When I add an object, it is invisible. When I change to evee, it turns black, and when I move around the view, it takes away from the black and makes it grey. Then when I turn to cycles it turns grey like in solid view. I've already tried rolling back my drivers, and it doesn't effect anything. I am using Blender 2.81. Everything was normal for a while, then all of the sudden I added an object and it was invisible. When I restarted my computer, it did the same thing but only when I was close. The origin and outline are still there. My clip start was set very low, but even when I change it it didn't change anything. One time when this was happening, I deleted everything, then added a new cube. It was displayed as one of the old cubes, but the outline was not in the right place. When I add more objects, some of them disappear and others display the wrong shape. This only happens in object mode, and going into edit mode fixes it. This only happens every so often, and seems to just magically be fixed.
I don't know why but now, when i go in material preview or rendered (in solid mode it works fine), my object is entirely gray, or invisible idk, because it has the same color of the background. I've been working for days on this project without having this issue, it suddenly showed up. I tried opening other old projects and suddenly they all do the same so i guess it is some blender program setting that applies for all the projects, but idk how it happened cuz i didn't change anything :/, anyone got any tips?
Ways to evaluate $\int \sec \theta \, \mathrm d \theta$ The standard approach for showing $\int \sec \theta \, \mathrm d \theta = \ln|\sec \theta + \tan \theta| + C$ is to multiply by $\dfrac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta}$ and then do a substitution with $u = \sec \theta + \tan \theta$. I like the fact that this trick leads to a fast and clean derivation, but I also find it unsatisfying: It's not very intuitive, nor does it seem to have applicability to any integration problem other than $\int \csc \theta \,\mathrm d \theta$.  Does anyone know of another way to evaluate $\int \sec \theta \, \mathrm d \theta$? I'm trying to typeset some python code, and I would like integer literals to be highlighted. I've come up, after about two hours of fighting with the documentation and various other questions here and on SO, with the following:  \lstset{ %     language=Python,     otherkeywords={1, 2, 3, 4, 5, 6, 7, 8, 9, 0},     morekeywords=[2]{1, 2, 3, 4, 5, 6, 7, 8, 9, 0},     keywordstyle=[2]{\color{orange}},     keywordstyle=\color{blue}\bfseries,     stringstyle=\color{red}, }   Unfortunately, that doesn't work. Numbers found in strings are also highlighted, which is rather annoying, and not only are they highlighted: they are highlighted in blue. Numerals in regular program text, on the other hand, are rightly highlighted in red.  Help?  EDIT: Thanks Peter Grill for your very nice solution. Is there a way to not break string colouring, though? When I use you code, the stringstyle=\color{red} doesn't seem to work anymore (strings are just plain black). Similarly, comments can be enclosed in triple quotes in python, and if I adapt your quote to detect """ blocks, then the corresponding  comments are not highlighted anymore. Norm and unit sphere How do you reopen a closed question? Can I clear an area of my prison for everyone BUT max security prisoners? Having problems with hair transparency using cycles. I ported over a model from daz studio (have done this a few times before with no issues).   Using a simple mix shader, transparent top, diffuse bottom, black/white texture in the fac.  Heres the blend  with images packed. ESM is available for 12.04, which enables Advantage customers to keep receiving kernel updates and updates for many user space packages. In our case it's very important to know if we'll be able tot use Ubuntu 14.04 safely after April '19, because it is not possible to upgrade our application and migrate it to say, 16.04. Wielding two weapons without "Two-Weapon Fighting" or "Dual Wielder"? We have a Barbarian on our team that very much likes to switch up his weapons; he goes from a Dwarven Hammer, to a pair of Hand-Axes, to occasionally just going bare-fisted. No issues there, he's fairly adept with all 3.  However, I have only just recently picked up on the way that he rolls his attacks for the two Hand-Axes: He rolls the first attack, adds his modifiers etc. Then he rolls the second attack, this time without modifiers, which is all well and good. But I'm not sure if this action uses his bonus action or not, or is considered a separate action (though, being a level 4 Barbarian, he only has one action per turn), or is merely part of the same action.  If we have a look at the Two-Weapon Fighting feature from the Fighter's fighting styles:     Two-Weapon Fighting   When you engage in two-weapon fighting, you can add your ability modifier to the damage of the second attack. (PHB5e p.72)   Similarly, the Dual Wielder feat also uses the Two-Weapon Fighting feat:     Dual Wielder   You master fighting with two weapons, gaining the following benefits:      ...         You can use two-weapon fighting even when the one-handed melee weapons you are wielding aren’t light. (PHB5e p.164, emphasis mine)      So; yes, this gives a bonus to the second weapon attack, but having asked a  to clarify which action this secondary weapon attack belongs to, it was determined that this uses your bonus action;     [...] the second attack requires a bonus action   So after that long winded explanation, here is the question: Without any additional feats, can you wield and use two weapons at once, and if so, how many actions does it require to make a full attack with both weapons in a turn? Is it possible to check where an alias was defined? The answer is, according to the book: $(m-1)!\cdot \binom{m+r-1}{r}$ I get why this is true. You arrange $m$ people in their seats in $(m-1)!$ ways and then you put $r$ empty spots inbetween them.I was thinking - choose $m$ seats out of $m+r$ to put the people on => $\binom{m+r}{m}$ Arrange them once you've chosen the seats $\to (m-1)!$ So in total: $\binom{m+r}{m} \cdot (m-1)!$ This is obviously not true. What's wrong with the way I'm thinking? Euclidean geometry in non-inertial frame What's the difference between the work function and ionisation energy? For a submerged object, buoyant force ($F_b$) is defined as:  $$F_b = V_{\text{submerged}} \times \rho \text{ (density)} \times g \text{ (gravitational constant)}$$   Conceptually, the buoyant force equation says that buoyant force exerted is equal to the weight of the volume of water a given object displaced. Why? I went online to   and found the following explanation, but it seems like non-sequitur to me:      Explanation: When an object is removed, the volume that the object occupied will fill with fluid. This volume of fluid must be supported by the pressure of the surrounding liquid since a fluid can not support itself. When no object is present, the net upward force on this volume of fluid must equal to its weight, i.e. the weight of the fluid displaced. When the object is present, this same upward force will act on the object. I want to show without using Sylow theorem that     Any group of order $2p$ has a subgroup of order $p~$($p$ being a prime)   My attempt: Since $|G|=2p,$ even $\exists~a\neq e\in G$ such that $a^{-1}=a.$ Then $H=\{e,a\}\leq G.$ Let $S$ be the set of all left cosets of $H$ in $G.$ Now $S$ forms a group w.r.t. the composition $g_1Hg_2H=(g_1g_2)H.$ The mapping $\sigma:S\to G:gH\mapsto g$ is an $1-1$ homomorphism. Then $S\simeq\sigma(S)\leq G.$ Thus $G$ has a subgroup of order $p.$  Please tell me whether I'm right!     I'm skecptical as I didn't use that $p$ is prime. Identify duplicate lines in a file without deleting them? I have my references as a text file with a long list of entries and each has two (or more) fields.   The first column is the reference's url; the second column is the title which may vary a bit depending on how the entry was made. Same for third field which may or may not be present.  I want to identify but not remove entries that have the first field (reference url) identical. I know about sort -k1,1 -u but that will automatically (non-interactively) remove all but the first hit. Is there a way to just let me know so I can choose which to retain?  In the extract below of three lines that have the same first field (http://unix.stackexchange.com/questions/49569/), I would like to keep line 2 because it has additional tags (sort, CLI) and delete lines #1 and #3:  http://unix.stackexchange.com/questions/49569/  unique-lines-based-on-the-first-field http://unix.stackexchange.com/questions/49569/  Unique lines based on the first field   sort, CLI http://unix.stackexchange.com/questions/49569/  Unique lines based on the first field   Is there a program to help identify such "duplicates"? Then, I can manually clean up by personally deleting lines #1 and #3? How can I easily clone my entire system? I am running an Ubuntu 14.04 LTS and I experienced some issues with my hard drive. I bought a new HDD of the same size (1TB), and I intend to move all my data (softs, work, pics, etc.) on the new HDD.  What is the easiest way to do that? Does $ \int_0^{\infty}\frac{\sin x}{x}dx $ have an improper Riemann integral or a Lebesgue integral? In  wikipedia article for improper integrals,  $$ \int_0^{\infty}\frac{\sin x}{x}dx $$ is given as an example for the integrals that have an improper Riemann integral but do not have a (proper) Lebesgue integral. Here are my questions:   Why does this one have an improper Riemann integral? (I don't see why $\int_0^a\frac{\sin x}{x}dx$ and $\int_a^{\infty}\frac{\sin x}{x}dx$ converge.) Why doesn't this integral have a Lebesgue integral? Is it because that $\frac{\sin x}{x}$ is unbounded on $(0,\infty)$ and Lebesgue integral doesn't deal with unbounded functions? CentOS 7 DVD ISO vs Everything ISO different Proving that an ideal in a PID is maximal if and only if it is generated by an irreducible I have been running Ubuntu 18.04 for a couple of weeks. I upgraded from 16.04 and all went smoothly.   I am not sure what I changed, but this morning I noticed that the "Settings" Icon that launches gnome-control-center is missing from "Show Applications" and the applications menu as well. When I searched from "Show Applications" for gnome-control-center or just settings, nothing showed up.   So, I figured I must have accidentally removed it and figured a reinstall of the package would resolve the issue. I have searched and found several postings where this worked. I tried:  sudo apt install gnome-control-center   And then rebooted - which reinstalled the gnome-control-center and I could then launch it from the command line. However, there was still no Settings Icon in any of the menus.   I also tried many other suggestions including:  sudo apt-get install --reinstall gnome-control-center sudo apt-get install ubuntu-desktop sudo apt install ubuntu-desktop --reinstall sudo apt-get install unity-control-center   However, none of these has been successful in restoring the icon into "Show Applications" or any applications menu.   Anyone have any suggestions on something I haven't tried? Would be much appreciated.  Or, is there a way to just manually recreate the icon?   Output of locate gnome-control-center.desktop:  /usr/share/applications/gnome-control-center.desktop  /home/dccrens/.local/share/applications/gnome-control-center.desktop /home/dccrens/.gnome/apps/gnome-control-center.desktop   I am running Ubuntu 18.04 desktop on kernel 4.15.0-45-generic.