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Anettt [7]
3 years ago
13

According to the given information, segment AB is parallel to segment DC and segment BC is parallel to segment AD.. Using a stra

ightedge, extend segment AB and place point P above point B. By the same reasoning, extend segment AD and place point T to the left of point A. Angles BCD and PBC are congruent by the Alternate Interior Angles Theorem. Angles PBC and BAD are congruent by the Corresponding Angles Theorem. By the __________ Property of Equality, angles BCD and BAD are congruent. Angles ABC and BAT are congruent by the Alternate Interior Theorem. Angles BAT and CDA are congruent by the Corresponding Angles Theorem. By the __________ Property of Equality, ∠ABC is congruent to ∠CDA. Consequently, opposite angles of parallelogram ABCD are congruent. What properties accurately complete the proof?

Mathematics
2 answers:
Eva8 [605]3 years ago
7 0
Pls. see attachment. 

svetoff [14.1K]3 years ago
7 0
1. transitive

2. transitive
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The two triangles can be made as shown below, the bigger triangle will be an equilateral triangle with each side measuring 10 units, while the smaller triangle will be an isosceles triangle and the measure of the same length side will be equal to the radius of the circle.

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Step-by-step explanation:

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Of 100 students,65 are members of a mathematics club and 40 are members of a physics club. if 10 are members of neither club the
alexandr1967 [171]

a. 15 students are members of both clubs

b. 50 students only are members of the mathematics club

c. 25 students only are members of the physics club

Step-by-step explanation:

The given is:

  • There are 100 students
  • 65 are members of a mathematics club
  • 40 are members of a physics club
  • 10 are members of neither club

We need to find how many students are members of

a. both clubs?

b. only mathematics club

c. only physics club

∵ The total number of students = 100

∵ 10 are members of neither club

- Subtract 10 from 100 to find the members of the mathematics

  club or the physics club

∵ 100 - 10 = 90

∴ There are 90 members of the mathematics club or physics club

∴ n(mathematics or physics) = 90

∵ 65 students are members of a mathematics club

∵ n(mathematics) = 65

∵ 40 students are members of a physics club

∴ n(physics) = 40

∵ n(mathematics or physics) = n(mathematics) + n(physics) - n(both)

∴ 90 = 65 + 40 - n(both)

∴ 90 = 105 - n(both)

- Add n(both) to each side

∴ n(both) + 90 = 105

- Subtract 90 from each side

∴ n(both) = 15

∴ 15 students are members of both clubs

a. 15 students are members of both clubs

∵ 65 students are members of the mathematics club

∵ 15 of them are members of the physics club

∴ n(mathematics only) = 65 - 15 = 50

∴ 50 students only are members of the mathematics club

b. 50 students only are members of the mathematics club

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∵ 15 of them are members of the mathematics club

∴ n(physics only) = 40 - 15 = 25

∴ 25 students only are members of the physics club

c. 25 students only are members of the physics club

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Let's do the math for this question.

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