I'm using Budgie Desktop currently, and I really enjoy it. It's got good battery life, customizability, stability and speed, but there's just one thing that bothers me; it doesn't have a real app switcher. Pressing Alt + Tab will switch apps, but no dialog appears, and it doesn't allow me to select an app I want to see; it just calls them all to front, on at a time.  Because of this, I've been looking for some third party app switcher that'll show a dialog and let me select the app to switch to instead of just pulling everything to the front. Right now, I'm using skippy-xd, which is an exposé-style switcher. It's OK, but I'd prefer the linear method that Unity and GNOME use.  Googling has turned up nothing. I think I'm just not doing it correctly.  Are there any standalone app/task switchers (besides skippy-xd) available for Ubuntu or Linux in general?
I have Ubuntu budgie 17.04 (budgie-desktop 10.2.9). While pressing Alt+Tab to switch between the windows, the windows are switched randomly and I am not able to select which window to shift.  This feature was available in Ubuntu 16.04 but not in Ubuntu-budgie-17.04.    How to resolve this?
Find all integer roots of: $x^2(y-1)+y^2(x-1)=1$   Obviously $(2,1)$ and $(1,2)$ are two answers. But I was unable to manipulate the equation algebraically giving a useful form for finding all other possible answers! I also tried to view it as a quadratic equation in terms of $x$, but forming the Delta of that equation didn't help much other than if $y=1$ then certainly Delta would be a square and equation would have answer for $x$... Travel to Australia with new passport (with visa in old passport) I was granted a visa to Australia few months ago. I’d like to travel to Australia in few days. However, I’ve just renewed my passport and I don’t have enough time to inform them about my passport renewal. Can I travel to Australia by carrying both my old and my new passports? How do I use the Chinese remainder theorem to find all the square roots of 11 in $\mathbb Z_{35}$? What process created the /var/lib/dpkg/lock-frontend file in Ubuntu? I have just booted my clean Ubuntu 18.04 LTS VM and tried to install some application with sudo apt install.  But got error message:     E: Could not get lock /var/lib/dpkg/lock-frontend - open (11: Resource temporarily unavailable)       E: Unable to acquire the dpkg frontend lock (/var/lib/dpkg/lock-frontend), is another process using it?   What process created the /var/lib/dpkg/lock-frontend file in Ubuntu? How to gently stop it and take manual control on APT? Two mice heterozygote and black are reproduced. Find the probability of getting the filium dark and heterozygote How to deploy wsp to single web application Recently, I am trying to create 3D models for my own game project. However, when I do some researches on game models, I come up with the term - game-ready models and I could not find any clear definition online. Could anyone tell me about what exactly is game-ready models and how to know if my 3D models (created by Blender mainly) is suitable for games.  From what I know by far, quad topology is preferred in Blender. However, for most of the game engine, triangle topology is preferred (Please correct me if I am wrong). Does it mean I have to manually change the models in Blender from quad topology to triangle topology before importing to the game engine?  Thanks for any helps. In my case I'm talking about Assassin's Stance, but I'd like to know the answer for stances in general.  I feel like I might be missing something incredibly obvious caused by missing a word or two (or a paragraph), but I can't find anything that explicitly states whether or not I can use the Tome of Battle stances with ranged weapons like the bow (or crossbow).  I realize that half the classes have "blade" in their names so using ranged weapons is not in the spirit of things, but I'm looking for an explicit answer (or as close to one as possible) from any WotC official resource. If this has already been answered I'm sorry, but I could not find the answer myself from searching. Windows INSERT key anti-functionality accidentally triggers; how to stop it permanently? Folland, "Real Analysis", Chapter 5.3, Exercise 36:     Let $\mathcal{X}$ be a separable Banach space and let $\mu$ be   counting measure on $\mathbf{N}$.  Suppose that   $\left\{x_n\right\}_1^\infty$ is a countable dense subset of the unit   ball of $\mathcal{X}$, and define $T \colon L^1(\mu) \to \mathcal{X}$   by $Tf = \sum_1^\infty f(n) x_n$.  (a)  $T$ is bounded.  (b)  $T$ is   surjective.   I have proved (a).  I would like help on (b).  Here are my ideas so far.    Say we want to show $y \in \mathcal{X}$ is in the image of $T$.  Reduce to the case $||y|| = 1$ and let $x_{n_m} \to y$.  My thought is to produce for each $N$ a function $f_N \in L^1(\mu)$ such that $||y - \sum_n^\infty f(n)x_n|| &lt; 1/N$ and also such that the $f_N$ converge in $L^1(\mu)$ to some $f \in L^1(\mu)$.  Then use the continuity of $T$ to conclude $T(f) = y$.    The way I have been setting up my inequalities is as follows. Pick $M_N$ such that for all $m &gt; M_N$, $||y - x_{n_m}|| &lt; 1/N$.  Then I want to define $f_N$ so that $$\sum_n^\infty f_N(n) = \sum_{m &gt; M_N}^\infty f_N(m) = 1, $$ hence $$||y - T(f_N)|| = ||\sum_{m &gt; M_N}^\infty f_N(m) y + \sum_n^\infty f_N(n)x_n|| \leq \sum_{m &gt; M_N}^\infty f_N(m) || y - x_{n_m} || &lt; \frac{1}{N}. $$  Presumably, if the $f_N$ are appropriately chosen, then I can find a dominating function and apply the Dominated Convergence Theorem to finish up.    However, getting all these hypotheses to hold simultaneously has been difficult.  In essence, the difficulty seems to be in requiring that the $f_N$ be "normalized" as well as possess a dominating function.  If this method is feasible, at least it seems to require a bit of analysis to show the desired series all converge.  Perhaps I am missing something cleaner.  If so, a simpler solution would be much appreciated.    -Thanks. We have some new hardware in our office which runs its own customized Linux OS.   How do I go about figuring which distro it's based on? How to connect to Android through SSH over 3G/4G public IP? Is this NPN transistor AND logic gate practical? Why is there no absolute maximum temperature? If temperature makes particles vibrate faster, and movement is limited by the speed of light, then I would assume that temperature must be limited as well. Why is there no limit? Evaluating indefinite integrals of the form $\int \frac{x^2 \,dx}{a x^5 + b}$ Evaluate the indefinite integral   $$\int \frac{x^2 \,dx}{a x^5 + b},$$   for real parameters $a, b \neq 0$.   No apparent substitutions simplify the expression (if the exponent of $x$ were an integral multiple of $3$, this would suggest the substitution $u = x^3$, $du = 3 x^2 \,dx$, but the exponent is not). Applying integration by parts with $u = a x^5 + b$, $dv = x^2 dx$ is straightforward, but it produces an integrand with a much larger degree in the denominator and so appears only to make the situation worse. Applying integration by parts instead with $dw = \frac{x^k dx}{a x^5 + b}$ results in integrating $\frac{x^k dx}{a x^5 + b}$, which except when $k \equiv 4 \pmod 5$ (which doesn't appear immediately fruitful), doesn't appear much easier than the given integral. Favorite questions in the Android app The  (ESC), states that for every natural number $n \geq 2$, there exists a set of natural numbers $a, b, c$ , such that the following equation is satisfied:  $$\frac{4}{n}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\tag{1}$$  The basic approach to solving this problem outlined by Mordell [Ref1] is described below  By defining $t$ and $m$ as positive integers greater than zero and $q$ a positive integer greater than one we can observe that  a) There is always a solution for even $n$, since if $n=2^qt$ we have the trivial solution $$\frac{4}{4t}=\frac{1}{t}$$  In the remaining case $n=2(2t+1)$, a solution in the form of two Egyptian fractions can always be found  e.g. $$\frac{4}{2(2t+1)}=\frac{2}{2t+1}=\frac{1}{t+1}+\frac{1}{(t+1)(2t+1)}$$  b) If $(1)$ is a solution for some particular prime $n$ then all composite numbers $mn$ divisible by $n$ are also solutions, thus  $$\frac{4}{mn}=\frac{1}{ma}+\frac{1}{mb}+\frac{1}{mc}$$  will also be a solution. This means that we can simplify the analysis to the cases where $n$ is a prime greater than 2.  Using Mordell's approach we have just shown that we only need to consider the cases where $n$ is prime and where $n \equiv 1 \pmod{2} \;\;[meaning \;\;n=2t+1]$  The argument continues...  Mordell goes on to show in turn that the search can be reduced further to the cases when $$n \equiv 1 \pmod{4} \;\;[meaning \;numbers \;\;n=4t+1]$$ $$n \equiv 1 \pmod{8} \;\;[meaning \;numbers \;\;n=8t+1]$$ $$n \equiv 1 \pmod{3} \;\;[meaning \;numbers \;\;n=3t+1]$$ $$n \equiv 1,2,4 \pmod{7} \;\;[meaning \;numbers \;\;n=7t+1,n=7t+2 \;or\;n=7t+4 ]$$ $$n \equiv 1,4 \pmod{5}  \;\;[meaning \;numbers \;\; n=5t+1 \;or\;n=5t+4]$$  Assembling these results together, Mordell showed that the conjecture can be proved in this context except for the cases when $$n \equiv 1,11^2,13^2,17^2,19^2,23^2 \pmod{840}$$  Mordell stated that since the first prime meeting this condition is 1009, this is proof that the conjecture holds for $n&lt;1009$.  This basic approach can be pursued further. Other workers have shown that the conjecture holds for much higher values of $n$ using similar methods as can be seen on the above Wikipedia page.   Note that other intermediate results can be constructed from the above congruence's, e.g. $n \equiv 1 \pmod{24}$.  The question is:  Are there any other elementary approaches to solving this problem than the one outlined by Mordell (and described above)?  [Ref1] Louis J. Mordell (1969) Diophantine Equations, Academic Press, London, pp. 287-290. Overnight charges for brokers holding stocks? I'm trying to learn about stock markets. I eventually want to invest a small amount over a long period. Yet, I notice on a lot of broker sites that they tend to charge an overnight fee for any stocks held over night. I presume this is every night thereafter?  I just wondered because it seems like a wall when you want to let your stock price grow, and an overnight charge eats into this. or, is this only the case for "CFD" stocks? I'm confused.  Any help?  Cheers Let $W^{1,p}(U)$ be the Sobolev space. Suppose that $U$ is connected bounded domain in $\mathbb{R}^n$ and $u \in W^{1,p}(U)$ satisfies $Du=0$ a.e. in $U$. How can I prove that $u$ is constant a.e. in $U$? Erase Windows Vista to Install Ubuntu new here. I have an old E-machine which as Windows Vista on it and I want to erase the Windows Vista, and install Ubuntu on it instead, but I am not sure where to start. Can I first install Ubuntu, and THEN erase Windows Vista, or is there a more sure method? I appreciate your help.