0
[ -2n ] is positive for all negative values of 'n' .
Wiki User
∙ 14y ago
Add your answer:
Earn +20 pts
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.
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 is the difference of a positive integer n and twice its absolute value?
Since n is positive, |n| = n, so you have 2n - n = n. The difference is n.
What is the answer to the sum of the squares of two consecutive positive integers is 3785 find the two integers?
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
Find two positive integers where one is 3 more than twice the other and their product is 90?
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.
What is the largest positive number that can be stored using 8 bits?
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.
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.
What is true of meiosis n to n n to 2n 2n to n or 2n to 2n?
it is 2n -n because in meiosis a diploid 2n becomes a haploid n.
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
Which or these is true of meiosis n n n 2n 2n n 2n 2n?
Meiosis produces haploid gametes which have the ' n ' symbol.
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.
Is 2n prime or composite?
2 is a prime number. 2 times anything but 1 is composite.
