A rocket is launched from a tower. The height of the rocket , y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the maximum height reached by the rocket, to the nearest tenth of a foot. Y=-16x^2+147x+118

Explanation:

The general equation

y = ax^2 + bx + c

has the vertex (h,k) such that

h = -b/(2a)

In this case, a = -16 and b = 147. This means,

h = -b/(2a)

h = -147/(2*(-16))

h = 4.59375

The x coordinate of the vertex is x = 4.59375

Plug this into the original equation to find the y coordinate of the vertex.

y = -16x^2+147x+118

y = -16(4.59375)^2+147(4.59375)+118

y = 455.640625

The vertex is located at (h,k) = (4.59375, 455.640625)

The max height of the rocket occurs at the vertex point. Therefore, the max height is y = 455.640625 feet which rounds to y = 455.6 feet

where's the diagram?

step-by-step explanation:

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %