the height of the pentagonal pyramid is 5. 70 meters
<h3>Volume of a regular pentagonal pyramid</h3>
The formula for determining the volume of a regular pentagonal pyramid is given as;
V=5/12tan(54°)ha^2
Where
- a is the base edge
- h is the height
We have the volume to be;
volume = 82. 5 cubic centers
height = h
a = 5m
Substitute the values
× 
× 
×
× 
× 
×
Make 'h' subject of formula
h = 5. 70 meters
The height of the pentagonal pyramid is 5. 70 meters
Thus, the height of the pentagonal pyramid is 5. 70 meters
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Answer:
[0,9]
Step-by-step explanation:
The domain is the first set of the ordered pair. When finding the domain you are supposed to find which x value it starts from to where it ends. The x value starts from 0 and ends and the 9th unit on the x-axis which would give us 0,9 now since both points are closed you would use these symbols [ ] to show that it is closed meaning the answer would be [0,9]
Note:
I'm sorry if this is wrong but this is what I believe the answer is.
Hi,
Perimeter AEDCB=3*10+15+20=65 (units)
Ratio of the similitude r*15=21==>r=21/15=1.4
Perimeter RQPTS=65*1.4=91(units)
Answer: 15) 1 16) 30
Step-by-step explanation:
15 ) Since 15 you are shown that the larger has 30 & 5 so, if it has shrunk until now that 30 has turned into 6 that means it has shrunk by 5 times the original bigger triangle has (30 divided by 6 = 5). 5 can go into 5 1 time so you get 1
16) We have an isosceles triangle so all the sides will be equal. So If one side is 30 then the rest will be 30. This is also shown with the smaller triangle, with both sides equaling 6.
Since the given figure is a trapezoid, here is how we are going to find for the value of x. Firstly, the sum of the bases of the trapezoid is always equal to twice of the median. So it would look like this. 2M = A + B.
Plug in the given values above.
2M = (<span>3x+1) + (7x+1)
2(10) = 10x + 2
20 = 10x + 2
20 - 2 = 10x
18 = 10x < divide both sides by 10 and we get,
1.8 = x
Therefore, the value of x in the given trapezoid is 1.8. Hope this is the answer that you are looking for.
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