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Rom4ik [11]
3 years ago
6

Help asap..............

Mathematics
1 answer:
MatroZZZ [7]3 years ago
3 0

The common denominator for 25 and 20 would be 100

25 x 4 = 100

20 x 5 = 100

Rewrite 14/25 as 14 x 4 / 25 x 4 = 56/100

Rewrite 11/20 as 11 x 5 / 20 x 5 = 55/100

14/25 is greater than 11/20

14/25 > 11/20

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Given:

Consider the given expression is

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To find:

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Solution:

We have,

(4x^3y^2)^{\frac{3}{10}}=(2^2)^{\frac{3}{10}}(x^3)^{\frac{3}{10}}(y^2)^{\frac{3}{10}}

(4x^3y^2)^{\frac{3}{10}}=(2)^{\frac{6}{10}}(x)^{\frac{9}{10}}(y)^{\frac{6}{10}}

(4x^3y^2)^{\frac{3}{10}}=(2)^{\frac{3}{5}}(x)^{\frac{9}{10}}(y)^{\frac{3}{5}}

(4x^3y^2)^{\frac{3}{10}}=\sqrt[5]{2^3}\sqrt[10]{x^9}\sqrt[5]{y^3}       [\because x^{\frac{1}{n}}=\sqrt[n]{x}]

(4x^3y^2)^{\frac{3}{10}}=\sqrt[5]{8y^3}\sqrt[10]{x^9}       [\because x^{\frac{1}{n}}=\sqrt[n]{x}]

Therefore, the required radical form is \sqrt[5]{8y^3}\sqrt[10]{x^9}.

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