That is the property of infinite density of rational numbers. If x and y are any two rational numbers then w = (x + y)/2 is a rational number between them. And then there is a rational number between x and w. This process can be continued without end.
½W - 3 and ½W + 3 (½W - 3) + (½W + 3) = ½W + ½W - 3 + 3 = W |(½W - 3) - (½W + 3)| = |½W - 3 - ½W - 3| = |½W - ½W - 3 - 3| = |-6| = 6 (The difference between two numbers can be calculated by subtracting one from the other and ignoring the sign; |n| is the value of n ignoring the sign.)
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Five 'w' could signify the five common question words:whowhatwhenwhywhere
As w and w are the same number, there are no numbers between w and w. However, if it is inclusive of the limits, (ie "between w and w inclusive"), then there is only 1 number w, which to be a multiple of 4 6 and 8 must be a multiple of their lowest common multiple (lcm) which is 24; ie all multiples of 24, namely w is one of: 24, 48, 72, 96, 120, 144, 168, ...
The number (w +W)/2. To find a halfway point (which is also the average), add the numbers together and divide by 2.
That is the property of infinite density of rational numbers. If x and y are any two rational numbers then w = (x + y)/2 is a rational number between them. And then there is a rational number between x and w. This process can be continued without end.
No.
½W - 3 and ½W + 3 (½W - 3) + (½W + 3) = ½W + ½W - 3 + 3 = W |(½W - 3) - (½W + 3)| = |½W - 3 - ½W - 3| = |½W - ½W - 3 - 3| = |-6| = 6 (The difference between two numbers can be calculated by subtracting one from the other and ignoring the sign; |n| is the value of n ignoring the sign.)
The formula for the perimeter of a rectangle is 2(L + W), where L is the length and W is the width. Then 16 = 2(L + W) : 8 = L + W. And L = W - 8, or W = L - 8 So, if you were dealing integers, you can choose any two numbers between 1 and 7 so that the numbers when added equal 8.
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Five 'w' could signify the five common question words:whowhatwhenwhywhere
There are infinite rational numbers between 2 and 3.Explanation:Let us write a few decimal numbers between 2 and 3: 2.01, 2.001, 2.0001,.., 2.4, 2.90 etc. Just change digits after the decimal point and this way we can write infinite decimal numbers between 2 and 3. And each decimal number can be expressed in the form of p/q(rational number)2.01 = 201/1002.001 = 2001/1000... 2.4 = 24/10 and so on.So there are infinitely many rational numbers b/w 2 and 3.
Oh, what a happy little math problem we have here! To factor w^2 + 8w + 12, we're looking for two numbers that multiply to 12 and add up to 8. Those numbers are 2 and 6. So, we can rewrite the equation as (w + 2)(w + 6). Just like that, we've created a beautiful factorization!
As w and w are the same number, there are no numbers between w and w. However, if it is inclusive of the limits, (ie "between w and w inclusive"), then there is only 1 number w, which to be a multiple of 4 6 and 8 must be a multiple of their lowest common multiple (lcm) which is 24; ie all multiples of 24, namely w is one of: 24, 48, 72, 96, 120, 144, 168, ...
Perimeter = 2*L + 2*W Area = L*W L = Area / W Perimeter = 2*Area / W + 2*W W * Perimeter = 2*Area + 2*W^2 Let A=area, and P= perimeter. 0=2W^2-PW+2A Quadratic formula: W= (P + root(P^2-16A))/4 or W= (P - root(P^2-16A))/4 Once you have W, L is simply A/W. Have fun, and maybe include some numbers to work with next time.
Suppose W/3 is the even integer X. Then the numbers are (X-2, X , X+2)If W/3 is not an even integer then there is no solution.