 
 <translate> Given: x≥−3x≥−3 and x≤2 x≤2
(x+3)≥0(x+3)≥0 and (x−2)≤0(x−2)≤0
(x+3)≥0(x+3)≥0 and (2−x)≥0(2−x)≥0
So, the answer is (D). <hau> Ganin: x≥−3x≥−3 da kuma x≤2 x≤2 (x+3) ≥0(x+3) ≥0 da (x−2) ≤0(x−2) ≤0 (x+3) ≥0(x+3) ≥0 da (2−x) ≥0(2−x) ≥0 Amsar ita ce (D). 