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Ali Fritz
23 January, 00:20
How do I evaluate (27x^3/8y^9) ^-5/3
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Leandro Bradley
23 January, 02:58
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First calculate 27 ^ - 5/3 = 1 / 27^5/3 = 1 / 243,
x^3 ^ - 5/3 = 1 / x^3 ^5/3 = 1 / x^5
8 ^-5/3 = 1 / 8^5/3 = 1/32
y^9 ^ - 5/3 = 1 / y^9^5/3 = 1 / y^15
so we have 1/243 x^5 * 32 * y^15
= 32y^15 / 243x^5 Answer
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