# Need help with a formula!

• Oct 18th 2013, 07:43 AM
Howlin1
Need help with a formula!
I'm not sure if this is the correct forum for this problem, but I am puzzled by this. I'm not the best with maths so I don't know how to solve this.
There are three numbers (below) that make three results (also below). Thing is I don't know if the 3rd number is required or not. There is a formula for each of the results. I know there is a formula for the result (each result 1 has the same formula, each result 2 has the same formula and each result 3 has the same formula), but I don't know if there is any other hidden variables to it or not. All I have figured out is that Number 2 is 25 times the value of number 1. Can anyone solve this with the information here?

Number 1: 76
Number 2: 1900
Number 3: 152
Result 1: 95.4
Result 2: 31.4
Result 3: 24.8

Number 1: 106
Number 2: 2650
Number 3: 212
Result 1: 192
Result 2: 63.3
Result 3: 49.8

Number 1: 112
Number 2: 2800
Number 3: 224
Result 1: 221.7
Result 2: 73.2
Result 3: 57.6

Number 1: 132
Number 2: 3300
Number 3: 264
Result 1: 361.5
Result 2: 119.2
Result 3: 93.8
• Oct 18th 2013, 08:46 AM
SlipEternal
Re: Need help with a formula!
For the function that takes Number 1 -> Result 1, the following quartic equation works:

$\displaystyle x^4 - \dfrac{1116459377}{2620800}x^3 + \dfrac{12588804713}{187200}x^2 - \dfrac{1525417666789}{327600}x + \dfrac{232244321233}{1950}$.

I don't have time to find quartics for the other two sets of numbers.
• Oct 18th 2013, 11:37 AM
Howlin1
Re: Need help with a formula!
I think you misunderstood my question (or I didn't clarify myself). Number 1 2 and (possibly) 3 are used to make result 1 and another formula to make result 2 and another one to make result 3
• Oct 18th 2013, 04:25 PM
Shakarri
Re: Need help with a formula!
Call number 1 x, number 2 y, number 3 z
As you have pointed out already $\displaystyle (\text{number }2)= 25\times(\text{number }1)$
i.e. $\displaystyle y=25x$

Similarly $\displaystyle (\text{number }3)= 2\times(\text{number }1)$
i.e. $\displaystyle z=2x$

If you had an equation like: $\displaystyle (\text{result 1})=x+3y-2z+xy+z^2$ you could use what you know about z and y to change this into
$\displaystyle (\text{result 1})=x+3(25x)-2(2x)+x(25x)+(2x)^2$

Which simplifies to $\displaystyle (\text{result 1})=301x+4x^2$
So whatever the formula is you can always simplify it so that it only depends on Number 1

The formula for result 1 is not simple. Approximately
$\displaystyle (\text{result 1})=0.0005429x^3-0.11158x^2+9.91578x-252.085$
There is some error because I used a computer which gave the answer as a rounded decimal instead of an exact fraction.