The depreciation of the value for a car is modeled by the equation y = 100,000(.85)^x for x years since 2000.

a. In what year was the value of the $61,412.50?

b. In what year will the value of the car reach 1/4 of its original value?

Step-by-step explanation:

The depreciation of the value for a car is modeled by the equation y = 100,000(.85)^x for x years since 2000

a) for the value to be $61,412.50, then

61412.50 = 100000(.85)^x

61412.50/100000 = 0.85^x

0.614125 = 0.85^x

We would take log to base 10 of both sides. It becomes

Log 0.614125 = log 0.85^x =

x log0.85

- 0.2117 = x × - 0.07058

- 0.2117 = - 0.07058x

x = - 0.2117/- 0.07058

x = 3 years

b) 1/4 × 100000 = 25000

Therefore ,

25000 = 100000(.85)^x

25000/100000 = 0.85^x

0.25 = 0.85^x

We would take log to base 10 of both sides. It becomes

Log 0.25 = log 0.85^x =

x log0.85

- 0.602 = x × - 0.07058

- 0.602 = - 0.07058x

x = - 0.602/- 0.07058

x = 9 years

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2