mn123 mn123
  • 19-04-2018
  • Mathematics
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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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