Answer:
Step-by-step explanation:
Our inequality is |125-u| ≤ 30. Let's separate this into two. Assuming that (125-u) is positive, we have 125-u ≤ 30, and if we assume that it's negative, we'd have -(125-u)≤30, or u-125≤30.
Therefore, we now have two inequalities to solve for:
125-u ≤ 30
u-125≤30
For the first one, we can subtract 125 and add u to both sides, resulting in
0 ≤ u-95, or 95≤u. Therefore, that is our first inequality.
The second one can be figured out by adding 125 to both sides, so u ≤ 155.
Remember that we took these two inequalities from an absolute value -- as a result, they BOTH must be true in order for the original inequality to be true. Therefore,
u ≥ 95
and
u ≤ 155
combine to be
95 ≤ u ≤ 155, or the 4th option
The answers to the questions
The domain of the graph is all real numbers because the value doesn’t have an end point.
Based on the picture, we know that ED is equal to DB. Therefore:
2 times ED=EB
2(10)=20
EB is 20