Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 2 - Graphs and Functions - 2.5 Equations of Lines and Linear Models - 2.5 Exercises - Page 245: 64

Answer

(a) See graph, linear. (b) $y=77x$, see graph. (c) $779$ megaparsec. (d) $12.34\ billion$ years. (e) from 9.5 to 19 billion years.

Work Step by Step

(a) See graph, the relationship appears to be linear. (b) Use points $(520,40000)$ and $(0,0)$, we have $m=\frac{40000}{520}\approx77$ and $y=77x$, see graph. (c) For $y=60000$, we have $x=60000/77\approx779$ megaparsec. (d) With $m=77$, use the formula, we have $A=\frac{9.5\times10^{11}}{77}\approx12.34\ billion$ years. (e) For $m_1=50$, we have $A_1=\frac{9.5\times10^{11}}{50}=19\ billion$ years. For $m_2=100$, we have $A_2=\frac{9.5\times10^{11}}{100}=9.5\ billion$ years. Thus the range is from 9.5 to 19 billion years.
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