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A car that originally cost $20,000 depreciates by 15% each year in about how many years will be car be worth half its value use the equation 10,000=20,000(0.85)^x to solve the p…
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A car that originally cost $20,000 depreciates by 15% each year in about how many years will be car be worth half its value use the equation 10,000=20,000(0.85)^x to solve the p…
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Jun 17, 2018
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A car that originally cost $20,000 depreciates by 15% each year in about how many years will be car be worth half its value use the equation 10,000=20,000(0.85)^x to solve the problem round your answer to the nearest dollar
Mathematics
high-school
Nitish
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0.5=0.85^x taking the natural log of both sides...
ln0.5=x ln0.85
x=ln0.5/(ln0.85)
x≈4.265
Then:
f=20000(0.85)^(4.265)
f=$10000.00 to nearest dollar
Hajin
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Jun 24, 2018
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