I don't know how to work out the equation below

1 answer:

8 0

3(x + 4) = 27

First, since our variable is already on one side by itself, we can start out by dividing each side by 3.
$x + 4 = \frac{27}{3}$

Second, we need to simplify our fraction we just got. To do so, think of it as this way: what times 3 equals 27? Or, (3 × x = 27). Either way, the answer is 9.
$x + 4 = 9$

Third, we can remove (subtract) 4 from each side of the problem. This will give us the answer after we subtract it.
$x = 9 - 4$

Fourth, our last step is to subtract 9 - 4 to get our answer. (9 - 4 = 5), meaning 5 is our answer.

Answer: $\fbox {x = 5}$

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Find the domain of the function y = 3 tan(23x)

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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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Mekhanik [1.2K]

Answer:

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