Pls. Help me to solve this math problem.

Mathematics

1 answer:

Tju [1.3M]4 years ago

4 0

$\bf (\stackrel{x_1}{x}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-6}~,~\stackrel{y_2}{8}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{8-(-7)}{-6-x}=\stackrel{\textit{slope}}{-\cfrac{15}{7}}\implies \cfrac{8+7}{-6-x}=-\cfrac{15}{7} \\\\\\ \cfrac{15}{-6-x}=-\cfrac{15}{7}\implies 105=(-6-x)(-15)\implies 105=90+15x \\\\\\ 15=15x\implies \cfrac{15}{15}=x\implies 1=x$

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3 years ago

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Bumek [7]

11. 2x+3x+1=21
5x+1 =21
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2x-8=x+17
2x=x+25
x=25
AB =42
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AC=84

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3x-31
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x+6=2x-37
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3 years ago

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Bingel [31]

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Answer:

Part A

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The coordinates of the vertices of the image, ΔL'M'N' are L'(-1, -1), and M'(-2, -2), and N'(-3, -3)

Please find attached the graph of triangle ΔLMN and ΔL'M'N'

Part B

The lines drawn through L and L' and through M and M' are colinear

Part C

A line is defined by two points, such as L and M. Rotation of a line through 180° about the origin will give an image location on the same path as the extension of the original line.

The third point, N, however, defines a plane, and the rotation of a plane by 180° will give an image which is turned upside down with regards to the preimage

Therefore, a trough in the preimage becomes a peak in the image and the lines drawn through N and N' crosses the colinear lines drawn throgh M and M' and L and L'

Step-by-step explanation:

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