I'm a brand new user and a first timer to this type of site - I like to use all the resources available when I have a question or a problem (research, google, bing, etc.) so I came here when I got my first real road block in my script where I couldn't find the answer anywhere, but I can't use many of the features (ex: chat) on this site until someone, somehow finds my question in the mist of many, answers it, or gives enough 1 ups... Why is that? How are we supposed to build Reputation when it seems questions by new users are a needle in a haystack to find because they have little or no Reputation? Just curious...       P.S. I saw "How does “Reputation” work?" and "How to gain rep (new users)?"     before posting my question and I didn't mean for it to appear that I was asking a      duplicate question, sorry)    In order to clarify, I just wondered if there was a way to improve this wonderful site to make it easier and more helpful on us newbies who know basic to little knowledge of JavaScript who design websites as hobby?   By the way great site before becoming a member the search engine I used always put this site at the top of their list and I was able to find many useful answers to solve my code issues. Thank-you.  My Question:
Seems like a stupid question, but I'm after a nice solid answer if possible.  I joined SO a while ago to ask the odd question, but recently I'm starting to use it more often.  I've been reviewing the 40 max items (20 first posts/20 late answers) per day, and editing anything I see wrong with them (lots of +2's)... flagging them... etc.  I've also been  per day.  But other than commenting where I can and answering the odd question, I'm only  (1k at the moment).  I might just be overeager, but I was hoping some high-rep users might be able to write a little something on how they earned their stripes?  I've , and to be honest I was hoping I'd get a hit, because it sounds like a stupid question... anyways, here you go!  Also, I'm not referring to how to gain rep on Meta, since it's a bit of a fluke I've earned the rep I have on here. I'm asking about main SO.
There has been talk of how to answer a negative question without ambiguity, most often with a qualifying phrase needed for clarification. (For example, "yes, I do"/"no, I don't.)  I've noticed that other languages have a single word to address a negative question on the contrary. In German, ja means yes and nein means no. Alternatively, doch is used to respond in the positive to a negative question, in essence meaning "No, I intend to do the opposite of what your question presumes."  Take for example the question, "You don't want to go to the movie then?" German speakers could simply reply, "nein" if they don't want to go or they would say "doch" if they did want to go; an unambiguously positive response to a negatively phrased question. No need for, "no, I actually would like to go" or "yes, I'll come actually". In French, the respective equivalents of ja, nein, and doch are oui, non, and si. I imagine many other languages contain this set as well.  I am wondering whether there was at one time an equivalent word in the English language that has since (unfortunately) disappeared. Any ideas? 3rd iterate of a continuous function equals identity function According with Mumford the answer is yes, but there are some obscure points in the proof.  We know that there exists a very ample line bundle on $A$ since every abelian variety is projective. Hence, via some corollaries of the Theorem of the Cube, we get an ample line bundle whose restriction to the kernel of $n_A$ is both the trivial bundle and very ample. This should imply that the dimension of $\ker n_A$ is $0$.  Furthermore, Mumford says that the fact that the dimension of the kernel is $0$ implies the surjectivity of $n_A$.  Is anybody able to clarify these two points to me?  Thanks in advance! I am visiting Vietnam in November 2015, I will entering the country on the 10th and my flight home is 14 days later, I will be leaving Vietnam and travelling to Cambodia in between the 2 weeks then travel back into Vietnam to get my flight home on the 24th November, do I need a visa? Can't get hold of anyone at Embassy and not clear about entering twice on the website. Number of pairs of strings satisfying the given condition We are provided with a string A (may contain repeated characters as well ). One needs to make all possible permutations of A , and find out how many pairs of strings chosen from these permutations are not similar.  Two strings are similar if we can make both of them exactly same by carrying out at most 1 swap in each of them .  Example :- if we are provided with string abcd  , then for permutation abcd , the permutations which are not similar to it are bcda , bdac , cadb , cdba , dabc , dcab . Similar is the case for other permuations as well. So overall number of desired pairs are 24*6=144 .  I tried solving it for each possible pair by approach mentioned here     But as the permutations can be large enough , I need a better approach of solving this. Check if file is already open PDF of a sum of exponential random variables Let $X_i$ for $i=1,2,...$ be a sequence of i.i.d exponential random variables with common parameter $\lambda$. Let $N$ be a geometric random variable with parameter $p$ that is independent of the sequence $X_i$. What is the pdf of the random variable $Y=\Sigma_{i=1}^N X_i$. Does $\int_a^bf(x) = 0$ imply $f(x) = 0$ for a Riemann Integral function? Let $a &lt; b$ and $f:[a, b] \rightarrow \mathbb R$ be a Riemann integral function. Suppose that $f(x) \geq 0$ for all $x \in [a, b]$. I want to show whether $\int_a^bf(x) = 0$ implies $f(x) = 0$ or not, but I'm not sure where to start. I don't know much about Riemann integral functions in general. I only know that $$\underline \int_a^b f= \overline \int_a^b f.$$ Also, I know the fact that if $f$ is continuous, then $f(x) = 0$ for all $x\in [a, b]$. Can anyone provide any hints for me, because I'm really confused about this whole thing. Fighting Spam - What can I do as an: Email Administrator, Domain Owner, or User? How to solve missing libmng.so.1 in Trusty (14.04) Package is no more available I would like to run SoulseekQT. However unfortunately one of the dependencies is libmng1, which is no more included in Trusty. sudo apt-get install libmng.so.1 or libmng1 or libmng1:i386 in Trusty (14.04) no more available. Do you have any idea how to solve this conflict? I mean is it really sure that libmng1 is excluded? Could I use libmng2 somehow instead? If not, could I somehow rebuild libmng1 for Trusty? Or do I have to downgrade back to Saucy (13.10)? "Can't find Project or Library" for standard VBA functions Is it possible for more than two particles to be entangled in a quantum way? So I know that two particles can be entangled in a quantum way, but is it possible that more than two particles be entangled in a quantum way? Most descriptions provide with two-particles cases, so I wonder. (It's hard to think of three particles entangled in spin, or so.) Are there naturally occurring lasers? Are there any open source alternative to Landscape? mathtools: \dddot cause the term to rise How to back up my entire system? If I want to make a backup of my entire OS (including but not just my home directory), how do I go about this?    Is it as simple as backing up everything in / and then when if I suffer a crash, just copying the files back over? Will this cover grub, and how do I actually do this when the system is inoperable? How do I add custom validations to Webform? I have created the country and state fields with the Webform module, and I want to add a field validation for the state field when user select US as country.   How can I achieve this? How can I customize different nodepart with different textures? My question is how can I customize hatch patterns(grid, vertical line, etc.) for different nodepart in a node?  The following is a failed try.  \documentclass[border=2mm,convert]{standalone} \usepackage{tikz} \usetikzlibrary{shapes.multipart,patterns}  \begin{document} \begin{tikzpicture} \node[rectangle split,rectangle split parts =2,rectangle split part fill={red!30,red!10}, rectangle split part pattern={grid,},draw]{one \nodepart{second}two}; \end{tikzpicture} \end{document} Coefficients change signs I have a dataset where there is a high degree of multicollinearity, with all variables correlating positively with each other and the dependent variable. However, on some of the models I run I get a couple of significant negative coefficients. Basically there are two coefficients that depending on what variables I include in the model, I can manipulate their signs.  My understanding is that if the variance-covariance matrix only contain positive values, then all coefficients should also be positive. Is this correct? How do I backup my Ubuntu Phone?