 
<translate> Nyere: x≥−3x≥−3 na x≤2 x≤2 (x+3)≥0(x+3)≥0 na (x−2)≤0(x−2)≤0 (x+3)≥0(x+3)≥0 na (2−x)≥0(2−x)≥0 Ya mere, azịza ya bụ (D). <eng> 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). 