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dangina [55]
3 years ago
8

Which equation best represents the model?

Mathematics
1 answer:
igor_vitrenko [27]3 years ago
5 0
C is the the answer. Hope this helped
You might be interested in
Please help.
denis23 [38]

9514 1404 393

Answer:

  1. Angle 1 = 139°
  2. Angle 2 = 41°
  3. x = 29; exterior angle = 131°

Step-by-step explanation:

These problems let you make use of the fact that the sum of the remote interior angles is equal to the exterior angle.

__

1. 53° +86° = ∠1

  139° = ∠1

__

2. ∠2 +92° = 133°

  ∠2 = 133° -92°

  ∠2 = 41°

__

3. (x +9)° +93° = (4x+15)°

  87 = 3x . . . . . . . . . . . . . . . . subtract x+15°

  29 = x . . . . . . . divide by 3

The exterior angle is ...

  (4x +15)° = (4·29 +15)° = 131° . . . exterior angle

3 0
3 years ago
Calculate the value of x
Katena32 [7]

Answer:

22⁰

Step-by-step explanation:

By angle sum property :

x + 127 + 31 = 180⁰

x = 180 - 158

x = 22⁰

3 0
3 years ago
Read 2 more answers
HURRY PLEASE!!!!!
RideAnS [48]
First we need to find the radius of the cylindrical tree trunk if the circumference C=4 feet: C=2rπ, r=4/(2π)=2/π. Now we find the volume V: V=r^2π*h, where h=20 feet is the height. So we input the numbers: V=(2/π)^2*π*20=(4/π²)*π*20. π in the nominator and the denominator cancel out: V=(4/π)*40=80/π. The volume of the cylindrical tree trunk is V=80/π feet^3
7 0
3 years ago
Use the following area model to multiply 29x97
alexandr402 [8]
The answer to 29×97 would be 2813
6 0
3 years ago
Consider the game of independently throwing three fair six-sided dice. There are six combi- nations in which the three resulting
Murrr4er [49]

Answer:

See explanation below.

Step-by-step explanation:

1) First let's take a look at the combinations that sum up 10:

  1. 1+3+ 6,
  2. 1+ 4+ 5,
  3. 2+2+6,
  4. 2+3+5,
  5. 2 + 4 + 4,
  6. 3+3+4

Notice that when we have 3 different numbers on the dice, we can permute them in 6 different ways. For example: Let's take 1 + 3 + 6, we can get this sum with these permutations:

1 + 3 + 6, 1 + 6 + 3, 3 + 6 + 1, 3 + 1 + 6, 6 + 1 + 3, 6 + 3 + 1.

And when we have two different numbers on the dice, we can permute them in 3 different ways:

2 + 2 + 6, 2 + 6 +2, 6 + 2 + 2.

So now we're going to write down the 6 combinations that sum up 10 but we're going to write down how many permutations of them we get:

  1. 1+3+ 6 : 6 permutations
  2. 1+ 4+ 5 : 6 permutations
  3. 2+2+6: 3 permutations
  4. 2+3+5: 6 permutations
  5. 2 + 4 + 4: 3 permutations
  6. 3+3+4: 3 permutations

Total of permutations: 6 + 6 + 3 + 6 + 3 + 3 =27.

Thus we have 27 different ways of getting a sum of 10.

2) Now we're going to take a look at the combinations that sum up 9 and we're going to proceed in a similar way:

  1. 1 + 2 + 6: 6 permutations
  2. 1+3+5: 6 permutations
  3. 1+4+4: 3 permutations
  4. 2+ 3+ 4: 6 permutations
  5. 2+2 +5: 3 permutations
  6. 3+3+3: 1 permutation.

Total of permutations: 6 + 6 + 3 + 6 +3 + 1 = 25.

Thus we have 25 different ways of getting a sum of 10

And we can conclude that the probability of getting a total of 10 is larger than the probability to get a total of 9.

5 0
3 years ago
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