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.

Select one:
a. x=6
b. x=18
c. x=27
d. x=36

Respuesta :

Answer:

a

Step-by-step explanation:

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

y = [tex]\frac{kx^2}{z^3}[/tex] ← k is the constant of variation

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

12 = [tex]\frac{k(4)^2}{2^3}[/tex] = [tex]\frac{16k}{8}[/tex] ( multiply both sides by 8 )

96 = 16k ( divide both sides by 16 )

6 = k

y = [tex]\frac{6x^2}{z^3}[/tex] ← equation of variation

when y = 1.728 and z = 5 , then

1.728 = [tex]\frac{6x^2}{5^3}[/tex] = [tex]\frac{6x^2}{125}[/tex] ( multiply both sides by 125 )

216 = 6x² ( divide both sides by 6 )

36 = x² ( take square root of both sides )

[tex]\sqrt{36}[/tex] = x , that is

x = 6