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const2013 [10]
3 years ago
12

A polygon with 7 sides has how many triangles​

Mathematics
1 answer:
vivado [14]3 years ago
3 0

Answer:

I THINK THE ANSWER IS 287

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Step-by-step explanation:

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sp2606 [1]

Answer:

I think that what you are trying to show is:  If s is irrational and r is rational, then r+s  is rational. If so, a proof can be as follows:

Step-by-step explanation:

Suppose that r+s is a rational number. Then r and r+s can be written as follows

r=\frac{p_{1}}{q_{1}}, \,p_{1}\in \mathbb{Z}, q_{1}\in \mathbb{Z}, q_{1}\neq 0

r+s=\frac{p_{2}}{q_{2}}, \,p_{2}\in \mathbb{Z}, q_{2}\in \mathbb{Z}, q_{2}\neq 0

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r+s=\frac{p_{1}}{q_{1}}+s=\frac{p_{2}}{q_{2}}

Then

s=\frac{p_{2}}{q_{2}}-\frac{p_{1}}{q_{1}}=\frac{p_{2}q_{1}-p_{1}q_{2}}{q_{1}q_{2}}\in \mathbb{Q}

This is a contradiction because we assumed that s is an irrational number.

Then r+s must be an irrational number.

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Step-by-step explanation:

This is the expanded form if this is what you wanted

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