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3 Exponents; Question 3.11

Algebra 2
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3 Exponents; Question 3.11

Postby Eigenvalue » Fri Oct 09, 2026 10:34 pm

3.11. Simplify:
a) (xy³)⁴(3x²y)³

b) [tex]\frac{(x²y³)³}{(2x³y⁴)²}[/tex]
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Re: 3 Exponents; Question 3.11

Postby Eigenvalue » Fri Oct 09, 2026 10:36 pm

a) (xy³)⁴(3x²y)³
=(x⁴y¹²)(27x⁶y³)
=27x¹⁰y¹⁵

b) [tex]\frac{(x²y³)³}{(2x³y⁴)²}[/tex]

=[tex]\frac{x⁶y⁹}{4x⁶y⁸}[/tex]

=[tex]\frac{y}{4}[/tex]

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