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# Edit detail for SandBoxReduce revision 20 of 36

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 Editor: crashcut Time: 2008/01/30 10:28:31 GMT-8 Note:

added:

From crashcut Wed Jan 30 10:28:27 -0800 2008
From: crashcut
Date: Wed, 30 Jan 2008 10:28:27 -0800
Subject:
Message-ID: <20080130102827-0800@axiom-wiki.newsynthesis.org>

\begin{reduce}
solve({c=-1/m*(1-(m*a+c1)^2)^(1/2)+c2, b=-1/m*(1-c1^2)^(1/2)+c2}, {c1,c2});
\end{reduce}


Try Reduce calculations here. For example:

  \begin{reduce}
solve({z=x*a+2},{z,x});
int(sqrt(1-sin(x)*cos(x)),x);
\end{reduce}


 solve({z=x*a+2},{z,x}); reduce
 int(sqrt(1-sin(x)*cos(x)),x); reduce

example from my daughter's college calc --pbwagner, Mon, 10 Sep 2007 12:58:25 -0500 reply
 int(log(log(x)),x); reduce

axiomintegrate(log(log(x)),x)
 (1)
Type: Union(Expression Integer,...)

 solve({c=-1/l*sqrt(1-(l*a+c1)^2)+c2, b=-1/l*sqrt(1-c1^2)+c2}, {c1, c2}) reduce

 solve({c-b=-1/l*sqrt(1-(l*a+c1)^2)+1/l*sqrt(1-c1^2)}, {c1, c2}) reduce

 solve({c-b=-1/l*(1-(l*a+c1)^2)^(1/2)+1/l*(1-c1^2)^(1/2)}, {c1, c2}) reduce

 solve({c-b=-1/l*(1-(l*a+c1)^2)^(1/2)+1/l*(1-c1^2)^(1/2)}, {c1}) reduce

 solve(c-b=-1/l*(1-(l*a+c1)^2)^(1/2)+1/l*(1-c1^2)^(1/2), c1) reduce

 solve(c-b=-1/l*(1-(l*a+c1)^2)^(1/2)+1/l*(1-c1^2)^(1/2), c1); reduce

 solve({c=-1/l*(1-(l*a+c1)^2)^(1/2)=c2, b=-1/l*(1-c1^2)^(1/2)+c2}, {c1,c2}); 2 2 2 - sqrt( - a *l - 2*a*c1*l - c1 + 1) ***** c=---------------------------------------- invalid as scalar l ***** Continuing with parsing only ...  reduce

 solve({c=-1/l*(1-(l*a+c1)^2)^(1/2)+c2, b=-1/l*(1-c1^2)^(1/2)+c2}, {c1,c2}); reduce

 solve({c=-1/m*(1-(m*a+c1)^2)^(1/2)+c2, b=-1/m*(1-c1^2)^(1/2)+c2}, {c1,c2}); reduce