I am going to examine the time taken for a whole tray of tomatoes Essay Example

I am going to examine the time taken for a whole tray of tomatoes Essay Example I am going to examine the time taken for a whole tray of tomatoes Essay I am going to examine the time taken for a whole tray of tomatoes Essay In this project I am going to examine the time taken for a whole tray of tomatoes to go bad when a single bad tomato is put in a particular position. I will see how this time changes when I vary the size of the tray and alter the starting position. I will start with a small tray and gradually the size of the tray will be larger and also the positions will move from corner to corner, side to side and so on. At the end of this project I want to be able to have a formula which will tell me how long it would take a bad tomato to spread over cover the whole tray when the first bad tomato is placed in a curtain position in the tray. Part 1 The diagram below represents the look of a tray with 16 tomatoes in it. The number 1 is there to show where the first bad tomato began. The other numbers 2,3,4,5 and 6 represent the number of hours that have gone. Therefore for example after two hours in the diagram below in total, which includes the 1 and all the 2s, 4 tomatoes have gone bad. 2 3 4 5 1 2 3 4 2 3 4 5 3 4 5 6 Hours (n) Total No. Of Bad Tomatoes 1st Difference 2nd Difference 1 1 3 2 4 1 4 3 8 0 4 4 12 -1 3 5 15 -2 1 6 16 The table on the previous page tells me what is involved in the nth term. The column labelled 1st Difference tells us the difference between the number of bad tomatoes in the first hour to the second hour and so on. The column labelled 2nd Difference is the difference between the figures in the 1st Difference column. When the differences become the same this is when I can stop. If there was a 2nd Difference column then that means there will be a 2 involved in the nth term. In total there are only 3 starting positions. They are one in the corner, one next to it on the side and one of the four squares in the middle. The tray above in a 4 x4 and in total it took 5 hours for all the tomatoes in the tray to go bad. The table just above this paragraph shows the total number of bad tomatoes. The columns towards the right hand side determine what the nth term will involve. If there are two differences that means the nth term will involve a 2. Now I am going to look at the same size tray with the bad tomato starting in another position. We can see now long it will take all the tomatoes in the tray to go bad. 4 3 2 1 5 4 3 2 6 5 4 3 7 6 5 4 Hours (n) Total No. Of Bad Tomatoes 1st Difference 2nd Difference 1 1 2 2 3 1 3 3 6 1 4 4 10 1 3 5 13 1 2 6 15 1 1 7 16 As it is possible to see from the table above there are two differences, which means that the nth term will involve 2. Now that I have found out that there is a 2 involved in the formula. I will now draw out a table in which I will find out the formula through trial and error. To begin with I will double the number and then 2 it. When I begin to see that there is something there I will 2 it and then double to try to get to the right number. n nà ¯Ã‚ ¿Ã‚ ½ nà ¯Ã‚ ¿Ã‚ ½-n nà ¯Ã‚ ¿Ã‚ ½+n (nà ¯Ã‚ ¿Ã‚ ½+n) / 2 t 1 1 0 1 2 4 2 3 3 9 6 6 4 16 12 10 5 25 20 13 6 36 30 15 The nth term is: Part 2 Now I am looking at different sizes of trays and different positions in the tray. The first size I am going to look at is: 10 x 10 and the bad tomato will be positioned in the corner. 1 2 3 4 5 6 7 8 9 10 2 3 4 5 6 7 8 9 10 3 4 5 6 7 8 9 10 4 5 6 7 8 9 10 5 6 7 8 9 10 6 7 8 9 10 7 8 9 10 8 9 10 9 10 10 Now I am going to find out how many differences are involved. Hours (n) Total No. Of Bad Tomatoes 1st Difference 2nd Difference 1 1 2 2 3 1 3 3 6 1 4 4 10 1 5 5 15 1 6 6 21 1 7 7 28 1 8 8 36 1 9 9 45 1 10 10 55 Now I know that there is a 2 involved in the formula as there are 2 differences. So I will now place the numbers in table to find out the formula. To begin with I will double the number and then 2 it. If that does not work I will try to use different ways such as 2ing and then misusing the n number. N nà ¯Ã‚ ¿Ã‚ ½ nà ¯Ã‚ ¿Ã‚ ½-n nà ¯Ã‚ ¿Ã‚ ½+n (nà ¯Ã‚ ¿Ã‚ ½+n) / 2 T 1 1 0 2 1 1 2 4 2 6 3 3 3 9 6 12 6 6 4 16 12 20 10 10 5 25 20 30 15 15 6 36 30 42 21 21 As it can be seen I found the formula in four steps and these steps I may use in other trays further in this project. The nth term is: nà ¯Ã‚ ¿Ã‚ ½+n 2 So by putting the numbers we know in the formula (hours) we can find out how many tomatoes go bad in particular number of hours in a 1010 tray with the bad tomato starting in a corner. Now I am going to look at another position on a 10 x 10 tray. The position I am going to look as is starting from the middle. 5 5 4 5 5 4 3 4 5 5 4 3 2 3 4 5 5 4 3 2 1 2 3 4 5 5 4 3 2 3 4 5 5 4 3 4 5 5 4 5 5 Now like previously in this project I am going to find out how many differences are involved so I can work out the formula. Hours (n) Total No. Of Bad Tomatoes 1st Difference 2nd Difference 1 1 4 2 5 4 8 3 13 4 12 4 25 4 16 5 41 6 61 7 85 8 113 9 145 10 181 As you can see from the table above there are two differences which means there is a 2 involved in the formula. The table on the next page will show my working outing out and how I got the formula. N nà ¯Ã‚ ¿Ã‚ ½ nà ¯Ã‚ ¿Ã‚ ½+n nà ¯Ã‚ ¿Ã‚ ½-n 2(nà ¯Ã‚ ¿Ã‚ ½-n) 2(nà ¯Ã‚ ¿Ã‚ ½-n)+1 t 1 1 2 0 0 1 1 2 4 6 2 4 5 5 3 9 12 6 12 13 13 4 16 20 12 24 25 25 5 25 30 20 40 41 41 The nth term is: 2(nà ¯Ã‚ ¿Ã‚ ½-n)+1 Now I am going to use the same size tray, which is 10 x 10, and place the bad tomato on the side. 5 5 4 5 5 4 3 4 5 5 4 3 2 3 4 5 5 4 3 2 1 2 3 4 5 Now like previously in this project I am going to find out how many differences are involved. Hours (n) Total No. Of Bad Tomatoes 1st Difference 2nd Difference 1 1 3 2 4 2 5 3 9 2 7 4 16 2 9 5 25 6 36 7 49 8 64 9 81 10 100 As it is possible to view, there are also two difference here just like the others we have seen so far. It is also possible to see from the table above that the formula is not going to be very complicated. N nà ¯Ã‚ ¿Ã‚ ½ t 1 1 1 2 4 4 3 9 9 4 16 16 5 25 25 6 36 36 The nth term is: nà ¯Ã‚ ¿Ã‚ ½ Conclusion: I have looked at different ways by which the bad tomatoes can spread. All of the nth terms above are not for a particular size tray. Even though I said I used a 10 x 10 tray that was only to start off with. In fact all the formulas above are really for large trays. I believe the main reason behind this investigation is to find out the a formula that by just knowing the size of the tray and exact position in the tray of the bad tomato will tell how long it will take for all the tomatoes to go bad. Before I write down the formula I will tell you how I arrived at it. The formula will tell you how long it will take to make all the tomatoes in the tray go bad. Firstly the formula should involved the size of the tray e.g. 6 x 8 and also must involved the starting position of the first bad tomato. Now that I have explained what have got to be in the formula, well, here it is. (a-x) + (b-y) a and b stand for the width and depth of the tray. x and y stand for the positioning of the first bad tomato. So, for example if we take a 8 x 6 tray with starting bad tomato at the position (4,4) the working to find how long it will take for the whole tray to go bad should look like: (a-x) + (b-y) (8-4) + (6-4) Answer: 6 hours

Displacement Reaction Definition and Examples

Displacement Reaction Definition and Examples A displacement reaction is a type of reaction where part of one reactant is replaced by another reactant.  A displacement reaction is also known as a replacement reaction or a metathesis reaction.  There are two types of displacement reactions: Single Displacement Reactions Single displacement reactions are reactions where one reactant replaces part of the other.AB C → AC B An example is the reaction between iron and copper sulfate to produce iron sulfate and copper: Fe CuSO4 → FeSO4 Cu Here, both iron and copper have the same valence. One metal cation takes the place of the other bonding to the sulfate anion. Double Displacement Reactions Double displacement reactions are reactions where the cations and anions in the reactants switch partners to form products.AB CD → AD CB An example is the reaction between silver nitrate and sodium chloride to form silver chloride and sodium nitrate: AgNO3 NaCl → AgCl NaNO3