Answer:
x = -5, x = -6
Step-by-step explanation:
After canceling common terms from numerator and denominator, there are two factors remaining in the denominator that can become zero. The vertical asymptotes are at those values of x.
![\displaystyle F(x)=\frac{x\frac{2x}{2}}{x(x+5)(x+6)}=\frac{x}{(x+5)(x+6)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F%28x%29%3D%5Cfrac%7Bx%5Cfrac%7B2x%7D%7B2%7D%7D%7Bx%28x%2B5%29%28x%2B6%29%7D%3D%5Cfrac%7Bx%7D%7B%28x%2B5%29%28x%2B6%29%7D)
The denominator will be zero when ...
x + 5 = 0 . . . . at x = -5
x + 6 = 0 . . . . at x = -6
Answer:
<h3>
6-16y</h3>
Step-by-step explanation:
Hello!
apply distributive law
![a(b - c) = ab - ac](https://tex.z-dn.net/?f=a%28b%20-%20c%29%20%3D%20ab%20-%20ac)
2(3-8y)=2.3-2.8y
=6-16y
Hope it helps!
Answer:
Step-by-step explanation:
the y intercept is the y value of ur points when ur line crosses the y axis. It is the initial value and is the output value when the input of a linear function is 0.
Consider the equation:
![\sqrt{x+2}+4 = 8](https://tex.z-dn.net/?f=%5Csqrt%7Bx%2B2%7D%2B4%20%3D%208)
Subtracting '4' from both the sides of the equation, we get as
![\sqrt{x+2}+4-4= 8-4](https://tex.z-dn.net/?f=%5Csqrt%7Bx%2B2%7D%2B4-4%3D%208-4)
![\sqrt{x+2}= 4](https://tex.z-dn.net/?f=%5Csqrt%7Bx%2B2%7D%3D%204)
Squaring on both the sides of the equation, we get
![(\sqrt{x+2})^2 = (4)^2](https://tex.z-dn.net/?f=%28%5Csqrt%7Bx%2B2%7D%29%5E2%20%3D%20%284%29%5E2)
![x+2 = 16](https://tex.z-dn.net/?f=x%2B2%20%3D%2016)
Subtracting '2' from both the sides of the equation, we get
![x+2-2=16-2](https://tex.z-dn.net/?f=x%2B2-2%3D16-2)
x=14
Since, An extraneous solution is a solution that arises from the solving process that is not really a solution at all. But, in this equation x=14 is the solution of the given equation.
Hence, it is not an extraneous solution.