The magnitude of $|-\sin t + \cos t|$ is.
Prove: $|a\sin x+b \cos x|\leq \sqrt{a^2+b^2}$
Many To Many Relationships I am working on a conceptual model for a database. I have the following entities:  Student (id, name, date of birth, department)  Subject (id, name)   This is a many-to-many relationship, one student may be enrolled in many courses, each course has many students. From what I understand, as I should break this many-to-many relationship by introducing a new into a entity, say 'registered' which allows a one-to-one relationship or one-to-many relationship?  Is this correct design decision? It would help everyone if we provided a mechanism that people could use to flag their post if English is language they are not comfortably fluent in. A lot of non-English speakers have to go to English forums for tech support and find it embarrassing to ask questions in broken Englis.   Worse, people reading a post often cannot tell if a badly-worded post results from a poor English or just laziness and sloppiness on the part of a native English speaker. Personally, I will take the time to puzzle out a poorly-worded post if I know the author is struggling with English, but if I think they're just lazy and careless I won't bother.   A simple check box could flag the author as a non-English speaker and allow everyone to adjust accordingly.   Edit01:  Some answers below have accused me of being 'elitist'. I'm not sure where they got the idea that I was sneering down at anyone because their abilities at either English or coding. That was not my intent.   When I said I would not bother with post that appeared to incoherent owing to laziness or carelessness, I am not being elitist. I'm giving priority in my time budget to people who are respectful enough of my time to spend the their time writing as clear a question as their grasp of English and coding allows. Why should I spend time answering a careless and slapdash post when I could spend that time helping posters who took more of their time writing their questions?   I am perfectly willing to burn a lot of time helping someone struggling to write code in their non-native language. I have a hard enough time coding myself in English that I can't even imagine what a dog's breakfast I would make of it if I had to code in, say, Korean.    I have seen many instances in which questions have been downvoted when a quick glance at the poster's profile gives a strong hint they are not native English speakers. I would like a system that would spare those posters from being downvoted or ignored by using an interface element that clearly communicates in a standardized way whether the poster natively speaks English or not.   I don't see what's elitist about that.    Edit02:   (original unedited) I'm talking about:     Hy, I have a childe view that put it   self over other brother views. The   problem is that when i touch it the   brother also reacts at the touch   event. How can I disable touch   propagation on the underlieng brother   views?   I'm not even exactly sure what he is describing. I can guess that he has a sibling view that extends over its siblings but I'm not sure. If I edited it, I might not end up asking the same question the poster intended.   To answer the question, I will either have to (1) first ask for more detail and then keep checking back until I get it or (2) try to answer what I think the question is and possibly mangle it then have to come back and edit my answer all the while risking a downvote because I answered the wrong question.   I will take the time and the risk of down vote if the poster is not a native English speaker (or is a child) and the confusing question represents their honest best effort. If they're just being lazy, I have other people to help.   I would like a mechanism that would help me decide how to allocate my time and would protect the non-English speaking posters from unjust downvotes. Uncertainty and wave-trains My textbook and the following extract from feynman's lectures present the same idea regarding wavetrains and uncertainty in their wavelengths. Why is it that a wavetrain confined to some space has an uncertainty in its wavelength or the wave number? Is not a confined wave-train equivalent to a burst of successive pulses which can have a definite wavelength according to their origin or nature of origin.   Next, i understand that the De Broglie relation relates the uncertainty in wavelength to the uncertainty in momentum but what links the finiteness of the wave-train to the uncertainty in position? nginx open reverse proxy? How can I restore configuration files? I would like to restore LightDM to its default state, because for some reason /etc/lightdm/unity-greeter.conf is now an empty file.  Deleting /etc/lightdm/unity-greeter.conf and then running sudo apt-get install --reinstall unity-greeter doesn't create a new config file as you might expect.   How can I restore a missing config file? I would like to replace the search submit button with a Font Awesome character like the Magnifier (Search Icon) character, and put Search the site inside the search box.  How can I achieve something like this? Bash/Korn shell script edited on Windows throws error '...^M: not found' wordcount with xesearch errors The function $f(z)=z+\frac{1}{z}$ maps $\Omega$ biholomorphically on the upper plane The question states:     Let $\Omega$ be the upper half plane minus the unit disk. Show that the map $f: z \mapsto z+1/z$ maps $\Omega$ biholomorphically on the upper plane. Describe the preimages of the lines $\mathrm{Im}(z)=\text{a constant}&gt;0$.   So how exactly does one show the mapping is biholomorphic. I'm not sure where I should start, so unfortunately I haven't made any progress Nonconstant polynomials do not generate maximal ideals in $\mathbb Z[x]$ Let $f$ be a nonconstant element of ring $\mathbb Z[x]$. Prove that $\langle f \rangle$ is not maximal in $\mathbb Z[x]$.   Let us assume $\langle f \rangle$ is maximal. Then $\mathbb Z[x] / \langle f \rangle$ would be a field. Let $a \in \mathbb{Z}$. Then $a + \langle f \rangle$ is a nonzero element of this field, hence a unit. Let $g + \langle f \rangle$ be its inverse. Then $a g - 1 \in \langle f \rangle$, hence $ag(x)-1 = f(x)h(x)$ for some $h \in Z[x]$, hence $ag(0) + f(0)h(0) = 1$, thus $(a,f(0))=1$ for all $a \in \Bbb Z$, contradiction, hence the proof.  Is my argument correct? Is there any other method? A field that is conservative must have a curl of zero everywhere. However, I was wondering whether the opposite holds for functions continuous everywhere: if the curl is zero, is the field conservative? Can someone please give me an intuitive explanation and insight into this, and if it is true, why? Also, please try to not be too rigorous (only in grade 9). Word for someone who repeats words Let $X$ be a locally compact metric space which is also $\sigma$-compact. Let $C_{c}(X)$ be the continuous functions on $f$ from $X$ to $\mathbb{R}$ with compact support. Is $C_{c}(X)$ separable?  My work so far:   If $X$ is a compact metric space, then by Urysohn's Lemma and Stone-Weierstrass, the continuous functions $C(X)$ on $X$ are separable and hence the result follows as $C_{c}(X) = C(X)$.  Suppose $X = \mathbb{R}$. Write $\mathbb{R} = \bigcup_{N  = 1}^{\infty}[-N, N]$. Let $f \in C_{c}(\mathbb{R})$. Then $f$ is supported on a compact set $K \subset [-N, N]$ for some $N$. Thus $C_{c}(\mathbb{R}) =\bigcup_{N = 1}^{\infty}C([-N, N])$. Each $C([-N, N])$ has a countable dense subset $\{\psi_{N, n}\}_{n = 1}^{\infty}$ and so $\bigcup_{N, n = 1}^{\infty}\{\psi_{N, n}\}$ is a countable dense subset of $C_{c}(\mathbb{R})$.  In the general case, $X = \bigcup_{i = 1}^{\infty}X_{i}$ where each $X_{i}$ is compact and $X_{1} \subset X_{2} \subset \cdots$. Let $f \in C_{c}(X)$. Then $f$ is supported on a compact set $K = \bigcup_{i = 1}^{\infty}K \cap X_{i}$. Is it still true that $C_{c}(X) = \bigcup_{i = 1}^{\infty}C(X_{i})$? Why do regex constructors need to be double escaped? jQuery: Uncheck other checkbox on one checked Automorphism group of the quaternion group How to install Anaconda on Ubuntu? How to install  on Ubuntu?   Is there a way to use apt-get install?   I only have command line access to my server. How do I install Anaconda on Ubuntu 14.04 from the command line? Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$ The following identity is true for any given $x \in [-1,1]$: $$\arcsin(x) + \arccos(x) = \frac{\pi}{2}$$  But I don't know how to explain it. I understand that the derivative of the equation is a truth clause, but why would the following be true, intuitively?  $$\int^{x}_{C1}\frac{1\cdot dx}{\sqrt{1-x^{2}}} + \int^{x}_{C2}\frac{-1 \cdot dx}{\sqrt{1-x^{2}}} =\\  \arcsin(x) - \arcsin(C1) + \arccos(x) - \arccos(C2) = 0 \\  \text{while } \arcsin(C1) + \arccos(C2) = \frac{\pi}{2}$$  I can't find the right words to explain why this is true?    Edit #1 (25 Jan, 20:10 UTC): The following is a truth clause: $$ \begin{array}{ll} \frac{d}{dx}(\arcsin(x) + \arccos(x)) = \frac{d}{dx}\frac{\pi}{2} \\ \\ \frac{1}{\sqrt{1-x^{2}}} + \frac{-1}{\sqrt{1-x^{2}}} = 0 \end{array} $$  By integrating the last equation, using the limits $k$ (a constant) and $x$ (variable), I get the following:  $$ \begin{array}{ll} \int^x_k\frac{1}{\sqrt{1-x^{2}}}dx + \int^x_k\frac{-1}{\sqrt{1-x^{2}}}dx = \int^x_k0 \\ \\ \arcsin(x) - \arcsin(k) + \arccos(x) - \arccos(k) = m  \text{ (m is a constant)}\\ \\ \arcsin(x) + \arccos(x) = m + \arcsin(k) + \arccos(k) \\ \\ \text{Assuming that } A = m + \arcsin(k) + \arccos(k) = \frac{\pi}{2} \text{ ,for } x \in [-1,1] \end{array} $$ Using Calculus, why is that true for every $x \in [-1,1]$?    Edit #2:  A big mistake of mine was to think that $\int^x_k0 = m \text{ (m is const.)}$, but that isn't true for definite integrals.  Thus the equations from "Edit #1" should be as follows: $$ \begin{array}{ll} \int^x_k\frac{1}{\sqrt{1-x^{2}}}dx + \int^x_k\frac{-1}{\sqrt{1-x^{2}}}dx = \int^x_k0 \\ \\ \arcsin(x) - \arcsin(k) + \arccos(x) - \arccos(k) = 0\\ \\ \arcsin(x) + \arccos(x) = \arcsin(k) + \arccos(k) \\ \\ A = \arcsin(k) + \arccos(k) = \frac{\pi}{2} \text{ ,for } x \in [-1,1] \end{array} $$ Allow people to answer protected questions if they have earned at least some amount of combined SE rep The purpose of protected questions is, according to the message presented on such a question, "to prevent 'thanks!', 'me too!', or spam answers by new users." The more verbose  agrees on the purpose. And the  says this is why protected questions exist:     Some questions are protected because they are expected to attract   either spam or users -- often new users -- who may mistake the site as   a traditional forum, posting "noisy" answers such as "Thank you" or   "This worked for me" or "I'm also having this problem".   Considering this, shouldn't a user be able to answer protected questions if they have gained at least 10 rep on any stack exchange site? If they've already demonstrated one site that they are familiar enough with the SE Q/A format to not be a drive-by useless answerer, shouldn't that confidence in them carry over to other SE sites?  Or maybe at some level of combined SE rep this restriction should be waived. I find it silly that at 15k+ combined network rep, I'm unable to answer a protected question on a SE site I've just started using. How can an object's instantaneous speed be zero and it's instantaneous acceleration be nonzero? I'm studying for my upcoming physics course and ran across this concept - I'd love an explanation.