Algebra help...?

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Solve and explain please...

7b^-2 / 7^2b^-8

2007-07-04 06:08:53 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

2 answers

7b^-2 / 7^2b^-8

it can be re written as 7/7^2 * b^-2/b^-8
= 1/7 * (b^ -2*b^8) coz wne raise to the power is taken from denominator to numerator the power sign changes.

=1/7 * b^6(as base is same powers r added)

=b^6/7

2007-07-04 06:58:59 · answer #1 · answered by sweet n simple 5 · 0⤊ 0⤋

this is not a solve, it's a simplify.

7b^-2 / 7Â²b^-8 = b^6 / 7
the 7/7Â² = 1/7, and for the b's, subtract the exponents: -2 - -8 = 6.

2007-07-04 06:17:49 · answer #2 · answered by Philo 7 · 0⤊ 0⤋