What should be the output of the Java code and why?
Is Java "pass-by-reference" or "pass-by-value"?
How to make all faces flip to the right/consistent direction? What's happening is that I have tried to flip the normals, inside and out, but there are always faces which have to be flipped!  Normals recalculated   Inside View    If I flip them again, all faces get flipped and end up horrible and, not working at all. So how can I make all of the faces flip to the right direction? Follow-up to question: Aut(G) for G = Klein 4-group is isomorphic to $S_3$ This is most likely a lack of understanding of wording on my part. I was considerind the Klein 4-group as the set of four permutations: the identity permutation, and three other permutations of four elements, where each of those is made up of two transposes, (i.e., 1 $\rightarrow$ 2, 2 $\rightarrow$ 1 and 3 $\rightarrow$ 4, 4 $\rightarrow$ 3) taken over the three possible such combinations of four elements.  Here, then, is my question. I am assuming (?) that Aut(G) in this case is the set of permutations of the four elements of the Klein 4-group - or the three non-identity ones for the purpose of showing isomorphic to $S_3$. If this is the case, what does it mean to have a permutation of these three permutations that I mentioned above? As always, thanks for your help. Is the Stack Exchange engine that powers Stack Overflow available to download or install internally for an enterprise or company?  I think Stack Exchange's engine is very great and could be very cool to use for internal enterprise patterns and practices, like the engine of Wikipedia. Background  Would like to center all tables within floats, without having to add \centering to each table.  Ideas  Tried the following:  \let\originaltabular\tabular \renewcommand\tabular{\centering\originaltabular}   It compiles, but the tables remain left-justified.  Question  How would you edit the Preamble of a LyX document to make all the floating tables be centred? Continuity of absolute value of a function Let $f(x)$ be a continuous function. Prove that $\left|f(x)\right|$ is also continuous.  Is it correct to say that, by the reverse triangle inequality, $\left|f(x)-f(c)\right| \geq \left|f(x)\right|-\left|f(c)\right|$ in all cases, so we will always have that for all $\left|x-c\right|&lt;\delta$ implies $\left|f(x)-f(c)\right|\leq \epsilon$?  I am not sure if my solution adequately uses the functional form, since I just used the equation and not my actual function. Please help with the proper solution! Extending from two classes lim sup inequality $\limsup ( a_n b_n ) \leq \limsup a_n \limsup b_n $ I´m not sure how to start with this proof, how can I do it? $$ \limsup ( a_n b_n ) \leqslant \limsup a_n \limsup b_n  $$ I also have to prove, if $ \lim a_n $ exists then: $$ \limsup ( a_n b_n  ) = \limsup a_n \limsup b_n  $$ Help please, it´s not a homework I want to learn. How can I split a beamer bibliography across two slides? I dug a large room to specifications, 3 blocks tall and 16 blocks wide as well as long. I dug it at 10 levels above bedrock, in a location {(-240, 0) -> (-224, 16)} specified by  slime finder app. I am definitely more than 24 blocks away and less than 32 away from the spawn room, as specified by the wiki .  Yet, no slimes. I have been sitting in my location patiently listening for the moist slapping sounds of slime. I have gone in to the room every time I heard a phantom slap but still nothing.  Am I just being dumb somehow or do the minecraft lords just hate me?  If it helps/anyone cares here's my seed: 6405460532872068998 Prove that the additive groups $\mathbb{R}$ and $\mathbb{Q}$ are not isomorphic. Is there a better (or other) way(s) to prove the following statement? Also, the same argument works for multiplicative groups $\mathbb{R}-\{0\}$ and $\mathbb{Q}-\{0\}$, right?       Problem Prove that the additive groups $\mathbb{R}$ and $\mathbb{Q}$ are not isomorphic.   Solution By cantor's diagonal argument, there is no possible bijection between $\mathbb{Q}$ and $\mathbb{R}$. Since an isomorphism needs to be a bijection, there is no possible isomorphism between the additive groups $\mathbb{R}$ and $\mathbb{Q}$.    Thanks How can I record my screen on Ubuntu?   The app I'm looking for has ideally all of these features:   Can record in a format that can be played back easily on any platform and/or accepted by YouTube or another popular video site Can record just a window (instead of the whole screen), possibly selecting it with a mouse click Can start recording after a configurable delay (e.g., I launch the app and have time to do arrangements to my desktop/window before actual recording starts) So  i have a scene with a few objects, and a water simulation in the scene. However I want to animate the camera turning around in the scene, while the water simulation is frozen in time. So something like a time freeze effect. But when ever I try to render an animation it renders the water simulation too. I just need one frame of the water simulation. Is there a way to do this? macOS Sierra restarting download instead of resuming divisibility of binomial coefficients by primes If $p$ is a prime number and $r = 1,2,3 \dots ,p -1$, how can I prove that the binomial coefficients $_nC_r(p,r)$ are divisible by $p$? By divisible I mean $_nC_r(p,r)/p$ leaves remainder $0$.  oivind Show $100+t$ is reducible in $\mathbb{Z}[[t]]$ $\mathbb{Z}[[t]]$ is the ring of formal power series with integer coefficients, and the problem asks to show that $100+t$ is reducible. This means I need to find two power series, $\sum_{i=0}^\infty a_i t^i$ and $\sum_{j=0}^\infty b_j t^j$ with $a_i,b_j\in\mathbb{Z}$, that when multiplied give $100+t$. Since the integers are an integral domain, I know that neither of these power series may be finite, but I'm having a really hard time figuring out how to come up with two series that will telescope properly. Any suggestions? I'm running a VPS of a german provider and want to switch from Ubuntu to CentOS. There I tried to install firewalld (was not preinstalled), but it always fails with the error  ERROR: Failed to load nf_conntrack module: modprobe: ERROR: could not find module by name='nf_conntrack' modprobe: ERROR: could not insert 'nf_conntrack': Function not implemented modprobe: ERROR: Error running install command for nf_conntrack modprobe: ERROR: could not insert 'nf_conntrack': Operation not permitted   Obviously the nf_conntrack kernel module was not found. Since this is a VPS, I cannot tinker with the kernel modules. On the support site of my provider I found a list of available kernel modules, for example:  ip_conntrack_netbios_ns ipt_conntrack    ip_conntrack    ip_conntrack_ftp    ip_conntrack_irc   Since nf_conntrack is not listed here, does that mean, that I cannot use firewalld? Is it possible to run it without this module?  I guess I could use iptables, if I wanted to (it seems to run correctly). But as I have more experience with firewalld, I would like to stick with it.    CentOS Linux release 7.7.1908 (Core)  uname -a: Linux xxxxxx.xxxxxxxx.xxx 3.10.0-042stab140.1 #1 SMP Thu Aug 15 13:32:22 MSK 2019 x86_64 x86_64 x86_64 GNU/Linux Can a Tourist from a VWP country carry (Rent/ Borrow) a gun in say Florida Looking to find out if it is legal to borrow or rent a gun from an individual(s) or company to use to either visit the various gun ranges or wildlife hunting in Florida.  I have a shotgun licence in the UK for hunting etc... but looking to do something similar in Florida without falling foul of any gun carrying laws in transit to any gun range or hunting area?  I know various friends in the US that would lend the guns while there but are they required to be with me while attending any event? Basically not looking to be tethered to any third party... Suppose $\int_{[a,b]}f=0,\text{ then }f(x)=0 \forall x\in[a,b]$ Let $a&lt;b$ be real numbers. Let $f:[a,b]\to\mathbb{R}$ be a continuous non-negative function. Suppose $\int_{[a,b]}f=0,\text{ then }f(x)=0 \forall x\in[a,b]$    Proof: Suppose for the sake of contradiction $\exists x_0\in[a,b] \text{ such that } f(x_0)= D&gt; 0$  Since the function is continuous at $x_0$, we have $$\forall \epsilon&gt;0, \exists \delta&gt;0 \text{ such that } |f(x)-f(x_0)|\leq\epsilon\text{ whenever } x\in[a,b]\cap(x_0-\delta,x_0+\delta)$$ choose $\delta=\delta_1$ such that $\epsilon=D/2$, then we can say: $$-D/2 \leq f(x)-f(x_0)\leq D/2$$ $$-D/2 \leq f(x) - D$$ $$f(x)\geq D/2&gt;0$$ This implies that $\int_{[a,b]\cap(x_0-\delta_1,x_0+\delta_1)}f &gt;0$  Now since $\int_{[a,b]} f = 0$. we have  $$\int_{[a,b]}f=\int_{[a,x_0-\delta_1]}f+ \int_{(x_0-\delta_1, x_0+\delta_1)} f + \int_{[x_0+\delta_1,b]}f=0$$ $$\int_{[a,x_0-\delta_1]}f+\int_{[x_0+\delta_1,b]}f&lt;-D/2,$$ but this is a contradiction since $f(x)\geq 0$ and hence $\int_{[a,x_0-\delta_1]}f+\int_{[x_0+\delta_1,b]}f\geq 0$  Is my proof correct? The annual student intake of Hogwarts seems about 100. Since in book 7 Hogwarts is declared mandatory for all children, this means that the total magical population of Britain cannot be much more than 10,000.  How do these people make a living?  All jobs mentioned in the books are non-productive occupations (government, stores, sports &amp;c) - not farmers or engineers. There are exceptions - makers of wands/robes/broomsticks - but where do they get the raw materials from?  Who and how "creates goods" in the wizarding world?  Note that if you say "wizards exchange their gold for Muggle money and buy stuff from Muggles (through intermediaries, if necessary)" then you would have to explain where they get their gold: they would either have to mine it, or produce something of value to the Muggles (in which case the separate financial system of galleons/sickles/knuts does not make sense).  PS. A simple issue of maintaining a steam locomotive for Hogwarts express is pretty much insurmountable for 10k people - without Magic, that is. So, I can imagine that the skilled tradesmen can transfigure dirt into raw materials for their trade - except for food, which "is the first of the five Principal Exceptions to Gamp's Law of Elemental Transfiguration". How do I disable Dunst and go back to notify-osd? I installed i3, which recommends dunst, but I'm not using i3 at the moment, and the dunst notifications don't really fit with Unity. I'd rather not uninstall dunst totally, since I might want to start using it at some point in the future; is there a way to switch back to notify-osd without uninstalling dunst?