Use a recursive function for the geometric sequence 2, -6, 18, -54,... to represent the 9th term.

Mathematics

2 answers:

WARRIOR [948]3 years ago

5 0

Answer:

The Answer is A. f(9) = f(8) • (−3)

Hope This Helps!

Jlenok [28]3 years ago

4 0

Answer:

f(9) = f(8)·(-3)

Step-by-step explanation:

A recursive function is one where each successive term is calculated using the previous term;

The clear pattern demonstrated in the sequence of terms shown is that each term is multiplied by -3 to give the next one;

Hence, the correct option is the first.

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Suppose integral [4th root(1/cos^2x - 1)]/sin(2x) dx = A<br>What is the value of the A^2?<br><br>

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$\large \mathbb{PROBLEM:}$

$\begin{array}{l} \textsf{Suppose }\displaystyle \sf \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx = A \\ \\ \textsf{What is the value of }\sf A^2? \end{array}$

$\large \mathbb{SOLUTION:}$

$\!\!\small \begin{array}{l} \displaystyle \sf A = \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx \\ \\ \textsf{Simplifying} \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\sec^2 x - 1}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\tan^2 x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\cdot \dfrac{\sqrt{\tan x}}{\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\tan x}{\sin 2x\ \sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{\sin x}{\cos x}}{2\sin x \cos x \sqrt{\tan x}}\ dx\:\:\because {\scriptsize \begin{cases}\:\sf \tan x = \frac{\sin x}{\cos x} \\ \: \sf \sin 2x = 2\sin x \cos x \end{cases}} \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{1}{\cos^2 x}}{2\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sec^2 x}{2\sqrt{\tan x}}\ dx, \quad\begin{aligned}\sf let\ u &=\sf \tan x \\ \sf du &=\sf \sec^2 x\ dx \end{aligned} \\ \\ \textsf{The integral becomes} \\ \\ \displaystyle \sf A = \dfrac{1}{2}\int \dfrac{du}{\sqrt{u}} \\ \\ \sf A= \dfrac{1}{2}\cdot \dfrac{u^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C = \sqrt{u} + C \\ \\ \sf A = \sqrt{\tan x} + C\ or\ \sqrt{|\tan x|} + C\textsf{ for restricted} \\ \qquad\qquad\qquad\qquad\qquad\qquad\quad \textsf{values of x} \\ \\ \therefore \boxed{\sf A^2 = (\sqrt{|\tan x|} + c)^2} \end{array}$