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3 years ago

The congruence theorem that can be used to prove triangle LON is congruent to triangle LMN is

Mathematics

2 answers:

USPshnik [31]3 years ago

8 0

Answer: SSS

Step-by-step explanation:

In the given picture we have two triangles $\triangle{LON}$ and $\triangle{LMN}$ in which

$\overline{LO}\cong\overline{LM}\ \ \ \text{[Given]}\\\\\overline{ON}\cong\overline{MN}\ \ \ \text{[Given]}\\\\\overline{LN}\cong\overlien{LN}\ \ \ \text{[Reflexive Property]}$

Thus by SSS congruence criteria we have,

$\triangle{LON}\cong\triangle{LMN}$

SSS congruence criteria says that if all three sides of a triangle are congruent to the corresponding sides of the other triangle then the triangle are congruent.

OlgaM077 [116]3 years ago

6 0

Sss, because LN=LN (reflexive property)

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How many different 5-letter radio station call letters can be made a. if the first letter must be Upper C comma Upper X comma U

Lady_Fox [76]

Answer:

a) 1,518,000

b) 2,284,880

c) 60,720

Step-by-step explanation:

a) a. if the first letter must be Upper C comma Upper X comma Upper T comma or Upper M and no letter may be repeated?

We draw 5 boxes, and based on that we will see the total possible cases. There are 26 alphabets

The first box should have C or X or T or M .No letter may be repeated.

Any Any Any Any Any

5 alphabets of the of the of the of the

C,X , T , M remaining remaining remaining remaining

25 alphabets 24 alphabets 23 alphabets 22 alphabets

Therefore; total possible call letters = 5 × 25 × 24 × 23 × 22 = 1,518,000

The first box should have C or X or T or M Repeats as allowed

Any Any Any Any Any

5 alphabets of the of the of the of the

C,X , T , M remaining remaining remaining remaining

26 alphabets 26 alphabets 26 alphabets 26 alphabets

Therefore Total possible call letters = 5 × 26 × 26 × 26 × 26 = 2,284,880

c) The first box should have C,X , T , M and end with S

So the last place if fixed, and we now have 25 alphabets. The first box can go in 5 ways. The next box then will have only 24 letters to choose from, as the first box has taken a letter and the last box already has S in it. Repetition not allowed

Any Any Any Any S

5 alphabets of the of the of the is fixed

C,X , T , M remaining remaining remaining here

24 alphabets 23 alphabets 22 alphabets

Therefore Total possible call letters = 5 × 24 × 23 × 22 × 1 = 60,720

3 0

3 years ago

Two girls had apples for 14 days. They had 512 in total. Each girl had an equal number of apples. How many apples did each girl

frosja888 [35]

You have to divide and 512/14=36 and 8/14 and that simplified is 36 and 4/7.

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4 years ago

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FinnZ [79.3K]

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Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$