Elementary Technical Mathematics

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

Chapter 1 - Section 1.9 - The U.S. System of Weights and Measures - Exercises - Page 54: 37


$2520\ in^2$

Work Step by Step

$3\frac{3}{4}ft\times4\frac{2}{3}ft=\frac{15}{4}ft\times\frac{14}{3}ft=\frac{/\!\!3\times5}{/\!\!2\times2}ft\times\frac{/\!\!2\times7}{/\!\!3}ft=\frac{35}{2}ft^2$ $\frac{35}{2}ft^2\times\frac{144in^2}{1ft^2}=\frac{5040}{2}in^2=2520in^2$
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