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JcAlmighty
The answer is E.4

The equation is: [tex]n ^{-2}= \frac{1}{16} [/tex]

Since [tex] x^{-a}= \frac{1}{ x^{a} } [/tex], then [tex] n^{-2} = \frac{1}{n^{2} } [/tex]

Therefore, the equation is:
[tex] \frac{1}{ n^{2} } = \frac{1}{16} [/tex]
⇒ [tex]n^{2} = 16[/tex]
    [tex]n= \sqrt{16}=4 [/tex]

Answer:

Option B and E are correct

n could be 4 and -4

Step-by-step explanation:

Given the following:

[tex]n^{-2} = \frac{1}{16}[/tex]

we know that:

[tex]a^{-n} = \frac{1}{a^n}[/tex]

Then;

[tex]\frac{1}{n^2}=\frac{1}{16}[/tex]

By cross multiply we have;

[tex]16=n^2[/tex]

or

[tex]n^2=16[/tex]

⇒[tex]n = \pm\sqrt{16}[/tex]

Simplify:

[tex]n =\pm 4[/tex]

⇒n = -4 and n = 4

Therefore, n could be 4 and -4

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