I believe that the answer to that one would be A.
Basically when the object hits the ground, f(t) is equal zero. So all you need to do is to solve the quadratic equation:
-5t^2 + 20t + 60 = 0
D = b^2 - 4ac = 20^2 - 4 × (-5) × 60 = 400 + 1200 = 1600
t1 = (-b + sqrt(D)) / 2a = (-20 + 40) / (-10) = -2
t2 = (-b - sqrt(D)) / 2a = (-20 - 40) / (-10) = 6
Since time can't be negative, our solution is t2.
So the object will hit the ground in 6 seconds.
So...the diameter is increasing at a rate if 2cm/minute, therefore the radius (1/2 the diameter) is increasing at half the rate. You will learn how to calculate the rate of change at a specific point in time in calculus.
Answer:
•Absolution is not a method to solve system of equation
A.Dose every point on the line
