z is directly proportional to the square of x and inversely proportional to the square root of y. If z=12 when x=2 and y=4, find z when x=6 and y=32


any help is greatly appreciated!

Guest Jun 3, 2017

We have  \(z=k\frac{x^2}{\sqrt{y}}\)   where k is a constant


Find k as  \(k=12\frac{\sqrt{4}}{2^2}\rightarrow6\)


So when x = 6 and y = 32 we have \(z=6\frac{6^2}{\sqrt{32}}\rightarrow27\sqrt2\)


Alan  Jun 3, 2017

13 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.