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6a2 + 5ab - 6b2 = (3a - 2b)(2a + 3b)
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Which is a factor of 6a2 5ab - 6b2?
(3a - 2b)(2a + 3b)
A closed box with square base has volume of 16000 cm3. Material for the top of the box costs 3 dollars per cm3 and the sides costs 1.50 cm3. Whats minimum costs.?
$6000 the box has a square base of 20 cm x 20 cm and a height of 40 cm. Its all in the calculus b = side of base Cost = 6b2 + (96,000/b) Derivative of Cost = 12b - (96,000/b2) Solve for b ______________________________ I respectfully disagree with your answer. Try using the cube root of 16000. Or, if you prefer rounder numbers, try length = 25, width = 25, height = 25.6. Of course, this is a crude approximation, but you still end up with a total cost of $5715, so I saved you at least $285. If you used exact numbers, you'd end up with $5714 and some change. Matt
Which is a factor of 6a2 5ab - 6b2?
(3a - 2b)(2a + 3b)
Which is a factor of 6a2 plus 5ab - 6b2?
Considering the minus sign between 5ab and 6b2 then we have the polynomial as 6a2 + 5ab - 6b2. The polynomial is a quadratic polynomial.Steps to factorize a quadratic polynomial:1 - Multiply first term by third term. 6a2 x (-6b2) = -36a2b22 - If possible break the second term into two terms such that they multiple to -36a2b2. If not then it is factorized by Sridharacharya's formula.5ab can be broken as 9ab + (-4ab).These two terms multiply to give -36a2b2.So we can write 6a2 + 5ab - 6b2 = 6a2 + 9ab + (-4ab) - 6b2.6a2 + 9ab - 4ab - 6b2 = 3a(2a + 3b) - 2b(2a + 3b) = (2a + 3b)(3a - 2b).So the factors are (2a + 3b) and (3a - 2b).
What is the factor polynomial 6b2-15b3?
-3b2(5b - 2)
How do you factor out 6b2-15b-6?
3(2b2 - 5b - 2)
How do you factor 6b2-7b-20?
(-3b - 4)(-2b + 5)
How do you factor the trinomial 6b2-13bs-63s2?
6b^2-13bs-63s^2 is factorised to (2b-9s)(3b+7s)
Solve for a a2-8a plus 12 a-2 -6b2 6 c -3 -4 d 3 4?
a2-8a plus 12, if done correctly according to calculus, results in the number 2.
Prove that square root of 6 is irrational number?
If a/b=sqrt(6), then a2=6b2 On the other hand, given integers ''a'' and ''b'', because the valuation (i.e., highest power of 2 dividing a number) of 6b2 is odd, while the valuation of a2 is even, they must be distinct integers. Contradiction.
What are all the factors of the monomial 6ab to the 2nd power?
They are 1, 2, 3, 4, 6, 9, 12, 18, 36,1a, 2a, 3a, 4a, 6a, 9a, 12a, 18a, 36a,1b, 2b, 3b, 4b, 6b, 9b, 12b, 18b, 36b,1a2, 2a2, 3a2, 4a2, 6a2, 9a2, 12a2, 18a2, 36a2,1ab, 2ab, 3ab, 4ab, 6ab, 9ab, 12ab, 18ab, 36ab,1b2, 2b2, 3b2, 4b2, 6b2, 9b2, 12b2, 18b2, 36b2.
Solve by factoring 6b2-13b 6 equals 0?
The question is ambiguous because there is no sign shown before the final 6. Assume it is +6 since -6 would make factorisation very difficult. Thus the equation is 6b2 - 13b + 6 = 0 Then 6b2 - 4b - 9b + 6 = 0 or 2b(3b - 2) - 3(3b - 2) = 0 ie (3b - 2)(2b - 3) = 0 then 3b -2 = 0 or 2b - 3 = 0 so 3b =2 or 2b = 3 ie b = 2/3 or b = 3/2
How do you factor 6ab - 6ac plus 6b2 - 6bc?
Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "times", "equals", "squared", "cubed" etc. Furthermore, it is not possible to solve one linear equation in two variables. You can only express one variable in terms of the other.
Why is the square root of 24 irrational?
If it were rational, then the square root of 6 would be also since sqrt 24 is sqrt(4x6)=2xsqrt(6) and here is a proof the sqrt(6) is irrational So let's assume by contradiction that the square root of 6 is rational. By definition, that means there are two integers a and b with no common divisors where: a/b = square root of 6. So let's multiply both sides by themselves: (a/b)(a/b) = (square root of 6)(square root of 6) a2/b2 = 6 a2 = 6b2 But this last statement means the RHS (right hand side) is even, because it is a product of integers and one of those integers (at least) is even. So a2 must be even. But any odd number times itself is odd, so if a2 is even, then a is even. Since a is even, there is some integer c that is half of a, or in other words: 2c = a. Now let's replace a with 2c: a2 = 6b2 (2c)2 = (2)(3)b2 2c2 = 3b2 But now we can argue the same thing for b, because the LHS is even, so the RHS must be even and that means b is even. Now this is the contradiction: if a is even and b is even, then they have a common divisor (2). Then our initial assumption must be false, so the square root of 6 cannot be rational.
