A line is drawn through the point $A(1,2)$ to cut the line $2y=3x-5$ in $P$ and the line $x+y=12$ in $Q$. If $AQ=2AP$, find the coordinates of $P$ and $Q$.  From: Mathematics, The Core Course for A-level, Bostock and Chandler. Chapter 4 Q15.  The answer is $(4,3.5)(7,5)$or$(0.4,-1.9)(2.2,9.8)$  I have no idea where to begin with this question.  I've tried letting $P=(x,12-x)$ and $Q=(x,\frac{3}{2}x-\frac{5}{2})$ and then used the distance formula to try and equate $|AQ|=2|AP|$ but didn't get anywhere.  How should I approach this problem?
A line is drawn through the point $A(1,2)$ to cut the line $2y=3x-5$ in $P$ and the line $x+y=12$ in $Q$. If $AQ=2AP$, find the coordinates of $P$ and $Q$.  I found the lengths of the lines $AQ$ and $AP$ in terms of $x$ and $y$ and used $AQ=2AP$ to get this equation $11x^2-40x-16=0$  Working: $AP=\sqrt{(x-1)^2+(\frac{3}{2}x-\frac{9}{2})^2}=\sqrt{\frac{1}{4}(13x^2-62x+85)}$  $AQ=\sqrt{(x-1)^2+(-x+10)^2}=\sqrt{2x^2-22x+101}$  $AQ=2AP\Rightarrow \sqrt{2x^2-22x+101}=2\sqrt{\frac{1}{4}(13x^2-62x+85)}$  $\Rightarrow 2x^2-22x+101=13x^2-62x+85\Rightarrow 11x^2-40x-16=0$  Using the quadratic formula I found $x=4,x=-\frac{4}{11}$  I then substituted these values into the two equations in the question to get the coordinates $(4,\frac{7}{2}),(-\frac{4}{11},-\frac{15}{11})$ or $(4,8),(-\frac{4}{11},\frac{136}{11})$  But the answers in the book are $(4,\frac{7}{2}),(7,5)$ or $(\frac{2}{5},-\frac{19}{10}),(\frac{11}{5},\frac{49}{5})$  I don't know where I went wrong in my method.  Edit: Found some errors in my method and calculation.  Using $x=4$ in $2y=3x-5$ gives $P(4,\frac{7}{2})$  Line with points $AP$ has equation $y-2=\frac{\frac{7}{2}-2}{3}(x-1)\Rightarrow y=\frac{1}{2}x+\frac{3}{2}$  $Q$ lies on the same line. Therefore, the intersection between $y=\frac{1}{2}x+\frac{3}{2}$ and $y+x=12$ is the point $Q$. Working: $\frac{1}{2}x+\frac{3}{2}=-x+12\Rightarrow x=7$ Therefore, $y=-7+12=5$. Point Q has coordinates $(7,5)$.  But $x=-\frac{4}{11}$ is not the other $x$ coordinate of $P$ according to the answers in the book. I have no idea where I went wrong at this point.
Echo from MySQL database with spaces and line breaks? Should we sign-then-encrypt, or encrypt-then-sign? Frequently, we want to send messages that are (a) encrypted, so passive attackers can't discover the plaintext of the message, and (b) signed with a private-key digital signature, so active attackers can't make Alice think that a message came from Bob when it didn't. Is it better to (a) generate the digital signature from the (hashed) plaintext, and then encrypt a file containing both the plaintext message and the digital signature? Or is it better to (b) encrypt the message first, and then generate a digital signature from the (hashed) encrypted file? Or (c) combine encryption and public-key digital signatures in some other way? A closely related earlier question (  ) seems to focus on symmetric-key MAC authentication. As , Do the same issues with (symmetric-key) MAC-then-encrypt apply to (public-key) sign-then-encrypt? How does the Ranger's Extra Attack feature work with Two-Weapon Fighting? I think I know the answer to this one but I'm double checking.  At the 5th level the Ranger gets Extra Attack, which lets them attack a second time during the Attack action. If they are fighting with 2 weapons, the second weapon normally uses their bonus action.  Does this mean when using the Extra Attack and using two weapons, a ranger can attack twice during the Attack action, and still make 1 extra attack as a bonus action, for a total of 3 attacks? Recently I downloaded Pokemon GO but when I try to sign up with google account,the option isn't there. Help please. I use a LG G2 mini Identify $E(X\mid X+Y)$ for $X$ and $Y$ independent and Poisson I am currently trying to understand how does an instance of the EM algorithm is derived when there are "hidden variables" in this old paper: .  I don't really understand how the expectation step is derived, and what bothers me the most, is that I cannot prove the following statement (right before 2.11):  $$ \mathbb{E}[X|X+Y] = \frac{(X+Y)\lambda_X}{\lambda_X+\lambda_Y} $$  where $X \sim Poisson\{\lambda_X\}$ and $Y \sim Poisson\{\lambda_Y\}$  To begin, I wrote the discrete expectation as follows: $$ \mathbb{E}[X|X+Y] = \sum_{k=1}^{X+Y} k P(X=k|X+Y)\\ = \sum_{k=0}^{X+Y} k \frac{P(X+Y|X=k)P(x)}{P(X+Y)} $$  Where I can replace $$ P(X+Y=k) = \frac{(\lambda_X+\lambda_Y)^ke^{-(\lambda_X+\lambda_Y)}}{k!}$$ $$ P(k=x) = \frac{\lambda_X^xe^{-\lambda_X}}{x!}$$ $$ P(X+Y=k|x) = P(Y=x-k) = \frac{\lambda_Y^{x-k}e^{-\lambda_Y}}{(x-k)!}$$  but I dont really know how to mix that together, because I don't really see what does the event $X+Y$ stands for, and how does it rely on the x in the integral ?  Thank you very much for you help Limit of a sequence defined by recursive relation : $ a_n = \sqrt{a_{n-1}a_{n-2}}$ Does borrowing from my 401(k) make sense in my specific circumstance? I know the standard answer and understand that I will be losing out on that compound interest, but I am young and only need to borrow a few thousand. here are some more details:   Need to borrow $6500 to pay off a car and two student loans ($300/month in payments combined) I am 22 and work full time, making about $70K annually I contribute 8% of my income with a 4% company match I increase that by 2% every year The loan would be over 48 months ($140/month) I live very frugally and have almost no discretionary spending that can be cut (don't eat out all the time, no gym membership/ Netflix/ Apple music etc.) I'm cash-poor: 90% of my salary is going to mortgage, groceries, students loans, car payment, newborn baby medical expenses, and commute. These are not my only debts, just the ones I am seeking to pay off in the short term. I work two jobs, one full and one part time. I started a business, and my wife has two jobs. Working more is not a good answer to the original question.   It seems to me that I would be reducing middle to high interest debt while also saving myself $150 per month.   Loans:     Student loan 1: $1200 @ 5%   Student loan 2: $1100 @ 4.25%   Car Loan: $4000 @ 6% $\ker \phi = (x_1-a_1, ..., x_n-a_n)$ for a ring homomorphism $\phi: R[x_1, ..., x_n] \to R$ Let $R$ be a commutative ring, $a_1, ..., a_n$ its elements and  $\phi: R[x_1, ..., x_n] \to R$  defined by  $ \phi(f(x_1, ..., x_n)) = f(a_1, ... ,a_n)$  a ring homomorphism.  Prove: $\ker \phi = (x_1-a_1, ..., x_n-a_n)$  It is obvious that $(x_1 - a_1, ..., x_n -a_n) \subseteq \ker \phi$. I'm not sure how to prove the converse.  At this point I don't know any division algorithms for multivariable polynomials, only for the ones in $R[x]$(and the book from where I taken the exercise doesn't assume the reader to know something beyond basics of rings and ideals, and the division algorithm for $R[x]$). Though I know this could be solved for $i = 1$ by dividing by $x-a$:   Let $f(x) \in \ker \phi$, divide by $x-a: f(x) = q(x)(x-a) + r, f(a) = r = 0$, so $f(x) = q(x)(x-a) \in (x-a)$. Photon energy - momentum in matter $E = h\nu$ and $P = h\nu/c$ in vacuum. If a photon enters water, its frequency $\nu$ doesn't change. What are its energy and momentum: $h\nu$ and $h\nu/c$ ? Since part of its energy and momentum have been transferred to water, it should be less.  If water's refractive index is $n$, are the energy and momentum equal to $h\nu/n$ and $h\nu/c/n$ ? I'm completely new and lost in Ubuntu 12.04, I'm having trouble navigating to my Sublime folder which I know is in my /opt folder. Yet I've tried to no avail to navigate into it via Terminal window. The directory is colored in blue, no idea what this means...but I am able to access the folder using the GUI explorer...?  k@k-Ubuntu:~$ cd /opt k@k-Ubuntu:/opt$ ls Sublime Text 2 k@k-Ubuntu:/opt$ cd Sublime Text 2 bash: cd: Sublime: No such file or directory k@k-Ubuntu:/opt$ cd /Sublime Text 2 bash: cd: /Sublime: No such file or directory Did Voldemort refer to Snape as one of the missing Death Eaters at his rebirth? In Goblet of Fire, Voldemort says of the missing six Death Eaters:     “And here we have six missing Death Eaters… three dead in my service. One,   too cowardly to return… he will pay. One, who I believe has left me forever… he   will be killed, of course… and one, who remains my most faithful servant, and   who has already reentered my service."   The answer to  speculates about who the three dead Death Eaters were.  We know who the last one is. I’m guessing one of the remaining two is Karkaroff. Is the other Snape? If so, which one? Showing that if a subset of a complete metric space is closed, it is also complete Let $(X, d(x,y))$ be a complete metric space. Prove that if $A\subseteq X$ is a closed set, then $A$ is also complete.  My attempt: I tried to prove that every Cauchy sequence $(b_n)$ of points of $A$ converges to a point $b\in A$. However could not figure out the exit way. Maybe I am on the wrong track. Could you please help me?  edit: More from my attempt:  Suppose $A$ is a closed set and let $(x_n)$ be a sequence of points $A$ such that $\lim_n x_n\to b$.  Suppose now that $A$ has the property that $b\in A$, whenever $x_n$ converges to $b$. We know that every element of $x_n$ which is convergent in $X$ also converges to a point in $A$. Since $x_n$ is a Cauchy sequence in $A$, it must converge to a point $y\in A$. But the limit of a convergent sequence is unique.   Take $x\in A$ and select an appropriate $n$, which enables $x_n$ to converge to a point $x$ in $X$. Since the limit is unique, it must follow that $x=y$. Thus $x\in A$ and $A$ is closed.   If $A$ is a closed subset of $X$, then any Cauchy sequence of a point in $A$ is convergent in $X$ and hence converges to a point in $A$. Thus $A$ is complete. Find the image of (1,1,1) by T Does Earth's Rotation Affect Its Shape? The question I am working on is, &quot;Consider the following. (a) Find the angular speed of Earth's rotation about its axis. rad/s (b) How does this rotation affect the shape of Earth?&quot;  I am fully capable of solving part (a); however, I am not sure how to describe the effect earth's rotation on its shape. I tried to search my textbook for the answer, but could not find anything. Is there an actual effect on the shape? Prove that $Z(G)$ which is the center of $G$ is a subgroup of $G$ Strictly Upper Triangular $n\times n$ Matrices form a Nilpotent Lie Algebra over $\mathbb{C}$ for $n\geq 2$ I have been attempting this problem for a while, it is an assignment problem so I don't want somebody to just post the answer, I'm just looking for hints.  Let $\mathfrak{u}(n,\mathbb{C})$ be the Lie algebra of strictly upper triangular $n\times n$ matrices over $\mathbb{C}$.  I am trying to show that $\mathfrak{u}(n,\mathbb{C})$ is a nilpotent Lie algebra for $n\geq 2$.  This is what I have looked at so far.  I have done some explicit calculations (I shall omit the bulk of them and just summarise below) for $n=2,3,4$ and have come up with some ideas.  Let $L=\mathfrak{u}(n, \mathbb{C})$.  If $n=2$,  $L^1 = [L, L] = &lt;0&gt;_{\mathbb{C}}$  So $L$ is nilpotent.  If $n=3$,  $L^1 = [L, L] = &lt;e_{13}&gt;_{\mathbb{C}}$  $L^2 = [L, L^1] = &lt;0&gt;_{\mathbb{C}} = \left\{ 0 \right\}$  So, $L$ is nilpotent.  If $n=3$  $L^1 = [L,L] = &lt;e_{13},e_{14},e_{24}&gt;_\mathbb{C}$  $L^2 = [L,L^1] = &lt;e_{14}&gt;_{\mathbb{C}}$  $L^3 = [L, L^2] = &lt;0&gt;_{\mathbb{C}} = \left\{ 0 \right\}$  So, $L$ is nilpotent.  So my hypothesis so far is that for an arbitrary $m\in\mathbb{N}$, $L^{m-1}=\{0\}$  My instinct is to do this by induction. My base case, $n=2$ has already been shown. Now I assume that $\mathfrak{u}(n,\mathbb{C})$ is nilpotent for $n=k$ ($k\in\mathbb{N}$) and consider the case for n=k+1.  However, I have no idea how to link these two cases together. I have read in Introduction to Lie Algebras and Representation Theory - J.E. Humphreys that $L^k$ should be spanned by basis vectors $e_{ij}$ where $j-i=k+1$ - however in my calculations for $L^1$, I obtain an $e_{14}$ which does not satisfy 4-1=2.  Am I missing something completely obvious? Or have I made a mistake somewhere? (I have checked my calculations numerous times, but cannot find any errors!)  Thanks,  Andy. What I am up to is to write some exercises dealing with logical formulas for my students, like:     And the students should draw these formulas on . At the end of the lesson, I really would like to print the correct answer for them. I found a great resource on a forum thread at , which helped me a lot to make up some Venn diagrams with tikz, but have some problems with visualizing complements, like ~A.  A simple, modified version of the TeX file found on the forum linked above, can be seen below, which produces the following expression:    \documentclass{letter}   \usepackage{tikz}   \def\firstcircle{(90:1.75cm) circle (2.5cm)}   \def\secondcircle{(210:1.75cm) circle (2.5cm)}   \def\thirdcircle{(330:1.75cm) circle (2.5cm)}   \begin{document}     \begin{tikzpicture}       \begin{scope}     \clip \secondcircle;     \fill[cyan] \thirdcircle;       \end{scope}       \begin{scope}     \clip \firstcircle;     \fill[cyan] \thirdcircle;       \end{scope}       \draw \firstcircle node[text=black,above] {$A$};       \draw \secondcircle node [text=black,below left] {$B$};       \draw \thirdcircle node [text=black,below right] {$C$};     \end{tikzpicture}   \end{document}   Which looks like:    Could anyone please help me out plotting/defining some expressions dealing with complements? A nice example could be:    That should look like: (image from )    I do not insist on the red color :)  I would like to use the simplest possible solution, as I would like to mass generate the exercises with the help of . So any suggestion dealing with gnuplot, R or any other opensource packages is welcome. Thank you!    UPDATE (25/01/2011): added details based on answers.  Thank you @Leo Liu, you helped me a lot! I modified a bit the code you suggested to be able to color the area outside of the two circles also (in the H universe), but have no idea how to set a background to that polygon also. The code:  \begin{tikzpicture}[fill=gray] % left hand \scope \clip (-2,-2) rectangle (2,2)       (1,0) circle (1); \fill (0,0) circle (1); \endscope % right hand \scope \clip (-2,-2) rectangle (2,2)       (0,0) circle (1); \fill (1,0) circle (1); \endscope % outline \draw (0,0) circle (1) (0,1)  node [text=black,above] {$A$}       (1,0) circle (1) (1,1)  node [text=black,above] {$B$}       (-2,-2) rectangle (3,2) node [text=black,above] {$H$}; \end{tikzpicture}   And the image generated:    I will also look for even odd rule in the near future which does not make sense for me at the moment but looks really simple and promising! When to use ' (or quote) in Lisp? After making it through the major parts of an introductory Lisp book, I still couldn't understand what the special operator (quote) (or equivalent ') function does, yet this has been all over Lisp code that I've seen.    What does it do? Why does instanceof return false for some literals? &quot;foo&quot; instanceof String //=&gt; false &quot;foo&quot; instanceof Object //=&gt; false  true instanceof Boolean //=&gt; false true instanceof Object //=&gt; false false instanceof Boolean //=&gt; false false instanceof Object //=&gt; false  12.21 instanceof Number //=&gt; false /foo/ instanceof RegExp //=&gt; true  // the tests against Object really don't make sense  Array literals and Object literals match... [0,1] instanceof Array //=&gt; true {0:1} instanceof Object //=&gt; true  Why don't all of them? Or, why don't they all not? And, what are they an instance of, then? It's the same in FF3, IE7, Opera, and Chrome. So, at least it's consistent. The equation $X^{n} + Y^{n} = Z^{n}$ , where $ n \geq 3$ is a natural number, has no solutions at all where $X,Y,Z$ are intergers.