mn123
mn123
19-04-2018
Mathematics
contestada
How do I evaluate (27x^3/8y^9) ^-5/3
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Аноним
Аноним
19-04-2018
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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