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photoshop1234 [79]
3 years ago
10

In the given figure ABCD is a parallelogram, E is the midpoint of BC. DE produced meets AB produced at L. Prove that AB = B L an

d AL=2DC​

Mathematics
1 answer:
aliina [53]3 years ago
7 0

Answer:

<em>See Reasoning Below</em>

Step-by-step explanation:

To prove that AB = BL, or in other words AB ≅ BL, let us consider the triangles CED and BEL. If we were to prove they were congruent, then by CPCTC ( corresponding parts of congruent triangle are congruent ) DC ≅ BL. As AB ≅  DC by " Properties of Parallelogram " it would be that through transitivity, AB ≅ BL / AB = BL;

m

Now for " part 2 " we can consider that AB = DC, from part 1. If AB = BL, then AL = 2 ( AB ) by the Partition Postulate. AB = DC, so we can also say that AL = 2 ( DC ) - Proved

See attachment in statement reasoning form for part 1;

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Genuinely confused rn
frosja888 [35]

Using it's concept, it is found that the probability that he winds up wearing the white shirt and tan pants is of \frac{1}{8}.

<h3>What is a probability?</h3>

A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.

In this problem:

  • For the shirt, there is 4 outcomes, hence the probability of the white shirt is 1/4.
  • For the pair of pants, there are 2 outcomes, blue or tan, hence the probability of tan pants is 1/2.

Since the shirt and the pants are independent, the probability is given by:

p = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}.

More can be learned about probabilities at brainly.com/question/14398287

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8 0
1 year ago
In how many distinct ways can the letters of the word mathematics be arranged? (first, does the order matter?)
TEA [102]
Our current list has 11!/2!11!/2! arrangements which we must divide into equivalence classes just as before, only this time the classes contain arrangements where only the two As are arranged, following this logic requires us to divide by arrangement of the 2 As giving (11!/2!)/2!=11!/(2!2)(11!/2!)/2!=11!/(2!2).

Repeating the process one last time for equivalence classes for arrangements of only T's leads us to divide the list once again by 2
4 0
2 years ago
Identify the interval that is not equal to the other three 15-19 30-34 40-45 45-49
Lunna [17]

Answer:

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Step-by-step explanation:

The first interval is 15-19.

The width of this interval is: 19-15=4

The second interval is 30-34.

The width of this interval is: 34-30=4

The third interval is 40-45.

The width of this interval is: 45-40=5

The fourth interval is 45-49.

The width of this interval is: 49-45=4

<h3>Therefore the interval that is not equal to the other three is 45-40</h3>
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Setler [38]

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Step-by-step explanation:

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2 years ago
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kkurt [141]
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3 years ago
Read 2 more answers
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