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Alex
3 years ago
7

A. Triangle ABC and CDA by angle-side-angle

Mathematics
1 answer:
Debora [2.8K]3 years ago
4 0

Answer:

D. ABE and CDE by SAS Theorem (Side-Angle-Side)

Step-by-step explanation:

First lets establish that by the Vertical Angles Theorem, when two lines intersect, the vertical angles created are congruent. Thus, ∠AED is congruent to ∠CED. Now, we need to figure out which pair of triangles Craig is referring to. We know that to figure out if two triangles are congruent, they must have three congruent of either sides and/or angles. We know that Craig can't be referring to triangle AED and BEC because the angle of their vertex's are not congruent. We also know that Craig can't be referring to triangles ABC and CDA because although they share a similar side (Indicated by the one tic mark), they don't have anything else that is congruent.

Thus, this is why triangles ABE and CDE are the triangles Craig is referring to. Both of these triangles have congruent vertex because of the vertical angles theorem.  Both of these triangles also share the same sides as indicated by the one and two tic marks. Since there are two sides and an included angle, triangles ABE and CDE are congruent by the SAS theorem.

Finally, the way Craig can prove that AB is parallel to CD is by the CPCTC theorem. The CPCTC theorem states that corresponding parts of congruent triangles are congruent. Since we already established that the two triangles are congruent, we can safely say that line AB is parallel to CD.

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Answer:

9

Step-by-step explanation:

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9

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3 years ago
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Find the median, first quartile, third quartile, interquartile range, and any outliers for each set of data. 111, 68, 93, 88, 74
lilavasa [31]

Answer:

Q1 = 87 ;

Q2 = 102 ;

Q3 = 115 ;

IQR = 28

OUTLIER = None

Step-by-step explanation:

Given the data:

111, 68, 93, 88, 74, 152, 119, 87, 88, 105, 84, 102, 151, 115, 112

Ordered data : 68, 74, 84, 87, 88, 88, 93, 102, 105, 111, 112, 115, 119, 151, 152

Sample size, n = 15

The first quartile, Q1 = 1/4(n+1)th term

Q1 = 1/4(15+1)th term

Q1 = 1/4(16) = 4th term

Q1 = 87

The Median , Q2 = 1/2(n+1)th term

Q2 = 1/2(15+1)th term

Q2 = 1/2(16) = 8th term

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The third quartile, Q3 = 3/4(n+1)th term

Q3 = 1/4(15+1)th term

Q3 = 3/4(16) = 12th term

Q3 = 115

The interquartile range, IQR = Q3 - Q1

IQR = (115 - 87) = 28

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Q2 = 102 ;

Q3 = 115 ;

IQR = 28

OUTLIER = None

4 0
3 years ago
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Answer:

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Step-by-step explanation:

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Step 1: Distribute

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Step 2: Combine like terms

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Step-by-step explanation:

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