The value x = - 2 is not part of the domain of the function f(x) = √(x + 1).
<h3>Is a given point part of the domain of a radical function?</h3>
Radical functions are functions that involves root operators. We see here a radical function of the form f(x) = √(x + a), whose domain is x ∈ [- a, + ∞), that is, x ≥ - a. If we know that f(x) = √(x + 1), then a = 1 and the domain of the function is x ∈ [- 1, + ∞), that is, x ≥ - 1.
Hence, the value x = - 2 is not part of the domain of the function f(x) = √(x + 1).
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Answer:
A=592
Step-by-step explanation:
Answer:
y=4x+5
Step-by-step explanation:
You have your mx already: (mx is slope)
y=4x
And you have your b already: (b is y-intercept)
+5
Combine equations: (Solve)
y=4x+5
Hence, the correct answer is y=4x+5
Answer:
B. $58,931
Step-by-step explanation:
Hope it helps :)
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