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
Answer:
y=-6x+48
Step-by-step explanation:
Answer:
answer=30
Step-by-step explanation:
by taking the length as x
1 millimetre = 10 metres
3 millimetres = x
10×3= 30 metres
HOPE THIS HELPS
You'd find the vertical asymptotes by seeing where the denominator equals zero; you can do so by factoring the denominator.
In this case, you can factor the denominator into (x+3)(x+2), so if you set each of those equal to zero you can find the equations of the vertical asymptotes (x=-3 and x=-2).