Answer:supplementary
Step-by-step explanation:
Opposite angles in any quadrilateral inscribed in a circle are supplements of each other.
Problem One
Call the radius of the second can = r
Call the height of the second can = h
Then the radius of the first can = 1/3 r
The height of the first can = 3*h
A1 / A2 = (2*pi*(1/3r)*(3h)] / [2*pi * r * h]
Here's what will cancel. The twos on the right will cancel. The 3 and 1/3 will multiply to one. The 2 r's will cancel. The h's will cancel. Finally, the pis will cancel
Result A1 / A2 = 1/1
The labels will be shaped differently, but they will occupy the same area.
Problem Two
It seems like the writer of the problem put some lids on the new solid that were not implied by the question.
If I understand the problem correctly, looking at it from the top you are sweeping out a circle for the lid on top and bottom, plus the center core of the cylinder.
One lid would be pi r^2 = pi w^2 and so 2 of them would be 2 pi w^2
The region between the lids would be 2 pi r h for the surface area which is 2pi w h
Put the 2 regions together and you get
Area = 2 pi w^2 + 2 pi w h
Answer: Upper left corner <<<<< Answer
120/4=30. Remember its widthxlength. if you forget ask your teacher or look online for more help. Good luck.
Answer:
40 feet
Step-by-step explanation:
We are given a right isosceles triangle having lengths of two sides 12 feet and 16 feet.
<em>Since, the triangle is an isosceles triangle i.e. two sides of the triangle are equal.</em>
That is, the three sides of the triangle are 12 feet, 12 feet and 16 feet.
We know that, Perimeter of a triangle = Sum of the sides
Thus, Perimeter of the given triangle = 12 + 12 + 16 = 24 + 16 = 40 feet.
Hence, the total length of the fencing needed is 40 feet.
The answer would b be d 8 I think
