Question

i inserted a picture of question 12graph the solution to the inequality.

Answer 1

Let's first determine the solution of the given inequality,

$\text{ x}^2\text{ }<\text{ }16$

We get,

$\text{ x}^2\text{ }<\text{ }16$$\text{ }\sqrt{\text{x}^2}\text{ }<\text{ }\sqrt{16}$$\text{ x }<\text{ }\pm\text{4}$

Since +4 is the highest, we will be starting the mark at +4, because it will still include all numbers less than -4.

Graphing this in a number line will be,