56x7y5, if x+0 and y#.0?

Mathematics

2 answers:

Marianna [84]2 years ago

8 0

Answer:

A is the answer.

Step-by-step explanation:

We have

$\sqrt[3]{56 {x}^{7} {y}^{5} }$

Factor the radical by taking perfect squares of the real numbers and the exponents.

56 can be made up of perfect 8, x^7 can be made up of x^6 and y^5 can be made up of y^3. So we get

$\sqrt[3]{8 {x}^{6} {y}^{3} \times \sqrt{7xy {}^{2} } }$

Take the cube root of 8x^6y^3.

Which is 2x^2y.

$2 {x}^{2} y \sqrt[3]{7xy {}^{2} }$

A is the answer.

Scrat [10]2 years ago

8 0

B) answer is there

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