Revision history [back]
 | 1 | initial version |
Sorry, I tried to attach an image of it and I messed it up. Here is the code: a,b=var('a b') solve([7.0==asqrt(16)/(1+b/16.0),22==asqrt(4)/(1+b/4)],a,b) print a.n() The result the solve command gives is the ratios of integers.
| | 2 | No.2 Revision |
Sorry, I tried to attach an image of it and I messed it up. Here is the code:
a,b=var('a b')
b')
solve([7.0==asqrt(16)/(1+b/16.0),22==asqrt(4)/(1+b/4)],a,b)
sqrt(4)/(1+b/4)],a,b)
print a.n() The result the solve command gives is the ratios of integers.
| | 3 | No.3 Revision |
Sorry, I tried to attach an image of it and I messed it up. Here is the code: a,b=var('a b')
solve([7.0==asqrt(16)/(1+b/16.0),22==asqrt(4)/(1+b/4)],a,b)
print a.n()
a.n()
The result the solve command gives is the ratios of integers.
| | 4 | No.4 Revision |
Sorry, I tried to attach an image of it and I messed it up. Here is the code: a,b=var('a b')
solve([7.0==asqrt(16)/(1+b/16.0),22==asqrt(4)/(1+b/4)],a,b)
print a.n()
The result the solve command gives is the ratios of integers.
| | 5 | No.5 Revision |
Sorry, I tried to attach an image of it and I messed it up. Here is the code: a,b=var('a b')
solve([7.0==asqrt(16)/(1+b/16.0),22==asqrt(4)/(1+b/4)],a,b)
print a.n()
The result the solve command gives is the ratios of integers.
(I typed an asterisk between the a and the sqrt, but it doesn't show up...)
