Question

x varies jointly as y^3 and square root of z . If x = 7 when y = 2 and z = 4, find x correct to 2 decimal places, when y = 3 and z = 9.

Answer 1

x=35.44

Explanation:

x=ky^3 and√z

x=k(y^3)(√z)

Find k when

x=7

y=2

z=4

7=k(2^3)(√4)

7=k(8)(2)

7=k(16)

7=16k

k=7/16

Find x when y=3, z=9

x=k(y^3)(√z)

x=(7/16)(3^3)(√9)

x=(7/16)(27)(3)

=7/16(81)

=567/16

x=35.4375

Approximate to 2 decimal places

x=35.44