Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 8 - Review Exercises - Page 411: 76


$y = \frac{1}{x-4}$, where $x\geq 5$

Work Step by Step

$x = t^2+5$ $t = \sqrt{x-5}$ We can replace this value in the equation with $y$: $y = \frac{1}{t^2+1}$ $y = \frac{1}{(\sqrt{x-5})^2+1}$ $y = \frac{1}{(x-5)+1}$ $y = \frac{1}{x-4}$ Since $x = t^2+5$, then $x\geq 5$ $y = \frac{1}{x-4}$, where $x\geq 5$
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