How do you solve #(10^2-4*8)-:(8+9)#?

1 Answer
EZ as pi
Aug 10, 2016

#4#

Explanation:

Rather than just blindly following PEDMAS/PEMDA/BODMAS etc, just realise that the operations we do with numbers are not equally strong. The strongest operations are done first.

Powers and roots have the biggest effect on a number and are thetherefore the strongest, then comes multiplication and division, and the weakest operations are addition and subtraction.

If a weaker operation must be done first, parentheses are used to show this.

Always count the number of terms first. Each term must simplify to a single answer and these are added or subtracted in the LAST step.

#(10^2-4*8)-:(8+9)# has one term, but there are different operations.
Note that they can be done at the same time because they are independent from each other.

#(color(red)(10^2)color(blue)(-4xx8))-:color(magenta)((8+9))#

=#(color(red)(100)color(blue)(-32))-:color(magenta)((17))#

=#68 div color(magenta)17#

=#4#

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