CoreyHammond74901
09.06.2020 •
Mathematics
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.
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Ответ:
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
Ответ:
13 players
Step-by-step explanation:
65 x 0.2 = 13