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soldier1979 [14.2K]
3 years ago
12

Solve the following system of equations:

Mathematics
1 answer:
grin007 [14]3 years ago
7 0
<span>A) 4x − 2y − z = −5
B) x − 3y + 2z = 3
C) 3x + y − 2z = −5
Looking at ALL 4 solutions, x =0 in all of them.
So, let's say x = 0, eliminate the "x's" and just keep equations B and C
</span>
<span>B) -3y + 2z = 3
C) y − 2z = −5
</span>Add B) and C)
-2 y = -2
y = 1

<span>1 − 2z = −5
</span>-2z = -6

z = 3

and x = 0


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umka21 [38]

Answer:

Step-by-step explanation:

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8 0
3 years ago
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Mandarinka [93]
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7 0
3 years ago
Read 2 more answers
5. (09.06 MC)
polet [3.4K]

Answer:

12.566

Step-by-step explanation:

A = ∫1/2(r)^2dtheta from a - b = 0 to 2π

∫1/2(sin^2theta-4sintheta+4)dtheta

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when you solve this you get 14.137 without taking out the cos(2theta+3)

5 0
2 years ago
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