## College Algebra (10th Edition)

$8.25$ days
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