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Goshia [24]
3 years ago
14

PLEASE HELP! I HAVE A TIME LIMIT AND IM BEHIND

Mathematics
1 answer:
Kipish [7]3 years ago
3 0

Answer:

Part A none

Part B drawn

Step-by-step explanation:

Suppose the transversal intersects a pair of parallel lines it creates two pairs of alternate exterior angles.

Now if we look at the first figure we find that the alternate exterior angle 1 and 4 are equal . Also 3 and 2 are equal.

But <1 and < 2 are not equal . They are supplementary angles. Their sum is equal to 180 degrees.

Similarly angles 3 and4 are supplementary angles.

This situation does not support Ricky's claim. So you will select none in part A.

Now in part B we see that angle 5 &8 are equal and 6 & 7 are congruent.

This refutes Ricky's claims and this can be proved mathematically.

Angles 5 & 6 are supplementary angles. Similarly 7 & 8 are supplementary therefore they cannot be equal as they are unequal and a transversal is drawn not a perpendicular.

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Vitek1552 [10]

120º is 1/3 of a complete revolution of 360º. So the area of this sector should be 1/3 the area of the complete circle.

A circle with radius 9 has area 9^2 π = 81π.

So the sector has area 81π/3.

Put another way: The area <em>A</em> of a circular sector and its central angle <em>θ</em> (in degrees) occur in the same ratio as the area of the entire circle with radius <em>r</em> according to

<em>A</em> / <em>θ </em>º = (π <em>r </em>^2) / 360º

==>  <em>A</em> = π/360 <em>θ r</em> ^2

In this case, <em>r</em> = 9 and <em>θ</em> = 120º, so

<em>A</em> = π/360 * 120 * 81 = 81π/3

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3 years ago
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Answer:

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Amiraneli [1.4K]

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Hey!

-----------------------------------------------

Solution:

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Answer:

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