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kap26 [50]
2 years ago
7

Tonya and Leo each bought a cell phone at the same time. The trade-in values, in dollars, of the cell phones are modeled by the

given functions, where x is the number of months that each person has owned the phone. Tonya's Phone Leo's Phone f(x) =490(0.88)x x g(x) 0 480 2 360 4 270 phone has the greater initial trade-in value. During the first four months, the trade-in value of Tonya's phone decreases at an average rate the trade-in value of Leo's phone.
Mathematics
1 answer:
Paha777 [63]2 years ago
6 0

Answer:

The answer is Tonya's phone had the greater initial trade-in value.

Leo's phone decreases at an average rate slower than the trade in value of Tonya's phone.

Step-by-step explanation:

Given

Tonya

f(x) = 490 * 0.88^x

Leo

x \to g(x)

0 \to 480

2 \to 360

4 \to 470

Solving (a): The phone with greater initial value

The initial value is when x = 0. So, we have:

f(x) = 490 * 0.88^x

f(0) = 490 * 0.88^0

f(0) = 490 * 1

f(0) = 490

From Leo's table

g(0) = 480

By comparison;

f(0) > g(0)

i.e.

490 > 480

<em>So: Tonya's had the greater initial trade-in value</em>

Solving (b): The phone with lesser rate

An exponential function is:

y = ab^x

Where:

b \to rate

For Tonya

b = 0.88

For Leo, we have:

(x_1,y_1) = (0,480)

(x_2,y_2) = (2,360)

So, the equation becomes:

y = ab^x

480 = ab^0 and 360 = ab^2

Solving 480 = ab^0, we have:

480 = a * 1

480 = a

a= 480

360 = ab^2 becomes

360 = 480 * b^2

Divide both sides by 480

0.75 = b^2

Take square roots

0.87 = b

b=0.87 -- Leo's rate

By comparison; Leo's rate is slower i.e. 0.87 < 0.88

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