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Lostsunrise [7]
3 years ago
9

How is the pythogorean theorem used to determine the distance formula

Mathematics
1 answer:
Fiesta28 [93]3 years ago
5 0

Answer:

The distance formula uses the coordinates of points and the Pythagorean Theorem to calculate the distance between points. If A and B form the hypotenuse of a right triangle, then the length of AB can be found using this formula, leg2 + leg2 = hypotenuse2 (or you can use A squared + B squared = C squared)

Am not a real genius but I hope that answers your question! ^w^

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Word problem
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Answer:

p=2

Step-by-step explanation:

4.05p+14.40=4.50(p+3)          < equation

4.05p+14.40=4.50p+13.50     < multiply

14.40=.45p+13.50                   < subtract

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2=p                                          < divide

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3 years ago
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alex41 [277]

the correct answer should be (b) 6 is extraneous.

If you look at the second row of the solution you see (x-6) as a factor in one of the denominators. That's a problem: if 6 were a solution this denominator would become 0. But denominators being 0 is bad (undefined), so while 6 comes out of the algebraic manipulation as a solution, it is invalid, or extraneous.

-3 is a fine solution.

6 0
3 years ago
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Explain how an estimate helps you place the decimal point when multiplying 3.7×5.1
Cloud [144]
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3 0
3 years ago
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ohaa [14]

Answer:

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Step-by-step explanation:

3x^2-2x-1=0\\\\3(x^2-\frac{2}{3}x-\frac{1}{3})=0\\ \\3(x^2-\frac{2}{3}x-\frac{1}{3}+\frac{4}{9})=3(\frac{4}{9})\\ \\ 3(x^2-\frac{2}{3}x+\frac{1}{9})=\frac{4}{3}\\ \\ 3(x-\frac{1}{3})^2=\frac{4}{3}\\ \\(x-\frac{1}{3})^2=\frac{4}{9}\\ \\x-\frac{1}{3}=\pm\frac{2}{3}

x-\frac{1}{3}=\frac{2}{3}\\ \\x=\frac{3}{3}\\ \\x=1

x-\frac{1}{3}=-\frac{2}{3}\\ \\x=-\frac{1}{3}

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8 0
2 years ago
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