The answer for a x b x c is the product of the three numbers a, b, and c. When you multiply three numbers together, you are finding the total result of combining the values. This operation is associative, meaning you can multiply any two of the numbers first and then multiply the result by the third number without changing the final answer.
∫ 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
Yes. It's the graph of [ Y = f(X) ] described by (X/A)2 + (Y/B)2 = C2 A, B, and C are constants. If 'A' and 'B' are both '1', then the graph is a circle with radius 'C'.
A quadratic involving x and y is usually in the form 'y = ax2 + bx + c'. This form is y in terms of x, so we must rearrange it. y = ax2 + bx + c y/a = x2 + bx/a + c/a y/a = x2 + bx/a + d + e, where c/a = d + e, e = (b/a)2 y/a - e = x2 + bx/a + d y/a -e = (x + b/a)2 √(y/a - e) = x + b/a √(y/a - e) - b/a = x
a^2b^2c^2 ^2 is squared
Given the limitations of the browser through which you post questions, it is virtually impossible to be certain. From what I can see and what I can guess, the most likely candidate isa(x) = 1.5*0.85^x.
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.
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
(a x b) x c = a x (b x c)
∫ 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
X= (b-c)/a
ax - b = c Add 'b' to both sides ax = c + b Divide both sides by 'a' x = (c + b) / a The answer!!!!!