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stich3 [128]
3 years ago
14

4(3x^2y^4)^3 / (2x^3y^5)^4

Mathematics
1 answer:
tatyana61 [14]3 years ago
8 0

Solving the expressions  \frac{4(3x^2y^4)^3}{(2x^3y^5)^4} we get \frac{27}{4x^{6}y^{8}}

Step-by-step explanation:

We need to Solve the expression: \frac{4(3x^2y^4)^3}{(2x^3y^5)^4}

Solving:

\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}

=\frac{4(3^3x^6y^{12})}{(2^4x^{12}y^{20})}\\=\frac{4(27x^6y^{12})}{(16x^{12}y^{20})}\\=\frac{108x^6y^{12}}{16x^{12}y^{20}}

Applying exponent rule: \frac{x^a}{x^b}=x^{a-b}

=\frac{108x^{6-12}y^{12-20}}{16}\\=\frac{27x^{-6}y^{-8}}{4}

Another exponent rule says: x^{-a}=\frac{1}{x^a}=

=\frac{27}{4x^{6}y^{8}}

So, solving the expressions  \frac{4(3x^2y^4)^3}{(2x^3y^5)^4} we get \frac{27}{4x^{6}y^{8}}

Keywords: Solving Exponents  

Learn more about Solving Exponents at:

  • brainly.com/question/13174260
  • brainly.com/question/13174254
  • brainly.com/question/13174259

#learnwithBrainly

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