "J'ai éprouvé la même chose que toi maintenant"
"éprouver" means "to have the feeling that" and is more idiomatic than "avoir le sentiment". You could use "se sentir" but "se sentir" is less precise as it can refer to physical sensations as well as sentimental feelings.


Minimum Requirements of a microcontroller to run embedded Linux?

Question: I am looking for the cheapest microcontroller to run embedded Linux. what should I be looking at, which class of microcontrollers or microprocessors is compatible? is there a minimum system requirements (RAM, Flash, clock, etc)?
Comment: As the question is (re)worded it seem totally on-topic to me. Vote to reopen and +1 for the (modified!) question.
Comment: I've seen some TI SoCs with onboard GPUs for <$20 and some 300MHz atmel micros for <$10 (but no dedicated gpu), almost everyone has something that can run *some* version of linux. Do you want a gpu? hdmi? fast processor? dual core? full SoC (i.e. just add ram, flash and power and bingo) or do you need something that requires designing a full blown bios? Do you want something with onboard programmable logic? something that'll run Ubuntu? The application determines the part.
Comment: you need a microprocessor something like beagle Bone
Comment: Looks like a shopping question to me... a broad one, really need to know which distro. You can run Linux on an 8088, but you won't be able to do much. You also need some requirements on what you want to do. When designing a system, you first come up with a list of capabilities you need and requirements, then pick the distro and system.
Answer: That totally depends on what you want to do with it, really.
I've run uClinux on an MMUless ARM7TDMI with 8 MB RAM -- but that will only run a kernel, a minimalistic init and dedicated services. If you are going for volume, 8 or 16 MB RAM, 32 MB Flash and some small ARM core are the minimum required, but you will need a lot of time shrinking the system to fit, so the expense for that needs to be offset by cheaper hardware.
In the middle, 64 MB RAM and 512 MB Flash with something that has an MMU (ARM9 and up, basically) gives you enough space to run a standard distribution like Debian, but you still need to compile your own kernel and bootloader. Less software effort, and the hardware should be cheap enough so anything below a million units is probably cheapest here.
Last but not least, there are readymade Linux systems that come with full vendor support, and generally have lots of memory, lots of flash, an eMMC disk, an SD card slot, PCIe slots for wireless cards, HDMI outputs and so on. These tend to be fairly expensive, but you can usually cross-compile your Linux applications fairly easily, and get set up in a week.
There are a few standard boards that are produced in bulk, like the Raspberry Pi and the Odroid boards. These are fairly cheap for the kind of specs they have, but the boards are larger than necessary for most embedded applications, and certainly more expensive in bulk than a custom-made board with just the required peripherals and the smallest CPU, RAM and Flash you can get away with (especially as the board isn't friendly to being used as a building block -- all interfaces are either fixed connectors on the edge, or unpopulated headers that need to be soldered in.
For a prototype or one-off, I'd definitely go with the Pi or one of its clones (which often have additional interfaces, so comparing here might pay off). For mass production, build your own PCB with a mid-range CPU with MMU, enough memory to run a standard system, enough flash to keep it and allow updates using the package manager, and just the peripherals you actually need.
Comment: Odroid and RPi are certainly NOT the cheapest boards out there. Remember this [reddit post](https://www.reddit.com/r/raspberry_pi/comments/3j0rn6/15_orange_pi_pc_hacker_sbc_packs_16ghz_quadcore/)?
Comment: Dimitry, that link goes to a page where people are having a go at the banana pi and the orange pi for lacking decent gpu software support, the sale price is deceptive as it is not the *true price*, once you factor in the work required to get a system fully operational the numbers can shift dramatically.
Comment: @Tom - in the context of a legitimate design question, one has to separate the engineering cost from the unit cost.  Some of the options out there may indeed require substantial engineering effort, the thing is that you can make your own board in moderate quantities - the $5 pi zero is just a tease, as you can neither buy the boards in quantity nor readily make a custom derivative.
Comment: @ChrisStratton I agree completely, an extreme example would be Intel's 5th gen processors, the foundry retrofit for 14nm was over $10B I think, so if you wanted to make just *one*, it'd cost you $10B, but if you sold 1B units, the amortised costs would only add 10 bucks to the sale price. So yes, the engineering and sale costs are different things, which is why I'd advise people to use a cots system if they can (ideally one with a decent software ecosystem). Personally I'd choose a TI solution (e.g. beaglebone) over the Pi, the ecosystem's smaller, but you can actually *buy* the bare chips


Magento 2.2.2 Admin panel grids pushed to right

Question: At some point during developing my Magento 2 website, I flushed my cache and after a refresh, all the content on the admin panels are pushed to the right. I was working on a custom mass action under the orders tab, so I undid all my actions to the point before flushing the cache, but nothing seems to help.
See following screen:
Comment: Try removing var folder and check. It may help. Or run static content deploy only for Magento/backend
Comment: Hm doesn't seem to work. Same results. I've checked the css file, and for some reason the following css is enabled "width: calc( (100%) * 0.75 - 30px )". When changing the 0.75 to 1, it works fine. But I'm working on a module which needs to be distributed, so I don't get why this suddenly is causing the width to alter and this should not happen on other installations of magento when using the module..
Answer: Okay, I couldn't figure out the problem. I reinstalled everything and now it works again. Probably some typo in some .xml file, I'm not sure.


Smart way to copy multiple files from different paths using scp

Question: I would like to know an easy way to use scp to copy files and folders that are present in different paths on my file system. The SSH destination server requests a password and I cannot put this in configuration files. I know that scp doesn't have a password parameter that I could supply from a script, so for now I must copy each file or directory one by one, writing my password every time.
Answer: in addition to the already mentioned glob:
you can use {,} to define alternative paths/pathparts in one single statement
e.g.: scp user@host:/{PATH1,PATH2} DESTINATION
Comment: Note that you have to quote or escape the braces so the shell doesn't expand them if you want to avoid entering the password twice. e.g. `scp 'user@host:{PATH1,PATH2}' DESTINATION`
Comment: What about the target path? Is there a way to have something like `scp 'user@host:{PATH1,PATH2}' DIRECTORY/{FILE1,FILE2}` ?
Comment: What about the target path? Is there a way to have something like scp 'user@host:{PATH1,PATH2}' DIRECTORY/{FILE1,FILE2}
Answer: From this site:
Open the master
SSHSOCKET=~/.ssh/myUsername@targetServerName
ssh -M -f -N -o ControlPath=$SSHSOCKET myUsername@targetServerName
Open and close other connections without re-authenticating as you like
scp -o ControlPath=$SSHSOCKET myUsername@targetServerName:remoteFile.txt ./
Close the master connection
ssh -S $SSHSOCKET -O exit myUsername@targetServerName
It's intuitive, safer than creating a key pair, faster than creating a compressed file and worked for me!
Comment: This is a great answer! Useful when passwordless keys are not allowed.
Answer: If you can express all the names of the files you want to copy from the remote system using a single glob pattern, then you can do this in a single scp command. This usage will only support a single destination folder on the local system for all files though. For example:
scp 'RemoteHost:/tmp/[abc]*/*.tar.gz' .

copies all of the files from the remote system which are names (something).tar.gz and which are located in subdirectories of /tmp whose names begin with a, b, or c. The single quotes are to protect the glob pattern from being interpreted from the shell on the local system.
If you cannot express all the files you want to copy as a single glob pattern and you still want the copy to be done using a single command (and a single SSH connection which will ask for your passsword only once) then you can either:

Use a different command than scp, like sftp or rsync, or
Open an SSH master connection to the remote host and run several scp commands as slaves of that master. The slaves will piggyback on the master connection which stays open throughout and won't ask you for a password. Read up on master & slave connections in the ssh manpage.
Answer: create a key pair, copy the public key to the server side.
ssh-keygen -t rsa

Append content inside the file ~/.ssh/identity.pub to file ~/.ssh/authorized_keys2 of server side user. You need not to type password anymore.
However, be careful! anybody who can access your "local account" can "ssh" to the server without password as well.
Comment: Oh, I thought the OP has some reason for being unable to use SSH keys. I misread. There doesn't seem to be any such requirement. Keys are, as you suggest, what I would normally use!
Comment: I knew this method, but i didn't want to put my hands on common directories like .ssh.
Answer: Alternatively, if you cannot use public key authentication, you may add the following configuration to SSH (either to ~/.ssh/config or as the appropriate command-line arguments):
ControlMaster auto
ControlPath /tmp/ssh_mux_%h_%p_%r
ControlPersist 2m

With this config, the SSH connection will be kept open for 2 minutes so you'll only need to type the password the first time.
This post has more details on this feature.


Is it okay/safe to mention/display information about the example format of user Login ID on the website?

Question: Is it okay/safe to mention/display information about the example format of user Login ID (aka Username) on the website?
Will this become a source of information for hackers to figure out the Login ID format?
Comment: Safe against what? What specific security concerns do you have? Please edit your question to include more information.
Answer: The short answer is that yes, an attacker would benefit from knowing the format of your usernames if they are going to attempt to launch an online guessing attack of likely username/password pairs against your system.  But this may not be a risk you should worry about.
The long answer is that you need to judge the value this practice provides you against the risk.  Our industry has generally embraced the idea that while usernames should not be easily disclosed to attackers, knowledge of a username alone should not grant access to user accounts.  Accounts should primarily be protected by a password or other authentication factors. There is a faction that believes the username should be more secretive to further decrease risk, but this idea hasn't been embraced by most system owners.
So by disclosing the format of usernames you are merely giving an attacker seeking that knowledge a slight advantage.  They still have to take the info that usernames are in a certain format and apply it to their guessing process.  Hopefully the username format itself isn't risky (e.g. not 'user1', 'user2', etc.) and you aren't trying to obscure that fact to avoid attacks against it.
Presumably you ask because you or someone else sees a benefit in publishing this information on your site to begin with.  Whether it's to help users through the account creation process or something similar, you may need to communicate your standard to them.  If not, then you can easily decide to remove this info from your site.  But if so, you should weigh this benefit against the low risk incurred from the practice and decide if it is worth it.
I see similarities between this and disclosing your password policy.  Yes, knowing passwords must contain uppercase, lowercase, and numeric characters can help attackers.  But it doesn't tend to help them very much.
Personally, I wouldn't worry about disclosing the username format.


What is this house plant with bright pink flowers and dark green leaves? How should I take care of it?

Question: I bought this plant from a home repair store 3 days back, but I don't know what it is. Now it seems that its decaying. What is this plant and what do I need to do to take care of it?
Comment: If you bought it just 3 days ago, do you still have the receipt/tag on the plant? Are you watering the plant/have you watered since you bought it?
Comment: You could also go back to the store to see if they still have them on sale. it should, probably have had a care tag.
Comment: This is a begonia. Is it only the flowers that are browning, or are the leaves browning (first at the edges) and shriveling, too? Is there any rot on the stem or leaves? It's hard to judge from the photo. How often are you watering the plant, how much water do you give it and is it standing in direct sunlight? Answers to these questions would be a great help in diagnosing the problem. Thanks.
Comment: @yoda: thanks. yes, I've been watering since I bought it. Does matter where I put that plat? I've been moving it around since I found it loosing its shape. And I do not have the tag/receipt :(
Answer: It is a begonia, probably a Rieger begonia. Check this begonia care site for details.
Begonias commonly get gray mold (Botrytis). Since this fungus moves from dead tissue into living, pick off any dead flowers or leaves as soon as you can. Try to keep the foliage and flowers dry.
Answer: Does your pot have drain holes? The begonia is very sensitive to wet feet. Allow the plant to dry out somewhat in-between watering. In most areas this plant is an annual.  I'm not sure where you live, if you can and if you have a covered patio, this plant would thrive outside.
Cut flowers off as soon as you can.  Annuals have one purpose in life and that is to produce seed.  Thus lots of flower display.  But once it's made seed the whole plant just starts relaxing.  It has done its job. To help the plant remain vigorous and produce flowers all season, cut the flowers off as soon as possible.
All that energy going into making flowers (seed) is diverted back to the plant.  If it hasn't already been fertilized by a greenhouse or the nursery (something else you should ask if you go back to the store is when was it fertilized last, what kind of fertilizer) fertilize it with a low-nitrogen fertilizer.  If you use a higher percentage of Nitrogen in relation to phosphorus and potassium you will get lots of vegetative growth and little or no blooms.
Put it in indirect sunlight when indoors, but still a brightly lit area. For the summer it would love to go outside on the covered porch but protect from wind.  If you have winter, bring it back inside by gradually increasing the time inside.  It'll need to get used to the dry air again.  And keep an eye out for spider mite or other insects. Gray mold, too as Eric brought up!


Database to CSV(Done) trying to Save file to folder on server

Question: I've have been successful in exporting my database to csv as a downloadable file. However what I now need to do is instead of creating a straight .csv file that's downloaded I need it to just save to a folder called "csv" on the server. Here is my code for the current export. I need help in the saving it to server. I'm not saving the data correctly.
<code>// output headers so that the file is downloaded rather than displayed
header('Content-Type: text/csv; charset=utf-8');
header('Content-Disposition: attachment; filename=data.csv');

// create a file pointer connected to the output stream
$output = fopen('php://output', 'w');

// output the column headings
fputcsv($output, array('tax_class_id','_product_websites'));

// fetch the data
mysql_connect('localhost:3036', 'x', 'x');
mysql_select_db('lato');
$rows = mysql_query('SELECT taxclass,productwebsite FROM product');

// loop over the rows, outputting them
while ($row = mysql_fetch_assoc($rows))
fputcsv($output, $row);

$filename = "data.csv"; // Trying to save file in server
file_put_contents("download/" . $filename, "$header\n$rows");
</code>
Answer: Why do write in streams, read it and than try to save the content?
I would do it in a smaller way:
<code>//Open a file in write-mode (he creates it, if it not exists)
$fp = fopen('./my/path/on/server/data.csv', 'w');

// output the column headings
fputcsv($fp, array('tax_class_id','_product_websites'));

// fetch the data
mysql_connect('localhost:3036', 'x', 'x');
mysql_select_db('lato');
$rows = mysql_query('SELECT taxclass,productwebsite FROM product');

// loop over the rows, outputting them
while ($row = mysql_fetch_assoc($rows))
fputcsv($fp, $row);
//close the handler
fclose($fp);
</code>


difference between diplomatic and forthright?

Question: Can a person(a politician specifically) be described as diplomatic as well as forthright at the same time? Or do the two attributes contrast with each other?
Answer: In general the two attributes contrast one another. In other words a diplomatic statement would be a "flowery" forthright statement. But of course for a particular statement it depends. A particular statement could be diplomatic as well as forthright.
Answer: They have very different meanings. In some ways they are opposites.
Diplomatic (when it doesn't mean "of diplomacy") means carefully phrased to avoid offence.

When we inform the parents that their children will move to a low maths group, we will need to be very diplomatic.  Make sure we use langauge like "this group will ensure your child receives the proper support".

"Forthright" means direct and strongly expressed.

He made a forthright defence of the Brexit policy.

Being diplomatic suggests being subtle and indirect. The opposite of "forthright".
If something was forthright and diplomatic, it would suggest "strongly expressed" but "carefully phrased so as not to offend". That is possible, but unusual.

The ambassador's statement was worded with typical diplomatic charm. Yet he was forthright in expressing his government's rejection of the UN resolution


the number of the positive integer numbers $k$ that makes for the two quadratic equations $ \pm x^2 \pm px \pm k$ rational roots.

Question: If we assume that  $p=2^{24036583}-1$ is the greatest prime number until now .How to find the number of the positive integer numbers $k$ that makes for the two quadratic equations $ \pm x^2   \pm px \pm k$ rational roots.
Answer: (Assuming that the question is rather about $x^2-px+k$ and $x^2-px-k$ simultaneously having rational [=integer] solutions)
Let $a,p-a$ be the roots of $x^2-px-k$ and $b,p-b$ the roots of $x^2-px+k$. Then $a(p-a)=k=-b(p-b)$, i.e.,
$$\tag1 a^2+b^2=p(a+b).$$
Unless $p\mid a,b$, this means that $x^2+1\equiv 0\pmod p$ has a solution. But for $p\equiv -1\pmod 4$, no such solution exists. We conclude $p\mid a$, $p\mid b$, say $a=pu,b=pv$. Then $(1)$ becomes
$$\tag2u^2+v^2=u+v.$$
For $x\in\Bbb Z$ we have $x^2\ge x$ with equality iff $x\in\{0,1\}$.
Thus we concolude from $(2)$ that $a,b\in\{0,p\}$. At any rate, this implies that the  only possible value for $k$ si $k=a(p-a)=0$ (which is not positive).
Answer: $x^2+px+k=0$ will have rational (indeed, integer) roots if and only if $p^2-4k$ is a square, $p^2-4k=q^2$. Let's write this as $p^2-q^2=4k$. This works for every odd number $q$ less than $p$, so there are $(p-1)/2$ such numbers $k$.
I don't know what you mean by "the two quadratic equations." The equation $x^2+px+k=0$ is the same as $-x^2-px-q=0$. If you want to choose the signs independently, then you get four (pairs of inequivalent) equations, not two, but they can be handled by the same methods as the one I did.
Comment: @GerryMyerson, I think the OP is really asking, in effect, for the number of positive integers $k$ for which *both* $x^2-px+k$ *and* $x^2-px-k$ have integer roots.  The more recent question that you marked as a duplicate of this one states the question somewhat more clearly.  If I'm correct, the sought-for answer is $0$ (i.e., there are no postive $k$ for which both quadratics have integer roots), ultimately because $p\equiv-1$ mod $4$.
Comment: Can you explain how did you know that there are $(p-1)/2$,please ?
Comment: @Mohammed, that's how many odd numbers there are less than $p$.
Comment: How about $k$ ?
Comment: @Mohammed, $p^2-q^2=4k$ gives you a value of $k$ for every odd number $q$ less than $p$. There are $(p-1)/2$ odd numbers less than $p$.


Role of stakeholders in the control of execution stage

Question: What role stakeholders play  in the control of a project in execution stage of the project cycle.
Comment: Hi Usman, I went ahead and put this on hold, since it's not really clear what the problem is. Please feel free to [edit] with more details and we can look at reopening.
Answer: This question is far too broad to answer in any meaningful way.  Industry, location, and type of project would yield different answers.  So, too, would the breakdown and segmentation of the stakeholders.  Based on how their segmented, each group would play a different role.
So the only feasible answer one could give now is, yes, they play several of many roles during the execution stage of a project.
EDIT:  In general, stakeholders can participate in the design of the widget being built.  They certainly participate in setting requirements or discussing changes or compromises to the requirements.  Some groups would provide subject matter expertise that would be needed by the builders of the widget.  All groups would require staying informed so they would play a communication receiving role.  If user acceptance testing is required, then a segmented group would play in that QC activity.  On some project, a stakeholder group would be at what I call level zero, meaning they are side by side with the project team on a daily basis, even part of the project team.  I would consider them a stakeholder even though they are part of the team because I am typically in a seller/buyer project relationship where I am the seller.
Some stakeholder segments would be deployed as champions of change, meaning they would advocate to other segments about the change in order to reduce the likelihood of change resistance, a huge killer of project acceptance.  Other stakeholder segments may have projects going on where your project and their project have one or many touch points, so coordination between the groups to resolve dependencies would need to occur.
...just some general ideas.
Comment: Thank you Mr. David but could you please explain what roles they play in general?
Comment: My pleasure.  Good luck on your project.
Answer: Stake holders play following key roles and many other
1.Stating requirements clearly
2.Clarification of requirements in all the phases of project
3.Validating POC or prototypes if there are any
4.Change controls if there are any and budget approvals
5.User Acceptance testing
6.Sign off for the developed piece of software
7.Production issues if there are any after product went live and follow up to rsolve
8.Feedback how it worked after the product is stable or mean time too


Translations of「しっとり」for describing food

Question: On the television the other day, there were some people talking about how the word 「しっとり」 applied to food can't be translated exactly into other languages. It seems that 「しっとり」 could be translated into "moist" when talking about a cake for example, we say a cake is "moist". Are the TV people right that 「しっとり」 is untranslatable?
Answer: According to the search on cookpad, at the moment「しっとり」has the hit of 51,369 menus.
For instance, "Gateau chocolate" :「しっとり濃厚★ガトーショコラ」, "Chocolate cookie" :「 バレンタイン♡しっとり生チョコクッキー 」, "Chicken Ham" :「しっとり鳥胸ハム」, "Banana Cake" :「しっとりバナナケーキ」, "Tofu Donuts" :「しっとり簡単豆腐ドーナツ」, "Rice flour crepe" :「もちもち&しっとり米粉クレープ」, etc.
If the texture of food is "moist", you can safely use it to any kinds of food.
I am not sure why they said that the word is untranslatable. Unless the food is  the thing which you can not eat outside Japan or hard to find, it should be translatable.
Answer: In my opinion, calling しっとり untranslatable in the context of food is a rather flippant exaggeration. The English word 'moist' is a perfectly good translation in many cases. It is common to describe a cake as 'moist', invoking the idea of it being soft or luscious. しっとり has a similar sense of a high moisture content giving the food item a pleasing feeling in the mouth.
To be honest, I would say there are very few (if any) untranslatable words in any language. If you insist on having a one-word to one-word correspondence, then maybe you could argue there are words which have no counterpart in other languages. But who is insisting on that? Language is such a flexible thing that concepts can generally be rendered easily into other languages.
Comment: +1 more too for the comment about translatability -- any claim that XYZ word is untranslatable is usually made by someone who doesn't understand language very well, or who is using a very narrow definition for "translate".
Comment: But there is also しっとりチョコ, I'm not sure how to translate that into English.
Comment: @Ben What's wrong with 'moist chocolate'? Sounds fine to me. Instead, you could use something like 'velvety', 'oozy', 'soft', 'luscious', etc, depending on the specific product.
Comment: This: https://www.google.com/search?q=%E3%81%97%E3%81%A3%E3%81%A8%E3%82%8A%E3%83%81%E3%83%A7%E3%82%B3


If banks create money by lending it out, how are they taking any risk?

Question: When I lend money to someone and they don't pay it back, I lose my money.  I can only lend out money that I own (or borrow).  I understand that when a bank lends money, that money is created the moment they lend it out.  If that is so, how does the bank take any risk in lending it out, considering they didn't own it in the first place?
Comment: That's not how banking works
Comment: @EnergyNumbers I suppose there must be more to it, as Henry has excellently explained; there must be some liability to offset the loan even after B spends the money, and now I understand there indeed is.  I suppose there's a similar but slightly different story if A decides to take out the money in cash.
Answer: Try the magic of double entry bookkeeping to see that creating money does not immediately make a bank more valuable. Initially the loan is matched by a deposit, and this is creation of what is effectively money:

Bank A has an asset: a newly created loan to customer B
Bank A has a liability:  B's newly created deposit with Bank A, which B can spend

B then spends the amount with C, who banks with D, and the deposit is transferred on B's instruction to C's account. Now

Bank A has an asset: a loan to B
Bank A has a liability:  Bank A owes money to Bank D
Bank D has an asset: Bank A owes money to Bank D
Bank D has a liability:  C's deposit with Bank D

Bank A now needs to settle with Bank D: this can happen through clearing if other customers are moving money to Bank A, but it is easier to imagine that Bank A tells the central bank to give some of A's reserve deposits there to Bank D. So Bank A has an asset, namely a loan to B, but this has been offset by an increase in liabilities or a reduction in other assets.  Similarly Bank D has an increase in assets, offset by a liability to C. The created money is now C's deposit with D, which C can spend.
Suppose the loan to B then goes bad and is never likely to be repaid.  From Bank A's point of view, the value of that asset falls to zero, while the associated increase in liabilities or reduction in other assets remains, and Bank A has to take this as a loss (negative profit), with the side effect of destroying that amount of money.  Bank A is now worth less (net) than when it started the loan to B.  This is the risk for Bank A from creating new financial assets and liabilities (i.e. new money).
If instead nothing goes wrong, B makes some money selling to E, pays A some interest, and possibly repays the loan (in which case the total amount of money in the economy reduces again).  In this case Bank A is more valuable (net), not directly from the creation of the loan or from its repayment or from use or return of the corresponding deposit, but from receiving the interest on the loan.
When Bank A makes too many loans, it may run out of reserve deposits at the central bank after its customers spend the new money.  The solution to this is to take deposits from other customers, as Bank D does in the example above.  But these deposits may need interest to attract them, at a cost to the bank.
So Bank A's overall profit on loans and deposits is the difference between the interest it receives from the loans it makes and the interest it pays on deposits it takes, allowing for the cost of bad loans; it may also make money from fees and have to pay for staff and other expenses.  The creation and destruction of money by making the initial loan and it being repaid is not Bank A's particular concern, though it is for the macroeconomic policy of the central bank and the government; if the economy grows and more loans are made, then the money supply also rises and the hope it that will happen neither too quickly nor too slowly.  Ultimately Bank A cares about the risk from bad loans, the interest rate margin and its administrative costs.


Is it advisable to use SharePoint SQL DB instance to another application?

Question: We are having SP 2013 on-prem environment. We are planning to have few applications which are using SQL as a backend. Is it advisable to use the same SP SQL Server DB instance to create DB for other applications? Will it cause any performance issues?
Answer: As per the Best Practices from MSFT:
To ensure optimal performance for farm operations, we recommend that you install SQL Server 2008 R2 with SP1 or SQL Server 2012 on a dedicated server that does not run other farm roles and does not host databases for other applications.
https://technet.microsoft.com/en-us/library/hh292622.aspx?f=255&MSPPError=-2147217396
But all of this depends on your existing usage of your SharePoint databases as well as the future usage of these non-SharePoint databases.
Similar issue post for your reference:
Non sharepoint databases in same sharepoint sql server instance and issues?
Answer: It isn't best practice as per the documentation from MS but it can be done and as far as I could tell, it isn't unsupported. There are a few things to note that might help you decide if it is worth it.
SharePoint needs certain SQL Server settings (such as Max Degrees of Parallelism) and if they conflict with the other applications then you can't do it.
You also need to make sure you have sufficient resources.
We've been testing out putting our 2013 SP databases on a shared cluster. In this situation, our DBA doesn't want our SP accounts having the high permissions it normally gets by default when setting up the farm. This means we need to request temporary Sys admin privileges before installing patches or even setting up a new service application.
Another thing we ran into related to how SP monitors itself. There are timer jobs that help maintain the DBs but these had to be disabled so our DBA could handle that instead.
The upside we are hoping for is some better redundancy as well as saving on some SQL licensing.
I would probably recommend a separate server if possible. An in-between option might be a separate instance on a shared SQL server.
Hope that helps.


Best practices attaching Event Handlers in large systems

Question: I am starting to build bigger things on javascript, and I am wondering about coding style and performance. Now I am trying to figure out what is the best way to attach event handlers in this case: I have Item() constructed objects for which there are divs in the document, one for each, with correspondent ids. I want to have access from each div's event handlers to its item object representation. The items will be manipulated heavily, created, deleted, etc., and so with be their divs. You can look at the code here (any feedback in general coding style would be really appreciated).
I could do this:

By directly having all my event handlers in the Item constructor so
they have access to the object being constructed (var that = this).
I don't think it's such a good idea to have all this code here, but
I might be overthinking this.
Having the handlers stored in a separate handlerCollection object
and accessing the item object with an auxiliary function
getItemFromDOM(div) which reads the divs id and searches for the
corresponding item in an array where all items in the document are
stored.
Having somewhere separarte in the code a function
registerItemEvents(item) called in the Item constructor and used to
generate a closure over the item for the handlers, which would live
inside it.

I am guessing the most elegant is the third one, but I might be overrating tidiness turning down the first one. Also is there any chance, using the third, that if I start deleting items the closure over them will prevent memory from being released, ultimately slowing down the whole thing? Can I fix this by simply releasing the event handlers before deleting an item?
Answer: I think you should consider using delegates, if possible.  From what I know, adding and removing event handlers adds unnecessary overhead.
Regardless of how you choose to implement (jquery, yui, other, custom), this post from Yahoo! has some good info
http://www.yuiblog.com/blog/2009/11/13/event-delegation-3/


Starting a new campaign with basic weapons

Question: I completed Resident Evil 6 on normal mode, all 4 chapters (pretty poorly, I might add) and am now going back through the game to achieve S rank on all chapters. When I went to start a new game it asked if I wanted to start a new game using previous weapons. Having pressed 'yes', I'm now on Leon's chapter 1 and only have the Wingshooter and knife.. I've also bought the infinite shotgun skill but have no shotgun to use it :( any ideas?
Answer: Actually, that is not what it says. If you read it carefully then you will find out that you actually need to press "No" to start with all your previous weapons (I cannot remember the exact text at the moment).
I admit I found the text to be a bit misleading too, so I can see how it could easily be overlooked.
It's also worth mentioning that replaying the campaign with the "other" character, does not allow use of weapons collected from the first playthrough, ie. you have to collect each characters weapons individually.
Comment: thanks for answering. I have started a new campaign with chris and he seems to have all his old weapons. So, does this mean having already started with leon previously that he will only have his original weapons? Also, to use infinite shotgun do you have to achieve an S grade in a particular chapter in order to use it?
Comment: @nardo6890: I am not sure you can restart with leon and re-use weapons until you complete again. But as you have infinite shotgun ammo you should be ok as you get this weapon earlier in Leon's campaign. To enable it you need to purchase the skill, asign it to a skill slot, and make sure you have that skill slot selected in your game. you also MUST enable "infinite ammo" on the "network settings" screen when you load a save game
Comment: aw man, thats a gutter! thank you for your help, i'll get right on it ;)


harvard-style tag

Question: There are currently 4 questions with the tag harvard-referencing. Our other major bibliography style tags (apa-style, chicago-style, mla-style) all have the form <name>-style.  It would seem to make sense to change harvard-referencing to harvard-style.
Comment: Sounds fine to me
Comment: @Joseph Not being a retagging expert, do I just add the `harvard-style` tag to the existing questions and then a mod will delete the old tag?
Comment: That does tend to be the way, but I think we can do this from the mod panel directly.
Comment: @Joseph That would be great.  I'm not opposed to helping, but if you can do it with a click or two that would make more sense.
Answer: I now merged harvard-referencing to harvard-style but without creating a, IMHO unnecessary, synonym. I don't think the first form will come back. The auto-completion will show the new form when someone enters harvard.
Such tag renaming can be done by the moderator using one of the moderation tools and does not create new revisions for the tagged questions, which is the preferable way.

harvard-style tag