No. 3n isn't a factor of 3n + 7.
The GCF of 3n + 7 and 9n is 1.
No. 3n is not a factor of 3n + 7.
No. 3n is not a factor of 3n + 7.
The GCF is 3n3
6n² - 18n - 18 = 6, so 6n² - 18n - 24 = 0.Dividing by 6:n² - 3n - 4 = 0.What multiplies to -4 and adds to -3? The only possible factors are:(-1)4(-2)2(-4)1...with sums of 3, 0, and -3, respectively. We want that last one, son² - 3n - 4 = (n - 4)(n + 1).
Yes, here's the proof. Let's start out with the basic inequality 81 < 83 < 100. Now, we'll take the square root of this inequality: 9 < √83 < 10. If you subtract all numbers by 9, you get: 0 < √83 - 9 < 1. If √83 is rational, then it can be expressed as a fraction of two integers, m/n. This next part is the only remotely tricky part of this proof, so pay attention. We're going to assume that m/n is in its most reduced form; i.e., that the value for n is the smallest it can be and still be able to represent √83. Therefore, √83n must be an integer, and n must be the smallest multiple of √83 to make this true. If you don't understand this part, read it again, because this is the heart of the proof. Now, we're going to multiply √83n by (√83 - 9). This gives 83n - 9√83n. Well, 83n is an integer, and, as we explained above, √83n is also an integer, so 9√83n is an integer too; therefore, 83n - 9√83n is an integer as well. We're going to rearrange this expression to (√83n - 9n)√83 and then set the term (√83n - 9n) equal to p, for simplicity. This gives us the expression √83p, which is equal to 83n - 9√83n, and is an integer. Remember, from above, that 0 < √83 - 9 < 1. If we multiply this inequality by n, we get 0 < √83n - 9n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus √83p < √83n. We've already determined that both √83p and √83n are integers, but recall that we said n was the smallest multiple of √83 to yield an integer value. Thus, √83p < √83n is a contradiction; therefore √83 can't be rational and so must be irrational. Q.E.D.
No. 3n is not a factor of 3n + 7.
No. 3n is not a factor of 3n + 7.
2+3n+2+9n would equal 4+12n after adding all like numbers.
9n + 12y = 3(3n + 4y)
(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
3n x 3n = 9n if 9n = 36 then n = 36/9 = 4
s = 9n - 18.
3(3n - 1)(n + 1)
9n+6-3n
-6N - 7
The expression (3n^2 + 9n + 6) can be factored into simpler polynomial expressions, indicating that it has more than one factor. A polynomial of degree 2, like this one, generally has two factors when expressed in its factored form. Additionally, since it can be factored using methods such as grouping or the quadratic formula, it reveals multiple combinations of factors. Thus, it showcases the polynomial's ability to be expressed in various multiplicative forms.
-5 plus 9n plus 6 is 1 plus 9n.