How do you evaluate #4y ^ { 2} + 5# when #y=3#?

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Answer 1 · 2018-03-04T13:37:15

#41#

#"substitute y = 3 into the expression"#

#rArr(4xxcolor(red)(3)^2)+5#

#=(4xx9)+5=36+5=41#

Answer 2 · 2018-03-04T13:39:30

#= 41 #

The first thing is to use the knowledge of squaring:

#y^2 = y xx y #

Hence using "bidmas":

#4y^2 + 5 = (4xxyxxy) + 5 #

We can just replace #y# with #3#:

#= (4xx3xx3) + 5 #

#= (4xx9) +5 #

#= 36 + 5 #

#= 41 #