[ ( (3)^2 - 6^0 ) / 9^(-3/2) ] ^(2/3) . . . find the value?

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[ ( (3)^2 - 6^0 ) / 9^(-3/2) ] ^(2/3) . . . find the value?

2007-07-21 13:02:11 · 4 answers · asked by CPUcate 6 in Science & Mathematics ➔ Mathematics

4 answers

[ ( (3)^2 - 6^0 ) / 9^(-3/2) ] ^(2/3)

= [ ( (3)^2 - 6^0 ) * 9^(+3/2) ] ^(2/3)

= [ ( 9 - 1 ) * 27 ] ^(2/3) = (8 * 27) ^(2/3)

= 4 * 9 = 36

2007-07-21 13:09:51 · answer #1 · answered by Northstar 7 · 1⤊ 0⤋

3^2 = 9
6^0 = 1

[(9-1) / 9^(-3/2)] ^(2/3) = 8^(2/3) / [9^(-3/2)]^(2/3)

= 8^(2/3) / 9^(-1) = 8^(2/3) * 9^(1) = (Â³â 8^2) * 9

= (Â³â2^6) * 9 = 4*9 = 36

2007-07-21 20:13:33 · answer #2 · answered by Anonymous · 0⤊ 0⤋

4*9 = 36
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Ideas of mental math approach:
numerator = 8^(2/3) = 4
denominator = 9^(-1)

2007-07-21 20:05:30 · answer #3 · answered by sahsjing 7 · 0⤊ 0⤋

do your own homework son!

2007-07-21 20:05:31 · answer #4 · answered by 星戦争は最もよい 3 · 0⤊ 2⤋