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edieva2001
@edieva2001
July 2022
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Найти точки пересечения графика функции y=x+1 y=3-x^2
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sedinalana
Verified answer
X+1=3-x²
x²+x-2=0
x1+x2=-1 U x1*x2=-2
x1=-2⇒y1=-2+1=-1
x2=1⇒y2=1+1=2
(-2;-1);(1;2)
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Answers & Comments
Verified answer
X+1=3-x²x²+x-2=0
x1+x2=-1 U x1*x2=-2
x1=-2⇒y1=-2+1=-1
x2=1⇒y2=1+1=2
(-2;-1);(1;2)