An employment research company estimates that the value of a recent MBA graduate to an accounting company is

V = 4e2 + 4g3

where V is the value of the graduate, e is the number of years of prior business experience, and g is the graduate school grade point average. A company that currently employs graduates with a 2.9 average wishes to maintain a constant employee value of V = 300, but finds that the grade point average of its new employees is dropping at a rate of 0.4 per year. How fast must the experience of its new employees be growing in order to compensate for the decline in grade point average? (Give the answer to two decimal places.)

(de/dt) = 0.70929

Step-by-step explanation:

Given

V = 4*e² + 4*g³

where

If V = 300 (constant)

g = 2.9

dg/dt = - 0.4

de/dt = ?

we have to get e as follows

V = 4*e² + 4*g³ ⇒  (V/4) = e² + g³  ⇒ e = √((V/4) - g³)

⇒ e = √((300/4) - (2.9)³)

⇒ e = 7.1141

Then we apply

dV/dt = d(4*e² + 4*g³)/dt

d(300)/dt = 4*((2*e*de/dt) + 3*g²*(dg/dt))

0 = (2*e*de/dt) + 3*g²*(dg/dt)

⇒ 0 = 2*(7.1141)*(de/dt) + 3*(2.9)²*(- 0.4)

⇒   (de/dt) = 0.70929

none of the above.

for one of the equations to be used in the process, there has to be a term in the equation since squared is .

since there is only a term in the first equation, we cannot derive a from completing the square.

solving with completing the square

step-by-step explanation:

ok one would be money, and the other would be refreshments or something similar. if each bag of popcorn is 5.00 and each drink is half of that that means drinks are 2.50 the proper way to do that would to look and see how much of each you can buy without exceeding a limit of 30.00