Dakotajnorris
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Which expression is equivalent to (StartFraction (3 x y Superscript negative 5 Baseline) cubed Over (x Superscript negative 2 Baseline y squared) Superscript negative 4 Baseline) EndFraction) Superscript negative 2? Assume StartFraction x Superscript 10 Baseline y Superscript 14 Baseline Over 729 EndFraction.
StartFraction x Superscript 10 Baseline y Superscript 14 Baseline Over 729 EndFraction
StartFraction x Superscript 22 Baseline Over 18 y Superscript 46 Baseline EndFraction
StartFraction 729 Over x Superscript 10 Baseline y Superscript 14 Baseline EndFraction
StartFraction 18 y Superscript 46 Baseline Over x Superscript 22 Baseline EndFraction

Respuesta :

Option a: [tex]\frac{x^{10} y^{14}}{729}[/tex] is the correct answer.

Explanation:

The expression is [tex]\left(\frac{\left(3 x y^{-5}\right)^{3}}{\left(x^{-2} y^{2}\right)^{-4}}\right)^{-2}[/tex]

We shall simplify the expression, to determine the expression which is equivalent to [tex]\left(\frac{\left(3 x y^{-5}\right)^{3}}{\left(x^{-2} y^{2}\right)^{-4}}\right)^{-2}[/tex]

Multiplying the powers, we get,

[tex]\left(\frac{\left3^{3} x^{3} y^{-15}\right}{x^{8} y^{-8}}\right)^{-2}[/tex]

Again multiplying the powers, we get,

[tex]\left\frac{\left3^{-6} x^{-6} y^{30}\right}{x^{-16} y^{16}}\right[/tex]

Dividing the fractions, we have,

[tex]\left\frac{\left3^{-6} y^{14}\right}{x^{-10} }\right[/tex]

Applying the exponent rule, [tex]a^{-b}=\frac{1}{a^{b}}[/tex], we have,

[tex]\frac{x^{10} y^{14}}{3^{6} }[/tex]

Hence, the expression can be written as

[tex]\frac{x^{10} y^{14}}{729}[/tex]

Thus, the expression which is equivalent to [tex]\left(\frac{\left(3 x y^{-5}\right)^{3}}{\left(x^{-2} y^{2}\right)^{-4}}\right)^{-2}[/tex] is [tex]\frac{x^{10} y^{14}}{729}[/tex]

Hence, Option a is the correct answer.

MrRoyal

The equivalent expression of [tex](\frac{(3xy^{-5})^3}{(x^{-2}y^2)^{-4}})^{-2[/tex] is [tex]\frac{x^{10}y^{14}}{729}[/tex]

What are equivalent expressions?

Equivalent expressions are expressions that have the same value, when evaluated or simplified

The expression is given as;

[tex](\frac{(3xy^{-5})^3}{(x^{-2}y^2)^{-4}})^{-2[/tex]

Start by evaluating the exponents

[tex](\frac{(3xy^{-5})^3}{(x^{-2}y^2)^{-4}})^{-2} = (\frac{3^3x^3y^{-15}}{x^{-8}y^8})^{-2}[/tex]

[tex](\frac{(3xy^{-5})^3}{(x^{-2}y^2)^{-4}})^{-2} = (\frac{27x^3y^{-15}}{x^{8}y^{-8}})^{-2}[/tex]

Evaluate the exponents, again

[tex](\frac{(3xy^{-5})^3}{(x^{-2}y^2)^{-4}})^{-2} = \frac{27^{-2}x^{-6}y^{30}}{x^{-16}y^{16}}[/tex]

Apply the law of indices

[tex](\frac{(3xy^{-5})^3}{(x^{-2}y^2)^{-4}})^{-2} = 27^{-2}x^{16-6}y^{-16+30}[/tex]

Evaluate the exponents

[tex](\frac{(3xy^{-5})^3}{(x^{-2}y^2)^{-4}})^{-2} = 27^{-2}x^{10}y^{14}[/tex]

Evaluate 27^-2

[tex](\frac{(3xy^{-5})^3}{(x^{-2}y^2)^{-4}})^{-2} = \frac{x^{10}y^{14}}{729}[/tex]

Hence, the equivalent expression is [tex]\frac{x^{10}y^{14}}{729}[/tex]

Read more about equivalent expressions at:

https://brainly.com/question/15775046

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