it stays the same
I THINK ......
∫ 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
ax - b = c ax = b + c x = (b + c)/a
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)
(a x b) x c = a x (b x c)
X= (b-c)/a
∫ 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
ax - b = c Add 'b' to both sides ax = c + b Divide both sides by 'a' x = (c + b) / a The answer!!!!!