How can I explain to someone who is a beginner, the basic idea and intuition behind dy/dx? I'm introducing calculus to one of my friends and here are the questions I might encounter: Why dy/dx? What do I even need to find the slope? How does that relate to rate of change of y with respect to x? What could be some real life examples for a better understanding? What information does the slope at every point of a curve, which is just a line, give me?  I'd like to explain all of this giving an intuitive sense to the formal differential calculus. Thanks!
In the book Thomas's Calculus (11th edition) it is mentioned (Section 3.8 pg 225) that the derivative $dy/dx$ is not a ratio. Couldn't it be interpreted as a ratio, because according to the formula $dy = f'(x)dx$ we are able to plug in values for $dx$ and calculate a $dy$ (differential). Then if we rearrange we get $dy/dx$ which could be seen as a ratio.   I wonder if the author says this because $dx$ is an independent variable, and $dy$ is a dependent variable, for $dy/dx$ to be a ratio both variables need to be independent.. maybe?
I am looking for either an online tool or one that I can download to convert DD to DMS.  For example, I want to convert from:  41.590833, -93.620833 to 41° 35′ 27″ N, 93° 37′ 15″ W ...  Those are the coordinates, according to , for Des Moines. If $\int_0^x f \ dm$ is zero everywhere then $f$ is zero almost everywhere I have been thinking on and off about a problem for some time now. It is inspired by an exam problem which I solved but I wanted to find an alternative solution. The object was to prove that some sequence of functions converges weakly to zero in $L^2$.   I managed to show (with some help) that the limit $f$ (of a subsequence) satisfies $\int_0^x f \ dm=0$ for all $x&gt;0 $. From this I want to conclude that $f=0$ a.e. I can do this with the fundamental theorem of calculus in its Lebesgue version but there ought to be a more elementary proof.  Can someone here help me out? Javascript custom button stopped working, Synchronous XMLHttpRequest is deprecated We have a custom javascript button on the quotes object in our org that has been working for several months that someone has noticed today is not working.    I'm not sure if there was an update that broke it or what day exactly it stopped working.  I'm hoping someone has an idea of where I can even begin to look.   Here's the code:   javascript:location.href= 'mailto:emailaddress@ourdomain.com?SUBJECT='+ escape("{!Quote.Name}")+' - '+ escape("{!Quote.Contact_s_Account__c}") + '&amp;BODY=Please convert ' + escape(location.href)+ ' into a proposal and send it to the email address on the quote.  Thanks!';   Using developer tools in chrome, I see this error after I click the javascript button:   presence: onFailure in event parser ={"status":-5,"statusText":""}page load happens value=true           umps1w4.salesforce.com/umps/UMPSWidget_184/js/dist/umps-min.js?cflag=184.36:14 presence: onFailure in event parser ={"status":-5,"statusText":""}page load happens value=true   I changed my code to something simple:   javascript:location.href= 'mailto:emailaddress@ourdomain.com?SUBJECT='hello'    and got this error in the developer tools in chrome:   Synchronous XMLHttpRequest on the main thread is deprecated because of its detrimental effects to the end user's experience. For more help, check http://xhr.spec.whatwg.org/.   This lead me to the conclusion that I'm having XSS issues.  I do remember a SF update not too long about about protecting against XSS issues.  I guess I'll have to figure out a different way to implement this functionality. Show that $\lim_{z \rightarrow \infty} -z^xe^{-z} = 0$ Let $x&gt;0$ be a constant. Show that $\lim_{z \rightarrow \infty} -z^xe^{-z} = 0$.  Do I apply L'hopital's rule here? It doesn't seem to help, for example, after one round of differentiating the numerator and denominator, I get $\frac{d}{dz} z^x = xz^{x-1}$ and $\frac{d}{dz}e^z = e^z$. I am unsure on how to proceed. I saw this today, I checked in Mathematica and the integral comes out to $\pi$, but I have no idea how to solve it.     FREE Wi-Fi: The Wi-Fi password is the first $10$ digits of the answer.   $$\int_{-2}^2\left(x^3\cos\frac x2+\frac12\right)\sqrt{4-x^2}\ dx$$ Uncommitted work pending in unit test with trigger and queueable callout I have an after insert trigger that invokes queueable Apex which makes a callout.  The trigger and callout work fine in the UI and I don't get any errors in normal operation.  However, my unit test keeps failing with the "uncommitted work pending" error despite all my attempts to separate the callout.  I've used a testVisible variable to disable the callout from the trigger so that I can invoke the queueable job explicitly in a startTest/stopTest block after the DML that fires the trigger.  I've pored over the callout class to make sure it's not doing any DML beforehand.  Anyway, if it was doing DML, I should get the uncommitted work pending errors in the UI, right?  I've tried API version 37 and version 41 but the unit test still fails.   I've had a case open with Salesforce premier support for 6 weeks now but haven't made any progress.  Any suggestions?  The unit test:  @isTest static void testCallout() {     // Disable the callout while setting up data.     CCx_EHR_CMCT_Services.isQueueSendEnabled = false;      // Set up the data and execute the DML.     / ...     insert act;      // Set up the http mock     // ...      // Now enable and execute the callout     Test.startTest();     CCx_EHR_CMCT_Services.isQueueSendEnabled = true;     Test.setMock(HttpCalloutMock.class, mock);     System.enqueueJob(new QueueSendToRedox(new List&lt;ActivityHL__c&gt;{act}, desk)));     Test.stopTest();  }   Trigger:  trigger HL_ActivityHL_c on ActivityHL__c (before insert, before update, after insert) {     if (Trigger.isAfter) {         if (Trigger.isInsert) {             CCx_ClientDataDomain.sendEHRClinicalComm(Trigger.new);         }     } }   Trigger handler class:  public without sharing class CCx_ClientDataDomain {     public static void sendEHRClinicalComm(List&lt;ActivityHL__c&gt; triggerNew) {     // do some processing here.     // ...     // then queue the callout job.     System.enqueueJob(new CCx_EHR_CMCT_Services.QueueSendToRedox(             deskActivities.get((Id) desk.getDeskId()), desk));     } }   Queueable job class with callout method:  public with sharing class CCx_EHR_CMCT_Services {     // isQueueSendEnabled is an attempted workaround for unit tests     @testVisible     private static Boolean isQueueSendEnabled = true;      public class QueueSendToRedox implements Queueable, Database.AllowsCallouts {         private List&lt;ActivityHL__c&gt; activities;         private CCx_DeskHierarchySelector.Desk desk;         private Boolean isEnabled;          public QueueSendToRedox(List&lt;ActivityHL__c&gt; activities, CCx_DeskHierarchySelector.Desk desk) {             this.activities = activities;             this.desk = desk;             this.isEnabled = isQueueSendEnabled;         }          public void execute (QueueableContext context) {             if(this.isEnabled) sendClinicalCommToRedox(activities, desk);         }     }      public static void sendClinicalCommToRedox(List&lt;ActivityHL__c&gt; activities, CCx_DeskHierarchySelector.Desk desk) {         // Set up data for the callout.  No DML actions are executed before the callout.         HttpRequest req = buildRequest();         // ...          // Callout         Http http = new Http();         HttpResponse resp = http.send(req);     } } I am having an error compiling the following code with xelatex and lualatex:  ** WARNING ** "seac" character deprecated in Type 2 charstring. ** ERROR ** Type2 Charstring Parser: Parsing charstring failed: (status=-1, stack=5)  Output file removed.  \documentclass{article}  \usepackage{xunicode} \usepackage{fontspec}  \setmainfont[Mapping=tex-text,Ligatures={Common}, Numbers={OldStyle}]{IM FELL English PRO} \begin{document}  Buongiorno a tutti. Trallallà.  \end{document}   The error disappears when I omit the accented à. It also doesn't happen when I use other fonts.  Could you suggest any workaround?  Thank you, A Cropping Shapefiles in QGIS? Make all footnotes numbered subsequently, but one with a symbol How to load a .NET assembly for reflection operations and subsequently unload it? Exit from both root and user with one command I know about "not using sudo su -" etc. But let's be honest, almost all of us do it.   So, here is the issue. We can't have root logins enabled, so we have to ssh in as our user then su to root.  Here is the process tree:       1  7897  7826  7826 ?           -1 S     1000   0:00 sshd: josh@pts/0  7897  7898  7898  7898 pts/0     8182 Ss    1000   0:00  \_ -bash  7898  7990  7990  7898 pts/0     8182 S        0   0:00      \_ sudo su -  7990  7991  7990  7898 pts/0     8182 S        0   0:00          \_ su -  7991  7992  7992  7898 pts/0     8182 S        0   0:00              \_  -su  7992  8182  8182  7898 pts/0     8182 R+       0   0:00                  \_ ps axjf   I would like to exit from root, then from my user with one command. Is there a way to do this?  BTW exit &amp;&amp; exit does not work because it exits the shell and doesnt process the rest of the command  josh@ubuntu:~$ sudo su - root@ubuntu:~# exit &amp;&amp; exit logout josh@ubuntu:~$ Are there any distinct finite simple groups with the same order? In finite group theory, it's often quite easy to show that there are no simple groups of a given order $n$.  My question is different: is there some natural number $n$ such that there are two non-isomorphic simple groups of order $n$?  The two easiest families of finite simple groups - $\mathbb{Z}_p$ for $p$ a prime and $A_n$ for $n\ge5$ - clearly don't yield any examples.    It should be possible to check this using the , but I don't know enough about the fabled 'sixteen infinite families of groups of Lie type' to do the check myself.  Also, if the answer is 'no', I'd be much more interested to see an elementary(ish) proof than I would be to see a proof that relies on over 3,000 pages of non-stop hard group theory. Requires installation of untrusted packages? error: file '/grub/i386-pc/normal.mod' not found error: file '/grub/i386-pc/normal.mod' not found. grub rescue&gt;   What can I do? I just sit and stare at it.   I found my old netbook (Dell Inspiron 1010) which I have not used for about four years. I replaced Windows XP with Ubuntu 12.10. I used my bootable USB drive. I installed and rebooted. I got the message that normal.mod is not found.   What should I do? Type exit, reboot, or quit? Should I re-install? After upgrading to Firefox 57 Quantum many website have rendering issues, such as empty frames, missing text or fonts, etc. I use Kubuntu 16.04 64 bit.  So far, I've tried:   Restarting Firefox in safe mode Disabling "hardware acceleration"   Wiping out my profile Purging Firefox and reinstalling Running Firefox from a clean user account   But nothing helped so far.  Any tips on how to solve this are much appreciated Firstly, I apologise if my thinking is completely incorrect - I am by no means a physicist (yet!), I have included layman illustrations to hopefully assist in explaining my thinking.  I am imagining the big bang to be, for lack of a better word, an explosion, thus the universe to be progressively expanding:    Therefore, to me it seems logical to suggest that there is an edge.  If this is the case, then is it reasonable to suggest that explosion + edge = 'bright edge'? Where light projects in all directions at the edge, not just outwards into the 'nothingness'.     So, this brings me to my question: why, when someone looks deeply enough 'at the edge' do they see an absence of light?   Again, apologies if I'm jumping to irrational conclusions &amp; am way off. Changing units in QGIS: degrees to meters I have a global map with CRS WGS84-EPSG:4326 and I want to calculate the area of my polygons. From what I read, to change the unit from degrees to meters I need to change my CRS. However, there is no global system for this. Is there any solution to change the unit from degrees to meters for a global map? Word that means "outdated name" Is there a word that means "outdated name"? For example:   Record, although very little music is on vinyl Film, although most movies are digital Horsepower, although no one uses horses as a metric anymore Phone, although it's mostly used to access the web   and so on. Proving that $\left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} \right) \lvert f(z) \rvert^2 = 4 \lvert f'(z) \rvert^2$ Given $f$ entire show that $$ \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} \right) \lvert f(z) \rvert^2 = 4 \lvert f'(z) \rvert^2 $$ I've come close to getting the exact answer by writing $f(z)$ as $u(x,y)+iv(x,y)$ and realizing the Laplacian of f is equal to zero. This leads to a lot of cancellations but I'm still making mistakes in my computations and I can't figure out where. Any help is appreciated. Clarify an idiom in a political text Please carefully read the text below:     On 24 November, 1993, a meeting of Leftist intellectuals occurred in   London under the auspices of the Institute for Public Policy   Research (IPPR), which is a Labour-leaning think-tank.  A short document   was circulated in advance of the said meeting to clarify its   purpose.        Among other things, the document declared that the task of the   IPPR was: "To do what the Right did in the seventies, namely to break   through the prevailing parameters of debate and offer a new   perspective on contemporary British politics."    The explanatory document also said, "Our concern is not to engage in a   philosophical debate about foundations of socialism."      If this meant that those foundations were not the appropriate thing to   talk about at the 24 November meeting, then that might have been   right; not every thing has to be discussed at every meeting.  But if   what was meant was that discussion of philosophical foundations is not   what the Left now needs, then I disagree, and if that indeed is what   was meant, then it is curious that the breakthrough by the Right should   have been invoked as an achievement for the left to emulate, for if   there is a lesson for the Left in the Right's breakthrough, it is that   the Left must repossess itself of its traditional foundations on   pain of continuing along its present, politically feeble, reactive   course.  If the Left turns its back on its foundations, it will be   unable to make statements that are truly its own.   Since I've heard only one definition of the part identified in bold, I am at a dead end.  As you know, and according to Oxford's dictionary, the verb "repossess" means: To take back property or goods from SB who has arranged to buy them and can not pay.  Here, the object of verb is a thing not a person, but in the text, first,itself comes before of and its traditional foundations. As a result, I thought that this phrase might be an idiom and so has a different meaning with the original verb.  Please explain the meaning of this phrase.