SOLUTION: IF A PIECE OF REAL ESTATE PURCHASED FOR $75,000 IN 1998 APPRECIATES AT THE RATE OF 6% PER YEAR, THEN ITS VALUE t YEARS AFTER THE PURCHASE WILL BE f(t)=75,000(1.06^t). ACCORDING TO

Algebra ->  Points-lines-and-rays -> SOLUTION: IF A PIECE OF REAL ESTATE PURCHASED FOR $75,000 IN 1998 APPRECIATES AT THE RATE OF 6% PER YEAR, THEN ITS VALUE t YEARS AFTER THE PURCHASE WILL BE f(t)=75,000(1.06^t). ACCORDING TO      Log On


   

Question 64215: IF A PIECE OF REAL ESTATE PURCHASED FOR $75,000 IN 1998 APPRECIATES AT THE RATE OF 6% PER YEAR, THEN ITS VALUE t YEARS AFTER THE PURCHASE WILL BE f(t)=75,000(1.06^t). ACCORDING TO THIS MODEL, BY HOW MUCH WILL THE VALUE OF THIS PIECE OF PROPERTY INCREASE BETWEEN THE YEARS 2005 AND 2008?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
IF A PIECE OF REAL ESTATE PURCHASED FOR $75,000 IN 1998 APPRECIATES AT THE RATE OF 6% PER YEAR, THEN ITS VALUE t YEARS AFTER THE PURCHASE WILL BE f(t)=75,000(1.06^t). ACCORDING TO THIS MODEL, BY HOW MUCH WILL THE VALUE OF THIS PIECE OF PROPERTY INCREASE BETWEEN THE YEARS 2005 AND 2008?
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f(t)=75,000(1.06^t)
f(7)=75000(1.06)^7=112,772.27 (Value in 2005)
f(10)=75000(1.06)^10= 134313.58 (Value in 2008)
The property will increase $21,541.31 in those
three years.
Cheers,
Stan H.