Question

Find an equation that meets the conditions p varies inversely as d, and p = 2 when d = 3

Answer 1

pd = 6

Explanation:

p ∝ 1/d

change ∝ to = by adding constant k or c

p = k * 1/d

p = k/d

Cross multiply

pd = k

k = pd

k = 2 * 3

k = 6

From pd = k

pd = 6