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3 years ago

Translation 5 units right and 1 unit up

Mathematics

1 answer:

dsp733 years ago

8 0

Answer:

hope this helps

Step-by-step explanation:

just have to move the points 5 times to the right but don't stop there!you gotta move 1 unit up then plot your final points

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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

Help me

dybincka [34]

Answer:

First you have to multiply negative four in parentheses, then add minus two

-4×(-8+3x)=32-12x

2+32-12x=34-12x

34-12x=82

-12x=48

x=-4

6 0

2 years ago

Is 2.0 a solution to the equation?<br><br> X-2y=0 , 2x-3y=1

Travka [436]

No, because (2,0) is a coordinate. x=2 and y=0. So just plug in the numbers where there's x or y with the appropriate number, (2 or 0). So in the first equation, x-2y=0, when you pug in the numbers, 2-2(0)=0, you know it's wrong because 2-0=0 isn't correct. So no. the point (2,0) is not a solution to the first equation. Now plug in the numbers for the second coordinate. You get 2(2)-3(0)=1. So 4-0=1. This is once again false no no. (2,0) satisfies neither equations.

8 0

3 years ago

What is the approximate area of a circle with a radius of 11 m?

Katarina [22]

Answer:

380.13

Step-by-step explanation:

area= (pi)x(r^2)

pi times r squared

8 0

3 years ago

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