How do you solve #(6^ { 2} - 2^ { 2} ) \div ( 4^ { 1} - 1^ { 2} + 1) \div ( 8+ 10)#?

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Answer 1 · 2016-12-07T21:06:19

#4/9#

#(6^2-2^2)=32#

We then do the second set of parenthesis:

#(4^1-1^2+1)=4#

Repeat this step for the last pair of parenthesis:

#(8+10)=18#

Now solve:

#32/4/18#

Now simplify the fraction:

#32/4=8# therefore the original fraction is #8/18#.

Since these two numbers both are factorable by #2# we can do the following:

#8/2/18/2=4/9#

Therefore the simplified factored version is:

Answer: #4/9#