SOLUTION: A computer that was purchased for $6000 depreciates at a rate of 13.4% per year. An equation modeling this is, f(x)=6000(0.866)x, where x is the number of years.
What is the com
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-> SOLUTION: A computer that was purchased for $6000 depreciates at a rate of 13.4% per year. An equation modeling this is, f(x)=6000(0.866)x, where x is the number of years.
What is the com
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Question 1199046: A computer that was purchased for $6000 depreciates at a rate of 13.4% per year. An equation modeling this is, f(x)=6000(0.866)x, where x is the number of years.
What is the computer worth after 2 years?
You can put this solution on YOUR website! after two years, the computer will be worth 6000 * .866 * .866 = 6000 * .866^2 = 4499.736.
to find how long it takes for the computer to be equal o 2530, use the following formula.
2530 = 6000 * .866 ^ n
n is the number of years.
divide both sides of the equation by 6000 to get:
2530/6000 = .866 ^ n
take the log of both sides of the equation to get:
log(2530/6000) = log(.866 ^ n)
by log rules, this becomes:
log(2530/6000) = n * log(.866)
solve for n to get:
n = log(2530/6000) / log(.866) = 6.002209934.
confirm by replacing n in the original equation to get:
2530 = 6000 * .866 ^ 6.002209934.
this becomes 2530 = 2530, confirming that vallue of n is correct.
your solution is that the value of the computer will be 2530 in 6.002209934 years.
round as required.
