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kirill [66]
3 years ago
6

Which statement about the function is true?

Mathematics
1 answer:
krok68 [10]3 years ago
7 0

Answer:

The function is negative for all real values of x where -6< x < -2

Step-by-step explanation:

Given f(x) = (x + 2)(x + 6)

The graph of the function is as shown in the attached figure.

As shown, we can deduce the following:

The function is positive for all real values of x where x > -2 and x < -6

The function is zero at x = -2 and x = -6

The function is negative for all real values of x where -6 < x <-2

Compare the observations to the given statements:

So, The true statement is The function is negative for all real values of x where -6 < x <-2

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What are the domain and range of the function?
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  see below

Step-by-step explanation:

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The output (range) is from negative infinity to zero: (-∞, 0].

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A graph of the function is included for your convenience.

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Step-by-step explanation:

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3 years ago
Perform the indicated operations below and simplify ((a+b)/4) + ((2a-b)/5).
ArbitrLikvidat [17]

Answer:

\frac{(a+b)}{4} +\frac{(2a-b)}{5}=\frac{13a}{20} +\frac{b}{20}

Step-by-step explanation:

We have:

\frac{(a+b)}{4} +\frac{(2a-b)}{5}

We can use <em>common denominator</em>.

Observation:

If you have, \frac{a}{b} +\frac{c}{d}=\frac{(a*d)+(c*b)}{b*d}

Then,

\frac{(a+b)}{4} +\frac{(2a-b)}{5}=\frac{5(a+b)+4(2a-b)}{4*5}

Using <em>distributive property</em>:

Observation:

c(a+b)=ca+cb

\frac{5(a+b)+4(2a-b)}{4*5}=\frac{5a+5b+8a-4b}{20}=\frac{(5a+8a)+(5b-4b)}{20}

Finally,

\frac{(5a+8a)+(5b-4b)}{20} =\frac{13a+b}{20}=\frac{13a}{20} +\frac{b}{20}

The answer then is:

\frac{(a+b)}{4} +\frac{(2a-b)}{5}=\frac{13a}{20} +\frac{b}{20}

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