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The approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.
Depreciation is to decrease in the value of a product in a period of time. This can be given as,
Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.
Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.
Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is . Thus, by the above formula for the first car,
Take log both the sides as,
Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.
Thus, the depreciation price of the car is 0.6y. Let the number of year is . Thus, by the above formula for the second car,
Take log both the sides as,
The difference in the ages of the two cars is,
Thus, the approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.
Learn more about the depreciation here;
The slope m of a line
through points 
and 
is given by :<span>
Thus, the slope of the line passing through points </span><span>(−1, 7) and (2, 10) is
</span>
<span>
The equation of a line with slope m passing through a point P(a, b) is given by
(y-b)=m(x-a).
We can consider any of the points (-1, 7), or (2, 10). Let's choose (2, 10):
y-10=1(x-2)
y-10=x-2
y-x=-2+10
y-x=8
Answer: </span><span>C. −x+y=8</span>
Thus, the slope of the line passing through points </span><span>(−1, 7) and (2, 10) is
</span>
<span>
The equation of a line with slope m passing through a point P(a, b) is given by
(y-b)=m(x-a).
We can consider any of the points (-1, 7), or (2, 10). Let's choose (2, 10):
y-10=1(x-2)
y-10=x-2
y-x=-2+10
y-x=8
Answer: </span><span>C. −x+y=8</span>