Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Appendix A - Review - A.1 Algebra Essentials - A.1 Assess Your Understanding - Page A12: 72

Answer

All values must be included.

Work Step by Step

The given expression is $=\frac{x}{x^2+9} $ (a) Substitute $x=3$. $=\frac{3}{3^2+9} $ Simplify. $=\frac{3}{9+9} $ $=\frac{3}{18} $ $=\frac{1}{6} $ (b) Substitute $x=1$. $=\frac{1}{1^2+9} $ Simplify. $=\frac{1}{1+9} $ $=\frac{1}{10} $ (c) Substitute $x=0$. $=\frac{0}{0^2+9} $ Simplify. $=\frac{0}{0+9} $ $=\frac{0}{9} $ $=0 $ (d) Substitute $x=-1$. $=\frac{-1}{(-1)^2+9} $ Simplify. $=\frac{-1}{1+9} $ $=\frac{-1}{10} $ $=-\frac{1}{10} $ Hence, all values must be included.
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