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Sam launched a ball vertically upward from the ground at the same time a bird was on a linear path toward the ground. The equation that models the approximate position, in meters, of the ball is p = - 5t ^ 2 + 18t ; the equation that models the path of the bird is p=-2t+15.At

what position should the ball first cross the path of the bird?

I will mark brainliest Sam launched a ball vertically upward from the ground at the same time a bird was on a linear path toward the ground The equation that mo class=

Respuesta :

Answer:

[tex](1,13)[/tex]

Step-by-step explanation:

[tex]p=-5t^2+18t[/tex]

[tex]p=-2t+15[/tex]

[tex]-5t^2+18t=-2t+15[/tex]

[tex]-5t^2+20t=15[/tex]

[tex]-5t^2+20t-15=0[/tex]

[tex]-5(t^2-4t+3)=0[/tex]

[tex]t^2-4t+3=0[/tex]

[tex](t-1)(t-3)=0[/tex]

[tex]t_1=1[/tex], [tex]t_2=3[/tex]

Since we want the position to be when the ball FIRST crosses the path of the bird, we will need to choose [tex]t_1=1[/tex] as our solution.

Therefore, the position would be:

[tex]p=-2t+15[/tex]

[tex]p=-2(1)+15[/tex]

[tex]p=-2+15[/tex]

[tex]p=13[/tex]

[tex](1,13)[/tex]

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