Math (Pre-Cal) (Exponential Functions)

A car was valued at \$38,000 in the year 2003. The value depreciated to \$11,000 by the year 2009. Assume that the car value continues to drop by the same percentage.
-What will the value be in the year 2013?

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1. let the rate of depreciation be r
38000(1-r)^6 = 11000
(1-r)^6 = 11/38
1-r = (11/38)^(1/6)
1-r = .8133326
r = .186667

rate of depreciation is 18.67 %

so in 2012 it will be
38000(.8133326)^10 = \$4813.55

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2. A new car was valued at \$43,000, and it's value depreciated to \$15,000 over the next 6 years. Use the exponential equation for depreciation to answer the following questions.
) What was the annual percent rate of depreciation, to 2 decimal places? Your answer will be a positive number.

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3. ctm

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