b+b+b+c+c+c+c =3b+4c
b=2 a=1 c=3 so b plus a =c
2b + 2c or 2(b + c)
And how does this relate to coins?
A+c= 2a+b
b+b+b+c+c+c+c =3b+4c
b + b + b + c + c + c + c = 3b + 4c
b=2 a=1 c=3 so b plus a =c
a+b+c+d+e = 30 (i) c+e = 14 (ii) d+b = 1 (iii) a = 2b-1 (iv) a+c = 10 (v) By (i)-(ii)-(iii): a = 15 then, by (iv): b = 8 and by (v): c = -5 Also, b = 8 so by (iii): d = -7 and then by (i), e = 19
It is impossible to give any decimal/numeric value if we are not given the values of at least one variable, so the answer is B + B + B + C + C + C.
2b + 2c or 2(b + c)
And how does this relate to coins?
a= (+a) or a= (-) b= 2a b= 2a c= (-a) c= (+a)
A+c= 2a+b
If a + b + c + d + 80 + 90 = 100, then a + b + c + d = -70.
∫ f(x)/[(f(x) + b)(f(x) + c)] dx = [b/(b - c)] ∫ 1/(f(x) + b) dx - [c/(b - c)] ∫ 1/(f(x) + c) dx b ≠c
If: a = b+c+d Then: c = a-b-d