The slope of the line that contains the point (13,-2) and (3,-2) is 0
<em><u>Solution:</u></em>
Given that we have to find the slope of the line
The line contains the point (13,-2) and (3,-2)
<em><u>The slope of line is given as:</u></em>

Where, "m" is the slope of line
Here given points are (13,-2) and (3,-2)

<em><u>Substituting the values in formula, we get,</u></em>

Thus the slope of line is 0
Answer: There are two solutions and they are
theta = 135
theta = 225
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Explanation:
Recall that x = cos(theta). Since the given cosine value is negative, this indicates x < 0. Theta is somewhere to the left of the y axis, placing it in quadrant 2 or quadrant 3.
It turns out there are two solutions, with one solution per quadrant mentioned above. Use the unit circle to find that the two solutions are:
theta = 135
theta = 225
You're looking for points on the unit circle that have x coordinate equal to x = -sqrt(2)/2. Those two points correspond to the angles of 135 and 225, which are in quadrants 2 and 3 respectively.
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I recommend using your calculator to note that
-sqrt(2)/2 = -0.70710678
cos(135) = -0.70710678
cos(225) = -0.70710678
The decimal values are approximate. Make sure your calculator is in degree mode. Because those three results are the same decimal approximation, this indicates that cos(135) = cos(225) = -sqrt(2)/2.
Answer:
i)D:
ii)R: 
iii) Y-int:(0,-1)
Step-by-step explanation:
i) The given absolute value function is;
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The absolute value function is defined for all real values of x.
The domain is all real numbers.
ii) The range is all y-values that will make x defined.
The given function,
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has vertex at, (-2,-3) and opens upwards.
This implies that, the minimum y-value is -3.
The range is 
iii) To find the y-intercept substitute x=0 in to the function.
.
.
.
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The y-intercept is (0,-1)
See attachment for graph.