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}\\)
check answer

Explanation:

Step1: Multiply numerators and denominators

Multiply the numerators \(5x^3 \cdot 48xy^4\) and denominators \(12xy^2 \cdot x^2y^3\).
Numerator: \(5x^3 \cdot 48xy^4 = 240x^{3 + 1}y^4 = 240x^4y^4\)
Denominator: \(12xy^2 \cdot x^2y^3 = 12x^{1 + 2}y^{2 + 3} = 12x^3y^5\)
So we have \(\frac{240x^4y^4}{12x^3y^5}\).

Step2: Simplify coefficients and exponents

Simplify the coefficient \(\frac{240}{12} = 20\).
For \(x\): \(x^{4 - 3} = x^1 = x\) (using \(a^m / a^n = a^{m - n}\)).
For \(y\): \(y^{4 - 5} = y^{-1}\), but we need nonnegative exponents, so \(y^{-1} = \frac{1}{y}\)? Wait, no, wait—wait, \(4 - 5 = -1\), but maybe I made a mistake. Wait, no, let's re - check the exponents. Wait, numerator \(y\) exponent is 4, denominator is 5. So \(y^{4-5}=y^{-1}\), but we can also simplify before combining. Wait, maybe I should simplify the fractions before multiplying. Let's try that approach.

Alternative Step1: Simplify before multiplying.
\(\frac{5x^3}{12xy^2} \cdot \frac{48xy^4}{x^2y^3}\)
Simplify the coefficients: \(\frac{5}{12} \cdot \frac{48}{1} = 5\times4 = 20\) (since \(48\div12 = 4\)).
For \(x\): \(\frac{x^3}{x} \cdot \frac{x}{x^2}=\frac{x^{3 - 1 + 1}}{x^2}=\frac{x^3}{x^2}=x^{3 - 2}=x\) (using \(a^m \cdot a^n=a^{m + n}\) and \(a^m/a^n=a^{m - n}\)).
For \(y\): \(\frac{1}{y^2} \cdot \frac{y^4}{y^3}=\frac{y^4}{y^{2+3}}=\frac{y^4}{y^5}=y^{4 - 5}=y^{-1}\)? No, wait, that's not right. Wait, no, \(\frac{y^4}{y^2 \cdot y^3}=\frac{y^4}{y^{5}}=y^{4 - 5}=y^{-1}\), but we need nonnegative exponents. Wait, no, I think I messed up. Wait, \(y^4\) divided by \(y^2\) is \(y^{4 - 2}=y^2\), then times \(y^4\) over \(y^3\)? No, original is \(\frac{5x^3}{12xy^2}\times\frac{48xy^4}{x^2y^3}\). So when multiplying, it's \(\frac{5x^3\times48xy^4}{12xy^2\times x^2y^3}\). So \(x\) terms: \(x^3\times x = x^{4}\), denominator \(x\times x^2=x^3\), so \(x^{4}/x^3=x\). \(y\) terms: \(y^4\) in numerator, \(y^2\times y^3 = y^5\) in denominator, so \(y^4/y^5=y^{-1}\), but we can write it as \(\frac{1}{y}\), but then multiply by 20x. Wait, no, that can't be. Wait, no, wait \(4 - 5=-1\), but maybe I made a mistake in the exponent subtraction. Wait, no, \(a^m\div a^n=a^{m - n}\). So \(y^4\div y^5 = y^{4 - 5}=y^{-1}=\frac{1}{y}\). But the problem says "nonnegative exponents". Wait, maybe I made a mistake in the exponent calculation. Wait, numerator \(y\) is \(y^4\), denominator \(y\) is \(y^2\times y^3=y^5\). So \(y^4 - y^5\) exponent is \(4-5=-1\). But maybe there's a miscalculation. Wait, no, let's do it again.

Wait, the original expression: \(\frac{5x^3}{12xy^2}\times\frac{48xy^4}{x^2y^3}\)

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

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

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

Simplify the coefficient: \(240\div12 = 20\)

For \(x\): \(x^{4-3}=x^1=x\)

For \(y\): \(y^{4 - 5}=y^{-1}\), but we can rewrite \(y^{-1}\) as \(\frac{1}{y}\), but then the expression is \(20x\times\frac{1}{y}=\frac{20x}{y}\)? Wait, no, that seems odd. Wait, maybe I messed up the \(y\) exponents. Wait, numerator \(y\) is 4, denominator \(y\) is \(2 + 3 = 5\). So \(4-5=-1\). But the problem says "nonnegative exponents". Wait, maybe there's a mistake in my calculation. Wait, no, let's check the multiplication again.

Wait, \(\frac{5x^3}{12xy^2}\times\frac{48xy^4}{x^2y^3}=\frac{5\times48\times x^3\times x\times y^4}{12\times x\times x^2\times y^2\times y^3}\)

Simplify the coefficients: \(5\times48 = 240\), \(240\div12 = 20\)

Simplify \(x\) terms: \(x^3\times x=x^{4}\), \(x\times x^2=x^3\), so \(x^{4}\div x^3=x\)

Simplify \(y\) terms: \(y^4\div(y^2\times y^3)=y^4\div y^5=y^{4 - 5}=y^{-1}=\frac{1}{y}\)

But then the expression is \(20\times x\times\frac{1}{y}=\frac{20x}{y}\). Wait, but that has a positive exponent on \(x\) and a positive exponent in the denominator for \(y\), which is allowed (since the exponent in the denominator is positive, and the overall expression has nonnegative exponents in the numerator). Wait, the problem says "nonnegative exponents"—so exponents in the numerator should be nonnegative, and exponents in the denominator can be positive (since \(y^1\) in the denominator is \(y^1\), which is nonnegative exponent). Wait, actually, \(y^{-1}=\frac{1}{y^1}\), so the expression \(\frac{20x}{y}\) has \(x\) with exponent 1 (nonnegative) and \(y\) with exponent 1 in the denominator (which is a positive, hence nonnegative, exponent in the denominator's term). So that's acceptable.

Wait, but let's check with another approach. Let's simplify the fractions before multiplying.

\(\frac{5x^3}{12xy^2}\times\frac{48xy^4}{x^2y^3}\)

Simplify \(\frac{5}{12}\times\frac{48}{1}=5\times4 = 20\) (since \(48\div12 = 4\))

For \(x\): \(\frac{x^3}{x}\times\frac{x}{x^2}=\frac{x^{3-1}\times x}{x^2}=\frac{x^2\times x}{x^2}=\frac{x^3}{x^2}=x\) (using \(a^m\div a^n=a^{m - n}\))

For \(y\): \(\frac{1}{y^2}\times\frac{y^4}{y^3}=\frac{y^4}{y^{2 + 3}}=\frac{y^4}{y^5}=y^{4 - 5}=y^{-1}=\frac{1}{y}\)

So putting it all together: \(20\times x\times\frac{1}{y}=\frac{20x}{y}\)

Answer:

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