Any number in the form of a+-bi, where a and b are real numbers and b is not equal to 0 is considered a pure imaginary number
2 answers:
Answer:
<h2>False.</h2>Step-by-step explanation:
<em>A pure imaginary number is a complex number that doesn't have a real part.</em>
So, if a is a real number, and it doesn't specify that a is only equal to zero, then the expression a+bi is not a pure imaginary number, it's only a complex number. Examples of imaginary numbers are where
In this case, , that is, <em>a </em>can be
<em>Therefore the statement is </em><em>false</em><em>, because a can take any real value.</em>
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Answer:
<em>See attached</em>
<u>Numbers are either negative or greater tan 5:</u>
- x < 0 and x > 5
- or
- x = (-∞, 0) ∪ (5, +∞)
The domain of the composite function is given as follows:
[–3, 6) ∪ (6, ∞)
<h3>What is the composite function of f(x) and g(x)?</h3>The composite function of f(x) and g(x) is given as follows:
In this problem, the functions are:
.
The composite function is of the given functions f(x) and g(x) is:
The square root has to be non-negative, hence the restriction relative to the square root is found as follows:
The denominator cannot be zero, hence the restriction relative to the denominator is found as follows:
Hence, from the restrictions above, of functions f(x), g(x) and the composite function, the domain is:
[–3, 6) ∪ (6, ∞)
More can be learned about composite functions at brainly.com/question/13502804
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