## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 10 - Exponents and Radicals - Review Exercises: Chapter 10 - Page 694: 50

#### Answer

$\sqrt{101}\approx10.050 \text{ units}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the Distance Formula to find the distance between the given points $\left( -7,-2 \right)$ and $\left( 3,-1 \right)$. $\bf{\text{Solution Details:}}$ With the given points, then $x_1= -7 ,$ $x_2= 3 ,$ $y_1= -2 ,$ and $y_2= -1 .$ Using the Distance Formula which is given by $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} ,$ then \begin{array}{l}\require{cancel} d=\sqrt{(-7-3)^2+(-2-(-1))^2} \\\\ d=\sqrt{(-7-3)^2+(-2+1)^2} \\\\ d=\sqrt{(-10)^2+(-1)^2} \\\\ d=\sqrt{100+1} \\\\ d=\sqrt{101} .\end{array} Hence, the distance is $\sqrt{101}\approx10.050 \text{ units} .$

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