(6(x^2-1))*((6x-1)/(6(x+1))
(6((x+1)(x-1)))((6x-1)/(6(x+1))
(6(x-1))*(6x-1)/(6)
(x-1)(6x-1)
6x^2-x-6x+1
6x^2-7x+1
Answer:
The answer to your question is:
Step-by-step explanation:
1.-
2sec
2.-
sec²x - tanxsecx
I don’t know if you need to show work but here is the answer 2(x-2)/x-3
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:
x= -3-sqr30/7
Step-by-step explanation:
