How do you solve $3- 1+ 6+ ( - 6)^{2} + | 3\cdot 2|$ ?

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Answer 1 · 2016-09-23T10:16:00

$50$

We have: $3 - 1 + 6 + (- 6)^(2) + abs(3 cdot 2)$

$= 3 - 1 + 6 + 36 + abs(3 cdot 2)$

$= 3 - 1 + 6 + 36 + abs(6)$

$= 3 - 1 + 6 + 36 + 6$

$= 3 + 6 + 36 + 6 - 1$

$= 9 + 42 - 1$

$= 51 - 1$

$= 50$

How do you solve 3- 1+ 6+ ( - 6)^{2} + | 3\cdot 2|3- 1+ 6+ ( - 6)^{2} + | 3\cdot 2|?