Answer:
6.74 s
Step-by-step explanation:
y = ½ at² + vt + h
Given a = -32, v = 80, h = 188:
y = ½ (-32) t² + (80) t + (188)
y = -16t² + 80t + 188
When y = 0:
0 = -16t² + 80t + 188
0 = 4t² − 20t − 47
Solve with quadratic formula:
t = [ -(-20) ± √((-20)² − 4(4)(-47)) ] / 2(4)
t = [ 20 ± √(400 + 752) ] / 8
t = (20 ± √1152) / 8
t > 0, so:
t = (20 + √1152) / 8
t ≈ 6.74
Step 1: Find f'(x):
f'(x) = -6x^2 + 6x
Step 2: Evaluate f'(2) to find the slope of the tangent line at x=2:
f'(2) = -6(2)^2 + 6(2) = -24 + 12 = -12
Step 3: Find f(2), so you have a point on y=f(x):
f(2) = -2·(2)^3 + 3·(2)^2 = -16 + 12 = -4
So, you have the point (2,-4) and the slope of -12.
Step 4: Find the equation of your tangent line:
Using point-slope form you'd have: y + 4 = -12 (x - 2)
That is the equation of the tangent line.
If your teacher is picky and wants slope-intercept, solve that for y to get:
y = -12 x + 20
The outlier is the number that’s farther away from the others, in this case it’s 9. The mean is the average of all the data points so it is 77/20 or 3.85. Finally, the range is the highest number minus the lowest number, so the range is 8.
