If I intend to earn on the side doing different category of things should I do it as an individual or as a small business? If I have a few areas of expertise (that can't fit a classification of services/products) and I'd like to make use of those and provide products or services based on those, should I do it as an individual or should I register a business?  Examples of things I have in mind:   Gardening work Teaching Computer Programming in community colleges or private institutes Selling art work of mine and possibly of some friends and family Marketing for other businesses   Examples of concerns:   Tax complications Liability (in case someone decides to sue me) Other benefits or issues I may not be thinking about   So instead of providing the above services as a person, could I provide it under LLC/LLP/Inc where I am the sole contributor. There are somewhat similar questions but here I don't have a fixed business category and it is also a mix of selling things and services.  Another point that I want to focus on is if there is a point beyond which establishing a small business (instead of an individual provider) doesn't make financial sense - tax (esp when the business may make losses on consecutive years).
When should I start an LLC for my side work? I am doing quite a bit of sidework this year and wondering at what point would it be a good idea to "start a business" and bill through that.
Is it possible to switch from one area of graduate study to another in US universities? For example, suppose someone has enrolled in a computer science phd program. Can he switch over to math(or physics) phd program in the same school later?(or say from applied mathematics to pure mathematics?)  What are the steps for doing for doing that? Fill the area determined by two pgfplots graphs I need to fill the region determined by two (Gaussian) curves and the right red line (the region is brushed in green in the figure below). I also wonder what is the best way to draw vertical line which goes through the intersection of both graphs (the left red line on the figure).  Any pointers would be greatly appreciated.    My initial attemt is pasted below.  \documentclass{standalone} \usepackage{pgfplots}  \begin{document}  \pgfmathdeclarefunction{dnorm}{2}{%   \pgfmathparse{1/(#2*sqrt(2*pi))*exp(-((x-#1)^2)/(2*#2^2))}% }  \begin{tikzpicture} \begin{axis}[domain=0:12, samples=100, height=5cm, width=10cm] % Fill aread under the curves \addplot [fill=red!20, draw=none, domain=0:6] {dnorm(6.5,1.5)} \closedcycle; \addplot [fill=blue!20, draw=none, domain=6:10] {dnorm(4,1)} \closedcycle; % Draw curves \addplot [thick] {dnorm(4,1)}; \addplot [thick] {dnorm(6.5,1.5)}; \end{axis} \end{tikzpicture}  \end{document}   My solution  A picture is worth a thousand words. Code is pasted below.   \documentclass{article} \usepackage{tkz-fct} \usetikzlibrary{intersections}  \begin{document}  \tikzset{    name plot/.style={every path/.style={name path global=#1}} }  % Extract coordinates for point X \makeatletter \newcommand{\gettikzxy}[3]{%   \tikz@scan@one@point\pgfutil@firstofone#1\relax   \edef#2{\the\pgf@x}%   \edef#3{\the\pgf@y}% } \makeatother  % Dimlines \def\Dimline[#1][#2][#3][#4]{ \begin{scope}[thin, &gt;=stealth'] % redefine as flechas \draw let \p1=#1, \p2=#2, \n0={veclen(\x2-\x1,\y2-\y1)} in [|&lt;-&gt;|, decoration={markings,mark=at position .5 with {\node[#3] at (0,0) {#4};}, }, postaction=decorate] #1 -- #2 ; \end{scope} }  \begin{tikzpicture}[scale=1,font=\small] \tkzInit[xmin=0,xmax=12,ymin=0,ymax=.3,ystep=.05] % Draw coordinates \draw[&gt;=stealth', &lt;-&gt;] (0,6) node[above] {$y$} -- (0,0) --  (12.5,0) node[right] {$x$}; % Draw functions and areas \tkzFct[name plot=A,thick,color=red,domain=0:12]{1/(1.5*sqrt(2*pi))*exp(-((x-4.5)**2)/(2*1.5**1))} \tkzDrawArea[opacity=.3,color=blue,domain = 7:12] \tkzFct[name plot=B,thick,color=blue,domain=0:12]{1/(2*sqrt(2*pi))*exp(-((x-7)**2)/(2*2**1))} \tkzDrawArea[opacity=.3,color=red,domain=0:7] \tkzDrawAreafg[between=b and a,opacity=.3,color=green,domain = 0:7] % Intersection between curves \path [name intersections={of=A and B,by=C}]; % Extract coordinates of C \gettikzxy{(C)}{\cx}{\cy} % Vertical lines \draw [thick,dashed, black] (\cx,0) -- (\cx,5.5) node [above] {$x_{0}$}; \draw [thick,dashed, black] (7,0) -- (7,5.5) node [above] {$\hat{x}$}; % Define regions \Dimline[($(0,0)+(0,-.6)$)][($(7,0)+(0,-.6)$)][above,black][$\mathcal{R}_{1}$]; \Dimline[($(7,0)+(0,-.6)$)][($(12,0)+(0,-.6)$)][above, black][$\mathcal{R}_{2}$]; \end{tikzpicture}  \end{document} Solution af a system of 2 quadratic equations I have a system of two quadratic equations with unknowns $x$ and $y$:  $$a_{1 1} x y + a_{1 2} x^2 + a_{1 3} y^2 + a_{1 4} x + a_{1 5} y + a_{1 6} = 0,\\ a_{2 1} x y + a_{2 2} x^2 + a_{2 3} y^2 + a_{2 4} x + a_{2 5} y + a_{2 6} = 0,$$  where $a_{i j}$ are arbitrary scalars.  Is there an algebraic solution of the above system? In this MWE, I want to obtain bold letter for the letter "H" (compare the two "H") :  \documentclass[10 pt]{beamer} \usetheme{default} \usepackage{lmodern} % load a font with all the characters \usepackage{fontspec} \usepackage{cmbright} \SetSymbolFont{largesymbols}{normal}{OMX}{iwona}{m}{n} \usepackage{amsmath}%loads amsmath that should go before unicode-math \usefonttheme{professionalfonts} \defaultfontfeatures{Mapping=tex-text}  % This seems to be important for mapping glyphs properly  \begin{document} \begin{frame} \[ \mbox{First bold H : }\|x\|_{\mathbf{H}},\; \mbox{Second bold H : }\mathbf{H}  \] \end{frame} \end{document}   but I don't know how doing it. THe "H" produced by this MWE is tiny but no bold effect. If I use      \SetSymbolFont{largesymbols}{normal}{OMX}{iwona}{m}{n}   I obtain bold tiny "H" but the font used still not the same. Hide deleted answers with certain characteristics Today I exceeded 10K reputation on Stack Overflow, and I see deleted answers.  I am now faced with pages like  where half the answers are "deleted."    Among these are many non-answers deleted by diamond moderators, and a self-deleted answer deleted on Nov 25 '09.  I find this very distracting, especially in my chosen sort order of "oldest" where these deleted answers are NOT displayed at the bottom.  I wish to be able to hide deleted answers based on combinations of characteristics such as:   age since deletion deleted-by-owner deleted by diamond moderators zero undelete votes   This would be a setting every user could enable for himself.    Here are HTML excerpts that may be useful for userscript filtering.  This shows the code that appears for a moderator deletion, as well as the date stamp.  &lt;div class="deleted-answer-info"&gt;deleted by &lt;a href="/users/10936/dori"&gt;Dori&lt;/a&gt;&lt;span class="mod-flair" title="moderator"&gt;♦&lt;/span&gt; &lt;span title="2011-06-26 09:13:34Z" class="relativetime"&gt;Jun 26 at 9:13&lt;/span&gt;&lt;/div&gt;   Here is the same class for deleted by owner:  &lt;div class="deleted-answer-info"&gt;deleted by owner &lt;span title="2009-11-25 19:12:11Z" class="relativetime"&gt;Nov 25 '09 at 19:12&lt;/span&gt;&lt;/div&gt; Determine if two rectangles overlap each other? I am trying to write a C++ program that takes the following inputs from the user to construct rectangles (between 2 and 5): height, width, x-pos, y-pos. All of these rectangles will exist parallel to the x and the y axis, that is all of their edges will have slopes of 0 or infinity.  I've tried to implement what is mentioned in  question but I am not having very much luck.  My current implementation does the following:  // Gets all the vertices for Rectangle 1 and stores them in an array -&gt; arrRect1 // point 1 x: arrRect1[0], point 1 y: arrRect1[1] and so on... // Gets all the vertices for Rectangle 2 and stores them in an array -&gt; arrRect2  // rotated edge of point a, rect 1 int rot_x, rot_y; rot_x = -arrRect1[3]; rot_y = arrRect1[2]; // point on rotated edge int pnt_x, pnt_y; pnt_x = arrRect1[2];  pnt_y = arrRect1[3]; // test point, a from rect 2 int tst_x, tst_y; tst_x = arrRect2[0]; tst_y = arrRect2[1];  int value; value = (rot_x * (tst_x - pnt_x)) + (rot_y * (tst_y - pnt_y)); cout &lt;&lt; "Value: " &lt;&lt; value;     However I'm not quite sure if (a) I've implemented the algorithm I linked to correctly, or if I did exactly how to interpret this?  Any suggestions? What is the difference between == and ===?  How exactly does the loosely == comparison work? How exactly does the strict === comparison work?  What would be some useful examples? I didn't know how to solve this problem.  How to prove $$\sum_{n=0}^{\infty}\sin(e\pi n!)$$ converges? Adding Layer using esriRequest Response in ArcGIS API for JavaScript? For various reasons I need to add layers to a map object using a response returned by an esriRequest to an ArcGIS Server. One reason being the data will change constantly and hard coding FeatureLayer URLs would prove inefficient.  I have successfully been able to return a JSON object containing attribution that appears to be the same data one would see on the layers REST node.     maps = [];  require([     "esri/map", "esri/layers/FeatureLayer",      "esri/dijit/Print", "esri/tasks/PrintTemplate", "esri/request",      "esri/config","esri/dijit/BasemapToggle",      "dojo/_base/array", "dojo/dom", "dojo/parser",  "dojo/domReady!"  ], function (Map, FeatureLayer,Print, PrintTemplate, esriRequest,      esriConfig,BasemapToggle,arrayUtils, dom, parser){  parser.parse();  console.log("Map Start");  esriConfig.defaults.io.proxyUrl = "http://........./proxy.ashx";  maps.addRespLyrs = function addRespLyrs (layer) {     if (layer.type == "Feature layer") {         console.log(layer.name);         maps.map.addLayer(layer);     } };   maps.handleMapInfo = function handleMapInfo (resp) {     console.log("handleMapInfo called");     maps.layers = resp.layers;     console.log(maps.layers);     for (i in maps.layers) {         var layer = maps.layers[i];         maps.addRespLyrs(layer);     } };  maps.layerRequest = function layerRequest () {     var mapInfo = esriRequest({         "url": "https://........./Application Name/MapServer/layers",         "content" : {"f":"json"},         "handleAs" : "json",         "callbackParamName" : "callback"     });     mapInfo.then(maps.handleMapInfo); };  maps.map = new Map("cpCentCent", {     basemap:"topo",     center: [-98.5795, 39.8282],     zoom: 5 });   console.log("Map End");   When attempting to add the data using the addLayer() method I recieve an error thrown from deep with in esri's API.      init.js:199 TypeError: a.id.match is not a function(…) "TypeError: a.id.match is not a function   Any thoughts as to what could be going wrong here or how to create a layer from my JSON? I wonder if it is possible to consider $\mathbb R$ as a Cartesian product $X\times X$ for some set $X$. From the point of view of the dimensionality, there are spaces with a Hausdorff dimension $1/2$ (sort of Cantor sets), but I guess there are other problems in this construction  unrelated to the dimension.  Edited: replying on the comment by Asaf. I want that for all $r\in\mathbb R$ there exists a unique representation $r=\langle x,y\rangle$ where $x\in X$ and $y\in X$. Also, what if we want to have $X$ to be a topological space and $h:X\times X\to\mathbb R$ to be a homeomorphism? Activate Rigid Body Physics at certain time? I've been trying to make a wrecking ball smash through a figure of sorts, not really sure what exactly but I made an example. The problem I'm having is that the figure does indeed fall apart, but I would like it to stay put together until the ball hits it. I achieved this by using Cell Fracture on the figure and just using Rigid Body. Again, I just need a way for the figure to stay still until hit by the ball, and then have all of its pieces come apart. Why isn’t the definite article used in "The commission commanded that work [...] should cease"? In this sentence:     The commission commanded that work on the building should cease.   Why isn't the used before the word work? It seems that the work is specified by on the building.  Another example is,     Construction of the building will start next week.   I understand construction here is something like putting up the building, and so the construction isn’t general in meaning. But the definite article isn’t used here, either.  I’ve asked about this several times, but none gave me a good answer. I've been trying to change the language of my LaTeX document by replacing \usepackage[swedish]{babel} with \usepackage[english]{babel} but I only get this message:  ! Package babel Error: You haven't loaded the option swedish yet.   I've read in an other question at this forum that compiling enough would solve the problem, but that doesn't work for me. Does anyone know what I should do? I’m able to copy-paste from any application to any application. I’m able to copy from a comment or from a question or from an answer. I’m able to paste to a question or answer. However, if I attempt to paste into a comment, nothing happens. I am able to subsequently paste into another application, confirming that the text is actually on the clipboard.   However, copying SE links will paste them successfully, but all other text - including other links - still fails. This leads me to believe that it’s a SE issue and not an Apple one.   This has been an issue since 11.0 beta and persists through iOS 13 release. Consider the examples:    "My name is Ricky."   He said his name is Ricky.    The tense is not backshifted in #2 because his name is and will always be Ricky. Also consider:    Ricky says: "I live in Paris."   Ricky said that he lives in Paris.    The tense is not backshifted in #4 because he still lives in Paris.  For examples #2 and #4, it is possible to not backshift the tense because the reporting words are still true at the time of reporting, but what will you say about the following examples:        I asked what his name is.   I asked where he lives.      Here, is it correct for #5 and #6 to not backshift the tense?  If #5 and #6 are correct, then under what conditions is it correct to not backshift the verb like that? Show $f$ can't be irreducible over a finite field if $f^\prime$ is the zero polynomial. Apparently this is a  and should hopefully be fixed in an upcoming update.    I have installed SQL Server Management Studio v17.9 on a clean Windows 10 installation.  This associated all .sql files with SSMS.  I want to associate .sql files with a plain text editor (eg: Sublime) but I am unable to change the default application.  I have tried:   Right click .sql file -> Open With -> Sublime + "Always use this program". Settings -> Default Applications by File Type -> Find .sql -> Change to Sublime Right click .sql file -> Properties -> "Change" button -> Sublime. Using ftype via elevated command prompt as seen    In all 3 cases, my selection is not reflected in any way, anywhere.  The icon remains SSMS, and the file opens with SSMS.  Is there some kind of "File Association retention" setting within SSMS I am unaware of?  Edit: I can change the default application for non-SSMS-associated extensions without issue.  Edit: after deleting registry key HKCU\Software\Microsoft\Windows\CurrentVersion\Explorer\FileExts\.sql, .sql files show a VSCode icon:    If I then double-click New Users.sql, it opens with SSMS and a refresh of regedit shows the .sql entry recreated with ssms.exe associated (the icon remains VSCode):    If I then right-click -> open with -> Choose Another App -> Select Sublime and hit "Always use this to open .sql files", it opens with Sublime and a refresh of regedit shows a new "b" entry created for Sublime.  The icon remains VSCode, and double-clicking opens it in SSMS:    If I then delete the a entry for ssms.exe and rename b -> a such that only sublime exists, then double-click the .sql file, it opens with SSMS and generates a new b entry for itself: Cartographic techniques for symbolizing routing data Hi I'm traveling to Uzbekistan and have 12 hours layover in Seoul via Korean Airlines. Do I need a visa to go out of the airport for sightseeing? I'm a Philippine passport holder. How to create a Facebook page without a personal account In the past, Facebook used to let business create pages or apps without a personal account. This account created in the page/app creation process had limited capacities and could be switch to a full personal account after—if needed (irreversible).  But today, if I want to create a page for a new brand, without having a Facebook account, it seems I can’t anymore. And I can’t find any communication from Facebook regarding this change.   Anybody has any info? Is it still possible (in a new way maybe)?