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Vlad [161]
3 years ago
15

Graph the system of equations. −x+2y=−8 3x−y=−6

Mathematics
2 answers:
BARSIC [14]3 years ago
7 0

Answer: x = - 4

y = - 6

Step-by-step explanation:

The given system of linear equations is expressed as

- x + 2y = - 8- - - - - - - - - - - - -1

3x - y = - 6 - - - - - - - - - - - -2

We would eliminate x by multiplying equation 1 by 3 and equation 2 by 1. It becomes

- 3x + 6y = - 24- - - - - - - - - - -3

3x - y = - 6- - - - - - - - - - - -4

Adding equation 3 to equation 4, it becomes

5y = - 30

Dividing the left hand side and the right hand side of the equation by 3, it becomes

5y/5 = - 30/5

y = - 6

Substituting y = - 6 into equation 2, it becomes

3x - - 6 = - 6

3x + 6 =- 6

Subtracting 6 from the left hand side and the right hand side of the equation, it becomes

3x + 6 - 6 = - 6 - 6

3x = - 12

Dividing the left hand side and the right hand side of the equation by 3, it becomes

3x/3 = - 12/3

x = - 4

Evgen [1.6K]3 years ago
6 0

Answer:

x = -4, y = -6

Step-by-step explanation:

-x+2y = -8

3x-y = -6

x = 2y+8

3(2y+8) - y = -6

6y + 24 -y = -6

5y = -30

y = -6

x = 2(-6)+8 = -12+8 = -4

x = -4, y = -6

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Answer:

Arc length MK = 15.45 units (nearest hundredth)

Arc measure = 58.24°

Step-by-step explanation:

Calculate the measure of the angle KLN (as this equals m∠KLM which is the measure of arc MK)

ΔKNL is a right triangle, so we can use the cos trig ratio to find ∠KLM:

\sf \cos(\theta)=\dfrac{A}{H}

where:

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Given:

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\implies \sf \angle KLM=\cos^{-1}\left(\dfrac{8}{15.2}\right)

\implies \sf \angle KLM=58.24313614^{\circ}

Therefore, the measure of arc MK = 58.24° (nearest hundredth)

\textsf{Arc length}=2 \pi r\left(\dfrac{\theta}{360^{\circ}}\right) \quad \textsf{(where r is the radius and}\:\theta\:{\textsf{is the angle)}

Given:

  • r = 15.2
  • ∠KLM = 58.24313614°

\implies \textsf{Arc length MK}=2 \pi (15.2)\left(\dfrac{\sf \angle KLM}{360^{\circ}}\right)

\implies \textsf{Arc length MK}=\sf 15.45132428\:units

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