Question

simplify. assume that no variable equals 0.\\(\frac{(3x^{-2}y^{3})(5xy^{-8})}{(x^{-3})^{4}y^{-2}}\\)\\(\bigcirc\frac{15x^{11}}{y^{3}}\\)\\(\bigcirc\frac{15}{y^{3}x^{11}}\\)\\(\bigcirc15x^{11}y^{3}\\)\\(\bigcirc\frac{15x^{12}}{y^{2}}\\)

Explanation:

Step1: Simplify the numerator

Multiply the coefficients and use the product rule for exponents ($a^m \cdot a^n = a^{m + n}$) for $x$ and $y$ terms.
For the coefficients: $3 \times 5 = 15$
For the $x$ terms: $x^{-2} \cdot x = x^{-2 + 1} = x^{-1}$
For the $y$ terms: $y^3 \cdot y^{-8} = y^{3 + (-8)} = y^{-5}$
So the numerator becomes $15x^{-1}y^{-5}$

Step2: Simplify the denominator

First, simplify $(x^{-3})^4$ using the power rule for exponents ($(a^m)^n = a^{m \cdot n}$), so $(x^{-3})^4 = x^{-3 \times 4} = x^{-12}$
Then the denominator is $x^{-12}y^{-2}$

Step3: Divide the numerator by the denominator

Use the quotient rule for exponents ($\frac{a^m}{a^n} = a^{m - n}$)
For the coefficients: $\frac{15}{1} = 15$
For the $x$ terms: $\frac{x^{-1}}{x^{-12}} = x^{-1 - (-12)} = x^{11}$
For the $y$ terms: $\frac{y^{-5}}{y^{-2}} = y^{-5 - (-2)} = y^{-3} = \frac{1}{y^3}$ (but we can write it as $\frac{15x^{11}}{y^3}$ by combining)

Answer:

$\frac{15x^{11}}{y^3}$ (corresponding to the first option: $\boldsymbol{\frac{15x^{11}}{y^3}}$)