Asked by BERNIE
repost sorry about that
(k^1/8)^-4 oer(k^7)^1/2
(k^1/8)^-4 oer(k^7)^1/2
Answers
Answered by
Reiny
(k^1/8)^-4 over (k^7)^1/2
= k^(-1/2) / k^(7/2)
= k^-4 or 1/k^4
= k^(-1/2) / k^(7/2)
= k^-4 or 1/k^4
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