a2+2ab+b2+2ac+2bc+c2+2ad+2ae+2bd+2be+2cd+2ce+d2+2de+e2
{ int a,b; { a=a^2; } { b=b^2; } { c=a^2+b^2+2*a*b; print f("%d%d%d",&c); get ch(); } ]
a= (+a) or a= (-) b= 2a b= 2a c= (-a) c= (+a)
If a + b + c + d + 80 + 90 = 100, then a + b + c + d = -70.
int x = 2 * (a + b);
You add 2 fractions with the same denominator [c], so the sum is the sum of the numerators divided by the denominator: a/c + b/c = (a+b)/c
2b + 2c or 2(b + c)
The Pythagoream Thereom is a^2 + b^2 = c^2. Written out it is a squared plus b squared equals c squared.
sqrt[(a + b)2*(c + d)/pi] = (a + b)*sqrt[(c + d)/pi]
b+c^2 There is no way to simplify that.
Because it is mathematically incorrect. a^2 + b^2 = c^2 Take square root of both sides. SQRT (a^2 + b^2) = c So you see, it is not a plus b equal c.
b+b+b+c+c+c+c =3b+4c
b + b + b + c + c + c + c = 3b + 4c
{ int a,b; { a=a^2; } { b=b^2; } { c=a^2+b^2+2*a*b; print f("%d%d%d",&c); get ch(); } ]
b=2 a=1 c=3 so b plus a =c
a3 + b3 + c3 + 2(a2)b + 2(b2)c + 2(a2)c + 2ab2 + 2(c2)b +abc
The equality 6+3+2 = 6+2+3 is an example of the commutative property of addition. When using only addition, the order of the values does not change their sum. Since b+c = c+b then a+(b+c) = a+(c+b)
2a + 2b = 26 2b + c = 22 ===> b = 11 - c/2 4c + 2b = 46 ===> b = 23 - 2c From the 2nd and 3rd equations: 11 - c/2 = 23 - 2c 22 - c = 46 - 4c 22 - 46 = -4c + c ===> -24 = -3c ===> c = 8 Substitute c=8 into the 3rd equation: b = 23 - 16 = 7 Substitute b=7 into the 1st equation: 2a + 14 = 26 2a = 12 ===> a = 6