The First is sec^2(x). Second is 2sec^2tanx. From that you get 2[2sec^2xtan^x+sec^4x]
9514 1404 393
Answer:
16 miles
Step-by-step explanation:
The problem can be modeled by a right triangle with one angle of 7° and the side opposite being 10,000 ft. The distance needed is the hypotenuse of the triangle, so the relevant trig relation is ...
Sin = Opposite/Hypotenuse
Hypotenuse = Opposite/Sin
air distance = (10,000 ft)/sin(7°) ≈ 82,055 ft
At 5,280 ft per mile, that is ...
(82,055 ft)/(5,280 ft/mi) ≈ 15.54 mi
The plane's air distance to the airport is about 16 miles.
I think $8.50 is how much 5 pounds is
Answer: $445.83
Step-by-step explanation:
