How do you evaluate $2 \cdot 6 + 3 \cdot 8 - 3$ $2\cdot 6+ 3\cdot 8- 3$ ?

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Answer 1 · 2017-03-29T15:32:18

Use PEMDAS to get to $33$ $33$ .

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Using PEMDAS, our first step $P$ $P$ , is parentheses. As that's not part of the question, we can ignore it. Same with exponents, the $E$ $E$ .

The next step is $M$ $M$ and $D$ $D$ , multiplication and division.

We would have to multiply $2 \cdot 6$ $2*6$ , to get $12$ $12$ . We'd then multiply $3 \cdot 8$ $3*8$ , to get $24$ $24$ .

Our current equation is $12 + 24 - 3$ $12 + 24 -3$ .

Our last step is $A$ $A$ and $S$ $S$ , addition and subtraction.

We'd just have to combine the last three terms.

$12 + 24 = 36$ $12+24 = 36$ , and $36 - 3 = 33$ $36-3 = 33$ . Our final answer is $33$ $33$ .

How do you evaluate 2\cdot 6+ 3\cdot 8- 32⋅6+3⋅8−32\cdot 6+ 3\cdot 8- 3?