Question

which of the following expressions are equivalent to \\(\frac{4^{-2}x^{0}}{2x^{-2}}\\). choose two.\\(\frac{2x^{2}}{16}\\)\\(\frac{x^{2}}{32}\\)\\(\frac{4^{-2}x^{2}}{2}\\)\\(\frac{-16x^{2}}{2}\\)

Explanation:

Step1: Simplify $x^0$ and $x^{-2}$

Any non-zero number to the 0 power is 1, and $x^{-n}=\frac{1}{x^n}$, so $x^0=1$ and $x^{-2}=\frac{1}{x^2}$. The expression becomes:
$$\frac{4^{-2} \cdot 1}{2 \cdot \frac{1}{x^2}} = \frac{4^{-2}x^2}{2}$$

Step2: Simplify $4^{-2}$

A negative exponent means reciprocal: $4^{-2}=\frac{1}{4^2}=\frac{1}{16}$. Substitute this in:
$$\frac{\frac{1}{16}x^2}{2} = \frac{x^2}{16 \cdot 2} = \frac{x^2}{32}$$

Step3: Match with options

Compare the simplified form to the given options:

1. $\frac{2x^2}{16}=\frac{x^2}{8}$ (not equivalent)
2. $\frac{x^2}{32}$ (equivalent)
3. $\frac{4^{-2}x^2}{2}$ (matches the result from Step1, equivalent)
4. $\frac{-16x^2}{2}=-8x^2$ (not equivalent)

Answer:

$\frac{x^2}{32}$, $\frac{4^{-2}x^2}{2}$