Average = (7 + 14 + 12) / 3Average = 33 / 3 = 11
7 X 2 = 14 6 X 2 = 12 12 + 14 = 26 ----
12=)Improved Answer:-It has 14 because the formula of 0.5*(49-27) = 14 diagonals
From the Pythagorean Theorem: c^2 = a^2 + b^2. So,c = √(a^2 + b^2) substitute the given values:c = √(4^2 + 3^2)c = √(16 + 9)c = √25c = 5 (since the length is always positive)One of the Pythagorean triples is 3,4,5. So, if you know all the Pythagorean triples, you don't need to do the computations above.The Pythagorean triple: A set of three positive integers a, b, and c such that a^2 + b^2 = c^2. Pythagorean triples that have greatest common divisor equal to 1 include the following: {3, 4, 5}, {5,12, 13}, {8, 15, 17}, {7, 24, 25}, and {20, 21, 29}.
There are more than four, but so I'll give you the ones that I know: 3-4-5. 5-12-13. 8-15-17, 7-24-25. 12-35-37, and 9-40-41. Multiples of these triplets work as well.
No.
The combination 5, 12 and 13 is a Pythagorean triple.
7
There are probably tens of such sets. Some are: 5, 6, 7, 7, 8, 11, 13, 14, 15, 16, 17, 18 5, 6, 7, 7, 7, 12, 12, 14, 15, 16, 17, 18 5, 6, 7, 7, 7, 12, 12, 14, 14, 16, 17, 18 5, 6, 7, 7, 7, 12, 12, 14, 14, 15, 17, 18 5, 6, 7, 7, 7, 12, 12, 14, 15, 15, 17, 18 5, 6, 7, 7, 7, 12, 12, 14, 15, 16, 16, 18 5, 6, 7, 7, 7, 12, 12, 14, 14, 16, 16, 18
There are infinitely many combinations. Using any pair of integers m and n such that m > n > 0, the integers m2 - n2, 2mn and m2 + n2 will form a Pythagorean triple.
Good old Arthur. There is a Pythagorean triple (5-12-13) lurking in this question. Diagonal is 26 cm making x = 7
12/14 in its simplest form is 6/7
6/7 = 12/141/2 = 7/14(6/7) - (1/2) = (12/14) - (7/14) = 5/14
Average = (7 + 14 + 12) / 3Average = 33 / 3 = 11
7
A Pythagorean triple is a group of three whole number which can be the sides of a right angled triangle, eg: (3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17), (9, 40, 41), (11, 60, 61), (12, 35, 37), (13, 84, 85), (16, 63, 65), (20, 21, 29), (28, 45, 53), (33, 56, 65), (36, 77, 85), (39, 80, 89), (48, 55, 73), (65, 72, 97)
( 3 , 4 , 5 ) ( 5, 12, 13) ( 7, 24, 25)