For
, we have
So for
to be continuous at
, we require that the limit as
is equal to 4.
The absolute value inequality can be decomposed into two simpler ones.
x < 0
x > -8
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Which two inequalities can be used?</h3>
Here we start with the inequality:
3|x + 4| - 5 < 7
First we need to isolate the absolute value part:
3|x + 4| < 7 + 5
|x + 4| < (7 + 5)/3
|x + 4| < 12/3
|x + 4| < 4
The absolute value inequality can now be decomposed into two simpler ones:
x + 4 < 4
x + 4 > - 4
Solving both of these we get:
x < 4 - 4
x > -4 - 4
x < 0
x > -8
These are the two inequalities.
Learn more about inequalities:
brainly.com/question/24372553
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Answer:
8-2x I think
Step-by-step explanation:
You would first divide 39 by 3, ( 2 + 1 = 3) and get 13.
You would next take 5 away from 13 and get 8.
Then you have your answer, because you cannot simplify it more because both the numbers do not have the same variable.
