Answer:
break it down by using pmedas
It depends if the one side is in between the angles or not.
Step-by-step explanation:
f(x) = -0.01x² + 0.7x + 6.1
a) f(x) is a downward facing parabola, so the maximum height is at the vertex. The vertex of a parabola can be found using x = -b/(2a).
x = -0.7 / (2 × -0.01)
x = 35
f(35) = -0.01(35)² + 0.7(35) + 6.1
f(35) = 18.35
The maximum height is 18.35 feet.
b) The maximum horizontal distance is when the ball lands, or when f(x) = 0.
0 = -0.01x² + 0.7x + 6.1
0 = x² − 70x − 610
Solve with quadratic formula:
x = [ -b ± √(b² − 4ac) ] / 2a
x = [ -(-70) ± √((-70)² − 4(1)(-610)) ] / 2(1)
x = (70 ± √7340) / 2
x = 35 ± √1835
x = -7.84, 77.84
x can't be negative, so x = 77.84. The ball's maximum horizontal distance is 77.84 feet.
c) When the ball is first launched, x = 0. The height at that position is:
f(0) = 6.1
The ball is launched from an initial height of 6.1 feet.
Answer: 2factor : f(x)=x3
Step-by-step explanation: Graph.
Note: My bad if this is incorrect. I'm rlly sure.
