Solver Problem 'Maybe'

01042017, 05:17 PM
Post: #1




Solver Problem 'Maybe'
Below is the equation for a Circle.
R^2=(XH)^2+(YK)^2. The solver will solve for R=7.2801. X=4, H=2, Y=6 and K=1 R^2=53. You can solve any of the variables except 'X'. Solving for X returns '0'. Rearranging the formulae to solve for X. Note: SQRT below instead squareRoot sign on calculator. X=H+SQRT(K^22*Y*K(R^2Y^2)). Note: 'H+' Now I can solve for any variables including 'X'. 2+2=4 Of course using X=HSQRT(K^22*Y*K(R^2Y^2)). Note: 'H' cannot solve for X either, since SQRT(K^22*Y*K(R^2Y^2))=2. 22=0 Which may be a clue????? Its seems that the Solver precedes the SQRT sign with H instead of H+ and returns '0' instead of '4'. The other option is, maybe I've had too much to drink over the Holidays. Thanks BM 

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Messages In This Thread 
Solver Problem 'Maybe'  BERNARD MICHAUD  01042017 05:17 PM
RE: Solver Problem 'Maybe'  roadrunner  01052017, 02:36 PM
RE: Solver Problem 'Maybe'  BERNARD MICHAUD  01052017, 05:11 PM
RE: Solver Problem 'Maybe'  Nigel (UK)  01052017, 06:27 PM
RE: Solver Problem 'Maybe'  BERNARD MICHAUD  01052017, 07:36 PM

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