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disa [49]
3 years ago
12

Please show your work and solve :)

Mathematics
2 answers:
pickupchik [31]3 years ago
8 0
A/-3 -6=-1
-a/3 -6=-1
-a-18=-3
-a=-3+18
-a=15
A=-15
Amiraneli [1.4K]3 years ago
5 0

Answer:

a = -15

Step-by-step explanation:

a/-3 -6 = -1

add 6 to both sides

a/-3 = 5

now multiply -3 to both sides

a= -15

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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 />

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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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