## Intermediate Algebra for College Students (7th Edition)

$d = \sqrt\frac{kP_{1}P_{2}}{C}$
$C = \frac{kP_{1}P_{2}}{d^2}$. Solve for $d$. $Cd^2 = kP_{1}P_{2}$ $d^2 = \frac{kP_{1}P_{2}}{C}$ $d = \sqrt\frac{kP_{1}P_{2}}{C}$