#### OctangularAddict

##### New member

- Joined
- Jul 3, 2019

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190 -> 42

180 -> 37

160 -> 24

140 -> 16

122 -> 10

154 -> ?

Thank you for your help.

- Thread starter OctangularAddict
- Start date

- Joined
- Jul 3, 2019

- Messages
- 2

190 -> 42

180 -> 37

160 -> 24

140 -> 16

122 -> 10

154 -> ?

Thank you for your help.

- Joined
- Jun 18, 2007

- Messages
- 18,470

Please follow guidelines to posting in this forum. The guidelines are enunciated at:

190 -> 42

180 -> 37

160 -> 24

140 -> 16

122 -> 10

154 -> ?

Thank you for your help.

Welcome to our tutoring boards! :) This page summarizes some main points from our posting guidelines. As our name implies, we provide math help (primarily to students with homework). We do not generally post immediate answers or step-by-step solutions. We prefer to help by having a...

www.freemathhelp.com

This problem can be attempted in many ways. What methods have you been taught in your class.

- Joined
- Jul 3, 2019

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- Joined
- Jun 18, 2007

- Messages
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Okay .... then .... what are the topics in math you had covered last academic session (semester)?

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Is this the entire problem, word for word? If not, please quote it exactly.Hello, I can't figure out the answer to this exercise. It says:

190 -> 42

180 -> 37

160 -> 24

140 -> 16

122 -> 10

154 -> ?

But if it is, then it can't be answered!

It isn't really about a sequence at all, but a function: determining an "output" for a particular "input". And there is no one function that produces these outputs; in particular, if you graph these pairs, you'll see that they don't lie on a straight line or any other "nice" curve. The best you could do would be to "fit" the data to a particular kind of function (e.g. linear) to estimate the requested value. Or you could use what is called linear interpolation. Neither of those can claim to give "the correct answer"; but you could give reasons why your answer makes sense as a good guess.

- Joined
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Any such polynomial can be written \(\displaystyle y= ax^4+ bx^3+ cx^2+ dx+ e\). Putting (x, y) from your data into that equation gives you five equations to solve for a, b, c, d, and e.

A variation of this same idea is to use "Newton's divided difference formula". A polynomial satisfying this data is \(\displaystyle 42\frac{(x- 180)(x- 160)(x- 140)(x- 122)}{(190- 180)(190- 160)(190- 140)(190- 122)}\)\(\displaystyle + 37\frac{(x- 190)(x- 160)(x- 140)(x- 122)}{(180- 190)(180- 160)(180- 140)(190- 122)}\)\(\displaystyle + 24\frac{(x- 180)(x- 190)(x- 140)(x- 122)}{(160- 190)(160- 180)(160- 140)(160- 122)}\)\(\displaystyle + 16\frac{(x- 190)(x- 180)(x- 160)(x- 122)}{(140- 190)(140- 180)(140- 160)(140- 122)}\)\(\displaystyle + 10\frac{(x- 190)(x- 180)(x- 160)(x- 140)}{(122- 190)(122- 180)(122- 160)(122- 140)}\).