Mathematics, 23.07.2019 09:00 sammielucero

Select all angles that are coterminal with an angle of rotation of 300 degrees a. -420 degrees b. 2100 degrees c. -900 degrees d. -π/3 e. 23π/3 f. -7π/3

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Answer from: ike9264

A. B. D. E. F. are all coterminal angles to 300°.

Step-by-step explanation:

A 360 degree angle is a complete turn of a circle, if you keep going you can reset the counter or keep it and it will still be valid, therefore a 361° angle is the same as a 1° angle. With that in mind you can keep substracting 360 to answer B. to check if it is valid:

$2100-360=1740\quad 1740-360=1380\quad 1380-360=1020\quad 1020-360=660\quad 660-360=300$

so answer B. gives the same angle (coterminal) to 300°.

In case of a negative angle you sweep the angle clockwise, instead of the normal positive counterclockwise. In that order of ideas, a -1° angle is the same a 359°. So you can add 360° to see if it reaches the desired angle.

In case of angle A.

$-420+360=-60\quad -60+360=300$

It is valid. Let's look at angle C.

$-900+360=-540\quad -540+360=-180\quad -180+360=180$

It is not coterminal with 300°.

To the angles in radians, remember that a pi/3 angle is just a 60° angle. Thus, answer D. is valid because a -60° is coterminal with a 300° angle.

Answer E. is examined taking into account that a 360° angle is just a 2 pi angle, or a 6 pi/3 angle. So

$\frac{23 \pi}{3} -\frac{6 \pi}{3} =\frac{17 \pi}{3} \quad \frac{17 \pi}{3} -\frac{6 \pi}{3} =\frac{11 \pi}{3}\quad \frac{11 \pi}{3} -\frac{6 \pi}{3} =\frac{5 \pi}{3}$

Which is exactly a 300° angle (check the image attached, look for the blue lines).

Answer F is another valid answer since

$-\frac{7 \pi}{3} +\frac{6 \pi}{3} =-\frac{\pi}{3}$

Select all angles that are coterminal with an angle of rotation of 300 degrees a. -420 degrees b. 2

Answer from: Quest

(2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

step-by-step explanation:

simplify the following:

x/x^2 + 3 x + 2 - (x (x + 1))/(x + 2)

combine powers. x/x^2 = x^(1 - 2):

x^(1 - 2) + 3 x + 2 - (x (x + 1))/(x + 2)

1 - 2 = -1:

x^(-1) + 3 x + 2 - (x (x + 1))/(x + 2)

put each term in 1/x + 3 x + 2 - (x (x + 1))/(x + 2) over the common denominator x (x + 2): 1/x + 3 x + 2 - (x (x + 1))/(x + 2) = (x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)):

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2))

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)) = ((x + 2) + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2)):

(2 + x + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

3 x^2 (x + 2) = 3 x^3 + 6 x^2:

(2 + x + 3 x^3 + 6 x^2 + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

2 x (x + 2) = 2 x^2 + 4 x:

(2 + x + 6 x^2 + 3 x^3 + 2 x^2 + 4 x - x^2 (x + 1))/(x (x + 2))

-x^2 (x + 1) = -x^2 - x^3:

(-x^2 - x^3 + 3 x^3 + 6 x^2 + 2 x^2 + 4 x + x + 2)/(x (x + 2))

grouping like terms, 3 x^3 - x^3 + 6 x^2 + 2 x^2 - x^2 + 4 x + x + 2 = (3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2:

((3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

3 x^3 - x^3 = 2 x^3:

(2 x^3 + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

6 x^2 + 2 x^2 - x^2 = 7 x^2: