How do you evaluate #(2d-a)/(b-2c)# when #a=13, b=5, =-1# and #d=-4#?
1 Answer
Jul 22, 2017
Explanation:
#"substitute the given values into the expression"#
#rArr((2xx-4)-13)/(5-(-1))#
#=(-8-13)/(5+1)#
#=(-21)/6larr" cancel using 3"#
#=-7/2#
