Equation to Model Situation

A particular type of cell triples in number every hour. Which function can be used to find the number of cells present at the end of h hours if there are initially 6 of these cells?

• f(h) = 3(6)^h

• f(h) = 6x^h

• f(h) = 6(3)^h

• f(h) = 3(6h)

The amount of cola in a soda machine decreases by a rate of 5% each hour. The amount of cola in the machine was originally 75 gallons. Which function models the amount of cola in gallons after h hours?

• f(h) = 75(-5)^h

• f(h) = -5(75)^h

• f(h) = .95(75)^h

• f(h) = 75(.95)^h

Write an exponential function to model the situation. Then solve.

The cost of tuition at a college is $12,000 and is increasing at a rate of 6% per year. Which equations represents this situation?

• C(t) = 1.06(12000)^t

• C(t) = 12000(.06)^t

• C(t) = 12000(.06)^t

• C(t) = 12000(1.06)^t

Write an exponential function to model the situation. You invest $1500. Your investment increases at a rate of 3.5%.

• B(t) = 1500(0.965)^t

• B(t) = 1500(1.035)^t

• B(t) = 1500(.035)^t

• B(t) = 1500(1.965)^t

• A

• B

• C

• D

A linear equation is used to model the bank balance, B, as a function of time, T for two people after monthly withdrawals.

----- ----- ----- -----

Amber = -800t + 10,000

Timothy = -500t + 15,000

----- ----- ----- -----

Complete the sentences:

​ (a)   spends money faster

​ (b)   has more money to start

• Amber

• Tim

Write an exponential function to model the situation. The value of a car is $18000 and is depreciating at a rate of 12% per year.

• V(t) = .88(18000)^t

• V(t) = .12(18000)^t

• V(t) = 18000(.88)^t

• V(t) = 18000(.12)^t

The value of a car is $15,000 and depreciates at a rate of 8% per year. What is the exponential equation that models this situation?

• y = 8(15,000)^x

• y = 15,000(1.08)^x

• y = 15,000(0.92)^x

• y = 15,000(0.08)^x

The equation f(x) = 300(1.06)^x is being used to calculate the amount of money in a savings account.  What does the 1.06 represent?

• 6% growth

• 6% decay

• 1.06% growth

• .06% decay

Suppose a culture of bacteria begins with 5000 cells and dies by 30% each year. Write an equation that represents this situation.

• y=5000(0.7)^x

• y=30(5000)^x

• y=5000(1.3)^x

• y=5000x^x