Tell me please, how calculate this expression:  $$ \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} $$  The result should be a number.  I try this:  $$ \frac{\left(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}\right)\left(\sqrt[3]{\left(2 + \sqrt{5}\right)^2} - \sqrt[3]{\left(2 + \sqrt{5}\right)\left(2 - \sqrt{5}\right)} + \sqrt[3]{\left(2 - \sqrt{5}\right)^2}\right)}{\left(\sqrt[3]{\left(2 + \sqrt{5}\right)^2} - \sqrt[3]{\left(2 + \sqrt{5}\right)\left(2 - \sqrt{5}\right)} + \sqrt[3]{\left(2 - \sqrt{5}\right)^2}\right)} = $$  $$ = \frac{2 + \sqrt{5} + 2 - \sqrt{5}}{\sqrt[3]{\left(2 + \sqrt{5}\right)^2} + 1 + \sqrt[3]{\left(2 - \sqrt{5}\right)^2}} $$  what next?
Why $\sqrt[3]{{2 + \sqrt 5 }} + \sqrt[3]{{2 - \sqrt 5 }}$ is a rational number?
Show that $a^{61} \equiv a\pmod{1001}$ for every $a \in \mathbb{N}$ Remove Meta Privilege for MSE (only) Maximum and Minimum values for ints For a project into balloon simulation I'd like to know how the force on a balloon changes with altitude:  I know that the Buoyant force on a balloon is:  $F = (\rho_{air} - \rho_{helium})gV$  Using changing air density from the international standard atmosphere and changes in gravitational acceleration I can calculate the buoyant force and add this to the (negative) force of gravity mg.  However I know that it isn't this simple, the balloon changes volume with altitude as the air pressure outside the balloon changes.  I'd like to know how the balloon inflates - this would be a great help!  I think it may be something to do with ideal gas laws -  I can calculate air temperature and pressure from the international standard atmosphere but do I use pressure as the air pressure? - or air pressure minus balloon pressure? How to render a background image in 2.8? Will self-citation be viewed as self-promotion in academia? As we know, citation counts are important to judge one's research activity. Is it good to cite one's previous works? Will it be viewed as an act of advertisement or self-promotion? Changing the style of the first *typeset* line of a paragraph Change the URL in the browser without loading the new page using JavaScript Reflection of sound waves from the open end of an organ pipe & relationship b/w nodes & pressure We know that standing waves are created when any wave traveling along the medium will reflect back when they reach the end. But in an open organ pipe, there is nothing to oppose the wave and reflect it back. So then how are standing waves created in such a case? In standing waves where is the pressure greater, at nodes or anti-nodes? How do I split a forest tree into two trees with proper vertical alignment? \documentclass[oneside,12pt]{article} \usepackage{geometry} \usepackage{microtype} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage{lmodern}  \usepackage{enumitem} \setlist[enumerate,1]{label=\bfseries\Alph*,align=left,leftmargin=*,   labelsep=1.5em} \setlist[enumerate,2]{label=\arabic*.,labelindent=1em,labelsep=1.5em,   leftmargin=*}  \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsthm} \theoremstyle{definition} \newtheorem*{sol}{Solution}  \usepackage{forest} \forestset{   ass/.append style={     before computing xy={l=\baselineskip},     no edge   },   prooftree/.style={     for tree={       child anchor=north,       parent anchor=south     }   } }  \begin{document} \subsection*{Exercises 29} \begin{enumerate} \item Use QL trees to evaluate the entailment claims (1) to (10) in Exercises   28A.   \begin{enumerate}   \item $\forall x(Fx\supset Gx)\vDash\forall x(Gx\supset Fx)$      \begin{forest}       prooftree       [$\forall x(Fx\supset Gx)$       [$\neg\forall x(Gx\supset Fx)\checkmark$,ass       [$\exists x\neg(Gx\supset Fx)\checkmark$,ass       [$\neg(Ga\supset Fa)\checkmark$,ass       [$Ga$,ass       [$\neg Fa$,ass       [$(Fa\supset Ga)\checkmark$,ass       [$\neg Fa$] [$Ga$]]]]]]]]     \end{forest}      $\neg Fa$ and $Ga$ make the premisses and the negation of conclusion true.     Thus, the entailment is invalid.   \item $\forall x(Fx\supset Gx)\vDash\forall x(\neg Gx\supset\neg Fx)$      \begin{forest}       prooftree       [$\forall x(Fx\supset Gx)$       [$\neg\forall x(\neg Gx\supset\neg Fx)\checkmark$,ass       [$\exists x\neg(\neg Gx\supset\neg Fx)\checkmark$,ass       [$\neg(\neg Ga\supset\neg Fa)\checkmark$,ass       [$\neg Ga$,ass       [$\neg\neg Fa$,ass       [$(Fa\supset Ga)\checkmark$,ass       [$\neg Fa$ [*,ass]] [$Ga$ [*,ass]]]]]]]]]     \end{forest}      Since the tree closes, the entailment is valid.   \item $\forall x\exists yLxy\vDash\forall y\exists xLyx$      \begin{forest}       prooftree       [$\forall x\exists yLxy$       [$\neg\forall y\exists xLyx\checkmark$,ass       [$\exists y\neg\exists xLyx\checkmark$,ass       [$\neg\exists xLax\checkmark$,ass       [$\forall x\neg Lax$,ass       [$\exists yLay\checkmark$,ass       [$Lab$,ass]]]]]]]     \end{forest}      \begin{forest}       prooftree       [$\neg Lab$,ass [*,ass]]     \end{forest}   \item $\forall x((Fx\wedge Gx)\supset Hx)\vDash\forall     x(Fx\supset(Gx\supset Hx))$   \item $(\forall xFx\vee\forall xGx)\vDash\forall(Fx\vee Gx)$   \item $\forall x(Fx\supset Gx),\forall x(\neg Gx\supset Hx)\vDash     \forall x(Fx\supset\neg Hx)$   \item $\exists x(Fx\wedge Gx),\forall x(\neg Hx\supset\neg Gx)\vDash     \exists x(Fx\wedge Hx)$   \item $\forall x\exists y(Fy\supset Gx)\vDash\forall y\exists x(Gx\supset     Fy)$   \item $\forall x\forall y(Lxy\supset Lyx)\vDash\forall xLxx$   \item $\forall x(\exists yLxy\supset\forall zLzx)\vDash\forall x\forall y     (Lxy\supset Lyx)$   \end{enumerate} \end{enumerate} \end{document}   Usually, I have to manually split a forest tree into two to emulate page breaking in a forest tree. However, the second tree doesn't vertically align with the first as in the picture below. How can I vertically align them together? @cfr I just need to check the reasoning in my proof is correct, I think it is valid although I'm not totally convinced because I can't follow the logic; does proving that $x$ is an integer prove that $ac|bd$?  Theorem: Let $a$, $b$, $c$, $d$ be integers. If $a|b$ and $c|d$, then $ac|bd$.  Proof: $aj=b$ and $ck=d$ for integers $j$, $k$. Then, $ac|bd$ implies $acx=ajck$ and thus $x=jk$ for some $x$. Since the product $jk$ is an integer, $x$ is an integer and thus $acx=ajck$ and thus $ac|bd$. I need to display math in my report, but is it possible to remove the vertical space around align*?  A little example is the following (where I just want the align* to have the same vertical space as a \\  \documentclass[10pt,danish,a4paper,oneside,fleqn]{report}  \usepackage[english]{babel} \usepackage{amsmath}  \begin{document}  ssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss \begin{align*}   A{\cap}B\ &amp; =\ {\left\{{b,d,e}\right\}}{\;}{\cap}{\;}{\left\{{a,b,f,g}\right\}}   \\   \nonumber\ &amp; =\ {\left\{{b}\right\}} \end{align*} ssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss \\ ssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss  \end{document} Practically speaking, how do people handle ANOVA when the data doesn't quite meet assumptions? This isn't a strictly stats question--I can read all the textbooks about ANOVA assumptions--I'm trying to figure out how actual working analysts handle data that doesn't quite meet the assumptions. I've gone through a lot of questions on this site looking for answers and I keep finding posts about when not to use ANOVA (in an abstract, idealized mathematical context) or how to do some of the things I describe below in R. I'm really trying to figure out what decisions people actually make and why.   I'm running analysis on grouped data from trees (actual trees, not statistical trees) in four groups. I've got data for about 35 attributes for each tree and I'm going through each attribute to determine if the groups differ significantly on that attribute. However, in a couple of cases, the ANOVA assumptions are slightly violated because the variances aren't equal (according to a Levene's test, using alpha=.05).   As I see it, my options are to: 1. Power transform the data and see if it changes the Levene p-val. 2. Use a non-parametric test like a Wilcoxon (if so, which one?). 3. Do some kind of correction to the ANOVA result, like a Bonferroni (I'm not actually sure if something like this exists?). I've tried the first two options and gotten slightly different results--in some cases one approach is significant and the other is not. I'm afraid of falling into the p-value fishing trap, and I'm looking for advice that will help me justify which approach to use.  I've also read some things that suggest that heteroscedasticity isn't really that big of a problem for ANOVA unless the means and variances are correlated (i.e. they both increase together), so perhaps I can just ignore the Levene's result unless I see a pattern like that? If so, is there a test for this?  Finally, I should add that I'm doing this analysis for publication in a peer-reviewed journal, so whatever approach I settle on has to pass muster with reviewers. So, if anyone can provide links to similar, published examples that would be fantastic. How do I calculate the resistor value for a simple LED circuit? I have a simple circuit:    The max rating (current) of the LED is 30mA.  How can I work out what the resistance of the resistor needs to be? By using ohm's law I found it to be \$3V/0.03A = 100 \Omega\$. However using software called , and trial and error I got the minimum possible resistance to be 36&Omega;. However, if I use a 35&Omega; resistor, then the LED breaks. Is the software wrong, or (more likely) am I doing something wrong? My mom and I are having a dispute on much more easy versus much easier. For example, consider the sentence:     It's [much more easy]/[much easier] to do action X than action Y.   I say that much easier is correct and that much more easy is grammatically incorrect, while she says that both are okay. I understand that both are acceptable colloquially, but which is grammatically correct here? If you could explain why, that would be even better! Every time I try to do something seemingly-simple in CSS, it doesn't work.  I have a content div that contains a 460x160 image. All I want to do is position the image at the bottom-right corner and wrap my text around it.  &lt;div id="contents"&gt;     &lt;img src="..." /&gt;     text text text text text text ... &lt;/div&gt;   So I want it to look like:  ------------------ | text text text | | text text text | | text text -----| | text text |    | ------------------   Doing it with a top-left or top-right image is cake:  #contents img { float:right; }  ------------------ | text text |    | | text text -----| | text text text | | text text text | ------------------   Now how do I push that to the bottom? The best I've come up with so far are:  #contents img { float:right; margin-top: 90%} // really needs to be 100%-160px  ------------------ | text text      | | text text      | | text text -----| | text text |    | ------------------   In this case the text does not print in the margin, so there is white space above the image.  #contents { position:relative; } #contents img { position:absolute; right:0; bottom:0; }  -or-  // move the img tag under the text in the html and: #contents img { float:right; margin-top:-160; }  ------------------ | text text text | | text text text | | text text -----| | text text | te | ------------------   In this case the text prints over or under the image.  So... how can I accomplish this? Use the definition of derivative to prove $\lim_{x\to 0}\ln(x+1)/x =1$ How would I use the definition of derivative to prove  $$\lim_{x\to 0} \frac{\ln(1+x)}{x} = 1$$  I got to $$\frac{\frac{\ln(1+x+h)}{(x+h)} - \frac{\ln(1+x)}{x}}{h}$$ but have no idea where to go from here.   On another site I found someones answer where they stated the following: $$ \lim_{x\to 0} \frac{\ln(1+x) - \ln(1+0)}{x-0}  = [\ln(1+x)]'\rvert_x = 0 $$ but I am unsure why the $x$ in the $x-0$ is removed. Can someone please explain? If $A$ an integral domain contains a field $K$ and $A$ over $K$ is a finite-dimensional vector space, then $A$ is a field. Parallel.ForEach Slower than ForEach Just curious whether the process takes a while, and whether it scrapes through history to determine if it was once valid or not. Currently, I have 6 0-vote accepted answers and 20 1+-vote answers (), and that makes it > 5 and > 20% but not for long if voting goes up or down from here. ;-P  Update:  Special thanks to Grace Note for pointing out the  to follow. I upgraded this to a bug because it's been 48 hours during which the criteria supposedly were met. 6 0-votes, 20-23 1+ votes during the whole period. The criteria only last so long, so it may go out of scope soon. If it only checks every 3 days, I'm not sure if that's intentional or a bug.