Answer:
- 7 faces
- 15 edges
- 10 vertices
Step-by-step explanation:
This is a counting problem. As with many counting problems, it is helpful to adopt a strategy that helps ensure you count everything only once.
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<h3>Faces</h3>
There are two pentagonal faces and 5 rectangular faces for a total of ...
7 faces
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<h3>Edges</h3>
There are 5 edges around each of the pentagonal faces, and 5 edges connecting the top face to the bottom faces, for a total of ...
15 edges
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<h3>Vertices</h3>
There are 5 vertices on the top face, and 5 on the bottom face, for a total of ...
10 vertices
The perimeter = 20 and AC = 8
Now as it is not mentioned which sides are equal of the isosceles triangle ABC,
We have two possible situations.
1)
If AC is the base
In that case AB = BC
Now AC = 8, AB = x , BC = x
So x + x + 8 = 20
2x + 8 = 20
2x = 12
x = 6
AB = BC = 6
2)
IF AC is not the base,
Then
AC = BC or AC = AB
So BC = 8 or AB = 8
If AB = AC = 8
Then
BC + 8 + 8 = 20
BC = 4
So there are two possible lengths of BC
Either it is BC = 8 or BC = 6 or BC = 4
The figure is attached for your reference.
Filling in the given values in point-slope equation
y = m(x -x₁) +y1
you have
y = 1(x -5) +3
This simplifies to ...
A. y = x - 2
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<h3>Whats your question?</h3>