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

8 0

3 years ago

Find real or imaginary solutions of x^4-10x^2=-9

malfutka [58]

This resembles a quadratic and we can factor it using that pattern.

$(x^2)^2-(10x^2)+9=0$

$(x^2-9)(x^2-1)$

$(x+3)(x-3)(x+1)(x-1)=0$

6 0

3 years ago