An object was thrown upward from the top of an 80ft tower. the height h of the object after t seconds is represented by the quadratic equation h=-16t^2 + 64t + 80. after how many seconds will the object hit the ground?

After 5 seconds object will hit the ground.

Step-by-step explanation:

An object was thrown upwards form an 80 feet tower.

The height h after t seconds is represented by the equation h = -16t²+ 64t + 80

We have to find the time taken by the object to hit the ground.

At h = 0, object will hit the ground.

Now we solve the equation for h = 0

-16t² + 64t + 80 = 0

-16(t² - 4t - 5) = 0

t² - 4t - 5 = 0

t² - 5t + t - 5 = 0

t(t - 5) + 1(t - 5) = 0

(t - 5)(t + 1) = 0

t = 5, (-1)

Since time can not be negative so t = (-1) can not be the answer.

So t = 5 seconds will be the answer.

(n^3 - n)

= n(n-1)(n+1)

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18

because 1.2 isnt half of 6 duh