2n x 2n x 2n = 8n^3
5-2n=3 -2n=3-5 -2n=-2 n=1
It is a statement that is equivalent to the equation:(2n+1)*(2n+3) = (2n+1)+(2n+3) + 23.
2n - 3 = 5Add 3 to each side:2n = 8Divide each side by 2:n = 4
2n/3 = 4 2n = 3*4 2n = 12 n = 6 :)
2n x 2n x 2n = 8n^3
5-2n=3 -2n=3-5 -2n=-2 n=1
Solve for n. 2n+3=0 2n=-3 n=-3/2
2n + 3 = 352n = 35 -32n = 32n = 32/2n = 16
3=2n
Answer is 3, 5, 7, 9detailsassume the numbers are2n-3, 2n-1, 2n+1, 2n+3 ......................... (1)(2n-3) 2 + (2n-1) 2 + (2n+1) 2 + (2n+3) 2 = 16416n 2 + 20 = 16416n 2 = 144n2 = 9n = 3Substitute in eq 1 we get the answer above
-2n +7 when n=3?
No. of subsets = 2n - 1 3 = 2n - 1 3 + 1 = 2n - 1 + 1 4 = 2n 4/2 = 2n/2 2/1 = 1n/1 2 = n n = 2elements
It is a statement that is equivalent to the equation:(2n+1)*(2n+3) = (2n+1)+(2n+3) + 23.
2n - 3 = 5Add 3 to each side:2n = 8Divide each side by 2:n = 4
6n -3=2n+9 6n-2n=9+3 4n=12 n=12/4 n=3
Let the first odd number be 2n-5 (n ≥ 3), then the 6 consecutive odd numbers are: 2n-5, 2n-3, 2n-1, 2n+1, 2n+3, 2n+5 And their sum is: 2n-5 + 2n-3 + 2n-1 + 2n+1 + 2n+3 + 2n+5 = 12n The greatest common factor for all n(≥ 3) of 12n is 12. Thus 12 is the greatest whole number that MUST be a factor of the sum of any six consecutive positive odd numbers.