How to bevel edges evenly
First time using bevel tool, doesn't bevel evenly. What am i doing wrong?
Using Illustrator CS6 to create shape out of multiple paths I am very new to Illustrator, in fact I am only experimenting with it in an attempt to help my friend develop a logo for a new business venture. I managed to draw out the shell of the logo using the line tool and modifying the lines with the anchor tool to get the shape I wanted.  However, I would like to somehow take all of these paths and create one object, or symbol which I can then apply a drop shadow, fill and various other effects too. Currently when I try to do this I get unwanted effects...  I'd appreciate any help on the subject. Could someone please help me with this proof?     Suppose that $a, b \in N$, and $d = \gcd(a, b)$. Since $d$ divides $a$, we have $a = de$ for some integer $e,$ and similarly $b = df$ for some integer $f$. Prove that $\gcd(e, f) = 1$.   I understand why it works. Since d is all the common factors of $a$ and $b, e$ and $f$ had no common factors, therefore the $\gcd(e,f) = 1$. But how do I prove this?  Thanks in advance. Until recently  had one single reputation point, you can see it in this old screenshot:    Now it has over 2K reputation all of a sudden.. what has caused this? Has it affected "real" members as well? Which configuration of armor enchantments gives maximum protection? How can I check the version of the available package in the Ubuntu repositories without installing it? C# Change A Button's Background Color QCompleter Custom Completion Rules How to render as only wireframe in Cycles? How to disk copy with unreadable sectors? I want to clone a whole disk bytewise, something like  dd if=/dev/$SRC of=/dev/$DST bs=65536 count=$count   There are better tools for the job, which understand the file system (Windows) and work faster (I don't care), but they fail because of unreadable sectors on the old source disk.  There's actually a single invalid block I have to skip. So I thought about using dd multiple times like  dd if=/dev/$SRC of=/dev/$DST bs=65536 count=... dd if=/dev/zero of=/dev/$DST bs=65536 count=... skip=... seek=... dd if=/dev/$SRC of=/dev/$DST bs=65536 count=... skip=... seek=...   with replacing the broken sector data by zeros. Is this a valid approach? I found this  which confuses me. I'm self-studying Guillemin and Pollack, but I'm stuck on Problem 3 of section 2. It says that if $V$ is a vector subspace of $\mathbb{R}^N$, then $T_x(V)=V$ if $x\in V$.  If $x\in V$, then since $V$ is a manifold, there is a local parametrization $\phi: U\to V$ where $U$ is open in $\mathbb{R}^k$. Without loss of generality, we can require $\phi(0)=x$. Then $T_x(V)$ is defined to be the image of $d\phi_0$ on $\mathbb{R}^k$.   An arbitrary element of $T_x(V)$ looks like $$ d\phi_0(v)=\lim_{t\to 0}\frac{\phi(0+tv)-\phi(0)}{t}=\lim_{t\to 0}\frac{\phi(tv)-x}{t} $$  but this doesn't seem very useful to show $T_x(V)=V$. What is the right approach? Does the alternating sum of prime reciprocals converge? $$\sum_{n = 1}^\infty \frac{(-1)^n}{p_n}$$ where $p_n$ is the $n$th prime.  I have computed this to 10000 rather than infinity. My results suggest that convergence does happen but it's very slow. But I can't even be sure about the first few digits: $-0.26959$?  I have looked at the "questions that may already have your answer" and some of the "similar questions," but they involve somewhat different formulas. Ohm's law states that  electric current is directly proportional to voltage provided that physical conditions like temperature remain constant i.e. $$V = IR$$  On the other hand,  $$\text{Power = Voltage} \times \text{Current}$$  So here it seems that the higher the voltage, the lower the current, provided that the power remains constant (i.e. current is inversely proportional to the voltage here which is against Ohm's Law.).   Now my question is how do physicists explain this apparent contradiction? Or maybe this not a contradiction because I am analysing things incorrectly?   P.S.: I am a tenth grade student so please refrain from the usage of highly complicated terminologies in your answers. How do you make a deep copy of an object? What benefit do plasterboard/drywall screws have? I have some plasterboard screws like this:     However, without a drill I have a job getting them into the wall. Am I safe to use normal screws on hangings that aren't going to be taking a great deal of weight? In this case it is a rack of coat hooks. What do you call the functions that make a composite function.  Example : $e^{\sin x} $ is made up of $\sin x$ in the argument of $e^x$  So what are $\sin x$ and $e^x$ called here in this context? Basic functions or something?   (Please don't tell me the former is trig and latter is exponential function, I know that, but that's not what I'm asking here, please try to understand)  Hope I made myself clear. \includegraphics optional argument not resolved if macro How do I switch sides in the Civil War? \textcite macro produces "Author (year)" correctly but colours first paren with hyperref After reinstalling TeXLive distribution (install-tl-20160318, Ubuntu 14.04), a issue occurs with the updated version of the titlesec package: the section numbering is lost.  What could it be missing? Is there a modification I have not seen?  Below is a MWE comparison of the two versions. The prevtitlesec package is a reinstall of the previous version (2011/12/15).  \documentclass{book} \usepackage[margin=2cm,a4paper]{geometry} %\usepackage{prevtitlesec}% 2011/12/15 version \usepackage{titlesec}% 2016/03/15 version \usepackage{lipsum}  \begin{document}  \chapter{Title of the first chapter} \lipsum[1] \section{A test section} \lipsum[4] \subsection{A test subsection} \lipsum[4] \subsection{A test subsection} \lipsum[4]  \end{document}   titlesec package 2011/12/15:    titlesec package 2016/03/15: Showing that $\displaystyle\lim_{s \to{1+}}{(s-1)\zeta(s)}=1$ I need prove the following: ($\zeta(s)$ is the Riemann zeta function)  $\displaystyle\lim_{s \to{1+}}{(s-1)\zeta(s)}=1$  I really don't know, I have tried hard, but have obtained nothing for now.