Example research essay topic: Acute P 20 - 3,156 words

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Introduction A woman only has 10 and 20 pence coins to use in a phone box. I have to investigate the number of different ways she could use 10 and 20 pence coins in a phone box. I am also investigating a man who only has 10 and 50 pence coins. I will start off with the woman. 10 pence call 10 = 10 p = 1 way 20 pence call 20 = 10 p+ 10 p 20 = 20 p = 2 ways 30 pence call 30 = 10 p+ 10 p+ 10 p 30 = 20 p+ 10 p = 3 ways 30 = 10 p+ 20 p 40 pence call 40 = 10 p+ 10 p+ 10 p+ 10 p 40 = 20 p+ 10 p+ 10 p 40 = 10 p+ 20 p+ 10 p = 5 ways 40 = 10 p+ 10 p+ 20 p 40 = 20 p+ 20 p 50 pence call 50 = 10 p+ 10 p+ 10 p+ 10 p+ 10 p 50 = 20 p+ 10 p+ 10 p+ 10 p 50 = 10 p+ 20 p+ 10 p+ 10 p 50 = 10 p+ 10 p+ 20 p+ 10 p = 8 ways 50 = 10 p+ 10 p+ 10 p+ 20 p 50 = 20 p+ 20 p+ 10 p 50 = 20 p+ 10 p+ 20 p 50 = 10 p+ 20 p+ 20 p Pascal´ s triangle theory This should help me to work out long and write out long equations like the ones I am doing E. g. 50 5 10 p = 1 1 20 p and 3 10 p = 4 = 8 ways 2 20 p and 1 10 p = 3 To get this I used the calculator button which has a big c you put the total number of units at the top then you put the number of 20 pence coins at the bottom. You use the biggest one at the bottom I use 20 pence because it is relevant.

The way that Pascal made up this theory was he got a triangle and started with 1 at the top and started adding them up as he went down like this: 1 1 1 3 3 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 He did this for the rest of the way down. Instead of a calculator Pascal would have used the number of units added up to see how far he had to go down and the number of units in the biggest value to see how far he would of gone across. E. g. 50 5 10 p = 1 1 20 p and 3 10 p = 4 down and 1 across 2 20 p and 1 10 p = 3 down and 2 across You then look at where the numbers would be on the triangle and add them up that would come to 1 + 4 + 3 which would equal 8. Investigation I am now going to use Pascal´ s triangle in my investigation to shorten the length of my equations. 60 pence call 60 = 6 10 p = 1 60 = 1 20 p and 4 10 p = 5 60 = 2 20 p and 2 20 p = 6 60 = 3 20 p = 1 This would equal 13 different combinations 70 pence call 70 = 7 10 p = 1 70 = 1 20 p and 5 10 p = 6 70 = 2 20 p and 3 10 p = 10 70 = 3 20 p and 1 10 p = 4 This would equal 21 different combinations 80 pence call 80 = 8 10 p = 1 80 = 1 20 p and 6 10 p = 7 80 = 2 20 p and 4 10 p = 15 80 = 3 20 p and 2 10 p = 10 80 = 4 20 p = 1 This would equal 34 different combinations Results This is a table of my results and a prediction for this far into this investigation Cost of phone call 10 pence 20 pence 30 pence 40 pence 50 pence 60 pence 70 pence 80 pence number of combinations 12358132134 By these results I would say this was the Fibinachi series which means to work out the next one you have to add the two previous terms to work out the next term. You can right it as the equation: Tn = (Tn- 1) + (Tn- 2) Tn means term number Prediction I would say using the Fibinachi series the next number for a 90 pence call would be 21 + 34 so it would be 55 90 pence call 90 = 9 10 p = 1 90 = 1 20 p and 7 10 p = 8 90 = 2 20 p and 5 10 p = 21 90 = 3 20 p and 3 10 p = 20 90 = 4 20 p and 1 10 p = 5 This would equal 55 different combinations Conclusion My prediction was right that a 90 pence call would make 55 different combinations I would say that the rest of the combinations would follow the Fibinachi series Investigation 2 In this investigation I am investigating a man who only has 10 and 50 pence coins to use in a phone box.

I am going to use Pascal´ s triangle to help me in this investigation as well. 10 pence call 10 = 1 10 p = This would equal 1 different combinations 20 pence call 20 = 2 10 p = This would equal 1 different combinations 30 pence call 30 = 3 10 p = This would equal 1 different combinations 40 pence call 40 = 4 10 p = This would equal 1 different combinations Results From these results so far it looks like they are going up by the number 1 so I predict that for a 50 pence call there should be 1 combination. I will now test out this theory by working out how many combinations there are for a 50 pence call. 50 pence call 50 = 5 10 p = 1 50 = 1 50 p = 1 This would equal 2 different combinations This is wrong from my prediction so my prediction is wrong. I will now test 60 - 90 pence calls then look for any pattern. 60 pence call 60 = 6 10 p = 1 60 = 1 50 p and 1 10 p = 2 This would equal 3 different combinations 70 pence call 70 = 7 10 p = 1 70 = 1 50 p and 2 10 p = 3 This would equal 4 different combinations 80 pence call 80 = 8 10 p = 1 80 = 1 50 p and 3 10 p = 4 This would equal 5 different combinations 80 pence call 80 = 8 10 p = 1 80 = 1 50 p and 3 10 p = 4 This would equal 5 different combinations 90 pence call 90 = 9 10 p = 1 90 = 1 50 p and 4 10 p = 5 This would equal 6 different combinations Results By these results I can see that if you take a 70 pence call you add the last one and the one 5 back to make it out. I will check this theory on the 100 pence call this should be the 90 pence call (6) + the 50 pence call (2), which would be 6 + 2. My prediction for the 100 pence call should be 8. 100 pence call 100 = 10 10 p = 1 100 = 1 50 p and 5 10 p = 6 100 = 2 50 p = 1 This would equal 8 different combinations My prediction is right and there and there was 8 combinations for a 100 pence call. This means to get the next term you have to plus the last term and the number five terms back.

You can put this into an equation: Tn = (Tn- 1) + (Tn- 5) Tn = term number Comparison between different call charges with one 10 p and another one with a multiple of 10 p First I will look at 10 p and 20 p, 10 and 50 p and how you find out the next term. 10 and 20 pence For this one you add the one behind it and the one 2 places back to get the next term. 10 and 50 pence For this one you add the one behind it and the one 5 place back to get the next term. You can put these two in a bar chart: Prediction I think from looking at these results I would say that as the money you were adding to 10 pence the amount you were going back to add on to the one before to make the next term would go up by 1. I will test out my prediction on making a call with 10 p and 40 p. I will use Pascal´ s triangle to help me. 10 pence call 10 p = 1 10 p = 1 This would equal 1 different combinations 20 pence call 20 p = 2 10 p = 1 This would equal 1 different combinations 30 pence call 30 p = 3 10 p = 1 This would equal 1 different combinations 40 pence call 40 p = 4 10 p = 1 40 p = 1 40 p = 1 This would equal 2 different combinations 50 pence call 50 p = 5 10 p = 1 50 p = 1 40 p and 1 10 p = 2 This would equal 3 different combinations 60 pence call 60 p = 6 10 p = 1 60 p = 1 40 p and 2 10 p = 3 This would equal 4 different combinations 70 pence call 70 p = 7 10 p = 1 70 p = 1 40 p and 3 10 p = 4 This would equal 5 different combinations Results I would say from these results that you have to go 3 place back plus the one before you are going to add up to make the next term. So the term for eighty pence call will be a seventy pence call plus a forty pence call which should mak 80 pence call 80 p = 8 10 p = 1 80 p = 1 40 p and 4 10 p = 5 80 = 2 40 p = 1 This would equal 5 different combinations The equation for 10 p and 30 p is Tn = (Tn- 1) + (Tn- 3) Tn = term number This proves my theory for both the 10 and 30 pence way to find the next term and how to predict how to get the terms.

This is what the term from 10 p with 10 p to 60 p with 10 p should look like in a line graph. Conclusion I think that the reason for the results being like this is because of the bigger the multiple of ten you are using with a ten pence in a phone call the further the two are apart. This means with a 100 pence coin it would be a lot longer before you added the 100 pence coin so you would be using the ten pence. This would mean there would be a lot more ones at the start of you then only have a 100 pence coin and a 10 pence coin for a long time until another 100 pence coin is added.

So the higher the coin value of a multiple of ten + ten pence the higher the higher the difference. You can work out what the difference by taking the number you are adding to ten and take the zero off it and that is how far you have to go back. The general causes is that for every 10 p you add on to the ten pence you have to go back 1 more than the previous one like if you just worked out 60 p and 10 p then it is one further back to find the next one which would be 70 p and 10 p. If you are just told to work it out and you haven´ t worked out the previous one you just divide the one you are doing by ten and that is how far you have to go back and add the one first back from the one you are adding up. You can also turn it into a graph where the plot would be y = x Here is what the graph should look like: Conclusion You can also work out how far you have to go back with an equation the first two equations were: Tn = (Tn- 1) + (Tn- 2) This was for the 10 and 20 pence coins Tn = (Tn- 1) + (Tn- 5) This was for the 10 and 50 pence coins Then I did the 10 and 30 pence one, which came out as: Tn = (Tn- 1) + (Tn- 3) The only variable in the equations is this bit in the second brackets after the Tn- that is the variable. Also the variable is always the one you are adding on to the 10 pence divided by 10.

Formula 1 Tn = (Tn- 1) + (Tn-v) Tn = term number V = variable E. g. 60 pence call 60 divided by 10 = 6 Tn = (Tn- 1) + (Tn- 6) I have also seen form my investigations that the series does not start until the term after two. If you are adding a number to ten pence there is an easy formula to find which term you can start off using the equations. First you add the 10 p and the one you are trying to solve together, you then divide that by ten that is how far you have to go forward until you can use formula 1. We will cal this formula 2 Formula 2 H = 10 p+v/ 10 H = how far you have to go back until you can use formula 1 V = variable (one you are adding to ten) / = Divide My theory for two variables I predict that for two variables I could use the same I can adapted my formulas. I think that where there was a one in formula 1 there would be the smallest number and where a v would be there would be a biggest number.

Here is what the equation should look like. Formula 3 Tn = (Tn-V 1 / 10) + (Tn-V 2 / 10) Tn = term number V = variable / = Divide I will test my theory on 20 and 30 pence coins in a phone box here is what the equation should look like: Tn = (Tn-V 1 / 10) + (Tn-V 2 / 10) Tn = term number V = variable / = Divide Tn = (Tn- 20 / 10) + (Tn- 30 / 10) Tn = (Tn- 2) + (Tn- 3) 10 pence call 10 p = 0 There are 0 different combinations 20 pence call 20 p = 20 = 1 There are 1 different combinations 30 pence call 30 p = 30 = 1 There are 1 different combinations 40 pence call 40 p = 20 + 20 p = 1 There are 1 different combinations 20 and 30 pence investigation continued 50 pence call 50 p = 30 p+ 20 p = 1 50 p = 20 p+ 30 p = 1 There are 2 different combinations 60 pence call 60 p = 20 p+ 20 p+ 20 p = 1 60 p = 30 p+ 30 p = 1 There are 2 different combinations 70 pence call 70 p = 30 p+ 20 p+ 20 p = 1 70 p = 20 p+ 30 p+ 20 p = 1 70 p = 20 p+ 20 p+ 30 p = 1 There are 3 different combinations 80 pence call 80 p = 20 p+ 20 p+ 20 p+ 20 p = 1 80 p = 30 p+ 30 p+ 20 p = 1 80 p = 30 p+ 20 p+ 30 p = 1 80 p = 20 p+ 30 p+ 30 p = 1 There are 4 different combinations 90 pence call 90 p = 30 p+ 30 p+ 30 p = 1 90 p = 30 p+ 20 p+ 20 p+ 20 p = 1 90 p = 20 p+ 30 p+ 20 p+ 20 p = 1 90 p = 20 p+ 20 p+ 30 p+ 20 p = 1 90 p = 20 p+ 20 p+ 20 p+ 30 p = 1 There are 4 different combinations Results From these if you look my theory only comes in after the 5 term (the 50 pence call). If my theory is correct a 100 pence call should have 7 different combinations. I will now test my theory for how many combinations there are in a 100 pence call 100 pence call 100 p = 20 p+ 20 p+ 20 p+ 20 p+ 20 p = 1 100 p = 30 p+ 30 p+ 20 p+ 20 p = 1 100 p = 20 p+ 30 p+ 20 p+ 30 p = 1 100 p = 30 p+ 20 p+ 30 p+ 20 p = 1 100 p = 20 p+ 20 p+ 30 p+ 30 p = 1 100 p = 30 p+ 20 p+ 20 p+ 30 p = 1 100 p+ 20 p+ 30 p+ 30 p+ 20 p = 1 There are 7 different combinations My prediction was right and there were 7 different combinations for a 100 pence call so my formula is right for two variables: Tn = (Tn-V 1 / 10) + (Tn-V 2 / 10) Tn = term number V = variable / = Divide I think that I can adapt my how far you have to go forward until you use the formula I think it would be: Formula 4 H = V 1 / 10 +V 2 / 10 H = how far you have to go back until you can use a formula V = variable / = Divide E. g. 20 and 30 pence H = V 1 / 10 +V 2 / 10 H = 20 / 10 + 30 / 10 H = 2 + 3 H = 5 Conclusion If you use formula 3 and formula 4 together you can work out how far forward you can use your formula and what is the next term.

Research essay sample on Acute P 20