Pyhtagorean Theorem
Pythagoras' theorem states that a²+b²=c² As you can see, a is the longest side, b is the middle side and c is the longest side (hypotenuse). The point of this coursework is to find b when a is an odd number and all of the sides are positive integers. Then after that I will go looking when a is a positive number. The numbers 3, 4 and 5 work in Pythagoras' theorem, 3²+4²=5² because 3²=3×3=9 4²=4×4=16 5²=5×5=25 and so 3²+4²=9+16=25=5² The numbers 5, 12 and 13 also work, 5²+12²=13² because 5²=5×5=25 12²=12×12=144 13²=13×13=169 and so 5²+12²=25+144=169=13² The numbers 7, 24 and 25 also work, 7²+24²=25² because 7²=7×7=49 24²=24×24=576 25²=25×25=625 and so 7²+24²=49+576=625=25² 3, 4 and 5 Perimeter= 3+4+5=12 Area= 1/2 ×3×4=6 5, 12 and 13 Perimeter= 5+12+13=30 Area= 1/2 ×5×12=30 7, 24 and 25 Perimeter= 7+24+25=56 Area= 1/2 ×7×24=84 From the first three terms I have realised that: - * a increases by 2 each time * a is equal to the formula ×2+1 from n * b is always even * c is always odd * c is always +1 of b * b=(a×n) + n I have also added 2 more terms using what I think are my formulae. 'n' 'a' 'b' 'c' Perimeter Area 3 4 5 2 6 2 5 2 3 30 30 3 7 24 25 56 84 4 9 40 41 90 80 5 1 60 61 32 330 Here are my formulas. They are formulas on how to get from n to all of the
The Die Investigation.
3 People 'A, B & C' are playing a game of die. A wins the game if he throws a '1', B wins the game if he throws a '2' or a '3' and C wins the game if he throws a '4', '5' or a '6'. Once someone has thrown their number that's it, the game is over and they've won. A goes first, B second and C third. If none of those throw their number then A has another go and then if he doesn't get it, it goes to B and so on until someone wins. * What is the probability of each person winning? * How many goes do you reckon it should take until someone wins? I have decided to draw a tree diagram to see what I am dealing with and then put together in a chart the probability of each person winning on their first, second and third attempts. Then from that see if there are any patterns evolving. Then progress forward until I may get the right answer. I first set up a tree diagram to show what is happening. A B C W 1/6 * W 10/36 W 60/216 L L L # W - Wins L - Loses At # the diagram then starts again at * and follows this pattern continuously except the final probabilities will be different. Next I wrote down in a table the probabilities of A, B and C winning on their first, second, third and fourth attempts. Multiplying down through the probability tree's branches did this. Attempt 1 Attempt 2 Attempt 3 Attempt 4 A /6
Investigation into a driving test.
Investigation into a driving test By Declan Gervin I downloaded the following database from the Internet. Number of Number of Gender hour lessons minor mistakes Instructor Day Time M 8 3 A Mon 7.00 M 9 9 A Wed 9.00 M 4 0 A Fri 3.00 F 5 1 A Mon 9.00 F 5 5 A Tue 5.00 F 5 2 A Mon 4.00 F 20 8 A Wed 0.00 F 20 4 A Fri 9.00 F 20 9 A Mon 3.00 M 20 1 A Fri 9.00 M 20 2 A Thur 4.00 M 20 9 A Fri 9.00 M 20 8 A Mon 0.00 F 8 0 A Wed 0.00 F 0 A Thur 9.00 F 0 4 A Thur 0.00 F 1 3 A Fri 3.00 F 3 4 A Wed 3.00 M 5 4 A Mon 5.00 M 5 9 A Fri 0.00 M 6 4 A Wed 7.00 M 7 0 A Fri 1.00 F 23 5 A Mon 9.00 F 25 6 A Mon 5.00 F 27 5 A Fri 4.00 M 8 6 A Fri 3.00 M 9 9 A Thur 0.00 M 9 3 A Mon 2.00 F 5 1 A Wed 9.00 F 5 9 A Mon 6.00 F 9 3 A Fri 6.00 F 22 4 A Thur 6.00 M 24 5 A Mon 7.00 M 25 4 A Fri 0.00 M 26 3 A Wed 7.00 M 27 2 A Tue 7.00 F 6 5 A Mon 2.00 F 7 6 A Wed 0.00 M 21 8 A Thur 6.00 M 22 7 A Thur 7.00 F 5 27 A Fri 4.00 F 30 2 A Fri 0.00 F 1 7 A Mon 6.00 M 7 8 A Tue 5.00 F 2 3 A Thur 7.00 M 8 31 A Wed 0.00 M 6 23 A Mon 1.00 F 23 4 A Wed 4.00 M 9 9 A Thur 2.00 F 6 7 A Fri 9.00 M 4 8 A Fri 7.00 M 9
Beyond Pythagoras
Beyond Pythagoras In this piece of coursework I will be exploring Pythagorean triples which are beyond Pythagoras. Pythagoras is where you have a right angled triangle and you know the values of sides a + b but you don't know what side c is (hypotenuse). To calculate the hypotenuse you can square sides a + b and add the two answers together as one total. This total is equal to c2 so all you have to do now is find the square root of the total and you have worked out side c. A Pythagorean triple is where the first two sides of a triangle (a + b) suit this equation: a2 + b2 = c2. For example lets say a=1, b=2, c=3. Now let's try this in a Pythagoras equation: a2 + b2 = c2 (12 + 22 = 32) this is not correct!!! 12 + 22 = 5 we should know that 32 = 9!!!! From this we should acknowledge that the 3 numbers I used don't fit the equation a2 + b2 = c2, 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 make a Pythagorean triple. I have been given 3 sets of Pythagorean triples from family 1 to analyse: I've been told that family 1 of the Pythagorean triples has the following features: ?smallest side is odd ?the longest side is one more than the middle side ?on the middle side you add 4 more on than the last time If I
Beyond Pythagoras.
Mathematics Coursework - Beyond Pythagoras Mathematics Coursework BEYOND PYTHAGORAS By Asif Azam ) The numbers 3, 4, and 5 satisfy the condition 3² + 4² = 5² because 3² = 3x3 =9 4² = 4x4 = 16 5² = 5x5 = 25 and so 3² + 4² = 9 + 16 = 25 = 5² I will now have to find out if the following sets of numbers satisfy a similar condition of (smallest number) ² + (middle number) ² = (largest number) ². a) 5, 12, 13 5² + 12² = 25 + 144 = 169 = 13² b) 7, 24, 25 7² + 24² = 49 + 576 = 625 = 25² 2) Perimeter 5 5 + 12 + 13 = 20 2 studentcentral.co.uk 7 7 + 24 + 25 = 56 24 wwfe few stfefeud efe fent cfe enfetral fecofe uk: b) Nth term Length of shortest side Length of middle side wwdd ddw stddddud edd ddnt cdd enddtral ddcodd uk. Length of longest side Perimeter Area +2 +1 wwbf bfw stbfbfud ebf bfnt cbf enbftral bfcobf uk. 3 4 5 2 6 2 5 2 3 30 30 +2 3 7 24 25 56 84 4 wwef efw stefefud eef efnt cef eneftral efcoef uk; 9 40 41 90 80 +2 5 1 60 61 32 330 +2 6 3 84 Tochv8Ey Visit studentcentral cb co cb uk cb for more cb Do not cb redistribute Tochv8Ey 85 82 546 7 5 12 13 240 840 +2 8 7 44 45 306 224 9 9 80 81 380 710 wwdd ddw stddddud edd ddnt cdd enddtral ddcodd uk. 0 21 220 221 462 2310 I looked at the table and noticed that there was only 1 difference between the length of the
Beyond Pythagoras
Beyond Pythagoras I have been asked to investigate Pythagorean triplets where the shortest side is an odd number and all the three sides are positive integers. A pythagorean triple is a set of integers (a,b,c) that specifies the lengths of a right angle triangle a²+b²=c² in which 'a' is the shortest side 'b' is the middle side and 'c' is the hypotenuse. The first set of triples (3,4,5) which has already been proved to satisfy Pythagoras's theory. I have also been given two other pythagorean triples (5,12,13) and (7,24,25) I will now prove these to satisfy Pythagoras's theory a² = 5² =25 b² = 12² =144 c² = 13² =169 a²+b² =25+144 =169 a²+b²=c² so Pythagoras's theory holds for (5,12,13) because they satisfy the condition of a²+b²=c² in a right angled triangle. a² = 7² =49 b² = 24² =576 c² = 25² =625 a²+b² =49+576 =625 a²+b²=c² so Pythagoras's theory holds for (7,24,25) because they satisfy the condition of a²+b²=c² in a right angled triangle. I am now going to put my results in to a table so that I can predict more values: a b c 3 4 5 5 2 3 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 triplet I predict the next two 'a' values to be 9 and 11
Mathematics Coursework - Beyond Pythagoras
Mathematics Coursework Beyond Pythagoras The numbers 3, 4 and 5 satisfy the condition 3 +4 =5 Because 3 = 3x3 =9 4 = 4x4 =16 5 = 5x5 =25 So 3 +4 =9+16 =25 =5 I now have to find out whether the following sets of numbers satisfy a similar condition of (smallest number) + (middle number) = (largest number) a) 5, 12, 13 5 +12 = 25+144 = 169 = 13 b) 7, 24, 25 7 +24 = 49+576 = 625 =25 I looked at the table of results and noticed that there was only a difference of 1 between the length of the middle side and the length of the longest side. I already know that the (smallest number) + (middle number) = (largest number) . So I think that there might be a connection between the numbers. The problem is that this is not completely correct. (Middle number) + (largest number) = (smallest number) Because 12 + 13 = 144+169 = 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 now try 2 sides being squared. (middle number ) + largest number = (smallest number) = 12 + 13 = 52 = 144 + 13 = 25 = 157 = 25 This does not work. There is also no point in squaring the largest and the smallest or the middle number and the largest number. I will now try 1 side being squared. 12 + 13 = 5 This couldn't work
Towers of Hanoi.
Year 10 GCSE coursework Towers of Hanoi Introduction The aim of this piece of coursework is to complete different investigations. The name of these investigations is the Towers Of Hanoi. I will need to be patient and enthusiastic to complete these testing challenges. Basically I have 4 discs of decreasing radii and 3 towers named A, B and C. I am allowed to move only one disc at a time and I cannot place a larger disc on top of a smaller disc. I have to complete the challenges within a certain amount of goes. I will do 6 investigations using 1, 2, 3, 4, 5 and 6 discs. After I have completed these investigations I will compare them and try to find patterns etc. I will be required to show diagrams, graphs, tables of results and rules. I will also include a conclusion. Investigating some challenges Now I am going to show my the 6 investigations and try to find patterns and rules afterwards Investigation 1 In my first investigation I will attempt to successfully move 4 discs to tower C in the least number of moves. I will now confirm that with four discs it is possible to get from the start (A) to the finish (B) or (C) in a minimum of 15 moves. This is the position I will begin my challenge from. ) 2) 3) 4) 5) 6) 7) 8) 9) 10) 1) 12) 3) 14) 5) Moves: 1-B 2-C 1-C 3-B 1-A 2-B 1-B 4-C 1-C 2-A 1-A 3-C 1-B 2-C 1-C As I confirmed, it is possible to