Well, we know the cyclist left the western part going eastwards, at the same time the car left the eastern part going westwards
the distance between them is 476 miles, and they met 8.5hrs later
let's say after 8.5hrs, the cyclist has travelled "d" miles, whilst the car has travelled the slack, or 476-d, in the same 8.5hrs
we know the rate of the car is faster... so if the cyclist rate is say "r", then the car's rate is r+33.2
thus

solve for "r", to see how fast the cyclist was going
what about the car? well, the car's rate is r + 33.2
A) 6(4)^1-6(4)^0 = 24-6 = 18
6(4)^3-6(4)^2 = 384-96 = 288
b) It is 27.11 times difference. This is because g(x) is an exponential function and so as x increases, g(x) increases faster.
Answer:
194
Step-by-step explanation:
t63 = t1 + (n - 1)(d)
= 8 + (63 - 1)(3)
= 8 + (62)(3)
= 8 + 186
= 194
-3 (-6=82)+8
im pretty sure like 80%
Answer:
31°
Step-by-step explanation:
The upper and lower triangles are similar, because their bases are parallel.
This implies that their correspondent angles have the same measure: the correspondent angle of x is the 31° angle.