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(biblatex-apa) How to make a long quote (>40 words)

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I want to make a long quote for my thesis on latex, using biblatex with apa, but I can't find the way to do that. My preamble is:

documentclass[12pt]{report}

usepackage[utf8]{inputenc}
usepackage[spanish]{babel} % pone el idioma en español

usepackage{geometry} % margenes del documento
geometry{letterpaper, top = {2,5cm}, bottom = {2,5cm}, inner = {2,5cm},
outer = {2,5cm}, head = {1,1cm}, foot = {1,1cm}, bindingoffset = {1,0cm}}
% Si se va a usar dos paginas por hoja, agregar la opción twoside

usepackage{helvet} % Arial
renewcommand{familydefault}{sfdefault} % Arial

usepackage{titlesec}
titleformat{chapter}[display] % cambiar modo de ver capitulos y secciones
{largebfseriescenter}{MakeUppercasechaptertitlename thechapter}{1em}{MakeUppercase}{} % Cambios en Capitulo
titleformat{section}{largebfseries}{thesection}{1em}{MakeUppercase}

titleformat{subsection}{normalsizebfseries}{thesubsection}{1em}{MakeUppercase} % Cambios en Sección
titlespacing*{chapter}{0pt}{-2.5em}{3em} % Cambia el espaciamiento del capitulo de 50pt a lo pedido
titlespacing*{section}{0pt}{3em}{2em} % Cambia el espaciamiento de la sección a 3 espacios en blanco antes y 2 despues.

renewcommand*{theenumi}{thechapter.arabic{enumi}} % numero de capitulo en listas
renewcommand*{theenumii}{theenumi.arabic{enumi}} % añade lo anterior a sublistas

usepackage{titletoc}
 titlecontents{chapter}[0em]
    {vspace{0em}}
    {normalfontnormalsizecontentslabel[thecontentslabel]{2em}uppercase}
    {hspace*{-2em}uppercase}
    {titlerule*[.75em]{.}contentspage}
    
titlecontents{section}[2em]
    {vspace{0em}}
    {normalfontnormalsizecontentslabel[thecontentslabel]{2em}uppercase}
    {hspace*{-2em}uppercase}
    {titlerule*[.75em]{.}contentspage}
    
titlecontents{subsection}[5em]
    {vspace{0em}}
    {normalfontnormalsizecontentslabel[thecontentslabel]{2.9em}uppercase}
    {hspace*{-2em}uppercase}
    {titlerule*[.75em]{.}contentspage}

usepackage{setspace}
renewcommand{baselinestretch}{1.5} % interlineado de 1.5

usepackage[skip=6pt, indent=1cm, tocskip]{parskip} % configurar espacio entre parrafos y sangría izquierda. tocskip genera el salto entre parrafos en el indice

% paquetes necesarios para citar
usepackage{csquotes}
usepackage[style=apa, backend=biber, sortcites, url=true]{biblatex}
addbibresource{bibliografia.bib}

usepackage{etoolbox}
patchcmd{chapter}{thispagestyle{plain}}{thispagestyle{fancy}}{}{} % necesario para poner encabezado en paginas de capítulos

% paquete para encabezado y pie de pagina 
usepackage{fancyhdr}
pagestyle{fancy}
renewcommand{headrulewidth}{0pt}
renewcommand{footrulewidth}{0pt}

% - - - - - COMANDOS - - - - -


% ------- PORTADA ------- 

title{PORTADA}
author{}
date{}

begin{document}

maketitle

tableofcontents

fancyhead{}
fancyfoot{}

fancyfoot[R]{thepage}
fancyheadoffset[lh]{2cm}

% ------- RESUMEN EJECUTIVO ------- 

include{1-resumen-ejecutivo.tex}

% ------- CAP 1 INTRODUCCIÓN ------- 

include{2-introduccion.tex}

% ------- CAP 2 XXXX ------- 

include{3-capitulo-1.tex}


end{document}

According to APA style, long quotes should:

— Add +1.27 cm indent for the entire text.

— First word without indent.

— Double-spacing.

— No additional space above or below.

— If quote have 2 or more paragraphs, the first line (for each paragraph) should have an exta 1.27 cm indent.

I want to make something like this (sorry for Spanish example):
40 word longer APA quote


   
Quote

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$begingroup$

enter image description here

Hi, I am deriving the log likelihood of the observed data in part a. However, I am not sure if I am deriving it the right way.

Below is my solution:

Yi ~ N(mu, sigma^2) = f(Yi = xi) = ∅ (xi; mu, sigma^2) (for ri =1)

Yi ~ N(mu, sigma^2) = f(Yi = xi) = Φ (xi; mu, sigma^2) (for ri =0)

Then I multiply these two terms

f(xi,ri) = ∅ (xi; mu, sigma^2)^ri*Φ (xi; mu, sigma^2)^1-ri

and then I form the likelihood equation:

L(mu,sigma^2|x,r) = n i=1 prod ∅ (xi; mu, sigma^2)^ri*Φ (xi; mu, sigma^2)^1-ri

Is this the correct way to derive the equation in part a?
I have also attached the picture of the question.

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I’m trying do looping zonal statistics to many polygons in a folder, and rasters in aonther folder. Each polygon from a specific folder need to get the mean values from a many rasters from another folders, like slope, standard deviation, etc. But when I run this code the error is:

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File “D:/ZONAL1.py”, line 10, in
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What this means?
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$begingroup$

There are two quite different, but useful, theorems related to 3D rotation that are referred to as “Euler’s Rotation Theorem“.

The first definition is given as a quotation in the book by Kuipers

  • Any two independent orthonormal coordinate frames can be related by a sequence of rotations (not more than three) about coordinate axes, where no two successive rotations may be about the same axis.

There is a similar, but more pithy definition, from Wolfram MathWorld

  • An arbitrary rotation may be described by only three parameters.

This is related to the three-angle representation of an arbitrary rotation by Euler, roll-pitch-yaw, Tati-Bryan angles etc.

The second definition is from Wikipedia

  • When a sphere is moved around its centre it is always possible to find a diameter whose direction in the displaced position is the same as in the initial position. (Also given in latin)

This related to the angle-axis representation of an arbitrary rotation. This version of the Rotation Theorem is found in many other places as well, including here on Math Overflow.

Both definitions are related to rotation, and the second at least has a reference to Euler 1776. The first is related to Euler axes and Euler angles, seems like the sort of thing Euler might have figured out, but Kuipers does not provide a reference.

Euler was notorious for not publishing much. Somebody, post Euler, has declared these theorems as Euler’s Rotation Theorem. Which one is it? Is there a more nuanced way to refer to them?

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So i found a riddle on reddit which is a consequence of the axiom of choice.

There is a house with 100 rooms, and each room contains countably many boxes indexed with the natural numbers. Each box contains a random real number, which is the same over all the rooms (that is, box n contains the same real number in every room).

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In order to win, 99 of them need to guess correctly.

The mathematicians can discuss a strategy beforehand, but after they go into their respective rooms, no more communication is allowed. What is a 100% winning strategy for this seemingly impossible task?

The solution is here: https://www.reddit.com/r/math/comments/rpbhos/predicting_random_real_numbers_with_the_axiom_of/?utm_source=share&utm_medium=ios_app&utm_name=iossmf

My question is why can the mathematicians agree on a specific representative for each equivalence class? I thought that the axiom of choice guaranteed a choice function existed, but didnt specify it. But for the solution to the riddle, they would need to know what the choice function is. Can someone explain?

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