Adding symbols to the end of a line
Showing arrowheads in line end using QGIS?
Spinning bike's pedals are way bigger than the cleat pedals I have. What size do I need? I've shown the existence of unique square roots of all positive rational numbers, so now I want to prove that the square root is monotone:   $0&lt;a&lt;b$ if and only if $\sqrt{a} &lt; \sqrt{b}$ How do you start Windows 10 in safe mode? I've been through the Windows 10 upgrade, and am experiencing a recurring "INTERNAL_POWER_ERROR", which is stopping me from logging in.  I'd like to be able to try to boot to safe mode, but it appears that pressing F8 during boot is no longer enough, so how can I boot in safe mode (or any of the other recovery options) on a Windows 10 PC that doesn't let me log on? How do I create an invisible character? I cannot remember how to create an invisible character in LaTeX, i.e. I want to put a space that has the width of a particular character, say `M'. I know there is a command for this, and this is a really dumb question, but my Google-fu has failed me. HP laptop F1-12 keys inverted with Fn I recently bought an HP laptop. I'm pleased with it, save for one thing.  All the F1-12 keys are inverted. I mean the following:  All the F-keys have a symbol on it. Normally you would press Fn + your to use it. On this laptop, it's reversed. The symbol is what you get when you press the button normally and you have to press Fn + F-key to get the usual action.  So, if I want to refresh a webpage, instead of just F5, I have to do Fn + F5  I want to change this to regular behavior. Anybody know how I can do this?  Running Win7 Home premium, 64-bit. How to turn off chrome's annoying auto-fill? Why girth of Petersen Graph is five? (Proof) I want to know, what is the mathematical proof that shows the shortest cycle in Petersen Graph has length five? Merge two unadjacent partitions I have three partitions on my drive, arranged something like the following:   ___________ | 1 |  2 | 3| -------------   I wish to merge partitions 1 and 3.  Is there a way to do this, or do I have to manually "slide" partition 2 over to partition 3, and then merge the newly open space where partition 2 used to be?  Note: I do not have the space to copy partition 2 to partition 3 (2 is 200GB and 3 is 65GB). In other words, why do we need to connect the battery positive to the negative to get electron flow? As far as I know, voltage difference is what drives current flow.  From what I understand, there's a surplus of negative charge (electrons) in the positive end of a battery (weird I know, but I guess they do it for mathematical reasons). Between the positive and negative ends of the battery is some kind of wall that prevents the electrons from diffusing, so they have to go the long way (through a wire to the other end of the battery) to diffuse and reach the receiver atoms/molecules.  So why not just have the surplus of electrons on one end go to ground to equalize (e.g. 5 volts to 0 volts should produce current)? Why does it have to be the other end of the battery? I'm sure there's atoms/molecules in the ground that could accept the extra electrons (receivers atoms/molecules for the donors in the battery). Create a table with LaTeX How to deal with duplicate answers posted long after first answer(s) I've been seeing this for a while now, especially on the  tag.  A user will ask a question (usually very simple, and mostly a duplicate of older questions. But that's for another day).  The user will get one or more answers that successfully answer the question. A little bit later, usually around 5-10 minutes but can be more, another user or users will come along (mostly low rep users) and post the exact same solution, sometimes changing the wording a bit, but sometimes being lifted word for word from the other answer(s).  This is slightly different from the Fastest gun in the west problem. I have nothing against answers posted in very quick succession that solve the problem. The ones I'm talking about are answers posted long after the first correct solution, that say the same thing.  A good example of this is available . All three answers say the same thing. One offers code + explanation (first answer), the second one (three minutes later) offers only code and the third one (a full eighteen minutes later) offers only a text answer.  How should answers like this be dealt with? They are adding nothing of value to the community, as the problem has already been answered. The person posting the answer could not have run into the "The other and I posted the answer so close to each other we didn't know about duplicates until after posting" problem, as there is an eighteen minute gap.   is another example. Find $\min(\operatorname{trace}(AA^T))$ for invertible $A_{n\times n}$ For invertible $A_{n\times n}$ find $\min(\operatorname{trace}(AA^T))$  (a) $0$ (b) $1$ (c) $n$ (d) $n^2$  Clearly for $A=I$, it is $n$, and I am unable to get any lower values, but how do I prove it. I find myself needing to draw lots of elegantly curved paths in TikZ. Ideally, I'd just specify a series of points, and TikZ would calculate the extra data itself to draw a nice series of curves passing smoothly through these points, perhaps with an optional "looseness" parameter that I could specify. But the only way I can find to draw nice curves is by explicitly giving control points, or by manually specifying in and out angles.  I can think up a simple algorithm to do this, which would certainly be within TikZ's power to perform: just choose the in and out angles in a simple fashion based on the relative angles between each adjacent pair of line segments.  Is something like this already built-in? Or can someone cook something up that does the job?  Edit: Jake has provided an answer using the plot [smooth] functionality. This is almost perfect! But it can't do what I need, because it doesn't let me specify tangent angles manually where needed, which is especially important at the beginning and end of the curve. I would have thought this would be a natural and straightforward addition to the existing plot [smooth] algorithm: for every coordinate, an optional angle should be able to be specified as an argument, which if supplied is treated as the tangent angle for the curve at that point. And while we're at it, it wouldn't hurt to also allow the tension to be modified along the path.  A minimal extension to the algorithm would just accept two optional parameters, for the curve tangent at the beginning and end. Subjunctive mood: 'was' usage after 'I wish' and 'if only' had declared the sentence below as follow: 1) STAR TURN, 2) BEST LINES.      I wish Bernard Cribbins was my grandad. Perhaps he would be willing to   adopt?     Rattigan: “If only that was possible.” The Doctor: “If only that were   possible. Conditional clause.”    Is 'was' after 'I wish', 'if only' and 'if' (in conditional clauses) colloquial?  Does it be used in formal speech and writing? Give warning when [super method] is not called Cardinality of the Irrationals Possible Duplicate:          We previously proved that $\mathbb{Q}$, the set of rational numbers, is countable and $\mathbb{R}$, the set of real numbers, is uncountable. What can you say, then, about the cardinality of the irrational numbers?  I would say that the cardinality of irrational numbers is uncountable. That would be because the set $\mathbb{Q}$ is a subset of $\mathbb{R}$. The irrationals is also a subset of $\mathbb{R}$. Basically, my thought is that the continuum is composed of irrationals and rationals (only two subsets of reals). Because we can prove that the rationals are countable by Cantor's proof, then the remaining subsets of $\mathbb{R}$, the irrationals, has to be uncountable for $\mathbb{R}$ to be uncountable. My concern is how to go about the prove this (rigorous, intro level). Consider numbers of the type $n=2^m+1$. Prove that such an n is prime only if $n=F_K$ for some $ k \in N$, where $F_k$ is a Fermat Prime. Consider numbers of the type $n=2^m+1$. Prove that such an $n$ is prime only if $n=F_K$ for some $ k \in N$, where $F_k$ is a Fermat Prime. Consider numbers of the type $n=a^m-1$ where $a&gt;1$ is a natural number. Prove that such an $n$ is prime only if $n=M_p$ for p-prime, where $M_p$ is a Mersennt Prime.   I do not understand what the question is asking me for. Can I get some clarification? Referee says the proof is wrong, but it is not, what to do? About 3 months ago, my co-author and I submitted a mathematical paper to a journal. Yesterday, we got the editor's response and the paper has been rejected. The reason is that the referee thinks that our results are wrong or trivial.  However, we carefully read the referee's report and we conclude that the "errors" pointed out by the referee are not errors at all and that our proofs are correct. In all modesty, we think that the referee understood very poorly even the statements of our theorems.  Personally, this is the first time that such a situation occurs to me. I already have four papers published in such journal and the reports I got from the referees have been always really accurate.  What do you think is the best thing to do? Forget about and try another journal? Write an e-mail to the editor explaining why we disagree with the decision? Other ideas?  EDIT: Thank you very much for all your answers. At the end, we submitted our paper to another journal with only some slight edits we have already planned to do independently by the referee's report. How to recover a removed file under Linux? By accident, I used rm on a file I didn't want to delete. Is there a way that I can get it back under Linux? Creating your own Tinyurl style uid I'm writing a small article on humanly readable alternatives to Guids/UIDs, for example those used on TinyURL for the url hashes (which are often printed in magazines, so need to be short).  The simple uid I'm generating is - 6 characters: either a lowercase letter (a-z) or 0-9.   "According to my calculations captain", that's 6 mutually exclusive events, although calculating the probability of a clash gets a little harder than P(A or B) = P(A) + P(B), as obviously it includes numbers and from the code below, you can see it works out whether to use a number or letter using 50/50.  I'm interested in the clash rate and if the code below is a realistic simulation of anticipated clash rate you'd get from generating a hash. On average I get 40-50 clashes per million, however bare in mind the uid wouldn't be generated a million times at once, but probably only around 10-1000 times a minute.  What is the probability of a clash each time, and can anyone suggest a better way of doing it?  static Random _random = new Random();  public static void main() {     // Size of the key, 6     HashSet&lt;string&gt; set = new HashSet&lt;string&gt;();     int clashes = 0;     for (int n=0;n &lt; 1000000;n++)     {         StringBuilder builder = new StringBuilder();          for (int i =0;i &lt; 7;i++)         {             if (_random.NextDouble() &gt; 0.5)             {                 builder.Append((char)_random.Next(97,123));             }             else             {                 builder.Append(_random.Next(0,9).ToString());             }         }          if (set.Contains(builder.ToString()))         {             clashes++;             Console.WriteLine("clash: (" +n+ ")" +builder.ToString());         }          set.Add(builder.ToString());         _random.Next();         //Console.Write(builder.ToString());     }      Console.WriteLine("Clashes: " +clashes);     Console.ReadLine(); }   UPDATE:   from this question  I really asked two questions here so I was cheating. The answer I was after was rcar's, however Sklivvz's is also the answer to the 2nd part (an alternative). Is it possible to make a custom unique id generator in a database, or would it be client side (which would be 2 possible reads first)?  The general idea I was after was using Ids in databases or other stores that can be used by phone or printed material, not a giant 16 byte guid.  UPDATE 2: I put the formula for two mutually exclusive events above instead of 2 independent ones (as getting an 'a' the first time doesn't mean you can't get an 'a' the second time). Should've been P(A and B) = P(A) x P(B)