Question

What is the equation of this relation? Explain please...my math lesson won't
A = {(n, D): (4,2), (5, 5), (6, 9), ...}

Answer 1

+3 +4 ... ?

Answer 2

is it spose to fit a quadratic? or just a pattern for a sequence?

Answer 3

I don't think so, I'm doing a lesson in precalculus about ordered pairs and relations. :?

Answer 4

A(n) = A(n-1) + 2(n-1) maybe? ; where A(1) = 2

Answer 5

that aint it

Answer 6

x: 4 5 6
y: 2 5 9

just aint enough to go on to me

Answer 7

sorry :( I guess I'm just gonna have to figure out my cryptic math textbook. :P Thanks for trying tho

Answer 8

2(x-5)(x-6)/[(4-5)(5-6)]
+
5(x-4)(x-6)/[(5-4)(5-6)]
+
9(x-4)(x-5)/[(6-4)(6-5)]

Answer 9

multiply and combine like terms if you want

Answer 10

After (4,2), the next pair (5,5) can be found by taking the n-1 term (which is 2) and adding the n-1 domain value (4 in this case) minus 1, eg, (5,5)=(5,2+(4-1))

Answer 11

The third pair may be found similarly: (6,9)=(6,5+(5-1))

Answer 12

Thus, the fourth pair in the sequence would be (7,14)=(7,9+(6-1))

Answer 13

I am unabel to reason an explicit definition. I find it difficult to belive that slidewhistle's problem expects a recursive definition. Then there is the problem of the notation (if you think of this function as a sequence). The first coordinate is uaually though tof as the position in the sequence, but in this case the first term a(1) is a(4), . . . So neglecting notation (cont)

Answer 14

A recursive defintion would be: (i) the first term is D=2; (ii) the next term is the previous term D(n-1) + the previous "position value" n-1. Does that make sense?

Answer 15

n=4, D(1)=2; n=5, D(2)=2+(4-1); etc.

Answer 16

you need joemath

Answer 17

looks like you are adding successive integers starting at 2 right? 2, 2+3,2+3+4, ...

Answer 18

if it was
1,1+2,1+2+3,1+2+3+4, you could use the well known
\[\frac{n(n+1)}{2}\] but here you can subtract 1 to get the answer, then replace n by n-3 to adjust for starting at 4 instead of 1

Answer 19

no that is wrong maybe n - 2 for n

Answer 20

yup that one works

Answer 21

i get
\[f(n)=\frac{(n-2)^2+n-4}{2}\]

Answer 22

probably a snappier way to do it, but it works for 4, 5 and 6

Answer 23

Very nice satellite; it works beyond 6 too.

I came back just to write a better recursive definition:\[a _{4}=2\]and\[a _{n+1}=a _{n}+n-1\]