I though it was British vs. American but watching Star Trek I've noticed the same character using two different pronunciations. Then I noticed other characters doing the same.  Is there any rule for using one or another?  I've searched online and found two definitions, one singular and another plural, both uncountable.  How can something uncountable be plural?
My non-native English speaking friend just asked me: "Data is..." or "Data are..."?  I said both but that's because I've been desensitized from reading/writing both (especially from writing code and adding quick comments).  My question: Is it acceptable to utilize either for a university paper? Or is one safer than the other (when confronted with stickler professors)?    Related questions:
How does one know when to use a gerund or an infinitive? As a native speaker of English, the gerund version of this sentence sounds better:  infinitive:     When used together in chains,   extension methods are an unprecedented   tool to produce extremely concise   code.   gerund:     When used together in chains,   extension methods are an unprecedented   tool for producing extremely concise   code.   But how can I explain to someone learning English how to decide in situations like these whether to use the infinitive or the gerund? Can this identity be derived from the binomial theorem?  $$n(n+1)2^{n-2} = \sum_{i=1}^{n}  i^2 \binom{n}{i}$$    I tried starting from $2^n = \displaystyle\sum_{i=0}^{n} \binom{n}{i}$ and dividing it by $4$ in order to get $2^{n-2}$. I need to run Firefox browser from terminal. Browser works fine, but terminal still waits for Firefox to return and if I do CTRL+C, it terminates Firefox. How to run command to make Firefox work independently from terminal? Voldemort couldn't kill Harry as long as he called the Dursleys home. Why didn't he just murder the Dursleys? This would prevent Harry from being able to live there. Trouble downloading packages list due to a "Hash sum mismatch" error When I check for updates, I get a "Failed To Download Repository Information" error.   This is what comes up under details:   W: Failed to fetch gzip:/var/lib/apt/lists/partial/us.archive.ubuntu.com_ubuntu_dists_natty_main_source_Sources  Hash Sum mismatch,  E: Some index files failed to download. They have been ignored, or old ones used instead. Product of path connected spaces is path connected Will product of path connected topological spaces be necessarily path connected? Why or why not?  Give me some hints. Thank you. How do I size a UITextView to its content? What happens when a creature with "Whenever a creature dies" dies simultaneously with multiple other creatures? has the following ability:     Whenever a creature you control with a +1/+1 counter on it dies, return that card to the battlefield under your control at the beginning of the next end step.   I had a number of creatures on the battlefield, including Marchesa herself, with one or more +1/+1 counters on them when  was played, destroying each creature on the board.  When Marchesa and several other creatures would be killed simultaneously,   Does Marchesa's ability trigger for herself? Does Marchesa's ability trigger for the other creatures under my control with +1/+1 counters?   In attempting to resolve the issue, we tried to look things up, leading to , which cites Rule 603.6d as proving that Marchesa's ability should trigger both for herself and for each other creature with +1/+1 counters.  Unfortunately, none of us could actually understand Rule 603.6d, aside from this example:     Two creatures are on the battlefield along with an artifact that has the ability Whenever a creature dies, you gain 1 life. Someone plays a spell that destroys all artifacts, creatures, and enchantments. The artifacts ability triggers twice, even though the artifact goes to its owners graveyard at the same time as the creatures.   The example certainly seems analogous to me, but without understanding the actual rule, rather than just the example, we weren't too sure.  A complete answer has to not only answer the question, but back up that answer with citation and explanation of the official rules on the matter. How to select even or odd elements based on class name if you create html layout like so  &lt;ul&gt;      &lt;li class='a'&gt;&lt;/li&gt;      &lt;li class='b'&gt;&lt;/li&gt;      &lt;li class='a'&gt;&lt;/li&gt;      &lt;li class='b'&gt;&lt;/li&gt;      &lt;li class='a'&gt;&lt;/li&gt;      &lt;li class='b'&gt;&lt;/li&gt;      &lt;li class='a'&gt;&lt;/li&gt;      &lt;li class='b'&gt;&lt;/li&gt; &lt;/ul&gt;   and try to select odd elements with 'a' class like so $$('.a:odd') you will get empty array, and if you do $$('.a:even') you will get all four li elements with 'a' class.. It really strange.. But im new to mootools, maybe im doing something wrong..    So my question is how to select first and third li elements with a class. I know that i can do it with function like this  $$('.a').filter(function(item, index) { return index%2; }  but its too complicated for such small task as selecting odd or even elements.. g++ undefined reference with a shared library What does 'condescending' exactly mean? I'm really confused about the meaning of the word 'condescending'. Online dictionaries define it as bossy/disdainful, which fits with...    But I find other usages where being condescending is obviously considered a good thing, perhaps more like being humble, or equal, rather than superior...    (and  of "condescending and friendly" in Google Books)  What's going on? Does this word really have two almost opposite meanings? I want to include some Cyrillic with a TeX Gyre Pagella text and math font document. I'm using TeXLive 2019.  As indicated in , running pdflatex on the following source works just fine:  \documentclass{article}  \usepackage[T1,T2A]{fontenc} \usepackage[russian,main=english]{babel} \usepackage{paratype} \usepackage{tgpagella} \usepackage{substitutefont} \substitutefont{T2A}{\rmdefault}{PTSerif-TLF}  \usepackage{amsmath}  \usepackage[math]{blindtext}  \begin{document}  The topologist is {\Russian Алекс\'{а}ндров}.  \blindmathpaper  \end{document}   However, running xelatex on the following surce produces output from which the indicated Cyrillic just disappears:  \documentclass{article}  \usepackage[T2A,T1]{fontenc} \usepackage[russian,main=english]{babel} \usepackage{amsmath} % load here to avoid \dddot clash \usepackage{fontspec} \usepackage{paratype} \setmainfont[Ligatures=TeX]{TeX Gyre Pagella} \usepackage{substitutefont}% \substitutefont{T2A}{\rmdefault}{PTSerif-TLF}%  \usepackage{unicode-math} \setmathfont{TeX Gyre Pagella Math}  \usepackage[math]{blindtext}  \begin{document}  The topologist is {\Russian Алекс\'{а}ндров}.  \blindmathpaper  \end{document}   Question: How should this second source be fixed so as to allow the Cyrillic to appear, too?  (I did put Map paratype-truetype.map into my local updmap.cfg file and did run updmap-user.) Union of subgroups is a subgroup if and only if one subgroup is a subset of the other Let $H$ and $K$ denote two subgroups of a group $G$.  Prove that the union $H \cup K$ is a subgroup of $G$ if and only if $H \subset K$ or $K \subset H$. Neat series convergence result This is a nice problem on series convergence that I recently stumbled upon. Given a non-negative sequence of real numbers $(a_n)$ such that $$\sum_{n=1}^\infty a_n &lt; \infty,$$ show that there exists a non-decreasing sequence of non-negative numbers $b_n$ such that $$b_n \to \infty \quad\text{ and } \quad \sum_{n=1}^\infty a_n b_n &lt; \infty.$$ In other words, for every convergent series with non-negative terms, there is another convergent series with "substantially larger" terms.  I have a solution (see below), but maybe someone else has a different, simpler, and/or more elegant solution. Anyone on the Android Jelly Bean OS and not able to buy Apps/Books through wallet? I have just bought the Samsung GalaxyNote 800 and am facing this issue.   The error is (RPC:S-7:AEC-0).   What is the solution? Let $m$ be Lebesgue measure and $a \in R$. Suppose that $f : R \to R$ is integrable, and $\int_a^xf(y) \, dy = 0$ for all $x$. Then $f = 0$ a.e. This is a corollary to a proof in Bass, but I don't understand why it follows from the proof he gives. I follow everything up until the last statement. Why is it that proving that the integral is $0$ for all Borel measurable sets guarantees it is true for all Lebesgue measurable sets? Am I missing something vital about the relationship between Lebesgue measurable sets and Borel measurable sets? Statement: Let $m$ be Lebesgue measure and $a \in\mathbb R$. Suppose that $f : \mathbb R \to\mathbb R$ is integrable, and $\int_a^xf(y) \, dy = 0$ for all $x$. Then $f = 0$ a.e. Proof: We prove the conclusion first for intervals, then for unions of intervals, then for open sets, then for Borel measurable sets. For any interval $(c,d)$, we have $$\int_c^df = \int_a^d f - \int^c_a f = 0.$$ By linearity, if $G$ is the finite union of disjoint open intervals, then $\int_G f = 0$. Now, if $G$ is open, then $G$ is the countable union of disjoint, open intervals, $G = \bigcup^n_{i=1}I_i$. We have $$\int_G f = \int \lim_{n\to\infty} f \cdot \chi_{\bigcup^n_{i=1} I_i}$$ and since $f_n$ is integrable for all $n$, and $|f_n| \le f$ for all $n$, by the dominated convergence theorem, $$\int_G f = \lim_{n\to\infty} \int_{\bigcup^n_{i=1}I_i} f =0.$$ Now, if $G_n$ is a decreasing sequence of open sets converging to $H$, since $|f \cdot \chi_{G_n}| \le f \cdot \chi_{G_1}$ for all $n$, we again have by the DCT that $\int_H f = 0 = \lim_{n\to\infty} \int_{G_n}f = 0$. Finally, if $G$ is a Borel measurable set, then we know that there exists a sequence $G_n$ of decreasing, open sets, that converge to some set $H$, where $H \setminus G$ is a null set. Thus, $$\int_G f = \int_H f = 0.$$ We have found that for every Borel measurable set $G$, $\int_G f =0$. Since $f$ is real valued and measurable, $f = 0$ a.e. Calling future method from batch class? I have an external HD which used to be my TimeMachine drive. I got a different drive for that, so I want the original drive back as an external on my Windows box, but it has this random 200 MB partition labeled as EFI System Partition. How do I get rid of this? No Wifi-Adapter found on new Lenovo Yoga C930 with Ubuntu 18.10 jQuery/JavaScript: accessing contents of an iframe