When you add the digits of any number that is divisible by three, that sum of those digits also appears to be divisible by three (with no remainder).  For example a number (which I randomly grab from the top of my head):  289752  whose digits sum to 33 (2+8+9+7+5+2=33)  That sum 33 is divisible by three and so is the original number 289752.  This is not the case when dividing by 2, for example 12 is divisible by two but when its digits are summed (1+2=3) you receive 3 which is NOT divisible by 2.  I have yet to be able to find a counter example for this phenomenon of division by 3.  Why does this happen?

The divisibility rule for $3$ is well-known: if you add up the digits of $n$ and the sum is divisible by $3$, then $n$ is divisible by three.  This is quite helpful for determining if really large numbers are multiples of three, because we can recursively apply this rule:  $$1212582439 \rightarrow 37 \rightarrow 10\rightarrow 1 \implies 3\not\mid 1212582439$$ $$124524 \rightarrow 18 \rightarrow 9 \implies 3\mid 124524$$  This works for as many numbers as I've tried.  However, I'm not sure how this may be proven.  Thus, my question is:     Given a positive integer $n$ and that $3\mid\text{(the sum of the digits of $n$)}$, how may we prove that $3\mid n$?


Let's say I want to buy some shares on the stock market. I want to predict whether the price will go up or go down so I look at 3 indicators on my chart. 1 indicators says that there's a 55% chance the price will go up. 1 says there's a 65% chance it will go up and one says there's a 60% chance of a price rise. all indicators are equally reliable.  What can I then say about my total chance of the price going up? Can I treat these indicators as independent? (I don't think so because they all are in relation with the price.) But does that mean I just have to take the average of the 3? so in this case 60%?   But what about 1000 indicators all saying the chance of a price rise is 60%? Intuitively it seems to me that would mean the total odds are bigger than 60%, is that right?   And if there's no mathematically 'correct' way to calculate this, how could I approach a solution then?

How do I solve this probability question?   let's say there are 3 totally independent weather forecast websites. They use totally different ways to predict the weahter.  Website 1 says: there's a 55% chance it's gonna rain, Website 2 says: there's a 60% chance it's gonna rain, Website 3 says: there's a 65% chance it's gonna rain.  What would be a correct estimate of the total probability it will rain? You could argue it's 60% because that's the average, but intuitively that sounds wrong to me. Because for example if 1000 websites claim there's a 60% chance it's gonna rain, isn't that prediction more valuable than if 1 website would claim that? What am I missing here?


In the TNG episode Heart of Glory, Worf states that once a Klingon has died their body becomes an empty shell, and should be treated as such.  What exactly does this mean for the crew? Organic waste can be recycled into the replicator system or ejected from an airlock, but would they feel comfortable doing that? Would they respect Klingon tradition, and if so what does that entail?

In the ST:TNG episode "Heart of Glory", an Enterprise away team rescues three Klingons from a battle-damaged, derelict freighter.  One of the Klingons soon dies from his injuries, prompting the two other Klingons and Worf to howl heaven-ward in the Klingon Death Ritual.  This exchange then happens:     Crusher: "Is there any special arrangement you'd like for the body?"      Klingon: "It is only an empty shell now.  Please treat it as such."      Crusher nods in agreement and understanding   What would she do with the body?


At least how many numbers are needed to be taken to be sure that there are at least 11 numbers among these numbers where the difference between any two is divisible by 7?

Prove that, for any $n+1$ integers $a_{1},a_{2},....,a_{n+1}$, there exist two of the integers $a_{i}$ and $a_{j}$ with $i \neq j$, such that $a_{i} - a_{j}$ is divisible by $n$.  Please help me about this problem,I know there is lots of answers can find from Google, but I am not really understand them. I hope someone can give me more details explanation. I would really appreciate your help.


Let $a \in \Bbb Z$ be such that $a = b^2 + c^2,$ where $b,c \in \Bbb Z \setminus \{0\}.$ Then $a$ cannot be written as $:$   $(1)$ $p d^2,$ where $d \in \Bbb Z$ and $p$ is a prime with $p \equiv 1\ \left (\text {mod}\ 4 \right ).$   $(2)$ $p q d^2,$ where $d \in \Bbb Z$ and $p,q$ are distinct primes with $p,q \equiv 3\ \left (\text {mod}\ 4 \right ).$  $(1)$ is false because $2^2 + 1^2 = 5 = 5 \cdot 1^2,$ where $d = 1 \in \Bbb Z$ and $5$ is a prime with $5 \equiv 1\ \left (\text {mod}\ 4 \right ).$ How do I prove or disprove the other option? Any help in this regard will be appreciated. Thanks for your time.

Let $a\in \mathbb{Z}$ be such that $a=b^2+c^2$ where, $b,c \in \mathbb{Z}-\{0\}$. Then $a$ cannot be written as    $pd^2$ where $d \in \mathbb{Z}$ and $p$ is prime with $p \equiv 1~(mod~4)$ $pd^2$ where $d \in \mathbb{Z}$ and $p$ is prime with $p \equiv 3~(mod~4)$ $pqd^2$ where $d \in \mathbb{Z}$ and $p,q$ are primes with $p \equiv 1~(mod~4)$ and $q \equiv 3~(mod~4)$ $pqd^2$ where $d \in \mathbb{Z}$ and $p,q$ are primes with $p,q \equiv 3~(mod~4)$   The similar question with some part missing is asked , but I want to discuss additional options too.  Here, I got 1) is false Since $13=3^2+2^2$ For other options What should I do


Are both options correct? If so, are there some subtle differences between those?  An increase of serotonin   An increase in serotonin  Thank you!

I was in chemistry class when this question struck me, which sentence is more accurate?  A decrease in the mass of a reactant.  or  A decrease of the mass of a reactant.


Until now, by every boot my NTFS partition (sda3) was mounted for reading and writing and I am the owner, not root. Suddenly, I can't write to it. Here is what fstab looks like, and mount. Could it be that doual-boot Win did something? Fastboot, some flags by NTFS?    And here is mount cmd where it says 'ro' and permissions in Dolphion     Edit: It is not a duplicate of , because there was a problem with ntfs-3g missing which was 2011. thing.

Until recently both my Windows Partition (C:) and my Data Partition (E:) would mount as read/write. Now both seem to only mount as read-only.    GParted Resize (Error)  ntfsresize v2.0.0 (libntfs 10:0:0) Device name : /dev/sda4 NTFS volume version: 3.1 Cluster size : 4096 bytes Current volume size: 86207623680 bytes (86208 MB) Current device size: 86207627264 bytes (86208 MB) New volume size : 78662066688 bytes (78663 MB) Checking filesystem consistency ... Accounting clusters ... Space in use : 62126 MB (72.1%) Collecting resizing constraints ... Needed relocations : 1606868 (6582 MB) Schedule chkdsk for NTFS consistency check at Windows boot time ... Resetting $LogFile ... (this might take a while) Relocating needed data ... ERROR: Extended record needed (1032 > 1024), not yet supported! Please try to free less space.   chkdsk Output:     Checking file system on E: The type of the file system is NTFS. Volume label is Data.  A disk check has been scheduled. Windows will now check the disk.                            99482 file records processed.   495 large file records processed.                        0 bad file records processed.                            0 EA records processed.                                  0 reparse records processed.                             104412 index entries processed.                          0 unindexed files processed.                             99482 security descriptors processed.                    2466 data files processed.                              CHKDSK is verifying Usn Journal...   4479208 USN bytes processed.                                      Usn Journal verification completed. Windows has checked the file system and found no problems.  84187132 KB total disk space.   60479688 KB in 69969 files.      15520 KB in 2467 indexes.          0 KB in bad sectors.     174336 KB in use by the system.      65536 KB occupied by the log file.   23517588 KB available on disk.       4096 bytes in each allocation unit.   21046783 total allocation units on disk.    5879397 allocation units available on disk.  Internal Info: 9a 84 01 00 00 1b 01 00 fc 1a 01 00 00 00 00 00  ................ 01 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00  ................ 42 00 00 00 52 73 5c 77 b0 e7 1d 00 b0 df 1d 00  B...Rs\w........   fstab  # /etc/fstab: static file system information. # # Use 'blkid' to print the universally unique identifier for a # device; this may be used with UUID= as a more robust way to name devices # that works even if disks are added and removed. See fstab(5). # # <file system> <mount point>   <type>  <options>       <dump>  <pass> proc            /proc           proc    nodev,noexec,nosuid 0       0 # / was on /dev/sda5 during installation UUID=8213cc89-2438-41b9-899f-13b4c5b299af /               ext4    errors=remount-ro 0       1 # swap was on /dev/sda6 during installation UUID=45873767-d9ae-4f0c-baab-ffc2726ac24e none            swap    sw              0       0   Checked the drive in GParted, GSmartControl and with ntfsfix, all completed successfully, no change.  Partitions  dev/sda1 NTFS WinRE dev/sda2 NTFS Windows C: dev/sda3 Extended dev/sda4 NTFS Data E: dev/sda5 EXT4 Ubuntu dev/sda6 linux-swap    I am looking for an explanation of:    Internal Info: 9a 84 01 00 00 1b 01 00 fc 1a 01 00 00 00 00 00  ................ 01 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00  ................ 42 00 00 00 52 73 5c 77 b0 e7 1d 00 b0 df 1d 00  B...Rs\w........   and    Resetting $LogFile ... (this might take a while) Relocating needed data ... ERROR: Extended record needed (1032 > 1024), not yet supported! Please try to free less space.


I use a javscript slider called LayerSlider. In Google Webmaster tools it is telling me I have a 404 error with the following link: http://www.example.com/js/layerslider-5/skins/. It says that my home page links to that file path. When I check my home page I have code as follows:   $("#layerslider").layerSlider({   //Other LayerSlider Options   skinsPath: '/js/layerslider-5/skins/' });   This is the JS used to call LayerSlider. The skinsPath tells the script where to find the appropriate skin and apply it. Evidently Google is seeing this as a broken link.   How do I tell Google that this isn't a broken link? I can mark the URL as fixed but it just reappears.   On site there are no errors or problems, just in the webmaster tools.   Will this negatively effect the site in anyway?

This is my first question on here so I hope I'm not breaking any rules, but I have a problem:  I have a site that uses an Adobe Site Catalyst for its online analytics. The suite supports tagging through JavaScript and when implemented, it looks something like this that is placed on http://www.mydomain.com:  <script language="JavaScript" type="text/javascript"> s.channel="/my-tracking-tag/" </script>`   The problem is that Google is reading this JavaScript and interpreting the s.channel string as a relative location on my website and is crawling it. I then get a 404 error in Google Webmaster Tools saying that the URL http://www.mydomain.com/my-tracking-tag/ does not exist, and that it was linked to by the URL that the JavaScript tag code was on.  Is there any to avoid this behavior? I thought about adding URL with a 404 error to my robots.txt file but tracking campaigns are constantly updating and it would be burdensome to update the text file with the constantly updating tags. I also though about structuring the string in s.channel as something like 's.channel= "/" + "my-tracking-tag" + "/" but I'm not sure if that would due the trick, and would also require implementation guidelines to be developed for the future.


This question is from an assignment in complex analysis which couldn't be discussed due to pendamic.  Let $ f\in H(U) , f(U) \subset U, f(0) =0 , f(1/2) =0 $. Show that $ |f'(0) |\leq 1/2 .$  It seemed to me that Schwartz lemma might be useful but from it I can only deduce that |f'(0) |$\leq$ 1 not 1/2 . So, can anyone please tell me how should I deduce the conclusion. I am not able to think about any other result useful in this condition. Thank you.

There is a holomorphic function $f$ from the unit disc to itself s.t $f(0)=f(i/2)=0$. How can you prove the estimates $|f'(0)|\leq 1/2$ and $|f(-i/2)|\leq 2/5$?   Applying Schwarz's lemma we can get $|f(z)|\leq |z|$ but this gives an estimate too large. Cauchy's integral formula seems of no use since we don't know anything about other values of the function.


I am wanting to build a product catalog template that can take XML and import it through InDesign to automatically style text and place product photos. The problem is I don't have a product database of photos, descriptions or prices to work from. Does anyone have ideas on where I could go to find such content? I could write the text myself but at the very least I would like the photos to be sharp and professional.

I'm always on the hunt for free stock images to use while I design, what are some good resources for images that are under a a royalty free or creative commons licence?


I seek a simple explanation about the term "process substitution" aimed for general audience / non professional sysadmins, which do server environment maintenance for personal projects once in a while and not daily for customers or companies, if its okay to ask for it here.  What problem/s this concept was invented to solve What process is being substituted and with what (the name might mislead)? Is it also named "preserving stdin" and if so why? If the stdin device holds standard input which can be piped to another command, what is there to "preserve" further than that?  This might help me to understand the following command: bash <(wget -O - URL)  Note for newcomers: -O - tells wget to write data to stdout;

I was wondering how to understand the following:     Piping the stdout of a command into the stdin of another is a powerful   technique. But, what if you need to pipe the stdout of multiple   commands? This is where process substitution comes in.   In other words, can process substitution  do whatever pipe can do?   What can process substitution do, but pipe cannot?


I have the following regex pattern:  ^[A-Za-z][A-Za-z0-9_-]+$`    It is used to match; alphanumeric characters, underscores and dashes, with the first character being alphabetical.  This works as expected, but I also need it to be able to match single characters. A conditions of a fails.  How can I modify the pattern to make a single alphabetical character pass?

I don't really understand regular expressions.  Can you explain them to me in an easy-to-follow manner?  If there are any online tools or books, could you also link to them?


I tried snap run youtube-dl -F https://www.youtube.com/watch?v=g0TwYwqiQ9E  Got [youtube] g0TwYwqiQ9E: Downloading webpage [info] Available formats for g0TwYwqiQ9E: format code  extension  resolution note 249          webm       audio only tiny   62k , opus @ 50k (48000Hz), 6.35MiB 250          webm       audio only tiny   80k , opus @ 70k (48000Hz), 8.43MiB 140          m4a        audio only tiny  128k , m4a_dash container, mp4a.40.2@128k (44100Hz), 16.98MiB 251          webm       audio only tiny  148k , opus @160k (48000Hz), 16.91MiB 278          webm       256x144    144p   99k , webm container, vp9, 30fps, video only, 11.84MiB 160          mp4        256x144    144p  112k , avc1.4d400c, 30fps, video only, 6.44MiB 133          mp4        426x240    240p  215k , avc1.4d4015, 30fps, video only, 11.62MiB 242          webm       426x240    240p  226k , vp9, 30fps, video only, 18.12MiB 243          webm       640x360    360p  413k , vp9, 30fps, video only, 31.77MiB 134          mp4        640x360    360p  580k , avc1.4d401e, 30fps, video only, 27.32MiB 244          webm       854x480    480p  765k , vp9, 30fps, video only, 51.62MiB 135          mp4        854x480    480p 1160k , avc1.4d401f, 30fps, video only, 51.69MiB 247          webm       1280x720   720p 1525k , vp9, 30fps, video only, 103.96MiB 136          mp4        1280x720   720p 2313k , avc1.4d401f, 30fps, video only, 103.16MiB 18           mp4        640x360    360p  433k , avc1.42001E, 30fps, mp4a.40.2@ 96k (44100Hz), 57.93MiB 22           mp4        1280x720   720p  898k , avc1.64001F, 30fps, mp4a.40.2@192k (44100Hz) (best)  There is no download. How to specify format code?

I use youtube-dl to download files from YouTube. I have tried downloading multiple files, but it is not working out for me. I have to open multiple terminals and start afresh every time I want to download a video.   Can you help me to download multiple files with a single terminal window by just mentioning all the URLs at once? I use Ubuntu 12.04 64-bit.


I’m an American student looking to visit my family in the UK this summer prior to starting my course in September. My course is less than 6 months so I do no need a student visa, simply just enter with my student documents a week before my course begins. My question is if I could visit my family for the summer (approximately 2.5 months) on a tourist visa, which grants me a 6 month stay. I would leave the UK for approximately a month to visit some other family in Europe before returning and reentering now with my student documents a week before my course. Will this be a problem?

I planned a trip to Europe the month before my student visa starts but wanted to stop in the UK for 2 weeks before I continue traveling through Europe. Would I be allowed to enter the UK as a tourist for 2 weeks, leave the UK to go to France (etc.) and the enter later on my Student Visa?


From what I understand, Ubuntu 16.04 and Ubuntu Gnome 16.04 are 2 different OS.  My question is, is it possible to convert my installation of Ubuntu 16.04 to Ubuntu Gnome 16.04? Or maybe I got it all wrong and gnome is just a desktop environment that I can switch to from unity? Maybe?  Thanks in advance.  EDIT: Apparently the forum found a possible duplicate so lemme clear that up :). I am not asking what the differences between Unity and Gnome DE are, I'm asking about the entire OS flavors if I can convert one to the other without a complete reinstall.

I no longer want Unity, and I want to switch to Gnome, I read  that you have to install gnome-session-fallback package.  To do this, however, when running:  sudo apt-get install gnome-session-fallback   I've got this output:  Reading package lists... Done Building dependency tree        Reading state information... Done Package gnome-session-fallback is not available, but is referred to by another package. This may mean that the package is missing, has been obsoleted, or is only available from another source However the following packages replace it:   gnome-session-flashback  E: Package 'gnome-session-fallback' has no installation candidate   So does this mean that I should install gnome-session-flashback instead of gnome-session-fallback to get Gnome? Also as I want to get the latest version of Gnome, which would be Gnome 3 and not 2, is this the correct package to install?    OS Information:  Description:    Ubuntu 15.04 Release:    15.04


Problem:find the residue Res$(f(z),\infty)$ for $f(z)$ as follows $f(z)=\frac{e^z}{z^2-1}$ I tried two methods to solve the problem,but obtained two different results. The first method was like this:  Because Res$(f(z),\infty)$+Res$(f(z),1)$+Res$(f(z),-1)$=0,then I get the answer: Res$(f(z),\infty)=-\sinh1$ The second method was like this: Res$(f(z),\infty)$=-Res$(\frac{1}{z^2}f(1/z),0)$ $$ \frac{1}{z^2}f(1/z)=\frac{e^{1/z}}{1-z^2}=(1+z^2+z^4+z^6+...)(1+1/z+1/(2!z^2)+1/(3!z^3)+...) $$ Then $b_1=1$ so Res$(f(z),\infty)=-1$ I don't know where went wrong? I would appreciate it if you could help me find the reason.Thank you!

According to some reliable sources, the residue theorem can be stated as: the sum of residues of a function on the whole complex plane(including infinity) is zero.  Now consider $$f(x) = \frac{\csc(\pi/x)}{x^2}$$ which has singularities at $+1, -1, 0, \infty$.  By Laurent expansion, I found that the residue is $\frac{1}{\pi}$ for $1, -1$ and $0$, and the residue at infinity is $\frac{-1}{\pi}$.  Obviously, this contradicts the residue theorem.  What mistakes did I make?


I Googled how to do share my wifi connection from my Mac (new Macbook Pro)  to my iPhone (5S) and I tried their way. System Preferences > Sharing > Internet Sharing but then i don't have the same options they showed me.   I have share your connection from: Thunderbolt bridge, wi-fi, bluetooth PAN, and bluetooth DUN.    When I choose wifi, which is what I'm supposed to chose from what i've read, I only get two options in the to computers using: Thunderbolt bridge and bluetooth pan but no wifi and neither is working   All the other share you connection from options allow me to do so with computers using wifi thunderbolt bridge and bluetooth pan.   What should I do? Thanks

I'm using OSX 10.9. I'm trying to set up Internet Sharing to connect my Android to my Mac directly over Wi-Fi. Every tutorial I've come across (including ) mentions checking a box for "AirPort" in a "To computers using" list. However, the option is simply not available on my machine:    I have a few related questions:   Why isn't the AirPort option available? Is there anything I can do to make it available? If there isn't, how can I work around it to finish setting up Internet Sharing so I can connect my Android and my Mac directly over Wi-Fi?


re using 'better'  "You had better do that" (meaning you should do that) or "You better do that": are they both correct?  If so, what part of speech is the word 'better' in each case?

I'm interested to know why we use had better for recommendation. Technically we're speaking of an action that hasn't yet occurred. Like he had better leave a tip means he hasn't yet left a tip, but I suggest he do.    Why do we use the past tense had if it has nothing to do with the past? Why do we not use the subjunctive here? (He have better leave a tip.)


Starting with the following html code  <div ng-repeat="item in list">     <input type="text" class="texter"> </div>   I use jquery to listen on keypress on any input listed.  $('.texter').keypress(function(e){     console.log(this); }   but nothing happens. Even with  $('input').keypress(function(e){     console.log(this); }   Basically, I want to get the element where keypress event is fired.  Is there a way to get it with jquery?  Note that I already searched for an answer before asking but none seems to do what I'm expecting.

I have a bit of code where I am looping through all the select boxes on a page and binding a .hover event to them to do a bit of twiddling with their width on mouse on/off.  This happens on page ready and works just fine.  The problem I have is that any select boxes I add via Ajax or DOM after the initial loop won't have the event bound.  I have found this plugin (), but before I add another 5k to my pages with a plugin, I want to see if anyone knows a way to do this, either with jQuery directly or by another option.


I want to write a paper in latex but I don't know which program I should use. I know I can't use word. Sorry if this question is a duplicate but I can't find the answer.

What editors/IDEs are available for easing the process of writing TeX/LaTeX documents? Please state some useful features like code completion, spell checking, building final DVI or PDF files, etc.  This question is undergoing a systematic refurbishment, see  on Meta. If you’d like to see another editor feature covered here or to take care of an editor that’s not covered yet, share your ideas in that meta question.  List of editors Links to answers.


I'm using OSX 10.10 and its showing movies 4.03 GB in About This Mac>Storage. I don't have any movies in my mac. I tried to check all my files one by one but i can't find it. How to clear this occupied storage. Any idea ?

This is the disk storage info:    This is the Applications folder info:    Why is the storage usage so high and how can it be reduced? This is on a MacBook Pro Retina 13" with OSX 10.9.5.  Edit: I had tried all the suggestions made by @user3439894 in his answer but didn't mention it before posting.


I'm pretty new to JS objects. Can anyone tell me why I am getting this result?     hey arnold   While my desired output is:     hey amy   Here's the code   function greet (person) {            if (person == { name: 'amy' }) {         console.log('hey amy')     } else {         console.log('hey arnold')     } } greet({ name: 'amy' })

A strict equality operator will tell you if two object types are equal. However, is there a way to tell if two objects are equal, much like the hash code value in Java?  Stack Overflow question  is similar to this question, but requires a more academic answer. The scenario above demonstrates why it would be necessary to have one, and I'm wondering if there is any equivalent solution.


At level 20, Druids get access to the following feature:  Archdruid At 20th level, you can use your Wild Shape an unlimited number of times. Additionally, you can ignore the verbal and somatic components of your druid spells, as well as any material components that lack a cost and aren’t consumed by a spell. You gain this benefit in both your normal shape and your beast shape from Wild Shape  The emphasis is on the fact that you can ignore verbal, somatic and non-consumed material components, which makes these spells imperceptible, as per Xanathar:  If the need for a spell’s components has been removed by a special ability, such as the sorcerer’s Subtle Spell feature or the Innate Spellcasting trait possessed by many creatures, the casting of the spell is imperceptible.  Counterspell requires you to be able to see a creature casting:  Casting Time: 1 reaction, which you take when you see a creature within 60 feet of you casting a spell.  As a result, it seems to me that it is impossible to ever counterspell a level 20 druid unless they use a material component that is consumed. Am I missing something? Is the level 20 Druid feature essentially 'become immune to counterspell' as a hidden bonus? Or is there still a way to counterspell them?

On p.85 of Xanathar’s Guide to Everything, there’s a section clarifying that you need to perceive a spell being cast (by either a verbal, somatic or material component) in order to counterspell it.  It then goes on saying this :     If the need for a spell’s components has been removed by a special ability, such as the sorcerer’s Subtle Spell feature or the Innate Spellcasting trait possessed by many creatures, the casting of the spell is imperceptible.   The thing is that many monsters’ Innate spellcasting specify that only material components are discarded (not verbal or somatic). So I am not sure whether you can actually counterspell innate spellcasting when the stat block only mentions removing material components.  There seems to be  supporting that such spells could be counterspelled (although it is confusing a bit).  It basically comes down to whether the above quote’s “many creatures” refers to all creatures with innate spellcasting, or only those that forgo all spell components to innanely cast. So, which is it ?


Since Aviation is in beta, I got access to the 10k tools at 2k. I've already handled the one flag, but the [1] persists on every page I browse to, hereafter.  Can the flag count be updated to only show the number of flags I haven't handled? (i.e. remove it at 0)

I have the moderator privileges on .  Even though it's quite rare that flags appear over there, it is still quite annoying that once I have done my job with a flag and therefore cannot do anything else with it, the flag in the top bar remains there. Why do I have to keep seeing that flag when in fact, I can't do anything about it?  Images, with free-hand circles:  A flag exists in the notification bar:    I have already done my part of the flag handling:    And yet the flag remains in the notification bar.


This isnt a duplicate, as the original answer did not help into implementing it into my raw input command interpreter. Credit to original question/answer where due.  So I'm working on a DOS style Operating System in python and I was wondering how or even if it is possible to execute a command that would let me switch between directories like a regular command prompt would let you do.   Im not sure how well I'm explaining myself but basically I would have the same command as cd in terminal so I could read files imbedded in folders, change between folders etc. I don't need any sort of limit on how complex it is but i would prefer if it isn't too long.  I have yet to find a clear solution to this. in a best case scenario, this is all you would need to input  cd folder1 cd folderinfolder1   this would be inputted into a raw input or a input and then executed so im assuming it would require splits so it knows that before the space there is cd and after the space there is the filename, such as folder I am also have an issue with using os.listdir only showing files, and not any folders.   it would be used like this  directory=raw_input("enter cd followed with a directory")    This might be a bit much to ask, not even sure if it is possible

cd is the shell command to change the working directory. How do I change the current working directory in Python?


How do I find the sum of the following infinite series: $$\left(\frac12+\frac14-\frac23\right)+\left(\frac15+\frac17-\frac26\right)+\left(\frac18+\frac{1}{10}-\frac29\right)+\cdots$$ The series definitely seems to be convergent.

Was given the following infinite sum in class as a question, while we were talking about Taylor series expansions of $\ln(1+x)$ and $\arctan(x)$: $$1+\frac12-\frac23+\frac14+\frac15-\frac26+\cdots$$  The question isn't homework or anything, just a thought tease. I tried for a long while but couldn't find anything remotely close. Thanks in advance for the help.


I've commonly seen references to a wheel user group online as well as when setting up my sudoers file.  Does naming a group wheel imply something special about the group or is it just a name for a generic group used in the same manner that foo and bar are thrown about?

The wheel group on *nix computers typically refers to the group with some sort of root-like access. I've heard that on some *nixes it's the group of users with the right to run su, but on Linux that seems to be anyone (although you need the root password, naturally). On Linux distributions I've used it seems to be the group that by default has the right to use sudo; there's an entry in sudoers for them:  %wheel ALL=(ALL) ALL     But that's all tangential; my actual question is: Why is this group called wheel? I've heard miscellaneous explanations for it before, but don't know if any of them are correct. Does anyone know the actual history of the term?


I have a list of arbitrary length, and I need to split it up into equal size chunks and operate on it. There are some obvious ways to do this, like keeping a counter and two lists, and when the second list fills up, add it to the first list and empty the second list for the next round of data, but this is potentially extremely expensive.  I was wondering if anyone had a good solution to this for lists of any length, e.g. using generators.  I was looking for something useful in itertools but I couldn't find anything obviously useful. Might've missed it, though.  Related question:


I have been residing in France for a year on a D type visa. Now, by the end of this year the visa will be over, and I would like to take advantage of being a Schengen visa free passport holder (annex II country, like Argentina, Australia or Singapore for example) to spend another 3 months as a tourist in Europe, immediately following my current stay.   Do you think I should exit upon visa's expiry and re-enter the Schengen again, or can I just stay for the extra 3 months without leaving? In case I do exit end re-enter, I thought of doing that through the border in Andorra (it's not Schengen, but not sure if they stamp the passport), or make my way south to cross the border to Gibraltar (also not Schengen - and they do stamp for sure).   What would you advise on the situation?

I am an Australian currently on a Netherlands working holiday visa that ends in May next year. I would like to know if when my visa ends I then get another 90 days just as a regular tourist visa as I would like to do some more travel within Europe when my visa ends before going home to Australia.


Need a batch renamer app for Linux.  I have 4 folders with 3 folders each. after the rename operation, I want to have  folder1   - file1.txt   - file2.txt   - file3.txt folder2   - file1.txt   - file2.txt   - file3.txt and so on.   When I try to use Krename => [$dirname]####, I get   folder1   - file1.txt   - file2.txt   - file3.txt folder2   - file4.txt   - file5.txt   - file6.txt and so on.   Any suggestion for a GUI app that can do it? Thanks

I have only spent 5 months with Ubuntu and it has been an awesome experience, I hardly boot into Windows now, but there are some things I miss, like the  I used to have in Windows.   I use the program a lot when I boot into Windows but I'd prefer if there was an alternative or something similar cause the program had a plethora of renaming options with a decent gui.  I'm aware of the rename command in terminal, but I've never had any luck with it.  A screenshot of what it looks like:   Questions   Is there an alternative of this program for Ubuntu? A great command line resource on how to get this done without a gui?


I understand that when beta decay occurs in the nucleus, the reason that the emergent electrons possess a distribution of energies is that varying amounts of the total energy release in the reaction get shared with the neutrinos, which carry it off stealthily. Hence the distribution, which represents the energy theft by the neutrinos. What is the physical mechanism behind the "sharing process" that furnishes the distribution? In other words, what physical process is at work in the reaction which kicks out the neutrino with a nonunique amount of energy (hence leaving both the recoiling nucleus and the electron with nonunique energies)- as opposed to other nuclear reactions where, for example, a gamma ray gets emitted, always with a certain very well-defined characteristic energy?

the spectrum of the Gamma and Alpha decays are both discrete, i.e. the $\alpha$-particles and the $\gamma$-rays take on only discrete values when emitted from a decaying nucleus.  Why is it then, that the $\beta^{\pm}$ can take on continuous values?  The main thing that distinguishes the beta decay from the other two is, that it is a three body problem, i.e. the nucleus does not only decay into an electron/positron, but also into a electron-neutrino/antineutrino. I don't see yet how this immediately implies that the spectrum of the electrons is continuous though.  The way I understand the two body decays is, that the initial nucleus spontaneously decays into the two bodies, i.e. a smaller nucleus and a gamma or alpha particle. As the energy levels in both the nuclei are quantized, only certain values for the energy of the photon and the Helium core are allowed, sine Energy and momentum need to stay conserved.  Does the third particle change things in the way that there basically are two things (i.e. the electron and the neutrino in the beta decay) that are not restricted by an inner energy level hierarchy in the way the nuclei are, thus allowing the energy given by the nuclei to be split arbitrarily (and continuously) between the electron and the neutrino?  If this is the wrong explanation, please correct me.


I tried to install updates via the update manager, so then I used terminal and got   dpkg: error: parsing file '/var/lib/dpkg/available' near line 0: field name `../../../../java-7-openjdk-common/jre/lib/ext/sunjce_provider.jar' must be followed by colon E: Sub-process /usr/bin/dpkg returned an error code (2)   so if anyone could help me i would appreciate that thank you

Been trying to install git-core package for almost more than half a day now. Every time I do, I get this error:  dpkg: error: parsing file '/var/lib/dpkg/available' near line 0: newline in field name `../../../../../share/pyshared/UpdateManager/Core/__init__.py'   I've also tried running sudo apt-get update and sudo apt-get install --fix-missing. Even tried rebooting and cleaning the apt-get cache, still doesnt work.


For a few years, I have been happily using  and recommend it highly.  I haven'f found anything better to help me debug my PDO SQL statements with bound parameters. It will show me the SQL statement which will be generated after bound parameters are substituted, which is A Good Thing.  I post the code below. My problem is that when I introduce a SELECT query using LIKE %gail.com, where $agencyEmailDomain == "gmail.com" and $expandedSqlCommand = PdoDebugger::show($sqlText, $parameters);  expanded to   SELECT count(*) AS count FROM recruiters WHERE recruiter_email LIKE "%'Gmail.com'"  Notice those single quotes around the domain name?  Is there something that I can do, like escaping the % sign?  My code is   $sqlText = 'SELECT count(*) AS count FROM recruiters              WHERE recruiter_email LIKE "%:agencyEmailDomain"';  $sqlParameters =  array('agencyEmailDomain' => $agencyEmailDomain);   and PDO-debugger's code is  /**  * Class PdoDebugger  *  * Emulates the PDO SQL statement in an extremely simple kind of way  */ class PdoDebugger {     /**      * Returns the emulated SQL string      *      * @param $raw_sql      * @param $parameters      * @return mixed      */     static public function show($raw_sql, $parameters)     {         $keys = array();         $values = array();          /*          * Get longest keys first, so that the regex replacement doesn't          * cut markers (ex : replace ":username" with "'joe'name"          * if we have a param name :user )          */         $isNamedMarkers = false;          if (count($parameters) && is_string(key($parameters)))          {             uksort($parameters, function($k1, $k2)              {                 return strlen($k2) - strlen($k1);             });              $isNamedMarkers = true;         }          foreach ($parameters as $key => $value)          {              // check if named parameters (':param') or anonymous parameters ('?') are used             if (is_string($key))              {                 $keys[] = '/:'.ltrim($key, ':').'/';             }              else              {                 $keys[] = '/[?]/';             }              // bring parameter into human-readable format             if (is_string($value))              {                 $values[] = "'" . addslashes($value) . "'";             }              elseif(is_int($value))              {                 $values[] = strval($value);             }              elseif (is_float($value))              {                 $values[] = strval($value);             }              elseif (is_array($value))              {                 $values[] = implode(',', $value);             }              elseif (is_null($value))              {                 $values[] = 'NULL';             }         }          if ($isNamedMarkers)          {             return preg_replace($keys, $values, $raw_sql);         }          else          {             return preg_replace($keys, $values, $raw_sql, 1, $count);         }     } }    How can I change to correctly handle that LIKE "%:agencyEmailDomain"?

Here's my attempt at it:  $query = $database->prepare('SELECT * FROM table WHERE column LIKE "?%"');  $query->execute(array('value'));  while ($results = $query->fetch())  {     echo $results['column']; }


I have a British passport, I am a citizen, lived there all my life.  Do I need a visa to stay in Denmark? If not, how long would I be allowed stay there (while not working or studying, just on holiday)? Could I stay there longer than 90 days if I am not working or studying, just travelling? And if so, for how long can I stay and would I need any kind of visa for this?   This is for someone not claiming any money or benefits, only for staying with a Danish family and travelling around for a period of time.  Very confused about what is needed/not needed as a British person. I used to think a UK person can just stay there for any period of time as long as they're not working or studying, but so many sites seem to say different things about all the laws, I am getting very confused.  Thanks for any information.  Edit: Thanks for the answers so far. Another question: What happens if you have stayed longer than 3 months but are, as said, NOT working/studying/claiming any benefits and intend to go back to home country at a certain point... But have not registered with any authorities in this time? Is there punishments for this and what to do about it if in that situation?

I'm travelling to Sweden for a few days on the 26th of November. I'm a UK resident and passport holder, will I need any other documents in order to travel?


I have a quasi-poisson regression model for analysing the correlation between academic prestige and bulic visibility.  I have a set of independent variables - continuous and dummies in this model. I would like to create scatter plots which describes this model and maybe indicate of an interaction effects. I understand that poisson models are log-linear models, so I tried to plot scatter plot in which the dependent variable is in log transformation. However,  I don't know which line to choose- should it be a linear regression line or  loess line (I'm using ggplot2 plots). for example, in the model, the economics department seem to be more visible then the sociology department. However, in the the plots it seem to be the opposite.      How can I make a reliable plot?

What diagnostic plots (and perhaps formal tests) do you find most informative for regressions where the outcome is a count variable?  I'm especially interested in Poisson and negative binomial models, as well as zero-inflated and hurdle counterparts of each. Most of the sources I've found simply plot the residuals vs. fitted values without discussion of what these plots "should" look like.  Wisdom and references greatly appreciated. The back story on why I'm asking this, if it's relevant, is .  Related discussions:


A thief does not ______ the door.   Is the correct expression to fill the gap 'knock at' or 'knock on'?

Which is more natural and why: 'knocking on the door' or 'knocking at the door'?  And which is grammatically correct?


Suppose $\mu$ is the Lebesgue measure defined on $\Bbb R^k$, I want to show that $\mu$ has some kind of linearity, which seems intuitively correct:      Suppose $A$ is a linear transformation on $\Bbb R^k$ (namely, a square matrix), then given any Lebesgue measurable subset $V$,    $$\mu(AV)=|\det A|\mu(V),$$   where $AV:=\{Av\mid v\in V\}$.   I'm almost sure of it. But I couldn't find a proof about this property in any literature at hand.   I don't think the proof should be hard, though, if the elementary case has been proved. But that's exactly where I get stuck: how am I supposed to, for example, prove the $V=[0,1]^k$ case? Now if this can be proved, then I may just use elementary sets (unions of $k$-cells each two of which share an at most measure zero set) to approach $V$ and the result follows.  Could anyone help me on this part or give me other insights about the proof? Thanks in advance.

I am trying to work my way through the proof of the change of variables theorem for Lebesgue integrals. A key lemma in this context is as follows:  If $T:\mathbb{R}^n \rightarrow \mathbb{R}^n$ is a linear map and $A \subset \mathbb{R}^n$ is Lebesgue measurable then $\lambda(T(A)) = |\det T \  | \cdot \lambda (A)$, where $\lambda(X)$ denotes the Lebesgue measure of $X$.  Can anyone provide a reference for a proof of this lemma that clearly references the facts from linear algebra that are necessary to effect the proof? The source I have for this lemma refers to German texts that I am incapable of reading and I have been unsuccessful at finding an alternate proof.  Added For the Benefit of Future Readers: In addition to the excellent references I received in response to this question, I have managed to find an additional reference that also provides a good proof of this fact: Aliprantis and Burkinshaw's Principles of Real Analysis, Third Edition, Lemma 40.4 pp 389-390.


I have the below data and I have to concatenate the long text column and make it as a single row. The challenge part is only one row has the notification number and other rows are null. You cannot group by the notification number.    I need the output as 2 rows  row number  Notification Number Plant   Creation Date   Language    Lineno  Tag Long Text 1           10014354914         A057    43466            EN            1     >X  aaabbbcccdddeeefffggghhhjjjkkklll 2           10014354915         A057    43466            EN            1     >X  aaabbbcccdddeeefffgggpppqqqrrrsss   I have used cursor for this. But it is taking much time.

I'm trying to migrate a MySQL-based app over to Microsoft SQL Server 2005 (not by choice, but that's life).  In the original app, we used almost entirely ANSI-SQL compliant statements, with one significant exception -- we used MySQL's group_concat function fairly frequently.  group_concat, by the way, does this: given a table of, say, employee names and projects...  SELECT empName, projID FROM project_members;   returns:  ANDY   |  A100 ANDY   |  B391 ANDY   |  X010 TOM    |  A100 TOM    |  A510   ... and here's what you get with group_concat:  SELECT      empName, group_concat(projID SEPARATOR ' / ')  FROM      project_members  GROUP BY      empName;   returns:  ANDY   |  A100 / B391 / X010 TOM    |  A100 / A510   So what I'd like to know is: Is it possible to write, say, a user-defined function in SQL Server which emulates the functionality of group_concat?   I have almost no experience using UDFs, stored procedures, or anything like that, just straight-up SQL, so please err on the side of too much explanation :)


My laptop is Thinkpad T400, and running 64 bit Ubuntu 14.04. I wonder if there is no limit of the RAM size that I can use in my laptop?  Now I am running Windows 8.1 in VirtualBox, which takes 2GB RAM, and if I open 80 Chrome tabs, My RAM (8GB, which is two 4GB RAM cards occupying two slots in the laptop, I am not sure if there are extra slots) will be fully used, and my swap (16GB) will be half full, and the speed is slow because of constant swapping.   I am thinking about upgrade my RAM to a bigger size (maybe 32GB or more). But not sure if there is some limitations on the RAM size, how many RAM cards I could buy and insert into the laptop, and other aspects that I should be aware of. So I appreciate your reply!

I've an  Notebook with an  CPU. The Asus site says that the maximum RAM is 8 GB, but Intel's says my CPU supports 16 GB. What is the maximum RAM I can use on this system under Ubuntu?


I had an little issue when loggin in with Safari on Mac OS X (El Capitan).   When entering the password in Safari, there is a key symbol that lets you choose among any password that is stored in your keychain. This button interferes with the "forgot?" button that Stack Exchange adds and you can't press the Safari button.     Please move this "forgot?" button below the text field, like most other sites implement this.

Safari 9 on OS X introduced a key indicator to allow you to select a password from the Keychain. However, the "Forgot" link inside of the text box makes it unusable. Attempting to click the key always results in the forgot link getting clicked.    It is occasionally useful to be able to use the key to select a password.


I have several power sources (up to 4) going into a circuit and I would like to be able to power it with "automatic" selection. I have looked up power OR-ing and diodes are dissipating more power than I can spare so it was suggested to use P MOSFETS because they have a lower Z.  The problem is that I can't seem to make them work even in simulation and I see a significant voltage drop. Can anyone spot the error?  EDIT: Additional information:   Input voltages might be 3.3V or 5V with +/- 5%. Current consumption from the sources can not exceed 5mA. Power source selection is done while the sources are plugged in and are not expected to change during the operation of the device. When the usage is done the sources are simply unplugged. The solution sought after is passive in order to be low cost. $2 is parts is fine, $5 is probably not. Output voltage should not be under 3V.          – Schematic created using   The main difference between my question and the one suggested as having a possible answer is that I have 4 power inputs, not 2.

I have got two power supply options on my board:   USB 5.0V DC Power Jack of 5.0V   My intention is to design a or-ing power supply option, as either of them will be used at a time. Possible solutions are shown below:    Using oring diodes   or using two p-mos    Possible issues observed with the above options:   Using diode is cheap and good option but an ordinary diode gives a drop of 0.7V and a schottky diode will give a drop of 0.16 to 0.20 volts, which is also not acceptable because my processor  recommended operating power supply is 5.0V and minimum supply voltage is 4.75. With a 0.2V with schottky diode the 5.0V comes down to 4.80V and the space between recommended and supply fed is very less. Then I thought to use p-mosfets which serves the purpose of oring along with reverse polarity protection, but assume a case when both the supplies are fed together by mistake and there is minor difference between USB supply and DC jack say 4.9V for USB and 5.10V for DC Jack. Then, the DC jack supply may pump the current in USB Jack, which may damage USB.   Is there any other better and cheap option I should use for or-ing supply, or is there something wrong with my analysis?


I hava Bootstrap model as below

Priority Queue Selection
           <button type="button" class="close" data-dismiss="modal" aria-label="Close">             &times;           </button>

           <button type="button" class="btn btn-danger cancel">Cancel</button>

   In this modal I am writing some dynamic buttons accoridng to the JSON response from server as below.  var scrn = '';           $.each(pqueueList, function(index, val)            {                scrn += '
';               scrn += '
';               scrn += '

'+val["pqName"]+'

'               scrn += '
';               scrn += '
';            });           $('#modal-text').html(scrn);           $("#PQModel").modal('show');         }   I am driving to capture the click event on each of the button as below.          $("#PQModel #modal-text div.btn").on('click',function(){           var value = $(this).attr('value');           alert("value "+ value);         });   The modal is coming as expected and buttons are coming as expected. But the click I am not able to capture button click event.I am not sure, where am I going wrong. Please help me solve it out.

I have a bit of code where I am looping through all the select boxes on a page and binding a .hover event to them to do a bit of twiddling with their width on mouse on/off.  This happens on page ready and works just fine.  The problem I have is that any select boxes I add via Ajax or DOM after the initial loop won't have the event bound.  I have found this plugin (), but before I add another 5k to my pages with a plugin, I want to see if anyone knows a way to do this, either with jQuery directly or by another option.


I have a pc with 512 mb ram and 1.6ghz and duron processor and 40gb hard disk. I have tried to install Ubuntu, Linux mint,lubuntu but everytime it shows error. Kernel exitted with error.  Only Linux puppy found to be running on it but every software which I tried to install example chrome, Firefox etc get crashed all the time. So what should I be doing now.

For a given hardware configuration, how do I find out if Ubuntu will run on it? What considerations should I take into account when choosing an Ubuntu version and  such as:    with a lighter desktop than the usual Gnome and Unity  with the even lighter LXDE desktop   Obviously Ubuntu does not run on some processor architectures. So how do I go about choosing the right version and derivate. How can I find out the minmal system requirements?


I have a situation like this... I am working on a project that at some point got bigger than I thought it would be. This is project contains basically 2 important parts that work together but I want to separate them.  the first part contains this structure:  -- app ---- templates ------- compiled ------- form-template-files --------- [some-files] ---- Formbuilder ------- Handlers --------- Default.php --------- DGSD.php ------- HandlerInterface.php ------- FormBuilder.php -- public ---- [some client files] ---- index.php -- vendor   This project grew and now I wish to separate FormBuilder directory from the app, into a new repository so I can add it into the packagist so after I could include FormBuilder as a package using composer  The question is how I do this? How to grab the folder with related commits and move them into a new repository, so I could later use this as a package?

I have a  repository which contains a number of subdirectories. Now I have found that one of the subdirectories is unrelated to the other and should be detached to a separate repository.  How can I do this while keeping the history of the files within the subdirectory?  I guess I could make a clone and remove the unwanted parts of each clone, but I suppose this would give me the complete tree when checking out an older revision etc. This might be acceptable, but I would prefer to be able to pretend that the two repositories doesn't have a shared history.  Just to make it clear, I have the following structure:  XYZ/     .git/     XY1/     ABC/     XY2/   But I would like this instead:  XYZ/     .git/     XY1/     XY2/ ABC/     .git/     ABC/


Is there a way to get a spacing which is half or quarter the length of \quad in mathmode?  Maybe there's an easy way to define a shortcut for such command?

I know that \: in LaTeX produces a space when rendered.  Are there any alternatives, because my LaTeX renderer doesn't support \: (it renders it as text), and there is no help / FAQ that I can find.


Could you tell me which is the right sentence please?   I'm pleased to hear that you passed the exam. I'm pleased to hear that you have passed the exam.   Thanks!

I read (or do I say "have read"?) many rules for when to use the present perfect. I found them complex and hard to understand (or do I say "have found"?). I am finding it hard to apply these rules in real sentences.  Is there a simple and clear way to explain the difference between past simple and present perfect? If there is a simple explanation, please tell it! If there is no simple explanation, please tell why it is so complicated. Maybe if I knew why there is no simple explanation, that would help me understand what work I need to do in order to learn the difference.


I'm a beginnner with Blender and I try to train myself watching videos tuts, but today I'm confronted with a small problem that I can't find the answer.  As the pictures describe better than the words, I made a small picture describing my problem.  My problem is that I can't get a UV Map with the right size and the correct deformation. I tried several things but I can't do it. Could you please help me to fix my problem and especially to understand how to do it.

I have tried to map this simple object, but the uv mapping does not correspond to what i see in the textured object. It cuts the image a bit on top and a bit on the bottom of the faces. It does not fill the faces with the texture, even though i have mapped them correctly. Does anyone know what could be happening? Thanks in advance.


For all $x,y\in\mathbb{R}$   define that $x\equiv y$   if $x^{2}=y^{2}$  . Then $\equiv$  is an equivalence relation on $\mathbb{R}$  , there are infinitely many equivalence classes, one of them consists of one element and the rest consist of two elements.  Solution  True. To show that \equiv  is reflexive we need to show that \forall x\in\mathbb{R}  :x=x.  Let x\in\mathbb{R}  , x\equiv\mbox{\ensuremath{x}}  if x^{2}=x^{2},   which is obvious.  [x]  ={y\in\mathbb{R}  |x\equiv y}  =[0]={y\in\mathbb{R}|0\equiv y}={0}  . Hence y^{2}  =0^{2}   =0   which implies that y=0  .

Please give me feedback on my answer to this question. Question:  For all $x,y\in R$ define that $x\equiv y$ if $x^{2}=y^{2}$. Then $\equiv$ is an equivalence relation on $R$, there are infinitely many equivalence classes, one of them consists of one element and the rest consist of two elements. Answer: False, since;  $x\equiv y$   if $x^{2}=y^{2}$  on $R$  To show $\equiv$   is reflexive, we need to show that ,  $\forall x\in R:X\equiv x.$  Let $x\in R$,   $X\equiv x$   if $X^{2}=x^{2}$  , which is obvious.  $[x]\triangleq:\{y\in R/x\equiv y\}$  $[0]=[y\in R/0\equiv y\},$  when $y=0$,  Then $y^{2}=0^{2}=0$  Thus; there are not many infinity many equivalence classes.


I want to make a plus sign for adding notes for a notes-app.  I made a circle that is position fixed, so that it is always visible.  I want to then make two lines that make a plus-sign, and center these lines.  This is the code:   #note-add {   position: fixed;   background: rgb(189, 28, 175);   width: 75px;   height: 75px;   bottom: 0;   right: 50%;   border-radius: 50%;   margin-bottom: 30px; }  .lines {   position: relative;   background: red;   width: 30px;   height: 5px;   margin: 5px 0px; }  #line2 {   transform: rotate(90deg); }

    Please tell me how to do center the plus sign to the fixed-positioned circle.  Thank You!

How can I horizontally center a
 within another
 using CSS?

Foo foo


:     In classical electrodynamics, light is considered as an electromagnetic wave, which is described by Maxwell's equations. Light waves incident on a material induce small oscillations of polarisation in the individual atoms (or oscillation of electrons, in metals), causing each particle to radiate a small secondary wave in all directions, like a dipole antenna. All these waves add up to give specular reflection and refraction, according to the Huygens–Fresnel principle.   My question is, how do all the waves add up to give the direction of angle-in equal to angle-out if they are going in all directions? Also, in reflection does the photon get absorbed and re-emitted or does it act like a dipole antenna?

The "equal angles" law of refection on a flat mirror is a macroscopic phenomenon. To put it in anthropomorphic terms, how do individual photons know the orientation of the mirror so as to bounce off in the correct direction?


I have below a cube type object that has very sharp faces. These faces contain 3 different colors (3 different types of purple) and I want to know if there is a way to transform these faces in a 2D texture so I could then delete the sharp faces and add the texture to a flat face, without loosing the aspect and forms of the colors. Is there a way to do that?

I created a interior scene and  i don't know how to bake a texture using cycles bake.


The problem asks:    I don't even know where to start. Do I need to find the width of the shaded areas, or something else completely?  UPDATE: Using the information given , I was able to act as if the circles were inside of a square and use those methods.  Thank You

A square with edge length 2 cm has semicircles drawn on each side. Find the total area of the shaded region.   Here is an image of the diagram shown :       Please show your work in pictures, numbers, words, anything. (Try to keep it to a Grade 8 level too)


I have a dependency problem on 18.04, libglib2.0-0 is a bit too old (I need 2.56.3 but only 2.56.1-2ubuntu1 is available).  When looking at the , I see that  Paquet libglib2.0-0 (...) bionic (18.04LTS) (libs): GLib library of C routines  2.56.2-0ubuntu0.18.04.2 [security]: amd64 i386  2.56.1-2ubuntu1 [ports]: arm64 armhf ppc64el s390x bionic-updates (libs): GLib library of C routines  2.56.3-0ubuntu0.18.04.1: amd64 arm64 armhf i386 ppc64el s390x (...)   bionic-updates fits the bill. How can I use it with apt?  I currently have APT sources set by default to   deb http://mirror.plusserver.com/ubuntu/ubuntu/ bionic main restricted universe multiverse deb-src http://mirror.plusserver.com/ubuntu/ubuntu/ bionic main restricted universe multiverse

I know there is a GUI in Ubuntu Software & Updates to enable the update channels   updates proposed backports security   as shown in this screenshot:    I am looking for an easy way to do this from within a terminal using commands such as    sudo apt-add-update enable updates sudo apt-add-update enable proposed sudo apt-add-update enable backports sudo apt-add-update enable security  sudo apt-add-update disable updates sudo apt-add-update disable proposed sudo apt-add-update disable backports sudo apt-add-update disable security   and an additional thing  sudo apt-add-update enable default  sudo apt-add-update disable default     Some examples for a better understanding   An empty sources.list   cat /etc/apt/sources.list  <empty>  sudo apt-add-update enable security  <empty>   One enabled repository (main)   cat /etc/apt/sources.list  deb http://archive.ubuntu.com/ubuntu wily main  sudo apt-add-update enable security  deb http://archive.ubuntu.com/ubuntu wily main deb http://archive.ubuntu.com/ubuntu wily-security main   Two or more enabled repositories in one or two lines   cat /etc/apt/sources.list  deb http://archive.ubuntu.com/ubuntu wily main universe   or   deb http://archive.ubuntu.com/ubuntu wily main deb http://archive.ubuntu.com/ubuntu wily universe  sudo apt-add-update enable security  deb http://archive.ubuntu.com/ubuntu wily main universe deb http://archive.ubuntu.com/ubuntu wily-security main universe   or  deb http://archive.ubuntu.com/ubuntu wily main deb http://archive.ubuntu.com/ubuntu wily-security main deb http://archive.ubuntu.com/ubuntu wily universe deb http://archive.ubuntu.com/ubuntu wily-security universe   With deb-src entries   cat /etc/apt/sources.list  deb http://archive.ubuntu.com/ubuntu wily main universe deb-src http://archive.ubuntu.com/ubuntu wily main universe  sudo apt-add-update enable security  deb http://archive.ubuntu.com/ubuntu wily main universe deb-src http://archive.ubuntu.com/ubuntu wily main universe deb http://archive.ubuntu.com/ubuntu wily-security main universe deb-src http://archive.ubuntu.com/ubuntu wily-security main universe   With inactive deb-src entries   cat /etc/apt/sources.list  deb http://archive.ubuntu.com/ubuntu wily main universe # deb-src http://archive.ubuntu.com/ubuntu wily main universe  sudo apt-add-update enable security  deb http://archive.ubuntu.com/ubuntu wily main universe # deb-src http://archive.ubuntu.com/ubuntu wily main universe deb http://archive.ubuntu.com/ubuntu wily-security main universe   The default thing    cat /etc/apt/sources.list  deb http://archive.ubuntu.com/ubuntu wily-security universe  sudo apt-add-update enable default  deb http://archive.ubuntu.com/ubuntu wily universe deb http://archive.ubuntu.com/ubuntu wily-security universe   Only one entry and the disable action   cat /etc/apt/sources.list  deb http://archive.ubuntu.com/ubuntu wily-security universe  sudo apt-add-update disable security  <empty>   Different or the same servers for different or the samerepositories, respect each server   cat /etc/apt/sources.list  deb http://archive.ubuntu.com/ubuntu wily universe deb http://us.archive.ubuntu.com/ubuntu wily main  sudo apt-add-update enable security  deb http://archive.ubuntu.com/ubuntu wily universe deb http://us.archive.ubuntu.com/ubuntu wily main deb http://archive.ubuntu.com/ubuntu wily-security universe deb http://us.archive.ubuntu.com/ubuntu wily-security main   Different Ubuntu releases for different repositories, respect each release   cat /etc/apt/sources.list  deb http://archive.ubuntu.com/ubuntu wily main universe deb http://archive.ubuntu.com/ubuntu trusty main  sudo apt-add-update enable security  deb http://archive.ubuntu.com/ubuntu wily main universe deb http://archive.ubuntu.com/ubuntu trusty main deb http://archive.ubuntu.com/ubuntu wily-security main universe deb http://archive.ubuntu.com/ubuntu trusty-security main   PPAs or other package sources (not-Canonical)  in the sources.list?  Ignore! Don't change the protocols, e.g. https, http, tor, ...


I don’t like you going there.   I don’t like you to go there.      Like can take both -ing and a to-infinitive as complement. But can the expression that the matrix subject and the subject of to-infinitive are different, as (2), be used?  says “We'd like you to come and visit us” is possible. But I’m not sure if (2) is possible.    added samples after consulting Angela Downing's English Grammar, BNC, and COCA:     And her mother did not like her to be out for too long. (Angela Downing, English Grammar, from BNC)   He did not like her to leave his side for more than a day. (T S Eliot. Ackroyd, Peter. London, from BNC)    Christina did not like them to receive such a bad impression of Crystal Springs. (BNC)    she did not like you to call them geraniums! (BNC)    However, schools did not like students to feel at home in arts and science. (BNC)    Mrs Mantini did not like Rachaela to sit and read in the shop. (BNC)    Dear mamma did not like me to use them. (COCA)    David McCutcheon talked with therapists about the one thing he did not like women to do. (COCA)    We did not like them to run free, leaving us at the mercy of their floods. (COCA)    Dad did not like kids to touch his desk. (COCA)    The woman did not like him to speak of dying, death, burial. (COCA)

This is what I read in an answer to a :     Verbs Followed by Either Gerund or Infinitive      Sometimes the meaning changes according to the verb used.         <…>   (dis)like   <…>      I’d like to ask a more specific question: What is that change in the case of (dis)like, exactly? What is the difference between, say, I like being married and (?) I like to be married, I like to eat paper and I like eating paper?


I would suspect that the answer is "yes". Of course any connected metrizable space is uncountable. And any such space has uncountably many open sets: For one can easily show by an inductive argument that any such space has a countably infinite, pairwise disjoint sequence of open sets.  Moreover, the space obviously has infinitely many spheres and each sphere is connected.

Moderator's note: I've copied the title of the post from a more recently posted version of this question by the OP. The text below is from the original question: I am not sure whether the motivations still applied, hence I left it untouched.   It is well known that any connected metrizable space with more than two points is uncountable. And any connected metrizable space has infinitely many open spheres; as each such sphere is connected, there are infinitely many connected subspaces. But must it be the case that any connected metrizable space has uncountably many connected subspaces?


How does one get from Rovaniemi, Finland, to Trondheim, Norway, without a car?   Finnish citizen residing permanently Norway, with pictures of relevant documents and a passport at hand. Not in a terrible hurry.   Some possibilities:   Can one fly from Helsinki to Oslo or Trondheim after taking a train to Helsinki? The flights are booked full for about a week, which is not ideal. Is there a way of getting to Troms/Tromsø/Tromssa and can one fly on (or even take a bus) from there? Is it possible to pass through Sweden? A ship from Vaasa/Vasa or some method the cross the border at Tornio/Torneå are possible ways of getting into Sweden, if it is possible. Is there a functional way of finding people travelling the same route or part of it by private car?

Due to the outbreak of nCoV-19 (Novel Coronavirus) many Countries have imposed several entry restrictions. But these are quite confusing for a lot of people. Some countries allow transit through parts of China (like Shanghai) but some restrict transit even through Hong Kong/Macau. Philippines restricted passengers coming from Taiwan too, but I think that that was later changed.  Where can one find a list of such restrictions for all the countries, ideally a source that is updated frequently


My friend came to visit me and asked  If I wanted something. I wanted to tell him to not bring anything. In this context what is the the right sentence to tell him among two of the following:   a) Don't bring nothing!  b) Don't bring anything!

In my native language, double negations "cancel out". I have been trying to learn English using movies and such things, but double negation confuses me. I see instances when double negation cancels out, and some when they do not.  For example, "I haven't got no money" is interpreted as "I have no money." "There wasn't anyone who didn't enjoy the party" is interpreted as "Everyone enjoyed the party."  Is this context dependent? If so, how does one decide which interpretation is correct?


If I push my hand against a wall, the wall doesn't move, so there is no displacement and so no work done. But energy is being used to generate the force, so what is happening to the energy? Does that mean the total energy is being lost as heat? If so how would you calculate this energy? I though of a way to do it, but I'm not sure if its correct, and would like to know that. Work done i.e. W = F x s can be converted to a function of time and so it  becomes W =  F^2 * s^2 / 2m. And then if I knew the force I applied, the duration for which I applied it, and the mass of my hand, I can calculate the total energy lost as heat, based on the potential distance my hand would have gone if there was no wall. Will this give an accurate result for the energy being transferred to heat?

When we push the wall, no work is being done. But we feel tired after some time. Since energy cannot be destroyed , the question is where does all the energy get transferred in this scenario. Note : It cannot be all heat as we feel more tired than standing still and if it is being transferred in heat then why it is faster than standing still if no work is being done?


Over the years I've often used two of the available ArcMap basemap layers (Imagery and USA Topo Maps) as part of my work.  I've noticed that ArcMap consistently takes 10-15 seconds for the screen to refresh each time I pan or zoom to a new mapextent.  This delay only occurs when I have a basemap layer turned on for display.  I find my productivity lagging as I stare at the screen, waiting for it to refresh.    Yet at the same time other programs (QGIS, Google Maps) refresh their basemap imagery almost instantaneously.  I thus conclude that the problem is with Arc, and not my internet connection speed (direct ethernet connection with typical download speeds > 200 Mbps) or computer power (currently a sixth-gen i7 processor with a dedicated GPU and SSD).  I'm using Arc 10.5, but I've noticed this slow basemap refresh problem ever since ESRI has provided the service (9.0?).  I've always had Hardware Acceleration enabled via the Arcmap Options.    If I return the map extent to a previous view, the imagery displays quickly, implying that Arc is correctly caching the imagery.  But the slow basemap refesh problem always occurs if I pan/zoom to a new area.    I'm not using any advanced cartographic features such as representations, and all of my work is on a local machine - no remote connections.  Is this slow basemap refresh typical?  If not, are there any tips to speed it up?

I'd like to know if there are ways (within ArcGIS itself) to increase the display performance of basemaps within ArcMap. The internet connection was recently upgraded, so it is much faster than before, and by that I mean instead of taking 1 minute to draw the Bing Imagery basemap, it now takes about 30 seconds.   I normally pan and zoom with the basemap off, and once I think I'm in the location I want, I turn the basemap on and take my hands away from the mouse (trying to do anything else guarantees me a spinning circle, and most likely a crash). From  on the ArcGIS forums, I've changed the location of my display cache from My Documents to my C drive, and I could see an improvement.  I then changed the location to our server, and it now takes about 5 seconds to draw when I pan and zoom. While I am quite happy with this, does the fact that I am using 10.1 make a difference? I changed the cache locations for my colleagues who are using 10.0, and their performance did not improve at all. I don't know what else to do, as I can't do anything about the antivirus/firewall part (that's under IT's control).  It's a bit frustrating, particularly when the basemaps in ESRI Maps For Office and ArcGIS for AutoCAD draw almost instantly, and I can zoom in without worrying my computer will implode. My current settings in ArcMap options are hardware acceleration enabled, and cache location changed.  I've looked at the help on , but my settings appear to be optimised already, unless I am missing something.


$$\frac{1}{\Gamma(r)}\int_{\mu}^{\infty}t^{r-1}e^{-t}dt=\sum_{x=0}^{r-1}\frac{e^{-\mu}\mu^x}{x!}$$ What I have tried- $$\frac{1}{\Gamma(r)}\int_{\mu}^{\infty}t^{r-1}e^{-t}dt=e^{-\mu}\sum_{x=0}^{r-1}\frac{\mu^x}{x!}$$ $$\frac{1}{\Gamma(r)}\int_{\mu}^{\infty}t^{r-1}\sum_{n=0}^{\infty}\frac{(-1)^nt^n}{n!}dt=e^{-\mu}\sum_{x=0}^{r-1}\frac{\mu^x}{x!}$$ But I tried interchanging the sum and the integral and it didn't make sense anymore. Any ideas?

Let $T_1, T_2, \dots$ be iid sequence of exponential random variables with parameter $\lambda$. The sum $S_n = T_1 + T_2 + \dots + T_n$ is a Gamma distribution. Now as I understand the Poisson distribution is defined by $N_t$ as follows:  $$N_t = \max\{k: S_k \le t\}$$  How do I formally show that $N_t$ is a Poisson random variable?   Any suggestions appreciated. I tried to work out a number of proofs but cannot get to the final equation.  References


Is this always true? Is it always true for any consecutive integers (change 5 and 6 for any consecutive numbers).

I did some mathematical induction problems on divisibility   $9^n$ $-$ $2^n$ is divisible by 7. $4^n$ $-$ $1$ is divisible by 3. $9^n$ $-$ $4^n$ is divisible by 5.   Can these be generalized as $a^n$ $-$ $b^n$$ = (a-b)N$, where N is an integer? But why is $a^n$ $-$ $b^n$$ = (a-b)N$ ?  I also see that $6^n$ $- 5n + 4$ is divisible by $5$ which is $6-5+4$ and $7^n$$+3n + 8$ is divisible by $9$ which is $7+3+8=18=9\cdot2$.  Are they just a coincidence or is there a theory behind?  Is it about modular arithmetic?


I trying edit the python script archive through Sublime Text Editor instead through Blender Native Editor, because it give not a good support how Sublime Text Editor gives me.  I just want do this, when i edit the code through Sublime Text Editor the Blender detect that has one change in the same archive but not make nothing in script inside of blender editor without i need to click in warning button to refresh the code.  Somebody can give me a solution without need to develop so much code?(I want not be a Blender developer or study very much the Blender)

I have little experience scripting thus far but want to develop my skills further. I am particularly interested in autocomplete features and looking to script geometry within one context for Rhino, Blender, and Processing. Atom handles Rhino and processing, but not sure how to make it work with Blender.  Edit: Is it possible to create an atom package for a bpy/bmesh autocomplete library and send scripts directly to blender?    is how Atom can be used with Rhino Python:     Scripts can be executed within Atom.


We use many variants of Indian Keyboards - Devanagari Inscript (Default Indian), Hindi WX, Malayalam Inscript etc. In earlier Ubuntu versions, we could easily add these keyboards through the System Settings GUI dialogues. In recent 16.04 and 17.04, we could see default Indian (and Urdu) only. How to add other variants?  Do any package need to be installed? We searched the package repository but could not find any separate package describing these variants.  We suppose it needs additional configuration. It might be silly, but don't know how to.  We use multiple distros - Ubuntu 16.04.2, 17.04, Arch Xfce, Fedora 25 Xfce etc. and we could not solve in all those distros. If we could make it in one distro, it must be similar in others.

I've upgraded from Ubuntu 16.10 to 17.04 today. I'm from Bangladesh, and Bengali is my native language. There was an option to set keyboard layout to Bengali in all previous Ubuntu versions I've used. But there's no such option in 17.04, strangely. I can use Bengali by installing the package ibus-m17n,  so I thought maybe there was some problem during my upgrade. But then I checked my friend's PC which runs ubuntu 16.10 and found that ibus-m17n was not installed in 16.10 but there's option for Bengali keyboard layout. So is Bengali keyboard support really removed from 17.04, or was there a problem during my upgrade?


i might have completely missundestood how maven works, but i have the following issue. i created an application with maven in netbeans. it runs in netbeans fine, but when i make Clean and Build it just compiles my source files, and the maven dependendencies are not in the jar file, so cannot run in command window.  this is my pom file. it has build section, the dependencies are in the classpath of mainfest.mf  <?xml version="1.0" encoding="UTF-8"?> <project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd"> <modelVersion>4.0.0</modelVersion> <groupId>com.ibh</groupId> <artifactId>SafePassword</artifactId> <version>1.0-SNAPSHOT</version> <packaging>jar</packaging>  <build>     <plugins>         <plugin>             <groupId>org.apache.maven.plugins</groupId>             <artifactId>maven-jar-plugin</artifactId>             <version>2.4</version>             <configuration>                 <archive>                     <manifest>                         <addClasspath>true</addClasspath>                         <mainClass>com.ibh.safepassword.TestMain</mainClass>                     </manifest>                 </archive>                 <descriptorRefs>                   <descriptorRef>jar-with-dependencies</descriptorRef>                 </descriptorRefs>             </configuration>         </plugin>     </plugins> </build>  <dependencies>     <dependency>         <groupId>org.hibernate</groupId>         <artifactId>hibernate-entitymanager</artifactId>         <version>4.3.1.Final</version>     </dependency>     <dependency>         <groupId>org.hibernate.javax.persistence</groupId>         <artifactId>hibernate-jpa-2.1-api</artifactId>         <version>1.0.0.Final</version>     </dependency>     <dependency>         <groupId>com.h2database</groupId>         <artifactId>h2</artifactId>         <version>1.4.192</version>         <type>jar</type>     </dependency>     <dependency>         <groupId>org.hibernate</groupId>         <artifactId>hibernate-validator</artifactId>         <version>4.2.0.Final</version>         <type>jar</type>     </dependency>     <dependency>         <groupId>org.hibernate</groupId>         <artifactId>hibernate-validator-annotation-processor</artifactId>         <version>4.2.0.Final</version>         <type>jar</type>     </dependency> </dependencies> <properties>     <project.build.sourceEncoding>UTF-8</project.build.sourceEncoding>     <maven.compiler.source>1.8</maven.compiler.source>     <maven.compiler.target>1.8</maven.compiler.target> </properties>     i came from c# world, thought maven is similar than NuGet which is playing the game only in DEV phase, but at compile time everything is there ...  what is missing? how to execute the jar file to have the other dependencies in place and so the app can run?  thanks for your hints!  EDIT  i don't agree that this question is fully duplicated to the one you're referring to. as i can read these many solutions maven is quite complicated, there is no ONE solution for every issue

I want to package my project in a single executable JAR for distribution.  How can I make a Maven project package all dependency JARs into my output JAR?


I wanna convert the linked list to array. But when I print it it gives me the objects' hash code. How can I fix this? I create an array and then equalize it to the person object which is in the linked list.  Person [] personArray; personArray = new Person[phoneList.size()]; System.out.println(personArray);   Phone Book Class:  public class PhoneBook {  Scanner input = new Scanner(System.in); SLinkedList<Person> phoneList; Person [] personArray;  public PhoneBook(){     phoneList = new SLinkedList<Person>(); }  public void addPerson(){     System.out.println("Create new person.");     System.out.println("Name: ");     String name = input.next();      System.out.println("Surname: ");     String sn = input.next();      System.out.println("Address: ");     String a = input.next();      System.out.println("Cell Number: ");     int cell = input.nextInt();      System.out.println("Home Number: ");     int hn = input.nextInt();      System.out.println("Work Number: ");     int wn = input.nextInt();      Person per = new Person(name, sn, a, cell, hn, wn);      phoneList.addLast(per);     personArray = new Person[phoneList.size()];     System.out.println(personArray);  }   Person class :  public class Person {     private String name;     private String surname;     public  String address;     public int cell;     public int home;     public int work;       public Person(String name, String surname, String address, int cell, int home, int work){         this.name    = name;         this.surname = surname;         this.address = address;         this.cell    = cell;         this.home    = home;         this.work    = work;     }      // Accessor methods:     public String getName(){         return name;     }     public String getSurname(){         return surname;     }     public String getAddress(){         return address;     }     public int getCell(){         return cell;     }     public int getHome(){         return home;     }     public int getWork(){         return work;     }      // Modifier methods:     public  void setName(String name){         this.name = name;     }     public void setSurname(String surname){         this.surname = surname;     }    public void setAddress (String address){         this.address = address;     }     public void setCell (int cell){         this.cell = cell;     }     public void setHome (int home){         this.home = home;     }     public void setWork (int work){         this.work = work;     }      public String toString(){         return name + " " + surname + " " + address + " " + cell + " " + home + " " + work;     } }

In Java, arrays don't override toString(), so if you try to print one directly, you get the className + '@' + the hex of the  of the array, as defined by Object.toString():  int[] intArray = new int[] {1, 2, 3, 4, 5}; System.out.println(intArray);     // prints something like '[I@3343c8b3'   But usually, we'd actually want something more like [1, 2, 3, 4, 5]. What's the simplest way of doing that? Here are some example inputs and outputs:  // Array of primitives: int[] intArray = new int[] {1, 2, 3, 4, 5}; //output: [1, 2, 3, 4, 5]  // Array of object references: String[] strArray = new String[] {"John", "Mary", "Bob"}; //output: [John, Mary, Bob]


The full problem is as stated in the title. I am here to check if this is a valid proof. I thought it would be easiest using Linear Algebra.     Recall that an infinite cyclic group is isomorphic to $\mathbb{Z}$. We wish to show that we do not have an isomorphism between $\mathbb{ZxZ\;and\;Z}$. Note that $\mathbb{ZxZ}$ is an infinite group (under addition of course). Now, in order for there to even be potential for an isomorphism, two spaces must have equal dimension. Since the $\mathbb{dim(\mathbb{ZxZ})}=2>\mathbb{dim(\mathbb{Z})}=1$, we know that $\nexists$ an isomorphism between our spaces. Hence, $\mathbb{ZxZ}$ is not a cyclic group.   My question (besides a validity check): was there a better way to prove this? I just found this to be the easiest way.  EDIT Totally wrong with this one. Back to the cutting board.

Proof $\mathbb{Z} \times \mathbb{Z}$ is cyclic.   My intuition is that there are four generators in $\mathbb{Z} \times \mathbb{Z}: 1 \times 1, 1 \times -1, -1 \times 1, -1 \times -1$. And the group they generated are $(z,z)$ or $(z,-z)$, rather than two independent $(z_1, z_2)$, where $z_1, z_2 \in \mathbb{Z}$.  But is this the right way to think about it, and right way to write it? Thank you. :-)


I'm on Ubuntu Mate,  and I would like the system to run the command homebridgeat boot.  Do I just need to   sudo nano /etc/rc.local   and add this  #!/bin/sh -e # # rc.local # # This script is executed at the end of each multiuser runlevel. # Make sure that the script will "exit 0" on success or any other # value on error. # # In order to enable or disable this script just change the execution # bits. # # By default this script does nothing. homebridge exit 0

I need to run a bash script at boot-time as superuser in Xubuntu. I don't know where to start. I have to do something described in a Arch Linux guide. That guide says exactly:     Add following at the end of the file : /etc/rc.sysinit   # Do my action /etc/rc.d/do_my_action  /bin/dmesg >| /var/log/dmesg.log   But I do not have /etc/rc.sysinit and /etc/rc.d/ in Xubuntu. I'm sure there are equivalent ones, right?


I know for a fact that primitives are being passed by value in java. But what are references being passed by? By reference, or by value? I can't find a clear answer.

I always thought Java uses pass-by-reference. However, I've seen a couple of blog posts (for example, ) that claim that it isn't (the blog post says that Java uses pass-by-value). I don't think I understand the distinction they're making. What is the explanation?


I need to ask for your help again ;)  I am working at a 200-page-Document for our high school graduation, which is going to be printed as a book and contains a page for each student with a photo and a short text about him. After I created these ~135 pages by hand, I need to insert a few pages for students who sent their texts later. I used a master page which contained next to graphical elements a text frame for the students' text, I edited that text frame on each page by creating a copy with ctrl+shift+click.  But when inserting a single page somewhere, each left page after that page gets a right page and each right page gets a left page. Every object (except the text frame) changes it's position regarding the master page left / right site, , but since the text frame is rotated -6,25° on left pages and 6,25° on right pages, the text frame keeps its original rotation and is now rotated -6,25° on right pages and 6,25° on left pages (see screenshots).  page 16 and 17 before inserting a page:   page 16 and 17 after inserting a page after page 15:   page 18 and 19 after inserting a page after page 15:  and so on...  How can I get InDesign to understand that these textframes belong to their master frames and should be flipped as well when the pages change from left to right? Any help is much appreciated, since I don't want to edit each of these text frames by hand. Thanks in advance!

I have some designer that moved slightly a text-frame and in order to be managed again by the master page we need to remove the page and create a new one inserting the content in it.  Is there any different way to do it?  I would like to be able to restore the "power" of positioning/size to the master page because now we need to change the master-page to position the text-frame a little bit to the right and in the case above mentioned will not be updated.  Thanks


Hi i am trying to apply some bump maps and normal maps to a floor in my scene but when i uv unwrap it blender warns of non-uniform scale and i think because of that Vector Mapping is resulting in some wired shape

I've got a problem while unwrapping UVs. When I'm unwrapping UVs it showing me this message.     Why is this happening?


I tried to solve . Much of the obvious things to try are there but don't work. After several failed attempts I arrived at this (also failed?) argument:  Let $$f_n(x)=\frac{nx}{1+n^4x^2}$$ The goal is to prove that $$F_m(x)=\sum_{n=0}^{m}(-1)^nf_n(x)$$ converges uniformly in $[0,\infty)$. I've tried the following:  First of all, for every $n\in\mathbb{N}$ and every $x\in[0,\infty)$ we have $|f_n(x)|<\frac{1}{2n}$  Given $x_0\in[0,\infty)$ and taking $f_n(x_0)$ as a function of $n$ we get that the sequence $\{f_n(x)\}_{n\in\mathbb{N}}$ starts to be decrasing when $n^4\ge\frac{1}{3x_0^2}$ or $x_0\ge\frac{1}{n^2\sqrt{3}}$. Now, for $p>q$ $$|F_p(x_0)-F_q(x_0)|=\left|\sum_{n=q+1}^{p}(-1)^nf_n(x_0)\right|$$ and we separate the following cases:  If $x_0\ge\frac{1}{(q+1)^2\sqrt{3}}$ then $\{f_n(x_0)\}_{n\ge q+1}$ is decreasing so we can use "Leinbiz's criterion proof" to get $$|F_p(x_0)-F_q(x_0)|<\left|f_{q+1}(x_0)\right|<\frac{1}{2(q+1)}$$ If $x_0\le\frac{1}{p^3\sqrt{3}}$ then $x_0<\frac{1}{n^3}$ for every $n\le p$ and we have $$|F_p(x_0)-F_q(x_0)|=\left|\sum_{n=q+1}^{p}(-1)^nf_n(x_0)\right|<\sum_{n=q+1}^p\frac{1}{n^2}<\frac{1}{q}$$ which we can make small.  If $\frac{1}{p^3\sqrt{3}}\le x_0\le\frac{1}{(q+1)^2\sqrt{3}}$ then none of the bounds i came up with tend to zero. Does someone have an idea? If this worked somehow I think it would be a neat thing to try for other problems.  Edit: Please note that when I uploaded my question, the original one already had an accepted answer. I'm not looking for a solution of that exercise but rather for some way to complete my line of reasoning. The functions $f_n$ or $F_m$ don't interest me as much as the "dividing the real line" trick and if it works here or in general for alternating sums of functions  Thanks!

$\newcommand{\FUNC}[1]{#1 \frac{nx}{1+n^4x^2}}$  The following problem has bothered me for a very long time. I've been trying to solve this problem for 3 days long - and it feels I tried everything. To conclude - I absolutely need the help of the professionals; And that's why I came here.  I am given with the following function series:  $$S(x)\equiv\sum_{n=1}^{\infty}(-1)^nf_n(x)\equiv\sum_{n=1}^{\infty}\FUNC{(-1)^n}$$  My task is to decide whether the series $S(x)$ uniformly converges on $[0,\infty)$, or not.  Things I tried:  (1) I tried to use Leibniz's Alternating Series Test. I proved that $f_n(x)$ uniformly converges to $0$, but the problem is that $f_n(x)$ is not a decreasing sequence - for every $n\in\mathbb{N}$, there exists $0<x_0\in\mathbb{R}$, such that $f_n(x)$ is increasing on $[0,x_0]$, and decreasing on $[x_0,\infty)$. I've noticed that $\lim_{n\to\infty}x_0=0$, but that doesn't help - since it would never actually reach $0$ (that would only help to prove that the series uniformly converges on $[c,\infty)$ when $c>0$).  (2) I tried to use Weierstrass's M Test. I found that $f_n(x)\leq\frac{1}{2n}$ for every $n\in\mathbb{N}$, but that of course won't help (since $\sum_{n=1}^{\infty}\frac{1}{2n}$ diverges). The huge problem using the M test is that $\frac{1}{2n}$ is actually a tight bound.  (3) I proved that $\displaystyle \sum_{n=1}^{\infty}\FUNC{}$ uniformly converges on $[c,\infty)$ when $c>0$, but does not uniformly converges on $[0,\infty)$. That wasn't very helpful, since $S(x)$ might converges conditionally.  (4) Defining:  $$S_m(x)\equiv\sum_{n=1}^{m}\FUNC{(-1)^n}$$  It is sufficient to show that:  $$(*) \lim_{m\to\infty}\left(\sup_{[0,\infty)}\left|S_m(x)-S(x)\right|\right)\equiv\lim_{m\to\infty}\left(\sup_{[0,\infty)}\left|\sum_{n=m+1}^{\infty}\FUNC{(-1)^n}\right|\right)=0$$  in order to prove the series uniformly converges on $[0,\infty).$ Looks impossible and complicated - yet I was eager to try that too. But to no avail.  I will say that my intuition is that the series does uniformly converge on $[0,\infty)$, and that is because I used graphic calculators, and found that for every $n\in\mathbb{N}$:  $$\sup_{[0,\infty)}\left|\sum_{n=m+1}^{\infty}\FUNC{(-1)^n}\right|\leq\sum_{n=m+1}^{\infty}\frac{1}{n^2}$$  And since $\displaystyle \sum_{n=m+1}^{\infty}\frac{1}{n^2}$ converges, we can say that its limit when $m\to\infty$ is $0$, thus proving $(*)$. However - I couldn't prove this inequality. I tried using the triangle inequality, to distinguish between odd and even indices - nothing worked.  As you might see - I really need help. Thank you very much!


I am unable to add a repository via the command line on Ubuntu 14.04. Apt-get works perfectly fine, and I am able to update the system. I have configured my /etc/apt/apt.config file as follows:  Acquire::http::proxy "http://<username>:<password>@<proxy address>:<port>/";   For example, when trying to add the repository for y-ppa-manager, I receive:  add-apt-repository ppa:webupd8team/y-ppa-manager Cannot add PPA: 'ppa:webupd8team/y-ppa-manager'. Please check that the PPA name or format is correct.   If you need any more info., feel free to ask.

I'm trying to follow the  (on my Ubuntu 10.04 VM).  The first step is:  sudo add-apt-repository ppa:linaro-maintainers/toolchain   But when I run that I get:  Error reading https://launchpad.net/api/1.0/~linaro-maintainers/+archive/toolchain: <urlopen error [Errno 110] Connection timed out>   I'm guessing the problem is with my configuration for my employer's proxy server, particularly for HTTPS.  When I opened  in Firefox, I got "The connection has timed out".  I then went into Firefox Preferences -> Advanced -> Network -> Settings... and set HTTP Proxy to "gateway" and Port to 8080, and checked "Use this proxy server for all protocols".  Then the page loaded.  This supports my theory.  I've tried setting the http_proxy and https_proxy environment variables (both lower and upper case):  $ set | grep -i proxy HTTPS_PROXY=https://gateway:8080 HTTP_PROXY=http://gateway:8080 http_proxy=http://gateway:8080 https_proxy=https://gateway:8080   I also tried changing them to:  $ set | grep -i proxy HTTPS_PROXY=gateway:8080 HTTP_PROXY=gateway:8080 http_proxy=gateway:8080 https_proxy=gateway:8080   And I tried adding this second line to /etc/apt/apt.conf:  Acquire::http::proxy "http://gateway:8080/"; Acquire::https::proxy "https://gateway:8080/";   But I continue to get the same error.  How do I troubleshoot this?  Update: I followed the instructions in the  to  and got further:  $ sudo add-apt-repository ppa:linaro-maintainers/toolchain Executing: gpg --ignore-time-conflict --no-options --no-default-keyring --secret-keyring /etc/apt/secring.gpg --trustdb-name /etc/apt/trustdb.gpg --keyring /etc/apt/trusted.gpg --primary-keyring /etc/apt/trusted.gpg --keyserver keyserver.ubuntu.com --recv 1DD749B890A6F66D050D985CF1FCBACA7BE1F97B gpg: requesting key 7BE1F97B from hkp server keyserver.ubuntu.com gpgkeys: HTTP fetch error 7: couldn't connect to host gpg: no valid OpenPGP data found. gpg: Total number processed: 0   I wonder if the "HTTP fetch error" means I also need to add "Defaults env_keep = http_proxy" to /etc/sudoers...  Update #2: I added:  Defaults env_keep="http_proxy" Defaults env_keep="https_proxy"   ...as in , but I still get the same "gpgkeys: HTTP fetch error 7: couldn't connect to host" error.  And this is curious:  $ sudo env | grep proxy https_proxy=https://gateway:8080   ...because http_proxy isn't there.  Any advice is appreciated.  Update #3: Since the host of my VM is a laptop, I took it home and tried there (with no proxies):  $ sudo add-apt-repository ppa:linaro-maintainers/toolchain Executing: gpg --ignore-time-conflict --no-options --no-default-keyring --secret-keyring /etc/apt/secring.gpg --trustdb-name /etc/apt/trustdb.gpg --keyring /etc/apt/trusted.gpg --primary-keyring /etc/apt/trusted.gpg --keyserver keyserver.ubuntu.com --recv 1DD749B890A6F66D050D985CF1FCBACA7BE1F97B gpg: requesting key 7BE1F97B from hkp server keyserver.ubuntu.com gpg: key 7BE1F97B: public key "Launchpad Linaro Overlay PPA" imported gpg: Total number processed: 1 gpg:               imported: 1  (RSA: 1)   It worked.  So problem is related to proxy configuration.  But I suppose I have a workaround.


First off, I understand .  Today I ran across this for the first time when I signed up on the musicians' SE site. I was confused as to why I got all this free rep. After looking into it, I see that there is a good rationale behind starting you off right on a new site, as explained by this paragraph:     If you are an experienced Stack Exchange network user with 200 or more reputation on at least one site, you will receive a starting +100 reputation bonus to get you past basic new user restrictions. This will happen automatically on all current Stack Exchange sites where you have an account, and on any other Stack Exchange sites at the time you log in.   That's very clear. It explains how it works. I'm not asking "how" it works, but why. I have read  as well as several others, and they only explain what it is, and that it is by design. My issue is exactly the same as described in , but again, I understand why it happened--it is by design--I just don't understand why it is by design.  Why do I receive this rep bonus on my first SE account?   It makes sense I should get the bonus on a new account, as it says above, to get past some new user restrictions. But if I already have 200 rep on another site (this is the reason I'm getting it on the new site), why give me 100 rep on the old one as well?

Not sure if it is a bug or not, but it feels like a bug to me. When I got to 200 rep on SO the other day I got +100 rep to all my accounts including SO. So my rep jumped up to 300 on SO.  Considering there are others on SO with rep between 200 and 300 I think it is possible that the +100 was supposed to apply only to the linked accounts, not to the account with 200+.


I just installed an application that has the bad habit of automatically scanning all my folders for files to edit.  The app has no available options to disable this behavior. I cannot just remove read/list permissions for those files because the app runs with my user permissions, and I still want other applications to be able to access them.  Is there a security mechanism to prevent the application from being able to read some folders? I tried using apparmor, but since it is "default deny" there doesn't seem to be a way to say "allow the app to do everything it wants EXCEPT access these files".  Thanks.

I hear a lot of talk about apparmor, I want to know the following:    What is apparmor? How does apparmor work?