Find cotθ if θ terminates in Quadrant III and secθ = 2 a. - (sqrt 3)/3
b. (sqrt 3)/3
c. - sqrt 3
d. sqrt 3

we have

secθ = 2

=2

Cosθ =

b=1

h=2

p=

again

Cot θ=

Cot θ=

Cot θ=

It lies in Quadrant III cot is positive

Cot θ=

b. (sqrt 3)/3

b. (sqrt 3)/3 ✓

~ʆᵒŕ∂ཇꜱꜹⱽẻⱮë

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)

answer: you need the equation in the question so i can answer it

step-by-step explanation: