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2 years ago

Can someone help me with this proof

Mathematics

1 answer:

maksim [4K]2 years ago

8 0

Step-by-step explanation:

Given :

Given that lines a and b are parallel, angles 1 and 5 are congruent because they are corresponding angles, and angles 1 and 4 are congruent because they are vertical angles

To find : by which property are angles 4 and 5 congruent

Solution :

We know that if two parallel lines are cut by a transversal, then the pairs of alternate interior angles are congruent.

Also, we know that if two things are equal to the same thing then they are equal to each other . In this case, we can say that if two angles are congruent to a third angle, then they are congruent to each other. As angles 4 and 5 are both congruent to angle 1, they are congruent to each other but angles 4 and 5 are alternate interior angles. So, if parallel lines have a transversal, alternate interior angles are congruent.

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Which algebraic expression is a polynomial with a degree of 5

anygoal [31]

Answer: For example, the polynomial which can also be expressed as has three terms. The first term has a degree of 5 (the sum of the powers 2 and 3), the second term has a degree of 1, and the last term has a degree of 0. Therefore, the polynomial has a degree of 5, which is the highest degree of any term.

Step-by-step explanation: (If this helps)

8 0

3 years ago

Read 2 more answers

There are 14 juniors and 16 seniors in a chess club. a) From the 30 members, how many ways are there to arrange 5 members of the

Dmitrij [34]

Answer:

a) 17,100,720

b) 4,717,440

c) 10,920

d) 2821

Step-by-step explanation:

14 juniors and 16 seniors = 30 people

a) From the 30 members, how many ways are there to arrange 5 members of the club in a line?

As it is a ordered arrangement

30.29.28.27.26 = 17,100,720

b) How many ways are there to arrange 5 members of the club in a line if there must be a senior at the beginning of the line and at the end of the line?

16.28.27.26.15 = 4,717,440

c) If the club sends 2 juniors and 2 seniors to the tournament, how many possible groupings are there?

Not ordered arrangement. And means that we need to multiply the results.

C₁₄,₂ * C₁₆,₂

C₁₄,₂ = 14.13.12! = 14.13 = 91

12! 2! 2

C₁₆,₂ = 16.15.14! = 16.15 = 120

14! 2! 2

C₁₄,₂ * C₁₆,₂ = 91.120 = 10,920

d) If the club sends either 4 juniors or 4 seniors, how many possible groupings are there?

Or means that we need to sum the results.

C₁₄,₄ + C₁₆,₄

C₁₄,₄ = 14.13.12.11.10! = 14.13.12.11 = 1001

10! 4! 4.3.2.1

C₁₆,₄ = 16.15.14.13.12! = 16.15.14.13 = 1820

12! 4! 4.3.2.1

C₁₄,₄ + C₁₆,₄ = 1001 + 1820 = 2821

7 0

3 years ago

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Digiron [165]

Here is what I got:(maybe wrong) 12*8*3 = 288, which is how big the box is. Then, you remove 1.5 inches from the height and get 10.5*8*3, which is 252. Subtract 252 from 288 to get the answer, 36.

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3 years ago

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irakobra [83]

Answer: $\frac{10}{3}$