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What number is both odd and even?
There is no number that is both even (divisible by 2) and odd (not divisible by 2). Even numbers are of the form 2m for all integer m; Odd numbers are of the form 2n + 1 for all integer n; Assume there is an even number is also an odd number, then for some integer m and n: 2m = 2n + 1 � 2m - 2n = 1 � 2(m - n) = 1 � m - n = 1/2 But as m and n are both integers, their difference cannot be a fraction. Thus there are no integer m and n that satisfy 2m = 2n + 1, which means that the original assumption that there is an even number that is also an odd number is false. Thus there is no number that is both even and odd. This question is actually a riddle. The answer is 6 or 9, since flipping either number will give you the other.
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
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.
What is the procedure calculate the value of 'm' in m-derived filters?
m={1-[fc/f(infinity)]}^(1/2)fc={1/[Lc]^(1/2)} f(infinity)={fc/[(1-m^2)^(1/2)]}
How do you simplify the expression m-2 plus 1-2m plus 1?
m-2+1-2m+1 When simplified: -m
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 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 number is both odd and even?
There is no number that is both even (divisible by 2) and odd (not divisible by 2). Even numbers are of the form 2m for all integer m; Odd numbers are of the form 2n + 1 for all integer n; Assume there is an even number is also an odd number, then for some integer m and n: 2m = 2n + 1 � 2m - 2n = 1 � 2(m - n) = 1 � m - n = 1/2 But as m and n are both integers, their difference cannot be a fraction. Thus there are no integer m and n that satisfy 2m = 2n + 1, which means that the original assumption that there is an even number that is also an odd number is false. Thus there is no number that is both even and odd. This question is actually a riddle. The answer is 6 or 9, since flipping either number will give you the other.
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
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 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
If m is not equal to -1 then m divided by m plus 1 is less than 1?
Not true, here's a case where is doesn't work:m = -2m / (m+1) = -2 / (-2+1) = -2 / -1 = 2
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.
What is half of a number m?
m/2
What is the procedure calculate the value of 'm' in m-derived filters?
m={1-[fc/f(infinity)]}^(1/2)fc={1/[Lc]^(1/2)} f(infinity)={fc/[(1-m^2)^(1/2)]}
What is the probability of rolling a prime number?
total numbers taht can be occure=1,2,3,4,5,6 prime number taht can occure in a roll= 2, 3,5 total sample space= n=6 total faverable cases= m=3 P(prime number)=m/n=3/6=1/2 probability of rolling a prime number is 1/2
