Question

1. your starting salary at a new company is $34,000 and it increase by 2.5% each year.

a. write an exponential growth function to represent this situation
b. what will you salary be in 5 years? round your answer to the nearest dollar.

1. in 2010 an item cost $9.00. the price increase by 1.5% each year.

a. write an exponential growth function to represent this situation.
b. how much will it cost in 2030? round your answer to the nearest cent.

1. the yearly profits of a company is $25,000. the profits have been decreasing by 6% per year.

a. write an exponential decay function to represent this situation.
b. what will be the profits in 8 years? round your answer to the nearest dollar.

1. you bought $2000 worth of stocks in 2012. the value of the stocks has been decreasing by 10% each year.

Explanation:

Step1: Define growth function (Q11A)

The exponential growth formula is $Y = a(1+r)^t$, where $a=\$34000$, $r=0.025$, $t$ = years.
Expression: $Y = 34000(1+0.025)^t$

Step2: Calculate salary in 5 years (Q11B)

Substitute $t=5$ into the growth function.
Expression: $Y = 34000(1.025)^5$
Calculate $(1.025)^5 \approx 1.13140821$, then $Y \approx 34000 \times 1.13140821 \approx 38467.88$

Step3: Define growth function (Q12A)

Use exponential growth formula with $a=\$9.00$, $r=0.015$, $t$ = years after 2010.
Expression: $Y = 9(1+0.015)^t$

Step4: Calculate 2030 cost (Q12B)

Find $t=2030-2010=20$, substitute into the function.
Expression: $Y = 9(1.015)^{20}$
Calculate $(1.015)^{20} \approx 1.34685501$, then $Y \approx 9 \times 1.34685501 \approx 12.12$

Step5: Define decay function (Q13A)

Use exponential decay formula $Y = a(1-r)^t$, where $a=\$25000$, $r=0.06$, $t$ = years.
Expression: $Y = 25000(1-0.06)^t = 25000(0.94)^t$

Step6: Calculate profit in 8 years (Q13B)

Substitute $t=8$ into the decay function.
Expression: $Y = 25000(0.94)^8$
Calculate $(0.94)^8 \approx 0.60956893$, then $Y \approx 25000 \times 0.60956893 \approx 15239$

Step7: Define decay function (Q14A)

Use exponential decay formula with $a=\$2000$, $r=0.10$, $t$ = years after 2012.
Expression: $Y = 2000(1-0.10)^t = 2000(0.90)^t$

Answer:

11.
A. $Y = 34000(1.025)^t$
B. $\$38468$
12.
A. $Y = 9(1.015)^t$
B. $\$12.12$
13.
A. $Y = 25000(0.94)^t$
B. $\$15239$
14.
A. $Y = 2000(0.90)^t$