144 Formula: c= (a2)+(ab)
(a3 + b3)/(a + b) = (a + b)*(a2 - ab + b2)/(a + b) = (a2 - ab + b2)
Commutativity.
The existence of the additive inverse (of ab).
ab=1a+1b a is equal to either 0 or two, and b is equal to a
144 Formula: c= (a2)+(ab)
(a3 + b3)/(a + b) = (a + b)*(a2 - ab + b2)/(a + b) = (a2 - ab + b2)
yes because ab plus bc is ac
2ab
a2 + b2 + c2 - ab - bc - ca = 0 => 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca = 0 Rearranging, a2 - 2ab + b2 + b2 - 2bc + c2 + c2 - 2ca + a2 = 0 => (a2 - 2ab + b2) + (b2 - 2bc + c2) + (c2 - 2ca + a2) = 0 or (a - b)2 + (b - c)2 + (c - a)2 = 0 so a - b = 0, b - c = 0 and c - a = 0 (since each square is >=0) that is, a = b = c
Commutativity.
The existence of the additive inverse (of ab).
associative property
the midpoint of AB.
C minus B equals AB
This is the formula: (a3)+(b3)=(a+b)(a2-ab+b2)
ab=1a+1b a is equal to either 0 or two, and b is equal to a