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).
(m + 2n)(m - 6n)
(m - 6n)(m + 2n)
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.
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.
It's clear that the first set has 2m subsets and second one has 2n subests. so we have to solve 2m - 2n = 56 (m,n are positive integers) Its also clear that m>n let m = k+n so 2k+n- 2n= 56 2n(2k- 1)= 23*7 clearly 2k-1 is odd so 2n= 23, and n = 3 and 2k-1= 7 so k =3 so m = 3+3 = 6 and n = 3
(m + 2n)(m - 6n)
m2 - 4mn - 12n2 = m2 - 6mn + 2mn - 12n2 = m(m - 6n) + 2n(m - 6n) = (m + 2n)(m - 6n)
(m - 6n)(m + 2n)
They are 1, 2, 4, m, 2m, 4m, n, 2n, 4n, mn, 2mn and 4mn.
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.
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.
2n + 4m - 2n + m = 5m
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.
It's clear that the first set has 2m subsets and second one has 2n subests. so we have to solve 2m - 2n = 56 (m,n are positive integers) Its also clear that m>n let m = k+n so 2k+n- 2n= 56 2n(2k- 1)= 23*7 clearly 2k-1 is odd so 2n= 23, and n = 3 and 2k-1= 7 so k =3 so m = 3+3 = 6 and n = 3
0
The expression m^2n^3/p^3 divided by mp/n^2 can be simplified as (m^2n^3/p^3) / (mp/n^2). This simplifies to (m^3n^5)/(p^3n) = m^3n^4/p^3.
Let's do some algebra. Assume that "m" and "n" are any integers. An even number is divisible by 2, so 2m or 2n would be even. An odd number is one that is not divisible by 2, so 2m + 1, or 2n + 1, are odd numbers.Multiply those two odd numbers together: (2m+1)(2n+1) = 4mn + 2m + 2n + 1. Since the first three parts are even, the added 1 at the end makes the result odd.