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Ugo [173]
3 years ago
11

PLEASE HELP I WILL GIVE BRAINLIEST TO THE FIRST PERSON TO ANSWER

Mathematics
1 answer:
jekas [21]3 years ago
4 0

Answer:

1. 2x = 60

x = 30

y = x - 5 = 30 - 5 = 25

2. x= 11

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In triangle RST, m∠R > m∠S + m∠T. Which must be true of triangle RST? Check all that apply.
solmaris [256]

Answer:

1. m∠R > 90°

2. m∠S + m∠T < 90°

4. m∠R > m∠T

5. m∠R > m∠S

Step-by-step explanation:

<h3>General strategy</h3>
  1. prove the statement starting from known facts, or
  2. disprove the statement by finding a counterexample

Helpful fact:  Recall that the Triangle Sum Theorem states that m∠R + m∠S + m∠T = 180°.

<u>Option 1.  m∠R > 90°</u>

Start with m∠R > m∠S + m∠T.

Adding m∠R to both sides of the inequality...

m∠R + m∠R > m∠R + m∠S + m∠T

There are two things to note here:

  1. The left side of this inequality is 2*m∠R
  2. The right side of the inequality is exactly equal to the Triangle Sum Theorem expression

2* m∠R > 180°

Dividing both sides of the inequality by 2...

m∠R > 90°

So, the first option must be true.

<u>Option 2.  m∠S + m∠T < 90°</u>

Start with m∠R > m∠S + m∠T.

Adding (m∠S + m∠T) to both sides of the inequality...

m∠R + (m∠S + m∠T) >  m∠S + m∠T + (m∠S + m∠T)

There are two things to note here:

  1. The left side of this inequality is exactly equal to the Triangle Sum Theorem expression
  2. The right side of the inequality is 2*(m∠S+m∠T)

Substituting

180° > 2* (m∠S+m∠T)

Dividing both sides of the inequality by 2...

90° > m∠S+m∠T

So, the second option must be true.

<u>Option 3.  m∠S = m∠T</u>

Not necessarily.  While m∠S could equal m∠T, it doesn't have to.  

Example 1:  m∠S = m∠T = 10°;  By the triangle sum Theorem, m∠R = 160°, and the angles satisfy the original inequality.

Example 2:  m∠S = 15°, and m∠T = 10°;  By the triangle sum Theorem, m∠R = 155°, and the angles still satisfy the original inequality.

So, option 3 does NOT have to be true.

<u>Option 4.  m∠R > m∠T</u>

Start with the fact that ∠S is an angle of a triangle, so m∠S cannot be zero or negative, and thus m∠S > 0.

Add m∠T to both sides.

(m∠S) + m∠T > (0) + m∠T

m∠S + m∠T > m∠T

Recall that m∠R > m∠S + m∠T.

By the transitive property of inequalities, m∠R > m∠T.

So, option 4 must be true.

<u>Option 5.  m∠R > m∠S</u>

Start with the fact that ∠T is an angle of a triangle, so m∠T cannot be zero or negative, and thus m∠T > 0.

Add m∠S to both sides.

m∠S + (m∠T) > m∠S + (0)

m∠S + m∠T > m∠S

Recall that m∠R > m∠S + m∠T.

By the transitive property of inequalities, m∠R > m∠S.

So, option 5 must be true.

<u>Option 6.  m∠S > m∠T</u>

Not necessarily.  While m∠S could be greater than m∠T, it doesn't have to be.  (See examples 1 and 2 from option 3.)

So, option 6 does NOT have to be true.

4 0
2 years ago
Parallel lines r and s are cut by two transversals, parallel lines t and u.
Fiesta28 [93]

Answer:

5 and 13

Step-by-step explanation:

4 0
3 years ago
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Please find the square roots of the following complex number.
Nana76 [90]

Hey there,

First, find the square root of the given equation.

\sqrt{z}=\sqrt{81(\text{cos}(\frac{4\pi}{9})+i\text{sin}(\frac{4\pi}{9})}\\=9\text{cos}\frac{1}{2}(2k\pi+(\frac{4\pi}{9}))+i\text{sin}\frac{1}{2} (2k\pi+(\frac{4\pi}{9}))

Keep in mind that this is solved for k = 0 and 1.

Second, simplify.

=9\text{cos}(\frac{2\pi}{9})+i\text{sin}(\frac{2\pi}{9})~\text{or}~9\text{cos}(\frac{11\pi}{9})+i\text{sin}(\frac{11\pi}{9})

Best of Luck!

6 0
3 years ago
Find x if segment A B is perpendicular to segment C D in the figure below.
ki77a [65]
If segment AB is perpendicular to segment CD, then a right angle (90°) is formed.


7x + 27 = 90

7x = 63

x = 9
3 0
4 years ago
Read 2 more answers
4x what is close to 97
Hitman42 [59]

Lets try  96

4 x  ?= 96

? = 96/4

= 24

So we see that 4 times 24 is close to 97.

7 0
3 years ago
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