How do you evaluate this expression #\frac { 4( 3h - 6) } { 1+ h }, #for #h = -2#?

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Answer 1 · 2018-01-10T07:35:43

Plug in #h=-2# for the expression #(4(3h-6))/(1+h)#.

When #h=-2#,

#:.=(4(3*-2-6))/(1+(-2)#

#=(4(-6-6))/-1#

#=(4*-12)/-1#

#=(-48)/-1#

#=48#

Answer 2 · 2018-01-10T07:46:46

#48#

#"substitute h = - 2 into the expression"#

#rArr(4((3xxcolor(red)(-2))-6))/(1+color(red)(-2))#

#=(4(-6-6))/(1-2)#

#=(4xx-12)/(-1)=(-48)/(-1)=48#