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vivado [14]
3 years ago
5

HURRY PLEASE!

Mathematics
2 answers:
Goshia [24]3 years ago
6 0

Answer:

Option b.

4x^3y^2\sqrt[3]{4xy}

Step-by-step explanation:

we have the expression

\sqrt[3]{256x^{10}y^{7}}

Remember these properties

\sqrt[n]{x^m} =x^{\frac{m}{n}}

(x^{m})^{n} =x^{m*n}

(x^{m})(x^{n})=x^{m+n}

so

\sqrt[3]{256x^{10}y^{7}}=(256x^{10}y^{7})^{\frac{1}{3}}=(256^{\frac{1}{3}})(x^{\frac{10}{3}})(y^{\frac{7}{3}})

Rewrite the expression

256=(4^3)(2^2)

x^{10}=(x^9)(x)

y^7=(y^6)(y)

substitute

(256x^{10}y^{7})^{\frac{1}{3}}=((4^3)(2^2)(x^9)(x)(y^6)(y))^{\frac{1}{3}}

Applying properties of exponents

((4^3)(2^2)(x^9)(x)(y^6)(y))^{\frac{1}{3}}=(4^3)^{\frac{1}{3}}(2^2)^{\frac{1}{3}}(x^9)^{\frac{1}{3}}(x)^{\frac{1}{3}}(y^6)^{\frac{1}{3}}(y)^{\frac{1}{3}}

simplify

(4)^{\frac{3}{3}}(2)^{\frac{2}{3}}(x)^{\frac{9}{3}}(x)^{\frac{1}{3}}(y)^{\frac{6}{3}}(y)^{\frac{1}{3}}

(4)(2)^{\frac{2}{3}}(x)^{3}(x)^{\frac{1}{3}}(y)^{2}(y)^{\frac{1}{3}}

4x^3y^2\sqrt[3]{4xy}

castortr0y [4]3 years ago
4 0

Answer:

hello :3 your answer is B

Step-by-step explanation:

B for e2020

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Answer:

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Step-by-step explanation:

In order to find this, first find the greatest common factor of the coefficients. Since 3 goes in evenly to both 15 and -18, then we know that it is a common factor.

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8 0
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