0
2ab-2 = 0
What else can I help you with?
How do you calucate a2 plus b2?
a2+2a2b+2ab2+b2
How do you cube a binomial?
let binomial be (a + b)now (a+b)3 will be (a+b)(a+b)2 = (a+b)(a2 + 2ab+ b2) = a(a2+ 2ab+ b2) + b(a2 + 2ab+ b2) = a3+ 2a2b+ ab2 + a2b + 2ab2 + b3 = a3+ 2a2b+ ab2 + a2b + 2ab2 + b3 = a3 +3a2b + 3ab2 +b3 hope it helped... :D
What is the formula of a plus b plus c whole cube?
a3 + b3 + c3 + 2(a2)b + 2(b2)c + 2(a2)c + 2ab2 + 2(c2)b +abc
What is min minus superscript 1?
For those of us who know non-Lygerian mathematics ---- there is no unique solution but it can be approximated with the series sol = 1 + ab x + 2ab2/c3 x + 3ab3/c5 x + 4ab4/c7 x + ..... where x is either -ln(min) or +ln(min)
The diagonals of square abcd intersect at e if ae equals 2 find to perimeter of abcd?
First of all we work out the length of a sides ab, bc, CD, & ad. We know that ab = bc = CD = ad also ae = ac/2 If a to e = 2 then ac = 4 so ab2 + bc2 = ac2 2ab2 = 16 ab2 = 8 ab = 2.8284271247461900976033774484194 so the perimeter = ab * 4 = 11.31
How do you compare half of a add b and a add b dividing 2ab?
Algebraically:((a+b) / 2) / ((a+b) / (2ab)) = 2ab(a + b) / 2(a + b) = (2a2b + 2ab2) / (2a + 2b) = abAs an example, let a = 2 and b = 3 then:i) (a + b) / 2 = (2 + 3) / 2 = 5/2ii) (a + b) / (2 * a * b) = (2 + 3) / (2 * 2 * 3) = 5 / 12Therefore i) is ab times larger than ii). (for these specific example numbers it will be 6 times larger)
2a square b square plus 5ab square plus 8a square b square - 3ab square?
To simplify the expression 2a^2b^2 + 5ab^2 + 8a^2b^2 - 3ab^2, first combine like terms. The terms with a^2b^2 are 2a^2b^2 + 8a^2b^2 = 10a^2b^2. The terms with ab^2 are 5ab^2 - 3ab^2 = 2ab^2. Therefore, the simplified expression is 10a^2b^2 + 2ab^2.
How do you factor 8a to the 3 plus 4a to the 2b-2ab to the 2-b to the 3?
Your notation is confusing. It looks like: 8a3 + 4a2b - 2ab2 - b3. So if you look at it, notice 23 = 8 & 22 = 4, so take c = (2*a), and you have c3 + c2b - cb2 - b3, Now note that if c = b, the polynomial is zero, so (c-b) or (b-c) is a factor. So divide by (c-b) using long division: (c - b)(c2 + 2cb + b2) {the second polynomial is a perfect square: (c+b)2}, so (c-b)(c+b)2, then substitute c = 2a, and get (2a-b)(2a+b)2 finally, multiply it out to make sure you get the original polynomial.
