Not sure if I'm just missing something in the UI or if there's a Mod that might enable this functionality, but it seems like it would be really useful to be able to leave one of the views active while you're laying out roads / zones / districts.
While building things I want to see the overlays from other modes so I can make better decisions. Can I lock an overlay somehow?  For example: When building roads I want to see the natural resources overlay. When zoning I want to see the wind overlay
Context Around this Code: I am making a schedulable class run each Sunday to track a couple of metrics important to the team.   Question: I am looking to consolidate DML operations and a big part of this is understanding pass by reference vs pass by value.  I want to pass the salesOps user object to the method SDREfficiencyLastWeek. I made the return type void, because I figured that I will be passing this by reference and I can make all the updates in the method without having to anything.   Why I am confused:    I read the above article and am thrown off by the following statement:  "Non-primitive data type arguments, such as sObjects, are also passed into methods by value. "   If the object is indeed passed by reference, then the query for the salesOps user in my code should be sufficient enough to edit all fields on that object without actually having to state each field in the query.  global class SDREfficientMetricsSchedulable implements Schedulable  {     global void execute(SchedulableContext sc)      {         User salesOps = [SELECT ID FROM USER WHERE id = '0056A000002Z3p9QAC'];         List&lt;String&gt; queries = new List&lt;String&gt;();          String contactsWorked_lastWeek  = 'select count(id) results from contact where LastActivityDate = last_week';         queries.add(contactsWorked_lastWeek);          String demosSet_lastWeek            = 'select count(id) results from form__c where CreatedDate = last_week';         queries.add(demosSet_lastWeek);          String demosCompleted_lastWeek  = 'select count(id) results from opportunity where (stagename = \'Unqualified Demo Completed\'  OR stagename = \'Qualified Demo Completed\' )  and RecordTypeId = \'0126A000000yudDQAQ\' and CloseDate = last_week';         queries.add(demosCompleted_lastWeek);          String qualifiedDemos_lastWeek  = 'select count(id) results  from opportunity where RecordTypeId = \'0126A0000004Ao6QAE\' and isclosed = false and  SDR_Owner__c != null and CreatedDate = last_week';         queries.add(qualifiedDemos_lastWeek);          String closedWon_lastWeek       = 'select count(id) results  from opportunity where RecordTypeId = \'0126A0000004Ao6QAE\' and iswon = true and  SDR_Owner__c != null and CloseDate = last_week';         queries.add(qualifiedDemos_lastWeek);            SDREfficiencyLastWeek(queries, salesOps);          update salesOps;     }       public void SDREfficiencyLastWeek(List&lt;String&gt; queries, User salesOps )     {        }  } Windows 7 / 8 stop modal dialogue flash on parent focus Is there some setting in Windows where I can change the default behavior that causes the border and title of a modal dialogue box to flash like 6 or 7 times when I click on the parent window?  Simply focusing it would be plenty to let me know what I did was wrong.   I don't feel the need for all the sirens and bull horns and strobe lights to let me know I clicked in the wrong place. Proof: in $\mathbb{R}$, $((0,1),|\cdot|)$ is not compact. Let $(M,d)$ be a metric space, and $A\subset M$. By definition, $A$ is said to be compact if every open cover of $A$ contains a finite subcover.  What is wrong with saying that, in $\mathbb{R}$, if $I=(0,1)$, we can choose $G=\{(0,\frac{3}{4}), (\frac{1}{4}, 1)\}$, which satisfies $I \subset \bigcup_{U\in G} U$, but we can't extract a finite subcover, so $I$ is not compact. Is $G$ a finite subcover of $G$, so it is not a valid cover for proving this? I would take $\cup_{n\in\mathbb{N}} (\frac{1}{n},1)$ in order to prove this, can we conclude that every open cover is necessarily a infinite union of open sets $\neq \emptyset$? UMP of product functor The scenario:  In a bash script, I have to check if a password given by a user is a valid user password.   I.e suppose I have a user A with password PA.. In the script I asked user A to enter his password, So how to check if the string entered is really his password?... Total number paths between two nodes in a complete graph Suppose the mean noon-time temperature for September days in San Diego is 24∘ and the standard deviation is 4.9. (Temperature in this problem is measured in degrees celsius)  On September 26, 1963, the all-time record of noon-time temperature in San Diego of 44∘ was hit. Assume the temperature distribution is symmetric around the mean, what is the Chebyshev bound for the probability of breaking (or tieing) this record?  I am having a hard time understanding this. Could someone explain to me how to do this? $(a,b) \mathbin\# (c,d)=(a+c,b+d)$ and $(a,b) \mathbin\&(c,d)=(ac-bd(r^2+s^2), ad+bc+2rbd)$. Multiplicative inverse? Let $r\in \mathbb{R}$ and let $0\neq s \in \mathbb{R}$. Define operations $\#$ and $\&amp;$ on $\mathbb{R}$ x $\mathbb{R}$ by $(a,b) \mathbin\#(c,d)=(a+c,b+d)$ and $(a,b) \mathbin\&amp;(c,d)=(ac-bd(r^2+s^2), ad+bc+2rbd)$ as addition and multiplication, respectively. Do these operations form a field?  I have shown that every axiom of the field is satisfied except for the "existence of multiplicative inverses" axiom. This is where I need help. We either need to show that this axiom fails and so we don't have a field or that the axiom $aa^{-1}=1$ is satisfied for every $a\neq 0 \in \mathbb{R}$  The multiplicative identity in this problem is $(1,0)$.  Any solutions or hints are greatly appreciated. Why was the snaphat not awarded? Or why was it awarded incorrectly? There have been a few cases in which the  was not awarded.   There have also been other cases in which it was awarded incorrectly, on old questions.   What gives? Classification of groups of order 12 I have a lot of difficulty understanding this proof we went over in class about classification of groups of order 12.  Let $G$ be a finite group of order $n=p^rm$ where $m \nshortmid p$.  Denote $Syl_p(G)$ as the set of all sylow p-subgroups, and $n_p(G)$ as the cardinality of $Syl_p(G)$.  (1) G has subgroups of order $p^r$. (How do we know this??)  (2) We know all sylow p-groups are conjugate and their number $n_p(G) | m$, by the 2nd and 3rd Sylow Theorems.  (3) We have that $n_p(G) \equiv 1(p). $  Then $n=12=2^2*3^1$ so $n_2(G)|3$ gives $n_2(G) \in \{1,3\}$ and $n_3(G)|4$ so $n_3(G) \in \{1,4\}$ since $2 \not\equiv 1(3)$.  Case 1: $n_3(G) = 4.$ Then the action of G by conjugation on $Syl_3(G)$ gives a homomorphism $f: G \rightarrow S_{Syl_3(G)} \simeq S_4$. $Ker(f)$ consists of $g \in G$ that normalizes all 3-Sylow subgroups. (Why?). Let $P_3$ be a 3-Sylow subgroup. Then the order of $P_3$ is 3. (I don't get that either), and $[G:N_G(P_3)] = 4.$ Thus $P_3 \subset N_G(P_3)$ gives $P_3 = N_G(P_3)$. (I'm lost here...) So $Ker(f) = \cap Syl_3(G) = \{e\}$ (the intersection of the sylow 3-subgroups is trivial. (How did we arrive here...) Thus we conclude that $G$ is isomorphic to a subgroup of $S_4$ of order 12.  To show $G$ is isomorphic to $A_4$, we have that $G$ has 8 elements of order 3: 4 3-sylow subgroups each has 3-1=2 elements of order 3. And $S_4$ has 8 3-cycles meaning f(G) contains all 3-cycles and hence the group they generate, which is $A_4$, so $A_4 \subset f(G)$. But since $|A_4| = 12 = |f(G)$ then $f(G) = A_4$ and $G \simeq A_4$. (Not understanding the first half of the proof, I do not get this part either).  Case 2: $n_3(G) = 1.$ The scope of the question is too large, I will need to post separately to understand the second half, unless the organizational rules of math.stackexchange would require that I post here. In that case, I'll edit the question.  I basically cannot follow the proof because it seems to skip too many steps; it would be helpful if anyone could elaborate case 1 with more reasoning so that I can follow it. Any help would be appreciated; I've spent hours trying to look up and decipher the proof but without any success. Is there a way to make a true toon shader material in Cycles?  Here's, in order of importance, what would be ideal to find in such a shader:   Must use Cycles Must be an actual shader Must not rely on lighting Must be independent of camera position Must react directly to lighting rather than manually having to change it to emulate it Should be easily made into a node-group Should have the following configurability:   Main color Darkness color Shadow depth (how far the shadow goes from the unlit side)    Below, I'll be listing some already-found methods and their flaws:    Cycles's Toon BDSF:  Source: Blender  Result:    Node setup:    Cons:   Not a clear-cut two-color shading Reacts to diffusion of surrounding objects      Mask &amp; Ramp:  Source:   Result:    Node setup (Compositor):    Cons:   Very grainy Does not support colour textures Not easy to change shadow depth in a nodegroup Not a material, requires compositor Requires use of object indexes     Shadeless Fresnel:  Source:   Result:    Node setup:    Cons:   Uses camera's direction Ignores lighting Not easy to change shadow depth in a nodegroup     Shadeless Normal:  Source:   Result:    Node setup:    Cons:   Ignores lighting Direction of shadow is manually changed Not easy to change shadow depth in a nodegroup   However, this one is easily the best one. It's pretty great, and would work still images very well.    That's all I've got. Does anyone else have any methods that fall under most of the criteria I listed above? Also note that OSL is totally fine, if it works. Second derivative "formula derivation" I've been trying to understand how the second order derivative "formula" works:  $$\lim_{h\to0} \frac{f(x+h) - 2f(x) + f(x-h)}{h^2}$$  So, the rate of change of the rate of change for an arbitrary continuous function. It basically feels right, since it samples "the after $x+h$ and the before $x-h$" and the $h^2$ is there (due to the expected /h/h -> /h*h), but I'm having trouble finding the equation on my own.  It's is basically a derivative of a derivative, right? Newtonian notation declares as $f''$ and Leibniz's as $\frac{\partial^2{y}}{\partial{x}^2}$ which dissolves into:  $$(f')'$$ and $$\frac{\partial{}}{\partial{x}}\frac{\partial{f}}{\partial{x}}$$  So, first derivation shows the rate of change of a function's value relative to input. The second derivative shows the rate of change of the actual rate of change, suggesting information relating to how frequenly it changes.  The original one is rather straightforward:  $$\frac{\Delta y}{\Delta x} = \lim_{h\to0} \frac{f(x+h) - f(x)}{x + h - x} = \lim_{h\to0} \frac{f(x+h) - f(x)}{h}$$  And can easily be shown that $f'(x) = nx^{n-1} + \dots$ is correct for the more forthcoming of polynomial functions. So, my logic suggests that to get the derivative of a derivative, one only needs to send the derivative function as input to finding the new derivative. I'll drop the $\lim_{h\to0}$ for simplicity:  $$f'(x) = \frac{f(x+h) - f(x)}{h}$$  So, the derivative of the derivative should be:  $$f''(x) = \lim_{h\to0} \frac{f'(x+h) - f'(x)}{h}$$  $$f''(x) = \lim_{h\to0} \frac{  \frac{ f(x+2h) - f(x+h)}{h} - \frac{ f(x+h) - f(x)}{h}  }{h}$$  $$f''(x) = \lim_{h\to0} \frac{  \frac{ f(x+2h) - f(x+h) - f(x+h) + f(x)}{h}  }{h}$$  $$f''(x) = \lim_{h\to0} \frac{  f(x+2h) - f(x+h) - f(x+h) + f(x)  }{h^2}$$  $$f''(x) = \lim_{h\to0} \frac{  f(x+2h) - 2f(x+h) + f(x)  }{h^2}$$  What am I doing wrong? Perhaps it is the mess of it all, but I just can't see it. Please help. $x^{p-1} + ... + x^2 + x + 1$ is irreducible using Eisenstein's criterion? Eisenstein's criterion says that for a prime $p$ if the following conditions are satisfied for a primitive polynomial, $f(x)$, then that polynomial is irreducible in $\mathbb{Z}[x]$  $p \mid a_0, a_0, ..., a_{n-1}$   $p \not\mid a_n$  $p^2 \not\mid a_0^2$  Now I am almost certain I read somewhere that this lemma can be used to prove that the polynomial $x^{p-1} + x^{p-2} + ... + x^2 + x + 1$ is irreducible.  All coefficients are $1$ though and we have have an awkward amount of terms, $p - 1$ terms as opposed to $p$ terms - so how can Eisenstein's criterion be applied here? Suppose I have a question in my mind that I think is answerable but because I don't already know the answer, I'm not sure how to express in a way that makes it clear and definitively answerable. I suspect that someone more knowledgable than me would understand what I'm trying to get at though and could help me word the question properly.  Is there a good place to find this sort of help? I want to put a link to my SO profile on my homepage. I remember a codesnippet i saw somewhere that would basically display my Username and my reputation. Where was that? Eigenvectors of a circulant matrix How to test API integration in marketing cloud? I have created a API package under Setup -> Installed Packages.   I have created API Integration component for the package .I have the below basic details for the component.   Client Id Client Secret Authentication Base URI REST Base URI   I want to test the created API package using any tool( Postman or other API testing tools).  I was able to find API integration testing links for SFDC but not for the Marketing Cloud.   What is the redirect URL to be used in the API package? I have created an API integration package of Integration type ( Web APP)   In web app properties, what is the redirect URL to be provided? Why does flag marking as helpful/declined not always correlate with moderator action? I think there's an unintended motivation for users to post questions to higher-traffic Stack Exchange sites rather than their more specific or appropriate lower-traffic ones.  I think I have a solution.  What motivated this suggestion for me was that I am confident that if users were to post Android design questions to  beta and/or , that they would likely not be noticed by the appropriate audience compared to the vast army of Android experts here on Stack Overflow.    There are plenty of Android questions on SO that really should be on one or both of the aforementioned sites.  Presently there is little motivation for users to start questions like this on the appropriate site.  Is there something we could do to cross-pollinate related sites?  For instance, what if we let certain tags span certain sites?  Obviously the selection would be controlled by high-reputation users.   There could also be a feature to show (or explicitly hide) certain questions from other sites regardless of the tags.  This could be controlled by lower-reputation users than the global tag since it wouldn't have much statistical impact.  For instance, the android-layout tag could by default span sites, but OPs and other users could eliminate from the Graphic Design site questions that are of a purely technical nature.  Users interested in Android tagged questions would see related questions on design and UE sites--whether they are searching, posting, or viewing a question through the related items.  This would help eliminate cross-site duplicates, and encourage people to post on the correct site in the first place.   For related question lists and search results, there could even be a specific cross-site icon  next to an off-site question to show that it's from UX, for example. Embedding C++ Objects into QML with Context Properties According to Qt5 documentation:  all public slots of C++ classes that inherit from QObject, are accessible from QML Here what I did:  C++  class MyClass : public QObject {     Q_OBJECT  public slots:     void doStuffFromQmlSlot()     {         qDebug() &lt;&lt; Q_FUNC_INFO;     }  public:     MyClass()     {         qDebug() &lt;&lt; Q_FUNC_INFO;     } };   my main.cpp contains:  MyClass myClass; QQmlEngine engine; engine.rootContext()-&gt;setContextProperty( "myclass", &amp;myClass ); QQmlComponent component( &amp;engine, QUrl::fromLocalFile("qml/qtquick-01/main.qml") ); component.create();   QML  import QtQuick 2.0  Rectangle {     width: 360     height: 360     Text {         text: qsTr("Hello World")         anchors.centerIn: parent     }     MouseArea {         anchors.fill: parent         onClicked: {             myclass.doStuffFromQmlSlot();              Qt.quit();         }     } }   Actually the QtCreator seems to recognixe the exposed myclass object into the QML as it enables auto-completion of both class name (myclass) and the public slot doStuffFromQmlSlot(). When I run the application I unfortunately got the following error:  ReferenceError: myclass is not defined  Any clue on what I'm doing wrong?