Updating ubuntu 13.10 Problem on Dell E6410 When ever I try to update, I get this message:  W:Failed to fetch cdrom://Ubuntu-GNOME 13.10 _Saucy Salamander_ - Release amd64 (20131017)/dists/saucy/main/binary-amd64/Packages Please use apt-cdrom to make this CD-ROM recognized by APT. apt-get update cannot be used to add new CD-ROMs W:Failed to fetch cdrom://Ubuntu-GNOME 13.10 _Saucy Salamander_ - Release amd64 (20131017)/dists/saucy/multiverse/binary-amd64/Packages Please use apt-cdrom to make this CD-ROM recognized by APT. apt-get update cannot be used to add new CD-ROMs W:Failed to fetch cdrom://Ubuntu-GNOME 13.10 _Saucy Salamander_ - Release amd64 (20131017)/dists/saucy/restricted/binary-amd64/Packages Please use apt-cdrom to make this CD-ROM recognized by APT. apt-get update cannot be used to add new CD-ROMs W:Failed to fetch cdrom://Ubuntu-GNOME 13.10 _Saucy Salamander_ - Release amd64 (20131017)/dists/saucy/universe/binary-amd64/Packages Please use apt-cdrom to make this CD-ROM recognized by APT. apt-get update cannot be used to add new CD-ROMs W:Failed to fetch cdrom://Ubuntu-GNOME 13.10 _Saucy Salamander_ - Release amd64 (20131017)/dists/saucy/main/binary-i386/Packages Please use apt-cdrom to make this CD-ROM recognized by APT. apt-get update cannot be used to add new CD-ROMs W:Failed to fetch cdrom://Ubuntu-GNOME 13.10 _Saucy Salamander_ - Release amd64 (20131017)/dists/saucy/multiverse/binary-i386/Packages Please use apt-cdrom to make this CD-ROM recognized by APT. apt-get update cannot be used to add new CD-ROMs W:Failed to fetch cdrom://Ubuntu-GNOME 13.10 _Saucy Salamander_ - Release amd64 (20131017)/dists/saucy/restricted/binary-i386/Packages Please use apt-cdrom to make this CD-ROM recognized by APT. apt-get update cannot be used to add new CD-ROMs W:Failed to fetch cdrom://Ubuntu-GNOME 13.10 _Saucy Salamander_ - Release amd64 (20131017)/dists/saucy/universe/binary-i386/Packages Please use apt-cdrom to make this CD-ROM recognized by APT. apt-get update cannot be used to add new CD-ROMs W:Failed to fetch http://ppa.launchpad.net/tualatrix/ppa/ubuntu/dists/trusty/main/binary-amd64/Packages  404  Not Found W:Failed to fetch http://ppa.launchpad.net/tualatrix/ppa/ubuntu/dists/trusty/main/binary-i386/Packages  404  Not Found E:Some index files failed to download. They have been ignored, or old ones used instead.   Though i have internet connection. What would be the problem?
Failed to download repository information due to missing CDROM When updating I got this error message:  W:Failed to fetch cdrom://Ubuntu 11.10 _Oneiric Ocelot_ - Release amd64 (20111012)/dists/oneiric/main/binary-i386/Packages  Please use apt-cdrom to make this CD-ROM recognized by APT. apt-get update cannot be used to add new CD-ROMs W:Failed to fetch http://ppa.launchpad.net/synce/ubuntu/dists/your_ubuntu_version/main/source/Sources  404  Not Found   What to do?
Derivative of $x^{x^{\cdot^{\cdot}}}$? Using image sequence of medical scans as volume data in cycles I know it's possible to do this in BI, but I need your help to do this in cycles. For now, I have a setup to use an image sequence as volume but it only shows 1 slice/frame. I'm using generated texture coordinates for easy mapping (0-1 on each axis). The images are mapped on XY and I would like to use normalized frame# as Z coordinate to map the sequence as a volume.       Images sequence can be found  (I'm using 16-bit TIF files of MRBrain in this exemple).    Any solution is welcome but try to keep it simple and user friendly :).  Final Edit: I had a hard time choosing between Lemon's and PGmath's answers. Lemon's is more user friendly and flexible, and uses the image sequence without any change inside blender. PGmath's uses a 3rd party converter to make the sequence 1 single image, but allows GPU rendering. Well, ImageMagick's command line is : "convert +append *.tif anim.tif" from the folder where the sequence is unzipped. Not so hard. That's why I accepted PGmath's answer.      Now, i played with it, and here's the result.  The "volume" object contains the volume shader (playing with the color ramp gives funny things) and the boolean modifiers (turn them off to see the complete volume, but reduce the final density multiplier for better view). The "Slice Control" arrow empty controls the slice position/rotation. The "Texture Mapping" object controls...the mapping, if it need some tunning (position, rotation, scale, and then tiling if necessary?). How does the four card limit work? I'm new to the game and trying to make my first constructed deck. My understanding of the card rule is I can only have four cards of one kind.  What does this precisely mean?  For instance, does this mean I can only have four rat cards total or four rat cards of several kinds? If two different cards do the same thing (like  and ), can I have four of each? Prove if $ p$ is a prime of the form $3n+1$, then $p$ is also of the form $6m+1$ Prove that if $p$ is a prime of the form $3n+1$, $p$ is also of the form $6m+1$.  I guess one could use the division algorithm for this; but, what's the best method? Can a sql server table have two identity columns? I need to have one column as the primary key and another to auto increment an order number field. Is this possible?  EDIT: I think I'll just use a composite number as the order number. Thanks anyways. Limit of $\cos(x)/x$ as $x$ approaches $0$ As the title says, I want to show that the limit of  $$\lim_{x\to 0} \frac{\cos(x)}{x}$$  doesn't exist.  Now for that I'd like to show in a formally correct way that   $$\lim_{x\to 0^+} \frac{\cos(x)}{x} = +\infty$$  I'm sure this is right since $\displaystyle\lim_{x\to 0^+} \cos(x) = 1$ and $\displaystyle\lim_{x\to 0^+} x = 0$, but since $\displaystyle\lim_{x\to 0^+} x = 0$ I can't just say:  $$\lim_{x\to 0^+} \frac{\cos(x)}{x} = \frac{1}{\displaystyle\lim_{x\to 0^+} x}$$   Things I tried: expand the fraction and use L'Hospital's rule (This didn't seem to yield results even though I tried quite a few things, should it? If it does, should it always work? It worked for similar problems) and going by the definition of the limit which didn't work for me either.  I'm grateful for any hints for doing this problem and similar problems especially. Prove that for for all $p,q\in \mathbb{Z}$, $q&gt;0$ we have: $$ \left| \sqrt{2}-\frac{p}{q} \right| &gt; \frac{1}{3q^2}. $$  To be honest, I do not know where to start - any help would be appreciated. How do I use method overriding in this case? Understanding my electrical panel I just purchased .  I'll be installing it in my RV.  Current will be supplied by a 12VDC to 120VAC, 3000W inverter.  I'll be using Single-Pole Type BR breakers.  Here's a picture of the box:    My current understanding is that I must install a grounding bar to the box.  This is my current understanding of how to connect the breaker to the main supply.:    Here is my understanding of how I would wire in new circuits:    Does this look right?  I would really appreciate your help -- I've been beating myself over the head about this. How do I write a mathematical equation for ARIMA (2,1,0) x (0,2,2) period 12 I would appreciate if someone could help me write the mathematical equation for the seasonal ARIMA (2,1,0) x (0,2,2) period 12. I'm a little confused with how to go about this. I would prefer an equation involving $Y_t , e_t, θ$ and $Θ$. Prove there's a simple path of length $k$ in a simple graph $G$ where all the vertices have degree of at least $k$.   Relevant definitions:  $G$ is a simple graph that consists of a vertex set $V(G) = \{v_1, v_2, ..., v_n\}$ and an edge set $E(G) = \{e_1, e_2, ..., e_m\}$ where each edge is an ordered pair of vertices. The edge $\{u,v\}$ is denoted $uv$.  A walk of length $k$ is a sequence $v_0,e_1,v_1,e_2,...,e_k,v_k$ of vertices and edges such that $e_i=v_{i-1}v_i$ for all $i$. A path is a walk with no repeated vertex.   My attempt:  Induction, for $k=1$ it's obvious.  Suppose for $k-1$ and we'll prove for $k$.  Let $G$ be a simple graph such that $\forall v \in V : d(v)\ge k$.  We can assume that there are at least $k+1$ vertices since there are $k$ neighbours to every vertex.  Let $v_0$ be some vertex, it has $k$ neighbours, we'll move to one of its neighbours say $v_1$ and remove $v_0$. So now there are $k$ vertices with a degree of at least $k-1$ and from the induction hypothesis we'll have a path of length $k$.  Is using only $k$ vertices from $n$ right? Does it keep the generality when using only all the neighbours of $v_0$? How prove this $\sum_{k=1}^{2^{n-1}}\sigma{(2^n-2k+1)}\sigma{(2k-1)}=8^{n-1}$ Given the positive integer numbers $n$,prove that $$\sum_{k=1}^{2^{n-1}}\sigma{(2^n-2k+1)}\sigma{(2k-1)}=8^{n-1}$$  where $\sigma(n)$ is defined as $$\sigma{(N)}=\sum_{d|N}d$$  This problem I think is very interesting, and have nice methods?  Thank you everyone.  and I find some of this     or can see: How to get the current date without the time? Difference between $\Delta$, $d$ and $\delta$ Objective-C Blocks and variable scope Challenge room locked though I have full health I thought these opened if you have full health. I actually had full + 2 blue hearts before dropping a bomb on myself before taking the screenshot.   But even here I have full health... right? How to show only date after the date of today in JCalendar In the book , at page 27 we find the exercise asserting that       Exercise I.XXXVI.   The underlying space of a zero-dimensional scheme is discrete; if the scheme is Noetherian, it is finite.     Thoughts I thought that a zero-dimensional scheme is one such that, in every local ring $\mathscr O_{X,p}$, the only prime ideal is its unique maximal ideal $\mathfrak m_{X,p}.$ Hence, supposing that $X=\operatorname{Spec} R$ is affine, we deduce that every prime ideal is maximal, so each ponit in $X$ is closed. But how does this imply the discreteness? It only implies the Hausdorff-ness of $X$, right? Furthermore, I cannot perceive what the "Noetherianity" has to do with the finiteness claimed here. In the affine case, it seems to be claiming that there are only a finite number of prime ideals in a Noetherian ring, if its scheme is zero-dimensional? For example, $\mathbb Z$ is a Dedekind domain, hence its scheme is zero-dimensional, but it has infinitely many primes: does this contradict the statement?    Thanks in advance for any reference or hint, and point out any inappropriate point if any is presented. Edit I see why my example cannot work now: it is one-dimensinoal: don't forget the pirme $(0)$. But I am still wondering why the statement is true... Looking for a powershell script that can pull a file from a set of PC's and FTP I'm looking to write a script (preferably powershell) that will essentially copy a file from a bunch of PC's and FTP it to a server.   So the structure of the environment is that we have a file on multiple PC's (around 50 or so) that need to placed on a server. Sometimes one of the PC's may be turned off so the script would first need to ensure the PC is up and running (maybe a ping result), then it would need to go into a directory on that PC, pull a file off of it, rename the file, place into a source directory, then remove the file. Naming convention doesn't matter, but date/time stamp would be easiest. Ideally, it would be best to first move all the files to a source directory to save on FTP bandwidth, but since the files will be named the same, the files must be renamed during the move process. Move not copy because the directory needs to be empty so the file can be re-created the next day. So once moved to the source directory, now all the files need to be FTP'd to a server for processing.   After all of this, we need to know which PC's on the list did not respond so we can manually retrieve the file so the script should output a file (txt is fine) that will show which PC's were offline.   Everything is one domain and script will be run from an server with admin creds.   Thank you! The nautilus-open-terminal package adds a menu/right-click option to open a terminal in the current directory.  I want to change this from opening gnome-terminal to opening terminator. , however these instructions do not appear to be valid any more, and I can't find any place in the system settings where the default emulator can be changed.  I also tried:  $ sudo update-alternatives --config x-terminal-emulator   choosing  * 0            /usr/bin/terminator               50        auto mode   But this does not appear to have any effect. Any idea how I can get this to work?