Growth and Decay II

The value of a car is $21,500. It loses 12% of its value every year. Write an equation.

• 21500^x(1+.12)

• 21500(1+.12)^x

• 21500 + .12^x

• 21500(1-.12)^x

The value of a car is $21,500. It loses 12% of its value every year is represented by:

21,500(1 − 0.12)t = 21,500(0.88)t

What is the initial value

• 21500

• 0.12

• 0.88

• t

The value of a car is $21,500. It loses 12% of its value every year is represented by:

21,500(1 − 0.12)t = 21,500(0.88)t

What is the rate of decay?

• 21500

• 0.88

• 0.12

• 12

The value of a car is $21,500. It loses 12% of its value every year is represented by:

21,500(1 − 0.12)t = 21,500(0.88)t

What is the decay factor?

• 21500

• 0.88

• 0.12

• 12

Give rate of decay y = 350(1 - 0.75)

• 0.75

• 1.75

• .25

What is the rate of growth? 10(1 + 0.4)t

• 0.4

• 1.4

• .004

y = 12(1.05)t

Growth or decay

• growth

• decay

y = 12(1.05)t

growth rate is

• .05

• 0.95

y = 12(0.05)t

rate is

• .05

• .095

g(t) = 240(0.72)t

initial value

• 72

• 240

• .72

• 28