Question

please help i dont get this at all

Answer 1

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:

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

Answer 2

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)  $\frac{20y^2a^2}{41x^{3}b}$.