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ozzi
3 years ago
15

In Euclidean (standard) geometry, prove: If two lines share a common perpendicular, then the lines are parallel. (You do not nee

d to use axioms of Euclidean geometry in this exercsise, you can use all the standard knowledge about geometry.)

Mathematics
1 answer:
Misha Larkins [42]3 years ago
6 0

Answer:

Step-by-step explanation:

Standard knowledge about geometry.

Lines can be described as y=mx+b where m is the slope of the line...(1)

Perpendicular lines have perpendicular slopes. Perpendicular slopes are negative reciprocals of each other...(2)

Two lines are parallel ⇔ they have the same slope...(3)

Check the image to read the proof........

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Which expression is equivalent to (picture attached)
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Its the last one a^2/b.

Step-by-step explanation:

(a^6 b^-3)^1/3

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3 years ago
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Answer:

Step-by-step explanation:

2\sqrt{3} sin^{2} \alpha -cos\alpha =0\\2\sqrt{3} (1-cos ^2 \alpha )-cos \alpha =0\\2\sqrt{3} -2\sqrt{3} cos^2 \alpha -cos \alpha =0\\2\sqrt{3} cos^2 \alpha +cos \alpha -2\sqrt{3} =0\\cos \alpha =\frac{-1 \pm\sqrt{1^2-4*2\sqrt{3}*(-2\sqrt{3})  } }{2*2\sqrt{3} } \\=\frac{-1 \pm\sqrt{1+48} }{4\sqrt{3} } \\=\frac{-1\pm7}{4\sqrt{3} } \\either~cos \alpha =\frac{6}{4\sqrt{3} }=\frac{\sqrt{3} }{2} \\=cos \frac{\pi }{6} ,cos(2\pi -\frac{\pi }{6} )\\=cos \frac{\pi}{6} ,cos \frac{11\pi }{6}

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3 years ago
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Kruka [31]

Answer:

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Step-by-step explanation:

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4 0
3 years ago
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Find the four terms of the sequence given by the following expression
soldi70 [24.7K]

Answer:

47, 40, 33, 26 are the first four terms of the sequence.

Step-by-step explanation:

Expression representing the sequence is,

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For n = 3,

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For n = 4,

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Therefore, first four terms of the sequence are 47, 40, 33 and 26.

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3 years ago
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