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Q4. L.T.: I can solve exponential decay equations. LaBrain buys a house for $246,800. The house appreciates at a rate of 5.28% annually for 7 years and then depreciates at a rate of 3.79% annually for the next 11 years, How much is the house worth after these 18 years have passed? Round to the nearest cent. (Note: A = P(1 +/-r)¹) a. $353,806.89 b. $623,118.91 c. $231,306.93 d. $254,183.65

Q4. L.T.: I can solve exponential decay equations. LaBrain buys a house for $246,800. The house appreciates at a rate of 5.28% annually for 7 years and then depreciates at a rate of 3.79% annually for the next 11 years, How much is the house worth after these 18 years have passed? Round to the nearest cent. (Note: A = P(1 +/-r)¹) a. $353,806.89 b. $623,118.91 c. $231,306.93 d. $254,183.65

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Q4. L.T.: I can solve exponential decay equations. LaBrain buys a house for $246,800. The house appreciates at a rate of 5.28% annually for 7 years and then depreciates at a rate of 3.79% annually for the next 11 years, How much is the house worth after these 18 years have passed? Round to the nearest cent. (Note: A = P(1 +/-r)¹) a. $353,806.89 b. $623,118.91 c. $231,306.93 d. $254,183.65
Transcribed Image Text:Q4. L.T.: I can solve exponential decay equations. LaBrain buys a house for $246,800. The house appreciates at a rate of 5.28% annually for 7 years and then depreciates at a rate of 3.79% annually for the next 11 years, How much is the house worth after these 18 years have passed? Round to the nearest cent. (Note: A = P(1 +/-r)¹) a. $353,806.89 b. $623,118.91 c. $231,306.93 d. $254,183.65
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