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user100 [1]
3 years ago
12

I need to use the power rule (if m and n are real numbers and x doesn't equal 0, then (x^m)^n = x^mn... ) to simplify this fract

ion: (2x^2 y^-2 / y^5)^-3
Mathematics
1 answer:
Keith_Richards [23]3 years ago
3 0

Answer:

= 2x^{-6}y ^{21}\\\\

Step-by-step explanation:

Given the indicinal expression  (2x^2 y^-2 / y^5)^-3

In indices

(x^m)^n = x^{mn

Applying to solve the question

(2x^2 y^{-2} / y^5)^{-3}\\\\= \frac{2x^{-6}y^6}{y^{-15}}\\=  2x^{-6} * \frac{y^6}{y^{-15}} \\= 2x^{-6} * y ^{6-(-15)}\\= 2x^{-6} * y ^{6+15}\\= 2x^{-6} * y ^{21}\\= 2x^{-6}y ^{21}\\\\

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