PLEASE HELP FOR REAL In exercises 22 24 , describe how the change affects the surface area of the right prism or right cylinder .

  22. The surface area is 1/9 of its original value

  24. The surface area is 120 m^2 less than its original value.

Step-by-step explanation:

22. For similar figures, the surface area is proportional to the square of the linear dimension scale factor. Multiplying all linear dimensions by 1/3 will result in reducing the surface area to (1/3)^2 = 1/9 of its original value.

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24. As above, the scaled base surface area will be (1/4)^2 = 1/16 of its original value. Since one dimension of the lateral area is multiplied by 4 and the other dimension is multiplied by 1/4, the net effect on that area is multiplication by (1/4)(4) = 1. That is, the lateral area will remain unchanged.

The overall surface area will be reduced by 15/16 of the area of the bases. The area of the bases is (2)(1/2)(8 m)(16 m) = 128 m^2. The changed figure will have a surface area that is 120 m^2 smaller.

the figures below are made out of circles, semicircles, quarter … find the area and the perimeter of each figure and give your answers as a completely simplified exact value in terms of π (no approximations). remember that we'll also have to cut whatever areas we get in half, since we're only dealing im pretty sure that is the right answer

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i can you with this sweetie!

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