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


$8.25$ 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
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