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)
Factors aremultiplicativeterms (i.e., they multiply into all the terms). The easiest way to see this is by example.Example 1:a(b+cd)a is a factor and (b+cd) is a factorExample 2:x2 - 9 = (x-3)(x+3)(x-3) is a factor and (x+3) is a factorHopefully you are beginning to notice you can always wrap parenthesis around a factor without altering the order of operations.Example 3:2a(b)(9n-x)+8This is a trick one, but you can factor out a 2:2a(b)(9n-x)+8 = (1)(2)a(b)(9n-x)+(4)(2) = (2)(ab(9n-x)+4)The factors are: 2 and (ab(9n-x)+4)Example 4:If that +8 wasn't on example 3, what would the factors be?2a(b)(9n-x)They would be:2, a, b, and (9n-x)