Answer:

<em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em><em>you</em><em>.</em><em>.</em><em>.</em><em>.</em>
Answer:
Im 100% sure that it’s (A)
Hello,
Answer D -3<x<27
Indeed:
1)
if x-12≥0 then x≥12 and
|x-12|=x-12x-12<15 → x<27
so 12≤x<27
2) if x-12<0 then
x<12 and|x-12|=-(x-12)<15
-x+12<15-x<3 → x>-3
So -3<x<12
Thus -3<x<12 U 12≤x<27 ⇔ -3<x<27
You should try using photomath.
Answer:
D. It would be D.
Step-by-step explanation:
Because anything divided by zero is zero, and I suppose zero is neither negative nor positive. or vise versa