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navik [9.2K]
3 years ago
13

Which expression is equivalent to ? Assume .

Mathematics
2 answers:
Ymorist [56]3 years ago
8 0

Answer: (x^{\dfrac{2}{7}})(y^{-\dfrac{3}{5}})

Step-by-step explanation:

The given expression : \dfrac{\sqrt[7]{x^2}}{\sqrt[5]{y^3} }

Law of radicals :-

\sqrt[n]{a}=a^{\frac{1}{n}}\\\\\sqrt[n]{a^m}=a^{\frac{m}{n}}

Law of exponent:

\dfrac{1}{a^n}=a^{-n}

Using the above law of radicals and law of exponent we have,

\dfrac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}\\\\=(x^{\frac{2}{7}})(y^{-\frac{3}{5}})

hichkok12 [17]3 years ago
5 0

\text{Use}\ \sqrt[n]{a^m}=a^{\frac{m}{n}}\ \text{and}\ a^{-1}=\dfac{1}{a}\\\\\dfrac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(x^{\frac{2}{7}\right)\left(\dfrac{1}{y^\frac{3}{5}}\right)=\left(x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)

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Answer:

see explanation

Step-by-step explanation:

Using the exact values of the trigonometric ratios

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Using the sine ratio on the right triangle on the left

sin60° = \frac{opposite}{hypotenuse} = \frac{a}{4\sqrt{3} } = \frac{\sqrt{3} }{2}

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Using the cosine ratio on the same right triangle

cos60° = \frac{adjacent}{hypotenuse} = \frac{c}{4\sqrt{3} } = \frac{1}{2}

Cross- multiply

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c = 2\sqrt{3}

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Using the sine/cosine ratios on the right triangle on the right

sin45° = \frac{a}{b} = \frac{6}{b} = \frac{1}{\sqrt{2} }

Cross- multiply

b = 6\sqrt{2}

cos45° = \frac{d}{b} = \frac{d}{6\sqrt{2} } = \frac{1}{\sqrt{2} }

Cross- multiply

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a = 6, b = 6\sqrt{2}, c = 2\sqrt{3}, d = 6

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3 years ago
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Step-by-step explanation:

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Answer:

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