Acute $\triangle{ABC}$, point $D$ is on side $AC$ so that $AD=BC$, $CF$ bisects $\angle{ACB}$. Point $E$ on side $AB$ so that $DE \parallel CF$ and $AE=CD$. Prove: $\angle{ADB}=3\angle{BAC}$
See below in acute triangle $ABC$, $D$ is on $AC$ such that $AD=BC$. $CF$ is the angle bisector of $\angle ACB$. $DE \parallel CF$. $E$ is on $AB$. $AE = CD$.  Prove that $\angle ABC = 2 \angle BAC$. In addition, find the condition in which $\angle ADB = 3 \angle BAC$  I struggle to find ways to use two conditions $AD = BC$ and $AE = CD$
Can $n!$ be a perfect square when $n$ is an integer greater than $1$? Can $n!$ be a perfect square when $n$ is an integer greater than $1$?  Clearly, when $n$ is prime, $n!$ is not a perfect square because the exponent of $n$ in $n!$ is $1$.  The same goes when $n-1$ is prime, by considering the exponent of $n-1$. What is the answer for a general value of $n$?  (And is it possible, to prove without Bertrand's postulate. Because Bertrands postulate is quite a strong result.) Theorem. Let $(X,\mathcal B,\mu)$ a finite measure space, where $\mu$ is a positive measure. Let $\mathcal A\subset \mathcal B$ an algebra generating $\cal B$. Then for all $B\in\cal B$ and $\varepsilon&gt;0$, we can find $A\in\cal A$ such that $$\mu(A\Delta B)=\mu(A\cup B)-\mu(A\cap B)&lt;\varepsilon.$$  I don't think there is a proof in this site. It's a useful result for several reasons:  We know what the algebra generated by a collection of sets is, but not what the generated $\sigma$-algebra is. The map $\rho\colon \cal B\times\cal B\to \Bbb R_+$, $\rho(A,A')=\mu(A\Delta A')$ gives a pseudo-metric on $\cal B$. This makes a link between generating for an algebra and dense for the pseudo-metric. We say that a $\sigma$-algebra is separable if it's generated by a countable class of sets. In this case, the algebra generated by this class is countable. An with the mentioned result, we can show that $L^p(\mu)$ is separable for $1\leq p&lt;\infty$, which makes a link between the two notions. In ergodic theory, we have to test mixing conditions only an a generating algebra, not on all the $\sigma$-algebra. Some prestigious universities like Caltech, Stanford, MIT and Berkeley never disclose their information regarding their cut off major and cumulative GPA for M.Sc and PhD programs in engineering. I also aware of the fact that graduate schools will look at the whole package and make their decision based on sets of factors like letter of recommendation, statement of purpose as well as GPA.   However, since application fee is pretty high it is not worth it to apply to schools that you know you have a slight chance of getting into it so, I would like to know what would be the acceptable GPA (Both major and Cumulative) for applicants to be competitive for top five schools in the world? nodes near coords with logarithmic axis: Wrong values I created a horizontal bar plot to compare benchmark times. Since one of the tested components is really slow, I need a logarithmic y axis for the other values to be distinguishable. Combined with "nodes near coords" this results in the calculated values showing, instead of the original ones. Is there any way to fix this?  What it looks like:   What the values(!) should look like:   Clearly, I'm overlooking something terribly obvious. I'm just a beginner at PGFPlots, so please don't kill me. :)  MWE:  \documentclass[fontsize=12pt]{scrartcl}  \usepackage{pgfplots} \pgfplotsset{compat=1.8}  \begin{document}   \begin{tikzpicture}   \begin{axis}[      width=12.5cm,     xbar=0.1cm,     xmin=0,     y=0.6cm,     bar width=0.4cm,     xlabel={Time (ms)},     xmode=log,     major x tick style = transparent,     major y tick style = transparent,     symbolic y coords={       {SpreadsheetGear},       {Aspose.Cells},       {NPOI},       {C1.Excel},       {Spire.XLS},       {GemBox.Spreadsheet},       {SmartXLS},     },     ytick=data,     y dir = reverse,      nodes near coords,     nodes near coords align={horizontal},   ]    \addplot coordinates {       (3078.6,{Aspose.Cells})       (10059.2,{C1.Excel})       (28794.8,{GemBox.Spreadsheet})       (7437079.8,{SmartXLS})       (19975.2,{Spire.XLS})       (573.2,{SpreadsheetGear})       (10025.2,{NPOI})};   \end{axis}   \end{tikzpicture} \end{document} Migrate from a virtual machine (VM) to a physical system As you can see in object mode there are 4 vertices that have separated the front of the cartridge into multiple faces when I use boolean modifiers on cartridge.001. How can I fix this?  I imagine it has to do with good topology.  Here's the cartridge in object mode showing faces:    Here is the cartridge with the unwanted vertices:    Attached is the .blend For any prime $p > 3$, why is $p^2-1$ always divisible by 24? If $|G/H|=4$ then $G$ is union of three proper subgroups Let $G$ be a finite group and $N$ a normal subgroup of $G$ of index $4$. Prove that $G$ has a subgroup of index $2$ (hence normal) and if $G/H$ is not cyclic then $G$ is equal to the union of three proper subgroups.   I understand that $G/H$ is isomorphic to $\mathbb{Z}_4$ or $\mathbb{Z}_2\times \mathbb{Z}_2$. Let $K^*$ be a subgroup of $G/H$ of order $2$. Then $K^*\cong K/H$ for some $K$ such that $H$ is normal in $K$ and hence $[G:K]=[G/H:K/H]=4/2=2$.  If $G/H$ is not cyclic then $G/H\cong \mathbb{Z}_2\times \mathbb{Z}_2$. I don't know how to prove $G$ is the union of three subgroups. Could it be true that as $G/H$ is the union of three proper subgroups $K_1^*,K_2^*,K_3^*$ then the corresponding groups $K_1,K_2,K_3$ have union equal to $G$? Would anyone give me a hint to prove this?  Thank you. Pick an urn and pick two balls from the urn There are 5 urns, the nth urn contains n − 1 red balls and 5 − n green balls (in total 4 ineach urn). We pick an urn at random and take 2 balls out of it without replacement.  (a) What is the probability the second ball is green?  (b) What is the probability the second ball is green given the ﬁrst is green?  (c) What is the probability we picked either the ﬁrst urn or second urn given both balls are green?  for (a), I did:   P(2nd ball is green) = 5*[{P(i th urn picked) * P(first ball drawn red/ ith urn picked) P(2nd ball drwan green/first ball red)}+ {P(i th urn picked) * P(first ball drawn green/ ith urn picked) P(2nd ball drwan green/first ball green)}]  = 5*[[{1/5 * (n-1)/4 * (5-n)/3 } + {1/5 * (5-n)/4 * (4-n)/3}] = 5* [{(n-1)(5-n)/60} + {(5-n)(4-n)/ 60}]  =[(n-1)(5-n) + (5-n)(4-n)]/12 = (5-n)(n-1+4-n)/12 = (5-n)*3/12 = (5-n)/4  And for b): 5*{P(i th urn picked) * P(first ball drawn green/ ith urn picked) P(2nd ball drwan green/first ball green)} = 5*{1/5 * (5-n)/4 * (4-n)/3} =(4-n)(5-n)/12  c), P(1st urn or 2nd urn/ both are green) = [ [P(1st or 2nd]* (4-n)(5-n)/6 ] / [{P(1st or 2nd]* (4-n)(5-n)/6} + {{P(3rd or 4th]* (4-n)(5-n)/6}] = {(4-n)(5-n)/15} /  2*{(4-n)(5-n)/15} = 1/2  But I am really uncertain about my solutions. If anyone has different solution or easier way to get the answer please let me know! Thanks a lot! Sine $\theta$ = opposite/hypotenuse   Cosine $\theta$  = adjacent/hypotenuse   Tangent $\theta$  = opposite/adjacent      In order to calculate the sine or the cosine or the tangent I need to know $3$ sides of a right triangle. $2$ for each corresponding trigonometric function. How does a calculator calculate the sine, cosine, tangent of a number (that is actually an angle ?) without knowing any sides? I am having some trouble with the following problem:     Let $K\subseteq L$ be fields. Suppose $\alpha,\beta\in L$ are algebraic elements over $K$ of degrees $p,q$ respectively, where $p$ and $q$ are distinct primes. Show that $\alpha+\beta$ is algebraic over $K$ with degree $pq$.   I have shown that $\alpha+\beta$ is algebraic over $K$, but I am having trouble showing that the degree is $pq$. I have previously shown that $[K(\alpha,\beta):K]=pq$, so I have been trying to use this result. I thought that if I could show that $K(\alpha,\beta)=K(\alpha+\beta)$ then I'd be done. To show this I first noted that clearly $K(\alpha+\beta)\subseteq K(\alpha,\beta)$. Then we have $$pq=[K(\alpha,\beta):K]=[K(\alpha,\beta):K(\alpha+\beta)][K(\alpha+\beta):K]$$ so that $[K(\alpha+\beta):K]=1,p,q,pq$. I tried looking at the cases when it is equal to $1,p,q$ and deriving a contradiction, but have been unsuccessful.   Another way I thought of solving this is to show that $\alpha,\beta\in K(\alpha+\beta)$, which would allow me to conclude that $K(\alpha+\beta)=K(\alpha,\beta)$, but I am not sure how to complete this either.    I am looking for some assistance to show that $[K(\alpha+\beta):K]\neq 1,p,q$, or that $\alpha,\beta\in K(\alpha,\beta)$. If you have any other solutions to this problem I would like to see those too. I guess nobody really knows the true nature of black holes, however, based on everything I know about black holes, there is a "singularity" at their center, which has finite mass but is infinitely small and thus infinitely dense.  So a black hole is really just a point mass that dramatically warps the space-time around it so that an event horizon is achieved.  What doesn't make sense to me is how a true mathematical "point" can rotate, or at least how it is even possible to determine if a point is rotating (or if it should even make a difference about whether or not it is rotating, for the same reason it would be impossible to tell if a perfect, featureless sphere were rotating (I know it would have higher rotational energy but lets assume you can't measure that because a true "point" can't attain rotational energy since it has no radius)).  The only conceptual solution I have to describe a rotating black hole would be to compare it to a rotating whirlpool, where the surface of the water is the fabric of space-time... but this is different because there is no "singularity" in a whirlpool that causes the water around it to rotate, whirlpools form from flow of water around a point (put into motion perhaps by some kid swirling water in a 2-liter bottle... it's due to an external cause), but the only thing that could cause rotation around a black hole would be the singularity itself, which doesn't make sense because, once again, how can we know if a true point is actually rotating?  Aside from everything I've said, here is what I am actually asking:  Why/how do some black holes rotate... what is the mechanism of their rotation (is analogous to a spinning point or a 3-d whirlpool, or something different)? Easiest way to read from and write to files Why is the automatic login disabled for users with encrypted home? I've tried to setup an autologin, but that isn't possible, because my home is encrypted. What are the reasons for this?  PS. Here I have described my problem: . How can I split a polygon by network line in QGIS? I have a road network as line layer and a land cover as polygon layer, I want to find number of subpolygon for each polygon that divided by line layer in QGIS?  I used union function in Geoprocessing Tools in Vector menu of QGIS and then polygonize function to convert union result to polygon but in table of new polygon there is NULL data. Split-like environment inside cases environment I'd like to have a cases environment with some cases that are too long to fit on one line. I tried this:  \begin{cases}   \begin{split}     long expression \\ second line of long expression   \end{split} &amp; condition \\   ...   but I get the warning     Package amsmath Warning: Cannot use 'split' here;   trying to recover with 'aligned' on input line 201.   It automatically replaces the split with an aligned environment, which looks like this:    Is there any way to get these long expressions to behave like they were in split or multline environments, with the top line flush left and the next line indented a little? How do I add swap after system installation? I did not create a swap partition during Ubuntu installation.  Later, I freed up some space and made a swap partition.  Now after each boot up, I am manually opening gparted to right-click the swap partition to turn on the 'swapon' option.  How can I automatically enable the swap partition at boot?  Also, if I had not created the swap partition, what are my other options? Which graphics cards are supported by the new AMDGPU driver in Ubuntu 16.04? The new open source AMDGPU driver pre-installed in Ubuntu 16.04 should support all the "Volcanic Islands" cards out-of-the-box. This means that AMD Radeon R9 285, R9 380/380X and R9 Fury/FuryX graphics cards will be supported out-of-the-box if Ubuntu 16.04 LTS is installed.  However I read that the open source AMDGPU driver has an experimental support for the "Sea Islands" graphics cards (HD7790, R7 260, R9 290, R7 360, R9 390). Is this support enabled by default on Ubuntu 16.04, or will these graphics cards use the Radeon driver?  Thanks If it is given that component of acceleration perpendicular to the velocity of a body has a constant, non-zero magnitude, how can we mathematically prove that the trajectory of the body will be circular?   I have basic knowledge of calculus, vectors and co-ordinate geometry. I can prove that the trajectory of a body in projectile motion is parabolic. I tried to follow a similar method to get an equation of circle but couldn't succeed. I tried forming parametric equations of x and y in terms of time t (the way we proceed when proving the parabolic path of a projectile) but could not reach the conclusion.  Edit: I have edited the question in order to improve it. However I still can not share my work on this question as I had asked the question almost 6 months ago and now I can not find where I tried solving it. Sincerely sorry for that. It  to disable internet search results in start menu searches before the Windows 10 Creators Update, but I've just clean installed a new Windows 10 Pro image and that setting is gone. Is there some way to disable it through editing the registry/etc?