If 180° < <em>θ</em> < 270°, then 90° < <em>θ</em>/2 < 135°, which places <em>θ</em>/2 in the second quadrant so that sin(<em>θ</em>/2) > 0 and cos(<em>θ</em>/2) < 0.
Recall that
cos²(<em>θ</em>/2) = (1 + cos(<em>θ</em>))/2
==> cos(<em>θ</em>/2) = -√[(1 + (-15/17))/2] = -1/√17
and
sin²(<em>θ</em>/2) = (1 - cos(<em>θ</em>))/2
==> sin(<em>θ</em>/2) = +√[(1 - (-15/17))/2] = 4/√17
Then
tan(<em>θ</em>/2) = sin(<em>θ</em>/2) / cos(<em>θ</em>/2)
… = (4/√17) / (-1/√17)
… = -4
1. 31e+16
hope that helps
Answer:
-3x-8
Step-by-step explanation:
To my best knowledge the slop that is increased is number 3 it is moving up in a slop and it is sloped.
To create this inequality graph, you will have an open circle at 50 and an arrow pointing to the right.
Since the bag must be over 50, we draw an open circle at fifty.
Since the fee goes towards the bags over 50, we draw an arrow pointing to the right.
