How to Undo Deny Access on Keychain
How can I undo "always allow" in keychain item
No internet connection after ubuntu server 20.04 install, ifconfig not available I recently installed Ubuntu Server 20.04 LTS on my desktop, and found out there was no internet connection even though I had the LAN cable connected. ifconfig and iwconfig failed saying the corresponding packages net-tools and wireless-tools were not installed.   EDIT  Here is the info requested. Please find the image How do favored classes work in Pathfinder? I'm working on creating a half-elf paladin character for a Pathfinder game.  Full disclosure: this is the first time I've tried creating a character; I've used ones other people created before.  In the text, it mentions something about "favored classes" I believe.  What is this? Computer doesn't boot up, shows PXE error and/or "operating system not found", "no bootable device", "insert boot media", or other similar error My computer no longer boots Windows or the installed operating system. Instead, I may get this error:  PXE-E61: Media test failure, check cable -or- Check cable connection!  PXE-M0F: Exiting PXE ROM.   and/or one of the following messages or similar:  Operating System not found  No boot device available- No bootable devices--strike F1 to retry boot, F2 for setup utility  No bootable device -- insert boot disk and press any key  No Boot Device Found. Press any key to reboot the machine  Default Boot Device Missing or Boot Failed  Reboot and Select proper Boot device or Insert Boot Media in selected Boot device   What does this mean and what can I do?     This question comes up often and the answers are usually the same. This post is meant to provide a definitive, canonical answer for this problem. Feel free to edit the answer to add additional details. Why is it "come" rather than "go" ? I've got a repo that I contribute to within an Organization that I administer. I'm trying to add it to careers, but it won't allow it. Is this on purpose? I don't really want to fork it to my personal account, since I won't be maintaining it from there. A friend of mine asked me the following question:      Suppose you have a basket in which there is a coin. The coin is marked   with the number one. At noon less one minute, someone takes the coin   number one and put into the basket two coins: the number two and the   number three. At noon less $\frac{1}{2}$ minute, he takes the coin   number two and put into the basket the coins number $4$ and the number   $5$ and so on. At noon less $\frac{1}{k}$ he takes the coin number $k$   and put two other coins.       The question is: How many coins I can find in the basket at noon?   Intuitively the answer is $\infty$ because I added one coin for infinite times. But the correct answer seems to be $0$ because every coin numbered $k$ has been removed at noon less $\frac{1}{k}$ of minute. What is the correct answer? Thanks. Mouse Pointer Locator Due to a stroke, I have had some vision impairment in that I sometimes have a hard time locating my mouse pointer on the screen.  I have tried making the mouse pointer larger which helps but is still not the solution.  I am new to the Mac world and would appreciate your help.  In Windows, there is an option where pressing the CTRL key will help to identify the mouse pointer using a temporary display of several radiating concentric circles around the mouse pointer.  Is there something like this on OS X? python multiprocessing on windows, if __name__ == "__main__" Prove that the sum of digits of $(999...9)^{3}$ (cube of integer with $n$ digits $9$) is $18n$ Someone had posted a question on this site as to what would be sum of digits of $999999999999^3$ (twelve $9s$ ) equal to?   I did some computation and found the pattern that sum of digits of  $9^3 = 18$,  $99^3 = 36$,  $999^3 = 54$ and so on. So, I had replied that sum of digits of  $999999999999^3 = 12 \cdot 18 = 216$.  Can anybody help me prove this, that sum of digits of  $(\underbrace{999\dots9}_{n\text{ times}})^3 = 18n.$ 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); Extract subtitle from mkv files Is it possible to make the default scene fill two monitors at startup? I know how to set up the default scene by pressing Ctrl+U, but whenever I startup Blender it always fills only one of my screens, even if I set it to fill both before setting the default scene. I end up having to resize it every time I start it up. Is there a way to set this to happen by default? There are several ideas on hard determinism, but the core seems to be that in an interconnected existence, there can be no "actions" (free will), but only "reactions" (determinism). An action of free will would have to be a separate entity outside of existence, so we'd say it can't occur.   Now fast forward to Joe Schmo the robbing/murdering/rapist.  Joe wasn't born in some infinite bliss utopia, then decided one day to invent the concept of scumbaggery.  He's a product of an incomprehensible number of variables - maybe some are apparent like his dad was a violent alcoholic or whatever, but most likely it's far more complicated than that.    So that's fine - I get all that - but if that's the case, then how can you ethically deal with people like that?  You can't hardly let them do what they please at the expense of society, but then how can you get medieval on them when they're basically riding a wave?  I think that's a big reason why people fight concepts of determinism that mitigate or eradicate the concept of free will - now everyone has a license to do what they want and how can you punish them, it's not their fault!  How would/could/should you consolidate hard determinism with responding to criminal acts? How to add an edge ring around a hole What is the easiest way to add/extrude an edge loop around the hole in the first image to make it like the second image? I have an application which is a relatively old. Through some minor changes, it builds nearly perfectly with Visual C++ 2008. One thing that I've noticed is that my "debug console" isn't quite working right. Basically in the past, I've use AllocConsole() to create a console for my debug output to go to. Then I would use freopen to redirect stdout to it. This worked perfectly with both C and C++ style IO.  Now, it seems that it will only work with C style IO. What is the proper way to redirect things like cout to a console allocated with AllocConsole()?  Here's the code which used to work:  if(AllocConsole()) {     freopen("CONOUT$", "wt", stdout);     SetConsoleTitle("Debug Console");     SetConsoleTextAttribute(GetStdHandle(STD_OUTPUT_HANDLE), FOREGROUND_GREEN | FOREGROUND_BLUE | FOREGROUND_RED); }   EDIT: one thing which occurred to me is that I could make a custom streambuf whose overflow method writes using C style IO and replace std::cout's default stream buffer with it. But that seems like a cop-out. Is there a proper way to do this in 2008? Or is this perhaps something that MS overlooked?  EDIT2: OK, so I've made an implementaiton of the idea I spelled out above. Basically it looks like this:  class outbuf : public std::streambuf { public:     outbuf() {         setp(0, 0);     }      virtual int_type overflow(int_type c = traits_type::eof()) {         return fputc(c, stdout) == EOF ? traits_type::eof() : c;     } };  int APIENTRY WinMain(HINSTANCE hInstance, HINSTANCE hPrevInstance, LPTSTR lpCmdLine, int nCmdShow) {     // create the console     if(AllocConsole()) {         freopen("CONOUT$", "w", stdout);         SetConsoleTitle("Debug Console");         SetConsoleTextAttribute(GetStdHandle(STD_OUTPUT_HANDLE), FOREGROUND_GREEN | FOREGROUND_BLUE | FOREGROUND_RED);       }      // set std::cout to use my custom streambuf     outbuf ob;     std::streambuf *sb = std::cout.rdbuf(&amp;ob);      // do some work here      // make sure to restore the original so we don't get a crash on close!     std::cout.rdbuf(sb);     return 0; }   Anyone have a better/cleaner solution than just forcing std::cout to be a glorified fputc? Removing unnecessary vertices? I have over sampled polygons that I need to simplify, without changing the shape. basically, I want to reduce the number of vertices, by eliminating the ones that are useless and have the polygons represented with the significant vertices only. I used to do that with ET GeoWizards Plug-in, but I don't have a license anymore. My table has 500.000 features, water polygons.   Any ideas? Please help me to prove this trigonometric equation.  $\cos \left( 36^\circ \right)-\cos \left( 72^\circ \right) = \frac{1}{2}$  Thank you. "Z-level" in TikZ Antiresonance of multiple parallel decoupling capacitors: use same value or multiple values? My quesiton: Is it better practice to put same-valued capacitors in parallel of capacitors of different values to decouple the high-frequency noise caused by digital ICs? Background Digital IC need a decoupling capacitor close to their supply pins to ensure a stable voltage during power transients and to deal with noise (mostly to prevent noise generated by the IC to affect neighboring circuitry). It seems sensible to place a bulk capacitor (say 10-100uF), to act as an energy reservoir, and several smaller capacitors to deal with higher frequencies. The reason to place several small capacitors instead of just one is to deal with their Equivalent Series Inductance (ESL), which in practice, causes them to behave like an LC circuit. The effect of anti-resonance Yet, here is where best design practices and electronic myth seem to get mixed up and confusing to me. Most electronic engineers I have met like placing several decoupling capacitors of different values in parallel (with the smaller capacitors closer to the IC). The logic behind it is that the each capacitor takes care of a different noise frequency as depicted in Figure 1.  Figure 1: Impedance over frequency of three different value capacitors in parallel (cyan) vs their individual contribution (brown, blue, red). Image taken from . Note the small anti-resonance peak. It seams no major trouble , and the overall behavior of the three different capacitors in parallel is vastly superior to their individual decoupling capabilities. However, I have read in  by [Henry W. Ott] that placing capacitors of different values may cause a much greater antiresonance-peak which can be very harmful for our designs (see Figure 2). In fact, it amplifies any noise that falls into the anti-resoance frequency range, which is corroborated by .  Figure 2: from Electromagnetic Compatibility Engineering, by Henry W. Ott, section 11.4.4. The 15nH inductance makes reference to the capacitors ESL. What is the rough estimate of salary value for a taxpayer to pay AMT? For US federal tax, whether a taxpayer is subject to AMT is determined by a complicated way in Instructions for Form 1040, Line 45, which involves an incomprehensible "worksheet to see if you should fill in form 6251".  By simplification and approximation, is there a rule of thumb for an estimate of a salary value of a taxpayer, so that a salary above the value might cause the taxpayer to pay AMT with a big possibility, and a salary below the value might not with a big possibility?  Why do the following two sources seem to differ greatly in estimates of the boundary salary value:    says that  the possible boundary salary value for a single taxpayer  to pay AMT is $53,900.  says that      Taxpayers with small incomes fall within the AMT exemption limits.   That leaves the upper middle class to face the tax, especially those   with incomes between $200,000 and $500,000    Thanks.