How do you evaluate #h(x ) = \frac { 1} { 4} x - 2# when # x = 8#?
1 Answer
Dec 11, 2017
Explanation:
#"substitute x = 8 into "h(x)#
#rArrh(color(red)(8))=1/cancel(4)^1xxcancel(color(red)(8))^2-2#
#color(white)(rArrh(8))=2-2=0#
