I have HP Pavilion 15 Notebook PC preinstalled with Windows 8. I installed ubuntu 14.04 (dual boot) , followed by an upgrade to Ubuntu 15.04. I also upgraded to Windows 10. Recently, due to some reasons, Windows OS directly loads without showing me the Grub menu and letting me select the OS I want to load. I have to press Esc key followed by F9 to see the boot options and select the OS from there. Is there any solution to this problem, without reinstalling Ubuntu?  I want to select the OS I want to load directly from the Grub Menu which should show up on system startup. None of the previous answers related to this question solved my problem.
I installed Ubuntu alongside Windows but I did not disable secure boot as someone told me it could cause problems with boot loader. So when installation finished and I rebooted, the system went straight into Windows. I went to Ubuntu IRC Support and they told me to turn off the secure boot and reinstall Ubuntu. I did that but my system still boot into windows.  This is a picture of partitions on my system:    This is my first Ubuntu install on my new laptop. I followed everdaylinux guide to install it and I used something else option.
How do I easily switch between windows rather than applications in GNOME Shell? How to find a program which is preventing sleeping? I am noticing that recently my laptop does not go to sleep/lockout after its configured X minutes of inactivity like it did in the past. I suspect some program (like media player does) is preventing it from going to sleep. I do not have any media player running, so how can I find which program is preventing it from sleeping?  It does go to sleep if I close the lid. But I would also like for it to go to sleep after X minutes of inactivity.  Update: Upgraded to Ubuntu 18.04 with default Gnome and Wayland enabled. Proving if $a$ is not a multiple of prime $p$, then an integer $b$ exists such that $p^b - 1$ is a multiple of $a$ I tried many approaches but none of them really worked I treated $p^b-1$ as a Geometric progression but it didn't work and that is about as far as I have been able to go I have no clue how to move forward IP address of host of Wi-Fi ad-hoc network? I often create a wireless ad-hoc network on my laptop running Windows 7. Now I'd like to connect to it using standard HTTP from my . I've tried 192.168.1.1 and that doesn't work. What should the address be so I can connect? I am calculating the class group of $\mathbb Q(\sqrt 6)$. My working is as follows:  The Minkowski bound is $\lambda(6)=\sqrt 6&lt;3$ so we only need to look at prime ideals of norm $2$. $2$ divides the discriminant, $24$, of $\mathbb Q(\sqrt 6)$, so $2$ ramifies, and so for any prime ideal $\mathfrak p$ of norm $2$, $\mathfrak p^2$ is principal. But $(2, \sqrt 6)^2=(2)$, so $\mathfrak p=(2, \sqrt 6)$ has norm $2$. If I can show $\mathfrak p$ is not principal, then I can conclude that $Cl(\mathbb Q(\sqrt 6))\cong C_2.$ But if there was an element of $\mathcal O_{\mathbb Q(\sqrt 6)}$ of norm 2, this would correspond to a solution of the diophantine equation $x^2-6y^2=2$. Clearly $x$ is even, so writing $x=2n$ we get $2n^2-3y^2=1$. Now reducing modulo $3$, we get $2n^2\equiv 1 \pmod 3$ which has no soultions. Hence $\mathfrak p$ is not principal and $Cl (\mathbb Q(\sqrt 6)) \cong C_2.$  However,  says that the class group is trivial. Where have I gone wrong? What am I misunderstanding? Is $(2, \sqrt 6)$ principal? tcolorbox manual tip box, how to clone? Show that a number has a terminating decimal expansion if and only if, it is rational and when in lowest terms, its denominator is coprime to all primes other than $2$ and $5$.   This is an unsolved question in my lecture notes. I can only seem to prove the converse direction for this. Would appreciate a solution for the other direction.     For converse direction:   Let the number, in lowest terms, be given by $\frac{l}{m}$ where $m = 2^ \alpha 5^\beta$, for some positive integers, $\alpha$ and $\beta$.    If $\alpha &gt; \beta$, let $k = 5^{(\alpha-\beta)}$. If $\alpha &lt; \beta$, let $k = 2^{(\beta-\alpha)}$.   Then $\frac{l}{m} = \frac{kl}{k2^\alpha5^\beta}=\frac{kl}{10^q}$ where $q = \max(\alpha,\beta)$.   Hence $\frac{l}{m}$ is a terminating decimal. Inequality Regarding e Difficulty in finding a counterexample I got a positive definite matrix $B,$ that is, $V(x) = x^TBx &gt; 0$ for any vector $ x \neq 0.$  I want to show that $ \lambda_\min \|x\|_2^2 \leq V(x) \leq \lambda_\max \|x\|_2^2$ for any $x \neq 0,$ where $\lambda_\min$ and $ \lambda_\max$ are defined by $$ \lambda_\min = \min \lbrace | \lambda|: \lambda \text{ is an eigenvalue of } B \rbrace$$ and  $$ \lambda_\max = \max \lbrace | \lambda|: \lambda \text{ is an eigenvalue of } B  \rbrace$$  Any hint please? The anime Elfen Lied is notorious for its one of a kind OP, featuring the mesmerizing theme song Lilium composed by MOKA and performed by Kumiko Noma, as well as sequences of art porn, a homage to Austrian symbolist painter Gustav Klimt. The idea to pose the characters as in Klimt's artworks came, as far as I know, from director Mamoru Kanbe, who continued this idea in the OP of Sora no Woto. (In my humble opinion, the OP of Sora no Woto is rather bland, and seriously, Klimt with Kalafina?)  This is a self-answered question and a modest attempt at unraveling the sources of and the symbolism represented in Elfen Lied's OP scenes. Other answers and/or suggestions are welcome. I'd like to provide template implementation in my module and allow themes to override it. Basically, I add a suggestion by this simplified code:  function attach_preprocess_node(&amp;$vars) {   $vars['theme_hook_suggestions'][] = 'node__test'; }   (I don't want to use hook_theme to add a new theme because I want to reuse preprocess node functions. The theme name are awkward but I don't want to write node_attach_% to avoid confusion with node type.)  Then I use hook_theme_registry_alter() to add module path:  function attach_theme_registry_alter(&amp;$theme_registry) {   $path = drupal_get_path('module', 'attach') . '/themes';   $theme_registry_copy = $theme_registry;   _theme_process_registry($theme_registry_copy, 'phptemplate', 'theme_engine', 'node', drupal_get_path('module', 'node'));   $theme_registry += array_diff_key($theme_registry_copy, $theme_registry);   if (!isset($theme_registry['node']['theme paths'])) {     $theme_registry['node']['theme paths'] = array();   }   if (!isset($theme_registry['node']['theme paths'])) {     $first_element = array_shift($theme_registry['node']['theme paths']);     if ($first_element) {       array_unshift($theme_registry['node']['theme paths'], $first_element, $path);     }     else {       array_unshift($theme_registry['node']['theme paths'], $path);     }   } }   However it does not work. That means: the file themes/node--super.tpl.php is not used. It is used only if I copied it to the theme folder. I live &amp; work in the UK.  Having pretty much exhausted all of the avenues I can find for saving (none of which have great returns anyway at the moment, obviously), I have recently started looking to buy my first house.  I have a reasonable sum saved up for a deposit, but I'm wondering what to do with that in the mean time, as a fair amount of it is not currently earning any interest. For the moment, I have put any surplus, that I'm not earning interest on into premium bonds, just in the off-chance that I win something with it, but this is obviously more of a lottery than an investment, and I'm not 'expecting' to win anything noteworthy with it.  Ideally, I would like to buy a house within the next year or so, so investing in the stock market is not really an option, as I understand the commonly accepted advice is that you invest with a minimum 5 year horizon.  What other options do I have for any surplus cash while I look for a house? $\nabla U=0 \implies U=\mathrm{constant}$ only if $U$ is defined on a connected set? Hill Cipher known plaintext attack I know a plaintext - ciphertext couple of length 6 for a hill cipher where its key is a [3x3] matrix.  Based on what I've read and learned, to attack and crack keys of [n x n], if we know a plaintext - ciphertext duo of length $n^2$ then we have our set of n equations with n variables, and this is generally solvable. However in my case there is only length of $6$ instead of $n^2 = 9$ .   My question is, how will I solve this problem and find the key? Or, since this is an obvious homework question, what should be my way of thinking? I cannot think of anything else but matrix multiplications and inverses but they do not help me at all. Add chat rooms for Android app Being able to join chat rooms from the Android app would be a plus! It would truly make us connected on the go, instead of having to sit down at our desks and open a browser window. This will allow more users to fully utilize the chat system and join events no matter where they are. Dictionaries on iOS 8 not working How to create a new partition from the existing partition? I wanna create a new partition from the existing root partition.My root partition is about 250GB.I wanna create a new partition say 150 GB taking space from this root partition.Can anyone help me with the procedure on doing this?  output: fdisk -l  karthick@karthick:~$ sudo fdisk -l [sudo] password for karthick:   Disk /dev/sda: 250.0 GB, 250000000000 bytes 255 heads, 63 sectors/track, 30394 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x28219744  Device Boot      Start         End      Blocks   Id  System /dev/sda1   *           1          61      487424   83  Linux Partition 1 does not end on cylinder boundary. /dev/sda2              61       30395   243650561    5  Extended /dev/sda5              61         124      498688   82  Linux swap / Solaris /dev/sda6             124         186      498688   82  Linux swap / Solaris /dev/sda7             186       30395   242651136   83  Linux Apply a string replacement on the citation as returned by \cite I have biblatex with options  \usepackage[backend=bibtex, style=authoryear]{biblatex}   The citations made with \cite look like this: Smith, 2019 or Smith and Baker, 2019 and so on. I would like to define a command that automatically replaces the occurrence of my name with my initials. Namely I need  \cite{SmithsPaper}  % --&gt; Smith, 2019 \cite{MyPaper}      % --&gt; Manny, 2019 %%% \newcite{SmithsPaper}  % --&gt; Smith, 2019 \newcite{MyPaper}      % --&gt; MC, 2019   I have tried with the package xstring to do something like  \newcommand{\newcite}[1]{%   \begingroup   \edef\@temp{\cite{#1}}%   \StrSubstitute{\@temp}{Manny}{MC}   \endgroup }   The problem is that I can't figure out how to tell LaTeX to expand \cite{#1}. As a result the citation comes out without the replacement. If I change \cite{#1} by simply #1 and I call \newcite{bla Manny bla} it gives bla MC bla as expected.  It shouldn't be relevant but I'm using this within the beamer documentclass. Do I have to buy or need a 3D-enabled phone to use Google Cardboard? Google cardboard comes with 3D glasses. Does that mean that we will be able to watch 3D videos using a smartphone like the Moto-G (1st edition) which is not 3D enabled?