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

the answer is (2x+8)²= 16

correct answer is infinitely many solutions

step-by-step explanation:

2(2x + 3) = -4 + 2(2x + 5)

solving the above equation:

4x + 6 = -4 +4x +10

combining like terms

4x -4x = -4+10-6

0=0

since both sides are 0=0, we end up into true statement it means it has infinite many solutions.

so correct answer is infinitely many solutions