A new car was valued at $28,000, and it's value depreciated to $14,000 over the next 3 years. Use the exponential equation for d
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I find it convenient to look at the differences and the rate at which those differences are made up.
8. Jim is closing the $150 gap at the rate of $7.50 per week. He will catch up in
... 150/(7.5/week) = 20 weeks
9. At noon, the price of Stock A has increased by 0.05×3 = 0.15, so is now $15.90, which is $0.63 more than Stock B at that time. The prices are closing the gap at the rate of $0.05 +0.13 = $0.18 per hour, so will be the same after
... $0.63/($0.18/hour) = 3.5 hours . . . . after noon, at 3:30 pm
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You can also write and solve equations for the prices of the stocks. Or you can use a graphing calculator to tell you the solution. When equations are involved, I like to solve them the simplest possible way: let technology do it.
You are given the value at a time, and the rate of change of that value, so the equations are easily written in point-slope form. You will note that the common price at 3:30 pm (15.5 hours after midnight) is one that is not a whole number of cents. (That's usually OK for when trading stocks.)
