X- (input) Y- (output) oot Is this mapping a function and how can you tell

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

solmaris [256]

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

4 0

3 years ago

(03.01 LC)

vampirchik [111]

Answer:

√7 units, √98 units

Step-by-step explanation:

Pythagorean theorem

a^2+b^2=c^2. So 9+b^2=16. So b^2=7. So b=√7

a^2+b^2=c^2. SO 49+49=c^2. So 98=c^2. So c=√98.

4 0

3 years ago

Read 2 more answers

A particle moves along a line with a velocity v(t)=t2−t−6, measured in meters per second. Find the total distance the particle t

ahrayia [7]

Answer:

10.67 m

Step-by-step explanation:

Velocity is given as;

v(t) = t² − t − 6

To get the distance, we will integrate the velocity equation.

x(t) = ∫v(t) = (t³/3) - (t²/2) - 6t

Thus, from t=0 seconds to t=4 seconds., we have;

(t³/3) - (t²/2) - 6t between 0 and 4.

Thus;

Total Distance = [(4³/3) - (4²/2) - 6(4)] - [(0³/3) - (0²/2) - 6(0)]

Total distance = (64/3) - 8 - 24 = -10.67 m

Distance can't be negative and so we take the absolute value which is 10.67 m

7 0

3 years ago

Select the correct answer.

mr Goodwill [35]

I believe its c because that is the suitable answer

6 0

2 years ago

Solve the equation : -3 • ( 2 - x ) + 4 = 2 • ( 1 - 2x) + 3 thanks :)

kakasveta [241]

Isolate the variable by dividing each side by factors that don't contain the variable.

x = 1

-6+3x+4=2-4x+3

-8+7x+1=0

7x=7

x=1

4 0

3 years ago