Question

which of the following is equivalent to $36^{-\frac{1}{2}}$?
options:
-6
-18
$\frac{1}{6}$
$\frac{1}{18}$

Explanation:

Step1: Recall the negative exponent rule

The rule for negative exponents is \(a^{-n}=\frac{1}{a^{n}}\) (where \(a
eq0\) and \(n\) is a positive integer). So, for \(36^{-\frac{1}{2}}\), we can rewrite it using this rule as \(\frac{1}{36^{\frac{1}{2}}}\).

Step2: Simplify the exponent \(\frac{1}{2}\)

The exponent \(\frac{1}{2}\) represents the square root. So, \(36^{\frac{1}{2}}=\sqrt{36}\). We know that \(\sqrt{36} = 6\) because \(6\times6 = 36\).

Step3: Substitute back

Substituting \(\sqrt{36}=6\) into \(\frac{1}{36^{\frac{1}{2}}}\), we get \(\frac{1}{6}\)? Wait, no, wait. Wait, let's check again. Wait, \(36^{-\frac{1}{2}}=\frac{1}{36^{\frac{1}{2}}}=\frac{1}{\sqrt{36}}=\frac{1}{6}\)? But wait, the options have \(\frac{1}{6}\)? Wait, no, the options: one of them is \(\frac{1}{6}\)? Wait, the user's image: let's re - examine. Wait, the problem is \(36^{-\frac{1}{2}}\). Let's do it again.

Wait, \(a^{-\frac{m}{n}}=\frac{1}{a^{\frac{m}{n}}}=\frac{1}{\sqrt[n]{a^{m}}}\). So for \(36^{-\frac{1}{2}}\), \(n = 2\), \(m = 1\), \(a=36\). So \(\frac{1}{\sqrt{36^{1}}}=\frac{1}{\sqrt{36}}=\frac{1}{6}\)? But wait, the options: looking at the image, one of the options is \(\frac{1}{6}\)? Wait, no, the options in the image: the first option (top left) is \(\frac{1}{18}\), top right is \(- 6\), middle is \(-18\), bottom is \(\frac{1}{6}\). Wait, so let's recalculate.

Wait, \(36^{-\frac{1}{2}}=\frac{1}{36^{\frac{1}{2}}}=\frac{1}{\sqrt{36}}=\frac{1}{6}\). So the equivalent value is \(\frac{1}{6}\).

Answer:

\(\frac{1}{6}\) (the bottom - most option in the given choices)