... side. I already know that the (smallest number) ² + (middle number) ² = (largest number) ². So I know that there will be a connection between the numbers written above. The problem is that it is obviously not: (Middle number) ²+ ...

... 1) ² 2n +1 2n 4n² 2n +1 2n 1 = 4n² + 4n + 1 Middle² = (2n² + 2n)² 2n² +2n 2n² 4n4 4n³ +2n 4n³ 4n² = 4n4 + 8n³ + 4n² Longest² = (2n² + 2n + 1)² 2n² +2n +1 2n² 4n4 4n³ 2n² +2n 4n³ 2n² 2n +1 2n² 2n 1 = 4n4 + 8n³ + 8n² +4n + 1 A + B = C 4n² + 4n + 1 ...

... 6 5, 12, 13 Perimeter = 5 + 12 + 13 = 30 Area = ? x 5 x 12 = 30 7, 24, 25 Perimeter = 7 + 24 + 25 = 56 Area = ? x 7 x 24 = 84 From the first three ...

... 6 5, 12, 13 Perimeter = 5 + 12 + 13 = 30 Area = ? x 5 x 12 = 30 7, 24, 25 Perimeter = 7 + 24 + 25 = 56 Area = ? x 7 x 24 = 84 From the first three ...

... of anywhere near 1. Part 1: Aim: To investigate the family of Pythagorean Triplets where the shortest side (a) is an odd number and all three sides are positive integers. By putting the triplets I am provided with in a table, along with ...

... I think it will become 4,8,12,16,20,24 So it will be 4,12,24, 40, 60 84. The difference is 4,8,12. Now I shall find the difference and it is n*4 Next I shall find the prediction of the longest side next. 5,13,25 It goes up by ...

... rule to get the formula in the middle of my investigation, where I have to make generalisation about the lengths of sides. From the first terms I have noticed the following: - * 'a' increases by +2 each term * ...

... values: a b c 3 4 5 5 12 13 7 24 25 I will now predict the next two values in the table so I can work out a general formula for this pythagorean family. By using the differencing method I can see that 'a' has a difference of 2 between each pythagorean ...

... this therefore means they are not Pythagorean triples!!!! So from this you should notice that a2 + b2 = c2, if a=3, b=4, and c=5 (32+44 = 52) we can see that 32 + 42= 52(25) this therefore means these numbers ...

... which matched with my others instead of something completely different. If we wish to work out a formula, a general rule and a relationship then I am going to use the triple which is most likely to help with that. ...

... I think it will become 4,8,12,16,20,24 So it will be 4,12,24, 40, 60 84. The difference is 4,8,12. Now I shall find the difference and it is n*4 Next I shall find the prediction of the longest side next. 5,13,25 It goes up by ...

... all this information as merely legends but, even if the reader treats it in this way, being such an early record it is of historical importance. Pythagoras's father was Mnesarchus, while his mother was Pythais and she was a native of ...

... 5² = 25 The difference between 25 and 313 is 288 which is far to big, so this means that the equation I want has nothing to do with 3 sides squared. I will now try 2 sides squared. M² ...

... Do not cb redistribute Tochv8Ey 85 182 546 7 15 112 113 240 840 +2 8 17 144 145 306 1224 9 19 180 181 380 1710 wwdd ddw stddddud edd ddnt cdd enddtral ddcodd uk. 10 21 220 221 462 2310 I looked at the table and noticed that there was only 1 difference between the length of the middle side and the length of the longest ...

... It is also used to determine certain trignometrical relationships such as sin² ? + cos² ? = 1. Pythagoras' theorem for right-angled triangles is likely to have been known long before the time of Pythagoras as it was probably used ...

... squared. I will now try 2 sides squared. (Middle)2 + Largest number = (smallest number)2 = 122 + 13 = 52 = 144 + 13 = 25 = 157 = 25 This does not work and neither will 132, because it is larger than 122. ...

... squared. I will now try 2 sides squared. (Middle)2 + Largest number = (smallest number)2 = 122 + 13 = 52 = 144 + 13 = 25 = 157 = 25 This does not work and neither will 132, because it is larger than 122. ...

... found that the formulas for B and C are . I know this because I know that A =B+C and that C=B+1. Then I simply worked out what A was equal to. I then found that with every example I ...

... 6 5, 12, 13 Perimeter = 5 + 12 + 13 = 30 Area = 1/2 x 5 x 12 = 30 7, 24, 25 Perimeter = 7 + 24 + 25 = 56 Area = 1/2 x 7 x 24 = 84 From the first three ...

... = 25² I will now work out the perimeter and area Perimeter = a + b + c Area = 1/2 x a x b 3, 4, 5 Perimeter = 12 Area = 6 5, 12, 13 Perimeter = 30 Area = 30 7, 24, 25 Perimeter = 56 Area = ...

... believe that the answer to this question could be 3 as it is the first opportunity for C to win and it has a 50% chance of winning every throw it has (3/6). Obviously A and more so B will ...

... were divisible by 18: 25/5 = 5...324/18 = 18...625/5 = 125...104976/18 = 5832. You will notice that when any of the probabilities are divided by 5/18 the result is the previous probability. This is because the probability of a player winning is ...

... - 5 18 P (LLL) = 13 18 I also found the probability of A, B and C winning in Round 1: P (A) wins = 1 6 P (B) wins = 5 x 1 = 5 6 2 18 P (C) wins = 5 ...

... now stratify the sample. We will just tale 80 pupils i.e. a third of each of the entries above. Doing it his way we will ensure we have proportional representation. Therefore we have the following numbers. Instructor A Instructor B Instructor C Instructor D Males 10 16 6 7 Females 10 17 7 7 We ...

... = 313 5 = 25 The difference between 25 and 313 is 288 which is far to big, so this means that the equation I need and want has nothing to do with 3 sides being squared. So I shall ...