How do you evaluate #6j+(23+k)/(l-3)#, when #j=3, k=7,# and #l=33#?
1 Answer
Sep 25, 2016
19
Explanation:
We must evaluate the division before adding. Substitute the given values into the expression.
#(6xx3)+(23+7)/(33-3)#
#=18+30/30=18+1#
#rArr18+1=19" is the evaluation"#
