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What are the first 4 mutpules of 2n?
I am assuming that 2n is an algebraic expression, and n is limited to positive integer values. The first 4 multiples of 2n are 0 (2n*0), 2n (2n*1), 4n (2n*2), and 6n (2n*3). If you are looking for non-0 multiples, you would also include 8n (2n*4).
What fractions are less than one-half.explain?
Any fraction with numerator = 1 and denominator > 2. Suppose numerator = 1 and denominator = n where n > 2 then 1/2 - 1/n = n/(2n) - 2/(2n) = (n-2)/(2n) Since n>2 the numerator of the above answer is positive nad the denominator (= 2n) is also positive so the difference between the two fractoins is positive. That is, 1/2 > 1/n or, equivalently, 1/n is less than 1/2.
Is the sum of two consecutive odd numbers is always divisble by 4?
Yes. Let the two consecutive odd numbers be (2n - 1) and (2n + 1) for all values of integer n (ie n ∈ ℤ) Then their sum is: sum = (2n - 1) + (2n + 1) = 2n + 2n = (2 + 2)n = 4n = a multiple of 4, so is divisible by 4.
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
What is n in the equation 41-2n equals 2 plus n?
We need to solve for "n" in 41-2n = 2 + n. Lets begin by getting the variable and constants on their own sides. 41-2n = 2 + n. Lets take the variable to the right side (since the value would be more positive) and the constant to the left. Lets subtract 2 from each side. (41-2) - 2n = 2+(-2) + n 39 - 2n = n Now lets bring the variable over to the left side by adding 2n. 39 - 2n + 2n = n + 2n 39 = 3n Finally lets solve for n. Since its being multiplied to n lets divide it. 39/3 = 3n/3 13 = n. So our answer is - 41 - 2n = 2 + n; N = 13.
