Question April's car was worth $9,000 at the beginning of 2000 and the value of the car decreased exponentially, decreasing by 21% each year. a. Write a function f that determines the value of April's car (in dollars) in terms of the number of years t since the beginning of 2000. f(t) 9000(1-21/100) Preview b. How many years after the beginning of 2000 was April's car worth $6,000? * years Preview c. How many years after the beginning of 2000 was April's car worth $2,400? * years Preview Submit

DBPCFA The Asker · Precalculus

Transcribed Image Text: April's car was worth $9,000 at the beginning of 2000 and the value of the car decreased exponentially, decreasing by 21% each year. a. Write a function f that determines the value of April's car (in dollars) in terms of the number of years t since the beginning of 2000. f(t) 9000(1-21/100) Preview b. How many years after the beginning of 2000 was April's car worth $6,000? * years Preview c. How many years after the beginning of 2000 was April's car worth $2,400? * years Preview Submit
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Transcribed Image Text: April's car was worth $9,000 at the beginning of 2000 and the value of the car decreased exponentially, decreasing by 21% each year. a. Write a function f that determines the value of April's car (in dollars) in terms of the number of years t since the beginning of 2000. f(t) 9000(1-21/100) Preview b. How many years after the beginning of 2000 was April's car worth $6,000? * years Preview c. How many years after the beginning of 2000 was April's car worth $2,400? * years Preview Submit
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The solution is given below We know that the formula for exponential decay or growth isf(t)=ab^(t)(1)Whe a is the initial value, b=1-r and r is the rate of decay.Here a=9000,quad r=21%=(21)/(100)=0.21And b=1-0.21=0.79Thus, the formula is f(t)=9000(0.79)^(t). (2)(b) Here f(t)=6000 is given, so from (2) 6000=9000(0.79)^(t)=>(0.79)^(t)=(6000)/(9000)=(2)/(3)Taking log on both sides, we get{:[t log(0.79)=log(2//3) ... See the full answer