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Colt1911 [192]
3 years ago
7

A region is bounded by semicircular arcs constructed on the side of a square whose sides measure 2/\pi, as shown. What is the pe

rimeter of this region?

Mathematics
1 answer:
Paladinen [302]3 years ago
4 0

Answer:

4

Step-by-step explanation:

The perimeter of the region is equal to the sum of the perimeter of the four semicircular arc. Since the semicircular arcs have the same measure, therefore:

Perimeter of region = 4 × Perimeter of semicircular arc.

The side of the square = diameter of the semicircle = 2/π.

The radius of semicircle = diameter/2 = \frac{2/\pi}{2} =\frac{1}{\pi}

The perimeter of a semicircle = perimeter of a circle ÷ 2

Perimeter\ of \ semicircle = \frac{perimeter\ of\ circle}{2}=\frac{2\pi r}{2}=\frac{2\pi (\frac{1}{\pi} )}{2} =1\\  Perimeter\ of\ region = 4*Perimeter\ of \ semicircle=4*1=4\

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<h2>Step-by-step explanation:</h2>

Given equations;

y₁ = 3x - 8               -------------------(i)

y₂ = 0.5x + 7          --------------------(ii)

To fill the table, substitute the values of x into equations (i) and (ii)

=> At x = 0

y₁ = 3(0) - 8 = -8

y₂ = 0.5(0) + 7 = 7

=> At x = 1

y₁ = 3(1) - 8 = -5

y₂ = 0.5(1) + 7 = 7.5

=> At x = 2

y₁ = 3(2) - 8 = -2

y₂ = 0.5(2) + 7 = 8

=> At x = 3

y₁ = 3(3) - 8 = 1

y₂ = 0.5(3) + 7 = 8.5

=> At x = 4

y₁ = 3(4) - 8 = 4

y₂ = 0.5(4) + 7 = 9

=> At x = 5

y₁ = 3(5) - 8 = 7

y₂ = 0.5(5) + 7 = 9.5

=> At x = 6

y₁ = 3(6) - 8 = 10

y₂ = 0.5(6) + 7 = 10

=> At x = 7

y₁ = 3(7) - 8 = 13

y₂ = 0.5(7) + 7 = 10.5

=> At x = 8

y₁ = 3(8) - 8 = 16

y₂ = 0.5(8) + 7 = 11

=> At x = 9

y₁ = 3(9) - 8 = 19

y₂ = 0.5(9) + 7 = 11.5

=> At x = 10

y₁ = 3(10) - 8 = 22

y₂ = 0.5(10) + 7 = 12

The complete table is attached to this response.

(ii) To find the solution of the system of equations using the table, we find the value of x for which y₁ and y₂ are the same.

As shown in the table, that value of <em>x = 6</em>. At this value of x, the values of y₁ and y₂ are both 10.

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