Tan θ = y over x tan 11 pi over 6 = the fraction negative 1 over 2 over the fraction square root 3 over 2 tan 11 pi over 6 = negative 1 over 2 times 2 over square root 3 tan 11 pi over 6 = negative 1 over square root 3 tan 11 pi over 6 = negative square root 3 over 3 this is an example on my algebra 2 of the tangent of a point on the unit circle. can anyone tell me how negative 1 over the square root of 3 simplifies to negative square root of 3 over 3?

answer: i'm not going to give you the answer, but some hints to work it out.

step-by-step explanation: so, if there is a straight line that means that the line angle is 180°. so we subtract the 2 angle measures provided, (15 and 38) from 180. so the answer is 180 - (38 + 15). (you have to solve the equation)

step-by-step explanation:

853.3

step-by-step explanation:

5x + 3y = - 4

step-by-step explanation:

the equation of a line in standard form is

ax + by = c ( a is a positive integer and b, c are integers )

express the line first of all in slope- intercept form

y = mx + c ( m is the slope and c the y-intercept )

to calculate m use the gradient formula

m = ( y₂ - y₁ ) / ( x₂ - x₁ )

with (x₁, y₁ ) = (7, - 3) and (x₂, y₂ ) = (4, - 8)

m = = -

y = - x + c ← is the partial equation

to find c substitute either of the 2 points into the partial equation

using (4, - 8 ), then

- 8 = - + c ⇒ c = -

y = - x - ← in slope-intercept form

multiply all terms by 3 to eliminate the fractions

3y = - 5x - 4 ( add 5x to both sides )

5x + 3y = - 4 ← in standard form