[ -2n ] is positive for all negative values of 'n' .
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∙ 14y agoI 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).
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.
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.
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
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.
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).
Since n is positive, |n| = n, so you have 2n - n = n. The difference is n.
The integers are 43 and 44. The answer can be found by representing the two consecutive numbers as "n" and "n+1". n2 + (n+1)2 = 3785 n2 + (n2 + 2n + 1) = 3785 2n2 + 2n + 1 = 3785 2n2 + 2n - 3784 = 0 n2 + n - 1892 = 0 (n + 44) (n-43) = 0 n = -44, +43 The answer is restricted to positive values, so n = 43 and n+1 = 44
Let one integer be n, then the other is 2n + 3 and n(2n + 3) = 90; solve this last equation for n: n(2n + 3) = 90 ⇒ 2n2 + 3n - 90 = 0 ⇒ (2n + 15)(n - 6) = 0 ⇒ n = 6 or n = -7.5 As n must be a (positive) integer, the solution n = -7.5 can be ignored, leaving n = 6, giving 2n + 3 = 15. Thus the two positive integers are 6 and 15.
An N-bit integer holds 2N different values.For an unsigned integer, the range of values is 0..2N-1 thus.For a signed integer using 2s complement, the range is -2N-1..+2N-1-1.Therefore, the largest positive number that can be stored using 8 bits is 255.
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.
Meiosis is a type of cell division that results in the formation of gametes (sperm and egg cells) in sexually-reproducing organisms. The process involves one round of DNA replication followed by two rounds of cell division, producing cells with half the chromosome number (n). This leads to genetic variation and ensures the correct chromosome number is maintained in the offspring.
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.
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
Meiosis produces haploid gametes which have the ' n ' symbol.
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.
2 is a prime number. 2 times anything but 1 is composite.