The table shows the value in dollars of a motorcycle at the end of x years. Motorcycle Number of Years, x & 0 & 1 & 2 & 3 Value, v(x) (dollars) & 9,000 & 8,100 & 7,290 & 6,561 Which exponential function models this situation? v(x)=9,000(1.1)^x v(x)=9,000(0.9)^x v(x)=8,100(1.1)^x v(x)=8,100(0.9)^x

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The table shows the value in dollars of a motorcycle at the end of x years. Motorcycle Number of Years, x & 0 & 1 & 2 & 3 Value, v(x) (dollars) & 9,000 & 8,100 & 7,290 & 6,561 Which exponential function models this situation? v(x)=9,000(1.1)^x v(x)=9,000(0.9)^x v(x)=8,100(1.1)^x v(x)=8,100(0.9)^x

Answer 1

B. \( v(x)=9,000(0.9)^{x} \)

Answer 2

Analyzing the data table, it's apparent that the value of the motorcycle decreases each year. Looking at the value from one year to the next, we see that the value drops to roughly 90% of the previous year's value each time. For example: The value dropped from 9,000 to 8,100 between 0 and 1 year which equates to a 10% decrease. This is found using the formula `(old value - new value) / old value * 100 = percentage decrease` ((9000 - 8100)/9000)*100 = 10%. If we calculate for the 2nd and 3rd years as well, we observe similar ~10% decreases. Using above formula: 1. for year 1-2 (8100 - 7290)/8100 = ~10% 2. for year 2-3 (7290 - 6561)/7290 = ~10%, That tells us each year the motorcycle is losing 10% of its previous years value. Maths convention dictates that a decreasing exponential function is modeled by a base that is between 0 and 1. So model comes out to be `value x (1 - rate/100)= newer_value` or `value x (0.9) = newer_value`. Accordingly, the correct formula from the choices given would be `v(x)=9000(0.9)^x`

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The table shows the value in dollars of a motorcycle at the end of x years. Motorcycle Number of Years, x & 0 & 1 & 2 & 3 Value, v(x) (dollars) & 9,000 & 8,100 & 7,290 & 6,561 Which exponential function models this situation? v(x)=9,000(1.1)^x v(x)=9,000(0.9)^x v(x)=8,100(1.1)^x v(x)=8,100(0.9)^x

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