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How do you solve L equals 2M plus 2N for N?
N=l-m
Is the sum of an even whole number and an odd whole number always odd?
Yes --------------------------------------------- Let n be an integer. Then 2n is an even number Let m be an integer. Then 2m is an even number and 2m + 1 is an odd number. Then: even + odd = (2n) + (2m + 1) = (2n + 2m) + 1 = 2(n + m) + 1 = 2k + 1 (where k = m + n) which is an odd number.
How do i factorise 49(m 2n)2 81(2m n)2?
Distribute the 49 and 81 first
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 the sum of 2 odd numbers always even?
An odd number can be written in the form '2n+1' where 'n' is an integer, and an even number can be written int the form '2n'. We can then write the sum of 2 odd numbers as: (2n+1) + (2m+1) * Combining and factoring out a 2, we arrive at: 2(n + m + 1) Since 'n' and 'm' are both integers, we know that the value contained int the '()' is also an integer. We can therefore rewrite this equation to be: 2n Which represents an even number as given previously. * 'm' is an integer
How do you solve L equals 2M plus 2N for N?
N=l-m
If two odd numbers are added together what is the result?
Two odd numbers added together will always be an even number. I'll show how: Let M and N be any two integers. 2M is even, and 2N is even. So (2M +1) is an odd number, and (2N + 1) is an odd number. (2M +1) + (2N +1) = 2M + 1 + 2N +1 = 2M + 2N + 2 = 2(M + N + 1), which is even.
Is the sum of an even whole number and an odd whole number always odd?
Yes --------------------------------------------- Let n be an integer. Then 2n is an even number Let m be an integer. Then 2m is an even number and 2m + 1 is an odd number. Then: even + odd = (2n) + (2m + 1) = (2n + 2m) + 1 = 2(n + m) + 1 = 2k + 1 (where k = m + n) which is an odd number.
Why odd number minus a odd number even?
Every odd number leaves a remainder of 1 when divided by 2. Therefore, every odd number is of the form 2k +1 where k is some integer.Suppose 2m + 1 and 2n + 1 are two odd numbers.Then (2m + 1) - (2n + 1) = 2m + 1 - 2n - 1 = 2m - 2n = 2*(m - n)By the closure property of integers under addition (and subtraction), m and n being integers implies than (m - n) is an integer. Therefore 2*(m - n) is an even integer.
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.
How do i factorise 49(m 2n)2 81(2m n)2?
Distribute the 49 and 81 first
Two finite sets have m and n elements If A has 56 more subsets than B then what are the values of m and n?
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
Why do you get an even number when adding two even numbers together?
The presentation of an even number is 2n (we can use any letter, such as 2m or 2r) Let the first even number be 2n and the second number be 2m. So we have, 2n + 2m = 2(n + m) let n + m = r, then by substituting r for n + m we have = 2r Since 2r represents an even number, then we can say that when are adding two even numbers we obtain another even number.
How do you prove by algebra that if you add two even numbers you get an even number?
You use the definition of even numbers which says it is a number of the form 2n for some integer n, and you take two of them, say 2n and 2m and add them and see that you have 2n+2m=2(n+m) and since n and m are integers so is n+m so lets call it p so the sum is 2p and this is even by definition of even numbers
How do you know that adding an even number to an even number results in an even number?
This is a simple proof of what you ask. Any even number can be thought of as 2n for some integer n. So let's take two even numbers 2n and 2m and add them. 2n+2m=2(n+m) and since n and m are integers so is n+m, call it p, then the sum is 2p and that is even since it is a multiple of 2.
Why is the sum of an even and odd number alsways odd?
A number a is even if there exists an integer n such that a = 2n A number b is odd if there exists an integer m such that b = 2m + 1. So: a+b = (2n) + (2m +1) = 2 (n+m) + 1 Since n and m are integers, n+m is also an integer. So a+b satisfies the definition of an odd number.
What are the factors of 4mn?
They are 1, 2, 4, m, 2m, 4m, n, 2n, 4n, mn, 2mn and 4mn.
