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Which expression is equivalent to the given expression? Assume the denominator does not equal zero. [tex]\frac{12 x^3 y^4}{6 z^2 y^3}[/tex]

A. [tex]2 x^3 y^2[/tex]
B. [tex]\frac{2}{x^6 y^2}[/tex]
C. [tex]\frac{2}{x^3 y^2}[/tex]
D. [tex]2 x^6 y^2[/tex]


Sagot :

To simplify the expression [tex]\(\frac{12 x^3 y^4}{6 z^2 y^3}\)[/tex], let's follow step-by-step:

1. Simplify the coefficients:

The coefficients in the fraction are [tex]\(12\)[/tex] (numerator) and [tex]\(6\)[/tex] (denominator). Simplifying these:
[tex]\[ \frac{12}{6} = 2 \][/tex]

2. Simplify the powers of [tex]\(x\)[/tex]:

The power of [tex]\(x\)[/tex] in the numerator is [tex]\(x^3\)[/tex] and in the denominator, there is no [tex]\(x\)[/tex].
[tex]\[ x^3 \][/tex]

3. Simplify the powers of [tex]\(y\)[/tex]:

The power of [tex]\(y\)[/tex] in the numerator is [tex]\(y^4\)[/tex] and in the denominator, it's [tex]\(y^3\)[/tex]. By applying the properties of exponents ([tex]\(a^m / a^n = a^{m-n}\)[/tex]):
[tex]\[ y^4 / y^3 = y^{4-3} = y^1 = y \][/tex]

4. Simplify the powers of [tex]\(z\)[/tex]:

The power of [tex]\(z\)[/tex] in the numerator is not present, but in the denominator, it's [tex]\(z^2\)[/tex]. Since the [tex]\(z\)[/tex] term is only in the denominator, it remains as [tex]\(z^2\)[/tex], and we represent it with a negative exponent for simplicity:
[tex]\[ z^2 = z^{-2} \][/tex]

Putting it all together, the simplified form of the expression is:
[tex]\[ 2 x^3 y z^{-2} \][/tex]

Or equivalently:
[tex]\[ \frac{2 x^3 y}{z^2} \][/tex]

Thus, the correct answer is:
[tex]\[ \boxed{\frac{2 x^3 y}{z^2}} \][/tex]

However, this specific form isn’t given as an option directly. Instead, our expression doesn't directly match the available choices A through D. After re-evaluating to see if there could have been common letters, it's evident each could have been over looked at it's simplest form while evaluating with the provided choices, matching correctly as other kind fraction simplification policies could validate clear result as:
\boxed{2 x^3 y \frac{1}{z^2}}