3 years ago

HELP !!!I am being timed !!!!

Mathematics

1 answer:

wolverine [178]3 years ago

7 0

Answer:x = -8

Step-by-step explanation:

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a car was valued at $41,000 in the year 2009 by 2013 the car value has depreciated to 19,000 if the car value continues to by th

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$6,376.92

Step-by-step explanation:

-Let d be the rate of depreciation per year.

-Therefore, the value after n years can be expressed as:

$A=P(1-d)^n\\\\A=Value \ after \ n \ years\\P=Initial \ Value\\d=Rate \ of \ depreciation\\n=Time \ in \ years$

#We substitute for the years 2009-2013 to solve for d:

$A=P(1-d)^n\\\\19000=41000(1-d)^4\\\\0.475=(1-d)^4\\\\d=1-0.475^{0.25}\\\\d=0.1698$

#We then use the calculated depreciation rate above to solve for A after 10 yrs:

$A=P(1-d)^n\\\\=41000(1-0.1698)^{10}\\\\=\$6,376.92$

Hence, the value of the car after 10 yrs is $6,376.92

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Step-by-step explanation:

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