In Minecraft Bedrock Edition, how would I specify larger, zoomed- out maps in a /clear command
Is it possible to use NBT tags in commands in Minecraft: Bedrock Edition?
Is a differentiable function always continuous? Continuous Functions are not Always Differentiable. But can we safely say that if a function f(x) is differentiable within range $(a,b)$ then it is continuous in the interval $[a,b]$ . If so , what is the logic behind it ? how do you make commands not show up on the chat I am making a server, and I put a comparitor clock down to constantly have clear weather. How do I silence the commands? How to insert page number in Beamer navigation symbols? Can I extend a class using more than 1 class in PHP? If I have several classes with functions that I need but want to store separately for organisation, can I extend a class to have both?  i.e. class a extends b extends c  edit: I know how to extend classes one at a time, but I'm looking for a method to instantly extend a class using multiple base classes - AFAIK you can't do this in PHP but there should be ways around it without resorting to class c extends b, class b extends a Increase progress bar each second -- error Identifying and solving javax.el.PropertyNotFoundException: Target Unreachable I have found that for only one page (so far) the spacing between the header and content is different from the spacing on all the other pages. You can see this behavior in the images below (from pages 2 and 3 of the document):     My .tex file contents are shown below.  \documentclass[a4paper]{report} \usepackage{fullpage} \usepackage{lipsum} \usepackage{fancyhdr} \pagestyle{fancy} \setlength{\headsep}{0.5cm}  \begin{document} \renewcommand{\headrulewidth}{0.4pt} \renewcommand{\chaptermark}[1]{\markboth{\thechapter. #1}{}} \fancyhead[L]{\bfseries \leftmark} \fancyhead[R]{\bfseries \rightmark}  \chapter{Version Control} \lipsum \lipsum \lipsum \lipsum  \end{document} How can I list all the printable ASCII characters in the terminal? Kinetic energy conservation in a collision Kinetic energy conservation in collision.  In any type of collision type of collision if we take the system to comprised of both bodies then the net external force is zero. So the work done by external force is also zero. By the relation $W=K_f-K_i$ as $W$ is zero hence kinetic energy is conserved. But this not the case.   Can you tell me what is the flaw in my argument? Some people say that the argument has neglected the fact that some kinetic energy is transformed into heat. But heat is created due to the interactions between bodies during collision which is thus an internal force and thus shouldn't affect kinetic energy. A and B disjoint, A compact, and B closed implies there is positive distance between both sets Intuitive explanation for formula of maximum length of a pipe moving around a corner? For one of my homework problems, we had to try and find the maximum possible length $L$ of a pipe (indicated in red) such that it can be moved around a corner with corridor lengths $A$ and $B$ (assuming everything is 2d, not 3d):    My professor walked us through how to derive a formula for the maximum possible length of the pipe, ultimately arriving at the equation $L = (A^{2/3} + B^{2/3})^{3/2}$.  The issue I have is understanding intuitively why this formula works, and exactly what it's doing. I understand the steps taken to get to this point, but there's an odd symmetry to the end result -- for example, is the fact that $\frac{2}{3}$ and its inverse are the only constants used just a coincidence, or indicative of some deeper relationship?  I also don't quite understand how the formula relates, geometrically, to the diagram. If I hadn't traced the steps myself, I would have never guessed that the formula was in any way related to the original problem.  If possible, can somebody give an intuitive explanation as to why this formula works, and how to interpret it geometrically?    Here's how he found the formula, if it's useful:  The formula is found by finding the maximum possible length of the pipe by expressing the length in terms of the angle $\theta$ formed between the pipe and the wall, and by taking the derivative to find when $\frac{dL}{d\theta} = 0$, which is the minimum of $\frac{dL}{d\theta}$ and is therefore when $L$ is the smallest:  $$ L = \min_{0 \leq \theta \leq \frac{\pi}{2}} \frac{A}{\cos{\theta}} + \frac{B}{\sin{\theta}} \\ 0 = \frac{dL}{d\theta} = \frac{A\sin{\theta}}{\cos^2{\theta}} - \frac{B\cos{\theta}}{\sin^2{\theta}} \\ 0 = \frac{A\sin^3{\theta} - B\cos^3{\theta}}{\sin^2{\theta}\cos^2{\theta}} \\ 0 = A\sin^3{\theta} - B\cos^3{\theta} \\ \frac{B}{A} = \tan^3{\theta} \\ \theta = \arctan{\left( \frac{B}{A} \right)^{\frac{1}{3}}} \\ $$  At this point, we can substitute $\theta$ back into the original equation for $L$ by interpreting $A^{1/3}$ and $B^{1/3}$ as sides of a triangle with angle $\theta$ and hypotenuse $\sqrt{A^{2/3} + B^{2/3} }$:  $$ \cos{\theta} = \frac{A^{1/3}}{ \sqrt{A^{2/3} + B^{2/3} }} \\ \sin{\theta} = \frac{B^{1/3}}{ \sqrt{A^{2/3} + B^{2/3} }} \\ \therefore L = A^{2/3} \sqrt{A^{2/3} + B^{2/3} } + B^{2/3} \sqrt{A^{2/3} + B^{2/3} } \\ L = (A^{2/3} + B^{2/3}) \sqrt{A^{2/3} + B^{2/3} } \\ L = (A^{2/3} + B^{2/3})^{3/2} \\ $$  The equation for the formula for the maximum length of the pipe is therefore $L = (A^{2/3} + B^{2/3})^{3/2}$. Let S be a finite set.Let F be a surjective function from S to S.  How do I prove that it is injective? I had given an cipher text (of JSON string), an IV and a key. I tried decrypting using 's method:      decryptWithManagedIV(algorithmName, privateKey, IVAndCipherText)     //Sample code   Blob key = Blob.valueOf(decryptionKey);   Blob encryptedDatafor = EncodingUtil.base64Decode(bodystring);   Blob decryptedData = Crypto.decryptwithManagedIV('AES128', key, encryptedDatafor);    cipher = decryptedData.toString();   System.debug('Encrypted BodyString: ' +cipher);   and I was able to decrypt the cipher successfully. But when I use the provided IV with the following method:      decrypt(algorithmName, privateKey, initializationVector, cipherText)    //Sample code  Blob exampleIV = Blob.valueOf('HSDlUkhrmfEo7SgM'); //provided IV   Blob key = Blob.valueOf(decryptionKey);  Blob encryptedDatafor = EncodingUtil.base64Decode(bodystring);  Blob decryptedDataIV = Crypto.decrypt('AES128', key, exampleIV, encryptedDatafor);   cipher = EncodingUtil.base64Encode(decryptedDataIV );  System.debug('Encrypted BodyString: ' +cipher);   I get some string which looks like, it is still in the encrypted form (different from the encrypted string I have passed).   Update:  I have tried the below code-snippet based on the suggestion from @identigral, but I am still unable to decrypt the cipher using the IV:  Blob exampleIV = Blob.valueOf('HSDlUkhrmfEo7SgM'); //provided IV  Blob key = Blob.valueOf(decryptionKey); Blob encryptedDatafor = EncodingUtil.base64Decode(bodystring);  String encodedCipherTextAndIV = EncodingUtil.convertToHex(encryptedDatafor); String encodedCipherText = encodedCipherTextAndIV.substring(32); String finCipher = encodedCipherText.difference(encodedCipherTextAndIV); Blob ciphertext = EncodingUtil.convertFromHex(finCipher);   Blob decryptedDataIV = Crypto.decrypt('AES128', key, exampleIV, ciphertext);  decryptStg = EncodingUtil.base64Encode(decryptedDataIV ); System.debug('Decrypted BodyString: ' +decryptStg); Explicit construction of a finite field with $p^n$ elements Eclipse doesn't start on Ubuntu 18.04 I installed eclipse via Ubuntu Software. When I start it, it take some seconds and then comes an error window showing the path of the error.  The following is the .log file:  !SESSION Wed May 02 12:53:28 BRT 2018 ------------------------------------------ !ENTRY org.eclipse.equinox.launcher 4 0 2018-05-02 12:53:28.587 !MESSAGE Exception launching the Eclipse Platform: !STACK java.lang.ClassNotFoundException: org.eclipse.core.runtime.adaptor.EclipseStarter     at java.base/java.net.URLClassLoader.findClass(URLClassLoader.java:466)     at java.base/java.lang.ClassLoader.loadClass(ClassLoader.java:566)     at java.base/java.lang.ClassLoader.loadClass(ClassLoader.java:499)     at org.eclipse.equinox.launcher.Main.invokeFramework(Main.java:626)     at org.eclipse.equinox.launcher.Main.basicRun(Main.java:584)     at org.eclipse.equinox.launcher.Main.run(Main.java:1438)     at org.eclipse.equinox.launcher.Main.main(Main.java:1414)   What is the problem and how can I fix it? Where does the term $|\psi\rangle\langle\psi|$ come from while calculating the expectation value? I would like to install Ubuntu 14.04.02 LTS on my machine which currently has Mageia 4.1.   So, I have the following partitions: One for /, one for swap and one for /home.   I have some important data on /home which I wouldn't like to lose (I have backed them up, but I would prefer to keep them) after a clean install. I plan to reformat partitions one and two as above and keep the /home one.   Will I be able to tell Ubuntu to do so during an installation? By this I mean: Will it be able to recognize the /home partition and keep all its data intact while formatting the rest? And one more question: Will it be able to keep the users as they are our will I be forced to change them too? (I have the bad feeling that it won't be able to keep the users as they are...) Permission for snap applications I have some trouble with the Ubuntu notepadqq package. After opening a file from my apache webroot I cannot see any content. Saving the file is also not possible. I thougt this would be a classic permission issue. Therefore, I changed the group of the webroot from root to www-data and added my user to this group. Permissions are 775, so in theory it should now be possible for me to edit files in my webroot. But nevertheless, when I open a file of this directory with notepadqq, it still only shows an empty file.  When using gksudo notepadqq I get the following error message:  No protocol specified QXcbConnection: Could not connect to display :0   I figured out that notepadqq is a so called snap application and I wonder if this is the reason for my problems.  Can someone give me a hint how I could fix this issue (without using 777 permissions)? I am looking for a way to get the output of a command when it is run from within a C++ program.  I have looked at using the system() function, but that will just execute a command. Here's an example of what I'm looking for:  std::string result = system("./some_command");   I need to run an arbitrary command and get its output. I've looked at , but I have not found anything that will give me what I need. How to allocate aligned memory only using the standard library?