a.
That a = ±b
(b + 2c)(b - c)
If we assume that the sqare root of 5 is a rational number, then we can write it as a/b in its simplest form, where a and b have no common factors. Therefore 5 = a2/b2 Therefore 5b2 = a2 Therefore a2 is divisible by 5, because b2 is an integer Therefore a is divisible by 5, because 5 is a prime number. Therefore 5 = 5c/b, where c is an integer Therefore 1 = c/b Therefore c = b Therefore sqrt 5 = 5c/c = 5, which is impossible. So sqrt5 cannot be expressed in the form a/b, and is irrational.
a^2 + b^2 + c^2 - ab - bc - ca = 0=> 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca = 0 => a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2ca + a^2 = 0 => (a - b)^2 + (b - c)^2 + (c - a)^2 = 0 Each term on the left hand side is a square and so it is non-negative. Since their sum is zero, each term must be zero. Therefore: a - b = 0 => a = b b - c = 0 => b = c.
b+b+b+c+c+c+c =3b+4c
b + b + b + c + c + c + c = 3b + 4c
312 is divisible by 2 b/c the 2 at the end of 312 is an even number312 is divisible by 3 b/c 3 + 1 + 2 = 6 which is divisible by 3312 is divisible by 4 b/c 12 is divisible by 4312 is not divisible by 5 b/c the one's place isn't 0 or 5312 is divisible by 6 b/c 312 is divisible by 2 and 3312 is not divisible by 7 b/c 2*7=14, 31-14=17 which is not divisible by 7312 is divisible by 8 b/c 312/8=39 (Explanation: A way to find out how 8 is divisible its that its first three digits [the ones, tens, and hundreds place] have to be divisible by 8)312 is not divisible by 9 b/c 3 + 1 + 2 = 6 which is not divisible by 9312 is not divisible by 10 b/c the ones place is not 0
Let A, B, C are digits. Number ABC is divisible by eleven if and only if A+C-B=0 or A+C-B=11
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.
If: a = b+c+d Then: c = a-b-d
Because there is no way to define the divisors, the equations cannot be evaluated.