At what point does the terminal side of the angle 5π6 in standard position intersect the unit circle? (-3√2, 12) (-12, 3√2) (12, −3√2) (3√2, −12) which point on the unit circle corresponds to −π6 ? (-12, 3√2) (3√2, −12) (12, −3√2) (-3√2, 12) what is the exact value of sin(5π4) ? −12 2√2 12 −2√2 what is the exact value of tan(−3π4) ? enter your answer in the box. in which quadrant does θ lie given that sinθ< 0 and cosθ> 0 ? quadrant i quadrant ii quadrant iii quadrant iv

1.) At what point does the terminal side of the angle 5π/6 in standard position intersect the unit circle?

First option: ( -√3/2, 1/2)

2. Which point on the unit circle corresponds to −π/6 ?

Second option: ( √3/2, -1/2)

3. What is the exact value of sin(5π/4) ?

Fourth option: −√2/2

4. What is the exact value of tan(−3π/4) ?

tan(−3π/4) = 1

5. In which quadrant does θ lie given that sinθ<0 and cosθ>0 ?

Fourth option: Quadrant IV

Step-by-step explanation:

1.) At what point does the terminal side of the angle 5π6 in standard position intersect the unit circle?

Point: P=(x,y)=?

x=cos 5π/6 → x = -√3/2

y=sin 5π/6 → y = 1/2

P=( -√3/2, 1/2)

2. Which point on the unit circle corresponds to −π/6 ?

Point: P=(x,y)=?

x=cos (-π/6) → x = √3/2

y=sin (-π/6) → y = -1/2

P=( √3/2, -1/2)

3. What is the exact value of sin(5π/4) ?

sin(5π/4) = −√2/2

4. What is the exact value of tan(−3π/4) ?

tan(−3π/4) = 1

5. In which quadrant does θ lie given that sinθ<0 and cosθ>0 ?

sinθ<0 (-) in Quadrants III and IV

cosθ>0 (+) in Quadrants I and IV

Then, sinθ<0 and cosθ>0 in Quadrant IV

need more information

step-by-step explanation:

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,