What is the solution to the trigonometric inequality sin(x) > cos(x) over the interval 0 ≤ x ≤ 2pi radians? Any explanation on how to find this would be helpful!

0 ≤ x < pi/4
0 ≤ x < pi/4 and 5pi/4 x ≤ 2pi
pi/4 < x < 5pi/4
pi/4 < x < 5pi/4 and 5pi/4 < x ≤ 2pi

f is the midpoint of

1/800 is the right answer

second option

step-by-step explanation:

the measures of a triangle equal 180. use that to find x.

2x+3x+4x=180

9x=180

9x/9=180/9

x=20

now you know x, you know angle a is 40°, angle b is 60°, angle c is 80°.

because angle a is 40°, the first option doesn't work.

because angle b is 60°, the second option does work.

when you add angle a and b, you get 100°. complementary means it equals 90°, so the third option doesn't work.

when you add angle a and c, you get 120°, which is not 100.

the height of the surface increases from the center out to the sides of the road.

step-by-step explanation:

as we move out we move away from the x-axis.

and as mentioned in the problem statement that height of a road increases when we move away from the center x.

hence proved that 3rd option

the height of the surface increases from the center out to the sides of the road.

is the correct option.