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What is the answer for question 10) ? http://postimg.org/image/5gm0wsol3/
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High School Teacher
The equation given is in the form V = a * b^y
where a = original value of the car, b = rate of depreciation, and
y number of years.
So, for the first question rate of depreciation is 0.85 or 85%.
For the second question, we replace the y by 5.
`V = 25000(0.85)^5 = 25000*0.4437053125 = $11092.63 `
` `Therefore, the value of the car after 5 years is $11.092.63.
For the last question, we will take the 10% of the original value.
`0.10 * 25000 = $2500`
So, our V = $2500, and we will solve for y.
`2500 = 25000(0.85)^y`
Divide both sides by 25000.
`0.10 = 0.85^y`
Take the natural logarithm of both sides.
`ln0.10 = ln0.85^y`
Use the property: lnu^v = vlnu on right side.
`ln0.10 = yln0.85`
Divide both sides by ln0.85.
`y = ln0.10/ln0.85 = 14.17 ` years or 14 years and 2 months.
Posted by violy on May 23, 2013 at 11:56 PM (Answer #1)
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