Question

Cami is comparing the growth rates in the value of two items in a collection. the value of a necklace increases by 3.2% per year. the value of a ring increases by 0.33% per month. write a function to represent the value of A of the necklace after t years, assuming the inital value of $1 then write an equivalent function with a power of 12t
B. which item is increasing in value at a faster rate? explain
function;
equivalent function;

Answer 1

Answer:

$A(n) = 1(1 + \frac{3.2}{100})^{t}$

$A(r) = 1(1 + \frac{0.33}{100} )^{12t}$

The value of the ring is increasing at a faster rate.

Step-by-step explanation:

It is given that the value of a necklace increases by 3.2% per year.

Therefore, the value of the necklace after t years will be  

$A(n) = 1(1 + \frac{3.2}{100})^{t}$ .......... (1)

{The initial value is given to be $1}

Again, the value of a ring increases by 0.33% per month.

Therefore, the value of the ring after t years will be  

$A(r) = 1(1 + \frac{0.33}{100} )^{12t}$ ............ (2)

{The initial value is given to be $1}

Therefore, from equation (1) the value of the necklace after 1 year will be  

A(n) = $1.032

And from equation (2) the value of the ring after 1 year will be  

$A(r) = 1(1 + \frac{0.33}{100} )^{12} = 1.04$ dollars.

Therefore, the ring will value more after 1 year.

Therefore, the value of the ring is increasing at a faster rate. (Answer)

Answer 2

Answer:

A(t) = 1.032^t

Ring

Step-by-step explanation:

A(t) = (1 + 3.2%)^t

A(t) = (1.032)^t

Ring:

(1 + 0.33%)^(12t)

(1 + 0.0033)^(12t)

(1.0033¹²)^t

1.040326705^t

Faster rate: ring