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An object is launched from a platform. Its height (in meters), xx seconds after the launch, is modeled by: h(x)=-5(x-4)^2+180 How many seconds after being launched will the object hit the ground?

Respuesta :

wegnerkolmp2741o

Answer:

x=10

Step-by-step explanation:

h(x)=-5(x-4)^2+180

Set y = 0 to find when it hits the ground

0 = -5(x-4)^2+180

Subtract 180 from each side

-180 =-5(x-4)^2+180-180

-180 =-5(x-4)^2

Divide by -5

-180/-5 =-5/-5(x-4)^2

36 = (x-4)^2

Take the square root of each side

±sqrt(36) = sqrt( (x-4)^2)

±6 =  (x-4)

Add 4  to each side

4±6 =  (x-4)+4

4±6=x

x = 10 or -2

But time cannot be negative so x=10

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