Answer:
![a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}](https://tex.z-dn.net/?f=a_n%20%3D%20128%5Cbigg%28%5Cdfrac%7B1%7D%7B2%7D%5Cbigg%29%5E%7Bn-1%7D)
Step-by-step explanation:
We are given the following in the question:
The numbers of teams remaining in each round follows a geometric sequence.
Let a be the first the of the geometric sequence and r be the common ration.
The
term of geometric sequence is given by:
![a_n = ar^{n-1}](https://tex.z-dn.net/?f=a_n%20%3D%20ar%5E%7Bn-1%7D)
![a_4 = 16 = ar^3\\a_6 = 4 = ar^5](https://tex.z-dn.net/?f=a_4%20%3D%2016%20%3D%20ar%5E3%5C%5Ca_6%20%3D%204%20%3D%20ar%5E5)
Dividing the two equations, we get,
![\dfrac{16}{4} = \dfrac{ar^3}{ar^5}\\\\4}=\dfrac{1}{r^2}\\\\\Rightarrow r^2 = \dfrac{1}{4}\\\Rightarrow r = \dfrac{1}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B16%7D%7B4%7D%20%3D%20%5Cdfrac%7Bar%5E3%7D%7Bar%5E5%7D%5C%5C%5C%5C4%7D%3D%5Cdfrac%7B1%7D%7Br%5E2%7D%5C%5C%5C%5C%5CRightarrow%20r%5E2%20%3D%20%5Cdfrac%7B1%7D%7B4%7D%5C%5C%5CRightarrow%20r%20%3D%20%5Cdfrac%7B1%7D%7B2%7D)
the first term can be calculated as:
![16=a(\dfrac{1}{2})^3\\\\a = 16\times 6\\a = 128](https://tex.z-dn.net/?f=16%3Da%28%5Cdfrac%7B1%7D%7B2%7D%29%5E3%5C%5C%5C%5Ca%20%3D%2016%5Ctimes%206%5C%5Ca%20%3D%20128)
Thus, the required geometric sequence is
![a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}](https://tex.z-dn.net/?f=a_n%20%3D%20128%5Cbigg%28%5Cdfrac%7B1%7D%7B2%7D%5Cbigg%29%5E%7Bn-1%7D)
K=19 which would leave the equation 181=181. The first step to solving this is subtracting 9k from each side.
after you do that it should be k-9=10
now you add 9 to both sides leave k=19. plug k into the equation and get 181=181
Answer:
This is guess of what the question is asking:
I had got 49x - 21y - 6.
Step-by-step explanation:
7 x 7 is 49 and 7 x 3 is 21.
f(x) = (x - 4)^2 - 5
Vertex (4 , -5)
This function opens upward and has min. value = -5
So range y >= - 5
So answer is A. -5 <= f(x) < ∞
Cause the it the oust side has blocks you count the blocks around it and you have your answer