9514 1404 393
Answer:
3433 m^2
Step-by-step explanation:
The total surface area is the sum of the end areas and the lateral area.
The end areas together are a circle of diameter 0.4 m and two rectangles that are 0.4 m by 0.55 m.
End areas = πr^2 + 2LW = 3.14(0.2 m)^2 + 2(0.4 m)(0.55 m)
= 0.1256 m^2 +0.44 m^2 = 0.5656 m^2
The perimeter of the end piece is half the circumference of a 0.4 m circle and three sides of a 0.4 m by 0.55 m rectangle.
P = (1/2)πd + 2L +W = 1/2·3.14·0.4 m + 2(0.55 m) + 0.4 m = 2.128 m
The lateral area of the mailbox is the product of its length and the perimeter of the end piece:
LA = PL = (2.128 m)(0.6 m) = 1.2768 m^2
Then the total surface area of the mailbox is ...
0.5656 m^2 + 1.2768 m^2 = 1.8424 m^2
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The area of aluminum required for 1863 mailboxes is ...
1863 × 1.8424 m^2 ≈ 3433 m^2
About 3433 square meters of aluminum are needed to make these mailboxes.
You make 2 2/7 an improper fraction (16/7) and then you subtract 8/7 from 16/7 and you get 1 1/7 (or 8/7) as your final answer
7.5 inches
the scale factor = 
= 3
hence height = 3 × 2.5 = 7.5 inches
