which is better?visibility property or display property of an element in html?
Performance differences between visibility:hidden and display:none
I installed Ubuntu 12.04 a month ago and am using it till now. I failed to notice that all this time there was no sound at all while running Ubuntu, even while playing a game in Wine. The weird thing is that only the startup sound comes when I log in (Indian/African drum tone), then comes the utter silence.  I tested both Digital Output (S/PDIF) and the speakers in the sound settings but can hear nothing.  Any help? Review Queues and Ignored Tags Steps to Recreate:   Have enough rep to use some of the review queues. Ignore a popular tag. Go to your profile-> prefs-> check 'hide ignored tags'. Open a review queue.   You will get questions from your ignored tag(s) in the review queue.  Personally, I ignored a tag on Movies.SE (and Scifi.SE) for a film that is out elsewhere in the world, but not locally.  It's a simple way to avoid being exposed to spoilers.  I then load up the First Posts review queue on Movies.SE, and I get an answer to a question about the end of the film.  Can the review queues be altered to skip posts on your ignored tags when you have the 'hide ignored tags' option enabled? 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? Probability of an even number of sixes We throw a fair die $n$ times, show that the probability that there are an even number of sixes is $\frac{1}{2}[1+(\frac{2}{3})^n]$. For the purpose of this question, 0 is even.   I tried doing this problem with induction, but I have problem with induction so I was wondering if my solution was correct:   The base case: For $n=0$, our formula gives us $\frac{1}{2}[1+(\frac{2}{3})^0] =1$. This is true, because if we throw the die zero times, we always get zero sixes.  Suppose it's true for $n=k$. Then the odds of an even number of sixes is $\frac{1}{2}[1+(\frac{2}{3})^n]$, and thus the odds of an odd number of sixes is $1 - \frac{1}{2}[1+(\frac{2}{3})^n]$.    For $n=k+1$, there are two ways the number of sixes are even:  a. The number of sixes for $n=k$ was even, and we do not throw a six for $n=k+1$: $ \frac{5}{6} \cdot  \frac{1}{2}[1+(\frac{2}{3})^n]$  b. The number of sixes for $n=k$ was odd, and we throw a six for $n=k+1$: $\frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$  So the probability $p$ for an even number of sixes at $n=k+1$ is $ \frac{5}{6} \cdot  \frac{1}{2}[1+(\frac{2}{3})^n] + \frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$  I have two questions   How do I get from $ \frac{5}{6} \cdot  \frac{1}{2}[1+(\frac{2}{3})^n] + \frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$ to $\frac{1}{2}[1+(\frac{2}{3})^n]$? I seem to have done something wrong, I can't get the algebra correct, I get $p = \frac{1}{3}[1+(\frac{2}{3})^n] + \dfrac{1}{6}$ Other than that, is my use of induction correct? Is it rigorous enough to prove the formula? Disk full, du tells different. How to further investigate? I have a SCSI disk in a server (hardware Raid 1), 32G, ext3 filesytem. df tells me that the disk is 100% full. If I delete 1G this is correctly shown.  However, if I run a du -h -x / then du tells me that only 12G are used (I use -x because of some Samba mounts).  So my question is not about subtle differences between the du and df commands but about how I can find out what causes this huge difference?  I rebooted the machine for a fsck that went w/out errors. Should I run badblocks? lsof shows me no open deleted files, lost+found is empty and there is no obvious warn/err/fail statement in the messages file.  Feel free to ask for further details of the setup. Is there any advantage in marketing my product for £99.99 instead of £100? Driver for Validity Sensors Fingerprint scanner prakhar@pr3d4tOr:~$ lsusb -s 001:003 -v  Bus 001 Device 003: ID 138a:0018 Validity Sensors, Inc. Fingerprint scanner Couldn't open device, some information will be missing Device Descriptor:   bLength                18   bDescriptorType         1   bcdUSB               1.10   bDeviceClass          255 Vendor Specific Class   bDeviceSubClass        17    bDeviceProtocol       255    bMaxPacketSize0         8   idVendor           0x138a Validity Sensors, Inc.   idProduct          0x0018 Fingerprint scanner   bcdDevice            0.78   iManufacturer           0    iProduct                0    iSerial                 1    bNumConfigurations      1   Configuration Descriptor:     bLength                 9     bDescriptorType         2     wTotalLength           46     bNumInterfaces          1     bConfigurationValue     1     iConfiguration          0      bmAttributes         0xa0       (Bus Powered)       Remote Wakeup     MaxPower              100mA     Interface Descriptor:       bLength                 9       bDescriptorType         4       bInterfaceNumber        0       bAlternateSetting       0       bNumEndpoints           4       bInterfaceClass       255 Vendor Specific Class       bInterfaceSubClass      0        bInterfaceProtocol      0        iInterface              0        Endpoint Descriptor:         bLength                 7         bDescriptorType         5         bEndpointAddress     0x01  EP 1 OUT         bmAttributes            2           Transfer Type            Bulk           Synch Type               None           Usage Type               Data         wMaxPacketSize     0x0040  1x 64 bytes         bInterval               0       Endpoint Descriptor:         bLength                 7         bDescriptorType         5         bEndpointAddress     0x81  EP 1 IN         bmAttributes            2           Transfer Type            Bulk           Synch Type               None           Usage Type               Data         wMaxPacketSize     0x0040  1x 64 bytes         bInterval               0       Endpoint Descriptor:         bLength                 7         bDescriptorType         5         bEndpointAddress     0x82  EP 2 IN         bmAttributes            2           Transfer Type            Bulk           Synch Type               None           Usage Type               Data         wMaxPacketSize     0x0040  1x 64 bytes         bInterval               0       Endpoint Descriptor:         bLength                 7         bDescriptorType         5         bEndpointAddress     0x83  EP 3 IN         bmAttributes            3           Transfer Type            Interrupt           Synch Type               None           Usage Type               Data         wMaxPacketSize     0x0008  1x 8 bytes         bInterval               4   I have searched high and low but I cannot get this to work. All questions related to 138a:0018 are either unanswered or old. Any help guys? Is the $ L^{p}$$[0,1]$ norm continuous in p? I ran into the following problem when I was doing my homework, and I have no thoughts on where I should start with:  (1) If $f\in L^{2}$, show that $\displaystyle \lim_{p \rightarrow 1^{+}}\int_{[0,1]}|f|^{p}=\int_{[0,1]}|f|$  (2) If $0&lt;p$, show that $\displaystyle \lim_{q\rightarrow p^{-}}||f||_{q}=||f||_{p}$  My first thought was Generalized LDCT, but it didn't seem to work. I also made some other attempts but none of them were successful... Can anybody give me some hints on how I should look at this question?  Also, I know if $p\rightarrow\infty$ then $||f||_{p}\rightarrow||f||_{\infty}$ on $[0,1]$, but does similar continuity in p holds for other $L^{p}[0,1]$ norms in general?  Thank you!  Edit:   Sorry if I did not make it clear enough in the question. All $L^{p}$ refers to $L^p[0,1]$.  The first question is found  (thanks to t.b.), but the second question remains, mainly because $f$ is not guaranteed to be in any $L^{p}$. Polygamma reflection formula Avoiding unused variables warnings when using assert() in a Release build This is a bit of a more unusual bug to report:    The screenshot above is unedited - the black box is rendered by the Stack Exchange application.  Phone:  running Android 4.0.4 (ICS) prove $\frac {ab}{a+b} \geq \sum ^n_{i=1} \frac{a_ib_i}{a_i+b_i}$ I've been using the biblatex package for a while, and since updating my TeX installation (post-el capitan upgrade) I have been experiencing the following.   If the following is typed:  Such and such has been said \citep[111]{citekey}.   The following text is displayed:  Such and such has been said (Author, 2016, p. 111).   And if the following is typed:  \citet[111]{citekey} has said such and such.   The following text is displayed:  Author, (2016, p. 111) has said such and such.   The citep command is working as I want it to, but I do not want a comma placed after the author's name when using the citet command. I would like it to display as follows:  Author (2016, p. 111) has said such and such.    If I use the \citeyearpar command, it is fine, but I would rather not have to type the author's name in manually.   Here is the command placed in the preamble:  \usepackage[backend=bibtex, style=authoryear-icomp, url=true, isbn=false, natbib=true, sortcites=true, block=space, pagetracker=false, loccittracker=constrict]{biblatex}   Here is an example of an entry in my bib file (made using BibDesk):  @article{poldrack2006, Author = {Poldrack, Russell}, Journal = {Trends in Cognitive Sciences}, Number = {2}, Pages = {59--63}, Title = {{Can cognitive processes be inferred from neuroimaging data?}}, Volume = {10}, Year = {2006}}   I have also tried changing the entry to the following:  @article{poldrack2006, Author = {Russell Poldrack}, Journal = {Trends in Cognitive Sciences}, Number = {2}, Pages = {59--63}, Title = {{Can cognitive processes be inferred from neuroimaging data?}}, Volume = {10}, Year = {2006}}   The bibliography entry appears as follows, and is fine:  Poldrack, Russell (2006). “Can cognitive processes be inferred from neuroimaging data?” In: Trends in Cognitive Sciences 10.2, pp. 59–63.  Thank you for any help that can be offered. Cost Distance Analyst In QGIS? Are there any tools, plugins or methods in QGIS which allows for an analysis of the best path between two points using least cost path method? As shown below ArcGIS has several distance tools. How does Lockable Suspension work? I recently asked a question about . There  told me about the “blow off” feature. Now, I am curious about how the whole system works.   So, can anyone tell me about how a lockable front suspension works, together with features like “remote-lock” and “blow-off”. Terminal method of formatting storage drive I want to know how to format a storage drive from the terminal. Helpful things to provide in the answer would be often used options for commands and base knowledge that one can use to extrapolate future uses. Specifically I would like to know how to format in the different file systems such as NTFS, FAT32, EXT4, etc. Information on how to partition the drive via terminal is also wanted.  I am trying to format a high capacity external hard drive (EHDD) into NTFS - from the terminal.   I know I can use gparted for this as well as other GUI programs, but I still want to now how to do it from the terminal. Prove: If $\gcd(a,b,c)=1$ then there exists $z$ such that $\gcd(az+b,c) = 1$ I can't crack this one.      Prove: If $\gcd(a,b,c)=1$ then there exists $z$ such that $\gcd(az+b,c) = 1$ (the only constraint is that  $a,b,c,z \in \mathbb{Z}$ and $c\neq 0)$ Can someone help me figure out the big-O for the recurrence relation $T(n) = T(n^{1/2}) + 1$?  I didn't think the master theorem would work since it requires $T(n) = T(n/b)$... to have $b$ as a constant. I was wondering if anyone could shed some light on a method to approach this problem? (not looking for a solution) Embedding images explodes generated PDF size How do I show apt-get package management history via command line? Is there a way to show the history of packages that were changed by apt-get via command line?