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Asked by BERNIE

repost sorry about that
(k^1/8)^-4 oer(k^7)^1/2
13 years ago

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Answered by Reiny
(k^1/8)^-4 over (k^7)^1/2

= k^(-1/2) / k^(7/2)
= k^-4 or 1/k^4
13 years ago
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repost sorry about that
(k^1/8)^-4 oer(k^7)^1/2

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