Arock is thrown straight up into the air from a height of 4 feet, the height of the rock above the ground , in feet, t seconds after it is thrown is given by -16t^2 + 56t + 4. for how many seconds will the height of the rock be at least 28 feet above the ground?

2.5 seconds.

Step-by-step explanation:

Find the value of t when the height is 28 feet:

-16t^2 + 56t + 4  = 28

-16t^2 +  56t - 24 = 0

-8(2t^2 - 7t + 3) = 0

-8(2t - 1)(t - 3) = 0

t = 0.5,  3 seconds.

So on the upward journey the rock is at 28 feet at 0.5 second after the throw and at 3 seconds it is at 28 feet again  while it is falling back.

Therefore the period when it is at least 28 feet above the ground is 3.0 - 0.5 = 2.5 seconds.

step-by-step explanation:

given perimeter: 28cm

length: 9cm

atp,

2(l+w)=p of the reactangle

2(9+x)=28

18+2x=28

2x=28-18

x=10/2

x=5

therefore width is 5 cm

d. 145*2 is 290 than the total angle is 360 then minus 290 then divide by two