iPhone - Strike out an NSString
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Do class spells take up spell slots? So I'm fairly new and ended up being the Cleric in my also fairly new group. I haven't played a caster before and the book says that at level 1 I have Bless and Cure Wounds. Do I need to assign these to a spell slot or do I have them inherently (not to be confused with a cantrip; I know I'd have to use a slot to cast them)? Lie bracket for Lie algebra of $SO(n,m)$ I want create my own IDE but I want a code editor that would syntax highlighting.  I want to embed it as a control into a WPF window.  Malcolm I tried to use my Virtual Box and this is the error I received - (I just updated to the new Ubuntu also) and I am VERY new at this Linux OS.     Kernel driver not installed (rc=-1908)      The VirtualBox Linux kernel driver (vboxdrv) is either not loaded or   there is a permission problem with /dev/vboxdrv. Please reinstall the   kernel module by executing      '/etc/init.d/vboxdrv setup'      as root. Users of Ubuntu, Fedora or Mandriva should install the DKMS   package first. This package keeps track of Linux kernel changes and   recompiles the vboxdrv kernel module if necessary. A characterisation of quadratic extensions contained in cyclic extensions of degree 4 Let $D$ be a squarefree integer. I am trying to prove that $\mathbb Q[\sqrt D]$ is contained in a Galois extension of $\mathbb Q$ with Galois group $\mathbb Z/4$ if and only if $D$ is the sum of two squares, $D = a^2 + b^2$ with $a,b\in \mathbb Q$.   The hint in the exercise in Dummit and Foote suggests considering the extension $\mathbb Q[\sqrt{s + s\sqrt D}]$ for the forward direction. However, I am unsure how to show that this extension is Galois, much less that its Galois group is cyclic of order 4.   For the reverse direction, I expressed the given Galois extension as $\mathbb Q[\sqrt{a+b\sqrt D}]$ for some $a,b\in \mathbb Q$. I showed that this extension has $\mathbb Q[\sqrt{c^2 - d^2 D}]$ as a subfield for some $c,d \in \mathbb Q$. From this, I deduced that the $\sqrt D$ and $\sqrt{c^2 - d^2 D}$ generate the same extension, so $\sqrt D = x \sqrt{c^2 - d^2 D}$ for some $x$, and hence that $D = \frac{x^2 c^2}{1 + x^2 d^2} $. It seems that there should be a way to decompose this as a sum of two squares, but I did not see how.   I did not tag this as homework as I am studying it independently. Firefox - Websites with custom fonts not displaying Let $C[0,1]$ denote the ring of continuous functions on $[0,1]$. Consider the maximal ideal $M_a =\{ f\in C[0,1] | f(a)=0 \}$ for $a$ $\in$ $[0,1]$. Is $M_a$ finitely generated?   Note: It may seem that $M_a = \langle x-a\rangle$ which is certainly not true because $ \vert x-a \vert $ is in $M_a$ but not in $\langle x-a\rangle$. Any ideas for solving this question? Is there a way to access QGIS plugins in Python?   I'd like to write a script to do some geoprocessing, and I'd like to use the clip, regular points and convex hull tools in fTools, as well as the Road graph plugin. The latter is probably a different can of worms because it is a C++ rather than Python plugin. Beamer: change header background color I have decided to switch from Powerpoint to Beamer, and for my first presentation I'd like to use the , and the same colors as in this goggle slide template:  .  I tried to do some research about it, the closest answers I found are the question  and the . I tried this code:  \documentclass{beamer} \usepackage[frenchb]{babel} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc}  \usetheme{metropolis}  \definecolor{bluegreen}{rgb}{3, 166, 155} \definecolor{pitchblack}{rgb}{0, 0, 0} \definecolor{lightbeige}{rgb}{255, 251, 241} \definecolor{mediumgray}{rgb}{183, 183, 183}  \setbeamercolor{background canvas}{bg=pitchblack} \setbeamercolor{normal text}{fg=lightbeige} \setbeamercolor{headline}{bg=bluegreen, fg=mediumgray}  \begin{document} \begin{frame}     \frametitle{The title}     Lorem ipsum... \end{frame} \end{document}   I does everything I want, except that the title bar's background color is lightbeige:    What did I do wrong? Or what argument should I pass to \setbeamercolor to get it right? Thanks in advance :-). Very large log files, what should I do? Sum of $k {n \choose k}$ is $n2^{n-1}$ Proof that $\suṃ̣_{k=1}^{n}k {n \choose k}$ for $n \in \mathbb N$ is equal to $n2^{n-1}$.  As a hint I got that $k {n \choose k} = n {n-1\choose k-1} $.  I tried solving this by induction but, in the inductive step I'm not arriving to the correct result. "I have never said" versus "I never said" Suppose Player 1 has a Dragon on the table and  in hand, while Player 2 has two Spiders on the table.  Player 1 chooses to attack with the Dragon and Player 2 chooses to block with one of the Spiders.  Before combat damage is to be dealt, Player 1 uses Unsummon on the blocking Spider.    Does Player 2 take any life damage from the attacking Dragon? Kinematics with non constant acceleration A particle experiences an acceleration described by  $$ a=kx^{-2} $$ where x is the displacement from the origin and k is an arbitrary constant.  To what value does the velocity v of the particle converge to as x approaches infinity if the particle starts at some point x0?  If I approach this problem with energy, then $$ W = \int F \mathrm{d}x $$ $$  = \int_{x_0}^\infty mkx^{-2} \mathrm{d}x $$ $$  = mk(-\infty^{-1}+{x_0}^{-1}) $$ $$ W = K = mk{x_0}^{-1} $$ $$ \frac{1}{2}mv^2 = mk{x_0}^{-1} $$ $$ v = \sqrt{2k{x_0}^{-1}} $$  How would I solve this problem with pure kinematics? (there appears to be some sort of cyclical dependency where acceleration affects velocity, velocity affects displacement, and displacement affects acceleration)  Likewise, two particles experience accelerations described by $$ a_1=k_1x^{-2} $$ $$and $$ $$ a_2=-k_2x^{-2} $$ where x is the distance between the two particles  What two velocities do the particles reach as x approaches infinity if the two particles are initially separated by some x0? I am looking for a short proof that $$\int_0^\infty \left(\frac{\sin x}{x}\right)^2 \mathrm dx=\frac{\pi}{2}.$$ What do you think?  It is kind of amazing that $$\int_0^\infty \frac{\sin x}{x} \mathrm dx$$ is also $\frac{\pi}{2}.$ Many proofs of this latter one are already in this . I finally got dragonrend and defeated Alduin, but I'm still too weak for fighting ancient dragons. Any certain weapon for killing them.  (I've got lvl 20 one handed weapons and lvl 46 for archery, so the Dragonbane is out of the question) I am stuck on the following problem that says:       a) Find a weak formulation for the partial differential equation $${\partial u\over\partial t\ }+ c{\partial u \over \partial x\ }=0$$    b) Show that $u=f(x-ct)$ is a generalized solution of $${\partial u\over\partial t\ }+ c{\partial u \over \partial x\ }=0$$ for any distribution $f$    My Attempt:   I know that in order to find a weak form of a pde, we need to multiply it by a test function, then integrate it. Also, to find a generalized solution, we need to find a weak solution and just multiply it by the Heaviside function.  Let's take any test function $\phi $, then we have (integrating by parts second part of the integral) $$\int_{\Omega } ( {\partial u\over\partial t\ }+ c{\partial u \over \partial x\ })*\phi(x) dx= \int_{\Omega } {\partial u\over \partial t}\phi(x) dx - c\int_{\Omega } u(x,t)\phi'(x)dx$$ where $\phi $ vanishes at boundaries. So, is it the final form or can we proceed further? And how am I supposed to find a generalized solution?  Can someone help me out? Thanks in advance for your time. Proving the snake lemma without a diagram chase Suppose we have two short exact sequences in an abelian category $$0 \to A \mathrel{\overset{f}{\to}} B \mathrel{\overset{g}{\to}} C \to 0 $$ $$0 \to A&#39; \mathrel{\overset{f&#39;}{\to}} B&#39; \mathrel{\overset{g&#39;}{\to}} C&#39; \to 0 $$ and morphisms $a : A \to A&#39;, b : B \to B&#39;, c : C \to C&#39;$ making the obvious diagram commute. The snake lemma states that there is then an exact sequence $$0 \to \ker a \to \ker b \to \ker c \to \operatorname{coker} a \to \operatorname{coker} b \to \operatorname{coker} c \to 0$$ where the morphisms between the kernels are induced by $f$ and $g$ while the maps between the cokernels are induced by $f&#39;$ and $g&#39;$.  It is not hard to show that the morphisms induced by $f, g, f&#39;, g&#39;$ exist, are unique, and that the sequence is exact at $\ker a, \ker b, \operatorname{coker} b, \operatorname{coker} c$. With the use of a somewhat large diagram shown , we can even construct the connecting morphism $d : \ker c \to \operatorname{coker} a$. However, I'm stuck showing exactness at $\ker c$ and $\operatorname{coker} a$. I thought Freyd might have had an element-free proof in his book, but it turns out he proves it by diagram chasing and invoking the Mitchell embedding theorem [pp. 98–99]. Is there a direct proof? How do I switch to first person? My HP Pavilion laptop has Ubuntu 18 version and I once tried to disable InBuilt webcam just for security reasons. But now it wasn't working and I reinstalled newer version of Ubuntu 20. But even after installing newer version the camera is still not working. When I open cheese it shows this screen. I assumed I might have deleted the drivers as I was doing package installation and removel that time. Please suggest me what should I do to get this working.