I don't know thats what I'm asking you!
you have 9n+17=-1 -17 -17 9n=-18 /9 /9 n= -2
Strangely enough, it is 9n + 1 for n = 1, 2, 3, ...
0.1 (recurring) = 1/9 Solve this type of problem as follows :- Let n = 0.1 (recurring) 10n = 1.1(recurring) 9n = 1 n = 1/9
37 - 9n
3(3n - 1)(n + 1)
9 more than the product of 1 and n
-5 plus 9n plus 6 is 1 plus 9n.
you have 9n+17=-1 -17 -17 9n=-18 /9 /9 n= -2
Strangely enough, it is 9n + 1 for n = 1, 2, 3, ...
If you mean: 9n = 3 then the value of n is 1/3 which is the solution to the equation
0.1 (recurring) = 1/9 Solve this type of problem as follows :- Let n = 0.1 (recurring) 10n = 1.1(recurring) 9n = 1 n = 1/9
No. 3n isn't a factor of 3n + 7. The GCF of 3n + 7 and 9n is 1.
(3n)(3n-1) = 3n * 3n - 3n * 1 Now, perform the multiplication: (3n * 3n) = 9n^2 (3n * 1) = 3n So, (3n)(3n-1) simplifies to: 9n^2 - 3n
37 - 9n
n = 1 7n + 6n - 5 = 4n + 4 13n - 5 = 4n + 4 13n - 4n - 5 = 4n - 4n + 4 9n - 5 = 4 9n -5 + 5 = 4 + 5 9n = 9 n = 1
3(3n - 1)(n + 1)
Yes you can, and you will write it as '1 million'.