I am trying to open Identify Results docked window using qgis plugin. How can i open the identify results window using feature id?  I am iterating through active layer, get a specific feature and then open the identify results window automatically within that plugin.
If I select a feature and run self.iface.actionIdentify().trigger() it will open the identify window but won't open the selected feature's data, how can I make it open the data of the selected feature?
What is a good set of naming conventions to use when developing on the Force.com platform? I've looked in the Apex developer's guide and a saw the   which has basically only has this:   We recommend following Java standards for naming, that is, classes start with a capital letter, methods start with a lowercase verb, and variable names should be meaningful.   I'm looking for something more in depth, such as end all Controllers with Controller and their tests with ControllerTest, etc.  What is a good set of naming conventions to use when developing on the Force.com platofrm?  It would be preferable if the answer has something that handles custom objects, classes, visualforce pages, and components and not just Apex classes. Delta-like symbol in LaTeX I'd like to write a majuscule delta-like symbol in LaTeX but I can't find it's syntax anywhere. You can see the symbol on equation (12) of the following paper:  "". PHP mail function doesn't complete sending of e-mail Join tables from two different server Layers with different directional drop shadows in Photoshop I am wondering how to have 2 different directional drop shadows in the one file with Photoshop. e.g. 2 different type layers — one has a drop shadow at 90 degrees, the other at -45 degrees.  Currently it seems Photoshop only lets me have 1 universal direction for the whole file, which I find really pointless.   I know how to apply different drop shadows and merge down layers to achieve the same thing, but its tedious and not exactly streamlined for making future changes to that layer.  How is this achieved? I read this book a few years ago which was part of a series, but I only found the first book and now I've forgotten the name.  The first book featured 3 children/teens from different places that end up on a quest to search for magic gems/stones together and the story takes place in a medieval kind fantasy world. One is a girl, while the other 2 are boys. One of the boys actually finds a stone/gem while he is on a school field trip (I think the gem/stone is red in colour), and when he picks it up, he gets transported to the fantasy land but ends up in some sort of military training ground/camp somewhere in a desert type area.  He isn't familiar with the fantasy world. The other 2 characters each start out on their own on a quest to collect some magical gems/stones which once collected, can be used to fight an evil ruler (I guess).  But halfway through the book, the girl and boy meet up and continue the quest together while the third boy(transported one) doesn't know that the stone he picked up is actually one of the magical stones others are looking for.  A few other details I remember:   The stone the boy finds is red in colour I think  The cover of the book had a border depicting the stones  It's part of a series or trilogy    Some books with similar storylines, but not the one I'm looking for:     Deltora Quest The Seven Towers Sword of Heaven series Westeria series Diadem series Let $A\neq I$ be a $5\times5$ matrix with real entries such that the sum of the entries in each row of $A$ is $1$. Then the sum of all the entries in $A^3$ is  1)$\space 3$ $\qquad $2)$\space 15$ $\qquad$ 3)$\space 5$ $\qquad$ 4)$\space 125$  Solution:  The answer is $5$ if $A = I$, but for $A \neq I$,  I found that vectors $\begin{pmatrix} 1\\ 1\\ 1\\ 1\\ 1\\ \end{pmatrix}$ and  $\begin{pmatrix} -1\\ -1\\ -1\\ -1\\ -1\\ \end{pmatrix}$  are eigen vectors for the matrix $A$ corresponding to eigen value $1$. From here, I dont know how to proceed, please help me solve the question..and if there are mistakes in my argument, please tell me.. I know this may be a tough or confusing question, but I have numerous existing materials I created and reuse for a project, which were all prepared for Blender Internal.  Now, I want to start the transition to Cycles, but I've hit a roadblock. How can I transition my materials without completely starting over? Or should I just forget it and start with fresh materials? Show that $a_n=\sin(n)$ does not converge     My idea:  Take two subsequences: $a_{n_k}=\sin(\frac {\pi k} 2)$ , $a_{n_l}=\sin(\frac {2\pi l} 3)$  So: $\forall n$ : $\lim_{n\to\infty} a_{n_k}=1$, $\lim_{n\to\infty} a_{n_l}=-1$  So two inifinite subsequence converge to different limits thus the sequence $a_n$ doesn't converge.   Is that correct ?  Edit:   The contrapositive of the defintion of a limit of a sequence:   $\exists\epsilon&gt;0 : \forall n\in N : \exists n&gt;N \Rightarrow |x_n-L|&gt;\epsilon$  Take $\epsilon =1$ and we know that $\sin(n)$ is bounded so lets take it's supermum: 1  $|1-1|&gt;\epsilon=1\Rightarrow 0&gt;1 \Rightarrow$ Contradiction. How does friction stop a car? I'm unable to understand how friction causes motion in cars and also is responsible for stopping them. When it starts the impending motion of tyres is in the backward direction that's why friction is in the forward direction and the car starts moving. But when the car engine turns off, the car stops after some time, but the condition of tyres are the same as in the previous case and friction must be in the forward direction. I am running the latest version of X3 Terran Conflict (3.1) and after about 30 min - 1 hour it's memory usage goes well above 2 GB.  I only have 4 GB of ram, running Windows 7 64 bit.  As a result of X3 using more ram than is available, my hard drive starts paging and makes the game and other applications become unusable.  I have to restart the game.  Is anyone aware of this issue or how to fix it and is there any way in Windows or through some tool, to limit how much memory is given to the game?  Is this a bug in the game? Can this even be fixed?  I'm using some mods but I doubt any of them are causing this as they are simple mods like adding a tractor beam etc. I just did a clean install of 20.04 and I noticed that no key shortcuts for files manipulation work when my focus in on Desktop (not in Nautilus).   When I have copied a file to clipboard, I cannot Ctrl+V it to Desktop, when I click on a file on Desktop, I cannot Ctrl+C it, or I cannot Delete it with keys. Only ways to do it is either navigate to the Desktop folder in Nautilus, or right-click the file on the Desktop and copy/move to trash/paste...  Sorry if I'm missing something obvious, all the key shortcuts worked normally for me in 18.04. $|A \cup B| = \mathfrak{c}$, then $A$ or $B$ has cardinality $\mathfrak{c}$. Is there an uncountable family of infinite parts of $\mathbb N$ with two to two intersection finite? Why particle hole symmetry and chiral symmetry are called symmetries? $PHP^{-1}=-H$ (particle-hole symmetry) and   $\Gamma H \Gamma^{-1}=-H$ (chiral symmetry)   I understand why we get the negative signs but im just a bit confused as to why such equalities mean $H$ is particle hole and chiral symmetric.   When we say $H$ is symmetric under some operation, like time reversal,  don't we usually mean that $H$ is invariant under the said transformation?  $THT^{-1}=H$ if $H$  is time reversal symmetric.  On the contrary, we say $H$ is symmetric under particle hole and chiral operations when $H$ picks up a negative sign. This seems to contradict our usual convention of what we mean by $H$ being symmetric(invariant). It is easy to find the minimum of $\|Ax-b\|_2$, when $A$ has full column rank. But how is the case when we add the constraint $\|x\|_2=1$? Or, to be explicit,  $$\min_{\|x\|_2=1}\|Ax-b\|_2=?$$  My idea is to construct the corresponding Lagrange function  $$L(x,\lambda) = (Ax-b)^T(Ax-b) + \lambda(x^T x - 1)$$  Differentiate the function with respect to $x,\lambda$, we have  $$A^T A x - A^T b + \lambda x = 0$$  $$x^T x=1$$  However, I find it hard to move forward to get the answer. Could anyone help me? Thanks a lot! Execute sudo without Password? Inspired by this ....  I am the sole person using my system with 12.04. Every time I issue a sudo command; the system asks for the user password (which is good in its own way).  However I was thinking; without activating the root account; how can I execute the sudo commands which will not ask for user password to authenticate.  NOTE: I want to execute sudo command without authenticating via password; only when they are executed via terminal. I don't want to remove this extra layer of security from other functions such a while using 'Ubuntu software center' or executing a bash script by drag-drop something.sh file to the terminal. I have $a,b,c\in\mathbb{Q}$ not all zero. ($a^2+b^2+c^2\ne 0$), I want to show that the following determinant is then non-zero. I failed to arrive at an appropriate form of the polynomial. Help please. $$\left|\begin{bmatrix} a &amp; 2c &amp; 2b\\b &amp; a &amp; 2c\\ c &amp; b &amp; a\end{bmatrix}\right| = a^3+2 b^3-6 a b c+4 c^3$$    Second question, what is the easiest way to argue that $\{1,\sqrt[3]{2},(\sqrt[3]{2})^2\}$ is linearly independent in $\mathbb{Q}$?    Motivation: Prove that $\mathbb{Q}[\sqrt[3]{2}] = \{a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2\;|\;a,b,c\in\mathbb{Q}\}$ forms a field.   Proof: Since $\mathbb{Q}[\sqrt[3]{2}] \subset \mathbb{R}$, we prove $\mathbb{Q}[\sqrt[3]{2}]$ is a subfield of $(\mathbb{R},+,\cdot)$    $\forall (a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2)\in \mathbb{Q}[\sqrt[3]{2}]\backslash\{0\}.$ We want to find $(a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2)$ such that    $(a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2)(d+e\sqrt[3]{2}+f(\sqrt[3]{2})^2) =$ $ (ad+2ec+2bf)+(ae+bd+2cf)\sqrt[3]{2}+(af+cd+be)(\sqrt[3]{2})^2 = 1$    Since $\{1,\sqrt[3]{2},(\sqrt[3]{2})^2\}$ is linearly independent (?) over $\mathbb{Q}$, we show there is unique solution to: $$\begin{bmatrix} a &amp; 2c &amp; 2b\\b &amp; a &amp; 2c\\ c &amp; b &amp; a\end{bmatrix} \cdot \left[\begin{array}{l l} d\\e\\f \end{array}\right] = \left[\begin{array}{l l} 1\\0\\0 \end{array}\right] $$ Which is equivalent in showing the determinant is non-zero $$\left|\begin{bmatrix} a &amp; 2c &amp; 2b\\b &amp; a &amp; 2c\\ c &amp; b &amp; a\end{bmatrix}\right| = a^3+2 b^3-6 a b c+4 c^3=(?)$$    By subfield test, 1)2)3)4) is enough to say that $(\mathbb{Q}[\sqrt[3]{2}],+,\cdot)$ is a subfield of $(\mathbb{R},+,\cdot)$ therefore a field.  EDIT: If you have shorter way that prove the proposition without touching my 2 questions, that is even better. I have a laptop which is connected to the internet with ADSL modem, and I would like to share the internet for any laptop connecting to wireless router from my laptop using hotspot.  This is what I want:  ┌────────────────────┐ │ ADSL modem         │ └───┬────────────────┘     │ ┌───┴───────────────────────┐ │ laptop's wired adaptor    │ │                           │ │ Hotspot software          │ │                           │ │ laptop's wireless adaptor │ └───┬───────────────────────┘     │ ┌───┴──────────────────────┐ │ wireless router(linksys) │ └───┬──────────────────────┘     │ ┌───┴─────────────────────────────┐ │ Laptop computers and cellphones │ └─────────────────────────────────┘   How can I do that? I'm not sure which of the following sentences is correct. Is the comma mandatory, optional or unnecessary? I've googled a bit about this topic, but I got confused.     Our mapping contains 2000 words, which map to more than one lemma.   or     Our mapping contains 2000 words which map to more than one lemma.   Context: In total, there are more than 2000 words.  I've read  . If I understand the answers correctly, my "which map to more than one lemma" is a defining clause and no comma is required. But is it wrong to add one?