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iris [78.8K]
3 years ago
14

Maggie constructs a triangle with sides lengths 7 centimeters long and 10 centimeters long.

Mathematics
1 answer:
MakcuM [25]3 years ago
7 0

7 centimeters is a possible length for the third side ⇒ B

Step-by-step explanation:

Let us revise the triangle Inequality Theorem

  • The sum of the lengths of any 2 sides of a triangle must be greater than the length of the third side
  • To prove that by easy way add the smallest two sides, if their sum greater than the third side,then the sides can form a triangle

Assume that the length of the third side is x cm

∵ The length of two sides are 7 cm and 10 cm

∵ The length of the third side is x cm

- Put the sum of x and 7 greater than 10 ( x and 7 are the smallest sides)

∴ x + 7 > 10

- Subtract 7 from both sides

∴ x > 3

- Put the sum of 7 and 10 greater than x (7 and 10 are the smallest sides)

∵ 7 + 10 > x

∴ 17 > x

∴ x < 17

- By using one inequality for x (combined the two inequalities in one)

∴ 3 < x < 17

That means the length of the third side is any number between 3 and 17

There is only one answer between 3 and 17

∵ 7 is between 3 and 17

∴ The length of the third side could be 7 cm

7 centimeters is a possible length for the third side

Learn more:

You can learn more about triangles in brainly.com/question/4599754

#LearnwithBrainly

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

<em>use </em><em>the </em><em>formula</em><em> for</em><em> </em><em>the</em><em> </em><em>nth </em><em>term </em><em>of </em><em>an </em><em>ap </em><em>Tn=</em><em>a+</em><em>(</em><em>n-1)</em><em>d</em>

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<em>T18=</em><em>1</em><em>3</em><em>.</em><em>5</em>

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<em>T12=</em><em>a+</em><em>(</em><em>1</em><em>2</em><em>-</em><em>1</em><em>)</em><em>d</em>

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<em>T18</em><em>=</em><em>a+</em><em>(</em><em>1</em><em>8</em><em>-</em><em>1</em><em>)</em><em>d</em>

<em>1</em><em>3</em><em>.</em><em>5</em><em>=</em><em>a+</em><em>1</em><em>7</em><em>d</em><em>(</em><em>2</em><em>n</em><em>d</em><em> </em><em>equation</em><em>)</em>

<em>then </em><em>solve </em><em>both </em><em>as </em><em>a </em><em>simultaneous</em><em> </em><em>equation</em>

<em>a+</em><em>1</em><em>1</em><em>d</em><em>=</em><em>1</em><em>0</em><em>.</em><em>5</em>

<em>a+</em><em>1</em><em>7</em><em>d</em><em>=</em><em>1</em><em>3</em><em>.</em><em>5</em>

<em> </em><em> </em><em> </em><em> </em><em>-6d/</em><em>-</em><em>6</em><em>=</em><em>-</em><em>3</em><em>/</em><em>-</em><em>6</em>

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<em>a=</em><em>1</em><em>0</em><em>.</em><em>5</em><em>-</em><em>5</em><em>.</em><em>5</em>

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<em>I </em><em>hope </em><em>this </em><em>helps</em>

<em>please </em><em>mark</em><em> </em><em>as </em><em>brainliest</em>

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