Answer:
its 2 i just did it now
Step-by-step explanation:
do it
A=-2 Im attaching a picture with the steps
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
Step-by-step explanation:
According to this trigonometric function, −C gives you the OPPOSITE terms of what they really are, so be EXTREMELY CAREFUL:
![\displaystyle Phase\:[Horisontal]\:Shift → \frac{0}{\frac{1}{7}} = 0 \\ Period → \frac{2}{1}π = 2π](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Phase%5C%3A%5BHorisontal%5D%5C%3AShift%20%E2%86%92%20%5Cfrac%7B0%7D%7B%5Cfrac%7B1%7D%7B7%7D%7D%20%3D%200%20%5C%5C%20Period%20%E2%86%92%20%5Cfrac%7B2%7D%7B1%7D%CF%80%20%3D%202%CF%80)
Therefore we have our answer.
Extended Information on the trigonometric function
![\displaystyle Vertical\:Shift → D \\ Phase\:[Horisontal]\:Shift → \frac{C}{B} \\ Period → \frac{2}{B}π \\ Amplitude → |A|](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Vertical%5C%3AShift%20%E2%86%92%20D%20%5C%5C%20Phase%5C%3A%5BHorisontal%5D%5C%3AShift%20%E2%86%92%20%5Cfrac%7BC%7D%7BB%7D%20%5C%5C%20Period%20%E2%86%92%20%5Cfrac%7B2%7D%7BB%7D%CF%80%20%5C%5C%20Amplitude%20%E2%86%92%20%7CA%7C)
NOTE: Sometimes, your <em>vertical shift</em> might tell you to shift your graph below or above the <em>midline</em> where the amplitude is.
I am joyous to assist you anytime.
