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Nina [5.8K]
2 years ago
10

MARKING BRAILIEST!!!! Select the statement that describes this expression: 13 + fraction 1 over 2 x (6 ÷ 2 + 6)

Mathematics
2 answers:
Sindrei [870]2 years ago
7 0

Answer:

a

Step-by-step explanation:

Sauron [17]2 years ago
3 0

Answer: D

You get the same answer from doing both!

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Answer:

\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

In other words, the x in f(x) = 3\, \tan(23\, x) could be any real number as long as x \ne \displaystyle \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) for all integer k (including negative integers.)

Step-by-step explanation:

The tangent function y = \tan(x) has a real value for real inputs x as long as the input x \ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

Hence, the domain of the original tangent function is \mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

On the other hand, in the function f(x) = 3\, \tan(23\, x), the input to the tangent function is replaced with (23\, x).

The transformed tangent function \tan(23\, x) would have a real value as long as its input (23\, x) ensures that 23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

In other words, \tan(23\, x) would have a real value as long as x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right).

Accordingly, the domain of f(x) = 3\, \tan(23\, x) would be \mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

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