Precalculus (6th Edition)

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

Chapter R - Review of Basic Concepts - R.2 Real Numbers and Their Properties - R.2 Exercises - Page 20: 46

Answer

$$\frac{1}{2}$$

Work Step by Step

$$\frac{-(q-6)^2-2p}{4-p}$$ $$\frac{-(8-6)^2-2(-4)}{4-(-4)}$$ $$\frac{-4+8}{8}$$ $$\frac{4}{8}$$ $$\frac{1}{2}$$
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