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andrew11 [14]
2 years ago
9

determine if it can be proven that the triangles are congruent if they are the choose a theorem that proves its congruent ​

Mathematics
1 answer:
Lostsunrise [7]2 years ago
7 0

Answer:

These triangles are congruent by SAS

Step-by-step explanation:

They share the same angle which is in the same angle which is the vertical angle with congruent sides on either side of the angle, therefore this is congruent by SAS.

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Three times b less than 100 is equal to the product of 6 and b
Irina-Kira [14]
Not exactly sure which equation you meant:

3(100-b)=6b
300-3b=6b
300=9b
b= about 33

100-3b=6b
100=9b
b=about 11
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The angle of 2 is 150 degrees
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jeka94

Answer:

to present another's viewpoint

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2. Check the boxes for the following sets that are closed under the given
son4ous [18]

The properties of the mathematical sequence allow us to find that the recurrence term is 1 and the operation for each sequence is

   a) Subtraction

   b) Addition

   c) AdditionSum

   d) in this case we have two possibilities

       * If we move to the right the addition

       * If we move to the left the subtraction

The sequence is a set of elements arranged one after another related by some mathematical relationship. The elements of the sequence are called terms.

The sequences shown can be defined by recurrence relations.

Let's analyze each sequence shown, the ellipsis indicates where the sequence advances.

a) ... -7, -6, -5, -4, -3

We can observe that each term has a difference of one unit; if we subtract 1 from the term to the right, we obtain the following term

        -3 -1 = -4

        -4 -1 = -5

        -7 -1 = -8

Therefore the mathematical operation is the subtraction.

b) 0. \sqrt{1}. \sqrt{4}, \sqrt{9}, \sqrt{16}, \sqrt{25}  ...

In this case we can see more clearly the sequence when writing in this way

      0, \sqrt{1^2}. \sqrt{2^2}, \sqrt{3^2 } . \sqrt{4^2} , \sqrt{5^2}

each term is found by adding 1 to the current term,

      \sqrt{(0+1)^2} = \sqrt{1^2} \\\sqrt{(1+1)^2} = \sqrt{2^2}\\\sqrt{(2+1)^2} = \sqrt{3^2}\\\sqrt{(5+1)^2} = \sqrt{6^2}

Therefore the mathematical operation is the addition

c)   ... \frac{-10}{2}. \frac{-8}{2}, \frac{-6}{2}, \frac{-4}{2}. \frac{-2}{2}. ...

      The recurrence term is unity, with the fact that the sequence extends to the right and to the left the operation is

  • To move to the right add 1

           -\frac{-10}{2} + 1 = \frac{-10}{2}  -   \frac{2}{2}  = \frac{-8}{2}\\\frac{-8}{2} + \frac{2}{2} = \frac{-6}{2}

  • To move left subtract 1

         \frac{-2}{2} - 1 = \frac{-4}{2}\\\frac{-4}{2} - \frac{2}{2} = \frac{-6}{2}

         

Using the properties the mathematical sequence we find that the recurrence term is 1 and the operation for each sequence is

   a) Subtraction

   b) Sum

   c) Sum

   d) This case we have two possibilities

  •  If we move to the right the sum
  •  If we move to the left we subtract

Learn more here: brainly.com/question/4626313

5 0
2 years ago
Find the value of the expression 12 ÷ v for v = 2
Zarrin [17]

Answer:

Step-by-step explanation:

12 / v....when v = 2

12 / 2 = 6 <==

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3 years ago
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