Question

HURRY PLEASE!

Which expression is equivalent to RootIndex 3 StartRoot 256 x Superscript 10 Baseline y Superscript 7 Baseline EndRoot?


a.4 x squared y (RootIndex 3 StartRoot x squared y cubed EndRoot)

b. 4 x cubed y squared (RootIndex 3 StartRoot 4 x y EndRoot)

c. 16 x cubed y squared (RootIndex 3 StartRoot x y EndRoot)

d. 16 x Superscript 5 Baseline y cubed (RootIndex 3 StartRoot y EndRoot)

Answer 1

hello :3 your answer is B

Explanation:

B for e2020

Answer 2

Option b.

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

Explanation:

we have the expression

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

Remember these properties

`\sqrt[n]{x^m} =x^{(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))^{(1)/(3)}=(256^{(1)/(3)})(x^{(10)/(3)})(y^{(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))^{(1)/(3)}=((4^3)(2^2)(x^9)(x)(y^6)(y))^{(1)/(3)}`

Applying properties of exponents

`((4^3)(2^2)(x^9)(x)(y^6)(y))^{(1)/(3)}=(4^3)^{(1)/(3)}(2^2)^{(1)/(3)}(x^9)^{(1)/(3)}(x)^{(1)/(3)}(y^6)^{(1)/(3)}(y)^{(1)/(3)}`

simplify

`(4)^{(3)/(3)}(2)^{(2)/(3)}(x)^{(9)/(3)}(x)^{(1)/(3)}(y)^{(6)/(3)}(y)^{(1)/(3)}`

`(4)(2)^{(2)/(3)}(x)^(3)(x)^{(1)/(3)}(y)^(2)(y)^{(1)/(3)}`

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