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Good catch!You can actually eliminate two variables to solve an equation in A (express all B's in terms of A, then substitute the C in the first equation by the third equation). You can then obtain expressions for B (and then C) trivially. But this equation which has lots of floor functions and exponents is probably going to be difficult/impossible to rearrange for A, so I'd say numerical is probably easiest here.
You get an exact value from plugging $B=18$ and $C=18$ into the first equation. There's no range. I get $A=1/67500$. How are you getting your range?Thanks.
However, B=18 and C=18 results in range A = 1.43000 * 10^(-5) to 1.72000 * 10^(-5)
Agree?
Sorry...was doing something different...not applicable here...How are you getting your range?