Question

Given: y varies directly as x squared and inversely as z cubed. If y = 12 when x = 4 and z = 2, find x when y = 1.728 and z = 5.

Answer 1

Answer:

a

Step-by-step explanation:

given y varies directly as x² and inversely as z³ then the equation relating them is

y = $\frac{kx^2}{z^3}$ ← k is the constant of variation

to find k use the condition y = 12 when x = 4 and z = 2 , then

12 = $\frac{k(4)^2}{2^3}$ = $\frac{16k}{8}$ ( multiply both sides by 8 )

96 = 16k ( divide both sides by 16 )

6 = k

y = $\frac{6x^2}{z^3}$ ← equation of variation

when y = 1.728 and z = 5 , then

1.728 = $\frac{6x^2}{5^3}$ = $\frac{6x^2}{125}$ ( multiply both sides by 125 )

216 = 6x² ( divide both sides by 6 )

36 = x² ( take square root of both sides )

$\sqrt{36}$ = x , that is

x = 6