do work son
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- May 22, 2010
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the original problem is 48÷2(9+3)
everything after the division sign is in the denominator. if you argue that the (9+3) is not in the denominator, then the problem would have been written (48÷2)(9+3)
everything after the division sign is in the denominator. if you argue that the (9+3) is not in the denominator, then the problem would have been written (48÷2)(9+3)