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How do you factor m²-4mn-12n²?
(m + 2n)(m - 6n)
How do you factor this polynomial m2-4mn-12n2?
(m - 6n)(m + 2n)
What is 4mn m (2n 3)?
The expression (4mn(2n + 3)) represents the product of (4mn) and the sum (2n + 3). To simplify, you can distribute (4mn) to both terms inside the parentheses: (4mn \cdot 2n + 4mn \cdot 3). This results in (8mn^2 + 12mn). Thus, the final simplified expression is (8mn^2 + 12mn).
How do you solve 4mn plus 8?
4mn + 8 is an expression: it is not an equation (or inequality) and therefore there is nothing that can be solved.
What are the factors of 4mn?
They are 1, 2, 4, m, 2m, 4m, n, 2n, 4n, mn, 2mn and 4mn.
What is the simplify 7bk-9mn plus 5mn plus 2bk?
The simplified form of 7bk-9mn + 5mn + 2bk is 9bk - 4mn.
What is 4mn to the third power over 10n to the second power in simplest form?
to find the answer, you have to find the prime factorization of the two monomials 4mn^3 2*2*m*n*n*n ---------- = ------------------ 10n^2 2*5*n*n then you cross out the common numbers (so in this case: 2, n, and n) the the ones that are left are your fraction in simplest form ANSWER: 2mn ------ 5
Why do you get an odd number when you multiply two odd numbers?
Let one odd number be "2m + 1", the other odd number "2n + 1" (where "m" and "n" are integers). All odd numbers have this form. If you multiply this out, you get (2m+1)(2n+1) = 4mn + 2m + 2n + 1. Since each of the first three parts is even, the "+1" at the ends converts the result into an odd number.
WHY IS AN ODD NUMBER TIMES AN EVEN NUMBER EVEN?
Suppose m and n are integers. Then 2m+1 is an odd number and 2n is an even number.(2m + 1) * 2n = 4mn + 2n = 2*(2mn + 1). Since m and n are integers, the closure of the set of integers under multiplication and addition implies that 2mn + 1 is an integer. Thus the product is a multiple of 2: that is, it is even.
What is the conjecture of the product of two odd numbers?
The conjecture about the product of two odd numbers is that the product will always be odd. This is because an odd number can be expressed in the form (2n + 1) (where (n) is an integer), and when two such expressions are multiplied, the resulting product simplifies to (4mn + 2m + 2n + 1), which is also of the form (2k + 1), confirming it is odd. Thus, the product of any two odd numbers is always odd.
Why is an odd number times an odd number an odd number?
Suppose m and n are integers. Then 2m + 1 and 2n +1 are odd integers.(2m + 1)*(2n + 1) = 4mn + 2m + 2n + 1 = 2*(2mn + m + n) + 1 Since m and n are integers, the closure of the set of integers under multiplication and addition implies that 2mn + m + n is an integer - say k. Then the product is 2k + 1 where k is an integer. That is, the product is an odd number.
If a and b are even then G(ab) must be 2?
If both ( a ) and ( b ) are even, they can be expressed as ( a = 2m ) and ( b = 2n ) for some integers ( m ) and ( n ). Thus, the product ( ab = (2m)(2n) = 4mn ), which is also even. The greatest common divisor ( G(ab) ) in terms of their even factors does not necessarily equal 2; it depends on the specific values of ( a ) and ( b ). Therefore, while ( ab ) is even, ( G(ab) ) can be greater than 2 if ( a ) and ( b ) share higher even factors.
