Originally posted by villa68-( (3/3) - 3) = 3
1 1 1=3
2 2 2=3
3 3 3=3
4 4 4=3
5 5 5=3
6 6 6=3
7 7 7=3
8 8 8=3
9 9 9=3
using just things from a scientific calculator make all of the above maths equations true
eg
1+1+1=3
2+(2/2)=3
(?4) / (?4) + (?4) = 3
-( cos(5!) * 5 + cos(5!) ) = 3
cos(6!) + cos(6!) + cos(6!) = 3
cos(7!) + cos(7!) + cos(7!) = 3
cos(8!) + cos(8!) + cos(8!) = 3
-( ( (?9) / (?9) ) - (?9) ) = 3 OR cos(9!) + cos(9!) + cos(9!) = 3
Is that what you were looking for?
Originally posted by CrowleyThe "?" are supposed to be the squareroot sign.
-( (3/3) - 3) = 3
(?4) / (?4) + (?4) = 3
-( cos(5!) * 5 + cos(5!) ) = 3
cos(6!) + cos(6!) + cos(6!) = 3
cos(7!) + cos(7!) + cos(7!) = 3
cos(8!) + cos(8!) + cos(8!) = 3
-( ( (?9) / (?9) ) - (?9) ) = 3 OR cos(9!) + cos(9!) + cos(9!) = 3
Is that what you were looking for?
Originally posted by villa68but where do all the zero's come from? I'd assume the only numbers allowed would be the three defined. by the way, for 9,9,9, there's
the simplist answer is
to put everything to the power of 0
cos everything to the power of 0 is 1
🙂
square_root( 9 * 9 ) / square_root( 9 )