I'm practising how to model my Kitchen radio. On the top of the model it has many holes, I've used subdivision modifier on my mesh. My first attempt was to deactivate the submodifier and make an array of holes, but once I turned on the subdivision modifier again, it was extremely unclean mesh.   So what would be the best way to achieve the holes on top of the mesh?  Thank you.
How would I be able to create a grill of sorts found on the front of some computer cases such as this one
Misleading chat usercard on SE 2.0 chat Trailing characters stripped from titles bypass length minimum Would a very small creature inside a sealed container carried by a character take damage from AOE spells? Assume character A is in the AOE of a damaging spell.  Character A is carrying a closed container inside his backpack, alongside several sheets of parchment.  Damaging spell description specifically states that items carried by a creature are not damaged even on a failed save. Unattended objects are damaged, though.  Inside said container there is a small creature. Creature has no problem living inside container for a long amount of time.  Does this creature take damage from the spell? If it does not, what mechanic prevents damage? pattern matching in vim Note: The legacy Skype for Linux application (v4.*) is now discontinued and was removed from Canonical’s partner repositories, so this question and its answers are obsolete. Please refer to  of our .    I am receiving the following message while I am trying to install Skype:  nidhin@ubuntu:~$ sudo apt-get update nidhin@ubuntu:~$ sudo apt-get install skype Reading package lists... Done Building dependency tree        Reading state information... Done Some packages could not be installed. This may mean that you have requested an impossible situation or if you are using the unstable distribution that some required packages have not yet been created or been moved out of Incoming. The following information may help to resolve the situation:  The following packages have unmet dependencies:  skype : Depends: skype-bin but it is not installable E: Unable to correct problems, you have held broken packages   How can I fix this? Closure of a Ball is Contained in Closed Ball (usual metric) Let $\overline{B_d (x, r)}$ be the closure of an open ball, and let $\hat B_d (x, r)$ be a closed ball. Show that in any metric space $(X, d)$, $\overline{B_d (x, r)} \subseteq \hat B_d (x, r)$ for all $x \in X$ and $r &gt; 0$.    I used the definition of closure: $\overline{B}$ = $B \bigcup B'$. If $x \in \overline{B}$, then $x \in B$ or $x \in \overline{B}$, or both. If $x \in B$, then clearly $x \in \hat B$ since $B \subseteq \hat B$. But, if $x \in B'$, then $x$ is an accumulation point. Since $\hat B$ is closed, it contains all accumulation points of $B$. Thus, $x \in \hat B$.  Is this approach correct, or at least in the right direction? Should I be using the closure definition, or try another one? Thank you. I just did a clean installation of Ubuntu 18.04.1 on a tablet which was previously running Ubuntu 17.10. Back then just after the installation the screen was rotated, but I was able to get it to work (I think it was running on Wayland).  Now with 18.04.1 the situation is similar: when Ubuntu first boots the login screen is upside-down, but that's not a problem. This time, after login, the screen is correctly oriented but the mouse cursor is upside down and mouse movements are inverted with respect to both axes. The problem is that when I try to click on something the click is registered on the position on the other side of the screen (symmetrically with respect to the center).   It almost seems like the screen (desktop, application, etc.) is correctly displayed, with also the mouse positioned correctly and moving as it should, but then the mouse cursor which is shown to the user is inverted, like if the screen had been inverted just before drawing the cursor.  I already tried the answers I could find, but none worked for me. Anyone have an idea of what's going on?  Thanks Detach (move) subdirectory into separate Git repository I have a  repository which contains a number of subdirectories. Now I have found that one of the subdirectories is unrelated to the other and should be detached to a separate repository.  How can I do this while keeping the history of the files within the subdirectory?  I guess I could make a clone and remove the unwanted parts of each clone, but I suppose this would give me the complete tree when checking out an older revision etc. This might be acceptable, but I would prefer to be able to pretend that the two repositories doesn't have a shared history.  Just to make it clear, I have the following structure:  XYZ/     .git/     XY1/     ABC/     XY2/   But I would like this instead:  XYZ/     .git/     XY1/     XY2/ ABC/     .git/     ABC/ Frobenius Inequality Rank I was looking for an answer for this problem in terms of matrices, but I really don't know how to prove this result. The proposition says that:  Let $A\in M_{m\times k}(\mathbb{C})$, $B\in M_{k\times p}(\mathbb{C})$ and $C\in M_{p\times n}(\mathbb{C})$, then $\textrm{rank}(AB)+\textrm{rank}(BC)\leq \textrm{rank}(B)+\textrm{rank}(ABC)$ Is there any matrix $2\times 2$ such that $A\neq I$ but $ A^3=I$? Is there any $2 \times 2$ matrix $A \neq I$ such that $A^3=I$?  In my opinion: no. Thank you very much. I’m a Filipino citizen. My flight goes from Vancouver to Manila via Tokyo Haneda. I will arrive in Japan at 6pm then depart at 9am the next day. I'm not planning to go out of the airport during my layover, so I don't see a need to have a transit visa. However, I called the Consulate-General of Japan here in Canada, and they asked me to get a transit visa.  Then I called the Ministry of Foreign Affairs in Japan and asked them the same question.. at first, they told me I don’t need a visa as long as I stay inside the airport, but when I told them that the Consulate-General of Japan in Canada asked me to get a visa, they also asked me to get one.  I'm so confused. I would like to know your knowledge or experience regarding this. I need to find a bottleneck and need to accurately as possible measure time.  Is the following code snippet the best way to measure the performance?  DateTime startTime = DateTime.Now;  // Some execution process  DateTime endTime = DateTime.Now; TimeSpan totalTimeTaken = endTime.Subtract(startTime); Which of those functions are PRFs? How to loop the particle simulation indefinitely? Orthogonal complement examples What damage will powering down instead of shutting down do? I am reading Velleman's "How to Prove it", and in one example he proves  $\neg \exists x P(x)  \implies \forall x \neg P(x)$    in the following way:    Suppose $\neg \exists x P(x)$. Let x be arbitrary. Suppose $P(x)$. Since we have a specific $x$ for which P(x) is true, it follows that $\exists x P(x)$, which is a contradiction. Therefore $\neg P(x)$. Since $x$ was arbitrary, we can conclude that $\forall x \neg P(x)$, so $\neg \exists x P(x)  \implies \forall x \neg P(x)$    What makes me confused is the line in bold text, that is "Suppose $P(x)$". The reason is that he writes that he is doing proof by contradiction, but $P(x) \not\iff \neg\forall x \neg P(x)$. He is not treating all possible cases where $\forall x \neg P(x)$ is false by assuming $P(x)$, does he? So how can this be a valid proof by contradiction?    Edit: I deleted my first example since it was clearly not representative for the problem.  Edit 2: So the thing that I am opposing I guess is that $\neg \forall x \neg P(x)$ is not $P(x)$. And from what I have undestood, a proof by contradiction should use the negation of the consequent which I do not think that Velleman does. Can someone explain Cayley's Theorem step by step? This is from Fraleigh's First Course in Abstract Algebra (page 82, Theorem 8.16) and I keep having hard time understanding its proof. I understand only until they mention the map $\lambda_x (g) = xg$. Can someone explain this proof step by step? Thank you so much! Here is the proof:  Let $G$ be a group. We show that $G$ is isomorphic to a subgroup of $S_G$. Define a one-to-one function $\phi: G \to S_G$ such that $\phi(xy)=\phi(x) \phi(y)$ for all $x,y \in G$. For $x \in G$, let $\lambda_x: G \to G$ be defined by $\lambda_x (g) = xg$ for all $g \in G$. The equation $\lambda_x (x^{-1} c) = x(x^{-1} c) = c$ for all $c \in G$ shows that $\lambda_x$ maps $G$ onto $G$. If $\lambda_x (a) = \lambda_x (b)$, then $xa=xb$ so $a=b$ by cancellation. Thus $\lambda_x$ is also one to one, and is a permutation of $G$. We now define $\phi: G \to S_G$ by defining $\phi(x) = \lambda_x$ for all $x \in G$. To show that $\phi$ is one to one, suppose that $\phi(x) = \phi(y)$. Then $\lambda_x = \lambda_y$ as functions mapping $G$ into $G$. In particular $\lambda_x (e) = \lambda_y (e)$, so $xe=ye$ and $x=y$. Thus $\phi$ is one to one. It only remains to show that $\phi(xy) = \phi(x) \phi(y)$, that is, that $\phi_{xy} = \lambda_x \lambda_y$. Now for any $g \in G$, we have $\lambda_{xy} (g) = (xy)g$. Permutation multiplication is function composition, so $(\lambda_x \lambda_y)(g) = \lambda_x (\lambda_y (g)) = \lambda_x (yg) = x(yg)$. Thus by associativity, $\lambda_{xy} = \lambda_x \lambda_y$. Is there any way to increase the drop rate of items? Some items in Terraria drop only very rarely from certain enemies. It's a pain having to farm so many enemies to try to get these rare items. Is there any way to make them drop more often? Sigma algebra and algebra difference An algebra is a collection of subsets closed under finite unions and intersections. A sigma algebra is a collection closed under countable unions and intersections.   Whats the difference between finite and countable unions and intersections? Does "countable" mean it implies there can be infinitely many unions and intersections?    Secondly, I was reading a definition  For an algebra on a set: By De Morgan's law, $A \cap B = (A^c \cup B^c)^c$, thus an algebra is a collection of subsets closed under finite unions and intersections.  What law are they using here to get $A \cap B = (A^c \cup B^c)^c$? I thought de morgan's law was $(A\cap B)^c = A^c \cup B^c$?  Finally, what exactly do they mean by "closed under finite unions and intersections?