Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 6 - Section 6.8 - Substituting Data into Formulas - Exercises - Page 258: 11

Answer

$$ \frac{L-\pi r_2-2d}{\pi}, \ \ 3.00 .$$

Work Step by Step

To solve the given equation for $r_1$, we move out the term $\pi r_2+2d$ and then divide both sides by $\pi$ as follows $$ L =\pi(r_1 + r_2) + 2d\Longrightarrow \pi r_1=L-\pi r_2-2d\\ \Longrightarrow r_1=\frac{L-\pi r_2-2d}{\pi} .$$ Now, by substitution, we have $$ r_1=\frac{L-\pi r_2-2d}{\pi}=\frac{37.68-5\pi -2(6.28)}{\pi}=3.00 .$$
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