Question

\\(\frac{a^5l^2}{3} \cdot \frac{b}{a^2}\\)
which expression is equivalent to the product for all \\(a > 0\\)?
choose 1 answer:
\\(\boldsymbol{\text{a}}\\) \\(\frac{bl^2}{3a^3}\\)
\\(\boldsymbol{\text{b}}\\) \\(\frac{a^7l^2}{3b}\\)
\\(\boldsymbol{\text{c}}\\) \\(\frac{bl^2}{3}\\)
\\(\boldsymbol{\text{d}}\\) \\(\frac{a^3bl^2}{3}\\)

Explanation:

Step1: Multiply the numerators and denominators

Multiply the numerators \(a^{5}b^{2} \cdot b\) and denominators \(3 \cdot a^{2}\), we get \(\frac{a^{5}b^{2} \cdot b}{3 \cdot a^{2}}\).

Step2: Simplify using exponent rules

For \(a\): \(a^{5} \div a^{2}=a^{5 - 2}=a^{3}\) (using \(a^{m}\div a^{n}=a^{m - n}\)).
For \(b\): \(b^{2}\cdot b = b^{2+1}=b^{3}\)? Wait, no, wait the original numerator for \(b\) is \(b^{2}\) and we multiply by \(b\) (which is \(b^{1}\)), so \(b^{2}\cdot b^{1}=b^{2 + 1}=b^{3}\)? Wait, no, wait the original expression is \(\frac{a^{5}b^{2}}{3}\cdot\frac{b}{a^{2}}\). So numerator: \(a^{5}b^{2}\times b=a^{5}b^{3}\), denominator: \(3\times a^{2}\). Then simplify \(a\): \(a^{5}/a^{2}=a^{3}\), \(b^{3}\)? Wait, no, maybe I misread the original problem. Wait the original problem is \(\frac{a^{5}l^{2}}{3}\cdot\frac{b}{a^{2}}\)? Wait, maybe it's a typo, maybe \(l\) is \(b\)? Wait no, the options have \(b\) and \(l\)? Wait, no, looking at the options, option D is \(\frac{a^{3}bl^{2}}{3}\)? Wait, maybe the original problem has \(l\) as a variable, but let's check again. Wait the first term is \(\frac{a^{5}l^{2}}{3}\), multiply by \(\frac{b}{a^{2}}\). So numerator: \(a^{5}l^{2}\times b=a^{5}bl^{2}\), denominator: \(3\times a^{2}\). Then simplify \(a\): \(a^{5}/a^{2}=a^{3}\), so we get \(\frac{a^{3}bl^{2}}{3}\), which is option D? Wait no, wait let's do it again.
Wait the expression is \(\frac{a^{5}l^{2}}{3} \times \frac{b}{a^{2}}\). Multiply numerators: \(a^{5}l^{2} \times b = a^{5}bl^{2}\). Multiply denominators: \(3 \times a^{2}\). Now, simplify the \(a\) terms: \(a^{5} / a^{2}=a^{5 - 2}=a^{3}\). So the expression becomes \(\frac{a^{3}bl^{2}}{3}\), which is option D? Wait no, wait option D is \(\frac{a^{3}bl^{2}}{3}\)? Wait the options:
A: \(\frac{bl^{2}}{3a^{3}}\)
B: \(\frac{a^{7}l^{2}}{3b}\)
C: \(\frac{bl^{2}}{3}\)
D: \(\frac{a^{3}bl^{2}}{3}\)

Ah, I see, so the first term is \(\frac{a^{5}l^{2}}{3}\), multiply by \(\frac{b}{a^{2}}\). So numerator: \(a^{5}l^{2} \times b = a^{5}bl^{2}\), denominator: \(3 \times a^{2}\). Then \(a^{5}/a^{2}=a^{3}\), so we have \(\frac{a^{3}bl^{2}}{3}\), which is option D.

Answer:

D. \(\frac{a^{3}bl^{2}}{3}\)