The censorship on Stackoverflow will kill the platform and it's elitist snakes will be haunted. Consider following words,  Its At That What  I often hear them as,  I/?/s Aa Tha/?/ Wha/?/  I'm interested to hear in which regional accents above pronunciations are used.
As I was taught in school and , the T ending sound of words is unvoiced and should be pronounced with air, but recently I met a friend from the US, those aired T sounds were missing from her speaking, like "hat", "hot", "right". She speaks like /hæ/, /hɑː/,/raɪ/.   So I wonder if it's common to miss those sounds for Americans?
Why $\sum_{k=0}^m x^k $ is equal to $\dfrac{x^{m+1}-1}{x-1} $? Who can help me proving this (but you can no longer use $\sum$ in the proof):  $$\sum_{k=0}^m x^k=\dfrac{x^{m+1}-1}{x-1}$$ Auto-fit TextView for Android Toy Line (Production) Run I am not sure about the meaning of this:          Transformers: Generation 1 (also known as Generation One or G1) was a children's toy line that ran from 1984 to 1991 and was produced by Hasbro.     Would the following rewrite be better?        Transformers: Generation 1 (also known as Generation One or G1) was a children's toy line the production of which ran from 1984 to 1991 and was produced by Hasbro. How to fix java.net.SocketException: Broken pipe? Joint distribution of two dependent variables Tiny lifeforms on planetoid that live at accelerated time rate A book I once owned and read back in late-80's or early-90's that I have not been able to identify title &amp; author for:  It was a SF story concerning tiny native lifeforms, living on either a planetoid or possibly a dead/neutron star, that evolve at a seemingly accelerated time rate. Most of their story is told in the few days of observation (from orbit) by a human expedition to the planet, during which time-frame the race below progresses from stone-age tech level to being more advanced than the human observers, and eventually flying up to meet the humans in small advanced craft powered by micro-singularities... one of which accidentally 'disintegrates' the tip of a human scientist's nose in the process. Let $a, b, \&amp; \, m$ be integers. Prove that if $2a + 3b \geq 12m + 1$, then $a \geq 3m + 1$ or $b \geq 2m + 1$.   My Attempt: I don't conceive apace how to contrive, from the one inequality in the antecedent, the two inequalities in the consequent. So a proof by contraposition may be more facile. The contrapositive is: $\text{If }  \color{Green}{a &lt; 3m + 1} \; \&amp; \; \color{#0073CF}{b &lt; 2m + 1}, \text{ then } 2a + 3b &lt; 12m + 1 \tag{*}.$  From $(*)$, $\color{Green}{2a} + \color{#0073CF}{3b} &lt; \color{Green}{6m + 2} \; + \;\color{#0073CF}{6m + 3} = 12m +5$. But this doesn't prove $(*)$.     Given Solution:  Assume that $a &lt; 3m+1$ and $b &lt; 2m+1.$ Since a and b are integers, $a ≤ 3m$ and $b ≤ 2m.$ Therefore, $2a + 3b ≤ 2(3m) + 3(2m) = 12m &lt; 12m+ 1,$ as desired. $\blacksquare$   What went wrong in my attempt? I understand the given solution but I persisted with the more natural and direct green and blue, instead of deviating to weak inequalities. Shouldn't both solutions work?  Source: Problem 4.19 on P102 (related to P90, Result 4.8) of Mathematical Proofs, 2nd ed by Chartrand et al.    $\large{\text{Supplement to Cameron Buie and pritam's Answers :}}$  What overriding, catholic lessons can be extrapolated from this example to generalise about inequalities?   Moreover, in order to circumvent the problem here (ie the loss of information from the weaker inequalities), should I always start with the strictest inequality? If so, should I always rework any weak inequality into the equivalent strict inequality?   For example, define $r \in \mathbb{R} $ and $n \in \mathbb{Z}$ and $g(...) \neq h(...)$. Then $ f(r) &lt; h(r) \require{enclose}      \enclose{updiagonalstrike}{\iff} f(r) \leq g(r).$ But $ f(n) &lt; h(n) \text{ MAY OR MAY NOT }\iff f(n) \leq g(n).$ I have this setup:   The mesh is only the visible top side, with a mirror modifier for the bottom part. Now I want to unwrap it like in the picture. I want the inner circle (red) to go to the top of the UV map...  The problem is that no matter what automatic unwrapping I use, it never works as intended.  The closest I've gotten was by using "follow quads", but that skewed the UV map a lot:   I used the result from the second image and manually adjusted the vertices with a lot of scaling and "align auto", but its very far from perfect as you can see in the first image.  Here's what I want exactly:   The uv vertices should be distributed evenly (horizontally) The middle line (yellow) should be on a "natural" y height (so there's no stretching)   I hope it's clear what I want to achieve. I'm sure there must be some automatic way to perfectly align and distribute the UVs in this way.  By the way: If you look closely you can see that I already placed a uv seam (aligned with the green y axis in the screenshot) but that didn't really help. I'm assuming that the answer to the question in the title is a resounding yes. Since Baryonic matter and dark matter interact via gravitational forces.  If this is the case how is information not lost if, as is presumed, dark matter does not interact via the electromagnetic force? We already had multiple questions about unbalanced data when using , , ,  and a number of other similar questions, what makes it a very popular topic! Unfortunately, each of the questions seems to be algorithm-specific and I didn't find any general guidelines for dealing with unbalanced data.  Quoting , dealing with unbalanced data     (...) heavily depends on the learning method. Most general purpose   approaches have one (or several) ways to deal with this.   But when exactly should we worry about unbalanced data? Which algorithms are mostly affected by it and which are able to deal with it? Which algorithms would need us to balance the data? I am aware that discussing each of the algorithms would be impossible on Q&amp;A site like this, I am rather looking for a general guidelines on when it could be a problem. Ubuntu 18.10 Cosmic not starting GNOME session on bootup Function $f(z)$ such that $|f(z)| > |z|$ for every $z\in\mathbb{C}, 1/f(z)$ $K\subseteq \mathbb{R}^n$ is a compact space iff every continuous function in $K$ is bounded. I need to prove that $K\subseteq \mathbb{R}^n$ is a compact space iff every continuous function  in $K$ is bounded.  One direction is obvious because of Weierstrass theorem. How can i prove the other direction? I tried to assume the opposite but it didn't work for me.  Thanks a lot. Why does GPS depend on relativity? In my preamble, you'll find:  \renewcommand{\baselinestretch}{1} \renewcommand{\familydefault}{\sfdefault} \usepackage[sfmath]{kpfonts} \usepackage[mathrmOrig]{sfmath} \usepackage{amsmath} \usepackage{amssymb} \usepackage{amsfonts} \usepackage{animate} \usepackage{anysize} \usepackage{arcs} \usepackage{array} \usepackage[scale=0]{background} \usepackage{blindtext} \usepackage{calc} \usepackage{cancel} \usepackage{chngcntr} \usepackage{CJK} \usepackage{CJKspace} \usepackage{CJKpunct} \usepackage{colortbl} \usepackage{enumitem} \usepackage{epstopdf} \usepackage{etoolbox} \usepackage{fancybox} \usepackage{fancyhdr} \usepackage{float} \usepackage{framed} \usepackage{graphicx} \usepackage{hyphenat} \usepackage{ifthen} \usepackage{latexsym} \usepackage{lmodern} \usepackage{multicol} \usepackage{multirow} \usepackage{mwe} \usepackage{pgffor} \usepackage{rotating} \usepackage{tabularx} \usepackage{textcomp} \usepackage{tgheros,tgtermes,tgcursor} \usepackage{tikz} \usepackage{times} \usepackage{tipa} \usepackage{varwidth} \usepackage{wasysym} \usepackage{wboiboites} \usepackage[abs]{overpic} \usepackage[breaklinks,colorlinks=true,linkcolor=red,citecolor=red, urlcolor=blue]{hyperref} \usepackage[framemethod=TikZ]{mdframed} \usepackage[breakable, theorems, skins]{tcolorbox,empheq} \usepackage[default]{gillius} \usepackage[document]{ragged2e} \usepackage[latin1]{inputenc} \usepackage[spanish]{babel} \usepackage[T1]{fontenc} \usepackage[total={18.59cm,26.5cm},left=1.5cm,top=2.5cm]{geometry} \usetikzlibrary{matrix,arrows, positioning,shadows,shadings,backgrounds,calc, shapes, tikzmark} \usetikzlibrary{shadows.blur} \usetikzlibrary{shapes.symbols} \usetikzlibrary{patterns,fadings} \tcbuselibrary{skins,breakable,listings,theorems}    I'm trying to draw an arrow for X axis and Y axis.   \begin{tikzpicture} \draw[yellow!10, fill=blue!5](-1, -1) rectangle (7, 5); \draw[dotted] (-0.75,-0.75) grid (6.75,4.75); \node[below] at (-0.25,0){$O$}; \draw[-&gt;,black, thick] (-0.75,0)--(6.75,0); \end{tikzpicture}   But I get this error:    Additionally, if I remove -> from the options of \draw the tikz picture compiles normally. I am running Wireshark on my linux machine and i have simple python web server running on localhost:8080  I am capturing only local port 8080. When i open browser on localhost:8080 wireshark captures huge amount of tcp communication (as page loads with all resources) but there`s no HTTP only TCP...   Why i cannot capture HTTP on custom port ? What is "offset null" in IC 741? What is Offset null in 1st and 5th pin in IC 741 (Op-Amp)? Why it is used, though it is not used in many circuits? Give me explanation regarding the offset null! Why offset voltage was formed in IC 741? Content of a polynomial Define the content of a polynomial (over an arbitrary commutative ring $A$) to be the ideal generated by its coefficients, denoted $c(f)$. I want to show that   $$c(fg) = c(f)c(g).$$      (I was told this is true.)    What I was able to show was that $c(fg) \subseteq c(f)c(g)$ (this is obvious), and that their radicals are the same. My reasoning for the latter was as follows: let $f = a_0 + \dotsb + a_nx^n,$ $g = b_0 + \cdots + a_mx^m,$ and consider the matrix  $$ \begin{pmatrix}  a_0 b_0 &amp; \cdots &amp; a_0b_m\\ \vdots &amp; \ddots &amp; \vdots\\ a_nb_0 &amp; \cdots &amp; a_nb_m \end{pmatrix} .$$  Then $c(f)c(g)$ is generated by the entries of this matrix, while $c(fg)$ is generated by the sums along the diagonals. Let $\mathfrak p$ be a prime ideal of $A$ not containing $c(f)$ or $c(g)$, and let $i,\,j$ be minimal such that $a_i,\, b_j \notin \mathfrak p$.  Then all the terms in the generator of $c(fg)$ corresponding to the coefficient of $x^{i+j}$ are in $\mathfrak p$ except for $a_ib_j$, so that $c(fg) \not\subset\mathfrak p$. It follows that $c(fg) \subseteq \mathfrak p \Longleftrightarrow c(f)c(g) \subseteq\mathfrak p$, and hence $\sqrt{c(fg)} = \sqrt{c(f)c(g)}$.     Can I go all the way and show that in fact $c(fg) = c(f) c(g)$? Uniform convergence for sequences of functions I just recently got an iPod Touch and wanted to download some free apps. The App Store wanted me to get an Apple ID, no problem. But now it's telling me to input a credit card. I don't have one, and anyway shouldn't need one to download free apps.   I did some searching here and on Google and found out that there should be a "None" option when selecting credit cards, but it was not there. I then read that it's disabled in some countried, so I tried telling iTunes that I'm in the US (I'm in Argentina), but still no "None" option. Later, I read that apparently this feature was disabled in iTunes 10, so I installed iTunes 9, but I still don't get the option.  How can I get the No Credit Card option to appear when downloading an app for the first time? It doesn't make sense that you should need a credit card for free stuff.  EDIT: To clarify, I already made an account in Apple's website, I'm looking for a way to be able to select "None" when buying an app through iTunes.