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Unformatted text preview: p 5 ( p 2 ) 2 p 6 = p 5 Â· p 2 Â· 2 p 6 by rule 3 = p 5 Â· 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. Â± 3 x 6 y1 Â²4 Solutions. The computation follows by Â± 3 x 6 y1 Â²4 = 1 Â± 3 x 6 y1 Â² 4 by rule 6 = 1 3 4 ( x 6 ) 4 ( y1 ) 4 by rule 4 = 1 81 x 6 Â· 4 y (1) Â· 4 by rule 3 = 1 81 x 24 y4 = 1 81 x 24 1 y 4 = 1 81 x 24 Â· y 4 1 by the rule of dividing by a fraction = y 4 81 x 24...
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 Fall '08
 HITCHCOCK
 Math, Decimal, Rover P4, positive exponents, p2Â·2 p6 p5

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