I have an ASUS N550JK laptop (BIOS version 204) running Ubuntu 14.04 (3.13.0-43-generic kernel).   My issue is that the battery indicator ALWAYS shows that it's using battery power - even when plugged in and fully charged. So even though I have selected to show the battery status in the menu bar only when charging or in use (in power settings), it is ALWAYS displayed. When it reaches a full charge, the indicator icon will either show just the percentage (usually 99% or 100%), or sometimes a very big number for time (like 22:14) along with the percentage. If I go to the power settings menu, it always says "Using battery power", with something like 40 hours remaining. This is all while plugged in and fully charged.  Are there any fixes less drastic than trying to update BIOS, or is a BIOS update necessary to fix this? The latest BIOS version is 208 according to the ASUS website, but the (very brief) descriptions of the updates don't mention anything about battery issues. If at all possible I'd like to avoid updating BIOS because of the risk involved.  I've already tried letting the battery drain all the way and then charging to full, with no luck.  Thanks in advance!
I have an ASUS N550JK laptop (BIOS version 204) running Ubuntu 14.04 (3.13.0-43-generic kernel).  My issue is that the battery indicator ALWAYS shows that it's using battery power - even when plugged in and fully charged. So even though I have selected to show the battery status in the menu bar only when charging or in use (in power settings), it is ALWAYS displayed. When it reaches a full charge, the indicator icon will either show just the percentage (usually 99% or 100%), or sometimes a very big number for time (like 22:14) along with the percentage. If I go to the power settings menu, it always says "Using battery power", with something like 40 hours remaining. This is all while plugged in and fully charged.  Are there any fixes less drastic than trying to update BIOS, or is a BIOS update necessary to fix this? The latest BIOS version is 208 according to the ASUS website, but the (very brief) descriptions of the updates don't mention anything about battery issues. If at all possible I'd like to avoid updating BIOS because of the risk involved.  I've already tried letting the battery drain all the way and then charging to full, with no luck.  Thanks in advance!
Do I need to take my weight off the bike to lock out the suspension? Tensor product in dual-space I am a bid confused regarding the notation for tensor products when going into dual-space  If $\left| \Psi \rangle \right. = \left| A \rangle \right. \left| B \rangle \right.$ is $ \left. \langle \Psi \right| = \left. \langle A \right| \left. \langle B \right|$ or $\left. \langle B  \right| \left. \langle A \right|$?  My guess is the last choice, since operators should act on   $\left| 0 \rangle \right. = \left( \begin{matrix}   1 \\   0  \end{matrix} \right) $  $\left| 1 \rangle \right. = \left( \begin{matrix}   0 \\   1  \end{matrix} \right) $  so  $\left| 01 \rangle \right. = \left| 0 \rangle \right. \left| 1 \rangle \right. = \left( \begin{matrix}   0 \\   1 \\   0 \\   0  \end{matrix} \right) $  But then, for me, it should be  $\left( \begin{matrix}   0 &amp; 1 &amp; 0 &amp; 0  \end{matrix} \right) = \left \langle 10 \right|$ But this is not correct if i calculate the tensor product, since   $\left \langle 1 \right| \left \langle 0 \right| = \left( \begin{matrix}   0 &amp; 0 &amp; 1 &amp; 0  \end{matrix} \right)$  For instance, shoulden't it be  $\left| 01 \rangle \right. \left \langle 10 \right| = \left( \begin{matrix}   0 &amp; 0 &amp; 0 &amp; 0\\   0 &amp; 1 &amp; 0 &amp; 0\\   0 &amp; 0 &amp; 0 &amp; 0\\   0 &amp; 0 &amp; 0 &amp; 0\\  \end{matrix} \right) $ ?  I hope I have made my confusion clear sed substitution with Bash variables Prove that if $(ab)^i = a^ib^i \forall a,b\in G$ for three consecutive integers $i$ then G is abelian I've been working on this problem listed in Herstein's Topics in Algebra (Chapter 2.3, problem 4):     If $G$ is a group such that $(ab)^i = a^ib^i$ for three consecutive integers $i$ for all $a, b\in G$, show that $G$ is abelian.   I managed to prove it, but I'm not very happy with my result (I think there's a neater way to prove this). Anyway, I'm just looking to see if there's a different approach to this.   My approach:  Let $j=i+1, k=i+2$ for some $i\in \mathbb{Z}$.  Then we have that $(ab)^i = a^ib^i$, $(ab)^j = a^jb^j$ and $(ab)^k = a^kb^k$.  If $(ab)^k = a^kb^k$, then $a^jb^jab =a^jab^jb$.   We cancel on the left and right and we have $b^ja = ab^j$, that is $b^iba = ab^j$.  Multiply both sides by $a^i$ on the left and we get $a^ib^iba = a^jb^j$, so $(ab)^iba = (ab)^j$. But that is $(ab)^iba = (ab)^iab$.   Cancelling on the left yields $ab=ba$, which holds for all $a,b \in G$, and therefore, $G$ is abelian.  Thanks! Page range truncation with imakeidx This is a followup question to egreg's answer to  The answer runs into a problem if an index topic occurs in more than one page range. The following MWE, based on egreg's answer:  \begin{filecontents*}{subjind.ist} delim_0 ", \\checkrange{" delim_t "}" \end{filecontents*}  \documentclass{book} \usepackage{xparse} \usepackage{imakeidx}  \makeindex[title=Subject Index,options=-s subjind]  \ExplSyntaxOn \NewDocumentCommand{\checkrange}{&gt;{\SplitArgument{1}{--}}m}  {   \formatrange#1  } \NewDocumentCommand{\formatrange}{mm}  {   \IfNoValueTF { #2 }    { % no --, it's not a range     #1    }    { % it's a range     \egreg_range:nn { #1 } { #2 }    }  }  \bool_new:N \l__egreg_range_compare_bool  \cs_new_protected:Nn \egreg_range:nn  {   % print the start number   #1--   \int_compare:nNnTF { \str_count:n { #1 } } = { \str_count:n { #2 } }    {% same number of digits, remove equal ones at the start     \__egreg_range_compare:nn { #1 } { #2 }    }    {% different number of digits, print both     #2    }  } \cs_new_protected:Nn \__egreg_range_compare:nn  {   \bool_set_true:N \l__egreg_range_compare_bool   % do a loop on the digits in the first number   \int_step_inline:nn { \str_count:n { #1 } }    {     \bool_if:NTF \l__egreg_range_compare_bool      {       % if the digits coincide, print nothing       \str_if_eq:eeF { \str_item:nn { #1 } { ##1 } } { \str_item:nn { #2 } { ##1 } }        {         \str_item:nn { #2 } { ##1 }         % at the first differing digit, don't compare any longer         \bool_set_false:N \l__egreg_range_compare_bool        }      }      { \str_item:nn { #2 } { ##1 } }    }  } \ExplSyntaxOff  \begin{document} \setcounter{page}{95}\index{Topic0|(}Text\newpage \setcounter{page}{101}\index{Topic0|)}Text\newpage \setcounter{page}{124}\index{Topic1|(}\index{Topic2|(}Text\newpage \setcounter{page}{126}\index{Topic1|)}Text\newpage \setcounter{page}{134}\index{Topic2|)}Text\newpage \setcounter{page}{141}\index{Topic0}Text\newpage \index{Topic0}Text\newpage \index{Topic0}Text \printindex \end{document}   yields the following error message:  ! LaTeX3 Error: Too many '--' tokens when trying to split argument.  For immediate help type H &lt;return&gt;.  ...                                                l.3   \item Topic0, \checkrange{95--101, 141--143}   Entering non-stop yields the following output:    which omits "-3" from the second index entry of Topic0. The question is how to get around this.  Edit following up on egreg's ingenious solution.  I don't know if I should ask a new follow up question, so I'm posing it first here in this edit.  egreg's solution doesn't seem to handle index sub-topics.  If one adds the following 4 lines  \setcounter{page}{135}\index{Topic0!sub0|(}Text\newpage \setcounter{page}{136}\index{Topic0!sub0!subsub0|(}Text\newpage \setcounter{page}{139}\index{Topic0!sub0!subsub0|)}Text\newpage \setcounter{page}{140}\index{Topic0!sub0|)}Text\newpage   after  \setcounter{page}{134}\index{Topic2|)}Text\newpage   in the solution, one gets the following errors:  Output written in mwe.ind. Transcript written in mwe.ilg.  (mwe.ind ! Extra }, or forgotten \endgroup. l.4     \subitem sub0, 135--140}  ? r OK, entering \nonstopmode... ! Extra }, or forgotten \endgroup. l.5       \subsubitem subsub0, 136--139}  ) [145] (mwe.aux) )   and this output:    Alas I don't understand egreg's code well enough to make the required changes. If $f,g : [a,b] \to (0,\infty)$ are continuous functions, prove that there exists $c \in (a,b)$ such that: $$\int_{a}^{c} f(x) \ \mathrm dx = \int_{c}^{b} g(x) \ \mathrm dx$$ How to zero out a drive? I'm currently running Ubuntu 12.04 and want to completely format my laptop so I can install Windows 7 (replace Ubuntu).   When I put in my bootable USB with the OS it just shows a black screen with a white blinking cursor on the top left corner. I wait for hours (to be specific 5 whole hours) waiting for something to happen but I get nothing, so I pull out the USB and my Ubuntu 12.04 loads up.  Repeated this several times with putting different boot priority options on top-est that relate to USB but the results are same. I go and visit some sites on the net like this  that say I have to zero out my hard drive.   My question is how?  Note: Erasing out all my data is no problem cause I have nothing important to backup.  If I have to lose my primary OS (Ubuntu 12.04) in the process I am ready to as my aim is just to install Windows 7 successfully. Please don't answer that there is something wrong with my USB or my USB reader/port/hardware or the content inside the USB cause they all works fine on my brothers PC as it boots up flawlessly. Is there a visual web editor based on bootstrap? Bootstrap is a great framework, and I want know if there is a tool that gives you the possibility of dragging html components, re-sizing and moving them, changing colours, all while generating the bootstrap html code.  I don't want to use Dreamweaver because you need to go and add classes and IDs to the html, which could be a pain sometimes. Is it a good practice to refer to countries, ships etc using the feminine form? I have read so many things already that I'm more confused than before. So decided to ask this by myself.  My paladin just got his Find Steed spell, and after some tweaking and talking, he decided to get a Worg and I was okay with it. This last session was his first time using it and I have few doubts when it comes to combat.  I know that while mounted, he uses the mounts movement and actions. But what about when unmounted? What if the combat began with him walking by his mount's side?   Does the mount share his initiative? Are the actions that the mount do, part of the Paladin's actions? Or are they both separately, having his Action, Movement and Bonus Action and then the mounts Action and movement?   We've stated that it can understand him but not talk back, he just telepathically receives feelings as a feedback to his commands, knowing that the mount understood what he needed. Last session he just said things to the mount like "I tell the mount to protect the ranger and the druid" and in case they got hit, it would attack them, but it didn't really do anything at all, as I didn't know when it would act and I had too many things already to remember about the mount. Why do different wavelength get impeded more or less when in different materials? Moving with the same speed, but a longer physical distance would imply that the fields oscillate less times in the material, but I don't know why a difference in the number of oscillations would impede the wave- I don't even know why things slow down in general. Why some electromagnetic wave would slow down just because it's entering other electromagnetic fields... It would seem to me that the only factor would be time taken to physically move some electron or something in the direction of the fields... But that seems to simple of an explanation to me. How to run firefox and firefox ESR simultaneously The main reason I want to use firefox ESR is to use a proprietary database tool for work that only allows oracle-java based web logins for linux (or a windows only thin client). I also want to use the full version of the outlook web application which only works in firefox. I don't want to run the NPAPI plugins for normal web browsing and email.   I have the repo version of firefox installed and firefox ESR installed in /opt/firefox and linked to /usr/local/firefox-esr  When I run firefox any attempt to run firefox-esr either brings up a window without the NPAPI plugins.   If I try running with the --new-instance or --no-remote options I receive the following error:  Firefox is already running, but is not responding. To open a new window, you must first close the existing Firefox process, or restart your system.   Regardless which one I start first, I can only open that one, and get the above error if I attempt to sidestep.  What makes this an especially annoying problem is that I often have to respond to emails that I need to access the database to deal with. So either I close and open many windows. View/edit ID3 data for MP3 files What's a quick and easy way to view and edit ID3 tags (artist, album, etc.) using C#? My friend had hosted a Minecraft server before and brought it back up today with my help. He and anyone in his LAN can connect. However we have had a few people try to connect, including me, and when we try to connect it gives the message:  Failed to connect to the server Connection timed out: connect  What does this exactly mean? How do we fix this? How do I change the screen resolution using Ubuntu command line? Question mark and colon in JavaScript I am trying to disable the touchscreen on my Asus S200E laptop.  In Ubuntu 16.04, I was able to do so with this:  Section "InputClass"     Identifier "evdev touchscreen catchall"     MatchIsTouchscreen "on"     MatchDevicePath "/dev/input/event*"     Driver "libinput" EndSection   How can this be done on Ubuntu 18.04? I have a vector space $V$ over a field of characteristic $0$. If $M_1,\dots,M_k$ are proper subspaces of $V$, and $N$ is a subspace of $V$ such that $N\subseteq M_1\cup\cdots\cup M_k$, how can you tell $N\subseteq M_i$ for some $i$?  I was first trying to show it just in the case $N\subseteq M_1\cup M_2$, and hoping to extend it to finitely many $M_i$. If either of the $M_i$ contains the other, the claim follows, so I suppose neither $M_i$ contains the other.   In hopes of a contradiction, I suppose $N\not\subseteq M_1$ and $N\not\subseteq M_2$, Picking $x_1\in N\setminus M_1$ and $x_2\in N\setminus M_2$, I'd have $x_1+x_2\in N\subseteq M_1\cup M_2$. The only thing I could conclude was that actually $x_1,x_2\notin M_1$ and $x_1,x_2\notin M_2$, which seems like a dead end. What's the right way to do this? Calculate the area of an irregular cyclic convex polygon Layover in Europe to the US - allowed? Many people have asked questions on this site regarding the rules for layovers, with no definitive answers. The rules are constantly in flux and it is very difficult to get a clear answer from government websites. I would like to buy a plane ticket for my mother-in-law (who has a visa) from Ukraine to the US. Most available flights have at least one layover in Europe. My understanding is that EU countries (minus some special cases and US citizens) are currently banned from travel to the US. If we buy a ticket with a layover in, say, Switzerland, can there be a problem? Is this allowed under current rules?