How do you evaluate #(1+4)times6#?

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Answer 1 · 2016-12-20T11:20:22

#30#

#(1+4) xx 6" "larr# brackets first

#= 5 xx 6" "larr# now multiply

#= 30#

You could also use the distributive law:

#(1xx6) + (4xx6)#

#= 6+24#

#=30#