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An object is launched from a platform.

Its height (in meters), xxx seconds after the launch, is modeled by:

h(x)=-5x^2+20x+60h(x)=−5x

2

+20x+60h, left parenthesis, x, right parenthesis, equals, minus, 5, x, squared, plus, 20, x, plus, 60

How many seconds after launch will the object land on the ground?

Respuesta :

abidemiokin

Answer:

6seconds

Step-by-step explanation:

Given the height of an object modeled by the expression

h(x)=-5x^2+20x+60

The height of the object on the ground is zero i.e h(x) = 0

On substituting

h(x)=-5x^2+20x+60

0 =-5x^2+20x+60

Multiply through by -1

5x^2-20x- 60 = 0

Divide through by 5

x^2 - 4x - 12 = 0

x^2 - 6x + 2x - 12 = 0

x(x-6)+2(x-6) = 0

(x-6)(x+2) = 0

x - 6 = 0 and x+2  = 0

x = 6 and -2

Since time cannot be negative, hence it takes 6seconds after launch for the object land to on the ground

jiaojason2020

Answer:

6

Step-by-step explanation:

It was correct on Khan.