Q: What is 4mn m (2n 3)?

Write your answer...

Submit

Still have questions?

Continue Learning about Math & Arithmetic

(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

Related questions

(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.

0

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

2n x 2n x 2n = 8n^3

5m+n = 37 m-2n = 14 => m = 2n+14 Substitute the value of m into the top equation: 5(2n+14)+n = 37 10n+70+n = 37 10n+n = 37-70 11n = -33 Divide both sides by 11 to find the value of n: n = -3 Substitute the value of n into the original equations to find the value of m: Therefore: n = -3 and m = 8