Example: a - b = b-a, So lets say a=2 b=3. 2-3=3-2 -1 =/= 1
There is no simple answer: in fact the search for the answer led to the study of elliptical integrals.C = 2*pi*a*{1 - [1/2]^2*e^2 + [(1*3)/(2*4)]^2*e^4/3 - [(1*3*5)/(2*4*6)]^2*e^6/5 + ... }wherea is the semi-major axis,b is the semi-minor axis, ande is the eccentricity = sqrt{(a^2 - b^2)/a^2}.The above infinite series converges very slowly.An approximation, suggested by Ramanujan, isC = pi*{3(a + b) - sqrt[(3a + b)*(a +3b)]} = pi*{3(a + b) - sqrt[10ab + 3(a^2 + b^2)]}
b/6
A-B-C as easy as 1-2-3 as simple as DO-RE-MEI A-B-C 1-2-3 baby you and me girl!! WOOOH!! :D
2 and 1 half = 2,5 2 thirds = 2/3 We use fraction flipping here, e.g.: a/(b/c) = a*(c/b) 2,5 / (2/3) = 2,5 * 3/2 = 7,5/2 = 3,75
a^(3) - b^(-3) = a^(3) - 1/b^(3) This factors to (a - 1/b)(a^2 + a/b + (1/b)^2))
Conmutative: a + b = b + a 5 + 2 = 2 + 5 (a)(b) = (b)(a) (2)(3) = (3)(2) Associative: (a + b) + c = a + (b + c) (1 + 2) + 3 = 1 + (2 + 3)
A B C + 1 2 3= 357
Example: a - b = b-a, So lets say a=2 b=3. 2-3=3-2 -1 =/= 1
3x2 - 2 is a polynomial of order 2. Therefore, dividing it by (x + 1) will result in a polynomial of order 1. Suppose the quotient is ax + b (where a is non-zero), and with the remainder c. Thus 3x2 - 2 = (x + 1)*(ax + b) + c = ax2 + ax + bx + b + c = ax2 + (a + b)x + (b + c) Comparing coefficients: 3 = a 0 = a + b => 0 = 3 + b => b = -3 -2 = b + c => -2 = -3 + c => c = 1 Therefore, (3x2 - 2)/(x + 1) = 3x - 3 = 3*(x - 1) and a remainder of 1.
a+b=6, or a=6-b a*b=1 substitute for a in the 2nd equation, using the first equation: (6-b)*b=1 -b2+6b-1=0 b2-6b+1=0 Using the quadratic equation, b = (6 +- (36-4).5)/2 = 3 +- 2(2).5 a = 6 - (3 +- 2(2).5) = 3 +- 2(2).5 (a,b) = (3+2(2).5,3-2(2).5) or (3-2(2).5,3+2(2).5)
5
It is not simple. The only systematic way is to find the prime factorisation of the number and write it in exponential form. So suppose n = (p1^r1)*(p2^r2)*...*(pk^rk) where p1, p2, ... pk are prime numbers and rk are the indices (or powers). Then the factors of n are (p1^s1)*(p2^s2)*...*(pk^sk) where 0 ≤ sk ≤ rk. And remember that anything raised to the power 0 is 1. Example: n = 72 = 2*2*2*3*3 = (2^3)*(3^2) so, the factors of n are (2^a)*(3^b) where a = 0, 1, 2 or 3 and b = 0, 1 or 2. When (a, b) = (0, 0) the factor is 1. (a, b) = (1, 0) the factor is 2. (a, b) = (2, 0) the factor is 4. (a, b) = (3, 0) the factor is 8. (a, b) = (0, 1) the factor is 3. (a, b) = (1, 1) the factor is 6. (a, b) = (2, 1) the factor is 12. (a, b) = (3, 1) the factor is 24. (a, b) = (0, 2) the factor is 9. (a, b) = (1, 2) the factor is 18. (a, b) = (2, 2) the factor is 36. (a, b) = (3, 2) the factor is 72.
1. Square of a binomial (a+b)^2 = a^2 + 2ab + b^2 carry the signs as you solve 2. Square of a Trinomial (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc carry the sings as you solve 3. Cube of a Binomial (a+b)^3 = a^3 + 3(a^2)b + 3a(b^2) + b^3 4. Product of sum and difference (a+b)(a-b) = a^2 - b^2 5. Product of a binomial and a special multinomial (a+b)(a^2 - ab + b^2) = a^3-b^3 (a-b)(a^2 + ab + b^2) = a^3-b^3
2.83 based on A=4, B=3, C=2, D=1. If each class has the exactly the same number of credits, then this would be a B+.
Well, isn't that just a happy little math problem! If A is less than B and B plus C equals 10, then it must be true that A plus C is less than 10. Just remember, in the world of numbers, everything adds up beautifully in the end.
In mathematics, special products are of the form:(a+b)(a-b) = a2 - b2 (Product of sum and difference of two terms) which can be used to quickly solve multiplicationsuch as:301 * 299 = (300 +1)(300-1) = 3002 - 12 = 90000 - 1 = 89999types1. Square of a binomial(a+b)^2 = a^2 + 2ab + b^2carry the signs as you solve2. Square of a Trinomial(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bccarry the sings as you solve3. Cube of a Binomial(a+b)^3 = a^3 + 3(a^2)b + 3a(b^2) + b^34. Product of sum and difference(a+b)(a-b) = a^2 - b^25. Product of a binomial and a special multinomial(a+b)(a^2 - ab + b^2) = a^3-b^3(a-b)(a^2 + ab + b^2) = a^3-b^3