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natita [175]
1 year ago
9

Given f of x is equal to 1 over the quantity x minus 3 end quantity and g of x is equal to the square root of the quantity x plu

s 3 end quantity comma what is the domain of f (g(x))?
[–3, ∞)
[–3, 6) ∪ (6, ∞)
(–∞, 3) ∪ (3, ∞)
ℝ
Mathematics
1 answer:
Lostsunrise [7]1 year ago
7 0

The domain of the composite function is given as follows:

[–3, 6) ∪ (6, ∞)

<h3>What is the composite function of f(x) and g(x)?</h3>

The composite function of f(x) and g(x) is given as follows:

(f \circ g)(x) = f(g(x))

In this problem, the functions are:

  • f(x) = \frac{1}{x - 3}.
  • g(x) = \sqrt{x + 3}

The composite function is of the given functions f(x) and g(x) is:

f(g(x)) = f(\sqrt{x + 3}) = \frac{1}{\sqrt{x + 3} - 3}

The square root has to be non-negative, hence the restriction relative to the square root is found as follows:

x + 3 \geq 0

x \geq -3

The denominator cannot be zero, hence the restriction relative to the denominator is found as follows:

\sqrt{x + 3} - 3 \neq 0

\sqrt{x + 3} \neq 3

(\sqrt{x + 3})^2 \neq 3^2

x + 3 \neq 9

x \neq 6

Hence, from the restrictions above, of functions f(x), g(x) and the composite function, the domain is:

[–3, 6) ∪ (6, ∞)

More can be learned about composite functions at brainly.com/question/13502804

#SPJ1

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Answer:

B(-2, -7)

Step-by-step explanation:

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A(x₁, y₁) B(x₂, y₂)

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Midpoint = (\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2} )\\\\Midpoint = (\frac{2+ x_{2} }{2}, \frac{1 + y_{2} }{2} )

(0, -3) = (\frac{2+ x_{2} }{2}, \frac{1 + y_{2} }{2} )

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              \frac{1 + y_{2} }{2} = -3\\\\1 + y_{2} = -6\ \ \ \ times\ both\ sides\ by\ 2\\y_{2} = -7\ \ \ \ subtract\ both\ sides\ by\ 1

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6 0
2 years ago
The length of a rectangle is 3 cm more than twice the width. The area of the rectangle is 65 square centimeters. Find the dimens
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Answer:

5 cm and 13 cm

Step-by-step explanation:

Let b be the width of the rectangle.

Length = 3+2b

The area of the rectangle is 65 cm²

We need to find the dimensions of the rectangle. The area of a rectangle is given by :

A = lb

65=(3+2b)b\\\\65=3b+2b^2\\\\3b+2b^2-65=0\\\\b=5\ cm,-6.5\ cm

Neglecting the negative value, the width of the rectangle is 5 cm.

Length = 3+2b

=3+2(5)

=3+10

=13 cm

Hence, the dimensions of the rectangle are 5 cm and 13 cm.

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