Answer:
36 feet.
Step-by-step explanation:
We have been given that a ball is thrown upward from ground level. Its height h, in feet, above the ground after t seconds is 
. We are asked to find the maximum height of the ball.
We can see that our given equation is a downward opening parabola, so its maximum height will be the vertex of the parabola.
To find the maximum height of the ball, we need to find y-coordinate of vertex of parabola.
Let us find x-coordinate of parabola using formula 
.
So, the x-coordinate of the parabola is 
. Now, we will substitute in our given equation to find y-coordinate of parabola.
Therefore, the maximum height of the ball is 36 feet.
The answer is zero. Oliver only has one. So one few would be zero
1444km(distance) divided by 110km(scale) = 13.13cm.
Hope this helped!
Answer:
The slant height of the pyramid is 3√2 ft, or to the nearest tenth ft,
4.2 ft
Step-by-step explanation:
The equation for the volume of a pyramid of base area B and height h is
V = (1/3)·B·h. Here, V = 432 ft³, B = (12 ft)² and h (height of the pyramid) is unknown. First we find the height of this pyramid, and then the slant height.
V = 432 ft³ = (144 ft²)·h, so h = (432 ft³) / (144 ft²) = 3 ft.
Now to find the slant height of this pyramid: That height is the length of the hypotenuse of a right triangle whose base length is half of 12 ft, that is, the base length is 6 ft, and the height is 3 ft (as found above).
Then hyp² = (3 ft)(6 ft) = 18 ft², and the hyp (which is also the desired slant height) is hyp = √18, or √9√2, or 3√2 ft.
The slant height of the pyramid is 3√2 ft, or to the nearest tenth ft,
4.2 ft
