Problems 1 and 2$2.69-$2.39 don't make it clear how many years but I will assume we subtract 2006-2003 = 3 years.
So the $200 has grown by $15 in the 3 year or $5/year.
Let x = number of years
200 + 0.05x = 230
Solve for x and add to year 2003.
$2.69-2.39 = 0.30 increase in 6 years (1995-1989) which is 0.05/year.
Let x = number of years.
2.39 + 0.05*x = 2.89
x = (2.89-2.39)/0.05 = 10 years and add that to 1989.
Problem 3 states how the years are to be counted; i.e., at the start of each year. The car is worth $11,000 at the start of 2003 and at the start of 2007 it was worth %9,600. That's 4 years so it is decreasing at the rate of $(11,000-9600)/4 = $350/year.
Let x = number of years.
11,000 - 350x = 8,900
x = 6 years; add that to 2003 so he must have sold the car at the start of the year
2009. Check my work.
Khiry has money in a savings account at the bank that increases by the same amount of interesr each year. In 2003 khiry had $200 in the bank. In 2006, Khiry had $215 in the bank.In how mnany years will Khiry hav $230?
The cost of gasoline per gallon in 1989 is $2.39. In 1995, the cost of gas per gallon is $2.69. In what year will the cost of gas be $2.89?
Mr. Jenssen's car decreased in value at a constant rate at the start of each successive year. The casr was worth $11,000 in 2003 and $9,600 in 2007. When the car was worth $8,900, he sold it, In what year did Mr. Jessen sell the car?
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