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Placing a question mark at the end of a list of expressions or numbers does not make it a sensible question.

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Q: 2ab 0 a 2 4a 2b 4?
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Who can factor 64a6n - b6n?

n(2a - b)(2a + b)(4a^2 - 2ab + b^2)(4a^2 + 2ab + b^2)


Who can factor 64a raised to 6n - b raised to 6n?

n(2a - b)(2a + b)(4a^2 - 2ab + b^2)(4a^2 + 2ab + b^2)


What is 2ab-2?

2ab-2 = 0


What is A B squared?

A^2-2ab+B^2 is actually (A+B)^2 AB squared is A^2B^2 or (AB)^2


12a2b plus 8a-5a2b plus 2ab - 4ab2 plus 6ab-2a-3ab?

Assuming that non-leading numbers are exponents, the expression becomes12a^2b + 8a - 5a^2b + 2ab - 4ab^2 + 6ab - 2a - 3ab= 7a^2b + 6a + 5ab - 4ab^2= a*(7ab + 6 + 5b - 4b^2)


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.


The quotient obtained by dividing 4a-2b by a plus b?

2


What is 9a plus 2b - 8a -3b?

The given terms can be simplified to: a -b


What are the values of a and b if 6 3 is the midpoint of the line joining 2a comma 2a-b to a-2b comma 4a plus 3b?

6 is the average of 2a and a - 2b so 12 = 2a + a - 2b = 3a - 2b3 is the average of 2a - b and 4a + 3b so 6 = 2a - b + 4a + 3b = 6a + 2bAdding the two equations gives: 18 = 9a so that a = 2Substituting for a in the first gives 12 = 6 - 2b so that b = -36 is the average of 2a and a - 2b so 12 = 2a + a - 2b = 3a - 2b3 is the average of 2a - b and 4a + 3b so 6 = 2a - b + 4a + 3b = 6a + 2bAdding the two equations gives: 18 = 9a so that a = 2Substituting for a in the first gives 12 = 6 - 2b so that b = -36 is the average of 2a and a - 2b so 12 = 2a + a - 2b = 3a - 2b3 is the average of 2a - b and 4a + 3b so 6 = 2a - b + 4a + 3b = 6a + 2bAdding the two equations gives: 18 = 9a so that a = 2Substituting for a in the first gives 12 = 6 - 2b so that b = -36 is the average of 2a and a - 2b so 12 = 2a + a - 2b = 3a - 2b3 is the average of 2a - b and 4a + 3b so 6 = 2a - b + 4a + 3b = 6a + 2bAdding the two equations gives: 18 = 9a so that a = 2Substituting for a in the first gives 12 = 6 - 2b so that b = -3


GIven Tx equals ax2 plus bx plus c find a b and c if T0 equals -4 T1 equals -2 and T2 equals 6?

T(x) = ax^2 + bx + cT(0) = -4T(1) = -2T(2) = 6Solution:Since T(0) = -4, then c = -4.So, when x = 1, we have:-2 = a(1)^2 + b(1) - 4-2 = a + b - 4-2 + 4 = a + b - 4 + 42 = a + b2 - a = a - a + b2 - a = bWhen x = 2, we have:6 = a(2)^2 + b(2) - 46 = 4a + 2b - 46 + 4 = 4a + 2b - 4 + 410 = 4a + 2b10/2 = 4a/2 + 2b/25 = 2a + b5 - 2a = 2a - 2a + b5 - 2a = b2 - a = 5 - 2a2 - 2 - a + 2a = 5 - 2 - 2a + 2aa = 32 - a = b2 - 3 = b-1 = bThus, a = 3, b = -1, and c = -4.Check:


What is 2ab when a -2 and b 7?

2ab = 2*(-2)*7 = -28


Asquare plus b square plus c square -ab -bc -ca equals 0 then show that a equals b equals c?

a^2 + b^2 + c^2 - ab - bc - ca = 0=> 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca = 0 => a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2ca + a^2 = 0 => (a - b)^2 + (b - c)^2 + (c - a)^2 = 0 Each term on the left hand side is a square and so it is non-negative. Since their sum is zero, each term must be zero. Therefore: a - b = 0 => a = b b - c = 0 => b = c.