z is jointly proportional to x^2 and y^2. If z = 410 when x = 6 and y = 5, find z when x = 2 and y = 4. (Round off your answer to the nearest hundredth.)

Question

can't get this one either!

Answer 1

SInce z is jointly proportional to x^2 and y^2; we get the equation: z=kx²y². Plug in the values to solve for k, we get k=410/(25*36). Reducing, k = 41/90. So our original equation with the k value in it is... z=(41*x²y²)/(90) Now plug in x=2 and y =4: z= 41(4)(16)/(90)=29.1555

When you round to the nearest hundredth becomes 29.16 since you have a 5 in the thousandths place you round up.

Answer 2

Hi Devan,

 

We can write z = kx^2y^2  where k is the constant of proportionality.

410 = k*6^2*5^2

410 = 900k

k = 41/90

 

when x=2 and y=4

z = (41/90)*2^2*4^2

= 29.16