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A car with an initial velocity of 25 miles per hour begins to accelerate at a constant rate. The table shows the elapsed time after the car begins to accelerate and thevelocity of the car at a particular time.Velocity25Elapsed Time024681045SS6575Which linear function best models the car's velocity.V(t), in terms of the elapsed time (t) after the car begins to accelerate?O V(t) - 5t + 25V(t) = 10t + 25O V(t) = 25t +5Vit) - 25t + 10

Respuesta :

[tex]y\text{ + 2 = }\frac{1}{3}(x\text{ - 4) (option D)}[/tex]

Explanation:

Given points: (4, -2) and ( -8, -6)

Applying the point slope formula:

[tex]y-y_1=m(x-x_1)[/tex]

We need to first find the slope (m)

[tex]slope=m\text{ = }\frac{y_2-y_1}{x_2-x_1}[/tex][tex]\begin{gathered} x_1=4,y_1=-2,x_2=-8,y_2\text{ = }-6 \\ m\text{ = }\frac{-6-(-2)}{-8-4}\text{ = }\frac{-6+2}{-12} \\ m\text{ = }\frac{-4}{-12}\text{ = 1/3} \end{gathered}[/tex][tex]\begin{gathered} u\sin g\text{ the slope and any two points, we get point slope} \\ point\text{ (}4,-2)=(x_1,y_1) \\ \\ y\text{ - }(-2)\text{ = }\frac{1}{3}(x\text{ - 4)} \end{gathered}[/tex][tex]y\text{ + 2 = }\frac{1}{3}(x\text{ - 4) (option D)}[/tex]

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