SOLUTION: -4a squared bto the sixth over two a to the fifth b to the fourth.
not certain how to solve- using prentice hall algebra i book, no examples
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Question 959201: -4a squared bto the sixth over two a to the fifth b to the fourth.
not certain how to solve- using prentice hall algebra i book, no examples
Found 2 solutions by stanbon, rothauserc:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
-4a squared b to the sixth over two a to the fifth b to the fourth.
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-4a^2*b^6/[2a^5*b^4]
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= (-4/2)(a^2/a^5)(b^6/b^4)
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= -2(1/a^3)(b^2/1)
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= -2b^2/a^3
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Cheers,
Stan H.
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Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
I get this from your statement of the problem
( (-4a)^2 )^6 / 2a^5b^4 = (16a^2)^6 / 2a^5b^4 = 16^6a^12 / 2a^5b^4 = 16777216a^12 / 2a^5b^4 = 8388608a^7 / b^4
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