The management of a chain of frozen yogurt stores believes that t days after the end of an advertising campaign, the rate at which the volume V (in dollars) of sales is changing is approximated by V ' ( t ) = − 26400 e − 0.49 t . On the day the advertising campaign ends ( t = 0 ), the sales volume is $ 170 , 000 . Find both V ' ( 6 ) and its integral V ( 6 ) . Round your answers to the nearest cent.

Step-by-step explanation:

Given the rate at which the volume V (in dollars) of sales is changing is approximated by the equation

V ' ( t ) = − 26400 e^− 0.49 t .

t = time (in days)

.v'(6) can be derived by simply substituting t = 6 into the modelled equation as shown:

V'(6) = − 26400 e− 0.49 (6)

V'(6) = -26400e-2.94

V'(6) = -26400×-0.2217

V'(6) = $5852.88

V'(6) = $5,853 to nearest dollars

V'(6) = 585300cents to nearest cent

To get v(6), we need to get v(t) first by integrating the given function as shown:

V(t) = ∫−26400 e− 0.49 t dt

V(t) = -26,400e-0.49t/-0.49

V(t) = 53,877.55e-0.49t + C.

When t = 0, V(t) = $170,000

170,000 = 53,877.55e-0 +C

170000 = 53,877.55(2.7183)+C

170,000 = 146,454.37+C

C = 170,000-146,454.37

C = 23545.64

V(6) = 53,877.55e-0.49(6)+ 23545.64

V(6) = -11,945.63+23545.64

V(6) = $11,600 (to the nearest dollars)

Since $1 = 100cents

$11,600 = 1,160,000cents

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q2 do

step-by-step explanation:

the expression (1.26)2t reveals the approximate rate of increases in the number of plants semiannually.

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