Answer:
a
Step-by-step explanation:
Answer:
For a point defined bt a radius R, and an angle θ measured from the positive x-axis (like the one in the image)
The transformation to rectangular coordinates is written as:
x = R*cos(θ)
y = R*sin(θ)
Here we are in the unit circle, so we have a radius equal to 1, so R = 1.
Then the exact coordinates of the point are:
(cos(θ), sin(θ))
2) We want to mark a point Q in the unit circle sch that the tangent has a value of 0.
Remember that:
tan(x) = sin(x)/cos(x)
So if sin(x) = 0, then:
tan(x) = sin(x)/cos(x) = 0/cos(x) = 0
So tan(x) is 0 in the points such that the sine function is zero.
These values are:
sin(0°) = 0
sin(180°) = 0
Then the two possible points where the tangent is zero are the ones drawn in the image below.
Y = -4x + 2
3x + 2y = 6 (or in y-intercept form y=-3/2x + 3)
2x - y = 7 equals y = 2x -7 in y-intercept form. This means that the line has a positive slope and therefore goes upwards. This cannot be a potential equation because the line shown is clearly pointing downwards.
y=5 indicates that the line is a horizontal line that neither points upwards or downwards. This cannot be a potential equation because the line is pointing downwards.
The only possible equations left are y= -4x + 2 and 3x + 2y = 6, both of which graph a line pointing downwards because their slopes are negative. Hope this helps!