Answer:
13 hours
Step-by-step explanation:
Average speed = Distance traveled / time taken
⇒ Distance = Average speed × Time
d = s × t
For the first trip;
Average speed = 280 mph
d₁ = 280t₁ ------(1)
where;
d₁ is the distance covered to get to the destination
t₁ is the time taken to get to the destination
For the second trip;
Average speed = 240 mph
d₂= 240t₂ ------(2)
where;
d₂ is the distance covered on the way back
t₂ is the time taken on the way back
The trip is the same distance to and fro. Therefore,
d₁ = d₂
Substituting the equation for d₁ and d₂
280t₁ = 240t₂ ------(3)
It took one hour less time to get there than it did to get back, then,
t₁ = t₂ - 1
t₂ = t₁ + 1 ------(4)
Substituting equation (4) into equation (3)
280t₁ = 240(t₁ + 1)
280t₁ = 240t₁ + 240
280t₁ - 240t₁ = 240
40t₁ = 240
t₁ = 240/40
t₁ = 6 hours
From equation (4)
t₂ = t₁ + 1
t₂ = 6 + 1
t₂ = 7 hours
The total time for the trip is t₁ + t₂ = 6 + 7
= 13 hours
8,020 divided by 10 is 802.
Hope this helps!
Y=4(x+1)
the form is y=mx+b, and they already gave you b so you just need to find m(slope) by rise over run
For the last part, you have to find where 
attains its maximum over 
. We have
so that
with critical points at 
such that
So either
or
where 
is any integer. We get 8 solutions over the given interval with 
from the first set of solutions, 
from the set of solutions where 
, and 
from the set of solutions where 
. They are approximately
