College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 2, Functions - Chapter 2 Review - Exercises - Page 269: 68

Answer

a) $3500$ and $6875$; the savings deposited in 1995 and 2010. b) 2025 c) $\$225/year$

Work Step by Step

a) Plugging in $t=0,15$ into the given function, we get: $D(0)=3500+15(0)^2=3500$ $D(15)=3500+15(15)^2=6875$ These values represent the savings deposited in 1995 and 2010. b) To find the year in which Ella deposits $\$17,000$, we plug in $\$17,000$ for $D$ and solve for $t$: $3500+15t^2=17000\\13500=15t^2\\900=t^2\\t=\pm30$ But $t$ must be positive, so we choose $t=+30$, which represents the year $2025$. c) The rate of change is: $\frac{f(y)-f(x)}{y-x}=\frac{6875-3500}{15-0}=\$225/year$ This value represents the average increase in deposit per.
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