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

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3x^2 - 11x -4 = 0

Factor:

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

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

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

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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?

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

height = f(11/6) = 14.083 ft

your question is incomplete, write all the numbers in the sequence that it tells you.  

step-by-step explanation:

the solution of the equation is 2.23 ⇒ answer b

step-by-step explanation:

* lets revise the meaning of exponential function

- the form of the exponential function is y = ab^x, where a ≠ 0, b > 0 ,  

  b ≠ 1, and x is any real number

- it has a constant base b

- it has a variable exponent x

- to solve an exponential equation, take the log or ln of both sides,  

  and solve for the variable

* lets solve the problem

∵

- the base is 7 and the exponent is x

- insert ㏑ in both sides

∴

- use the rule

∴ x ㏑(7) = ㏑(77)

- to find x divide both sides by ㏑(7)

∴

* the solution of the equation is 2.23 to the nearest 2 decimal places