A new car is purchased for 23900 dollars. The value of the car depreciates at 9.25% per year. To the nearest tenth of a year, ho
2 answers:
9514 1404 393
Answer:
14.2 years
Step-by-step explanation:
The value is described by the exponential function ...
y = (initial value)·(decay factor)^x
where y is the value after x years.
The decay factor is (1 - annual depreciation) = 1 - 0.0925 = 0.9075, so we want to find x when ...
6000 = 23900·0.9075^x
6000/23900 = 0.9075^x . . . . . . . . . . . . . . divide by 23900
log(6000/23900) = x·log(0.9075) . . . . . . . take logarithms
x = log(6000/23900)/log(0.9075) ≈ 14.2396
It will be about 14.2 years until the value of the car is $6000.
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For this case the area is given by:
A = (22 + x) * (28 + x) = 722
Rewriting we have:
616 + 22x + 28x + x ^ 2 = 722
x ^ 2 + 50x + 616 - 722 = 0
x ^ 2 + 50x - 106 = 0
Solving the polynomial we have:
x1 = 2.04
x2 = -52.04
We take the positive root:
x = 2.04 inches:
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The width of the border to the nearest inch is:
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PLEASE SEE ATTACHED IMAGE.
For this case the area is given by:
A = (22 + x) * (28 + x) = 722
Rewriting we have:
616 + 22x + 28x + x ^ 2 = 722
x ^ 2 + 50x + 616 - 722 = 0
x ^ 2 + 50x - 106 = 0
Solving the polynomial we have:
x1 = 2.04
x2 = -52.04
We take the positive root:
x = 2.04 inches:
Answer:
The width of the border to the nearest inch is:
x = 2inches