z varies directly as x^2 and inversely as y^2 If z = 142 when x = 4 and y = 5, find z if x = 10 and y = 6. (Round off your answer to the nearest hundredth.)

Question

z varies directly as x^2 and inversely as y^2 If z = 142 when x = 4 and y = 5, find z if x = 10 and y = 6

Answer 1

 

 

 Z = kX^2/y^2 where k is a fixed number constant.

 

 142 = k(4)^2/(5)^2 = k(16/25)

 

 Solving for k,  k = 142*25/16 = 221.875

 

 so z = 221.875X^2/y^2

 

 Then z = 221.875*(10)^2/(6^2) = 221.875*100/36 = 22187.5/36 = 616.3194444..... = 616.32 rounded to 2 digits

 

 

Answer 2

z = kx²/y²  to solve for k substitute the initial values of x , y and z and solve for k

 

142 = 16k/25

 

k = 221.875

 

z = (221.875)(100/36) = 616.32