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last edited 6 years ago by test int |
Edit detail for SandBoxReduce revision 20 of 36
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Editor: crashcut
Time: 2008/01/30 10:28:31 GMT-8 |
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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 |
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int(sqrt(1-sin(x)*cos(x)),x); | reduce |
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example from my daughter's college calc --pbwagner, Mon, 10 Sep 2007 12:58:25 -0500 reply
int(log(log(x)),x); | reduce |
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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=-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 |
