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

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - 10.7 The Distance Formula, the Midpoint Formula, and Other Applications - 10.7 Exercise Set: 64

Answer

$d\approx200.003 \text{ units} $

Work Step by Step

Using $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ or the Distance Formula, the distance, $d$, between the given points, $\left( 0.5,100 \right) \text{ and } \left( 1.5,-100 \right),$ is \begin{array}{l}\require{cancel} d=\sqrt{ \left(0.5-1.5 \right)^2+\left( 100-(-100) \right)^2} \\\\ d=\sqrt{ \left(-1 \right)^2+\left( 200 \right)^2} \\\\ d=\sqrt{ 1+40,000} \\\\ d=\sqrt{ 40,001} \\\\ d\approx200.003 \text{ units} .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.