turn off rendering in outliner view I'm interested in removing one object's geometry from another. I should be able to do this with a boolean modifier on the larger object (the "car") targeting another object (the "window"). I would then like the window not to show when rendering. I expected to find this option in the outliner view (based on ), but maybe this has changed in blender 2.80? In the outliner window, I see the "Hide in Viewport" option, but where is the option to hide it in the final render?
Where are the outliner toggles in 2.8? The selection toggle arrows in the Outliner disappeared. Is there some preference in the settings that has to be turned on?
I just saw now that a new build had rolled out in which the comment upvote and flag options are always visible. Previously, one had to mouse over the area in order to see them.  Why was this just changed? Is it to support touch screens? I'll admit, it was difficult to find the flag button using my touchscreen laptop running Microsoft Edge.  Cross-site duplicate on MSO: What should we do with old, very popular closed questions? Motivated by the deletion of     On one hand, I disagree with these closures, because:  These questions were (and still are) very popular They have some good information in them (or at least the first one does) and it feels like bad sportsmanship to remove good information from the internet, particularly when it was contributed by your fellow users  Number 2 in particular bothers me. A lot. On the other hand, I do agree that:  The community of 10k users should be able to decide which closed questions they wish to delete on their own without me intervening These questions should definitely remain closed, because they either have an absurd # of answers, or are duplicates (though some duplicates ). Which means they are eligible for deletion … forever. These questions wouldn't survive more than 10 minutes if asked today on Stack Overflow, so in some ways they are artifacts from the past, like dinosaurs in an era of mammals.  There is some related discussion in:   While I normally favor code solutions here, I am not sure I like the idea that once a question gets 400 upvotes it should become extremely difficult to delete. It feels like mob rule. Official policy on the blog: After installing Telegram Desktop on Ubuntu 16.04 by these commands:  sudo add-apt-repository ppa:atareao/telegram sudo apt-get update sudo apt-get install telegram   I get this error:  $ telegram telegram: Panic - Unable to find your tty (pts/2) in /var/run/utmp Where to place and how to read configuration resource files in servlet based application? iMessage doesn't say "delivered" or "not delivered" I sent an iMessage from my iPod touch to my friends iphone on Monday. It didn't say delivered or not delivered. It was just nothing. His phone is on (I know because my other friends are iMessaging him) and my friend said her messages don't say "delivered" at the bottom but they still talk. Mine always say delivered at the bottom. So is this an error, like it didn't go through, or is he just ignoring me? I have an iOS application - an alarm clock. I released it a little over a month ago. Initially it was paid, but when I saw there were virtually no purchases (from 0 to 4 a day), I changed it to be free for a period of time. When it become free, in less than 10 days it has been downloaded more than 30,000 times, and more importantly (I think) it moved much higher in the search results when the user searches "alarm" - it reached 20th place in the results. During that time I also received very good reviews in the AppStore. But today I changed it back to paid, and now when I search "alarm" my app is shown at the bottom of the 4TH PAGE. Given that one page contains 120 apps, this is very bad for me and I don't understand why this happened. Can anyone tell me how does AppStore ranks search results and is there a way to move my app higher in the rankings, please? :) Huge thanks! When I double-click on a script in Nautilus to run it, the script just opens in my text editor with no option to run it. Using Nautilus, how do I run executable text files and/or scripts? If a superconductor has zero resistance, does it have infinite amperage? Why doesn't LaTeX hyphenate words containing accents, e.g. "développement"? In this MWE, Latex doesn't hyphen the word "développement" and prefer to create a HUGE hbox.   \documentclass[a4paper,12pt,dvipsnames]{report} \usepackage[utf8]{inputenc} \usepackage[french]{babel} \usepackage{lmodern} \usepackage[a4paper, showframe]{geometry}  % Title Page \title{} \author{}   \begin{document} \section{Section}  \subsection{Subsection} \begin{description}  \item [Marhalie Toucse : ]  ingénieure de la plate-forme SORTATZ. Elle réalise   le développement du projet des tex écrit pour lipsum lagae legarc sgsi text text text text text text text text text.  \end{description} \end{document} I need to sync the Downloads folder to iCloud and then all the downloads should automatically sync with iCloud.   I tried to:   Create a new Downloads folder in the terminal of MacBook in the Documents directory. Tried to move the default Download folder to the newly created folder.   When I tried to use the mv command to move the default Downloads folder to the newly created Downloads folder in Documents in the terminal, it gives permission denied error.  I am trying to achieve what was suggested by user PhillipHolmes in the below thread, please. Thank you:    Thank you Maximum length of tags Possible Duplicate:        Is there a maximum limit on the length of tags? I noticed that the tag for SQL Server Management Studio is truncated to sql-server-management-stu.  It seems like this should be an easy thing to correct, if not to allow the creation of new longer tags, but to correct existing tags that have been truncated. Is this intended behavior (with plain XeTeX):  \input tikz \tikz\draw[dashed] node {$3T\over4$}; \bye   It renders as: !?  The fraction line is dashed, too! Is this intentional? Is there something I can do to change that? I know I can separate the fraction into its own node, but other than that? How can I prevent my Wi-Fi connection from glitching? Occassionally my Wi-Fi goes into this weird state where the phone "thinks" I'm connected, but the connection simply doesn't work.  Apps still seem to think I have connectivity, though they can't actually connect to anything, and the status bar still shows that I'm connected to Wi-Fi.  Turning Wi-Fi off and back on resolves the issue, and this occurs on multiple networks.  It seems sort of like whatever service manages the Wi-Fi connection gets killed occasionally.  It usually happens when I've been playing a game (that doesn't access the Internet) for a while or if my phone's been sitting for a while; it never happens while I'm actively using the connection.  I have my Wi-Fi timeout set to "Never" and the normal timeout isn't the issue since my phone regularly lasts several hours without the Wi-Fi dying.  Sometimes it dies a couple times in a day, and recently I went a week or so without it happening.  How can I fix this or prevent the service from being killed (if that's indeed what's happening)?  Shouldn't that be a system process that never gets killed?  (Does Android ever kill system processes other than when they crash?)  Could it be related to TouchWiz?  I have an SGS Vibrant running 2.2 (Bionix NextGen v1, but this has occurred on multiple ROMs). How to convert a one digit number to a two digit number I have a table that represents a timetable for bus, so each cell contains time of departure as hour followed by minute, both as two digits, but sometimes I don't have two digits,just one so I have to switch eg. from 12 4 -&gt; 12 04. How is it possible? Characterizing all mobius transformations from unit disk to itself. All answers to this problem involve the following process:   Pick a point $a$ in the unit disk such that $T(a)=0$. Then, $T(a^*)=\infty$ if $a^*$ is the symmetric point $a^*=\frac{1}{\bar{a}}$ of a. Clearly, such a transformation is of the form $T(z)=\frac{z-a}{1-z\bar{a}}$. We might as well place a constant C in there: $T(z)=C\frac{z-a}{1-z\bar{a}}$. Then, most answers magically jump to replacing $C=e^{i\theta}$ in the formula and claiming the job is done.  Why does a function built so that it sends a point inside the unit disk to zero, and its symmetric point to $\infty$, be the one that preserves the unit disk? What's the motivation behind approaching this problem in that manner? Could one proceed bruteforce: with $|\frac{az+b}{cz+d}|&lt;1$ and arrive at the same function? For some reason this seems to me like somebody pulled a rabbit out of a hat. Once given the formula, I understand why it does the job, but why would somebody think that choosing a point inside the disk and building a function that sends it to zero create a mobius transformation that does the job? Difference between "I'm home" vs "I'm at home" What's the difference between these two sentences? And which one's grammatically correct? Interpret geometrically the equation The function $f$ is convex when:  $$f(tx+ (1-t)x', ty + (1-t)y') ≤  tf(x,y) + (1-t)f(x', y')$$  for all numbers $0≤t≤1$ and all pairs of points $(x,y)$ and $(x',y')$ in the domain of $f$ (the whole of $R^2$)  How would I interpret this geometrically? Also what is an example of a convex function whose graph is not a plane? algebraic element over a field Let $F$ be a subfield of a field $K$, $a$ an element of $K$.  Prove that $a$ is algebraic over $F$ iff $F[a]=F(a)$.  For the first direction I've tried this: $a$ is algebraic over $F$ $\Rightarrow$ there is a polynomail $f$ over $F$ such that $f(a)=0$ $\Rightarrow$ $1,a,a^2,...,a^m$ are linearly dependent $\Rightarrow$ $\deg(F[a])&lt;m$.    but it contains $f(a)$ so they are equal..  is that correct? about the other direction.. i simply have no clue End Process vs End Process Tree What is the primary difference between the two? I always see this in the right click menu in windows. What happens if you end a process that is relied upon by another process? Is there a way to "kill" a process tree in linux or is the technology different? $A \oplus B = A \oplus C$ imply $B = C$? I don't quite yet understand how $\oplus$ (xor) works yet. I know that fundamentally in terms of truth tables it means only 1 value(p or q) can be true, but not both. But when it comes to solving problems with them or proving equalities I have no idea how to use $\oplus$.  For example: I'm trying to do a problem in which I have to prove or disprove with a counterexample whether or not $A \oplus B = A \oplus C$  implies $B = C$ is true.   I know that the venn diagram of $\oplus$ in this case includes the regions of A and B excluding the areas they overlap. And similarly it includes regions of A and C but not the areas they overlap. It would look something like this:    I feel the statement above would be true just by looking at the venn diagram since the area ABC is included in the $\oplus$, but I'm not sure if that's an adequate enough proof.  On the other hand, I could be completely wrong about my reasoning.    Also just for clarity's sake: Would $A\cup B = A \cup C$ and $A \cap B = A \cap C$ be proven in a similar way to show whether or not the conditions imply $B = C$? A counterexample/ proof of this would be appreciated as well.