Question

cube of y varise inversely as square of x and y=3, when x=16 if x=2 then what is the value of y​

Answer 1

Answer:

y^3=k(1/x)^2=k/(x^2) this is the equation where k is a constant. In this equation put y=3 and x=16

3^3=k/(16)^2

27=k/256

k=27×256.

Then the equation becomes

y^3=(27/256)(1/x^2)

put x=2 in this equation

y^3=(27/512)=3^3/8^3

y=3/8