Find the intersection of the intervals (-infinity, -10] and [-10, 8).

Step-by-step solution

1. 1

   Translate each interval into an inequality. $(-\infty, -10]$ is $\{x : x \le -10\}$ and $[-10, 8)$ is $\{x : -10 \le x < 8\}$. The bracket shapes matter: a square bracket includes the endpoint, a parenthesis excludes it.

2. 2

   Intersect the conditions. A number in both sets must satisfy $x \le -10$ and $x \ge -10$ simultaneously, which forces

   $x = -10$

   The third condition $x < 8$ is then automatic.

3. 3

   Confirm the endpoint really is included. $-10$ belongs to the first interval because its right end is closed, and to the second because its left end is closed. Had either bracket been a parenthesis the intersection would have been empty instead.

4. 4

   Write the answer both ways. As a set, $\{-10\}$; as an interval, the degenerate closed interval

   $[-10,\, -10]$

   It has length $0$ but is not empty — a distinction worth keeping straight.

5. 5

   Contrast with the union. The same two intervals have union $(-\infty, 8)$, since together they cover everything below $8$. Intersection collapses to a point precisely because the intervals meet end to end rather than overlapping.