Maths Presentation

If every pupil call individually early(a) on a mobile phone, how many calls lead thither be exclusively when there are 910 pupils? Ok. So if two heap rang individually other then it would look like this: =pupil 2 pupils=1 contacts =contact This shows that two people can single ring individually other once. If 3 people rang each other, then it would look like this: and so on and so forth... 3 pupils= 3 contacts 4 pupils= 6 contacts If I delegate all the combinations in up to six in a sequence then it would look like this: 2 3 4 5 6>>>>>>>>>> ordinal term 1 3 6 10 15>>>>>> ;>> sequence 2 3 4 5 >>>>>>>>>>>difference 1 1 1 To find the nth term an2+bn+c II. 4a+2b+c=1 troika. 9a+3b+c=3 IV. 16a+4b+c=6 directly I am going to cancel forbidden the c by taking away equivalence II from trinity: V.

(9a+3b+c=6) (4a+2b+c=1) = (5a+b=2) Then I forget do the same for equivalence III and IV: VI. (16a+4b+c=6) (9a+3b+c=3) = 7a+b=3 Now to work come out of the closet what a is, I will minus equation V from equation VI. VII . (7a+b=3) (5a+b=2) = 2a=1 ! a=0.5 upon purpose out that a equals 0.5, I can work out that... (5x0.5)+b=2 2.5+b=2 ...b=-0.5 So presently I will go vertebral column to equation II to work out what c is: 4x0.5+2x-0.5+c=1 2+ (-1)+c=1 C=0 So the formula is: 0.5n2+ (-0.5n)=y So question asks how many calls would there be if 910 students called each other once. Now I can work out what it is by replacing n for 910: 0.5n2+ (-0.5n)=n(n+1)2=910(910+1)2=414 505If you essential to get a honorable essay, order it on our website: OrderCustomPaper.com

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