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Why isnt subtraction commutative?
Example: a - b = b-a, So lets say a=2 b=3. 2-3=3-2 -1 =/= 1
Circumference of an oval?
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)]}
What is 1 over x 3 over a 2 over b?
b/6
What is the number of sides in an argument?
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
What is 2and 1 half divided by 2 thirds?
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
Algebraic expression a cubed minuis B-Cubed?
a^(3) - b^(-3) = a^(3) - 1/b^(3) This factors to (a - 1/b)(a^2 + a/b + (1/b)^2))
What are the similarities of commutative and associative properties?
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=?
A B C + 1 2 3= 357
Why isnt subtraction commutative?
Example: a - b = b-a, So lets say a=2 b=3. 2-3=3-2 -1 =/= 1
What is 3x2 - 2 divided by x plus 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.
What two numbers add up to six but multiply to 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)
What absolute value problem has the answer of 2 or -3?
5
How do you find all the factors for a number without listing them all?
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.
What are the different special product formulas?
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
What is the GPA for 2 b's 4 c's and 1 d?
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+.
If A is less than B and B plus C equals 10 and none of them equal zero then which of the following must be true?
You haven't provided any choices for the "which of the following" part of your question. Such questions are best avoided here. However, assuming a, b and c are all natural numbers, all of the following are true for a<b AND b+c=10: a=1, b=2, c=8 a=1, b=3, c=7 a=1, b=4, c=6 a=1, b=5, c=5 a=1, b=6, c=4 a=1, b=7, c=3 a=1, b=8, c=2 a=1, b=9, c=1 a=2, b=3, c=7 a=2, b=4, c=6 a=2, b=5, c=5 a=2, b=6, c=4 a=2, b=7, c=3 a=2, b=8, c=2 a=2, b=9, c=1 a=3, b=4, c=6 a=3, b=5, c=5 a=3, b=6, c=4 a=3, b=7, c=3 a=3, b=8, c=2 a=3, b=9, c=1 a=4, b=5, c=5 a=4, b=6, c=4 a=4, b=7, c=3 a=4, b=8, c=2 a=4, b=9, c=1 a=5, b=6, c=4 a=5, b=7, c=3 a=5, b=8, c=2 a=5, b=9, c=1 a=6, b=7, c=3 a=6, b=8, c=2 a=6, b=9, c=1 a=7, b=8, c=2 a=7, b=9, c=1 a=8, b=9, c=1
What are types of special product?
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
