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Unformatted text preview: 4 · 1)2 by b = 1 = a (4) · (2) = a 8 . 3. Simplify the expression. Express your answer using only positive exponents and evaluate whenever possible. ³ 2 x 3 y2 ´5 Math 111I Sec 001 Group Quiz 3 Solutions Solutions. The solution follows as ± 2 x 3 y2 ²5 = 1 (2 x 3 y2 ) 5 = 1 2 5 ( x 3 ) 5 ( y2 ) 5 = 1 32 x 15 y10 = 1 32 x 15 (1 /y 10 = 1 32 x 15 · y 10 = y 10 32 x 15 . 4. Simplify the expression. 4 p 16 x12 y 3 Solutions. The solution follows as 4 p 16 x12 y 3 = ± 16 x12 y 3 ² 1 / 4 = (16) 1 / 4 ( x12 ) 1 / 4 ( y 3 ) 1 / 4 = 2 x3 y 3 / 4 = 2 y 3 / 4 x 3 ....
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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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