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anonymous
 one year ago
Which line is perpendicular to 8x+3y=8
anonymous
 one year ago
Which line is perpendicular to 8x+3y=8

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marihelenh
 one year ago
Best ResponseYou've already chosen the best response.2I don't know if there are options, but if you find the slope of this line, just take the opposite reciprocal and that will give you the slope of the perpendicular line.

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1one approach is to put in the slope intercept form y = mx + b where m is the slope the product of the slopes for perpendicular lines = 1

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437439782206:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@marihelenh http://i1286.photobucket.com/albums/a605/kristinalucy/choices_zpsbbltuawp.png these are the choices

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@triciaal How wold you find the slope of 8x+3y=8 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0reverse it and put it into yintercept form y=MX+B (MX= SLOPE ) THEN YOU FLIP AND CHANGE THE SIGN :)

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.2This is how you would find the slope. The slope of the given line is 8/3. dw:1437440303973:dwdw:1437440334749:dw

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.2To have a perpendicular line, the slope of the other line has to be the opposite reciprocal, so the other slope has to be 3/8.

marihelenh
 one year ago
Best ResponseYou've already chosen the best response.2Now you can go through your answer choices and find the slopes of the equations given. If one is equal to 3/8, that is the perpendicular line.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you! Do you think you could possibly help me with another question?

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1sure open a new question
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