Text gets displayed during boot after Nvidia installtion
Enabling Nvidia driver messes up splash screen
How do the sets $\emptyset\times B,\ A\ \times \emptyset, \ \emptyset \times \emptyset $ look like? If we have a function $f:A \rightarrow B$, then one way to give meaning, I think, to this function, in terms of set theory, is to say, that $f$ is actually a binary relation $f=(A,B,G_f)$, where $G_f \subseteq A \times B$ is the graph of the function. Now my question is: what is $f$ if   $\bullet \ A=\emptyset, \ B\neq\emptyset$,?  $ \bullet \ B=\emptyset, \ A\neq\emptyset$ ?  $ \bullet \ B=\emptyset, \ A=\emptyset$ ?  (Another way to formulate this, I think, would be: How do the sets $\emptyset\times B,\ A\ \times \emptyset, \ \emptyset \times \emptyset $ look like? Are they all $\emptyset$ ?) How to get the last date of a particular month with JodaTime? Creating a selector from a method name with parameters I have a code sample that gets a SEL from the current object,   SEL callback = @selector(mymethod:parameter2);   And I have a method like    -(void)mymethod:(id)v1 parameter2;(NSString*)v2 { }   Now I need to move mymethod to another object, say myDelegate.  I have tried:  SEL callback = @selector(myDelegate, mymethod:parameter2);   but it won't compile. Random spawn location despite sleeping in a bed I am trying to make Minecraft harder, and one of the ways I wish to do this is to have a "pseudo-hardcore" mode where dying spawns you in a random location rather than completely locking you out of the world.  I intend to do this by setting the spawn radius to be very high.  However, I also want to make this "bed-proof" - that is, if someone places a bed and sleeps in it, I want the player to still be spawned in a random location, regardless of their personal spawn point.  How do I do this? Command blocks are fine, but data packs are not preferred. Is it possible to print a variable's type in standard C++? For example:  int a = 12; cout &lt;&lt; typeof(a) &lt;&lt; endl;   Expected output:  int Documents required by Indian citizens wishing to travel to Nepal (Originally  by AJAY.)  What is the geometric, physical, or other meaning of the third derivative of a function at a point?  If you have interesting things to say about the meaning of the first and second derivatives, please do so. Why do we need a diagonal matrix? Apart from simplifying matrix powers, why do want to diagonalize a matrix? Do they have any appealing application which can be used to motivate to study diagonal matrices.  Thanks for any answers. Does existence of magnetic monopole break covariant form of Maxwell’s equations for potentials? Absence of magnetic charges is reflected in one of Maxwell's fundamental equations: $$\operatorname{div} \vec B = 0. \tag1$$ This equation allows us to introducte concept of vector potential: $$\vec B = \operatorname{rot} \vec A.$$ Using this concept, it is possible to express Maxwell's equations in a graceful symmetric form: $$\nabla^2 \vec A - \frac{1}{c^2}\frac{\partial^2 \vec A}{\partial t^2} = - \frac{\vec j}{\epsilon_0 c^2}, \tag2$$ $$ \nabla^2 \phi -\frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2} = - \frac{\rho}{\epsilon_0}. \tag3 $$  Noticing, that vector $\vec A$ and scalar $\phi$ potentials, as well as electric current density $\vec j$ and charge density $\rho$, form a 4-vector in Minkovsky space-time. Therefore, Maxwell's equations can be expressed in a covariant form, using d'Alembertian: $$\nabla_{\mu}\nabla^{\mu} A_{\nu} = \frac{j_{\nu}}{\epsilon_0}. \tag4$$  If magnetic monopols exist, Maxwell's equation $(1)$ will look as: $$\operatorname{div} \vec B = \mu_0 c \rho_{\mathrm{magnet}}.$$  As the divergence of $\vec{B}$ isn't equal to zero, it impossible to introduce concept of vector potential. Thus, the equation in the form of $(4)$ will not be possible to express. How to add texture to fill colors in ggplot2 Evaluate $$\int \frac{2-3x}{2+3x}\sqrt\frac{1+x}{1-x}d x$$ What substitution should I use ? $\sqrt\frac{1+x}{1-x}$ suggests $x=cos2\theta $ but its not useful in $\frac{2-3x}{2+3x}$ Bash auto_completion with Xubuntu and xrdp from windows I'm using Xubuntu and connecting over RemoteDesktop with Windows 7.  Everything works fine, but the auto_completion in my bash doesn't work. I checked the .bashrc file in my home directory and also the /etc/bash.bashrc.  I uncommend the lines below in my local bashrc configuration and also in bash.bashrc, but nothing changes.  if ! shopt -oq posix; then   if [ -f /usr/share/bash-completion/bash_completion ]; then     . /usr/share/bash-completion/bash_completion   elif [ -f /etc/bash_completion ]; then     . /etc/bash_completion   fi fi How to set the fully qualified hostname on CentOS 7.0? How do I set the fully qualified hostname on CentOS 7.0?  I have seen a few posts online for example using:  $ sudo hostnamectl set-hostname nodename.domainname   However, running domainname returns nothing:  $ domainname (none)   Also:  $ hostname nodename.domainname   However,  $ hostname -f hostname: Name or service not known $ hostname -d hostname: Name or service not known   Some debug output:  $ cat /etc/hostname nodename.domainname  $ grep ^hosts /etc/nsswitch.conf hosts:      files dns quick selection of a random row from a large table in mysql I'm on a Sprint Samsung Galaxy S3 with Android 4.3.  I would love to become a power user of the dictation feature to dictate text messages and emails. More often than not, however, I have to hand-edit the dictated text not because of text comprehension issues but because of punctuation, capitalization and spacing problems.  These are the "commands" I've found to work. What is the full list?  "period"     --&gt; .   (works fairly reliably) "comma"      --&gt; ,   (works fairly reliably) "new line"   --&gt; \n  (works about 50% of the time)   I've not gotten capital or anything like that to work, or acronyms. If you know how it works, please share. Java FileInputStream ObjectInputStream reaches end of file EOF How to code around known MMa special-case failures? Function on $[a,b]$ that satisfies a Hölder condition of order $\alpha > 1 $ is constant I want to show that if a function $f:[a,b]\rightarrow \mathbb R$ satisfies a Hölder condition of order $\alpha &gt; 1 $ then it is constant.  The way I think of it is as follows:  $$|f(x) - f(y)| &lt; K|x-y|^\alpha$$  $$\frac{|f(x) - f(y)|} {|x-y]}  &lt; K|x-y|^{\alpha -1}$$  $$\lim_{y\rightarrow x} \frac{|f(x) - f(y)|} {|x-y]} \le \lim_{y\rightarrow x}  K|x-y|^{\alpha -1} =0 $$ As the limit is $0$, we can remove the modulus, so we get:  $$\lim_{y\rightarrow x} \frac{f(x) - f(y)} {x-y} = 0$$ So $f$ is derivable and $f'(x) = 0$ for all $x$ in $[a,b]$. Note that the reason we can add the limit $y\rightarrow x$ is because $[a,b]$ is closed in $\mathbb R$.  However, the question gives as a hint using the mean value theorem. I am not sure why one should do that. You would first have to prove that $f$ is derivable in a similar manner to what I did, and then prove that $f$ is constant. Or is there a simpler way to do it and I am missing it?  Also please inform me of any mistakes I did in the proof (if any)/ Thank you! I just noticed a weird thing in my reputation tab on StackOverflow.    That last question mentioned (but not linked, had to find it via search):   I must have amnesia as I have no memory of ever interacting with this post. I don't even know what xampp and pdftk are so I don't know why I would even be interested in such a question. 2 reputation points more or less is not a big deal, but the fact I got rep change from question I've never seen before is odd and I'm wondering what happened. Why avoid increment ("++") and decrement ("--") operators in JavaScript?