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Lorico [155]
3 years ago
11

2x^5+12x^4+16x^3-12x^2-18x+9=0​

Mathematics
1 answer:
irga5000 [103]3 years ago
4 0

Answer:

24702x + 9

Step-by-step explanation:

First, put everything you need to in parenthisis.

(2x^4) + (12x^4) + (16x^3) - (12x^2) - (18x + 9)

Then Simplifly

32x + 20736x + 4096x - 144x -18x + 9

Now Simplifly everything one by one

20768x + 4096x - 144x - 18x + 9

24864x - 144x - 18x + 9

24720x - 18x +9

24702x + 9

Now you can stop there, or solve for X by dividing each # by 24702

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The domain and range of a function are the possible <em>x and y values </em>of the function.

<em>The domain and the range of the function is: (a) </em>\mathbf{D:\{x \in R|x \ne -2,2\}}<em> and </em>\mathbf{R:\{(-\infty,1) \cup (2,\infty)\}}<em />

The functions are given as:

\mathbf{f(x) = \frac{4}{x^2 - 4}}

\mathbf{g(x) = x + 2}

(g o f)(x) is calculated as:

\mathbf{(g\ o\ f)(x) = g(f(x))}

So, we have:

\mathbf{(g\ o\ f)(x) = \frac{4}{x^2 - 4} + 2}

Take LCM

\mathbf{(g\ o\ f)(x) = \frac{4 + 2x^2 - 8}{x^2 - 4} }

\mathbf{(g\ o\ f)(x) = \frac{2x^2 - 4}{x^2 - 4} }

Represent the denominator as follows, to calculate the domain

\mathbf{x^2 - 4 \ne 0 }

Add 4 to both sides

\mathbf{x^2 \ne 4 }

Take square roots of bot sides

\mathbf{x \ne \±2 }

Hence, the domain of the function is:

\mathbf{D:\{x \in R|x \ne -2,2\}}

On the graph of \mathbf{(g\ o\ f)(x) = \frac{2x^2 - 4}{x^2 - 4} } (see attachment), the function does not have a value from <em>y = 1 to 2.</em>

Hence, the range is:

\mathbf{R:\{(-\infty,1) \cup (2,\infty)\}}

Read more about domain and range at:

brainly.com/question/1632425

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