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DIA [1.3K]
3 years ago
13

The figures below are based on semicircles and squares. Find the perimeter and the area of each shape. Give your answer as a com

pletely simplified exact value in terms of π (no approximations).

Mathematics
1 answer:
irina1246 [14]3 years ago
8 0

Answer:

Part 1) The area of the figure is A=48\ cm^2

Part 2) The perimeter of the figure is P=4(2+3\pi)\ cm

Step-by-step explanation:

Part 1) Find the area of the figure

we know that

The area of the figure is equal to the area of three squares

so

A=3[4^2]=48\ cm^2

Part 2) Find the perimeter of the figure

we know that

The perimeter of the figure is equal to two times the length side of the square, plus the circumference of six semicircles ( or the circumference of three circles)

Remember that

The circumference of a circle is equal to

C=\pi D

where

D is the diameter of the circle

In this problem

D=4\ cm

so

P=2(4)+3[\pi(4)]=(8+12\pi)\ cm

simplify

P=4(2+3\pi)\ cm

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Check whether the relation R on the set S = {1, 2, 3} is an equivalent
kozerog [31]

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R isn't an equivalence relation. It is reflexive but neither symmetric nor transitive.

Step-by-step explanation:

Let S denote a set of elements. S \times S would denote the set of all ordered pairs of elements of S\!.

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A relation R on S\! is a subset of S \times S. For any two elementsa,\, b \in S, a \sim b if and only if the ordered pair (a,\, b) is in R\!.

 

A relation R on set S is an equivalence relation if it satisfies the following:

  • Reflexivity: for any a \in S, the relation R needs to ensure that a \sim a (that is: (a,\, a) \in R.)
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  • Transitivity: for any a,\, b,\, c \in S, if a \sim b and b \sim c, then a \sim c. In other words, if (a,\, b) and (b,\, c) are both in R, then (a,\, c) also needs to be in R\!.

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Neither is R transitive. (3,\, 1) \in R and (1,\, 2) \in R. However, (3,\, 2) \not \in R. In other words, under relation R\!, 3 \sim 1 and 1 \sim 2 does not imply 3 \sim 2.

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