I'd be thankful if someone could explain to me how to get the result with minus sign in front in this "rationalize the denominator" task:  $\dfrac{\sqrt[3]2-\sqrt[3]3}{\sqrt[3]4-\sqrt[3]6+\sqrt[3]9}$ .  I get the solution with the numerator numbers switched (therefore the minus sign)...  Ehm, as soon as I posted this question, I realized that I'm actually right, I solved it correct, I thought I didn't get the minus. I guess it's late, just delete this...
Rationalize   $\dfrac{1}{\sqrt[3]{p^2}+\sqrt[3]{pq}+\sqrt[3]{q^2}}.$  How would I go about doing this without wading through lots of algebra?  Is there a trick similar to how you can multiply by $\dfrac{\sqrt a-\sqrt b}{\sqrt a-\sqrt b}$ with square roots?  Thanks in advance!
Prove that inequality is true for $x>0$: $(e^x-1)\ln(1+x) > x^2$ I was given a task to prove that inequality is true for x>0: $(e^x-1)\ln(1+x) &gt; x^2$. I've tried to use derivatives to show that the $f(x) = (e^x-1)\ln(1+x)-x^2$ is greater than zero, but has never succeeded.  Any help will be appreciated. How does Austria's suspension of Schengen affect travellers (both EU and non-EU)? I'm an Italian citizen. Next Sunday I'm going to Austria. I have a national ID card but no passport. Recently Austria suspended Schengen. So can I enter Austria? Is it possible to log back into a Pokemon Go account? How do I run a sudo command needing password input in the background? I have recently disabled  so that it now prompts me for a password every time.  And though this is good for security, it has caused one slight problem which I haven't been able to figure out a solution for, I am unable to run commands which go along the lines of:   sudo &lt;command&gt; &amp;   In the past I would have run a sudo command before that, it would have cached my password and allowed me to run sudo commands without prompt for the next few minutes, and thus it would allow me to run the command.  But when I run it now as there is no caching before hand and as it starts a new thread immediately and sudo doesn't even prompt me for a password, I am unable to run it in this way.  So unless I run sudo -i before it I am unable to run a command in this format which is becoming rather annoying. So I was wondering if there is any way to get around this and still run programs and commands in this way?  I am running Ubuntu GNOME 15.10 with GNOME 3.18, and specifically the program I want to run in this fashion is etherape if that makes any difference, but I would really like the solution to work for all programs and commands. Difference between pts and tty braces on the sides of an easybmat matrix I created the following matrix with , and I would like to add at the bottom (respectively right) something like an underbrace (resp. "rightbrace") under each of the two blocks, together with a number specifying the number of columns (resp. rows) of the block. This amounts to two braces on the bottom and two on the right, and each brace should have a number below (resp. at the right) of it.  I did not find anything about that on the easybmat manual. I think I see how to do the braces with the  package, but then I am not able to create the block structure of the matrix in the first place.  \documentclass{article} \usepackage{easybmat} \begin{document} \[  M(z)= \left[ \begin{BMAT}[5pt]{c|c}{c} \begin{BMAT}[5pt]{c:c:c}{ccccccc} &amp; &amp; \\ &amp; &amp; \\ &amp; &amp; \\ v_1 &amp; \dots &amp; v_m \\ &amp; &amp; \\ &amp; &amp; \\ &amp; &amp; \end{BMAT} &amp; \begin{BMAT}{c}{c|c} \begin{BMAT}[5pt]{c:c:c}{ccc} &amp; &amp; \\ w_1(a) &amp; \dots &amp; w_n(a) \\ &amp; &amp; \end{BMAT} \\ \begin{BMAT}[10pt]{c}{c} B(z) \end{BMAT} \end{BMAT} \end{BMAT}  \right] \] \end{document}     This is more or less what i wish: Who did Khan get his Starfleet insignia from? Any examples of life forms on Earth that could have survived on Mars or Moon? A former lecturer of mine once explained why, from a syntactic point of view, the English rule that negation and questions are formed with the auxiliary do follows from other syntactic facts about English.  More precisely, if you gave a good syntactician not familiar with English a corpus of English sentences containing everything but questions and negated phrases, then s/he could infer grammatical rules from it (such as headed-ness of phrases, what kind of arguments what word takes, etc.) From some of these rules, then, s/he could see that something specific would be wrong with the sentences "Gave you the present to Mary?" and "He went not to the cinema."  Can someone explain what exactly this something is and how it follows from other sentences?  FYI, my lecturer did this in Government &amp; Binding with X-Bar theory. This may be more suitable a question for the proposed  but as it's not online yet I'm trying it here first.   Edit: the beta is online now at . Does a card that affects 'each' creature affect creatures that enter the battlefield after it resolves? QGLWidget appears black in Windows 7 Proof that $\mathrm{ker}(f^*) = \mathrm{im}(f)^{\perp} $ Where $ f: V \rightarrow V  $ is a linear operator on an inner product space $V$, I'm trying to show that $$ \mathrm{ker}(f^*) = \mathrm{im}(f)^{\perp} $$  i.e. the Kernel of the adjoint is equal to the orthogonal complement of the image of f.  I taking the inner product of $f(v)$, where $v \in V$  and an arbitary element of the kernel, $k \in \mathrm{ker}(f^*) $  $$\langle f(v), k\rangle = \langle v, f^*(k)\rangle = \langle v, 0\rangle = 0$$  And then taking the inner product of $f(v)$ with $w$, where $w \in  \mathrm{im}(f)^{\perp}$  By definition  of the orthogonal complement to the image of f,   $$ \langle f(v), w\rangle = 0 $$  My question is, now that we have   $$ \langle f(v), k\rangle = \langle f(v), w\rangle = 0 $$  What can we conclude? I would like to conclude that because $w$ and $k$ are arbitary elements of their respective subspaces, that these must be equal, i.e. $ \mathrm{ker}(f^*) = \mathrm{im}(f)^{\perp} $  However, I am not sure this is mathematically sound, is that conclusion valid or are there any technical barriers or additional statements that would have to be made to make that watertight? How can I copy location info from geotagged phone images to images taken with my other camera around the same time? So i did quite a big trip recently, and took lots of pictures with my DSLR and occasionally Android phone. Only after coming back did i read about the whole GPX file, tracklog options in Lightroom. meaning i don't have such a tracklog file.  The android phone did geotag most photos it took. (at least, when it got a gps fix i guess)  So my question is, is there some software or way that will automatically copy the geotag from the android's photos to photos taken on the same time with my DSLR?  Or alternatively, generate a tracklog from the geotagged data in the android pictures? How can I pad an integer with zeros on the left? I've been battling the following two questions for more than a day today.  Write a regular expression (comprised of {a, b}) that contain at least two b and do not contain abb. Write a regular expression (comprised of {a, b}) that contain abba and do not contain baa.  For the first part I wrote a basic one to start off with that doesn't handle the must two-b: b* OR a* OR (a+ AND b)* OR a*  I ran a few &quot;truth tables&quot; inputs over it and it seems ok. But the moment I add the requirement, things go haywire. Mostly because when pushing in those &quot;b&quot;s, you can suddenly use the OR to break things up. Same thing on the second one.... Any help would be greatly appreciated! Norah Is there a way I can avoid imperial ships? Whenever I am traveling on foot, imperial ships will fly by and drop down troops for me to fight. In a normal situation, this is no big deal and I will fight or flee to continue with my travels.  However this will also occur when I am doing a quest that requires me to locate and pick up items. This is an inconvenience as I cannot pick up items while in combat and I have to kill them before I can continue.  Is there a way I can "hide" and allow the ships to by pass my location? The relativistic energy-momentum equation is:  $$E^2 = (pc)^2 + (mc^2)^2.$$ Also, we have $pc = Ev/c$, so we get: $$E = mc^2/(1-v^2/c^2)^{1/2}.$$  Now, accelerating a proton to near the speed of light, I get the following results for the energy of proton:  0.990000000000000   c =&gt;    0.0000000011    J =      0.01    TeV  0.999000000000000   c =&gt;    0.0000000034    J =      0.02    TeV  0.999900000000000   c =&gt;    0.0000000106    J =      0.07    TeV  0.999990000000000   c =&gt;    0.0000000336    J =      0.21    TeV  0.999999000000000   c =&gt;    0.0000001063    J =      0.66    TeV  0.999999900000000   c =&gt;    0.0000003361    J =      2.10    TeV  0.999999990000000   c =&gt;    0.0000010630    J =      6.64    TeV 0.999999999000000   c =&gt;    0.0000033614    J =      20.98   TeV  0.999999999900000   c =&gt;    0.0000106298    J =      66.35   TeV  0.999999999990000   c =&gt;    0.0000336143    J =      209.83      TeV  0.999999999999000   c =&gt;    0.0001062989    J =      663.54      TeV  0.999999999999900   c =&gt;    0.0003360908    J =      2,097.94    TeV  0.999999999999990   c =&gt;    0.0010634026    J =      6,637.97    TeV  0.999999999999999   c =&gt;    0.0033627744    J =      20,991.10   TeV    If the LHC is accelerating protons to $7 TeV$ it means they're traveling with a speed of $0.99999999c$.  Is everything above correct? I have an itemized list:  \begin{itemize}     \item Lorem     \item Ipsum     \item Sit     \item Dolor     \item Amet \end{itemize}   I want to create an accolade on the right side of 'Lorem' and 'Ipsum', and add a bit of text at the point of the accolade. The level of this text should be between the first and the second line. How can I get this done? I am trying to write a bash script for testing that takes a parameter and sends it through curl to web site.  I need to url encode the value to make sure that special characters are processed properly.  What is the best way to do this?    Here is my basic script so far:  #!/bin/bash host=${1:?'bad host'} value=$2 shift shift curl -v -d "param=${value}" http://${host}/somepath $@ How to derive variance-covariance matrix of coefficients in linear regression I am reading a book on linear regression and have some trouble understanding the variance-covariance matrix of $\mathbf{b}$:    The diagonal items are easy enough, but the off-diagonal ones are a bit more difficult, what puzzles me is that  $$ \sigma(b_0, b_1) = E(b_0 b_1) - E(b_0)E(b_1) = E(b_0 b_1) - \beta_0 \beta_1 $$  but there is no trace of $\beta_0$ and $\beta_1$ here.