The variable x varies directly as the cube of y, and y varies directly as the square root of z. If x equals 1 when z equals 4, what is the value of z when x equals 27?

Answer:[tex]z=36[/tex]Step-by-step explanation:According to the question,[tex]x[/tex]∝[tex]y^3[/tex]          .......(1)[tex]y[/tex]∝[tex]\sqrt{z}[/tex]    .......(2)From equation 1,2 let constant of proportionality be [tex]k1,k2[/tex] respectively.⇒[tex]x=k1(y^3)[/tex]            .......(3)⇒[tex]y=k2(\sqrt{z} )[/tex]    .......(4)From the above equations putting 4 into 3,[tex]x=k1((k2\sqrt{z})^3) =k1.k2^3.(\sqrt{z})^3[/tex]Let the new constant to the above equation be [tex]k3[/tex],[tex]x=k3(\sqrt{z})^3[/tex]Given,if x=1, when z=4[tex]1=k3(\sqrt{4} )^3=k3(8)[/tex]⇒[tex]k3=\frac{1}{8}[/tex]Now if x=27, then z=?[tex]27=\frac{1}{8} (\sqrt{z} )^3[/tex]⇒[tex](\sqrt{z} )^3=27(8)[/tex]⇒[tex]\sqrt{z}=3(2)=6[/tex][tex]z=36[/tex]