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No. 3n isn't a factor of 3n + 7.

The GCF of 3n + 7 and 9n is 1.

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Q: The gcf of the expressions 3n plus 7 and 9n is 3n?
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Is 3n the gcf of 3n plus 7 and 9n?

No. 3n is not a factor of 3n + 7.


Is 3n the gcf of 3n 7 and 9n?

No. 3n is not a factor of 3n + 7.


What is the GCF of 6n to the third power and 9n to the third power?

The GCF is 3n3


How do you factor 2n2 9n-81?

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).


Is the square root of 14 an irrational number?

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.