Backstory I have a Radeon RX 570 and decided to try out Blender and 3D modelling in general. But after about 2 months I started getting some strange graphical bugs. I did manage to fix a couple of them but apparently blender doesn't like AMD Users. Then I got the wireframe issue. Problem Wireframe viewpoint comes up with this now  and I have yet to fix it. Also for Cycles, It can't even render on the viewpoint and if I dare to render it as an image, my whole PC freezes up.  What I've tried For the wireframe viewpoint strangely breaking I've tried...  Reinstalling Blender to an older version Reinstalling Blender and install a portable install Going through each setting Updated any possible driver and software  For Cycles I've tried...  Changing Cycles rendering device to each possible selection Reinstalling Blender again  Each solution I've tried hasn't fixed anything
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?
How to log out of a Stack Exchange site? How to explain the long lifetime of Rydberg atoms with Fermi golden rule? Disable ClearType (blurry fonts) in Internet Explorer 10 on windows 7 I've installed recently the latest Internet Explorer 10 and found out again what was my issue with the previous upgrade -- blurry fonts aka ClearType! Unfortunately there are some apps that use IE engine and thus all content there is blurry too :/  I know about  but unfortunately it seems that it is only for IE9. Is there any other solution to disable CT in IE10?  (and for clarity - I don't like CT, I've tried tweaking it but for me it's just blurry and case pain in my eyes no matter what I do... it's probably related to the low dpi of my monitor) How is it possible for the Root Name Servers to handle all DNS requests? I was reading about DNS some days ago and learned how the requests are processed. If you surf to www.example.com, then a request will go to the Root Name Servers to see who owns that .com address, then another request will go to another, more local, DNS server to see who owns the example.com address and so on.  How is it technically possible that the 13 Root Name Servers can handle all requests done by earth's billions of Internet users simultaneosly without beeing ddos:ed? Rewiring a switch controlled receptacle to a ceiling light Is there a single word to describe a solution that hasn't been optimized? I am working with some code, and I would like to describe the difference in its performance when it is optimized versus when it is not optimized. Unfortunately, I can't find a word to describe the code before it is optimized; "sub-optimal" might work, but I want to emphasize that this is the first iteration of the code. If "unoptimized" were a word, I would use that, but I haven't found any references to it in any dictionary.  Perhaps "raw" or "original"? Holomorphic self map on the open unit disc Does there exist a holomorphic function $f:\mathbb{D}\longrightarrow\mathbb{D}$ such that $f(1/2)=-1/2$ and $f'(1/4)=1$, where $\mathbb{D}=\{z:|z|&lt;1\}$? Searching for an equation to find the center of a sphere given 4 points, one finds that taking the determinant of the four (non-coplanar) points together with the variables $x$, $y$, and $z$ arranged like so:  $\left|\begin{array}{ccccc} x^2+y^2+z^2 &amp; x &amp; y&amp;z&amp;1\\ x_1^2 + y_1^2 + z_1^2 &amp; x_1 &amp; y_1 &amp; z_1 &amp; 1\\ x_2^2 + y_2^2 + z_2^2 &amp; x_2 &amp; y_2 &amp; z_2 &amp; 1\\ x_3^2 + y_3^2 + z_3^2 &amp; x_3 &amp; y_3 &amp; z_3 &amp; 1\\ x_4^2 + y_4^2 + z_4^2 &amp; x_4 &amp; y_4 &amp; z_4 &amp; 1\\ \end{array}\right| = 0$   yields the equation for the sphere.  Then one need only re-arrange terms into the more familiar form to find the center and radius.  This works fine.  My question is why.    This same approach also works for one or two dimensions.  I'm guessing it also works for finding hyperspheres in higher-dimensional spaces as long as you have a corresponding number of points.  But where did that determinant form come from?  Is there an intuitive meaning for what that relationship is saying? What does this notation mean? F|U If $F$ is a function and $U$ is a set, then what does $F|U$ mean?   In , the Inverse Function Theorem, the notation is used  Also, the file in stating the Lemma 2.14 states that $A \in U$, what does this mean? A set is an element of another set? What is $A$?  EDIT: Could someone kindly explain to me what is going on in Lemma 2.16. How does $\delta$ come into play? I dont see the role in the lemma. I also don't see why this proves one-to-one I've installed Ubuntu 16.04 on my Asus X555LJ.  But I don't know how to install drivers or how to enable function keys to controls screen backlight brightness. So this homework question is in the context of $\mathbb{R}$ only, and we are using Lebesgue measure.  The sum $A+B$ is defined to be $A+B=\{a+b|a\in A,b\in B\}$. The question is: If $m(A),m(B)&gt;0$, is it necessarily the case that $A+B$ contains an open interval?  This question is quite annoying, since it is quite difficult to calculate the sum even after you construct an example. However, it seems like no matter how I constructed a set of positive measure, there is a region that have a lot of point in it (yet this region is not necessarily dense). This suggest that when summed up, the holes should be covered up somehow and thus there would be an open interval, and so the answer is yes. Sorry for a rather vague idea, but I'm just not sure what to do with this problem. I have a feeling that somehow topology would have a way to capture precisely the idea above about a region with a lot of point.  (also, I think my middle school teacher mentioned this once, and I remembered the answer to be yes; but that is quite a while ago though)  Thank you. Any help will be appreciated.  EDIT2: So Camilo's answer is gone. I am looking at the convolution method right now. Because the convolution is continuous, if it is nonzero I am done. But then there is no guarantee that the convolution is nonzero even if $x\in A+B$. I still need help with this direction. Can someone give me some hints?  EDIT: Camilo's answer below essentially suggest the method of getting 2 open interval from regularity theorem, one cover a subset of A and one cover a subset of B such that the sum of the measure of these two subset is bigger than the measure of the larger interval. user99680's answer point to a wiki which show a method that essentially simply acquire an open set. After some thinking, I believe that these method simply can't be made to work. I constructed this example in the hope that it would convince you that it is the case:  Let $I_{0},I_{1},\ldots$ be disjoint open interval such that the distance between any 2 of them is nonzero and $m(I_{n})=9^{-n}$ and all of them are inside $[0,2]$. Let $J_{0},J_{1},\ldots$ be disjoint open interval such that the distance between any 2 of them is at least $3$ and $m(J_{3n})=m(J_{3n+1})=m(J_{3n+2})=\frac{m(I_{n})}{3}$. Let $O=\bigcup I_{n}$ and $H=\bigcup J_{n}$. Construct $A\subset O$ such that $m(A\bigcap I_{n})=\frac{3m(I_{n})}{4}$. Construct $B\subset H$ such that $m(B\bigcap J_{n})=\frac{3m(J_{n})}{4}$. Now if you apply Camilo's method, it is possible that your $I$ would be one of the $I_{n}$ above, and your $J$ is one of the $J_{m}$ above. No matter which $I_{n},J_{m}$ you have, the argument does not work. Rewrite $m=3q+r$ where $0\leq r\leq 2$. If $q&lt;n$ then $m(J_{m})&gt;m(I_{n})$ so the longer interval have length $\frac{m(I_{q})}{3}$; $m(A\bigcap I_{n})$ is at most $\frac{3m(I_{q+1})}{4}=\frac{m(I_{q})}{12}$ and $m(B\bigcap J_{m})=\frac{3m(J_{m})}{4}=\frac{m(I_{q})}{4}$. Thus $m(A\bigcap I_{n})+m(B\bigcap J_{m})$ is no larger than the larger interval. If $n\geq q$ then $m(I_{n})&gt;m(J_{m})$ so the larger interval is of length $m(I_{n})$; $m(A\bigcap I_{n})=\frac{3m(I_{n})}{4}$ and $m(B\bigcap J_{m})$ is at most $\frac{3m(J_{3n})}{4}=\frac{m(I_{n})}{4}$. Once again $m(A\bigcap I_{n})+m(B\bigcap J_{m})$ is no larger than the larger interval. Hence the Camilo's argument does not work. user99680's method does not work either, since you get $O$ and $H$ which is even bigger than any interval. I have Ubuntu server 12.04 installed, so I have no GUI. When I do the command ifconfig, I cannot find my internal IP address. It says inet addr: 127.0.0.1. Here is the output of ifconfig -a: eth0   link encap:Ethernet  HWaddr 00:06:4f:4a:66:f0     BROADCAST MULTICAST  MTU:1500  Metric:1     RX packets:0 errors:0 dropped:0 overruns:0 frame:0     TX packets:0 errors:0 dropped:0 overruns:0 carrier:0     collisions:0 txqueuelen:1000     RX bytes:0 (0.0 B)  TX bytes:0 (0.0 B)  eth1   link encap:Ethernet  HWaddr 00:16:ec:05:c8:9c      BROADCAST MULTICAST  MTU:1500  Metric:1     RX packets:0 errors:0 dropped:0 overruns:0 frame:0     TX packets:0 errors:0 dropped:0 overruns:0 carrier:0     collisions:0 txqueuelen:1000     RX bytes:0 (0.0 B)  TX bytes:0 (0.0 B)  lo     Link encap:Local Loopback     inet addr 127.0.0.1  Mask:255.0.0.0     inet6 addr:  ::1/128 Scope:Host     UP LOOPBACK RUNNING MTU:65536  Metric:1     RX packets:1800 errors:0 dropped:0 overruns:0 frame:0     Tx packets:1800 errors:0 dropped:0 overruns:0 carrier:0     collisions:0 txqueuelen:0     RX bytes:143896 (143.b KB)  TX bytes:143896 (143.8 KB)  here are the contents of etc/network/interfaces: # The loopback network interface auto lo iface lo inet loopback  # The primary network interface auto eth0 iface eth0 inet dhcp  If someone could edit this for me, the contents of etc/network/interfaces should be on separate lines. The output of host askubuntu.com was: ;; connection timed out; no servers could be reached.  I set up owncloud and webmin a few months ago and was using them for a month with no problems. I think the power went off one day 2 months ago and I never turned the server back on until yesterday. I haven't done anything that would have affected the internet setup So i'm not sure why it doesn't work anymore. As far as my network topology goes, I have a pci-e network card for the pc. The ethernet line goes from the network card to a switch, and then to a modem/router from there. I've believed for a long time that every Lie group can be decomposed as the semidirect product of a connected Lie group and a discrete Lie group. However, in , it is mentioned that this is not true, and that there exist obstructions to this decomposition. Unfortunately, the level of complexity in the answers goes a bit over my head.  This type of decomposition is also addressed in .  With this loss of faith in my intuition, I have three questions:   Is it true that Lie groups do not necessarily decompose into the semidirect product of a connected group and a discrete group? If this decomposition doesn't always work, then is there a particularly nice argument as to why it doesn't? If this decomposition doesn't always work, then is there a nice example I can use to convince myself that my intuition is wrong?   Edit: As much as I love Qiaochu Yuan's answer, which is thorough and (as usual) a role model among Stack Exchange answers, this is not what I'm looking for. I'm aware of the general extension problem, and that not every short exact sequence splits, but my intuition has led me to think that each Lie group can be decomposed this way. I will certainly play around with his outline for a construction, but I would much prefer a more concrete answer. Once again, thank you!  Edit: As of yet, playing with the construction suggested by Qiaochu hasn't yielded anything interesting. This question is still open to answers! $\ker \phi = (x_1-a_1, ..., x_n-a_n)$ for a ring homomorphism $\phi: R[x_1, ..., x_n] \to R$ Let $R$ be a commutative ring, $a_1, ..., a_n$ its elements and  $\phi: R[x_1, ..., x_n] \to R$  defined by  $ \phi(f(x_1, ..., x_n)) = f(a_1, ... ,a_n)$  a ring homomorphism.  Prove: $\ker \phi = (x_1-a_1, ..., x_n-a_n)$  It is obvious that $(x_1 - a_1, ..., x_n -a_n) \subseteq \ker \phi$. I'm not sure how to prove the converse.  At this point I don't know any division algorithms for multivariable polynomials, only for the ones in $R[x]$(and the book from where I taken the exercise doesn't assume the reader to know something beyond basics of rings and ideals, and the division algorithm for $R[x]$). Though I know this could be solved for $i = 1$ by dividing by $x-a$:   Let $f(x) \in \ker \phi$, divide by $x-a: f(x) = q(x)(x-a) + r, f(a) = r = 0$, so $f(x) = q(x)(x-a) \in (x-a)$. When may adjectives precede determiners? (E.g. too difficult a task) The expression too difficult a task sounds a little pompous, but it doesn't sound ungrammatical. According to my folk-beliefs about English grammar, determiners precede adjectives. However, the expression too difficult a task seems to controvert that belief.  Additionally, the expression too difficult of a task seems grammatical too.    Under what conditions may determiners precede adjectives? Are there conditions under which too difficult of a task would be a more grammatical expression than too difficult a task would be?   (This question does not ask what the question “How big of a problem” vs. “how big a problem” asks. That question does not pertain to determiners.) Automorphisms of the complex plane How can it be shown that $$Aut(\mathbb{C})=\{f|f(z)=az+b,a\neq 0\},$$ where an automorphism of $\mathbb{C}$ is defined as a bijective entire function with entire inverse?  If $f$ is of the form $f(z)=az+b$, with $a\neq 0$, then obviously $f$ is in $Aut(\mathbb{C})$. How can I prove the converse? How do you copy the contents of an array to a std::vector in C++ without looping? Where can I find the source code for the Ubuntu Kernel? I'd like to build my own copy of the Ubuntu kernel. Where can I get the source code? Hi currently Nagios sends notifications from nagios@server.domain.com, How do I go about configuring this? Thanks We are in the early stages of development of the official Stack Exchange Android app and we need your help with testing. To set a few expectations: there will be bugs, crashes, and missing features, all of which will be fixed in .    We are initially targeting Android devices running  (Android 4.1 or higher) starting with phones (up to ~5 inch) and eventually moving our way to tablets.    To register for alpha testing .  After you register we will send you an invite to a private Google community where you can find a link to download our app directly from the  store. The invitations will be sent out in waves, so if you don’t get an invite immediately, be patient.  Crash reports will be automatically reported back to us using . Bugs or feature requests should be reported here on meta and tagged . Please do not use the Google community to report bugs.  As always, we welcome your feedback and are committed to building the best mobile app possible for browsing the Stack Exchange sites.