Not really sure whats up with that inequality but she can't, if she has to spend 1.5 hours in a lab that leaves her with 5 hours. 5/4 is 1.25, so no, she can only spend 1.25 hours with each student.
First, let's find the time where the ship arrives to point C
t=d/V, t= 45/30=1.5h
if the ship's departure time is at 1.5h, on the point C, let 's find its arrival to the point O, V=D/Dt, Dt=tfinal-tinitial
Dt=D/V= 20/30=0.6h=t final -t initial, implies t final=0.6+t initial=0.6+1.5=2.1h
so the ship will be on the radar at 2.1h,
Answer:
B
Step-by-step explanation:
It has a not as steep incline
