The variable c times the variable b simply equals cb. Just as the variable x times the variable y would equal xy, and so on.
b divided by 2
Associative: (a + b) + c = a + (b + c) (a x b) x c = a x (b x c)
suppose x is in B. there are two cases you have to consider. 1. x is in A. 2. x is not in A Case 1: x is in A. x is also in B. then x is in A intersection B. Since A intersection B = A intersection C, then this means x is in A intersection C. this implies that x is in C. Case 2: x is not in A. then x is in B. We know that x is in A union B. Since A union B = A union C, this means that x is in A or x is in C. since x is not in A, it follows that x is in C. We have shown that B is a subset of C. To show that C is subset of B, we do the same as above.
(a x b) x c = a x (b x c)
The Second Mean Value Theorem for Riemann integrals states that if ( f ) and ( g ) are continuous functions on the closed interval ([a, b]) and ( g ) is non-negative and integrable, then there exists a point ( c \in [a, b] ) such that: [ \int_a^b f(x) g(x) , dx = f(c) \int_a^b g(x) , dx. ] Proof: Define ( G(x) = \int_a^x g(t) , dt ). Since ( g ) is continuous, ( G ) is differentiable and ( G(a) = 0 ). By applying the Mean Value Theorem to ( G ) over ([a, b]), we find a ( c \in [a, b] ) such that: [ G(b) = G'(c)(b - a) = g(c)(b - a). ] Thus, we have: [ \int_a^b g(x) , dx = G(b) = g(c)(b - a), ] which leads to the conclusion that: [ \int_a^b f(x) g(x) , dx = f(c) \int_a^b g(x) , dx. ]
2 x b x t x a x c to the power 2
No. But multiplication is distributive over addition. This means that for any numbers A, B, and C A x (B + C) = (A x B) + (A x C). If addition were distributive over multiplication, that would mean that A + (B x C) = (A + B) x (A + C) which is not true.
b divided by 2
Associative: (a + b) + c = a + (b + c) (a x b) x c = a x (b x c)
Let the number be X, then B% = B/100 → B% of X = C → B/100 x X = C → X = C ÷ (B/100) = C x 100/B = 100C ÷ B So to find the number, divide C by B percent.
suppose x is in B. there are two cases you have to consider. 1. x is in A. 2. x is not in A Case 1: x is in A. x is also in B. then x is in A intersection B. Since A intersection B = A intersection C, then this means x is in A intersection C. this implies that x is in C. Case 2: x is not in A. then x is in B. We know that x is in A union B. Since A union B = A union C, this means that x is in A or x is in C. since x is not in A, it follows that x is in C. We have shown that B is a subset of C. To show that C is subset of B, we do the same as above.
Not sure what you mean by "missed" but the answer is 0.
Ax + B = Bx + C Ax - Bx = (C - B) x (A - B) = (C - B) x = (C - B) / (A - B)
(a x b) x c = a x (b x c)
x + y + z = 0 x = a - b, y = b - c, z = c - a, therefore a - b + b - c + c - a = ? a - a + b - b + c - c = 0
For any three numbers a, b, and c:a + b = b + a (commutative law)(a + b) + c = a + (b + c) (associative law)Both the commutative and associative laws are also valid for multiplication.a x (b + c) = (a x b) + (a x c) (distributive law)For any three numbers a, b, and c:a + b = b + a (commutative law)(a + b) + c = a + (b + c) (associative law)Both the commutative and associative laws are also valid for multiplication.a x (b + c) = (a x b) + (a x c) (distributive law)For any three numbers a, b, and c:a + b = b + a (commutative law)(a + b) + c = a + (b + c) (associative law)Both the commutative and associative laws are also valid for multiplication.a x (b + c) = (a x b) + (a x c) (distributive law)For any three numbers a, b, and c:a + b = b + a (commutative law)(a + b) + c = a + (b + c) (associative law)Both the commutative and associative laws are also valid for multiplication.a x (b + c) = (a x b) + (a x c) (distributive law)
ax - b = c ax = b + c x = (b + c)/a