I have a Lenovo Thinkpad X1 yoga, which has an OLED display. This means the brightness isn't set in the usual way, but luckily you can get around that with scripts using xrandr --brightness. Specifically, I have a  set up that listens for brightness changes and then uses xrandr to implement them. Unfortunately, this does not take into account color temperature, and so does not pick up changes from Gnome Night Light. Where does Night Light store its current settings, and how does one get access to them? Or better yet, is there a way to convince Night Light to use xrandr to set brightness?  NB: This is not a duplicate of . OLED displays do not respond to the usual way of setting brightness. So Night Light by itself only changes the color temperature, but not the brightness. On the other hand, I can set the brightness with xrandr, but this overrides the color temperature. Can I get Night Light to use xrandr to set brightness instead of the standard method? Or, failing that, how do I read out Night Light's settings to set the correct temperature and brightness myself?
I tried to use GNOME Nightlight in Ubuntu 17.04. The settings only allow me to turn it on and adjust the time. Is it possible to adjust the intensity or temperature of the hue? Currently, it's a bit too much.
Difference between sentences using 'was' and 'being'? What is the meaning of below sentences using was and being?     He committed suicide after he was severely beaten up by four teachers. He committed suicide after being severely beaten up by four teachers. How to fix sculpting ridges? I ran into a problem I have never seen before and that is when I try to sculpt on a mesh, it sculpts like this:  Instead of this: I am using dyntopo with these settings:   I feel like theres a very obvious solution to this, but I have no idea what it is. I can't add a book that I've written that's not on Amazon Is there any chance that there will be support to my Careers profile for adding books that aren't available to purchase through Amazon.com? My book was published through Smashing Magazine and it isn't available there. I have a document I am trying to copy to learn TeX. Here is what I have encountered: How can I have the "or" in the equation. This is what I have right now:  \[ f = ma; \]  But $a$ is the change in velocity,  \[ f = m \frac{dv}{dt};\] \[ f = m  \frac{d^2y}{dt^2};\]   Edit: Trying out Stefan's answer Some class features, monster traits, and feats allow you to take a −5 to your attack roll in order to gain a +10 to your damage roll. At first glance this looks really good, but when will it deal more damage than not using it on average?  For instance if you have a total attack bonus of +8 and you deal 1d12 damage. You have to roll a 7 to hit an AC of 15. If you take the −5 that makes it where you have to roll a 12. That's going to reduce your average damage by a lot. Cursor hitting "border" between differently sized monitors Are smaller p-values more convincing? How do I allow non-root access to /ttyUSB0? Since I updated from Ubuntu 10.04 (Lucid Lynx) to Ubuntu 12.04 (Precise Pangolin), I am unable to access the device connected in /ttyUSB0 if I don't have root access.  To be more specific, I use a USB-to-serial com port converter to access and configure Cisco equipment with software called . It happens that, since I upgraded to 12.04, PuTTY only sees the connected device when I run it as gksudo putty. If not, it says "unable to connect to port /ttyUSB0".  I have managed to change the permissions via chmod 666, and it works, but if I disconnect the USB device and reconnect, it goes back to root only. How do I permanently allow non-root access? I have a clue it might be via  rules, but I have no idea how to do it.  Obs.: Manufacturer and drivers are "PROLIFIC" When proving $\bar x$ and $S^2$ are independent in my noted it says that "functions of independent quantities are independent ". Can someone tell me how functions of independent quantities are independent happen?  Also let $X_1,X_2,X_3,\dots,X_n$ be a random sample.And suppose we want to estimate parameter $\theta$ using $T(x)$ as an estimator. In this I have a expression as $E\{ [  T(X)-E(T(X))][E(T(x)-\theta)]\}$.  I want to know if I can say that in $[T(X)-E(T(X))]$ since $E(T(X)$ (expected value is a one particular fixed value) and $T(X)$ are independent of one another. Also the expression $[E(T(x)-\theta)]$ is independent of $X$.  Hence the two expressions $[T(X)-E(T(X))]$ and $[E(T(x)-\theta)]$  are independent of one another. So that I can use if $a,b$ are independent $E(ab)=E(a)E(b)$ form What is the difference between the following commands:  su sudo -s sudo -i sudo bash   I know for su I need to know the root password, and for sudo I have to be in the sudoers file, but once executed what is difference?  I know there is a difference between su and sudo -s because my home directory is /root after I execute su, but my home directory is still /home/myname after sudo -s. But I suspect this is just a symptom of an underlying difference that I'm missing. I have a Samsung Galaxy S5 duos with stock Android Marshmallow, and I'm considering encrypting my SD card to prevent access to its contents in case I lose my phone. The main reason of not doing this is that I'm not sure if there's a way to mount an encrypted SD card on Windows (or other desktop OS's)   When I encrypt the SD card on my Samsung Galaxy S5, is it using a known implementation of encryption, or is it something that only works on Android (or perhaps Samsung) phones?   I do not want to use USB connection to my phone to access my SD card, it's terrible. Mounting the SD card is much easier and faster and less annoying.   EDIT: just to clarify, I am NOT talking about adoptable storage. I want a portable encrypted SD card that can both be read on PCs and Android phones (i.e. I want to be able to remove the SD card from my phone, plug it into my PC, provide a password to decrypt and read/write it like any other SD card) Query to find GUID's "not in" null containing subquery returns no results I want to hit enemy's axe to break it. How this is properly played in D&amp;D 5e? And what about shield or armor? Can I attack these items to break it? I want to block Facebook (the Facebook site) for concentrating purposes, but I don't want to break the share or like buttons on other pages (like a good article or a nice clip on YouTube I want to like or share). How can I do that? I use both Windows and Linux. Play a Sound with Python I read this book about 20-30 years ago. It's about a group of teenagers who wake up in a huge room where each is on a separate platform. As the book progresses, they go unconscious and the platforms change, move or become connected. The children are punished or rewarded through a variety of methods including food and pain. The goal is to get them to either hurt each other, or see if they band together, I can't quite remember. Please tell me if this sounds familiar. Is there a fairly simple way to divide a circle into equal parts using Adobe Illustrator CC? Let $a,b,c$ are non-negative numbers, such that $a+b+c = 3$.  Prove that $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$    Here's my idea:  $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(ab + bc + ca)$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - 2(ab + bc + ca) \ge 0$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - ((a+b+c)^2 - (a^2 + b^2 + c^2) \ge 0$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) - (a+b+c)^2 \ge 0$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$  And I'm stuck here.  I need to prove that:   $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$ or  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a+b+c)$, because $a+b+c = 3$  In the first case using Cauchy-Schwarz Inequality I prove that:  $(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$  $3(a^2 + b^2 + c^2) \ge (a+b+c)^2$  Now I need to prove that:   $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a^2 + b^2 + c^2)$   $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a^2 + b^2 + c^2)$   $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge a^2 + b^2 + c^2$   I need I don't know how to continue.  In the second case I tried proving:  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$ and  $a^2 + b^2 + c^2 \ge a+b+c$  Using Cauchy-Schwarz Inequality I proved:  $(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$  $(a^2 + b^2 + c^2)(a+b+c) \ge (a+b+c)^2$  $a^2 + b^2 + c^2 \ge a+b+c$  But I can't find a way to prove that $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$  So please help me with this problem.  P.S   My initial idea, which is proving:  $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$  maybe isn't the right way to prove this inequality. Trying to find a TV show about an event that leaves everyone dead apart from people who died at the moment of the event? This was a TV show aired in the UK probably sometime between 1990 and 1993, if memory serves it was only a one-off. The plot starts with the protagonist waking up in bed and being disappointed. He eventually goes outside and everyone is dead, whatever they were doing at the time they died doing it (driving, working, etc.) The protagonist is in a state of panic running around town (I seem to remember it being set in the UK) and he discovers that he's is 'alive' because he'd attempted suicide and died at the exact moment this mysterious event happened. I think he encounters others that 'died' at the time of the event and that's when he realises. I think the protagonist was a bald, white man, probably mid-thirties. Sorry for any vagaries, I was VERY young when it was on. Until recently I assumed it was The Quiet Earth but I've checked and it definitely isn't. It was in colour and was contemporary at the time so it would have been made around 1990. Every Artinian ring is isomorphic to a direct product of Artinian local rings Proposition. Let $R$ be commutative ring with $1_R$. We assume that $R$ is an Artinian ring and $M_1,\dots,M_n$ its maximal ideals. Then         $R/\mathrm{Jac}(R)\cong (R/M_1)\times \dotsb \times (R/M_n)$.   The ring $R$ is isomorphic to the direct product of a finite number of Artinian local rings.      Proof. 1. It's obvious that $M_i+M_j=R,\ \forall 1\leq i \neq j \leq n$. So, from Chinese Remainder Theorem, we have  $$R/\bigcap_{i=1}^{n}M_i = R/\mathrm{Jac}(R) \cong (R/M_1)\times \dotsb \times (R/M_n),$$ as we wanted.    Since $R$ is an Artinian ring, we have $\mathrm{Jac}(R)^m=\{0_R\}$, for some $m\in \Bbb Z^+$. But,  $$\{0_R\}\subseteq M_1^m\dotsb M_n^m=(M_1\dotsb M_n)^m\subseteq (\bigcap_{i=1}^{n}M_i=\mathrm{Jac}(R))^m=\{0_R\}.$$ So, if we apply CRT we will take  $$R\cong R/\{0_R\}\cong R/M_1^m\dotsb M_n^m\cong (R/M_1^m)\times \dots \times (R/M_n^m).$$   Questions.   1) Are these thoughts complete and correct?  2) Why $R/M_i^m$ are artinian local rings?  3) Could you please elaborate some examples as an application?