If $$ p=\frac{\sqrt{x+2}}{\sqrt{16-x^{2}}} $$, for which values of $$ x $$ will $$ p $$ be real?

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1. For the numerator $$ \sqrt{x+2} $$ to be real, the radicand must be non-negative:
   $$
   x+2 \ge 0 \quad \Longrightarrow \quad x \ge -2.
   $$

2. For the denominator $$ \sqrt{16-x^2} $$ to be real, the radicand must be non-negative:
   $$
   16-x^2 \ge 0 \quad \Longrightarrow \quad x^2 \le 16.
   $$
   This implies:
   $$
   -4 \le x \le 4.
   $$

3. Additionally, the denominator cannot be zero because division by zero is undefined. Thus, we require:
   $$
   \sqrt{16-x^2} \ne 0 \quad \Longrightarrow \quad 16-x^2 > 0 \quad \Longrightarrow \quad x^2 < 16,
   $$
   which gives:
   $$
   -4 < x < 4.
   $$

4. Combining the conditions from the numerator and the denominator, we need:
   $$
   x \ge -2 \quad \text{and} \quad -4 < x < 4.
   $$
   The intersection of these intervals is:
   $$
   -2 \le x < 4.
   $$

Hence, $$ p $$ is real for all $$ x $$ satisfying:
$$
-2 \le x < 4.
$$
$$ p $$ is real when $$ x $$ is between -2 and 4, including -2 but not including 4.

If , for which values of will be real?

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