Question 132367
(3m^2t^5)^2
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(2m^3t^-1)
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[(3m^2t^5)^2] / [(2m^3t^-1)]
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= [9m^4t^10] / [2 m^3/t]

Invert the denominator and multiply:

= [9m^4t^10] *[t/(2m^3]

= [9m^4t^11]/[2m^3]

= (9/2)*m*t^11

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Cheers,
Stan H.