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.

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1

step-by-step explanation:

there are 36 inches in a yard

25 : 36 = 0.69 yards (you can round to 0.70)