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What is the Maximum number of node at height h of a binary tree?

It is ((2^h) -1)/(2-1) generally for an m-tree is: ((m^h)-1)/(m-1)


What is the degree of freedom when the number of component and number of phase are both equal to 2?

df = (n-1)*(m-1) = (2-1)*(2-1) = 1*1 = 1


When was kenmore serial number M 31803091 built?

2 1/2 tons


What is m-2 over m plus 2 times m over m-1?

(m-2)/(m+2) * m/(m-1) = [(m-2)*m]/[(m+2)*(m-1)] = (m2 - 2m)/m2 + m - 2)


What is the average of all the integer's of 13 to 37?

To find the (mean) average, add all the numbers and divide by the number of numbers. The sum of a series of digits (in arithmetic progression, like 13, 14, 15, ... 37) is sum = (first + last) x number_of_digits / 2 So their average is: average = ((first + last) x number_of_digits / 2) / number_of_digits = (first + last) / 2 = average of first and last digits! So the average of the numbers 13, 14, 15, ..., 37 is: average = (13 + 37) / 2 = 50 / 2 = 25 To find the sum of n digits starting with m: Sum = m + (m+1) + ... + (m+n-2) + (m+n-1) Rewrite the sum in reverse order: Sum2 = Sum = (m+n-1) + (m+n-2) + ... + (m+1) + m Add the two sums, term by term: Sum + Sum2 = 2 Sum = (m + (m+n-1)) + (m + (m+n-1)) + ... + (m + (m+n-1)) + (m + (m+n-1)) There are n terms, all (m + (m+n-1)), so: 2 Sum = (m + m+n-1) x n Sum = (m + m+n-1) x n / 2 But n is the number of digits, m is the first number and (m+n-1) is the last, so: Sum = (first + last) x number_of_digits / 2


What 2 number you can mutiply to 80 but subtract to get -79?

mn = 80 m -n = -79 Substitute m - 80/m = -79 m^2 - 80 = -79m m^2 + 79m = 80 (NB Notice change of signs) Quadratic Eq'n m = { - 79 +/- sqrt[)79)^2 - 4(1)(-80)}] / 2(1) m = { - 79 +/- sqrt[6241 + 320}]/ 2 m = { -79 +/- sqrt[6561]} / 2 m = { - 79 +/- 81}/2 m = -160 / 2 = -80 or m = 2/2 = 1 Hence n = 1 + 79 = 80


What are the possible values of the quantum numbers n l m s for the second shell?

Possible values of quantum numbers in order of n,l,m,s in the second shell:2,0,0,-1/22,0,0,+1/22,1,-1,-1/22,1,-1,+1/22,1,0,-1/22,1,0,+1/22,1,1,-1/22,1,1,+1/2


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.


What is 1mm x 2 m?

1 mm = 1/1000 m → 1 mm × 2 m = 1 × 1/1000 m × 2 m = 0.002 m² Or 1 m = 1000 mm → 1 mm × 2 m = 1 mm × 2 × 1000 mm = 2000 mm²


What are 3 consecutive numbers?

They are numbers of the form m, m+1 and m+2 where m is an integer. However, sometimes it can be easier - particularly with an odd number of consecutive integers - to write them as n-1, n and n+1 where n is an integer (= m+1).


What are the factors of the quadratic function -m2 3m-2?

-m2+3m-2 -m2+2m+m-2 -m(m -2)+1(m-2) (-m+1)(m-2) or


What is the multiplicative inverse of a whole number?

Let m be a whole number, then the multiplicative inverse of m is a number n such that mn=1 since 1 is the multiplicative identity. There is only one choice for n, it is 1/m since m(1/m)=1