College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 4 - Section 4.1 - Properties of Linear Functions and Linear Models - 4.1 Assess Your Understanding - Page 282: 38

Answer

(a) The cost would be 18 dollars. (b) 64 minutes would've been used. (c) The maximum minutes I could talk is 173 minutes. (d) The implied domain is: $[0,43200)$

Work Step by Step

(a) Calculate C(50): $C(50)=0.26(50)+5$ $C(50)=13+5$ $C(50)=18$ dollars (b) Swap 21.64 for C(x) and solve for x: $21.64=0.26x+5$ $21.64-5=0.26x+5-5$ $16.64/0.26=0.26x/0.26$ $x=64$ minutes (c) Swap 50 for C(x) and solve for x: $50=0.26x+5$ $50-5=0.26x+5-5$ $45/0.26=0.26x/0.26$ $x\approx173$ minutes (d) There is no such thing as talking negative minutes, so we count from zero. Theoretically, one could make a 30-day straight call, so to find the implied domain, we must convert 30 days to minutes: $\dfrac{x}{30 \text{ days}}=\dfrac{1440 \text{ minutes}}{1\text{ day}}$ $x=43200$ minutes Thus, the implied domain is: $[0,43200)$
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