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Sagot :
Answer:
(0,14) and (2,11)
Step-by-step explanation:
points are on the graph of the line:-
[tex]\tt{3x+2y=28 }[/tex] ⠀
we get,
[tex]\rightsquigarrow[/tex] [tex]\tt{ 2y=28-3x }[/tex] ⠀
[tex]\rightsquigarrow[/tex] [tex]\tt{y=\dfrac{28-3x}{2} }[/tex] ⠀
In,(0,14)
If we put the value ,x=0
y=[tex]\tt{\dfrac{ 28-3×0}{2}=\dfrac{28}{2}=14 }[/tex]
So,we get one point,(0,14)✔
In,(2,11)
If we put ,x=2
y=[tex]\tt{\dfrac{28-3×2}{2}=\dfrac{28-6}{2}=\dfrac{22}{2}=11 }[/tex]
so,we get another point (2,11)✔
in other given ponts{(1,12),(4,9),(8,4)},It is impossible. ⠀ ⠀
we have to find the line which are in the line!!
points are on the graph of the line:-
3x+2y=28 ⠀
we get,
2y=28−3x ⠀
[tex]{y=\dfrac{28-3x}{2} }[/tex]
In,(0,14)
If we put the value ,x=0
[tex]y={\dfrac{ 28-3×0}{2}=\dfrac{28}{2}=14 }[/tex]
so one point is (0,14)
In,(2,11)
If we put ,x=2
[tex]y={\dfrac{28-3×2}{2}=\dfrac{28-6}{2}=\dfrac{22}{2}=11 }[/tex]
so,we get another point (2,11)
in other given ponts{(1,12),(4,9),(8,4)},It is impossible. ⠀ ⠀
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