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nordsb [41]
2 years ago
12

What is a necessary step for constructing perpendicular lines through a point off the line?

Mathematics
1 answer:
Nostrana [21]2 years ago
8 0

Answer:

Find another point on the perpendicular line.

Step-by-step explanation:

Given an original line "m", and a point off the line "Q", in order to construct a second line "p", meant to be perpendicular to "m" through the point "Q", fundamentally, the only truly necessary step to construct a perpendicular line through is to find another point on the yet-to-be-found perpendicular line.

Most often, this is accomplished by exploiting the fact that "p" is the set of all points that are equidistant from any pair of points that are symmetric about "p".

Since the symmetry must be about "p", and we don't even know where "p" is, one often finds two points on "m" that are equidistant from "Q".

This can be accomplished by adjusting a compass to a fixed radius (larger than the distance from "Q" to "m"), and making an arc that intersects "m" in two places.  Those two places will be equidistant from "Q", and are simultaneously on line "m".  Thus, these two points, "A" & "B" are symmetric about "p".

Since "A" & "B" are symmetric about "p", they are equidistant from "p", and are on "m".  One could try to find the point of intersection between "p" and "m" through construction, but this is unnecessary.  We need only find a second point (besides "Q") that is equidistant from "A" & "B", which will necessarily be a point on "p", to form the line perpendicular to "m".

To do this, fix the compass with any radius, and from "A" make a large arc generally in the direction of "B", and make the same radius arc from "B" in the direction of "A" such that the two arcs intersect at some point that isn't "Q".  This point of intersection we can call point "T", and the line QT is line "p", the line perpendicular to the original line, necessarily containing "Q".

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4 0
3 years ago
Ms. Patterson burns 236 calories riding her bike every hour. She wants to burn more than 590 calories riding her bike at the sam
Alekssandra [29.7K]

Ms. Patterson will take t>2.5 hours to burn 590 calories  riding her bike at the same rate which is 236 calories every hour.

<h3 /><h3>What amount of time Ms. Patterson  will take to burn more then 590 ?</h3>

Given that -

she burns 236 calories in 1 hour get = \dfrac{1}{236} hours

so time taken to burn 590 calories at same rate  = 590 \times\dfrac{1}{236}  hours

so time taken to burn 590 calories at same rate  = 2.5 hours

hence Ms. Patterson will take t>2.5 hours to burn 590 calories  riding her bike at the same rate which is 236 calories every hour.

To know more about  follow

brainly.com/question/14733838

3 0
2 years ago
How do you get the same base for the power
zhannawk [14.2K]
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10/3 x = 2x +4

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you can also use logarithms like so

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3x + 6 = 5x

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x = 3




7 0
3 years ago
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djyliett [7]

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Step-by-step explanation:

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Zinaida [17]

Answer:

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Step-by-step explanation:

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90 - 20 = 70

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3 years ago
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