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simplify the expression below. \\(\\frac{(-9x)^2}{x^4y}\\)

Question

simplify the expression below.
\\(\frac{(-9x)^2}{x^4y}\\)

Explanation:

Step1: Simplify the numerator

First, we simplify \((-9x)^2\). Using the power of a product rule \((ab)^n = a^n b^n\), we have \((-9x)^2 = (-9)^2 \cdot x^2 = 81x^2\).
So the expression becomes \(\frac{81x^2}{x^4y}\).

Step2: Simplify the variables with exponents

For the \(x\)-terms, we use the quotient rule of exponents \(\frac{a^m}{a^n}=a^{m - n}\). Here, \(m = 2\) and \(n = 4\) for the \(x\)-terms. So \(\frac{x^2}{x^4}=x^{2 - 4}=x^{-2}=\frac{1}{x^2}\) (we can also think of it as subtracting the exponents when dividing like bases).
Putting it all together, we have \(\frac{81x^2}{x^4y}=\frac{81}{x^{4 - 2}y}=\frac{81}{x^2y}\) (or using the negative exponent form and then rewriting it as a positive exponent in the denominator).

Answer:

\(\frac{81}{x^{2}y}\)