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 14: 43


a. $C=1500+1.50(n)$ for $0\le n\leq500$ b. 1875 dollars c. 333 CDs d. 500 CDs

Work Step by Step

a. C=cost, n=# of cds (up to 500) $C=1500+1.50(n)$ for $0\le n\le 500$ b. $C=1500+1.50(250)$ $C=1500+375$ $C=1875$ c. C=2000 $2000=1500+1.50(n)$ $2000-1500=1.50(n)$ $500=1.50(n)$ $500\div1.50=n$ $n=333$ d. C=3000 $3000=1500+1.50(n)$ $3000-1500=1.50(n)$ $1500=1.50(n)$ $1500\div1.50=n$ $n=1000$ but because the deal is only for up to 500, they can only get 500 CDs.
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