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There are 20 sets represented at least 1 of every five sets has more c’s. Given the total 3/4s=75%
Answer:
The distance of the overpass above the ground is approximately 26.795 ft
Step-by-step explanation:
The parameters given are;
The distance from the overpass the engineer stands before determining the angle of elevation of the overpass from his standing point = 100 ft
The angle of elevation of the overpass as determined by the engineer from 100 ft = 15°
By trigonometric ratios, we have;

The opposite side to the 15° angle of elevation in the above case is the distance of the overpass above the ground
The opposite side to the 15° is the distance of the engineer from the base of the overpass
Therefore;
Tan(15°) the height of the overpass=
length

The distance of the overpass above the ground = 100 × tan (15°) ≈ 26.795 ft.
9514 1404 393
Answer:
x ≈ {-2.80176, -0.339837}
Step-by-step explanation:
Write in terms of sine and cosine:
sec(x) -5tan(x) -3cos(x) = 0 . . . . . . given, subtract 3cos(x)
1/cos(x) -5sin(x)/cos(x) -3cos(x) = 0
Multiply by cos(x). (Note, cos(x) ≠ 0.)
1 -5sin(x) -3cos(x)² = 0
Use the trig identity to write in terms of sin(x).
1 -5sin(x) -3(1 -sin²(x)) = 0
3sin(x)² -5sin(x) -2 = 0 . . . . . . . . quadratic in sin(x)
(sin(x) -2)(3sin(x) +1) = 0 . . . . . . factor the quadratic
Values of sin(x) that make this true are ...
sin(x) = 2 . . . . . true only for complex values of x
sin(x) = -1/3
Then the possible values of x are ...
x = arcsin(-1/3), -π -arcsin(-1/3)
x ≈ {-2.80176, -0.339837} . . . . . rounded to 6 sf
C- Evolutionarily they have adapted to their surroundings (its lighter on the beach compared to inland)
hope the helped!