Testing divisibility by 7 of number in base 8
Casting nines and elevens in other bases (radix) and doing check sums for binary
How do I draw shapes inside a tikz node? I know how to draw different shapes using tikz, like this:  \documentclass{minimal} \usepackage{tikz} \begin{document}  \begin{tikzpicture} \draw (0,0) ellipse (1 and 1.5)       (0,2.5) circle (1)       (2.5,0) circle (1.5); \end{tikzpicture}  \end{document}     However, I would like to put a number of different shapes together inside a node, and I couldn't yet find a simple way to do that. I will make a little figure to include in a large diagram which is already made of several nodes.  If this is not possible, or not easy, then how do I include a JPG or EPS file in a node?  Thank you! Grub rescue - error: unknown filesystem I have a multiboot system set up. The system has three drives. Multiboot is configured with Windows XP, Windows 7, and Ubuntu - all on the first drive. I had a lot of unpartitioned space left on the drive and was reserving it for adding other OSes and for storing files there in the future.  One day I went ahead and downloaded Partition Wizard and created a logical NTFS partition from within Windows 7, still some unpartitioned space left over. Everything worked fine, until I rebooted the computer a few days later.  Now I'm getting:    error: unknown filesystem.   grub rescue&gt;   First of all I was surprised not to find any kind of help command, by trying:    help, ?, man, --help, -h, bash, cmd, etc.  Now I'm stuck with non-bootable system. I have started researching the issue and finding that people usually recommend to boot to a Live CD and fix the issue from there. Is there a way to fix this issue from within grub rescue without the need for Live CD?  UPDATE  By following the steps from , I was able to boot to initramfs prompt. But not anywhere further than that.  So far from reading the manual on , I was able to see my drives and partitions using ls command. For the first hard drive I see the following: (hd0) (hd0,msdos6) (hd0,msdos5) (hd0,msdos2) (hd0,msdos1)  I now know that (hd0,msdos6) contains Linux on it, since ls (hd0,msdos6)/ lists directories. Others will give "error: unknown filesystem."  UPDATE 2  After the following commands I am now getting to the boot menu and can boot into Windows 7 and Ubuntu, but upon reboot I have to repeat these steps.    ls ls (hd0,msdos6)/ set root=(hd0,msdos6) ls / set prefix=(hd0,msdos6)/boot/grub insmod /boot/grub/linux.mod normal   UPDATE 3  Thanks Shashank Singh, with your instructions I have simplified my steps to the following. I have learned from you that I can replace msdos6 with just a 6 and that I can just do insmod normal instead of insmod /boot/grub/linux.mod. Now I just need to figure out how to save this settings from within grub itself, without booting into any OS.  set root=(hd0,6) set prefix=(hd0,6)/boot/grub insmod normal normal   UPDATE 4  Well, it seems like it is a requirement to boot into Linux. After booting into Ubuntu I have performed the following steps described in the :  sudo update-grub sudo grub-install /dev/sda   This did not resolve the issue. I still get the grub rescue prompt. What do I need to do to permanently fix it?  I have also learned that drive numbers as in hd0 need to be translated to drive letters as in /dev/sda for some commands. hd1 would be sdb, hd2 would be sdc, and so on. Partitions listed in grub as (hd0,msdos6) would be translated to /dev/sda6.  UPDATE 5  I could not figure out why the following did not fix grub:  sudo update-grub sudo grub-install /dev/sda   So I downloaded  based on an answer from  post. That seemed to do the trick after I picked the "Recommended Repair (repairs most frequent problems)" option. Can I run two 4k (3840x2160) displays at 60hz on a Macbook Pro 13" 2015 I think the title says it all. Can I run two 4k (3840x2160) displays, both at 60hz on a Macbook Pro 13' 2015? I understand they might need to be SST and each connected to a Thunderbolt 2 port. But I cannot get a confirmation whether that setup would work.    says:     Simultaneously supports full native resolution on the built-in display and up to 3840 by 2160 pixels on up to two external displays, both at millions of colors.   …   Support for up to 5120-by-2880 resolution at 60Hz on a single external display (model with AMD Radeon R9 M370X only)   Unfortunately, it does not state at what frequency the two displays can be run.    says:     With OS X Yosemite v10.10.3 and later, most single-stream 4K (3840x2160) > displays are supported at 60Hz operation on the following Mac computers: … MacBook Pro (Retina, 13-inch, Early 2015) and later. MacBook Pro (Retina, 15-inch, Mid 2014) and later   Unfortunately it does not state whether two displays can be run at 3840 by 2160 at 60hz Deleting another user's directories from my own I am a non-root user, and have made a directory into which other users in my group can write. The directory is setgid, so files and directories within it have the same group.  I can delete files placed into this directory, but if a user creates a subdirectory with files in it, I can't seem to delete those. Is there something special I can do (other than, say, bothering the user in question or the sysadmin about it) to get rid of this subdirectory?  EDIT: Inevitably, of course, someone asks "why?"  The shared folder in question is a mercurial repository. We can't use a shared-account ssh-key-based server for this like mercurial-server, and we can't just host it externally for various reasons. But that part works fine. We use Redmine for project management, but it does not yet read the repo index for file contents - it reads the working directory. So the working directory must be kept up-to-date. crontab has been disabled by the sysadmin, so that's out. So I naïvely put a changeset hook in the shared repo that performed an "hg update". This seemed to work fine initially, until, of course, someone pushed a commit with a new directory. While I had vague thoughts of trying to make a setuid script for the hook, I'll probably just consider safer avenues. What's so special about standard deviation? Given a $k$-manifold $M$, such that $\partial M$ is a $(k-1)$-manifold, there is a standard way in which $\partial M$ inherits the orientation of $M$. So if $M$ is oriented by the form field $\omega$, then $\omega_\partial=\omega(v_{\text{out}},v_1,\dots,v_{k-1})$ orients $\partial M$, where the $v_1,\dots,v_{k-1}$ are basis vectors for a tangent space on a point of the boundary, and $v_\text{out}$ is an outward pointing vector (see here: , which follows the exposition of Hubbard &amp; Hubbard).  In practice I find this very difficult to interpret. How can one find this $v_\text{out}$ in practice?   For instance, take $M$ to be defined by $z=x^2+y^2$ for $x\ge0$ (technically a manifold with boundary, in fact a half-paraboloid). It's a fact that $\partial M$ is defined by the parabola $z=y^2$. If $M$ is oriented by $dx\wedge dy$, how does this induce an orientation on the boundary (in fact we should obtain a 1-form that induces an orientation on the boundary)?   Note: I find the whole discussion of boundary orientation to be rather clumsy in its exposition. In the case of surfaces inducing orientations on boundary curves, I was originally taught to imagine that the normal vector that induces the surface orientation is me walking on the surface. This normal should walk along the boundary such that the surface lies to its left. OK - but how does this translate to the $v_{\text{out}}$ interpretation, so that we can generalize to other dimensions easily?)  Edit: An ideal answer would merge the intuitive approach with the rigorous approach, if possible. I would grateful to anybody who could help order my (at present confused) understanding! Why are floating point numbers inaccurate? Prime number expressible as $2^2+3^3+5^5+7^7+11^{11}+\cdots$ $S^2 \times S^2$ is diffeomophic to $G_2(\mathbb{R}^4)$ $G_2(\mathbb{R}^4)$ is the Grassmannian manifold of two-dimensional subspaces of $\mathbb{R}^4$. I would like a detailed proof. Can it be done explicitly? I mean, showing the map and checking its differentiability using charts? Why are downloads destructive in the Star Trek universe? There are numerous places in various Star Trek episodes where references are made to the fact that "downloads" are a destructive operation: you download something from one place, and unless you explicitly made a copy beforehand, what you downloaded is gone from its original storage location and instead appears in some other location. Hence was born the term "copy and download".  There are of course out-of-universe, plot reasons for things happening that way which are exploited from time to time. There is also the out-of-universe explanation of how large media companies feel about people downloading copyrighted works, and a possible out-of-universe explanation that at the time, people in general may not have been intimitely familiar with the implications of the term "download", so they added "copy" to help describe what happens to at least portions of their viewership.  But is there a plausible in-universe explanation for why downloads would be destructive by default, and a specific copying operation would have to be made beforehand so as to not delete the data from its original location?  As for specific examples, these are two that I can think of right now (I'm quite sure there are more):   VOY "Twisted" (S2E06), right near the end, stated by Torres: "Our entire database has been copied and downloaded into somebody else's system." VOY "Message In A Bottle" (S4E14), a little more than five minutes into the episode, the EMH is "downloaded into the transceiver array" for transmission across an alien sensor network to the Prometheus when the latter is in the alpha quadrant, and some 20 minutes into the episode it's established that the EMH is not available in the sickbay and will return later. What is a complete list of the usual assumptions for linear regression? Evaluate the limit : $$\lim_{n \to \infty } {1 \over {\sqrt n }}\left( {1 + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }} + \cdots + {1 \over {\sqrt n }}} \right)$$  I can use the sandwich principle, certain convergence criteria, Cesaro mean theorem, limit arithmetic.. things around this area.  Any help would be greatly appreciated, thanks!  Sorry for not elaborating more at the beginning, rookie first-post mistake I suppose. :) My grandparents bought a flight on United Airlines that departs on Feb 10 2019, as the one and same ticket, from Toronto, to ORD with 1 55-minute layover, then Maui. Please ping me if I ought post their return flight on Feb 24's details.   advises:     There should be no customs and immigrations formalities at ORD, as YYZ is a , and CBP matters will be handled in Toronto.    I'm assuming that  persists during their vacation to Maui.   reported some TSA agents' refusal to work.  Will  refuse to work, or any other federal workers pertinent to travelling to and from the US? Prove that $ S=\{0\}\cup\left(\bigcup_{n=0}^{\infty} \{\frac{1}{n}\}\right)$ is a compact set in $\mathbb{R}$, but $\bigcup_{n=0}^{\infty} \{\frac{1}{n}\}$ is not a compact set.   (Can we use definition of compact set for this problem? ) Quoted vs unquoted string expansion How do I import a namespace in Razor View Page? Relationship between Partial Harmonic Sum and Logarithm. In English, I sometimes hear a phrase like green is the new black.  Literally, this makes absolutely no sense whatsoever.  What does it mean when someone says X is the new Y, when this makes no sense when taken literally?   gives plenty of examples:          I've found definitions online for specific examples, such as , but overall I don't understand what X is the new Y means, and I can't find it explained anywhere.  What does X is the new Y mean?  As far as I'm aware, this phrase exists only in English. Show that G is a group, if G is finite, the operation is associative, and cancellation law holds Let $G$ be a non-empty finite set with an associative binary operation so that cancellation law holds, i.e. $ab=ac$ or $ba=ca$ implies $b=c$, for any choices of $a,b,c$ in $G$. Assume that there is an identity element $e$ in $G$. Show that $G$ is a group.  Proof: To show $G$ is a group, conditions must hold. Suppose that $ab=ac$, then $b=eb=(a^{-1}a)b =a^{-1}(ab)=a^{-1}(ac) =(a^{-1}a)c=ec=c$. So left cancellation holds in $G$.  2) Suppose $ba=ca$, then $b=be=b(aa^{-1}) =(ba)a^{-1}=(ca)a^{-1} =c(aa^{-1})=ce=c$. So right cancellation holds in $G$.  3) If $a$ exists in $G$ then $aa^{-1}=a^{-1}a=e$, where $a$ is an inverse of $a^{-1}$. Since inverses are unique, $(a^{-1})^{-1}=a$.  4) Let $x$ be the inverse of $ab$. Then $(ab)x=e$. By associativity we have $a(bx)=aa^{-1}$. Through left cancellation we have  $bx=a^{-1}bx=ea^{-1}=b(b^{-1}a^{-1})$ and $x=b^{-1}a^{-1}$.  Thus $(ab)^{-1}=b^{-1}a^{-1}$. So all conditions hold, $G$ is a group.  Is this proof correct? I know to show $G$ is a group these conditions have to be met. This is all I have to show right? As seen at:           Update: I'm 101% sure the down votes were due to the lack of free hands.    Update 2: likes me, like me not, like me, likes me not... yeah !