College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 3 - Section 3.3 - Properties of Function - 3.3 Assess Your Understanding - Page 236: 106

Answer

$8.25$ days

Work Step by Step

We know that: $\displaystyle Shelf\ life=\frac{k}{Temperature}$ We plug in the given values to find $k$: $\displaystyle 33=\frac{k}{10}$ $k=330$ Thus: $\displaystyle Shelf\ life=\frac{330}{Temperature}$ So we find the shelf life at a temperature of $40$ degrees: $\displaystyle Shelf\ life=\frac{330}{40}=\frac{33}{4}=8.25$ days
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.