Question

1, Suppose z varies directly with z and inversely with the square of y. If z = 12 when I = 3 and y what is z when 3 = 8 and y = 8? 2 = Question Help: D Video

Answer 1

z = 1/2

Step-by-step explanation:
`\begin{gathered} z\text{ }\alpha\text{ x} \\ z\text{ }\alpha\text{ 1/}y^2 \end{gathered}`

combining both:

`\begin{gathered} z\text{ }\alpha x/y^2 \\ z\text{ = }(kx)/(y^2) \\ \text{where k = constant of proportionality} \\ \text{and }\alpha\text{ means varies } \end{gathered}`

when z = 12, x = 3, y = 1

Inserting the values, we get k:

`\begin{gathered} 12\text{ = }(k(3))/(1^2) \\ 12\text{= }(3k)/(1) \\ 12\text{ = 3k} \\ k\text{ = 12/3} \\ k\text{ = 4} \end{gathered}`
`z\text{ = }(4x)/(y^2)\text{ (relationship connnecting the variables)}`

when x = 8, y = 8, z= ?

`\begin{gathered} \text{Inserting into the relationship:} \\ z\text{ = }(4(8))/(8^2) \\ z\text{ = }(32)/(64) \\ z\text{ = 1/2} \end{gathered}`