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Whitepunk [10]
3 years ago
12

If you place 35 points on a piece of paper so that no three points are in a line, how many line segments are necessary to connec

t each point to all the others?
Mathematics
2 answers:
Ganezh [65]3 years ago
4 0
<span>Thank you for posting your Math question here at brainly. The line segments that are necessary to be connected to each point is 34. The total number of such connection is C(35,2) = 35 * 34/2. I hope the answer helps. </span>
bogdanovich [222]3 years ago
3 0

Answer:

595 line segments are necessary to connect each point to all the others.

Step-by-step explanation:

Consider the provided information.

The formula for the number of line segments between "n" non-collinear points is:

\frac{n\times ( n -1 )}{2} Segments

Here it is given that there are 35 points.

Substitute n = 35 in above formula.

\frac{35\times (35 -1 )}{2}

\frac{35\times 34}{2}

35\times 17

595 Segments

Hence, 595 line segments are necessary to connect each point to all the others.

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Answer:

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Step-by-step explanation:

The diagonals of a parallelogram bisect each other.  It was given that diagonals AC and BD intersect at point E.

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The circumference of the ellipse approximate. Which equation is the result of solving the formula of the circumference for b?
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Answer:

b = \sqrt{\frac{C^{2} }{2(\pi )^{2} }  -  a^{2}}

Step-by-step explanation:

Given - The circumference of the ellipse approximated by C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }where 2a and 2b are the lengths of 2 the axes of the ellipse.

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Squaring Both sides, we get

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∴ we get

b = \sqrt{\frac{C^{2} }{2(\pi )^{2} }  -  a^{2}}

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