## Algebra 1: Common Core (15th Edition)

Published by Prentice Hall

# Chapter 7 - Exponents and Exponential Functions - 7-1 Zero and Negative Exponents - Practice and Problem-Solving Exercises - Page 422: 46

#### Answer

For $d=-2$, There were 20 finches two decades before measurement. For $d=0$, There were $45$ finches at the time of measurement. For $d=1$, There will be about 63 finches a decade after the time of measurement.

#### Work Step by Step

For $d=-2$, $45\cdot 1.5^{-2}=\qquad$ apply: $\displaystyle \mathrm{a}^{-n}=\frac{\mathrm{l}}{\mathrm{a}^{n}}$ $=45\displaystyle \cdot\frac{1}{1.5^{2}}=\qquad$ evaluate $1.5^{2}$ $=\displaystyle \frac{45}{2.25}\times\frac{4}{4}= \qquad$ eliminate the decimals in the denominator $=\displaystyle \frac{45\times 4}{9}=5\times 4=20$ There were 20 finches two decades before measurement For $d=0$, $45\cdot 1.5^{0}=\qquad$ apply: $a^{0}=1$ $=45\cdot 1=45$ There were $45$ finches at the time of measurement For $d=1$, $45\cdot 1.5=62.5\approx 63$ There will be about 63 finches in a decade's time from the time of measurement

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