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Question 1022117: state the excluded value for each function
y = x-1/12x+36
y = x+1/ 2x +3
y= 1/5x-2
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the excluded value is the one that makes the denominator equal to 0 in all of these.
in the first one, the denominator is 12x + 36 as far as i can see.
set that equal to 0 and you get 12x + 36 = 0
subtract 36 from both sides to get 12x = -36
divide both sides by 12 to get x = -36/12 which becomes x = -3.
the excluded value is x = -3.
in the second one, the denominator is 2x + 3 as far as i can see.
set that equal to 0 and you get 2x + 3 = 0
subtract 3 from both sides to get 2x = -3
divide both sides by 2 to get x = -3/2.
the excluded value is -3/2.
in the third one, the denominator is 5x-2 as far as i can see.
set that equal to 0 to get 5x - 2 = 0
add 2 to both side to get 5x = 2
divide both sides by 5 to get x = 2/5.
the excluded value is 2/5.
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