If the -8 is under the square root, then...

OR
If the -8 is not under the square root, then...

Either way, we replace x with -3 and simplify.
For more information, refer to the direct substitution rule for limits.
Answer: 3/5
Step-by-step explanation: simplified completely
4h+9
We can plug 4 into h to result in this expression:
4*4+9
16+9=
25
The numbers would be -1, 0, 1, 2.
Hope this helps