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3 years ago

Can a graph be linear with a outlier

Mathematics

1 answer:

Makovka662 [10]3 years ago

7 0

Answer:

yes it can be.

Step-by-step explanation:

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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

2 years ago

I would like step by step explanation

Nezavi [6.7K]

Answer:

the number is 87

Step-by-step explanation:

the photo contains step by step

3 0

3 years ago

Read 2 more answers

What is the answer to the equation 6x-x/x+1

avanturin [10]

Answer:

5x/x+1

Step-by-step explanation:

Numerator: 6x minus an x equals 5x.

Denominator: Since x and 1 are not like terms, they can't be combined. You just leave them as they were.

Hope this helps!

6 0

3 years ago

$10,000 at an annual rate of 7%, compounded semi-annually, for 2 years

forsale [732]

Answer:

$\$13,107.96$

Step-by-step explanation:

Since interest is compounded semi-annually (half a year or 6 months), in a spawn of 2 years, the interest will have been compounded 4 times. As given in the problem, each time the interest is compounded, the new balance will be 107% or 1.07 times the amount of the old balance.

Therefore, we can set up the following equation to find the new balance after 2 years:

$\text{New balanace}=10,000\text{ (old balance)}\cdot 1.07\cdot 1.07\cdot 1.07\cdot 1.07,\\\text{New balanace}=10,000\cdot 1.07^4=\boxed{\$13,107.96}$

8 0

3 years ago

Ricky withdrew a total of $175 from his bank account over 5 days. He withdrew the same amount each day. By how much did the amou

kifflom [539]

Answer:

$35

Step-by-step explanation:

175/5 = 35

have a nice day! (maybe get this double checked in case I did it wrong :D)

8 0

2 years ago