Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 1 - Linear Functions - 1.1 Solving Linear Equations - 1.1 Exercises - Page 13: 40

Answer

a. 257.49 million b. 1997 c. 2009

Work Step by Step

a. t=1993-1990=3 $P=2.57(3)+249.78$ $P=7.71+249.78$ $P=257.49$ b. P=270 $270=2.57(t)+249.78$ $270-249.78=2.57(t)$ $20.22=2.57(t)$ $20.22\div2.57=t$ $t=7.86$ 1990+7.86=1997.86 or 1997 c. P=300 $300=2.57(t)+249.78$ $300-249.78=2.57(t)$ $50.22=2.57(t)$ $50.22\div2.57=t$ $t=19.54$ 1990+19.54=2009.54 or 2009
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