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Unformatted text preview: p 5 ( p 2 ) 2 p 6 = p 5 · p 2 · 2 p 6 by rule 3 = p 5 · p 4 p 6 = p 5+4 p 6 by rule 1 = p 9 p 6 = p 96 = p 3 by rule 2 . Math 111I Sec 001 Quiz 4 Solutions 4. (3 points) Simplify the expression. Express your answer using only positive exponents and evaluate whenever possible. ± 3 x 6 y1 ²4 Solutions. The computation follows by ± 3 x 6 y1 ²4 = 1 ± 3 x 6 y1 ² 4 by rule 6 = 1 3 4 ( x 6 ) 4 ( y1 ) 4 by rule 4 = 1 81 x 6 · 4 y (1) · 4 by rule 3 = 1 81 x 24 y4 = 1 81 x 24 1 y 4 = 1 81 x 24 · y 4 1 by the rule of dividing by a fraction = y 4 81 x 24...
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This note was uploaded on 02/16/2011 for the course MATH 111 taught by Professor Hitchcock during the Fall '08 term at South Carolina.
 Fall '08
 HITCHCOCK
 Math

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