## College Algebra (10th Edition)

$a.\displaystyle \quad D=\frac{429}{p}.$ $b.\quad 143$ bags sold.
If the demand $D$ is inversely related to the price $p$, then there is a nonzero constant $k$ such that $D=\displaystyle \frac{k}{p}$ a. If, when p=2.75, D=156, substituting, we find k: $156=\displaystyle \frac{k}{2.75}\quad/\times 2.75$ $k=429$ Thus, we write: $\displaystyle \quad D=\frac{429}{p}.$ b. If p=3, $D=\displaystyle \frac{429}{3}$ = $143$ bags sold.