Question

multiple-choice question what is $y^{\frac{9}{2}}$ in radical form? $\sqrt{y^{9}}$ $\sqrt{9}$ $\sqrt{9^{y}}$ $\sqrt9{y^{2}}$

Explanation:

Step1: Recall the exponent to radical rule

The rule for converting a rational exponent \(a^{\frac{m}{n}}\) to radical form is \(\sqrt[n]{a^m}\) (or equivalently \((\sqrt[n]{a})^m\)). For \(y^{\frac{9}{2}}\), here \(a = y\), \(m = 9\), and \(n = 2\).

Step2: Apply the rule

Using the rule \(\sqrt[n]{a^m}\) with \(n = 2\) and \(m = 9\) for \(a=y\), we get \(\sqrt[2]{y^9}\) which is written as \(\sqrt{y^9}\) (since the index 2 for square root is often omitted and written as \(\sqrt{\quad}\) instead of \(\sqrt[2]{\quad}\)).

Now let's analyze the other options:

• Option \(\sqrt{9}\): This is \(\sqrt{9}\) which is a constant (equal to 3) and has nothing to do with \(y^{\frac{9}{2}}\).
• Option \(\sqrt{9^y}\): This is \(\sqrt{9^y}\) which is of the form \(\sqrt{(3^2)^y}=\sqrt{3^{2y}}\) or \(3^y\) in simplified radical form (but not related to \(y^{\frac{9}{2}}\)).
• Option \(\sqrt[9]{y^2}\): This would correspond to \(y^{\frac{2}{9}}\) (using the exponent - radical rule in reverse: \(\sqrt[n]{a^m}=a^{\frac{m}{n}}\)), which is not equal to \(y^{\frac{9}{2}}\).

Answer:

A. \(\sqrt{y^9}\)