Answer:
The maximum height of the ball is 256m
Step-by-step explanation:
Given the equation of a pathway modelled as pathway can be modeled by h = -16t² + 128t
At maximum height, the velocity of the ball is zero.
velocity = dh/dt
velocity = -32t + 128
Since v = 0 at maximum height
0 = -32t+128
32t = 128
t = 128/32
t = 4seconds
The maximum height can be gotten by substituting t = 4 into the modelled equation.
h = -16t² + 128t
h = -16(4)²+128(4)
h = -16(16)+512
h = -256+512
h = 256m
(6x^3 + 4x-2) + (4x^2 + 3x^3 - 6)=
6x^3 + 4x-2 + 4x^2 + 3x^3 - 6=
9x^3+4x^2+4x-8
Answer:
1736
Step-by-step explanation:
Area= b x h
= 28 x 62
= 1736 m
