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2 years ago

How to prove these are congruent

Mathematics

1 answer:

pantera1 [17]2 years ago

4 0

Answer:

Yes, they are congruent by SAS postulate.

Step-by-step explanation:

Let us first name the vertices of both the triangles.

The labelled diagram is shown below.

Consider the triangles ABC and ACD

Statements Reasons

1. BC ≅ AC Given

2. ∠ ACB ≅ ∠ CAD Given

3. AC ≅ AC Common sides are congruent due to Reflexive property

As two corresponding sides and the included angles between them are congruent, therefore the two triangles are congruent by SAS postulate.

So, we can conclude that the given triangles are congruent.

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The editor of a textbook publishing company is deciding whether to publish a proposed textbook. Information on previous textbook

kogti [31]

Answer:

34.86% probability that it will be huge success

Step-by-step explanation:

Bayes Theorem:

Two events, A and B.

$P(B|A) = \frac{P(B)*P(A|B)}{P(A)}$

In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.

In this question:

Event A: Receiving a favorable review.

Event B: Being a huge success.

Information on previous textbooks published show that 20 % are huge successes

This means that $P(B) = 0.2$

99 % of the huge successes received favorable reviews

This means that $P(A|B) = 0.99$

Probability of receiving a favorable review:

20% are huge successes. Of those, 99% receive favorable reviews.

30% are modest successes. Of those, 70% receive favorable reviews.

30% break even. Of those, 40% receive favorable reviews.

20% are losers. Of those, 20% receive favorable reviews.

Then

$P(A) = 0.2*0.99 + 0.3*0.7 + 0.3*0.4 + 0.2*0.2 = 0.568$

Finally

$P(B|A) = \frac{P(B)*P(A|B)}{P(A)} = \frac{0.2*0.99}{0.568} = 0.3486$

34.86% probability that it will be huge success

4 0

2 years ago

Jose and Maris work for different car dealerships. Jose earns a monthly salary of $3,500 plus a 6% commision on his sales, x. Ma

patriot [66]

Answer:

x>$25000

Step-by-step explanation:

4 0

2 years ago

Grace has 1.35 pounds of strawberries, 1.4 pounds of bananas, and some apples. She has more pounds of apples than pounds of stra

satela [25.4K]

Answer:

Grace could have 1.36 pounds, 1.37 pounds, 1.38 pounds or 1.39 pounds of apples.

Step-by-step explanation:

Let a represent pounds of apples.

We have been given that Grace has 1.35 pounds of strawberries. She has more pounds of apples than pounds of strawberries.

This means that a is greater than 1.35. We can represent this information in an inequality as:

$a>1.35$

We are also told that Grace has 1.4 pounds of bananas. She has fewer pounds of apples than pounds of bananas. This means that a is less than 1.4. We can represent this information in an inequality as:

$a$

Upon combining both inequalities, we will get:

$1.35$

Therefore, Grace could have 1.36 pounds, 1.37 pounds, 1.38 pounds or 1.39 pounds of apples.

6 0

2 years ago

When a breeding group of animals is introduced into a restricted area such as a wildlife reserve, the population can be expected

jasenka [17]

Answer:

A. Initially, there were 12 deer.

B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. After 15 years, there will be 410 deer.

D. The deer population incresed by 30 specimens.

Step-by-step explanation:

$N=\frac{12.36}{0.03+0.55^t}$

The amount of deer that were initally in the reserve corresponds to the value of N when t=0

$N=\frac{12.36}{0.33+0.55^0}$

$N=\frac{12.36}{0.03+1} =\frac{12.36}{1.03} = 12$

A. Initially, there were 12 deer.

B. $N(10)=\frac{12.36}{0.03 + 0.55^t} =\frac{12.36}{0.03 + 0.0025}=\frac{12.36}{y}=380$

B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. $N(15)=\frac{12.36}{0.03+0.55^15}=\frac{12.36}{0.03 + 0.00013}=\frac{12.36}{0.03013}= 410$

C. After 15 years, there will be 410 deer.

D. The variation on the amount of deer from the 10th year to the 15th year is given by the next expression:

ΔN=N(15)-N(10)

ΔN=410 deer - 380 deer

ΔN= 30 deer.

D. The deer population incresed by 30 specimens.

8 0

2 years ago

What is the answer to 3^=55

kakasveta [241]

I think it’s 1.7444921e+26

8 0

2 years ago