The circle is open i think is the right answer i hope i got it right if not im very sorry :(
The y-intercept is -6, and the x-intercept is 18
Step-by-step explanation:
We will prove by contradiction. Assume that
is an odd prime but n is not a power of 2. Then, there exists an odd prime number p such that
. Then, for some integer
,
![n=p\times k.](https://tex.z-dn.net/?f=n%3Dp%5Ctimes%20k.)
Therefore
![2^n + 1=2^{p\times k} + 1=(2^{k})^p + 1^p.](https://tex.z-dn.net/?f=2%5En%20%2B%201%3D2%5E%7Bp%5Ctimes%20k%7D%20%2B%201%3D%282%5E%7Bk%7D%29%5Ep%20%2B%201%5Ep.)
Here we will use the formula for the sum of odd powers, which states that, for
and an odd positive number
,
![a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2-...+b^{n-1})](https://tex.z-dn.net/?f=a%5En%2Bb%5En%3D%28a%2Bb%29%28a%5E%7Bn-1%7D-a%5E%7Bn-2%7Db%2Ba%5E%7Bn-3%7Db%5E2-...%2Bb%5E%7Bn-1%7D%29)
Applying this formula in 1) we obtain that
.
Then, as
we have that
is not a prime number, which is a contradiction.
In conclusion, if
is an odd prime, then n must be a power of 2.
Answer:
the answer is B
Step-by-step explanation:
i just took the test, hope this can help?