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kari74 [83]
4 years ago
14

What is the area of the sector having a radius of 8 and a central angle of 5π3

Mathematics
2 answers:
olganol [36]4 years ago
5 0

Area of a full circle is PI x r^2

Area = 3.14 x 8^2 = 200.96

Area of a sector is area *  central angle / full circle

Area = 200.96 x 5PI/3 / 2PI = 160PI/3

Answer is C.

Shalnov [3]4 years ago
5 0

Answer:\frac{160\pi }{3} units^2

Step-by-step explanation:

Given

Angle subtended by sector is\frac{5\pi }{3}

radius =8 units

Area of sector=\frac{\theta }{2\pi }\pi r^2

A=\frac{20\times 8\pi }{3}

A=\frac{160\pi }{3}

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Answer:

(x (x - 4) (x - 1))/(2 (x + 4))

Step-by-step explanation:

Simplify the following:

((x^2 - 16) (x^3 - 2 x^2 + x))/((2 x + 8) (x^2 + 3 x - 4))

The factors of -4 that sum to 3 are 4 and -1. So, x^2 + 3 x - 4 = (x + 4) (x - 1):

((x^2 - 16) (x^3 - 2 x^2 + x))/((x + 4) (x - 1) (2 x + 8))

Factor 2 out of 2 x + 8:

((x^2 - 16) (x^3 - 2 x^2 + x))/(2 (x + 4) (x + 4) (x - 1))

x^2 - 16 = x^2 - 4^2:

((x^2 - 4^2) (x^3 - 2 x^2 + x))/(2 (x + 4) (x + 4) (x - 1))

Factor the difference of two squares. x^2 - 4^2 = (x - 4) (x + 4):

((x - 4) (x + 4) (x^3 - 2 x^2 + x))/(2 (x + 4) (x + 4) (x - 1))

Factor x out of x^3 - 2 x^2 + x:

(x (x^2 - 2 x + 1) (x - 4) (x + 4))/(2 (x + 4) (x + 4) (x - 1))

The factors of 1 that sum to -2 are -1 and -1. So, x^2 - 2 x + 1 = (x - 1) (x - 1):

(x (x - 1) (x - 1) (x - 4) (x + 4))/(2 (x + 4) (x + 4) (x - 1))

(x - 1) (x - 1) = (x - 1)^2:

(x (x - 1)^2 (x - 4) (x + 4))/(2 (x + 4) (x + 4) (x - 1))

((x - 4) (x + 4) x (x - 1)^2)/(2 (x + 4) (x + 4) (x - 1)) = (x + 4)/(x + 4)×((x - 4) x (x - 1)^2)/(2 (x + 4) (x - 1)) = ((x - 4) x (x - 1)^2)/(2 (x + 4) (x - 1)):

(x (x - 4) (x - 1)^2)/(2 (x + 4) (x - 1))

Cancel terms. ((x - 4) x (x - 1)^2)/(2 (x + 4) (x - 1)) = ((x - 4) x (x - 1)^(2 - 1))/(2 (x + 4)):

(x (x - 4) (x - 1)^(2 - 1))/(2 (x + 4))

2 - 1 = 1:

Answer: (x (x - 4) (x - 1))/(2 (x + 4))

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