Ask your own question, for FREE!
Mathematics 9 Online
OpenStudy (anonymous):

Can I get some help with this question: The price of products may increase due to inflation and decrease due to depreciation. Derek is studying the change in the price of two products, A and B, over time. The price f(x), in dollars, of product A after x years is represented by the function below. f(x) = 10250(0.63)^x Part A: Is the price of product A increasing or decreasing and by what percentage per year? Justify your answer. Part B: The table below shows the price f(t), in dollars, of product B after t years:

OpenStudy (anonymous):

Graph for Part B

OpenStudy (anonymous):

For Part A I said: "The price of product A is decreasing because the number inside the pretenses of the function is less than 1. It is decreasing by 37%"

OpenStudy (anonymous):

Looks good to me.

OpenStudy (anonymous):

Alright, now I'm just stuck at Part B

OpenStudy (anonymous):

What's the question?

OpenStudy (anonymous):

Oh oops here it is, its for Part B: Which product recorded a greater percentage change in price over the previous year? Justify your answer.

OpenStudy (anonymous):

To get the common ratio for Part B, divide the price from any year by the price 1 year EARLIER. What do you get?

OpenStudy (anonymous):

Do i have to use f(b)-f(a)/b-a?

OpenStudy (anonymous):

If I just divide the year by the year earlier I got .43

OpenStudy (anonymous):

Great. Now in Part A, this ratio was 0.63. So which one has the greatest percentage change? Remember, A changes by 1-0.63 = 37%

OpenStudy (anonymous):

B has a greater percent change because 1-0.43=57%

OpenStudy (anonymous):

So B recorded a greater percent change in price over the previous year

OpenStudy (anonymous):

Well done. Congrats.

OpenStudy (anonymous):

Yay all thanks to you!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!