What's the easiest way to unrar a file on Ubuntu 12.04
What's the easiest way to unrar a file?
Let, $X$ be a countable set. Which of the following are true?   There exists metric $d$ on $X$ so that $(X,d)$ is complete. There exists metric $d$ on $X$ so that $(X,d)$ is not complete. There exists metric $d$ on $X$ so that $(X,d)$ is compact. There exists metric $d$ on $X$ so that $(X,d)$ is not compact.   4 is true as we can take discrete metric. Then $\{\{x\}:x \in X\}$ is an open cover containing no finite subcover.  If there exists a compact metric then it would automatically become complete also. Hence if 3 is true, so is 1.  What about 2? Any other comment? Should it be obvious that independent quantum states are composed by taking the tensor product? I've read before that, by the Principle of Explosion, if a theory is inconsistent, then absolutely any statement can be proven within it.  Obviously, there are statements which are independent of ZFC (Continuum Hypothesis, etc). It seems to me that this proves that ZFC is incomplete. Why does it not then follow that ZFC is consistent? It seems to me that we could say "Assume ZFC is inconsistent. Then the the Continuum Hypothesis is provable in ZFC. But the Continuum Hypothesis is neither provable nor disprovable in ZFC. Therefore ZFC is consistent."  What am I missing here? How do I install Komodo Edit? New to linux, help me install this program Komodo-Edit-7.1.0-10496-linux-x86_64  I'm running Ubuntu 12.04 64bit. What damage can person/s do if they have your passport number and visa control number? I read  as well as  but it doesn't tell of what damages can be done. I have seen several sites on which people share/show the visa without the numbers and sometimes without name as well.   The only reason which many people perceive is identity theft but am unable to see how is that possible. I have to use public facilities such as printing, scanning etc. where it is easy for the operator to make copies without my consent. Also as shared, both these details are widely circulated when you stay in a hotel or anywhere else (when abroad) where identity information is needed.   Also particular to visas, as I'm given to understand, for short-stay visas (less than 3 months) which most tourist visas operate under, the window of doing anything nefarious activities (such as terrorism) would be extremely small. If I were a terrorist and I needed those bits to make work, then I would probably go for somebody who has a long-term stay as those credentials would look valid and I could jump from place to place under the radar under a fraudulent identity. Also as a terrorist, I would take quite a long view as any such attack takes years of planning, as shown in the investigation of various terrorist attacks in India as well as the U.S. Also, although I don't know (am speculating here) don't most airports might/should be checking on status quo of a passport to ensure issues such as showing 2 entries of a passport without exit (in some database where this information is collected and collated.)   Am I missing something in the logic here ? $n$ people attend the same meeting, what is the chance that two people share the same birthday? Given the first $b$ birthdays, the probability the next person doesn't share a birthday with any that went before is $(365-b)/365$. The probability that none share the same birthday is the following: $\Pi_{0}^{n-1}\frac{365-b}{365}$. How many people would have to attend a meeting so that there is at least a $50$% chance that two people share a birthday?   So I set $\Pi_{0}^{n-1}\frac{365-b}{365}=.5$ and from there I manipulated some algebra to get $\frac{364!}{(364-n)!365^{n}}=.5\iff (364-n)!365^{n}=364!/.5=.....$   There has to be an easier way of simplifying this. Is it (always) true that $$\mathrm{Var}\left(\sum\limits_{i=1}^m{X_i}\right) = \sum\limits_{i=1}^m{\mathrm{Var}(X_i)} \&gt;?$$ How to convert sshfs command to fstab entry? From another SO , I understand I should create an entry in fstab to permanently mount an access to a virtual machine.  I am not Linux expert. The magic command which allows me to perform this from a terminal is:  sudo sshfs -o idmap=user -o allow_other -o nonempty jverstrynge@devjverstrynge:/home/httpd /home/jverstrynge/httpd   When I check the above SO question, I see an entry looking like this:  /media/mybook/laptop_backup /export/laptop_backup none bind 0 0   Can someone explain how to transform the CLI command into an entry in fstab? Remove votes to close question X as a duplicate of Y if question Y is closed as a duplicate of X I just happened across , which has one close vote suggesting that it be closed as an exact duplicate of .  This suggestion is perfectly correct: these are exactly identical questions asked by the same user.  However, there's not much point in closing question A as a duplicate of B, since question B has already been closed as a duplicate of A.  Yet the close vote remains, doing nothing but wasting reviewers' time.  This is not the first time I've seen this happen, so I'd like to suggest that the Stack Exchange software should do something about it.  In particular, I'd like to suggest the following feature:     When a question Y is closed as an exact duplicate of another question X, any votes or flags to close X as a duplicate of Y should be automatically discarded.   Actually, I'm kind of surprised that the SE software doesn't do that already.  If anyone else thinks that this constitutes an actual bug, feel free to retag it. Prove that $E$ is disconnected iff there exists two open disjoint sets $A$,$B$ in $X$ Does the lens I just bought have a fungus? Make a computer act as a virtual USB device for other equipments Laggy viewport in 2.9 when model has a lot vertices App drawer icon gone from Cyanogenmod dock I have a Samsung Galaxy S2, running the latest CM10.1.  I lost the app drawer icon on my Dock; I must have deleted it. The app drawer icon lets you go to the app list, and the widget list.  How do I get the app drawer icon back? The icon looks like a circle with 6 dots inside it. I have recently bought a Logitech M510 as my main mouse switching over from a magic mouse. I didn't realize missing horizontal scrolling would be such an issue, but apparently it is.   Now a mac does support horizontal scrolling by holding shift + scrolling, this is however amazingly slow.   I have the Logitech Control Center installed, which offers a few extra options, but nothing regarding the horizontal scroll speed. The scrolling speed options offered there only affect vertical scrolling.    Can the vertical speed be one setting and the horizontal speed another? I wrote a script to create a big number of heat maps for a given scenario. I would like to save the result of every heat map as an PNG image. I have figured it out doing it with simple Map Rendering but the quality of the result is too low. Furthermore I would like to export to OpenLayers using the same code, which does not seem to be possible with simple rendering.   Now I am trying to do it with QgsComposition instead. Manually in the Map Composer you can just set the paper color to transparent and disable the background of the map canvas.   This is the code that I have so far. I just cannot figure out why it is not working. The result png-file always has a white opaque background.   My OS is Windows 7 and I'm using QGIS 2.6 Brighton.  I hope someone can help me out here.  # PARAMETERS mainPath = 'path/to/folder/' filename = 'filename' imageType = "png" imageWidth_mm = 400 imageHeight_mm = 160 dpi = 300  mapRenderer = iface.mapCanvas().mapRenderer()  c = QgsComposition(mapRenderer) c.setPlotStyle(QgsComposition.Print)  c.setPaperSize(400, 160)  x, y = 0, 0 w, h = c.paperWidth(), c.paperHeight() composerMap = QgsComposerMap(c, x ,y, w, h) c.addItem(composerMap)  # get all items and disable the backgrounds itemList = c.items() c.removeItem(itemList[2]) c.removeItem(itemList[2]) itemList[2].setBackgroundEnabled(False) itemList[3].setBackgroundEnabled(False) # add them to the composition c.addItem(itemList[2]) c.addItem(itemList[3])  c.refreshItems() c.refreshDataDefinedProperty(QgsComposerObject.AllProperties)  c.setPrintResolution(dpi) dpmm = dpi / 25.4 width = int(dpmm * c.paperWidth()) height = int(dpmm * c.paperHeight())  # create output image and initialize it # image = QImage(QSize(width, height), QImage.Format_ARGB32) image = QImage(QSize(width, height), QImage.Format_ARGB32_Premultiplied) image.setDotsPerMeterX(dpmm * 1000) image.setDotsPerMeterY(dpmm * 1000) #image.fill(0)  imagePainter = QPainter(image)  # This part does not seem to have any effect #brush = QBrush(Qt.BrushStyle(0)) # transparent brush # #imagePainter.setBackgroundMode(0) #c.setBackgroundBrush(brush)  c.renderPage( imagePainter, 0 ) imagePainter.end()   imageFilename =  mainPath + filename + '.' + imageType image.save(imageFilename, imageType) print 'image saved' print 'done' 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. Getting different answers when integrating using different techniques Question:  Is it possible to get multiple correct results when evaluating an indefinite integral?  If I use two different techniques to evaluate an integral, and I get two different answers, have I necessarily done something wrong?    Often, an indefinite integral can be evaluated using different techniques.  For example, an integrand might be simplified via partial fractions or other algebraic techniques before integration, or it might be amenable to a clever substitution.  These techniques give different results.  For example, looking over a few other questions on MSE:      From : evaluate $$ \int x(x^2+2)^4\,\mathrm{d}x. $$    Via the substitution $u = x^2+2$, this becomes $$ \int x(x^2+2)^4\,\mathrm{d}x       = \frac{1}{10}x^{10} + x^8 + 4x^6 + 8x^4 + 8x^2 + \frac{32}{5} + C.     $$ However, multiplying out the polynomial and integrating using the power rule gives $$ \int x(x^2+2)^4\,\mathrm{d}x       = \frac{1}{10}x^{10} + x^8 + 4x^6 + 8x^4 + 8x^2 + C    $$     From :  evaluate   $$ \int \frac{1-x}{(x+1)^2} \,\mathrm{d}x. $$    Simplifying the integrand using partial fractions then integrating gives $$ \int \frac{1-x}{(x+1)^2} \,\mathrm{d}x       = \frac{-2}{(x+1)} - \ln|x+1| + C.    $$ Via integration by parts, we get $$ \int \frac{1-x}{(x+1)^2} \,\mathrm{d}x       = \frac{x-1}{(x+1)} - \ln|x+1| + C.    $$     From : evaluate   $$ \int \frac {\tan(\pi x)\sec^2(\pi x)}2\,\mathrm{d}x. $$    Using the substitution $u = \sec(\pi x)$, this becomes $$ \int \frac {\tan(\pi x)\sec^2(\pi x)}2\,\mathrm{d}x       = \frac {\sec^2(\pi x)}{4\pi} + C.    $$ Using the substitution $u = \tan(\pi x)$, this becomes $$ \int \frac {\tan(\pi x)\sec^2(\pi x)}2\,\mathrm{d}x       = \frac {\tan^2(\pi x)}{4\pi} + C.    $$ Reset Search Index in Exchange 2013 Differences between hosting options for SharePoint apps