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pashok25 [27]
3 years ago
14

In △QRS, QR¯¯¯¯¯¯¯¯ has been bisected at point T, and TU¯¯¯¯¯¯¯ has been drawn such that U is the point of intersection of TU¯¯¯

¯¯¯¯ and RS¯¯¯¯¯¯¯. Which statement must be true in order to prove that TU=12QS?(1 point)
∠RTU≅∠RUT

∠TQS≅∠RUT

∠RTU≅∠TQS

∠TQS≅∠USQ
Mathematics
1 answer:
Allushta [10]3 years ago
5 0

9514 1404 393

Answer:

  ∠RTU≅∠TQS

Step-by-step explanation:

To show TU ║ QS, the angles made by transversal RQ with each of those segments must be shown to be congruent. That is, it must be true that ...

  ∠RTU≅∠TQS

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Please help me to solve this sum<br>please gI've correct answer​
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This question requires creating a few equations and working through them step-by-step. Now, first let's give each of the shapes a variable: let's say that the blue shape is a, the orange shape is b and the green shape is c.

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We can see in row 2 that there are three of the same shape (a) that add to 57. This makes it very simple to calculate the value of the shape.

Since 3a = 57

a = 57/3 = 19

2. Now we need to find a row or column that includes a and one other shape; we could choose either column 2 or 3, so let's go with column 2. Remembering that the blue shape is a and the orange shape is b:

2a + b = 50

Now, given that a = 19:

2(19) + b = 50

38 + b = 50

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3. We can now take any of the rows or columns that include the third shape (c) since we already know the values of the other two shapes. Let's take column 1:

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31 + c = 38

c = 38 - 31

c = 7

Thus, the value of the blue shape is 19, the value of the orange shape is 12 and the value of the green shape is 7.

3 0
3 years ago
What is the width of heng’s area model??
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Answer:

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3 years ago
There are 15 red, 14 green, and 9 yellow balls in a dark room. What is the minimum number of balls that Alicia needs to take so
Tems11 [23]

Answer:

32

Step-by-step explanation:

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2. If you take 30 balls, 9 could be yellow, 10 could be red and 11 could be green.  We want at least 12 of the same color, so 30 balls isn't a good minimum.

3. If you take 31 balls, 9 could be yellow, 11 could be red and 11 could be green.  We want at least 12 of the same color, so 31 balls isn't a good minimum.

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We want at least 12 of the same color, and we have it from the different samples of the 4th trial.

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No

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3 years ago
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