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.

x=35.44

Step-by-step 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