Answer:
Yes, the sequence is geometric
common ratio (r) = -3
Step-by-step explanation:
Given:
If a sequence is geometric, common ratio
:
![r=\dfrac{a_4}{a_3}=\dfrac{a_3}{a_2}=\dfrac{a_2}{a_1}](https://tex.z-dn.net/?f=r%3D%5Cdfrac%7Ba_4%7D%7Ba_3%7D%3D%5Cdfrac%7Ba_3%7D%7Ba_2%7D%3D%5Cdfrac%7Ba_2%7D%7Ba_1%7D)
Substituting the given values:
![\implies \dfrac{a_4}{a_3}=\dfrac{-27}{9}=-3](https://tex.z-dn.net/?f=%5Cimplies%20%5Cdfrac%7Ba_4%7D%7Ba_3%7D%3D%5Cdfrac%7B-27%7D%7B9%7D%3D-3)
![\implies \dfrac{a_3}{a_2}=\dfrac{9}{-3}=-3](https://tex.z-dn.net/?f=%5Cimplies%20%5Cdfrac%7Ba_3%7D%7Ba_2%7D%3D%5Cdfrac%7B9%7D%7B-3%7D%3D-3)
![\implies \dfrac{a_2}{a_1}=\dfrac{-3}{1}=-3](https://tex.z-dn.net/?f=%5Cimplies%20%5Cdfrac%7Ba_2%7D%7Ba_1%7D%3D%5Cdfrac%7B-3%7D%7B1%7D%3D-3)
Therefore, as
the sequence is geometric with common ratio of -3
Answer:
125y^5
Step-by-step explanation:
take the denominator to the numerator, then the exponents become positive. Then it's 5y^3 *(5y)^2 which is 125y^5
The answer is (B) polynomial of 4 terms.
Answer:
A
Step-by-step explanation:
mid point of AC=((2m+2p)/2,(2n+2r)/2)=(m+p,n+r)