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Viefleur [7K]
3 years ago
7

Plzz Help ASAP!! i posted the picture?

Mathematics
2 answers:
Andre45 [30]3 years ago
8 0

Hi!


So the original formula for a problem likes this or a quadratic is:


ax^2 + bx+ c


so a is the accelaration; b is the initial velocity; and c is the initial height


so none of those are the zeros, which means c is out


zeros are the x values when y is 0, that also means when the parabola crosses the x axis


so d is out too because the maximum height is the vertex which mean that there would probably be a y value. in some cases the vertex is the zero, but not here


B is out because x isn't time it is distance

your answer is A!


Hope this helps!

anastassius [24]3 years ago
5 0

You can write the function as


h = -2x^2 + 3x + 5


h is the height above the ground the shot reaches as a function of the horizontal distance it travels.


Solve -2x^2 + 3x + 5 = 0 for x.


-2x^2 + 3x + 5 = 0


2x^2 - 3x - 5 = 0


(2x - 5)(x + 1) = 0


2x - 5 = 0 or x + 1 = 0


2x = 5 or x = -1


x = 2.5 or x = -1


At x = 2.5 ft, the height was 0. That is when the shot fell on the ground.

From the zero of the function at x = 2.5 ft, we find out the shot traveled 2.5 ft horizontally.


Answer: A. the horizontal distance covered by the shot

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