To approximate the volume with 8 boxes, we have to split up the interval of integration for each variable into 2 subintervals, [0, 1] and [1, 2]. Each box will have midpoint 
that is one of all the possible 3-tuples with coordinates either 1/2 or 3/2. That is, we're sampling 
at the 8 points,
(1/2, 1/2, 1/2)
(1/2, 1/2, 3/2)
(1/2, 3/2, 1/2)
(3/2, 1/2, 1/2)
(1/2, 3/2, 3/2)
(3/2, 1/2, 3/2)
(3/2, 3/2, 1/2)
(3/2, 3/2, 3/2)
which are captured by the sequence
with each of 
being either 1 or 2.
Then the integral of 
over 
is approximated by the Riemann sum,
(compare to the actual value of about 4.159)
The ladder is the hypotenuse of a 45 45 90 triangle.
The legs of a 45 45 90 triangle = hypotenuse / sqroot (2)
15 / 1.414 = 10.608
(Incidentally it is the "bottom" of the ladder that is 10.608 feet from the wall. The "top" of the ladder is touching the wall. LOL)
If fog means f, then
answer: f(n)=15
Answer:
a) i) The drawing of Zeno's route is attached
ii) The bearing of Zeno's return journey is 241° from point C to point A
b) Yes, Zeno returns to Port A before 5.15pm
Step-by-step explanation:
a) i) Please find attached the drawing of Zeno's route
ii) From the attached diagram of Zeno's route created with Microsoft Whiteboard, we find the bearing of his return journey as 241° from point C to point A
b) The distance from point C to point A = 10·√5 km
The speed with which Zeno sails as he returns = 10 km/h
The time it takes Zeno to return t = Distance/Speed
∴ The time it takes Zeno to return t = 10·√5 km/(10 km/h) = √5 h ≈ 2.2361 hours ≈ 2 hours 14 minutes and 9.845 seconds
The time Zeno arrives at point A from point A ≈ 3.00 pm + 2 hours 14 minutes and 9.845 seconds = 5:14.1641 p.m. ≈ 5:14 pm.
Therefore, Zeno returns to Port A before 5.15pm.
Answer: I dont know the answer but I made this for you and i hope it will help
Step-by-step explanation: :D
