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Sveta_85 [38]
3 years ago
12

The sets of numbers 7, 24, 25 and 9, 40, 41 are pythagorean triples. use what you know about the pythagorean theorem and explain

or show why they are pythagorean triples. be sure to show your work for each set of triples! (5 points)
Mathematics
2 answers:
nadya68 [22]3 years ago
3 0
Pythagorean triples are sets of numbers that when plugged in to the pythagorean theorem they come out an whole number

7 squared plus 24 squared equals 25 squared
49 plus 576 equals 625
625 equals 625 
This would be a pythagorean triple

9 squared plus 40 squared equals 41 squared
81 plus 1600 equals 1681
1681 equals 1681
This would be a pythagorean triple

Hope this helps :)
Harman [31]3 years ago
3 0

Answer:

Check below.

Step-by-step explanation:

The Pitagoras' theorem says that for a right triangle is true that  

a^2+b^2 = c^2

where a and b are the legs and c the hypotenuse. Now, a pythagorean triple are three numbers that satisfies that rule. That is, we have that three integers x,y and n are a pytagorean triple if they satisfy that

x^2+y^2 = n^2.

For the case of 7,24 and 25:

7^2 + 24^2 = 49 + 576 = 625 and \sqrt{625} = 25, then 25^2 = 625, so

7^2 + 24^2 = 25^2

For the case of 9, 40 and 41:

9^2 + 40^2 = 81 + 1600 = 1681 and \sqrt{1681} = 41, then 41^2 = 1681, so

9^2 + 40^2 = 41^2.

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A farmer with 950 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to o
Gelneren [198K]

Answer:

22,562.5 ft²

Step-by-step explanation:

The largest possible area will be obtained when half the fencing is used for the long side of the 4 pens, so the dimension in that direction is (950 ft)/(2·2) = 237.5 ft.

The other half of the fencing will be used for the 2 ends and 3 partitions, each of which will be (950 ft)/(2·5) = 95 ft.

Then the overall area of the 4 pens is ...

... (237.5 ft)(95 ft) = 22,562.5 ft²

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<em>General Solution</em>

Suppose L is the length of fence available, and x is the length of the long side of the enclosed area. For n pens, the enclosed area will be ...

... A = x(L-2x)/(n+1)

For constant values of A, L, n, this describes a downward-opening parabola with zeros at x=0 and x=L/2. The vertex of the parabola (point of maximum area) will be halfway between these zeros, at x = (0 + L/2)/2 = L/4. That is, half the available fence is used in each of the orthogonal directions.

Note that adding partitions in the other direction replaces the 2 in the equation with (m+1), where m is the number of pens between the sides of length x. That is, if there are 4 pens in one direction by 3 pens in the other direction, the area will be

... A = x(L -(3+1)x)/(4+1)

and, once again, we find that half the fence is used in each of the orthogonal directions when we maximize the overall area.

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Answer:

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Step-by-step explanation:

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