Hence the required function of h i.e f(h) is expressed as
![f(x) = \frac{1}{x^2-3}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B1%7D%7Bx%5E2-3%7D)
Given the expression
![f(x) = \frac{1}{x^2-3}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B1%7D%7Bx%5E2-3%7D)
To get f(h), we will substitute f in place of x as shown;
![f(h) = \frac{1}{h^2-3}](https://tex.z-dn.net/?f=f%28h%29%20%3D%20%5Cfrac%7B1%7D%7Bh%5E2-3%7D)
Hence the required function of h i.e f(h) is expressed as
![f(x) = \frac{1}{x^2-3}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B1%7D%7Bx%5E2-3%7D)
Learn more: brainly.com/question/4357143
Answer:
![4x^2-10x+11+\frac{6}{2x+5}](https://tex.z-dn.net/?f=4x%5E2-10x%2B11%2B%5Cfrac%7B6%7D%7B2x%2B5%7D)
Step-by-step explanation:
The given functions are:
and ![g(x)=2x+5](https://tex.z-dn.net/?f=g%28x%29%3D2x%2B5)
By algebraic properties of functions;
![\frac{f(x)}{g(x)}=\frac{8x^3-28x+61}{2x+5}](https://tex.z-dn.net/?f=%5Cfrac%7Bf%28x%29%7D%7Bg%28x%29%7D%3D%5Cfrac%7B8x%5E3-28x%2B61%7D%7B2x%2B5%7D)
We perform the long division of the two polynomials as shown in the attachment.
Therefore:
![\frac{8x^3-28x+61}{2x+5}=4x^2-10x+11+\frac{6}{2x+5}](https://tex.z-dn.net/?f=%5Cfrac%7B8x%5E3-28x%2B61%7D%7B2x%2B5%7D%3D4x%5E2-10x%2B11%2B%5Cfrac%7B6%7D%7B2x%2B5%7D)
The last choice is correct
Answer:B
Step-by-step explanation:
Its b,i swear on my doggie
Answer:
So, I'll give it my best shot. Here it is factored:
Factor the polynomial.
−
(
64
c
^9 + 72
a
b
c − 27
a^
3 + 863
)
It's the same thing simplified, just without the parentheses.
Step-by-step explanation:
Hope this helps!
have a nice day
So, they are the same size because of the preminitor around. ^_^ Sorry if I'm wrong.