Equations with absolute value:

Where k is a positive number; if k is a negative number, the equation is impossible (absolute value is always positive).
How to solve:


Then:
1. |x+7|=12
x+7=12 V -x-7=12
x=5 V -x=19
x=5 V x=-19
{-19, 5}
2. |2x+4|=8
2x+4=8 V -2x-4=8
2x=4 V -2x=12
x=2 V x=-6
3. 3|3k|=27
3×3k=27 V 3×(-3k)=27
9k=27 V -9k=27
k=3 V k=-3
{-3, 3}
4. 5|b+8|=30
5×(b+8)=30 V 5×(-b-8)=30
5b+40=30 V -5b-40=30
5b=-10 V -5b=70
b=-2 V b=-14
{-14, -2}
5. |m+9|=5
m+9=5 V -m-9=5
m+9=5 V m+9=-5
Your answer would be “a”.
In order to receive a positive solution in the following equation, you must include a double negative to have them cancel each other out and come up with a positive value. Sorry if this is confusing it’s easier to explain this process in person rather than online but I hope this helps anyways :)
B is 26
c is 15
X is 80
g is 30
x is -5
W is -4
r is 45
k is 12
f is 50
Answer:
C. 4 radians
Step-by-step explanation:

Ctrl Angle = ∅
∅ = 
in radians:
∅ = 
Hope this helps