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Mathematics 18 Online
OpenStudy (anonymous):

Find the equation of the horizontal line through (5,3)

OpenStudy (mimi_x3):

y = 5 ( Not sure if im right though xD)

OpenStudy (anonymous):

How did you get that though?

OpenStudy (mimi_x3):

um , bcos x=5

OpenStudy (anonymous):

y=3

OpenStudy (anonymous):

for all values of x, y=3

OpenStudy (mimi_x3):

ohh im wrong then xD

OpenStudy (anonymous):

[here, jjwings22, y=-(5/3)+3

OpenStudy (anonymous):

SalaamPerc can you show me how you got -5/3?

OpenStudy (anonymous):

ok, a straight line graph is define by the th eequation... y=mx+c hop u understnd. where, m=gradient & c is the y-value when x=0, |dw:1314175903823:dw| therefore, y=mx+3, but the points are A(0,3) & B(5,0), Substitute B in the equation, which gives.. 0=m(5)+3, which gives m= -5/3, which is your gradient. Hence the equation y=-5/3m+3 is the equation of your graph, hop it helps ^_-

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