An object is launched from a platform. it's height in meters, x 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?

We are  given function: h(x)=-5(x-4)^2+180 for the height in meters, x seconds after the launch.

We need to find the time when it hit the ground.

The height of the object would be 0, when it hit the ground.

So, we need to set the given function equal to 0 and solve for x.

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

Subtracting both sides by 180.

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

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

Dividing both sides by -5, we get

(x-4)^2 = 36.

Taking square root on both sides, we get

x-4 = + - 6

x-4 =6 and x-4 =-6

Adding 4 on both sides in both equations, we get

x-4+4 =6+4 and x-4+6 =-6+4

x=10 and x=-2.

x represents time in seconds. So we can't take a negative value for time.

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5