## College Algebra (6th Edition)

Published by Pearson

# Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.1: 120

#### Answer

$-10$

#### Work Step by Step

Use the order of operations to simplify the expression:$$\frac{12\div3\times5|2^{2}+3^{2}|}{7+3-6^{2}}$$ As the fraction itself defines grouping symbols, the numerator and denominator are each individual groups. Within each group start with Grouped operations. Remember, absolute value bars are grouping symbols as well as parentheses. $$=\frac{12\div3\times5|4+9|}{7+3-6^{2}}$$ $$=\frac{12\div3\times5|13|}{7+3-6^{2}}$$ The absolute value is the distance of that number to zero. 13 is 13 spaces away from zero, so the absolute value of 13 is 13: $$=\frac{12\div3\times5(13)}{7+3-6^{2}}$$ Next complete any exponents, then multiplication and division in order left to right, and finally addition and subtraction left to right: $$=\frac{12\div3\times5(13)}{7+3-36}$$ $$=\frac{4\times5(13)}{7+3-36}$$ $$=\frac{20(13)}{7+3-36}$$ $$=\frac{260}{7+3-36}$$ $$=\frac{260}{10-36}$$ $$=\frac{260}{(-26)}$$ Finish with the division implied by the fraction or reduce the fraction: $$=-10$$

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