Question

1. $x^{-3}y^{0} cdot 2x^{-4}y^{2} =$

Explanation:

Step1: Simplify $y^0$ (any non-zero number to 0 power is 1)

$y^0 = 1$, so the expression becomes $x^{-3} \cdot 1 \cdot 2x^{-4}y^2 = 2x^{-3}x^{-4}y^2$

Step2: Combine like $x$-terms using exponent rule $a^m \cdot a^n = a^{m+n}$

$x^{-3} \cdot x^{-4} = x^{-3 + (-4)} = x^{-7}$, so the expression is $2x^{-7}y^2$

Step3: Rewrite negative exponent as reciprocal (optional standard form)

$x^{-7} = \frac{1}{x^7}$, so $2x^{-7}y^2 = \frac{2y^2}{x^7}$

Answer:

$\frac{2y^2}{x^7}$ (or $2x^{-7}y^2$)