The domain of the function is given by (–∞, –3) ∪ (–3, 2) ∪ (2, ∞)
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The function is given by:
![g(x) = \frac{x^2 + 3x}{x^2 + x - 6}, x < 3](https://tex.z-dn.net/?f=g%28x%29%20%3D%20%5Cfrac%7Bx%5E2%20%2B%203x%7D%7Bx%5E2%20%2B%20x%20-%206%7D%2C%20x%20%3C%203)
![g(x) = \log_{2}{x + 5}, x \geq 3](https://tex.z-dn.net/?f=g%28x%29%20%3D%20%5Clog_%7B2%7D%7Bx%20%2B%205%7D%2C%20x%20%5Cgeq%203)
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- For the second definition, the restriction is that x + 5 has to be positive. Since the definition is only valid for
, this will <u>always be true.</u> - In the second definition, the denominator cannot be zero, thus the <u>zeros of the denominator will be outside the domain.</u>
These zeros are found using Bhaskara, we have the quadratic function
, thus the coefficients are
.
![\Delta = b^2 - 4ac = (1)^2 - 4(1)(-6) = 25](https://tex.z-dn.net/?f=%5CDelta%20%3D%20b%5E2%20-%204ac%20%3D%20%281%29%5E2%20-%204%281%29%28-6%29%20%3D%2025)
![x_1 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{-1 + 5}{2} = 2](https://tex.z-dn.net/?f=x_1%20%3D%20%5Cfrac%7B-b%20%2B%20%5Csqrt%7B%5CDelta%7D%7D%7B2a%7D%20%3D%20%5Cfrac%7B-1%20%2B%205%7D%7B2%7D%20%3D%202)
![x_2 = \frac{-b - \sqrt{\Delta}}{2a} = \frac{-1 - 5}{2} = -3](https://tex.z-dn.net/?f=x_2%20%3D%20%5Cfrac%7B-b%20-%20%5Csqrt%7B%5CDelta%7D%7D%7B2a%7D%20%3D%20%5Cfrac%7B-1%20-%205%7D%7B2%7D%20%3D%20-3)
Thus, <u>x = 2 and x = -3 are outside the</u> domain, which is given by:
(–∞, –3) ∪ (–3, 2) ∪ (2, ∞)
A similar problem is given at brainly.com/question/13136492
This question mean no sense lol
Answer: Part A:
We observe that we have the following system of equations:
y = 4x
y = 2x-2
The solution to the system of equations is an ordered pair (x, y), that is, a point in common that both functions have.
Part B:
x y = 4x y = 2x-2
-3 -12 -8
-2 -8 -6
-1 -4 -4
0 0 -2
1 4 0
2 8 2
3 12 4
The solution is the ordered pair:
(x, y) = (-1, -4)
Part C:
See attached image
Step-by-step explanation:
Corresponding Angles Theorem:
2y = 58
Divide both side by 2
y = 29 degrees