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Oliga [24]
4 years ago
10

A new Neighborhood Activity Complex is being built in Safe Harbor. The perimeter of the rectangular playing field is 486 yards.

The length of the field is 2 yards less than quadruple the width. What are the dimensions of the playing field
Mathematics
1 answer:
siniylev [52]4 years ago
6 0

Answer:

the two lengths= 194 each side

teh two widths= 49 on each side

194+194+49+49= 486

Step-by-step explanation:

486= 2l+2w

l=4w-2

486-2w=2l -divide that by 2

243-w=l

243-w=4w-2

245=5w -divide that by 5

49=w

l=4(49)-2

l=194

w=49

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Solve the system of equation by guess sidle method
lilavasa [31]

Answer: The solution is,

x_1\approx 0.876

x_2\approx 0.419

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Step-by-step explanation:

Given equations are,

8x_1 + x_2 + x_3 = 8

2x_1 + 4x_2 + x_3 = 4

x_1 + 3x_2 + 5x_3 = 5,

From the above equations,

x_1=\frac{1}{8}(8-x_2-x_3)

x_2=\frac{1}{4}(4-2x_1-x_3)

x_3=\frac{1}{5}(5-x_1-3x_2)

First approximation,

x_1(1)=\frac{1}{8}(8-(0)-(0))=1

x_2(1)=\frac{1}{4}(4-2(1)-(0))=0.5

x_3(1)=\frac{1}{5}(5-1-3(0.5))=0.5

Second approximation,

x_1(2)=\frac{1}{8}(8-(0.5)-(0.5))=0.875

x_2(2)=\frac{1}{4}(4-2(0.875)-(0.5))=0.4375

x_3(2)=\frac{1}{5}((0.875)-3(0.4375))=0.5625

Third approximation,

x_1(3)=\frac{1}{8}(8-(0.4375)-(0.5625))=0.875

x_2(3)=\frac{1}{4}(4-2(0.875)-(0.5625))=0.421875

x_3(3)=\frac{1}{5}(5-(0.875)-3(0.421875))=0.571875

Fourth approximation,

x_1(4)=\frac{1}{8}(8-(0.421875)-(0.571875))=0.875781

x_2(4)=\frac{1}{4}(4-2(0.875781)-(0.571875))=0.419141

x_3(4)=\frac{1}{5}(5-(0.875781)-3(0.419141))=0.573359

Fifth approximation,

x_1(5)=\frac{1}{8}(8-(0.419141)-(0.573359))=0.875938

x_2(5)=\frac{1}{4}(4-2(0.875938)-(0.573359))=0.418691

x_3(5)=\frac{1}{5}(5-(0.875938)-3(0.418691))=0.573598

Hence, by the Gauss Seidel method the solution of the given system is,

x_1\approx 0.876

x_2\approx 0.419

x_3\approx 0.574

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