the path of a toy rocket is modelled by the equation y=-x^2+6x+2, where x is the horizontal distance, in metres, travelled and y is the height, in metres, of the toy rocket above the ground. what is the maximum height of the toy rocket? at what horizontal distance does the maximum height occur?

The path of a toy rocket is defined by the relation y=-3x^2+11x+4, where x is the horizontal distance, in meters, traveled and y is the height, in metres, above the ground.

Determine the zeros of the relation.

-3x^2 + 11x + 4 = 0

---

3x^2 - 11x -4 = 0

Factor:

3x^2 - 12x + x - 4 = 0

---

3x(x-4) + x-4 = 0

---

(x-4)(3x+1) = 0

---

x = 4 or x = -1/3

f(x) = -3x^2+11x+4

How far has the rocket traveled horizontally when it lands on the ground?

4 ft.

What is the maximum height of the rocket above the ground, to the nearest hundredth of a meter?

---

max occurs when x = -b/2a = -11/(2*-3) = 11/6

height = f(11/6) = 14.083 ft

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it

what am i looking for finish the question.

step-by-step explanation: