This is a curiosity question. Considering same mass black holes. As nothing can come out of either black hole, would the annihilation result into a black hole of energy? Assuming the two black holes do not repel one another - i.e. assuming antimatter does not have opposite gravity.   Could the big bang have been one such collision between black holes of unequal masses where black hole of normal matter was heavier?
Let's say we have one black hole that formed through the collapse of hydrogen gas and another that formed through the collapse of anti-hydrogen gas. What happens when they collide? Do they (1) coalesce into a single black hole or do they (2) "annihilate" into radiation?  One would expect (1) to be the case if the No Hair Theorem were to hold. So I guess what I'm really asking for is a modern understanding of this theorem and its applicability given what we know today.
Do I need a visa for Georgia? I am a Nigerian, holding a valid schengen visa issued by the French embassy, and a multiple entry visa for the United States. Will I be allowed to enter Tbilisi in Georgia? What is the analytic continuation of the Riemann Zeta Function animating an element's height using jquery Why are we using the squared residuals instead of the absolute residuals in OLS estimation?  My idea was that we use the square of the error values, so that residuals below the fitted line (which are then negative), would still have to be able to be added up to the positive errors. Otherwise, we could have an error of 0 simply because a huge positive error could cancel with a huge negative error.  So why do we square it, instead of just taking the absolute value? Is that because of the extra penalty for higher errors (instead of 2 being 2 times the error of 1, it is 4 times the error of 1 when we square it). I recently switched from Windows to Mac. Steam says I have fewer Mac specific games than total games and it should be an equal amount unless you can only download, install, and play once from your account. I thought I bought it so that it would work across all operating systems. If this is true, is there a work around or a discount for re-purchasing for the other OSes? While working together, how can I discuss my ideas and objections without killing the other person’s motivation?  I want to do a design project with a friend, for study purposes, but I experienced some obstacles I don’t know how to deal with:   We didn’t learn yet how to deal with objections or dislikes from the other. As a result our meetings are inefficient, either because we let the other do whatever they want, or because someone gets into a bad mood and instead of thinking, they try to put on an unaffected face. My friend has a habit of letting their attention stray away, and they do it at the most unexpected times.   I know that being sincere is needed and inevitable, but I would like to learn a method to keep the attention and motivation even in these situations. My problem is that I don’t know what to say or do to keep our task looking interesting and fun to do.  As a possible solution I thought of inserting a 5 minute period where we shouldn’t say any good things, not even a word, but have to point out the bad parts of our own work.  But I fear this one won’t be enough, that’s why I would like to hear ideas and tricks that could make our sessions going. I just installed my first Ubuntu Server (12.04) instance and immediately ran into the following problems:   Running apt-get install or update fails (see below) Running ifconfig produces peculiar IP addresses starting with 10.x.x.x and my router does not see the server at all in its "Attached Devices" page (its a consumer grade Netgear)   When I run apt-get I get 50+ of the following errors:  Failed to fetch &lt;some-http-url&gt;. Temporary failure resolving &lt;blah&gt;.ubuntu.com   I read up on this error for 12.04, and found some similar posts on this forum and on server fault that recommended going into /etc/resolv.conf and adding an entry:  nameserver 8.8.8.8   To my surprise, I found that /etc/resolv.conf was a symlink pointing to somewhere else  and I had to delete it first, then create a new one with the entry indicated above.  I restarted the server and tried running apt-get again, same results.  How do I start diagnosing the problem (I'm a relatively new Linux/Ubuntu user)?  Additional details that may help:   This is actually a virtual machine running as a VirtualBox guest OS The physical host that is running this VM is my laptop which has a wireless connection; I'm wondering if (somehow) the laptop is getting the network wirelessly, but perhaps either the VBox VM and/or the Ubuntu server OS isn't configured for wireless and so nothing is "getting through"?   Thanks in advance for any pointers!  Edit:  Here is a snapshot of my VM's Network settings (only Adapter 1 has any information populating its fields, so I assume my VM has only 1 adapter):    Here is the output when I run ifconfig from the terminal:    And here is the output when I run route from the terminal:    Running ping google.com: as of 4/2015 OS X Kindle App: Where are the books located? After I updated my Kindle for Mac app 2 mos. ago I can no longer find the azw book files on my Mac to convert them so I can read them on my KOBO ereader. I use to find them in a My Kindle Content file but that has disappeared. I have Ult. DRM &amp; Calibre but they won't work without the azw files, and I can't find them on my mac anymore. Can anyone help me? The answers on this site about this are years old &amp; do not apply to my problem. Let us consider that $u\in C^2(\Omega)\cap C^0(\Omega)$ and satisfying the following equation .  $\Delta u=u^3-u , x\in\Omega$ and   $u=0 $   in  $\partial \Omega$  $\Omega \subset \mathbb R^n$ and bounded .   I need hints to find out what possible value $u$ can take ?  Thank you for ur hints in advance . Most efficient way to see if an ArrayList contains an object in Java I have an ArrayList of objects in Java.  The objects have four fields, two of which I'd use to consider the object equal to another.  I'm looking for the most efficient way, given those two fields, to see if the array contains that object.  The wrench is that these classes are generated based on XSD objects, so I can't modify the classes themselves to overwrite the .equals.    Is there any better way than just looping through and manually comparing the two fields for each object and then breaking when found?  That just seems so messy, looking for a better way.  Edit: the ArrayList comes from a SOAP response that is unmarshalled into objects. Linear functional and a bounded norm I'm trying to work through some example questions for my class but there is no mention of "Piecewise affine functions" anywhere and I'm completely stumped on how to do this.  Any help would be appreciated please. Display Animated GIF Why do the less than symbol () appear wrong as upside down exclamation (¡) or question mark (¿)? Spice containers purchase  I'm planning on upgrading my spice storage. Reading some advice online (e.g. , and ) I came to the conclusion that spice containers should:   Not be exposed to direct sunlight and high temperatures Be airtight   Which means that either you store it inside of some closed cupboard/wardrobe/room or you have containers which do block light and are in a relatively shady place. And the containers need to be airtight.  Available solutions  I've been browsing for some containers available online in my country and the 2 most sensible options I've seen are these:       The problem  The first one is hermetic (it has some rubber around the lid) but it's clear glass. The other one does not have the rubber, just a regular twist-off lid, but it's amber glass.  It'll be hard for me to find any space inside of cupboards to store all of my spices (I need at least 30), so I'd rather buy amber glass container, which I believe will effectively block the sunlight, even when kept on the counter. But I can't find any hermetic amber-glass containers, only ones with regular lid.  The question  Is that regular twist-off lid enough to keep spices airtight and fresh? Or should I focus on buying a hermetic container (like the one in clear glass). I could then for example paint them black to block the sunlight. How can I upload files asynchronously with jQuery? I would like to upload a file asynchronously with jQuery.     $(document).ready(function () {     $("#uploadbutton").click(function () {         var filename = $("#file").val();          $.ajax({             type: "POST",             url: "addFile.do",             enctype: 'multipart/form-data',             data: {                 file: filename             },             success: function () {                 alert("Data Uploaded: ");             }         });     }); }); &lt;script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/2.2.0/jquery.min.js"&gt;&lt;/script&gt; &lt;span&gt;File&lt;/span&gt; &lt;input type="file" id="file" name="file" size="10"/&gt; &lt;input id="uploadbutton" type="button" value="Upload"/&gt;     Instead of the file being uploaded, I am only getting the filename. What can I do to fix this problem? Where to ask "subjective and argumentative" questions on programming? Recently  was "closed as subjective and argumentative".  I wonder, where would you recommend to ask such questions? If you willingly fall on your turn does it cost movement?  As a DM I'm thinking about how to handle things such as creatures who are immune to being prone, diving into the water from a height, or a flying creature diving off a cliff and taking a fall before taking flight. There are many circumstances where you can fall without taking damage or being knocked prone afterward, and I'm not sure how to determine movement following the dive.  For example, my character walks up to a ledge above a 10 foot drop to the ground below. After taking 5 feet of movement to step off the ledge, I allow myself to fall. When I hit the ground have I used 15 feet of movement or just 5? $(a^{2^n}+1,a^{2^m}+1)=1 or 2$ Prove that if $m\not =n,a$ are positive integers then $(a^{2^n}+1,a^{2^m}+1)$ is $1$ if $a$ is even and $2$ if $a$ is odd.    I solve the problem in the following way: I assume that $m&gt;n$ then  $$a^{2^m}+1=a^{2^m}-1+2$$ With this expression if I take a divisor prime $p$ of $a^{2^n}+1$ then if $p|a^{2^m}+1$ this prime must divide 2 and the result follows.   But, how can I prove this statement without use of primes? This exercise appears before the concept of prime is introduced then I prefer do not use it.    Any idea?   Thanks! What does string theory predict for the singularity inside a black hole? The usual explanation for what's going on inside a black hole goes something like "General Relativity predicts a singularity with infinite curvature, but when matters gets so tightly compressed we should expect quantum effects to become important, so the singularity will probably be blurred". Authors then proceed to say that we need a quantum theory of gravity to answer such things, and quickly abandon the subject.  Well, while we don't have an experimentally confirmed quantum theory of gravity, we do have a speculative one, string theory. What does it say about singularities? Are they smeared out like the books say? Do they remain points (or rings for a rotating black hole) with infinite curvature and density? Do we not know? E: Package 'libpng12-0' has no installation candidate [ubuntu 16.10 Gnome] I tried to install WPS Office on my laptop (ubuntu 16.10 Gnome, 64 bit) with  Gdebi installer but it said   Dependency is not satisfiable: libpng12-0   Then I tried to open this library as follows  sudo apt-get install libpng12-0   But I encountered an error:  Reading package lists... Done Building dependency tree        Reading state information... Done Package libpng12-0 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  E: Package 'libpng12-0' has no installation candidate   Can anyone help? Thanks a lot