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3 years ago

Solve system using eliminations 2x+4y=22 2x-2y=-8

1 answer:

5 0

$2x+4y=22 \\
2x-2y=-8 \ \ \ |\times (-1) \\ \\
2x+4y=22 \\
\underline{-2x+2y=8} \\
2x-2x+4y+2y=22+8 \\
6y=30 \ \ \ |\div 6 \\
y=5 \\ \\
2x-2y=-8 \\
2x-2 \times 5=-8 \\
2x-10=-8 \ \ \ |+10 \\
2x=2 \ \ \ |\div 2 \\
x=1 \\ \\
\boxed{(x,y)=(1,5)}$

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3 years ago

Solve for b:<br><br> 16b = 12<br> A. b = 28<br> B. b = 3/4<br> C. b = 4/3<br> D. b = 1/4

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3 years ago

Read 2 more answers

Lydia graphed ADEF at the coordinates D (-2,-1), E (-2, 2), and F (0,0). She thinks ADEF is a right triangle. Is Lydia's asserti

Setler79 [48]

The true statement is (c) No; the slopes of segment EF and segment DF are not opposite reciprocals.

<h3>Right triangles </h3>

Right triangles have a pair of perpendicular lines

Coordinates

The coordinates are given as:

D = (-2,-1)
E = (-2,2)
F = (0,0)

<h3>Slopes</h3>

Start by calculating the slopes of lines DF and EF using:

$m = \frac{y_2 -y_1}{x_2 -x_1}$

So, we have:

$m_{DF} = \frac{0 + 1}{0 +2}$

$m_{DF} = \frac{1}{2}$

Also, we have:

$m_{EF} = \frac{0 -2 }{0+2}$

$m_{EF} = \frac{-2 }{2}$

$m_{EF} = -1$

For the triangle to be a right triangle, then the calculated slopes must be opposite reciprocals.

i.e.

$m_1 = -\frac{1}{m_2}$

By comparison, the slopes of both lines are not opposite reciprocals.

Hence, the true statement is (c)