I would like to know how to color change a stroke in Illustrator or Photoshop like the Q in the picture:
Is it possible to paint only inside a stroke on illustrator? what i actually want to do is paint the same stroke with different colors in diferent segments of it but i can't do it...   Also... I've seen some tutorials that include editing a gradient but that would be very complicated and so would be to make a new stroke over this one, I just hope theres an easyer way XD  Edit per comment:  thats cool but not really what im looking for, what i want is lets say my stroke has the shape of a burger i want part of i to be bread color and part of it to be meat color... not one stroke over the other... i want something like this, imagine the jelly worm is my stroke:
Borel set preserved by continuous map Let $f:\mathbb{R}^m\rightarrow\mathbb{R}^n$ be a continuous map. Show that if $A$ is a Borel subset of $\mathbb{R}^n$, then $f^{-1}(A)$ is a Borel subset of $\mathbb{R}^m$.   I know that for $A$ open subset of $\mathbb{R}^n$, then $f^{-1}(A)$ is open. Likewise for closed. A Borel subset is a countable unions/intersections of these sets, done repeatedly. From those facts, combined with $f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$ and $f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$, the statement should be intuitively true.  But to do it rigorously, I want to write $A$ as open and closed sets. But since the countable union/intersection can be done repeatedly, I don't know how I can write it that way. Error while running Quickly Proving any number of the form $2^n$ can't be written as the sum of k consecutive numbers. Changing Quotes What is the term for changing a quote? For example, changing 'Who knows what evil lurks in the hearts of men?' to 'Who knows what mischief lurks in the hearts of children?' They agreed to stay connected for hours in their provisional support group, looking for answers in their counterparts.   I want to change the bold part to say that they all shared the same problem.  Might end up using one of the above, but I feel like I need some more options to weigh in.    Edit for context: they all share the same rare cognitive impairment How do I remove an OpenID or login? How do I remove login credentials (emails or OpenIDs) from my account?     For more information, see "" in the . Let be $u,v\in\mathbb{R}^n$, then $\det(I+uv^\intercal)=1+v^\intercal u $ where $I$ denotes the identity matrix of order $n$. How to prove this?   what I did:   let be $A=\{n\in\mathbb{N}: \forall u,v \in \mathbb{R^n}, \det(I+uv^\intercal)\neq1+v^\intercal u  \} $, and suppose $A\neq \varnothing $, then for the well-orderer-principle there exists $n_0\in A$ such $n_0\leq n,\;\forall n\in A$. since $1\notin A,\; n_0\neq 1 \rightarrow n_0-1\in\mathbb{N}\setminus A  $ so: $$\forall u,v\in\mathbb{R^{n_0-1}}: \det(I+uv^\intercal)=1+v^\intercal u $$ Then, let be $u,v\in\mathbb{R}^{n_0}$, so $$\det(I+uv^\intercal)= \displaystyle\sum_{j=1}^{n_0}(-)^{1+j}a_{1j}\det(A_{1j})$$ where $a_{ij}=\begin{cases} u_iv_j+1 &amp; i=j \\ u_iv_j &amp; i\neq j \end{cases}$, and $A_{1j}$ is the  submatrix of $I+uv^\intercal$ that results deleting the i-file and j-column. Then I try to open the term for $j=1$ and try to use the relation for the new matrix of $n_0-1$ order with the determinant because it holds the same form but I have other terms which difficult me the work. Do you know other method?.  Ps. , in page 2 I've found something similar but I don't understand what it means PS. I'm taking a course of numerical analisys Do I need to activate the firewall? I only use Ubuntu for a home desktop use? I'd like to find the files in the current directory that contain the text "chrome".  $ find . -exec grep chrome find: missing argument to `-exec'   What am I doing wrong? Asking a question: DO or ARE? When should you ask a question with &quot;do&quot;, and when with &quot;are&quot;?  In other words,  Are you going to Spain next week?   Do you like Chinese food?  What is the rule here?  We don't say  Do you go to Spain next week?   Are you liking Chinese food?  Why is that? Careers ads break the Android app Clicking any Careers (or Jobs) ad in the Android app feed results in an instant crash for the Android App version 1.0.89. One of my players has a +1 shortbow, and recently found some +1 arrows. If he fires a +1 arrow from a +1 bow, will the bonus to attack and damage be +1, or +2? Finding the expected value of a continuous random variable with a polynomial function. Update finding the integral. An insurance company's monthly claims are modeled by a continuous, positive random variable $X$ whose probability density function is proportional to $(x+1)^{-4}$ where $0 \lt x \lt \infty$   I need help understanding the reasoning behind the answer:  the PDF is given by:  $1 =\int_0^{\infty} \frac{k}{(1+x)^{-4}}$   Here is my first issue in the first step, where does K come from? We are given the pdf to be $\frac{1}{(x+1)^{4}}$ So to me it would stand that the function we are given is the PDF because it says thats the pdf, and we are then to find the expected value from here which would be equivalent to:  $\int_0^{\infty} x(1+x)^{-4}$ by definition of expected value  My guess would be that the words "proportional to" are key as to why we have a K introduced can someone explain this?  I understand everything except when taking the integral because and solving for 1 because I arrive at $\frac{3k}{(1+x)^4}$ rather than $\frac{-1}{3(1+x)^4}$  Taking the integral of $(1+x)^4$:  $u=1+x$ $du=1dx$ and   $\frac{u^{-4+1}}{-4+1} dx$  which gives $\frac{-3k}{(1+x)^3}$  Which gives me $k=\frac{1}{3}$ when solving $1=\frac{-3k}{(1+x)^4} \Big\vert_0^{\infty}$ rather than $k=3$ as the answer which is the correct answer. Can someone please explain this to me? Ubuntu 16.04 multitouch gestures Showing that $f$ is identically $0$ on $\mathbb{D}$ Here is the question: Let $f:\mathbb{D}\rightarrow\mathbb{C}$ be an analytic function.  Suppose that there exists $\epsilon&gt;0$ such that for every $\theta$, $|\theta|&lt;\epsilon$, and $\lim_{z\rightarrow e^{i\theta}}f(z)=0$.  Prove that $f$ is identically $0$ on $\mathbb{D}$. My thoughts:  Well, honestly, I am quite stumped here.  I was thinking I could let $f$ be analytic in some other region (can I even do this?), say the lower half plane and suppose $\lim_{z\rightarrow x}f(z)=0$ whenever $-1\leq x\leq1$, by assumption.  Maybe then I could show that $f$ is identically $0$ on the bigger region, and then consider $f$ on the intersection of those two regions, and then ultimately making my way towards using the uniqueness theorem to show what I want... but I am quite stuck on how to do that... any thoughts, suggestions, etc. are greatly appreciated!  Thank you. I need a CD or USB stick, that will install Ubuntu on a completely headless machine. Stick CD in and restart, no keyboard or screen should be involved.  There are a few obstacles:   The language selection menu when you first boot the CD. The fact that the CD menu waits. The installer's asking questions during installation. Timestamp in the  is going out of it's boundary.        AFAIK, the suggested edits timestamp was near to it's type (i.e. asked, proposed) like this:        Checked on Chrome (27.0.1453.110 m), Firefox (21.0) on Windows XP  The issue is in other Stack Exchange sites as well. E.g. Have a look at  on Physics.SE. Also in   Can we fix this? "Writ large", "Writ small", and others I have seen this used with other words besides "large" and "small".  What other words can be used? Can this structure be used more generally with other adjectives? Is there a c# equivalent of Java ConcurrentHashMap Texmaker generated pdf file not showing the position of cursor I am compiling LaTeX files with Texmaker on Ubuntu 12.04 system.   When I run pdflatex and then view the pdf file (internal), it does not show me the exact place where I am editing on Texmaker i.e. the generated pdf file does not show me the position of the cursor.   But some days before it was working fine. I don't know what causes this problem.  I already tried uninstalling and then reinstalling, I did update and upgrade too. But the problem still persists.