Ubuntu 16.04 dual boot Windows 7
How can I install Windows after I've installed Ubuntu?
What are the options for traveling to the airports in Paris? Paris has 3 airports (Orly, Roissy, Beauvais), what are the options to go to or to come from there?  (There is also Vatry (XCR) but there is very few flights and i see the shuttle system is currently off.) Combine and Minify Multiple CSS / JS Files I am from Bangladesh, and I have an Indian tourist visa, which I received from the Indian High Commission in Bangladesh. At this moment I am in Bhutan. Now I want to enter India by air. Is it possible? And will I face any problems when I come back from Kolkata to Dhaka? How do I bake multiple texture maps into 1 UV, In Blender Cycles? Uniform convergence for sequences of functions Given a string of JSON data, how can I safely turn that string into a JavaScript object?  Obviously I can do this unsafely with something like:  var obj = eval("(" + json + ')');   but that leaves me vulnerable to the JSON string containing other code, which it seems very dangerous to simply eval. Irreducible polynomial is always separable in char 0 field I can never figure out whether I should use who and whom. Most people use who for both colloquially, but some people say this is not correct.  What’s the rule for using who and whom correctly? Why does the wire-frame material fades across the object? Why does the wire-frame material fades across the object?  Here is what I did - I created a new material and I selected on "Wire" and I made it shadeless. Then, I created a new material and assigned the new material on the on the object. However, doing it that way, it allows me to render both materials - "Wire" and "Surface". Then I get a fading result like this -    But I want all of the edges to be shown, not a fading result. How can I make that?  Here is the blend file if anybody wants to check it out - Let $G$ be a connected planar graph with a planar embedding where every face boundary is a cycle of even length. Prove that $G$ is bipartite.  Any hints/tips will be greatly appreciated. Is this a correct demonstration for why elements above untriseptium cannot exist? With the confirmation that elements 113, 115, 117, and 118 are indeed fundamental elements that are now to be named on the periodic table, the next question is: what is the highest atomic number possible for an element? Feynman had a go at this years ago, and he derived (according to my limited understanding) that above element-137 (informally denoted 'Feynmanium') the electrons in the nearest orbit about the nucleus would be traveling at a velocity greater than the speed of light. (Note I am considering the Bohr model of the atom in this question.) I wished to demonstrate to my friends why this is and I came up with the following explanation.  According to Bohr's quantum condition, the angular momentum of an electron about the nucleus is given by $$ L=m_evr_n=n\hslash \implies r_n=\frac{n\hslash}{m_ev},$$ where $m_e$ is the mass of the electron, $v$ its velocity, $n$ an integer, and $r_n$ the radius of the $n$th possible orbit. Because we are concerned with the nearest orbit to the nucleus, $n=1$; and thus $$r=\frac{\hslash}{m_ev}.$$   Now, according to Coulomb's law, if the electron orbits about the nucleus the centripetal motion can be described by $$\frac{Ze^2}{4\pi\epsilon_or}=m_ev^2, $$ where $Z$ denotes the number of protons in the nucleus (the atomic number), and $e$ the elementary charge. Solving for $Z$ and substituting in $r$ from above yields $$Z=\frac{4\pi\epsilon_o \hslash v}{e^2}. $$ But what is $v$? Well, the maximum velocity an electron could ever have is the speed of light, and we wish to find the atomic number associated with an orbiting electron traveling at this speed, so we set $v=c$ and obtain our final result of $$Z=\frac{4\pi\epsilon_o\hslash c}{e^2}\approx 137.521,$$ which implies that for $Z&gt;137$, the electrons at a position of $n=1$ in the Bohr model would have a velocity $&gt;c$; and thus the highest atomic number achievable on the periodic table is 137.  Again, I just want to make sure this is a correct method for deriving element-137 before presenting. Perhaps one could explain how relativity plays a role here. I know Feynman used the Dirac equation to get this result...so could anyone (subsequently of course) expatiate on this in a simplistic manner? Thanks! How to install brackets.io from the command line? For some reason the gnome installer that handles .deb packages on my system is not working. I'm not concerned.  How would one install the Brackets.io program from the terminal? No Sound (Dummy Output) error Ubuntu 19.10 "We plan" versus "we are planning" Is there a semantic difference between the following two sentences?     In the future, we are planning to migrate our tool to the Z3 solver.         In the future, we plan to migrate our tool to the Z3 solver. Can I install items on Ubuntu, without using the command line? Everything I've googled so far has used the command line? Copyright of a set of colors? I often see color combinations on webpages, images, posters, products etc. that harmonize really well. Am I allowed to use those colors in my own design?   This is solely about the colors. I will not use images, logos or shapes from that source, and my content is not related to the original color source. I am also far from any commercial aspect here. I'm considering the ring $\mathbb{Z}[\sqrt{-n}]$, where $n\ge 3$ and square free. I want to see why it's not a UFD.   I defined a norm for the ring by $|a+b\sqrt{-n}|=a^2+nb^2$. Using this I was able to show that $2$, $\sqrt{-n}$ and $1+\sqrt{-n}$ are all irreducible. Is there someway to conclude that $\mathbb{Z}[\sqrt{-n}]$ is not a UFD based on this? Thanks. Number theory counting sums of squares modulus This is an extra question from an old examination paper:     VI. Let $n&gt;1$ in $\mathbf{Z}$ and let $r(n) = \#\{(a,b)\in \mathbf{Z}^{2};n=a^{2}+b^{2}\}$ Let also $n=n&#39;n&#39;&#39;n&#39;&#39;&#39;$ where $n&#39;,n&#39;&#39;,n&#39;&#39;&#39; \in \mathbf{N}$ and   $p|n&#39; \Longrightarrow p\equiv 1\mod 4 ; p|n&#39;&#39; \Longrightarrow p=2 ; p|n&#39;&#39;&#39; \Longrightarrow p\equiv 3 \mod 4$   for $p\in \mathbf{N}$ prime.   i) Show that if $n&#39;&#39;&#39;$ is not a square, then $r(n)=0$.  ii) Show that if $n&#39;&#39;&#39;$ is a square and $n&#39; = 1$, then $r(n)=4$  iii) Show that if $n&#39;&#39;&#39;$ is a square and $1&lt;n&#39;=p_{1}^{e_{1}}\ddots p_{k}^{e_{k}}$ where $p_{1},...,p_{k}$ are different primes in $\mathbf{N}$ and $e_{1},...,e_{k} \in \mathbf{N}$, then it holds that $r(n)=r(n&#39;)=4(e_{1}+1)\ddots (e_{k}+1)$    I am completely dumbstruck and can't see how to begin (and neither did an older student who took this exam and whom I asked for hints). Help is greatly appreciated. Is there any explanation for why the other 2 schools in the TriWizard tournament are both single-gender, unlike Hogwarts?  Was single-gender education normal in wizarding world with Hogwarts being the exception?  Was this some special intentional literary trick on Rowling's part or on movie creators'?    Here's the evidence from the movies:  Durmstrang Institute is boys-only:    Beauxbatons Academy of Magic seems to be an all-girl school: Prevent monitor from losing signal after screen saver / lock activates After the screen saver activates, or after I manually lock the screen via Ctrl+Alt+L, I see a blank black screen with just the date / time in the upper right corner. A few seconds after that, my HP flat screen monitor(s) lose the video signal and display ugly hopping "No Input Signal" boxes.  I'd like to keep the blank black screen (It's less distracting, consumes about as much power, and allows me to unlock faster, as the monitors don't have to re-sync), so Ubuntu should keep driving the video output while the screen is locked.  How can I configure this? I'm using Ubuntu 14.04.3 LTS with the Gnome Classic Fall-back. Graphics card is NVIDIA with the open source Nouveau driver, but I also see this with an external HP NL571AA USB graphics adapter.