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.
It could be 2.5+2.5=5 and then add that to 4 and that equals 9
Answer:
it is c
Step-by-step explanation:
because I said it is
Triangle MNP
Sorry if I’m wrong but I think that’s the one
For #3
You multiply 7 by 13
its 91, so you can't have that many 7 pointers
guess and check until you see the highest number of 7 point touch downs you can get is 8, in order to be even with three 5 pointers
7*8=56
5*3=15
56+15=71
