Okay, I know this might be a really stupid question to ask, but I have always been on the look out for the being to whole the title "most powerful" and when I was watching  I got curious as to who would fit the bill in the Marvel universe.  I have looked into characters such as , , , , , , and , all of which listed some way that they could be destroyed.  So, my question is two-fold. Who, in the Marvel Multi-verse, A) has the greatest power and B) the smallest (preferably non-existent) weakness?
There are many beings that have very big powers. There are beings that are worshipped as gods (Asgardians; and I would lump Azazel into that as well since Marvel seems to imply that he's - while a mutant - someone people consider to be "Devil"). There are mutants whose power would clearly have been considered as "god like" by people worshipping Greek or other polytheistic pantheons where the threshold to what made you a god was pretty low.  But are there any actual major "Gods" in Marvel universe?  Since there are many different definition of what a "god" is, the one that is pertinent to this question:   Power wise, being able to create Universe(s); e.g. gods more in a sense of Christianity/Judaism scale rather than "Pan" or "Hestia" in Greek pantheon who merely have some limited supernatural powers over small domain of natural events. Effectively immortal, at least excluding specific exotic attacks. Origin-wise, be "supernatural", not merely a human mutant or a powerful alien with advanced technology. Desire/demand/like followers' worship or prayers.
When I try to install a software using Ubuntu Software center I get:  Failed to download repository information Check Your Internet connection   When I try to do a apt-get install something, I get:  407  Proxy Authentication Required   I use a proxy server that requires a user-name and a password. I have set my systems proxy manually, by plugging in the required numbers in the Networks proxy and applied it system wide. I guess the problem now is plugging in my user-name and password.  When I use INTERNET via Mozilla, it specifically asks me for my user-name and password. A worker carries two bags. Each of the bags initially contains N nails. Whenever the worker needs a nail, he takes it from a bag picked at random. At some point the worker picks an empty bag. Find the probability that the other bag contains exactly m nails.  My reasoning:  The desired probability: $C_2^1 \cdot P(|A| = 0\ and\ |B| = m)$, where A, B - sets of nails in each bag. Then we can "divide" this probability (I suppose that I deal with independent events):  $P(|A| = 0\ and\ |B| = m) = P(|A| = 0) \cdot P(|B| = m);$  $P(|A| = 0) = (\frac{1}{2})^N$, where $|A| = |B| = N$ (initial condition).  $P(|B| = m) = (\frac{1}{2})^{N - m}$  Result: $C_2^1 \cdot P(|A| = 0\ and\ |B| = m)$ = $C_2^1 \cdot (\frac{1}{2})^{2N - m}$  Am I wrong or it is correct? How to attach my repo to heroku app Lenovo Legion Y720 laptop Keyboard backlight I just moved from Windows to Ubuntu, but I cannot get the keyboard backlight to work at all. On Windows there was a specific program for controlling the backlight. I have searched for days now without luck. Anyone got a solution?  I am using Ubuntu 18.04 I'm fitting a regression on the $\log(y)$.  Is it valid to back transform point estimates (and confidence/prediction intervals) by exponentiation?  I don't believe so, since $E[f(X)] \ne f(E[X])$ but wanted other's opinions.  My example below shows conflicts with back transforming (.239 vs .219).  set.seed(123)  a=-5 b=2  x=runif(100,0,1) y=exp(a*x+b+rnorm(100,0,.2)) # plot(x,y)  ### NLS Fit f &lt;- function(x,a,b) {exp(a*x+b)}  fit &lt;- nls(y ~ exp(a*x+b),  start = c(a=-10, b=15))  co=coef(fit) # curve(f(x=x, a=co[1], b=co[2]), add = TRUE,col=2,lwd=1.2)  predict(fit,newdata=data.frame(x=.7)) [1] 0.2393773  ### LM Fit # plot(x,log(y)) # abline(lm(log(y)~x),col=2) fit=lm(log(y)~x) temp=predict(fit,newdata=data.frame(x=.7),interval='prediction') exp(temp)         fit       lwr       upr 1 0.2199471 0.1492762 0.3240752 Do you really want to create a new tag? I'm entering the world of Linux and at work I'm using grep more and more.  By doing that I'm figuring out that sometimes it's not adequate for what I want.  I was struggling with grep a few days ago and a colleague of mine who is a senior Linux admin, told me to use awk. I was stunned by how fast I got a result.  So my question is when do you choose to use one over the other? What questions can I ask myself before going to work with grep and spending a lot of time, when I could have done it with awk and saved time? I'm having a heck of a time trying to disable the option  "Check for updates Automatically"  I ran composer to see what plist was being modified and it looks like ~/Library/Preferences/com.oracle.javadeployment.plist  &lt;?xml version="1.0" encoding="UTF-8"?&gt; &lt;!DOCTYPE plist PUBLIC "-//Apple//DTD PLIST 1.0//EN"    "http://www.apple.com/DTDs/PropertyList-1.0.dtd"&gt; &lt;plist version="1.0"&gt; &lt;dict&gt;     &lt;key&gt;/com/oracle/javadeployment/&lt;/key&gt;     &lt;dict&gt;         &lt;key&gt;deployment.macosx.check.update&lt;/key&gt;         &lt;string&gt;false&lt;/string&gt; &lt;/dict&gt; &lt;/dict&gt; &lt;/plist&gt;   So I experimented with changing that value to true or false and it would not effect the checkbox under the Java Control Panel > Update > Check for Updates Automatically  Your assistance would be much appreciated.  Initially I was looking @ /Library/Internet\ Plug-Ins/JavaAppletPlugin.plugin Finding multivariable limits for the function $\frac{3x^2y}{x^2+y^2}$ Ackermann function - how to calculate the number of times it calls itself Purely for my own amusement I've been playing around with the .  The Ackermann function is a non primitive recursive function defined on non-negative integers by:  $A(m,n) = n+1$, if $m=0$  $A(m,n) = A(m-1,1)$, if $m&gt;0$ and $n=0$  $A(m,n) = A(m-1,A(m,n-1))$, if $m&gt;0$ and $n&gt;0$  I'm not interested in values of the function itself, rather I'm interested in the number of recursions the function takes to arrive at a value. For example:  $A(1,2) = A(0,A(1,1)) = A(0,A(0,A(1,0))) = A(0,A(0,A(0,1))) = A(0,A(0,2)) = A(0,3) = 4$  That's 6 steps, so if we define $RA(m,n)$ as the number of recursions to calculate $A(m,n)$, then $RA(1,2) = 6$.  I made a short Excel macro to calculate $RA(m,n)$ for any given $m$ and $n$. The problem I have is that $RA(m,n)$ grows even faster than $A(m,n)$ and I quickly run out of stackspace. The table below shows $RA(m,n)$ for very small values of $m$ and $n$. The numbers shown for $RA(3,10)$ and $RA(4,1)$ are the number of steps before crashing!    A formula for $RA(1,n)$ is trivial, and it is relatively easy to see that $RA(2,n) = 5(n+1) + 4n(n+1)/2$.  So my questions: can anyone see an explicit formula for $RA(3,n)$? More generally, does a formula exist for $RA(m,n)$, in the same way that $A(m,n)$ can be expressed using Knuth arrows?   For the low values shown of $m$ and $n$, $RA(m,n)$ looks about order of magnitude larger than $A(m,n)$. Can we prove that $RA(m,n) &gt; A(m,n)$ for all $m,n$. Intuitively, it should be.  For bonus points, can $RA(m,n)$ be expressed in terms of $A(m,n)$? Why is it that when you add normally distributed random variables the variance gets larger but in the Central Limit Theorem it gets smaller? Proof: For all integers $x$ and $y$, if $x^3+x = y^3+y$ then $x = y$ "System" storage on macOS Sierra is 470GB! Demonstrate that if $f$ is surjective then $X = f(f^{-1}(X))$ I haven't been able to do this exercise:     Let $f: A \rightarrow B$ be any function. $f^{-1}(X)$ is the inverse   image of $X$. Demonstrate that if $f$ is surjective then $X = f(f^{-1}(X))$ where $X \subseteq B$.   Since $X \subseteq B$, all the elements in $X$ belong to the codomain of $f$.  Since $f$ is surjective, it means that all elements in the codomain $B$ have some preimage in $A$. Given that $X \subseteq B$, all elements in $X$ must also have a preimage in $A$.  Have $\triangle = f^{-1}(X)$, $\triangle$ is now a set containing the preimages of the elements in $X$. Because of this, $\triangle \subseteq A$.  If we evaluate $f(\triangle)$, we...... nope, I don't know what I'm doing now.    What do you think? Just recently, the right-click context menus in Windows 7 have been acting in a kind of left-handed way. That is, when possible, the initial menu appears to the left of the mouse pointer. When I hover over a submenu option, the next level of context menu appears to the left of the first and so-on.  Basically, what normally happens when you're right at the right-hand edge of the screen is happening all the time - except when I'm at the left-hand edge. Perhaps it's some kind of left-handed users option, but that doesn't seem likely - more like a right-to-left language option, except that I've never even tried to use a right-to-left language.  I haven't checked every app. for the issue - probably only Windows Explorer, definitely not in Firefox, for the moment that's as far as I've checked.  Scratch that - this affects menu bars as well as context menus. The menus dangle down to the left of the menu-bar item instead of the right, and submenus open to the left too. This seems to apply to most applications, but not too Firefox or Thunderbird.  I find this extremely annoying.  I have no idea how this occurred, though it isn't the only weird glitch I've had in Windows 7 in the last few days - see   I assume there's a setting somewhere that has got changed, but I can't find anything. There's also a chance that I've suffered some kind of registry corruption, so I might need to apply a fix in the registry - not something that scares me, but I need to know what to fix and how.  So - does anyone know what setting has been changed, and how I can fix it?  Also, beyond CCleaner (which I already have), are there any tips on making sure that the registry stays clean and sane?  Finally, both Microsoft Security Essentials and Malwarebytes Antimalware seem happy that I don't have a malware problem. Should my recent glitches make me doubt that? How do I rip audio from a DVD? I remember reading a long time ago how you can rip the audio (music) from a DVD and convert it to a usable/readable format.  I can't find that article any more. Can anybody help me? Comma usage with compound sentences {actually compound predicates as per Rob's answer} Rule 101 of compound sentences is that a coordinate conjunction must be accompanied by a comma most of the time when joining independent clauses, yet I stumble across sentences all that time that seem to abuse this rule. Normally, it's a case of the subject not being present in the second clause, as in the following example:      "They usually start out as an abbreviation, but turn into acronyms   when they become well known."   (taken from the Oxford Style Manual)   To me, this comma cannot be correct because the clause after the comma doesn't contain a subject and therefore cannot be independent (right?). As far as I'm concerned, it's not the 'Comma to set off a contrast rule because I can find plenty of other where 'and' is used and the subject's missing in the second clause.  what's the deal here?   Thanks in Advance for any responses I may receive! I'm a citizen of India and I'm travelling to Nepal by air. Is an Aadhaar Card a valid document used for travelling to Nepal by air? How are release codenames chosen? TexMaker - "File not found" when viewing as pdf At the moment I'm having a problem when trying to view my LaTeX file as a PDF. Whether I do it internal or external I both the error "File not found.".  I have no idea what file it means. I've read somewhere it has to with my Acrobat Reader, but I still can't get it fixed. And I think it's worth mentioning that I work on Windows 7, as I've also read that Texmaker on Windows has some bugs.  Any help will be appreciated!