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Aleks04 [339]
2 years ago
12

Solve for the value of w

Mathematics
1 answer:
Ivenika [448]2 years ago
4 0

Answer: 9°

Step-by-step explanation:

(3w + 7)° + 90° + (7w - 7)° = 180°

=> 3w + 7 + 90 + 7w - 7 = 180

=> 10w + 90 = 180

=> 10w = 180 - 90 = 90

=> w = 90 / 10 = 9°

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Solve the system of equations for x and y.<br> y = x + 5<br> y = 2x + 2
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Answer (x,y) (3, -2)

Explanation:
using the
substitution method
y
=
x
−
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→
(
1
)
y
=
−
2
x
+
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→
(
2
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since both equations are expressed in terms of x we

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⇒
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add 2x to both sides
2
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x
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2
x
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2
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⇒
3
x
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add 5 from both sides
3
x
+
5
−
5
=
4
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⇒
3
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9
divide both sides by 3
3
x
3
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3
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3
substitute this value in
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As a check
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Use matrices and elementary row to solve the following system:
LiRa [457]

I assume the first equation is supposed to be

5x-3y+2z=13

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5x-3x+2x=4x=13

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\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\4&-2&4&12\end{array}\right]

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\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\2&-1&2&6\end{array}\right]

Add -1(row 2) to row 3:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&5&5\end{array}\right]

Multiply through row 3 by 1/5:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&1&1\end{array}\right]

Add -2(row 3) to row 1, and add 3(row 3) to row 2:

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\left[\begin{array}{ccc|c}1&0&0&1\\2&-1&0&4\\0&0&1&1\end{array}\right]

Add -2(row 1) to row 2:

\left[\begin{array}{ccc|c}1&0&0&1\\0&-1&0&2\\0&0&1&1\end{array}\right]

Multipy through row 2 by -1:

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