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
Y=mx+b
so is 2
the last number of the equation
Answer:
5 terms
Step-by-step explanation:
nth term of the sequence =n^2 + 20
an= n^2 + 20
1st term when n= 1
1^2 + 20= 20
2nd term n= 2
2^2 + 20=24
3rd term when n= 3
3^2 + 20= 29
4th term when n= 4
4^2 + 20= 36
5th term when n= 5
5^2 + 20 =45
6th term when n= 6
6^2 + 20=56
Hence, terms in the sequence are less than 50 are first 5 terms
The answer is 500 becuase it is
