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:
The Amount initially deposited is $37046.64
Step-by-step explanation:
A = p (1+r/n)^(nt)
A= final amount= $5000
P = principal amount=
r = rate = 0.05
n = number of times compounded
= 6*12
= 48
t = years= 6
A = p (1+r/n)^(nt)
50000 = p (1+0.05/48)^(48*6)
50000= p(1.001041667)^288
50000= p (1.34965)
50000/1.34965= p
37046.64 = p
The Amount initially deposited is $37046.64
The nearest thousand is 547,000
Equal to 3n+p
i think thats it
