Problem  I had a problem with one bug in my Stack&nbsp;Overflow account user profile while I wanted to click on next badge.    Solution  Through Ctrl + Shift + I console I selected item boxes and deleted that panel which was before that blue one and after that I could lick on "Let us pick".  Versions/settings   I use Acer Aspire with Windows&nbsp;10 Pro, 64 bit - all up-to-date I opened Stack&nbsp;Overflow on extended monitor DELL(1280x1024) with correctly working VGA connector Google Chrome Version 69.0.3497.100 (Official Build) (64-bit) in Chrome I have only these extensions installed:   AdBlock Scopus Document Download Manager Google Scholar    Why did this GUI mistake happen? Is it a problem on my site or server site? Can I do something to not repeat it again?
When I visit  I see this:     The "Congratulations! You've earned (badge)" section has dropped down completely beneath the tabs under it. The "You've earned a new privilege!" section has a gap at the bottom:     Shog9 says in comments how to reproduce this:     Gotta have earned a badge, not dismissed the notification, and be looking at your own profile.
Blender 2.8 - Transparent background Problem with long chapter title and arsclassica I want to delete the record type which is default to the some profiles What does the complex electric field show? We have a complex electric field. Is there any definition for absolute and imaginary part of a complex electric field? What do they stand for? Show that if $a \neq b$ and a and b are positive then $\frac{a}{b}+\frac{b}{a}$ is never an integer Sometimes chicken breast can be thick and I never know how to tell when it's cooked! How can I be certain it is cooked all the way through? Finding roots/zeros for collision detection? For the most simple of 2D games, I have implemented a posteriori collision detection (overlapping rectangles) on the x/y Cartesian plane, but am now interested in understanding the basics of a priori collision detection...  In the link , I noticed the reference to Newton’s Method.  So, I want to better understand the link b/w finding zeros of an equation and detecting a posteriori collision.    Let's start with the most obvious case of finding roots of an equation:  If you have a projectile's trajectory modeled as a quadratic function, you can find the roots to predict at what time the height will equal zero (or any other constant height, actually, by just solving f(x)=C to be f(x)-C=0 and then finding roots of that expression).    So, as I understand, finding the root can tell me the time at which the object will be at any chosen height.  More broadly, we basically know the (x,y) location of the object for any given time.  But how does root finding, in particular, translate to detecting collisions with another object?  (The other object may be stationary, or be moving.  If the latter, do you also determine roots/location of this moving object as well? Calculate the (x,y) trajectory of both objects and determine where they will intersect?)  Again, the basic question is where does root finding come into play?  Any links to a basic primer on this topic, with a simple example would also be greatly appreciated. Which ligatures do the fonts used in my document offer? This is a follow-up question to:   In order to know which ligatures I might consider disabling (I have in mind microtype's \DisableLigatures command; some important notes about selective disabling of ligatures are found under ), knowledge about which ligatures might be applied will be very useful. Which ligatures do the fonts used in my document offer? Rules when using a Joker I had on my rack among other tiles, a black 9, 1 x blue 10, and a black 12.  On the table was a set of 3 x 10s including a Joker(no blue 10), and other sets of tiles including an available black 11.  Can I replace the joker with my blue 10 and then set out :  my black 9, the joker(representing a black 10), a black 11 from the table, and my black 12? Taking the second derivative of a parametric curve How come the year doesn't show here? Voldemort was said to be very interested in the four founders of Hogwarts. He created a Horcrux from the Ravenclaw diadem, the Hufflepuff cup, and the Slytherin locket. Did he try to find a Gryffindor relic to complete the set, or did he not bother?   I can't see him giving up on such a task, but maybe there was insufficient time.  Gryffindor's sword was hiding 'inside' the Sorting Hat, but a famous family usually has some other heirlooms too.  If there are no canon answers I'd be interested in your own speculation! I'm working on a small game and i'd like to know how can i write/read a user settings file in ubuntu touch? Weak solutions to the Neumann's problem (Evans PDE) Let $U$ be connected. A function $u \in H^1(U)$ is a weak solution   of Neumann's problem   \begin{equation} (*)\qquad\left\{ \begin{array}{rl} -\Delta = f &amp; \text{in } U \\ \frac{\partial u}{\partial \nu} = 0 &amp; \text{on } \partial U \end{array} \right. \end{equation}   if   $$ \int_U Du \cdot Dv \; dx = \int_Ufv \; dx $$   for all $v \in H^1(U)$. Let $f\in L^2(U)$. Prove $(*)$ has a weak solution   if and only if   $$ \int_U f \; dx =0. $$   For the only if part I set $v=1$. However I do not see where to start the if part. I was thinking of the Lax-Milgram theorem. That's where I'm now. At the end of a makefile I saw   rm -f *~ *.class   I understand the *.class, but what's *~? mkdir -p will create a directory; it will also make parent directories as needed.  Does a similar command exist for files, that will create a file and parent directories as needed? Java: Avoid inserting duplicate in arraylist Probability of an even number of sixes We throw a fair die $n$ times, show that the probability that there are an even number of sixes is $\frac{1}{2}[1+(\frac{2}{3})^n]$. For the purpose of this question, 0 is even.   I tried doing this problem with induction, but I have problem with induction so I was wondering if my solution was correct:   The base case: For $n=0$, our formula gives us $\frac{1}{2}[1+(\frac{2}{3})^0] =1$. This is true, because if we throw the die zero times, we always get zero sixes.  Suppose it's true for $n=k$. Then the odds of an even number of sixes is $\frac{1}{2}[1+(\frac{2}{3})^n]$, and thus the odds of an odd number of sixes is $1 - \frac{1}{2}[1+(\frac{2}{3})^n]$.    For $n=k+1$, there are two ways the number of sixes are even:  a. The number of sixes for $n=k$ was even, and we do not throw a six for $n=k+1$: $ \frac{5}{6} \cdot  \frac{1}{2}[1+(\frac{2}{3})^n]$  b. The number of sixes for $n=k$ was odd, and we throw a six for $n=k+1$: $\frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$  So the probability $p$ for an even number of sixes at $n=k+1$ is $ \frac{5}{6} \cdot  \frac{1}{2}[1+(\frac{2}{3})^n] + \frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$  I have two questions   How do I get from $ \frac{5}{6} \cdot  \frac{1}{2}[1+(\frac{2}{3})^n] + \frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$ to $\frac{1}{2}[1+(\frac{2}{3})^n]$? I seem to have done something wrong, I can't get the algebra correct, I get $p = \frac{1}{3}[1+(\frac{2}{3})^n] + \dfrac{1}{6}$ Other than that, is my use of induction correct? Is it rigorous enough to prove the formula? Nature of gravity: gravitons, curvature of space-time or both? General relativity tells us that what we perceive as gravity is curvature of space-time.   On the other hand (as I understand it) gravity can be understood as a force between objects which are exchanging (hypothetical) virtual particles called gravitons, similar to the way electromagnetic forces are due to objects exchanging virtual photons?  At least at first glance, the two concepts seem mutually exclusive. Is there a description of gravity which includes both, or is this contradiction one of the problems in combining GR with quantum mechanics? So I recently saw add comment (requires 50 reputation) on a deleted answer. After looking around a bit, it seems to be on all deleted posts now (only deleted posts, normal posts still say just add comment).  I'm sure it wasn't there before (I think it just said add comment (unclickable) or possible nothing).  It is also on , presumably the root cause is the same.  And yes, I do have more than 50 reputation (in case anyone was wondering).  Just to clarify, requires 50 reputation obviously shouldn't be there for users with 50 or more reputation. Choices:   Display nothing Unclickable add comment - probably not ideal Unclickable add comment (disabled for deleted posts) or something along those lines   FYI - You may be somewhat confused right about now if you can't  yet.