MathDB

5

Part of 1989 AMC 8

Problems(1)

1989 AJHSME Problem 5

Source:

6/25/2011
−15+9×(6÷3)=-15+9\times (6\div 3) =
(A) −48(B) −12(C) −3(D) 3(E) 12\text{(A)}\ -48 \qquad \text{(B)}\ -12 \qquad \text{(C)}\ -3 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 12