How do you evaluate $16 + [9 \times (3 - 1) + 8] \div 2$ $16+ [ 9\times ( 3- 1) + 8] \div 2$ ?

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Answer 1 · 2017-01-04T15:45:41

29

$16 + [9 \times (3 - 1) + 8] \div 2$ $16+[9 xx (3-1)+8]-:2$

$= 16 + [18 + 8] \div 2$ $=16+[18+8]-:2$

$= 16 + 26 \div 2$ $=16+26-:2$

$16 + 13$ $16+13$

$= 29$ $=29$

How do you evaluate 16+[9×(3−1)+8]÷216+ [ 9\times ( 3- 1) + 8] \div 2?