The height of the equilateral triangle is 41. 6 inches. Option C
<h3>How to determine the height</h3>
The formula for finding the height of an equilateral triangle is given as;
h = (a√3)/2
we have a = 48 inches
Let's substitute the value
Height, h = 
Height = 
Height = 
Height =
Inches
Thus, the height of the equilateral triangle is 41. 6 inches. Option C
Learn more about equilateral triangles here:
brainly.com/question/1399707
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The equation of line that passes through point (-3, 8) and has an undefined slope is x = -3
<u>Solution:</u>
Given that we have to find the equation of line that passes through point (-3, 8) and has an undefined slope
Since the line has an undefined slope. This informs us that it is a vertical line parallel to the y-axis and passing through all points in the plane with the same x-coordinate.
If the slope of the line is undefined, then, by definition the line is a vertical line.
For a vertical line, the value of x is the same for each and every value of y
Given point is (x, y) = (-3 , 8)
Because the value of x in the point provided in the problem is -3, we get
Equation of line is : x = -3
Answer:
5. Plane P
6. Line segment DB
7. Line CE
11. Ray BE (with an arrow on top from left to right) and Ray EC (with an arrow on top from left to right)
Answer:
Analyzed and Sketched.
Step-by-step explanation:
We are given 
To sketch the graph we need to find 2 components.
1) First derivative of y with respect to x to determine the interval where function increases and decreases.
2) Second derivative of y with respect to x to determine the interval where function is concave up and concave down.

is absolute maximum

is the point concavity changes from down to up.
Here, x = 0 is vertical asymptote and y = 0 is horizontal asymptote.
The graph is given in the attachment.