121. nina has just taken out a car loan for $12,000. she will pay an annual interest rate of 3% through a series of monthly payments for 60 months, which she pays at the end of each month. the amount of money she has left to pay on the loan at the end of the n th month can be modeled by the function f(n) = 86248 − 74248(1.0025)n for 0 ≤ n ≤ 60.
at the same time as her first payment (at the end of the first month), nina placed $100 into a separate investment account that earns 6% per year compounded monthly. she placed $100 into the account at the end of each month thereafter. the amount of money in her savings account at the end of the n thmonth can be modeled by the function g(n) = 2(1.005) n − 2 for n≥ 0.
a. use the functions f and g to write an equation whose solution could be used to determine when nina will have saved enough money to pay off the remaining balance on her car loan.

(1.0025ⁿ)² + 3.7124×(1.0025ⁿ) - 5.3124 = 0

Step-by-step explanation:

Nina will have saved the enough money to pay off the remaining balance on the loan when the functions are the same. So

f(x) = g(x)

86248 - 74248(1.0025)ⁿ = 20000(1.005)ⁿ - 20000

we sum 20000 on each side of the equation

106248 - 74248(1.0025)ⁿ = 20000(1.005)ⁿ

then we divide by 2000 each side of the equation

5.3124 - 3.7124(1.0025)ⁿ = 1.005ⁿ

the square root of 1.005 is aproximately 1.0025, so

1.005 ≈ 1.0025²

5.3124 - 3.7124(1.0025)ⁿ = (1.0025²)ⁿ

Since (1.0025²)ⁿ = 1.0025²ⁿ = (1.0025ⁿ)² we can write:

(1.0025ⁿ)² + 3.7124(1.0025ⁿ) - 5.3124 = 0

x equals 1

step-by-step explanation:

f(4)=f(×)

x=1

whole numbers.

step-by-step explanation:

we are given the average monthly rainfall (in millimeters) model r(m) for the years 1900-2009 as a function of the month number m.

as we can see m is taken for the number of months for years 1900-2009.

we know that number of months can't be a negative number or decimal number or a fraction.

so, the number of months could be only positive whole numbers.

therefore, whole numbers is more appropriate for the domain of r.