Only hole of function
is at <em>x=(-4)</em>
Step-by-step explanation:
Given the function is ![f(x) = \frac{x^{2}+7x+10 }{x^{2}+9x+20 }](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7Bx%5E%7B2%7D%2B7x%2B10%20%7D%7Bx%5E%7B2%7D%2B9x%2B20%20%7D)
In order to find holes of any function, you should find when function is becoming undefined or say " infinity"
Given function is polynomial function.
It will become undefined become denominator become zero
![x^{2}+9x+20=0](https://tex.z-dn.net/?f=x%5E%7B2%7D%2B9x%2B20%3D0)
Solving for x value when denominator become zero
![x^{2}+9x+20=0\\x^{2}+5x+4x+20=0\\x(x+5)+4(x+5)=0\\(x+4)(x+5)=0](https://tex.z-dn.net/?f=x%5E%7B2%7D%2B9x%2B20%3D0%5C%5Cx%5E%7B2%7D%2B5x%2B4x%2B20%3D0%5C%5Cx%28x%2B5%29%2B4%28x%2B5%29%3D0%5C%5C%28x%2B4%29%28x%2B5%29%3D0)
we get possible holes at x=(-4) and x=(-5)
Check whether you can eliminate any holes
Now, Solving for x value when numerator become zero
![x^{2}+7x+10=0\\x^{2}+5x+2x+10=0\\(x+5)(x+2)=0](https://tex.z-dn.net/?f=x%5E%7B2%7D%2B7x%2B10%3D0%5C%5Cx%5E%7B2%7D%2B5x%2B2x%2B10%3D0%5C%5C%28x%2B5%29%28x%2B2%29%3D0)
x=(-5) and x=(-2)
x=(-5) is common is both numerator and denominator.
So that, we can eliminate it.
![f(x) = \frac{(x+5)(x+2)}{(x+5)(x+4)}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B%28x%2B5%29%28x%2B2%29%7D%7B%28x%2B5%29%28x%2B4%29%7D)
![f(x) = \frac{(x+2)}{(x+4)}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B%28x%2B2%29%7D%7B%28x%2B4%29%7D)
Therefore, Only hole of function
is at x=(-4)
Answer:
The answer should be the third one y= x/2 +6
Step-by-step explanation:
Brainliest pls UnU
H3000 -you 67778 it’s equals to ninth five yes
5.6mile
7 x
-- = --
80 64
7×64=80x
448=80x
x=5.6
The answer is g=37. This is correct