Simplify both sides of your equation.
Subtract 4 from both sides.
Multiply both sides by 2(-1).
m = -12
Instantaneous rate of change = S'(r) = 8πr
S'(8) = 8π(8) = 64π
Therefore, the instantaneoud rate of change of the <span>surface area with respect to the radius r at r = 8</span> is 64π
Answer:
S(t) = -4.9t^2 + Vot + 282.24
Step-by-step explanation:
Since the rocket is launched from the ground, So = 0 and S(t) = 0
Using s(t)=gt^2+v0t+s0 to get time t
Where g acceleration due to gravity = -4.9m/s^2. and
initial velocity = 39.2 m/a
0 = -4.9t2 + 39.2t
4.9t = 39.2
t = 8s
Substitute t in the model equation
S(t) = -49(8^2) + 3.92(8) + So
Let S(t) =0
0 = - 313.6 + 31.36 + So
So = 282.24m
The equation that can be used to model the height of the rocket after t seconds will be:
S(t) = -4.9t^2 + Vot + 282.24
Answer: false
explanation;
