proyaop

The rectangle with these coordinates is centered at the origin, with side length 2. Note that any equation y = mx will bisect the area of the square, so the y-intercept is 0. (Try drawing it on graph paper too) 

f(x) = a{trig}(bx - c) + d

a is the amplitude

2pi/|b| = the period

c/b = the horizontal phase shift

d = midline

Following this generalization, the midline is y = 1.8

0.03 + 0.0014 repeating = 0.0314 repeating

Let 0.0014repeating = S, 100S = 0.141414repeating = S + 0.14

Solving, you have 99S = 0.14, S = 0.14/99 = 14/(99)(100) = 14/9900

0.03 + 7/4950 = 297/9900 + 14/9900 = 311/9900

A cubic polynomial, by the fundamental theorem of algebra, will have EXACTLY three roots (which generalizes to an nth degree polynomial to have n complex roots). Clearly, in this problme, there are four roots, so a cubic polynomial f does not exist, and all coefficients would be 0. Thus, the sum of the coefficients is 0.

Let the center of the small circle be O, and the point where circle O is tangent to the circle with diameter AC be D, and let the radius of circle O be r. Lastly, let the midpoints of AB and BC be points M and N, respectively.

Since B is the center of a semicircle, BD must also be the same length as AB (because they are the radius) = 2.

MO = 1 + r, and MB = 1. OMB forms a right triangle with angel OBM = 90 degrees, since O lies on the perpendicular bisector of AC.

This means that OB, by the pythagorean theorem, has length \(\sqrt{(1+r)^2-1}=\sqrt{r^2+2r}\).

Additionally, DO = r, and lines on the same line as OB. Recalling that DB = 2, we can set an equation: DO + OB = 2:

\(r + \sqrt{r^2+2r}=2\)

\(\sqrt{r^2+2r}=2-r\)

\(r^2+2r=(2-r)^2=r^2-4r+4\)

\(2r = -4r+4\)

\(6r=4\)

\(r={2\over3}\)

Because triangle ABC is a right triangle with B = 90 degrees, then we know that the pythagorean theorem is a possible strategy.

AB = 3, and BC = 4, and they are both the legs of the right triangle because they include the right angle, and so by pythagorean theorem:

\(AC^2=AB^2+BC^2=3^2+4^2=25\)

So, AC = 5 units.

Through construction, AB = AF + FB = 3.

AF = EF = x. Since ADEF is a square, DE is parallel to AB, and thus angle CED is congruent to angle CBA by corresponding angles, and ADE = angle EFB because they are both 90 degrees. Thus, triangle CDE is similar to EFB is similar to CAB, and CA : AB = 1 : 3, likewise, FE : FB = 1 : 3. Substituting, we have x : FB = 1 : 3, and therefore FB = 3x. Substituting this into the original equation, AB = x + 3x = 4x = 3. And thus, x = 3/4. The area of the square would be x^2 = 9/16.

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.