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Verizon [17]
3 years ago
8

Please help i dont get this at all

Mathematics
2 answers:
Tom [10]3 years ago
4 0

Answer:

\frac{20y^{2} a^{2} }{41bx^{3} }

Step-by-step explanation:

1) Flip the second fraction and change the operation to multiplication:

\frac{5x^{2}y^{3}  }{2a^{5}b^{4}  } *  \frac{8a^{7}b^{3}  }{41x^{5}y  }

2) Cross cancel factors:

\frac{5y^{2}   }{2b  } *  \frac{8a^{2}  }{41x^{3}  }

3) Multiply:

\frac{40y^{2} a^{2} }{82bx^{3} }

4) Simplify:

\frac{20y^{2} a^{2} }{41bx^{3} }

Leya [2.2K]3 years ago
3 0

Answer:

The correct option is D) \frac{20y^2a^2}{41x^{3}b}.

Step-by-step explanation:

Consider the provided expression.

\frac{5x^2y^3}{2a^5b^4}\div \frac{41x^5y}{8a^7b^3}

Apply the fraction rule: \frac{a}{b}\div \frac{c}{d}=\frac{a}{b}\times \frac{d}{c}

Change the expression by using the above rule:

\frac{5x^2y^3}{2a^5b^4}\times \frac{8a^7b^3}{41x^5y}

\frac{5x^2y^3\times \:8a^7b^3}{2a^5b^4\times \:41x^5y}

\frac{40a^7y^3b^3x^2}{82a^5x^5b^4y}

\frac{20a^7y^3b^3x^2}{41a^5x^5b^4y}

Apply the exponent rule: \frac{x^a}{x^b}=\frac{1}{x^{b-a}}

\frac{20a^{7-5}y^{3-1}b^{3-4}x^{2-5}}{41}

\frac{20a^2y^2b^{-1}x^{-3}}{41}

Again apply the exponent rule:

\frac{20y^2a^2}{41x^{3}b}

Hence, the correct option is D)  \frac{20y^2a^2}{41x^{3}b}.

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