Answer:
Starting with ΔABC, draw the dilation image of the triangle with a center at the origin and a scale factor of two. Notice that every coordinate of the original triangle has been multiplied by the scale factor (x2). Dilations involve multiplication! Dilation with scale factor 2, multiply by 2.
Answer:
n= 75
y= 75
z= 50
p= 105
other n across from p= 105
Step-by-step explanation:
n and y are the same since the triangle is isoseles.
p is supplemetary to n
z and 50 are opposites in a quadrilateral
They would be alternate exterior so they would have to to be equal
2k + 11 = 131
-11 -11
2k = 120
---- -----
2k 2k
k = 60
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
