
the denominator cannot be zero, because the division by zero is not defined, therefore:
![\begin{gathered} x^2-9=0 \\ \text{Solving for x:} \\ x^2=9 \\ \sqrt[]{x^2}=\sqrt[]{9} \\ x=\pm3 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%5E2-9%3D0%20%5C%5C%20%5Ctext%7BSolving%20for%20x%3A%7D%20%5C%5C%20x%5E2%3D9%20%5C%5C%20%5Csqrt%5B%5D%7Bx%5E2%7D%3D%5Csqrt%5B%5D%7B9%7D%20%5C%5C%20x%3D%5Cpm3%20%5Cend%7Bgathered%7D)
Therefore the domain of (f o g)(x) is:
Answer:
B
Step-by-step explanation:
To evaluate f(2) substitute x = 2 into f(x), that is
f(2) = - (2)³ + 2(2)² - 3 = - 8 +2(4) - 3 = - 8 + 8 - 3 = - 3
Answer:
5
Step-by-step explanation:
hope this helps
374.072
Step-by-step explanation:
Use a calculator
Answer:
20 icnhes
Step-by-step explanation:
not rlly