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1 vote
Help pleaseee!!!!!!!!

Help pleaseee!!!!!!!!-example-1
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User Jkalden
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2 Answers

5 votes

Answer:

1

Explanation:


((2x^3y^-3)^2)/(16x^7y^-2)\\ = (x^6y^-6)/(4x^7y^-2)\\ = (y^-4)/(4x)\\ = (1)/(4xy^4)

Value of numerator in final answer is 1 .

1 vote

Answer:


\textsf{Simplified:} \quad (1)/(4xy^4)

The value of the numerator is 1.

Explanation:

Given rational expression:


(\left(2x^3y^(-3)\right)^2)/(16x^7y^(-2))


\textsf{Apply the exponent rule} \;\;\boxed{(a^b)^c=a^(bc)}\;\;\textsf{to the numerator:}


\implies (2^2x^((3\cdot2))y^((-3\cdot 2)))/(16x^7y^(-2))


\implies (4x^(6)y^(-6))/(16x^7y^(-2))

Divide the numbers 4/16 = 1/4:


\implies (x^(6)y^(-6))/(4x^7y^(-2))


\textsf{Apply the exponent rule} \;\;\boxed{(1)/(a^n)=a^(-n)}:


\implies (x^(6)y^(-6)x^(-7)y^(2))/(4)

Collect like terms:


\implies (x^(6)x^(-7)y^(2)y^(-6))/(4)


\textsf{Apply the exponent rule} \;\;\boxed{a^b \cdot a^c=a^(b+c)}:


\implies (x^((6-7))y^((2-6)))/(4)


\implies (x^(-1)y^(-4))/(4)


\textsf{Apply the exponent rule} \;\;\boxed{a^(-n)=(1)/(a^n)}:


\implies (1)/(4xy^4)

Therefore, the value of the numerator is 1, and the value of the denominator is 4xy⁴.

answered
User Tyler Olthuizen
by
7.8k points

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