We just have to write one function inside the other:

Since we have a fraction with x in the denominator, we have to remove from the domain the value of that makes null the denominator. Thus,

Furthermore, we also have to remove the value of x that annuls the denominator of g(x), because that's the first function that we use. So:

Hence, the domain is:
![D((f\circ g)(x)) = \left]-\infty,-3\right[\cup \left]-3,\dfrac{22}{3}\right[\cup\left]\dfrac{22}{3},\infty\right[](https://tex.z-dn.net/?f=D%28%28f%5Ccirc%20g%29%28x%29%29%20%3D%20%5Cleft%5D-%5Cinfty%2C-3%5Cright%5B%5Ccup%20%5Cleft%5D-3%2C%5Cdfrac%7B22%7D%7B3%7D%5Cright%5B%5Ccup%5Cleft%5D%5Cdfrac%7B22%7D%7B3%7D%2C%5Cinfty%5Cright%5B)
Answer:
0.125
Step-by-step explanation:
Simply use a calculator
Answer:
C
Step-by-step explanation:
given the 2 equations
x + y = 0 → (1)
5x + 3y = - 8 → (2)
Rearrange (1) expressing x in terms of y by subtracting y from both sides
x = - y → (3)
Substitute x = - y into (2)
5(- y) + 3y = - 8, that is
- 5y + 3y = - 8
- 2y = - 8 ( divide both sides by - 2 )
y = 4
Substitute y = 4 into (3) for corresponding value of x
x = - y = - 4
Solution is (- 4, 4 ) → C
Answer:
≈ 8.9
Step-by-step explanation:
Assuming you are requiring the length of the unknown side
let the side ( hypotenuse ) be x
Using Pythagoras' identity
The square on the hypotenuse ie equal to the sum of the squares on the other 2 sides, that is
x² = 4² + 8² = 16 + 64 = 80 ( take square root of both sides )
x =
≈ 8.9 ( to 1 dec. place )