proyaop

If you construct a unit circle at a point, say (1,1), you will see that it satisfies the given constructions. 

The radius of a unit circle is just 1 unit.

If angle BAC = angle ABC = 60 degrees, then angle BCA = 60 degrees by triangle interior angle sum, and thus triangle ABC is equilateral. The angle bisector of the apex of an isosceles triangle (which includes the equilateral triangle) is also the height of the triangle. Since AD = 24, and angle A is being bisected, AD is also the height with length 24. By 30 - 60 - 90, the base of triangle ABC has length BC = 8sqrt(3), and the area is b*h/2 = 96sqrt(3) units^2.

\(f(x) = \left\lfloor\frac{2 - 3x}{x + 3 + 2x}\right\rfloor\)

f(1) = floor(-1/6) = -1

f(2) = floor(-4/9) = -1

f(3) = floor(-7/12) = -1

f(4) = floor(-10/15) = -1

...

f(10) = floor(-28/33) = -1

...

f(100) = floor(-298/303) = -1

... f(x) = floor((-3x + 2) / (3x + 3)), which the numerator will never be able to be -3x + 3, if it did it would be -1, so the floor'd always be -1.

Hence, f(1) + f(2) + ... f(1000) = -1 * 1000 = -1000

4) 35pi/4 = 35pi/4 - 8pi = 35pi/4 - 32pi/4 = 3pi/4. 3pi/4 = 135 degrees, which is located in Quadrant II.

3) 5pi/6 is > pi/2 or 90 degrees, but < pi or 180 degrees, so 5pi/6 lies in quadrant two.

-5pi/9 = -5pi/9 + 2pi (coterminal) = 13pi/9. 13pi/9 is greater than pi or 180 degrees, but is less than 3pi/2 = 13.5pi/9 or 270 degrees, so this terminal ray also lies in QUadrant III.

1) -21pi/8 = -21pi/8 + 4pi = 11pi/8. 11pi/8 is > than pi or 180 degrees, but is < 3pi/2 or 270 degrees, so the terminal ray lies in Quadrant III.

Coterminal angles means the same angle, just you rotated the terminal-line around the unit circle a multitude of times. Thus, all coterminal angles of 2pi/3 are in the form 2pi/3 + 2(k)pi, k is in the realm of integers: 2pi/3 - 4pi = -10pi/3, 2pi/3 - 2pi = -4pi/3, 2pi/3 + 2pi = 8pi/3, thus the FIRST, SECOND, and LAST are the only correct answers.