SOLUTION: At the beginning of 2000 Tyler's house was worth 250 thousand dollars and Conner's house was worth 112 thousand dollars. At the beginning of 2003, Tyler's house was worth 182 thous
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Question 1150148: At the beginning of 2000 Tyler's house was worth 250 thousand dollars and Conner's house was worth 112 thousand dollars. At the beginning of 2003, Tyler's house was worth 182 thousand dollars and Conner's house was worth 156 thousand dollars. Assume that the values of both houses vary at an exponential rate.
a) Write a function (f) that determines the value of Tyler's house (in thousands of dollars) in terms of the number of years (t) since the beginning of 2000.
b) Write a function (g) that determines the value of Conner's house (in thousands of dollars) in terms of the number of years (t) since the beginning of 2000. Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! formula to use is f = p * (1 + r) ^ n
f is the future value
p is the present value
r is the interest rate per time period
n is the number of time periods
for tyler, the formula becomes:
182 = 250 * (1 + r) ^ 3
divide both sides of the equation by 250 and then take the third root of both sides of the equation and then subtract 1 from both sides of the equation to get:
(182/250) ^ (1/3) - 1 = r
solve for r to get:
r = -.1004117109
for conner, the formula becomes:
156 = 112 * (1 + r) ^ 3
divide both sides by 112 and then take the third root of both sides of the equation and then subtract 1 from both sides of the equation to get:
(156/112) ^ (1/3) - 1 = r
solve for r to get:
r = +.1167831651
tyler's house is depreciating by about 10% per year.
connor's house is appreciating by about 11.7% per year.
the equation for tyler becomes f(t) = 250,000 * (1 - .100411709) ^ t
the equation for connor becomes g(t) = 112,000 * (1 + .1167831651) ^ t
these equations can be graphed as shown below:
the red line is tyler.
the blue line is connor.
at the given rates of growth, the values of their respective house will be the same in 3.713 years.
the equation for graphing of tyler becomes y = 250,000 * (1 - .100411709) ^ x.
the equation for graphing of connor becomes y = 112,000 * (1 + .1167831651) ^ x.
y replaces f(t) and g(t).
x replaces t.
it's the same equations with different names normally used for graphing.
