<h3>
Answer: Choice B</h3>
c^2-2cd+d^2 = (c-d)^2
We can verify this with the following steps
(c-d)^2 = (c-d)(c-d)
(c-d)^2 = c(c-d) - d(c-d)
(c-d)^2 = c^2-cd-cd+d^2
(c-d)^2 = c^2-2cd+d^2
Answer:
D)The range of f(x) includes values such that y ≥ 1, so the domain of f–1(x) includes values such that x ≥ 1.
Step-by-step explanation:
The missing tables are:
First table
x: 0 1 2
f(x): 1 10 100
Second table
x: 1000 100 10
f^-1(x): 3 2 1
Option A is not correct because f(x) has a y-intercept at (0, 1)
If f(x) has a y-intercept, then f^-1(x) has a x-intercept, which is located at (1, 0). Then option B is not correct
Option C is not correct because the domain of f^-1(x) is associated with x values.
Option D is correct because the domain of f(x) is the range of f^-1(x) and vice versa
