College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Are you Ready for College Algebra? - A. Diagnostic Test:: Real Numbers and Exponents: 1

Answer

a. $\frac{5}{6}$ b.$\frac{19}{12}$ c.$\frac{16}{3}$ d. $8$

Work Step by Step

a. $\frac{1}{3}$ + $\frac{1}{2}$ Multiply so the fractions have a common denominator. $\frac{1}{3}$ $\times$ $\frac{2}{2}$ = $\frac{2}{6}$ $\frac{1}{2}$ $\times$ $\frac{3}{3}$ = $\frac{3}{6}$ Perform addition. $\frac{2}{6}$ + $\frac{3}{6}$ = $\frac{5}{6}$ b. $2 - \frac{2}{3} + \frac{1}{4}$ Multiply so all terms have a common denominator. $\frac{2}{1} \times \frac{12}{12} = \frac{24}{12}$ $\frac{2}{3} \times \frac{4}{4} = \frac{8}{12}$ $\frac{1}{4} \times \frac{3}{3} = \frac{3}{12}$ Subtract then add. $\frac{24}{12} - \frac{8}{12} + \frac{3}{12} = \frac{19}{12}$ c. $4(2- \frac{2}{3})$ Multiply so the numbers in the parenthesis have a common denominator. $\frac{2}{1} \times \frac{3}{3} = \frac{6}{3}$ Perform the operation in the parenthesis. $\frac{6}{3} - \frac{2}{3} = \frac{4}{3}$ Multiply. $4(\frac{4}{3}) = \frac{16}{3}$ d. $\frac{12}{\frac{4}{3} + \frac{1}{6}}$ Multiply so the fractions in the denominator have a common denominator. $\frac{4}{3} \times \frac{2}{2} = \frac{8}{6}$ Add the fractions in the denominator. $\frac{8}{6} + \frac{1}{6} = \frac{9}{6}$ Multiply the numerator by the inverse of the denominator. $12 \times \frac{6}{9} = 8$
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