Solve the Math Puzzle: 66+935+1389

[Dai]

If
28+453+9012=81.534
91+712+1024=49.091
Then
66+935+1389=?

 

[topsquark]

Let
[math]\left [ \begin{matrix} a & b & c \\ a & b & c \end{matrix} \right ] ~ \left [ \begin{matrix} 28 & 91 \\ 453 & 712 \\ 9012 & 1024 \end{matrix} \right ] = \left [ \begin{matrix} 81.534 \\ 49.091 \end{matrix} \right ] [/math]

Letting a = 1 gives b = -0.07265695 and c = 0.009592496.

Thus
[math] \left [ \begin{matrix} a & b & c \end{matrix} \right ] ~ \left [ \begin{matrix} 66 \\ 935 \\ 1389 \end{matrix} \right ] = 11.389770769[/math]

But you probably didn't mean that. (Tongueout)

-Dan

 

[Dai]

Nope :)
But nice try tho

 

[topsquark]

The decimal values...do you have an exact expression for them or is your answer also going to be in decimal form?

And hey! It works!

-Dan

 

[topsquark]

[math]\left [ \begin{matrix} a & b & c \\ a & b & c \end{matrix} \right ] ~ \left [ \begin{matrix} 28 & 91 \\ 453 & 712 \\ 9012 & 1024 \end{matrix} \right ] = \left [ \begin{matrix} 81.534 \\ 49.091 \end{matrix} \right ] [/math]

My argument has been made, but I made a ghastly error in the way I wrote it. I am posting to correct this. I meant to say
[math]\left [ \begin{matrix} a & b & c \end{matrix} \right ] ~ \left [ \begin{matrix} 28 & 91 \\ 453 & 712 \\ 9012 & 1024 \end{matrix} \right ] = \left [ \begin{matrix} 81.534 & 49.091 \end{matrix} \right ] [/math]

-Dan

 

[topsquark]

Sooooooo... What's the solution? (Yes, I'm impatient! (Time) )

-Dan

 

[MarkFL]

Sooooooo... What's the solution? (Yes, I'm impatient! (Time) )

-Dan

Dan, in 10 hours it will have been a week...so at that time the OP should feel free to post their solution. (Yes)