Elementary Technical Mathematics

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

Chapter 6 - Review - Page 263: 30



Work Step by Step

Since we have $$C=\frac{5}{9}(F -32)$$ then by multiplying both sides by $\frac{9}{5}$, we get the formula $\frac{9}{5}C=F-32$ and hence we get $$F=\frac{9}{5} C +32^o=\frac{9}{5} ( 175^o)+32^o\\ =315^o+32^o=347^o$$ That is, $175^oC=347^oF.$
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