Is there any way to mirror an already rigged mesh and skeleton in Blender?
How can I copy one side of a rig to the other?
When is it appropriate to use the original pronunciation of a foreign word versus the English pronunciation? When reading to an audience, or speaking in conversation, when is it appropriate to use the original pronunciation of a foreign word versus the English pronunciation (assuming you know the appropriate pronunciation for it)? Is it considered rude, or condescending? Or is it considered a mark of being knowledgeable?  One of the things to consider are place names. When referring to Paris, France, should it be pronounced with the silent 'S' as the French would say it, or with the 'S'? Should Hiroshima be pronounced as a Japanese speaker would pronounce it slightly more emphasis on RO, rather than the SHI? Should Mount Pinatubo be pronounced as a Filipino speaker would pronounce it with shorter stronger vowels, or the longer vowels? (e.g. Pi as is 'pick' rather than 'pea'.) Should Wichita be pronounces as the original "shi" rather than the modern "chi"?  Pronounce pesos or sombrero as a Spanish speaker 'eh' or the English 'ay'?  (I can't think of other common words right now that aren't words taken from other languages like hurricane, boondocks, tornado, etc. which I think have (correctly) changed to English pronunciations.) Windows Metro apps have wrong icon in search UI This was posted on facebook and people are saying it is incorrect, it should be: "...as you and I"   Which is correct? Compensation for missed connection due to delay with same airline but separate bookings Long story short, I had to fly from A (in Spain) to C (in the UK) and I booked two separate flights with the same budget airline, so the actual route was A (Spain) -> B (UK) -> C (UK).  The first flight was delayed by 90 minutes, which means that I was unable to catch the second one (the layover time was around an hour and a half) and had to buy another ticket for the following day (with another airline).  I googled a lot and found mixed responses. Is there any chance I can get a refund for my second flight? I will obviously ask them but I'm unsure what my best course of action is. Why are only certain shutter speeds or apertures available? For the exposure control function DSLRs there are specific shutter speeds available. For example there are 1/4, 1/100, 1/500. Why, even in high end DSLRs, is there no option to use a custom shutter speed (such as 1/19)?   If the camera can do 1/4000, why can't it do 1/33?  Similarly, why are the aperture choices limited?  Why is it a technical challenge to add extra shutter speeds or aperture settings? I just want to know if it's possible to get Cain and abel installed in ubuntu, it would really help my work instead of dual booting back to my windows 7 just to use it. THanks! How to draw a crisp, opaque hairline in JavaFX 2.2? Is it possible to set the "urgent" flag in Gmail for outgoing mail? I'd like to be able to flag time-sensitive emails as such. Proof of a combinatorial identity: $\sum\limits_{i=0}^n {2i \choose i}{2(n-i)\choose n-i} = 4^n$ Possible Duplicate:          This was part of a homework assignment that I had, and I couldn't figure it out. Now it is bugging me. Can anyone help me? Although a proof would be nice, I wouldn't mind a push in the right direction.   $\displaystyle\sum_{i=0}^n {2i \choose i}{2(n-i)\choose n-i} = 4^n$ Why are there only derivatives to the first order in the Lagrangian? Why is the Lagrangian a function of the position and velocity (possibly also of time) and why are dependences on higher order derivatives (acceleration, jerk,...) excluded?  Is there a good reason for this or is it simply "because it works". What Nikon lens to buy to make photos of birds? I have been using Ubuntu 12.04 for quite a while and have never had a problem come at me out of the blue like this. I'm trying to install python 3.2 on my Ubuntu system and every time I run the code to compile the python source, I get this error:  'Command 'sudo' is available in '/usr/bin/sudo' The command could not be located because '/usr/bin' is not included in the PATH environment variable. sudo: command not found'   I've also tried running gksudo command to view what my environment shows but I pretty much get the same error. Ethernet Without Magnetics? How / can you connect Ethernet devices together without magnetics? How do you wire it up? What sorts of practical limitations are there on doing this?  To clarify, I'm not trying to connect two computers here. I'm thinking specifically of creating a small network of microcontrollers on a single board. Lets say the total length of a network segment is no longer than 6 inches. This would be for 10BaseT or at most 100BaseT. How can I redefine \int to put a negative space after the integral sign, but keep the limits properly set? I'd like to modify the \int command such that it adds an automatic negative space, e.g. \! after the integral sign. Preferentially, I'd like to be able to not edit any of my code, so cleverly redefining \int seems like my best option. Unfortunately, naive implementations get the limits wrong:  \documentclass{article} \usepackage[intlimits]{amsmath} \usepackage{ifthen}  \newcommand{\diff}[2][]{   \ifthenelse { \equal {#1} {} }   {\ensuremath{\mathop{\mathrm{d} #2}}}   {\ensuremath{\mathop{\mathrm{d}^#1 #2}}} }  \begin{document}  This is what I would want: \begin{equation}  \int_0^\infty \mspace{-4mu}\diff[3]{\boldsymbol r} f(\boldsymbol r), \end{equation} but with this ``user code", containing no negative spacing explicitly: \begin{verbatim}  \int_0^\infty \diff[3]{\boldsymbol r} f(\boldsymbol r), \end{verbatim} which for comparison with the above gives: \begin{equation} \int_0^\infty \diff[3]{\boldsymbol r} f(\boldsymbol r), \end{equation}  mostly because this gives me the option to remove it or change the spacing when I see fit. \end{document}   Which results in: Commas for multiple adjectives If the lines intersect the vertices of the square. The area of the square is $1$ and the lines also intersect the midpoints of the square lines. How to find the area of shaded region? Why is my ASCII art all wonky? Nothing looks right in the ASCII art.  It looks wrong and nothing is proportional, how do I fix this? I have tried fiddling with my font size, but nothing still looks right! How to set systemwide "XP style DPI scaling" in Windows 8.1? I run my ThinkPad with a system DPI setting of 140%. In Windows 7 and 8, I enabled the "Use Windows XP style DPI scaling" checkbox in the Custom DPI Setting dialog. When this setting is off, non-DPI-aware applications get DPI virtualization via bitmap stretching, which results in large but fuzzy text. Turning on the checkbox disables this bitmap stretching, so non-DPI-aware apps get smaller text but without the fuzzy stretching. I find most apps to be more readable with the smaller crisper text, so I enable XP style scaling (i.e. disable DPI virtualization).  Windows 8.1 RTM has removed this option completely. After upgrading, all of my non-DPI-aware apps are running with ugly bitmap stretching. I'm able to fix this by going into the Properties for each app and setting the "Disable diplay scaling on high DPI settings" box in the Compatibility tab, but that is a bit of a pain.  Is there any way to restore the systemwide XP-style DPI scaling in Windows 8.1? I'm using the beamer miniframes theme (with subsection=false) and the compress option. Currently the navigation shows the dots for each subsection side-by-side. I would prefer if there would be a horizontal space (e.g. with the width of a dot) between the dots of each subsection. Is this possible? Right now it looks like:  Section 1 oooooo  And I would prefer it to look like:  Section 1 ooo ooo  MWE: \documentclass[compress]{beamer}   \useoutertheme[subsection=false]{miniframes}   \begin{document}   \section{Sec 1}   \subsection{}   \begin{frame}[t]     hi   \end{frame} \subsection{}   \begin{frame}[t]     hi   \end{frame} \end{document}` Find the derived subgroup of $A_4$ Find the derived subgroup of $A_4$.   Since it is $A_4$, for a permutation $\sigma$ to be in $A_4$, $\sigma$ must have a cycle structure of $2$ cycles. Therefore, $\sigma=(ab)(cd)$.  The commutator of such elements, would be obviously another permutation of $2$-cycle (of length $2$).  Does it mean that $A'_4=A_4$?