Answer:
0.128rad/sec
Step-by-step explanation:
Let x represent the between the man and the point on the path
θ = the angle
dx/dt = 4 ft/s
dθ/dt = rate is the searchlight rotating when the man is 15 ft from the point on the path closest to the searchlight
tan θ = x/20 ft
Cross Multiply
20tan θ = x
dx/dt = 20sec² θ dθ/dt
dθ/dt = 1/20 × cos² θ dx/dt
dθ/dt= 1/20 × cos² θ × 4
dθ/dt = 1/5 × cos² θ
Note : cos θ = 4/5
dθ/dt = 1/5 × (4/5)²
dθ/dt = 16/125
dθ/dt = 0.128rad/sec
Only selections B and D give a maximum height of 13 at t=3. However, both of those functions have the height be -5 at t=0, meaning the ball was served from 5 ft below ground. This does not seem like an appropriate model.
We suspect ...
• the "correct" answers are probably B and D
• whoever wrote the problem wasn't paying attention.
Answer:
j=38
Step-by-step explanation:
j/-2 +7=-12, subtract 7 from both sides of the equation and you get j/-2 = -19, then you multiply j/-2 by -2/1 and multiply -19 by -2/1 to get j= 38
Answer:
Step-by-step explanation: