solve: a^2 · a^3 · a^-4

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solve: a^2 · a^3 · a^-4

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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do you know the answer?
use these formula ..a^-n = 1/a^n , a^n . a^m = a^(n+m) and a^n/a^m = a^(n-m) and you can get the answer quick
it's not a^9 though..

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Other answers:

ofcourse it's not, one of the power is negative...check that again :)
see that's what confuses me, but I thought when you flip the exponent to make it positive, you add it as a positive exponent?
would it be a^1?
correct
okay, So you're not supposed to flip it then? :S
it's two positive powers 2 and 3..you add..one negative power 4 ..you subtract
ty
yw :)

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