I have recently purchased my new laptop ASUS Vivobbok S14. i7, 8gb, 512gb version which comes with windows 10 pre installed. I am willing to install ubuntu 18 on this machine. Is there a support for my machine?? I have been reading many posts and came to know that there are backlight keyboard and wifi issue after the installation. And the battery backup is also reduced as compared to windows. I am a programmer, I need to upgrade only if there is full support/drivers available for my machine.
I have a device on which I would like to install Ubuntu. How do I find out if Ubuntu will work on it?
How to disable touchscreen permanently on ubuntu 18.04 I'm trying to install the drivers to my wireless adapter ():   I have already tried the following:  sudo add-apt-repository ppa:hanipouspilot/rtlwifi sudo apt update sudo apt install rtl8192eu-dkms   But I'm getting the following error:  E: Unable to locate package rtl8192eu-dkms   I have also tried manually downloading and installing it, but I'm getting this:  make make ARCH=x86_64 CROSS_COMPILE= -C /lib/modules/4.6.0-kali1-amd64/build M=/root/Desktop/20140812_rtl8192EU_linux_v4.3.1.1_11320  modules make[1]: *** /lib/modules/4.6.0-kali1-amd64/build: No such file or directory.  Stop. Makefile:1323: recipe for target 'modules' failed make: *** [modules] Error 2   edit: lsusb output  root@kali:~# lsusb Bus 002 Device 002: ID 8087:0024 Intel Corp. Integrated Rate Matching Hub Bus 002 Device 001: ID 1d6b:0002 Linux Foundation 2.0 root hub Bus 001 Device 004: ID 8087:07da Intel Corp.  Bus 001 Device 003: ID 1bcf:2c0e Sunplus Innovation Technology Inc.  Bus 001 Device 002: ID 8087:0024 Intel Corp. Integrated Rate Matching Hub Bus 001 Device 001: ID 1d6b:0002 Linux Foundation 2.0 root hub Bus 004 Device 001: ID 1d6b:0003 Linux Foundation 3.0 root hub **Bus 003 Device 002: ID 2357:0107**   Bus 003 Device 001: ID 1d6b:0002 Linux Foundation 2.0 root hub As the title asks, why didn't the basilisk bite received in the Chamber of Secrets destroy the Horcrux inside the bitten person?   Yes, I know Fawkes quite quickly healed the bite wound with his phoenix tears, but when Harry used the basilisk fang to destroy Tom Riddle's diary the destruction of that Horcrux was immediate. When Ron hit the locket with the Sword of Gryffindor, it was destroyed immediately. Although the destruction of the ring occurred off-page, Dumbledore used the Sword of Gryffindor to destroy it, so presumably, like the locket, it was gone right away. Nagini died straightaway when Neville decapitated her. The destruction of the Hufflepuff cup happens off-page. The diadem seems to have been destroyed by Fiendfyre immediately as well, as when Harry examined it outside the Room of Requirement it was already leaking a blood-like substance, and then it cracked in half. When Voldemort hit Harry with Avada Kedavra in the forest, Harry woke up in limbo with the last fragment of Voldemort's soul - that awful flayed baby-like being - separated from him immediately.   Is there a logical/canon-based explanation for why the fragment of Voldemort's soul residing in Harry wasn't destroyed upon Harry being bitten by the basilisk?  NOTE: I've read the HP Wiki's entry on Horcruxes; however, I find the HP Wiki to be inconsistent in its accuracy, so I'm hesitant to accept the Wiki's explanation on the matter. I'd prefer an explanation constructed from information in the books or a statement by JKR, or the like. I can't seem to find a satisfactory answer to this anywhere on the web. While  poses a related question, I don't think it sufficiently answers mine...  During the opening chapters of Jedi Search, Luke addresses the new Senate on the need to find new Jedi talent, or Force sensitive individuals. During this proposal, he cites that the most fruitful leads of Jedi potential are in the descendants of Jedi Knights: as Force sensitivity is hereditary. The book even discusses the notion that Leia should be comparatively well attuned to the Force, due to her being the child of Anakin Skywalker.  I understand that the concept of the Jedi as a totally celibate and monastic order was something that was retconned into the lore after Episode II: Attack of the Clones, and that there are examples of entire nuclear Jedi families exist (such as Andur Sunrider and Plo Koon)...  But if Force-manipulative ability is passed down genetically (as a potentially mutative property of their genetic makeup), why would the Jedi order be so numerous if they practice Chastity?  Even if procreation is not expressly forbidden (as in the pre-retcon Expanded Universe): why would Jedi need special permission to produce more Jedi potential children?  How are more Jedi created, if not through procreation? Has their profusion been deliberately retarded, and if so why? is there another explanation for their (pre-empirical) plenitude? Is there a cleaner proof of convergence for this almost-telescoping series I was wondering how the convergence of the series $$\sum_{k\geq1}\frac{(k+1)^y-k^y}{k^x},$$ depends on $x$ and $y$, where $x$ and $y$ are real numbers. I've been able to show that it converges if and only if $y\leq x$, by distinguishing nine cases; whether $x$ and $x-y$ are positive, negative or zero. If desired I can include a sketch of my proof. My question is whether there is a proof that works for all real $x$ and $y$? Or at least a proof that seems less artificial? What is the location of lock screen images on Windows 10? Where, or what format, are the images used by Windows 10 when logged out or in lock-screen?  I searched the entire drive for *.jpg, *.bmp, *.png, and *.tif but the one I was looking for did not appear.  At first glance, , although not exactly a duplicate, appeared to have the path I wanted.  But when I went there, the pre-login image was there, but the lock-screen image was not.  (And contrary to the claim in that answer, the files there did have extensions.)  UPDATE: I just did a lockscreen, and got an image that IS in that directory, but earlier today, it was consistently an image of an interesting footbridge that is not in that directory. If you draw two cards, what is the probability that the second card is a queen? How to change from csh to bash as default shell I have csh as my default shell, as shown by echo $SHELL. I want to switch to bash as my default shell. I tried the following approaches to no avail:   With chsh I get:  chsh: can only change local entries; use ypchsh instead.  With ypchsh I get:  ypchsh: yppasswdd not running on NIS master host ("dcsun2").    I only have .chsrc in my home directory and I cannot find any .profile files in /etc. How can I change my default shell to bash? Rename command not found when running a script How to stop event propagation with inline onclick attribute? Consider the following:  &lt;div onclick="alert('you clicked the header')" class="header"&gt;   &lt;span onclick="alert('you clicked inside the header');"&gt;something inside the header&lt;/span&gt; &lt;/div&gt;   How can I make it so that when the user clicks the span, it does not fire the div's click event? Why can a discontinuous function not be differentiable? I don't really understand why a discontinuous function cannot be differentiable.  In Stewart's Calculus, the definition of a function $f$ being differentiable at $a$ is that $f'(a)$ exists. Earlier it gives the definition of the derivative as $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$. It also has a theorem that if $f$ is differentiable at $a$, then $f$ is continuous at $a$. There is also a definition that a functions $f$ is continuous at a number $a$ if $\lim_{x\to a}f(x)=f(a)$.  Take for example the very simple function:  $$ f(x)=\begin{cases} x+1 &amp; x\geq0, \\ x &amp; x&lt;0.  \end{cases} $$  It is discontinuous at $x=0$ (the limit $\lim_{x\to 0}f(x)$ does not exist and so does not equal $f(0)$), but if I find the derivative using the limit above, I get the left and right limits to equal $1$. So therefore, the derivative exists.  According to the book, the function shouldn't be differentiable at $x=0$ as it has a discontinuity (continuity is a necessary condition of differentiability). What am I doing/understanding wrong? Using WCS layer in QGIS for processing gives FAILURE: Missing destination I have a GeoServer serving raster dem data out via WCS and can add the layer easily to QGIS (3.2) but when I try to use it as the input for an operation (e.g. HillShade) I get an error.  Processing algorithm… Algorithm 'Hillshade' starting… Input parameters: { 'ALTITUDE' : 45, 'AZIMUTH' : 315, 'BAND' : 1, 'COMBINED' : False, 'COMPUTE_EDGES' : False, 'INPUT' : 'dpiMode=7&amp;identifier=training:panorama&amp;url=http://localhost:8080/geoserver/ows?version%3D1.1.1', 'MULTIDIRECTIONAL' : False, 'OPTIONS' : '', 'OUTPUT' : 'C:/Users/ishare/AppData/Local/Temp/2/processing_868cc25180f24d9f8e193bc0cfedb6d3/7cf48f64d8b54ee694c3ee93ed17e903/OUTPUT.tif', 'SCALE' : 1, 'ZEVENBERGEN' : False, 'Z_FACTOR' : 5 }  GDAL command: gdaldem hillshade dpiMode=7&amp;identifier=training:panorama&amp;url=http://localhost:8080/geoserver/ows?version%3D1.1.1 C:/Users/ishare/AppData/Local/Temp/2/processing_868cc25180f24d9f8e193bc0cfedb6d3/7cf48f64d8b54ee694c3ee93ed17e903/OUTPUT.tif -of GTiff -b 1 -z 5.0 -s 1.0 -az 315.0 -alt 45.0 GDAL command output:   FAILURE: Missing destination.   Clearly the missing destination is a red herring, and it is failing to understand the WCS layer as an input. But GDAL is built with WCS support included so shouldn't this work?  Do I need to do something else or start working out where to hack the QGIS code? What is an "AppImage"? How do I install it? I just read . It talks about downloading an AppImage.     Yes, Linux; the Linux packages is distributed as an .appimage for 32-bit and 64-bit distributions, and should run across all major Linux distributions without any issues. The team currently have no plan to provide a native .deb (or .rpm) installer.   What are AppImages? How do they differ from snaps? Finding the limit $\displaystyle\lim_{x\to 0+} \left(\frac{\sin x}x\right)^{1/{x^2}}$ Find the following limit: $$\displaystyle\lim_{x\to 0+} \left(\frac{\sin x}x\right)^{1/{x^2}}$$   Well I tried to do the $\exp\left(\frac{ \ln\frac{\sin x}{x}}{x^2}\right)$ then apply LHR but I seem to get to endless dervivations...   There's got to be a more simple approach. Posts linked in deleted answers still show up under "Linked" E.g. see . The following is a geometry puzzle from a math school book. Even though it has been a long time since I finished school, I remember this puzzle quite well, and I don't have a nice solution to it.  So here is the puzzle:    The triangle $ABC$ is known to be isosceles, that is, $AC=BC$. The labelled angles are known to be $\alpha=\gamma=20°$, $\beta=30°$. The task is to find the angle labelled "?".  The only solution that I know of is to use the sine formula and cosine formula several times. From this one can obtain a numerical solution. Moreover this number can be algebraically shown to be correct (all sines and cosines are contained in the real subfield of the 36th cyclotomic field). So in this sense I solved the problem, but the solution is kind of a brute force attack (for example, some of the polynomials that show up in the computation have coefficients > 1000000). Since the puzzle originates from a book that deals only with elemetary geometry (and not even trigonometry if I remember correctly) there has to be a more elegant solution. Evaluating $\int^\pi_{-\pi} \cos^3(x) \cos(ax)~dx$ I need to evaluate $I = \int^\pi_{-\pi} \cos^3(x) \cos(ax)~dx$ where $a$ is some integer.   I get: $\dfrac{2a(a^2-7)\sin(\pi a)}{a^4 - 10a^2 + 9}$. However $\sin(\pi a)$ is $0$ for all $a$ so $I=0$. But as noted by an answer, there are answers for $a = 1,3,-1,-3$. On a plane, there are two congruent circles, $O_1, O_2$. One line intersects $O _1$ and A and B and $O_2$ at C and D such that AB=BC=CD=14. Another line intersects $O _1$ and E and F and $O_2$ at G and H such that EF=FG=GH=6. What is the radius of the circle.   The configuration was hard to draw. . You can reflect the lines over the centers, but it is still the same. Cleary, $\widehat{AB}=\widehat{CD}$ and $\widehat{EF}=\widehat{GH}$. I have no idea how to proceed. Can we say that $\frac{0}{0}$ is every number? Use the beamer TOC style elsewhere