Объяснение:
[tex](1;3)\ \ \ \ \ (-4;-1)\\\frac{x-x_1}{x_2-x_1} =\frac{y-y_1}{y_2-y_1}\\ \frac{x-1}{-4-1}=\frac{y-3}{-1-3} \\ \frac{x-1}{-5}=\frac{y-3}{-4}\ |*(-20)\\ 4*(x-1)=5*(y-3)\\4x-4=5y-15\\5y=4x+11\ |:5\\ y=0,8x+2,2\ \ \ \ \ \ \Rightarrow\\[/tex]
Ответ: B: 2,2.
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Объяснение:
[tex](1;3)\ \ \ \ \ (-4;-1)\\\frac{x-x_1}{x_2-x_1} =\frac{y-y_1}{y_2-y_1}\\ \frac{x-1}{-4-1}=\frac{y-3}{-1-3} \\ \frac{x-1}{-5}=\frac{y-3}{-4}\ |*(-20)\\ 4*(x-1)=5*(y-3)\\4x-4=5y-15\\5y=4x+11\ |:5\\ y=0,8x+2,2\ \ \ \ \ \ \Rightarrow\\[/tex]
Ответ: B: 2,2.
y=1÷x , x=4 , y=4 , y=0