How do you solve #4\times 8*2^{2}#?

1 Answer
Shwetank Mauria
Oct 3, 2016

#4xx8*2^2=128#

Explanation:

We can use PEMDAS, we should first start with Parentheses (none here), then Exponents, Multiplication, Division, Addition and Subtraction (last three too are not applicable here).

Hence #4xx8*2^2#

= #4xx8*4#

= #4xx32#

= #128#

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