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 only way to solve if it is equal to something
assuming that the teacher wanted you to make it equal to zero do
0=-3x^2-21x-54
remember if we can do
xy=0 then assume x and y=0
so factor
0=-3x^2-21x-54
undistribute the -3
0=-3(x^2+7x+18)
remember 0 times anything=0 so
x^2+7x+18 must equal zero
use quadratice formula which is
if you have
ax^2+bx+c=0 then
x=

x^2+7x+18
a=1
b=7
c=18
x=

x=

x=

i=√-1
x=

the zerose would be
x=

or
-6 5 -3 is the corrdante did those sir