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aliya0001 [1]
3 years ago
8

Solve the system by elimination. (Please explain how to do it)

Mathematics
1 answer:
Naddika [18.5K]3 years ago
3 0

Apply to Row 2 : Row 2 + Row 1

2x + 2y + 3z = 0

y + 4z = -3

2x + 3y + 3z = 5

Apply to Row 3: Row 3 - Row 1

2x + 2y + 3z = 0

y + 4z = -3

y = 5

Apply to Row 3: Row 3 - Row 2

2x + 2y + 3z = 0

y + 4z = -3

-4z = 8

Simplify rows

2x + 2y + 3z = 0

y + 4z = -3

z = -2

<em>Note that the matrix is in echelon form now. The next steps are for back substitution.</em>

Apply to Row 2: Row 2 - 4 Row 3

2x + 2y + 3z = 0

y = 5

z = -2

Apply to Row 1: Row 1 - 3 Row 3

2x + 2y = 6

y = 5

z = -2

Apply to Row 1: Row 1 - 2 Row 2

2x = -4

y = 5

z = 2

Simplify the rows

<u>x = -2</u>

<u>y = 5</u>

<u>z = -2</u>

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Read 2 more answers
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Answer:

A) The equation for circumference is C=2piR. So Filling in for circle A we have 28.26=2*pi*4.5 so we want to isolate pi which I'm gonna call x for it's easier for me xD. So we're gonna start by dividing 4.5 from each side which is gonna leave us with 6.28=2*x which gives you x(pi)= 3.14. For circle B we have 15.70=2*x*2.5 isolate x by first dividing 2.5 which leaves us again with 6.28=2x and x= 3.14.

B) The equation for area is A=piR^2. So again for circle A we have 63.585=x9^2. This one is harder but also are you sure that the area is 63.585 it's supposed to be 254.469 (We'll come back to this)

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hoped i helped :  )

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