Answer:
I can't see the work ?????!!!
Y = 1
Using y = mx + c.
Compare to y = 1, y = 0x + 1 ,
We can see that the slope m = 0 and the vertical intercept, c = 1.
For the line perpendicular to y = 1
Condition for perpendicularity m₁m₂ = -1
m₁ = 0, m₂ = ?
0*m₂ = -1
m₂ = -1/0 = Negative Infinite or Infinite
Slope of line perpendicular to y = 1, is = Infinite.
4

This is because there is an imaginary 1 in front of the first sqrt.
If we plot the data on the graph, we can see that the
data is skewed to the right (positive skew) and there is an outlier. In skewed
data and presence of outlier, the median is most commonly used measure of
central tendency. This is because a positive skew would result in a positive
bias to the mean. Meaning that it would be a lot larger than the median and not
really representing the actual central tendency. The median however is less
affected by the skew and outliers.
Answer: Median, because the data are skewed and there is
an outlier
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Answer:
Part 1) 
Part 2) 
Step-by-step explanation:
we have

The cos(A) is negative, that means that the angle A in the triangle ABC is an obtuse angle and the value of the sin(A) is positive
The angle A lie on the II Quadrant
step 1
Find the measure of angle A

using a calculator

step 2
Find the sin(A)
we know that

substitute the value of cos(A)




step 3
Find tan(A)
we know that

substitute the values

Simplify
