That factors to (a + 1)(a + b) a = -1, -b b = -a
(a3 + b3)/(a + b) = (a + b)*(a2 - ab + b2)/(a + b) = (a2 - ab + b2)
Commutativity.
7
If, as is normal, ab represents a times b, etc then ab + ab + cc = 2ab + c2 which is generally not the same as abc.
A = 1, B = 9
to solve ab=c+a for a:Divide both sides by b, so:a= (c+a)/b
That factors to (a + 1)(a + b) a = -1, -b b = -a
(a3 + b3)/(a + b) = (a + b)*(a2 - ab + b2)/(a + b) = (a2 - ab + b2)
yes because ab plus bc is ac
No - a pos can not donate blood to a negative
Commutativity.
2ab
3y=ab. Divide by 3 to get y=ab/3
7
then the baby will be O plus, or AB plus , I think .
If, as is normal, ab represents a times b, etc then ab + ab + cc = 2ab + c2 which is generally not the same as abc.