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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.

Respuesta :

MrRoyal

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

[tex]f(x) = 490 * 0.88^x[/tex]

Leo

[tex]x \to g(x)[/tex]

[tex]0 \to 480[/tex]

[tex]2 \to 360[/tex]

[tex]4 \to 470[/tex]

Solving (a): The phone with greater initial value

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

[tex]f(x) = 490 * 0.88^x[/tex]

[tex]f(0) = 490 * 0.88^0[/tex]

[tex]f(0) = 490 * 1[/tex]

[tex]f(0) = 490[/tex]

From Leo's table

[tex]g(0) = 480[/tex]

By comparison;

[tex]f(0) > g(0)[/tex]

i.e.

[tex]490 > 480[/tex]

So: Tonya's had the greater initial trade-in value

Solving (b): The phone with lesser rate

An exponential function is:

[tex]y = ab^x[/tex]

Where:

[tex]b \to rate[/tex]

For Tonya

[tex]b = 0.88[/tex]

For Leo, we have:

[tex](x_1,y_1) = (0,480)[/tex]

[tex](x_2,y_2) = (2,360)[/tex]

So, the equation becomes:

[tex]y = ab^x[/tex]

[tex]480 = ab^0[/tex] and [tex]360 = ab^2[/tex]

Solving [tex]480 = ab^0[/tex], we have:

[tex]480 = a * 1[/tex]

[tex]480 = a[/tex]

[tex]a= 480[/tex]

[tex]360 = ab^2[/tex] becomes

[tex]360 = 480 * b^2[/tex]

Divide both sides by 480

[tex]0.75 = b^2[/tex]

Take square roots

[tex]0.87 = b[/tex]

[tex]b=0.87[/tex] -- Leo's rate

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

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