Question

multiply. enter your answer as a single, completely simplified rational expression with nonnegative exponents. \\(\frac{5x^3}{12xy^2} \cdot \frac{48xy^4}{x^2y^3}\\)

Explanation:

Step1: Multiply numerators and denominators

Multiply the numerators \(5x^3\) and \(48xy^4\), and the denominators \(12xy^2\) and \(x^2y^3\) separately. So we get \(\frac{5x^3 \cdot 48xy^4}{12xy^2 \cdot x^2y^3}\).

Step2: Simplify coefficients and variables

For coefficients: \(5\times48 = 240\), and \(12\) remains. So coefficient part is \(\frac{240}{12}=20\). For \(x\) terms: use the rule \(a^m \cdot a^n=a^{m + n}\) and \(\frac{a^m}{a^n}=a^{m - n}\). So \(x^3\cdot x=x^{3 + 1}=x^4\), and \(x\cdot x^2=x^{1+2}=x^3\), then \(\frac{x^4}{x^3}=x^{4 - 3}=x\). For \(y\) terms: \(y^4\) in numerator and \(y^2\cdot y^3=y^{2 + 3}=y^5\) in denominator, so \(\frac{y^4}{y^5}=y^{4 - 5}=y^{-1}=\frac{1}{y}\) (but we can also handle it as \(y^{4-(2 + 3)}=y^{-1}\), but we need nonnegative exponents, so we can rewrite later or handle during simplification). Combining all: \(20\times x\times\frac{y^4}{y^5}=20x\times\frac{1}{y}=\frac{20x}{y}\)? Wait, no, wait let's re - check. Wait \(y\) terms: numerator \(y^4\), denominator \(y^2\times y^3=y^{5}\), so \(\frac{y^4}{y^5}=y^{-1}\), but let's do it step by step again. Wait the original multiplication:

Numerator: \(5x^3\times48xy^4=(5\times48)x^{3 + 1}y^4 = 240x^4y^4\)

Denominator: \(12xy^2\times x^2y^3=12x^{1+2}y^{2 + 3}=12x^3y^5\)

Now simplify \(\frac{240x^4y^4}{12x^3y^5}\)

Divide coefficients: \(240\div12 = 20\)

For \(x\): \(\frac{x^4}{x^3}=x^{4-3}=x\)

For \(y\): \(\frac{y^4}{y^5}=y^{4 - 5}=y^{-1}=\frac{1}{y}\), but we can write it as \(\frac{20x}{y}\) (since \(y^{-1}=\frac{1}{y}\)) or we can also think of it as \(20x y^{-1}\), but the problem says nonnegative exponents, so we use the form with positive exponents in the denominator. Wait, but let's check again. Wait \(y^4\) divided by \(y^5\) is \(y^{4-5}=y^{-1}=\frac{1}{y}\), so the entire expression is \(20\times x\times\frac{1}{y}=\frac{20x}{y}\)

Wait, but let's check the calculation of coefficients again. \(5\times48 = 240\), \(240\div12 = 20\). For \(x\): \(x^3\times x=x^4\), divided by \(x\times x^2=x^3\), so \(x^4\div x^3=x\). For \(y\): \(y^4\) divided by \(y^2\times y^3=y^5\), so \(y^4\div y^5=y^{-1}=\frac{1}{y}\). So combining all, we get \(\frac{20x}{y}\)

Wait, but let's do it in one step:

\(\frac{5x^3}{12xy^2}\cdot\frac{48xy^4}{x^2y^3}=\frac{5\times48x^{3 + 1}y^4}{12\times1x^{1+2}y^{2+3}}=\frac{240x^4y^4}{12x^3y^5}\)

Simplify fraction:

\(\frac{240}{12}=20\), \(\frac{x^4}{x^3}=x\), \(\frac{y^4}{y^5}=\frac{1}{y}\)

So the result is \(20\times x\times\frac{1}{y}=\frac{20x}{y}\)

Answer:

\(\frac{20x}{y}\)