Replacing list elements by other list elements
Join matrix based on conditions
How can I minimize Thunderbird 60.2.1 to tray after "MinimizeToTray revived" add-on stopped working? After update, my Thunderbird stopped getting minimized with the "MinimizeToTray revived" add-on. Is there any way to minimize my Thunderbird? Any other add-on that works with the new version? I added two torus objects, and when I click one of them, it also selects the other. They both share the same node thing. I want to delete one of them, but it keeps on deleting the other with it. I can't even undo placing the second torus without removing the other. Suppose that I have a &lt;div&gt; that I wish to center in the browser's display (viewport). To do so, I need to calculate the width and height of the &lt;div&gt; element.   What should I use? Please include information on browser compatibility. Disabling middle mouse button What does a leading zero do for a php int? Rounding Percentages SyntaxError: Non-ASCII character '\xa3' in file when function returns '£' I understand how to define include shared objects at linking/compile time. However, I still wonder how do executables look for the shared object (*.so libraries) at execution time.  For instance, my app a.out calls functions defined in the lib.so library. After compiling, I move lib.so to a new directory in my $HOME.  How can I tell a.out to go look for it there? find $\sum_{k=2}^{\infty}\frac{1}{k^2-1}$ Let be $u,v\in\mathbb{R}^n$, then $\det(I+uv^\intercal)=1+v^\intercal u $ where $I$ denotes the identity matrix of order $n$. How to prove this?   what I did:   let be $A=\{n\in\mathbb{N}: \forall u,v \in \mathbb{R^n}, \det(I+uv^\intercal)\neq1+v^\intercal u  \} $, and suppose $A\neq \varnothing $, then for the well-orderer-principle there exists $n_0\in A$ such $n_0\leq n,\;\forall n\in A$. since $1\notin A,\; n_0\neq 1 \rightarrow n_0-1\in\mathbb{N}\setminus A  $ so: $$\forall u,v\in\mathbb{R^{n_0-1}}: \det(I+uv^\intercal)=1+v^\intercal u $$ Then, let be $u,v\in\mathbb{R}^{n_0}$, so $$\det(I+uv^\intercal)= \displaystyle\sum_{j=1}^{n_0}(-)^{1+j}a_{1j}\det(A_{1j})$$ where $a_{ij}=\begin{cases} u_iv_j+1 &amp; i=j \\ u_iv_j &amp; i\neq j \end{cases}$, and $A_{1j}$ is the  submatrix of $I+uv^\intercal$ that results deleting the i-file and j-column. Then I try to open the term for $j=1$ and try to use the relation for the new matrix of $n_0-1$ order with the determinant because it holds the same form but I have other terms which difficult me the work. Do you know other method?.  Ps. , in page 2 I've found something similar but I don't understand what it means PS. I'm taking a course of numerical analisys Does Class need to implement IEnumerable to use Foreach This is in C#, I have a class that I am using from some else's DLL. It does not implement IEnumerable but has 2 methods that pass back a IEnumerator. Is there a way I can use a foreach loop on these. The class I am using is sealed. Calculating polygon area within polygons in QGIS? I am new to QGIS - I want to estimate the area of the reef under a polygon grid. My shapefiles are already in meters, however I am having a hard time calculating the area under the grid layer. Attached are the photos. I want to calculate the total area of the bold black polygons (reefs) under each grid square (image 2). How much reef is inside each square, for example, grid1: 25 sq.km, grid 1 sq.km, etc. Each reef polygon has its corresponding area in the layer attribute table.    ADDITIONAL PHOTO: RED IS THE REEF  THE ATTRIBUTES OF THE REEF LAYER A positive harmonic function on the punctured plane is constant Why has Steve Nash moved to the LA Lakers in sign-and-trade deal? Congratulations to Steve Nash on his move to Lakers !  His contract with the Suns has expired so I don't understand why he moved in a sign-and-trade deal.    Why resign and move in trade? Why didn't the Lakers sign Nash without any trade? Grub not showing on startup for Windows 8.1 Ubuntu 13.10 Dual boot Ok, so I'm such a newbie to Ubuntu and I bought a Windows 8 pre-installed laptop last month.  I updated to Windows 8.1 and then I thought about installing Ubuntu as a dual boot so I could mess around and learn more about it.   So I followed a Youtube tutorial (  ) and I got my stuff working fine.   The first few times I booted I got the GRUB menu instead of my default HP Boot OS Manager, and I was able to select my OS.   So I went to sleep and the next day I turned on my computer and the GRUB menu did not show up. I tried several times and it didn't automatically show up.  In order for me to see the GRUB menu I had to turn on my PC and on start had to press ESC to pause startup and press F9 to get boot options. Then from there I had to pick from OS Boot, Ubuntu, Ubuntu (Yes there were two Ubuntus available) and a default EFI file thingy.   When I click the first Ubuntu I get the GRUB Menu (I was too scared to try the second in case I screwed my laptop up) and I can safely load Ubuntu from there and use it (although I do have to increase my brightness every time I load Ubuntu because it somehow reduces my brightness to complete darkness on boot)  So my problem here is why isn't my GRUB showing on boot, after it worked on the first day?   My laptop is a HP TouchSmart j-078CA.  EDIT 1: FYI I just switched to Ubuntu LiveCD to check my Installation and found out that one of my partition is flagged as msftres. Here is a screenshot of my GParted on a LiveCD: I'm currently a beginner when it comes to 3D modelling but doing okay so far. I'm struggling a bit when it comes to the final renders of my scene, specifically on the camera view I want. Sometimes its to zoomed in and sometimes to zoomed out that it's gone out of the environment of the blender. I realized the settings for the camera on blender are for actual cameras / photography and I'm no expert on that too. A tip on how to control them precisely would be nice. Thank you. Prove that an infinite ring with finite quotient rings is an integral domain How can we show that if $R$ is an infinite commutative ring and $R/I$ is finite for every nonzero $I \unlhd R$, then $R$ is an integral domain?   I tried proceeding by contradiction: assume $a$,$b$ $\in R \backslash \{0\}$ and $ab=0$; then $R/(a)$ and $R/(b)$ must be finite, say $R/(a)=\{k_i + (a) : 1 \leq i \leq m\}$ and $R/(b)=\{l_j + (b) : 1 \leq j \leq n\}$. Does this mean $R$ must be finite? Or what about using the fact that $R/(a,b)$ finite?  Thanks for any help with this! java.sql.SQLException Parameter index out of range (1 > number of parameters, which is 0) Best answer versus best solution Travelling without a visa after passport was stolen