Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.4 Rates of Change - Exercises - Page 129: 12

Answer

$5.897$ $°C/hr$

Work Step by Step

The time interval from 8:36 AM to 9:34 AM has length 58 minutes The change in temperature over this time interval is $ΔT$ = $-42-(-47.7)$ = $5.7$ °C $\frac{ΔT}{Δt}$ = $\frac{5.7}{58}$ = $0.0983$ °C/min $\frac{ΔT}{Δt}$ = $0.0983\times60$ = $5.897$ °C/hr
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.

GradeSaver will pay $15 for your literature essays
GradeSaver will pay $25 for your college application essays
GradeSaver will pay $50 for your graduate school essays – Law, Business, or Medical
GradeSaver will pay $10 for your Community Note contributions
GradeSaver will pay $500 for your Textbook Answer contributions