What is the 6th term of the geometric sequence f(1) = 12, f(n) = -2/3f(n-1)

1 answer:

5 0

$\bf \begin{array}{ccll}
\stackrel{n^{th}}{term}&value\\\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\\\
1&12\\\\
2&-\frac{2}{3}f(2-1)\\\\
&-\frac{2}{3}f(1)\\\\
&-\frac{2}{3}(12)\\\\
&-8\\\\
3&-\frac{2}{3}f(3-1)\\\\
&-\frac{2}{3}f(2)\\\\
&-\frac{2}{3}(-8)\\\\
&\frac{16}{3}
\end{array}$

$\bf \begin{array}{cccl}
\qquad &\qquad \\
4&-\frac{2}{3}f(4-1)\\\\
&-\frac{2}{3}f(3)\\\\
&-\frac{2}{3}\left( \frac{16}{3} \right)\\\\
&-\frac{32}{9}\\\\
5&-\frac{2}{3}f(5-1)\\\\
&-\frac{2}{3}f(4)\\\\
&-\frac{2}{3}\left( -\frac{32}{9} \right)\\\\
&\frac{64}{27}\\\\
6&-\frac{2}{3}f(6-1)\\\\
&-\frac{2}{3}f(5)\\\\
&-\frac{2}{3}\left( \frac{64}{27}\right)
\end{array}$

You might be interested in

Evaluate 7C2. Explain how you got your answer.

Leviafan [203]

So, to evaluate a combination, there's a formula we use.

I don't remember the formula from the top of my head, lol, but this is how you solve them.

7 c 2

When doing combinations and permutations each number is always in a factorial. We always start with the number on the left.

7! That's the total amount. The number on the left divides into that.

7! / 2!

We're not done yet. Here's the tricky part. We also always divide the number on the left, in this case 7!, with the positive difference of both numbers given to us.

7 - 2 = 5

So, we have 7! / 5! / 2! = 21.

Hope that helped!

Let's work another one.
5 c 3

We have 5! / 3! ,but we need to also divide 5! by the positive difference of 5 and 3. We get 2.

So, 5! / 3! / 2! = 10.

If you have any questions then leave a comment. Good luck!

5 0

4 years ago

HELP!!

Liono4ka [1.6K]

Answer:

Part 1: The polygon ABCDE reflected across y-axis to get the polygon MNOPQ. So, the polygon ABCDE congruent to polygon MNOPQ.

Part 2: The vertices of polygon VWXYZ are V(1, 3), W(-1, 7), X(-7, 7), Y(-5, 3), and Z(-3, 1).

Step-by-step explanation:

Part 1:

The vertices of the polygon ABCDE are A(2, 8), B(4, 12), C(10, 12), D(8, 8), and E(6, 6).

The vertices of the polygon MNOPQ are M(-2, 8), N(-4, 12), O(-10, 12), P(-8, 8), and Q(-6, 6).

We need to find the transformation or sequence of transformations that can be performed on polygon ABCDE to show that it is congruent to polygon MNOPQ.

The relation between the vertices of ABCDE and MNOPQ are defined as

$(x,y)\rightarrow (-x,y)$