Answer:
T(3) = 13
Step-by-step explanation:
If we are trying to find the 3rd term of this <em>specific </em>sequence, then we simply plug in 3 as n.
T(3) = (3)² + 4
T(3) = 9 + 4
T(3) = 13
However, this isn't proper notation for an arithmetic or geometric sequence.
Answer:
-12
Step-by-step explanation:
In d ratio 5:9 which equals 5+9=14
the total female performers is 63
For men = 5/14*63=5/2*9
=45/2 =22.5
The men would be about 22.5 in the dance recital
Answer:
g(x) = (5x + 3)/ x - 7
Step-by-step explanation:
Let g(x) = y
y = (7x + 3)/x - 5
Make x the subject
xy - 5y = 7x + 3
xy - 7x = 5y + 3
x(y - 7) = 5y + 3
x = (5y + 3)/ y - 7
Therefore, the inverse of the function = (5x + 3)/ x - 7
Answer:
(a) 0
(b) f(x) = g(x)
(c) See below.
Step-by-step explanation:
Given rational function:
![f(x)=\dfrac{x^2+2x+1}{x^2-1}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cdfrac%7Bx%5E2%2B2x%2B1%7D%7Bx%5E2-1%7D)
<u>Part (a)</u>
Factor the <u>numerator</u> and <u>denominator</u> of the given rational function:
![\begin{aligned} \implies f(x) & = \dfrac{x^2+2x+1}{x^2-1} \\\\& = \dfrac{(x+1)^2}{(x+1)(x-1)}\\\\& = \dfrac{x+1}{x-1}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20%5Cimplies%20f%28x%29%20%26%20%3D%20%5Cdfrac%7Bx%5E2%2B2x%2B1%7D%7Bx%5E2-1%7D%20%5C%5C%5C%5C%26%20%3D%20%5Cdfrac%7B%28x%2B1%29%5E2%7D%7B%28x%2B1%29%28x-1%29%7D%5C%5C%5C%5C%26%20%3D%20%5Cdfrac%7Bx%2B1%7D%7Bx-1%7D%5Cend%7Baligned%7D)
Substitute x = -1 to find the limit:
![\displaystyle \lim_{x \to -1}f(x)=\dfrac{-1+1}{-1-1}=\dfrac{0}{-2}=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%20-1%7Df%28x%29%3D%5Cdfrac%7B-1%2B1%7D%7B-1-1%7D%3D%5Cdfrac%7B0%7D%7B-2%7D%3D0)
Therefore:
![\displaystyle \lim_{x \to -1}f(x)=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%20-1%7Df%28x%29%3D0)
<u>Part (b)</u>
From part (a), we can see that the simplified function f(x) is the same as the given function g(x). Therefore, f(x) = g(x).
<u>Part (c)</u>
As x = 1 is approached from the right side of 1, the numerator of the function is positive and approaches 2 whilst the denominator of the function is positive and gets smaller and smaller (approaching zero). Therefore, the quotient approaches infinity.
![\displaystyle \lim_{x \to 1^+} f(x)=\dfrac{\to 2^+}{\to 0^+}=\infty](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%201%5E%2B%7D%20f%28x%29%3D%5Cdfrac%7B%5Cto%202%5E%2B%7D%7B%5Cto%200%5E%2B%7D%3D%5Cinfty)