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3 years ago

What are the domain restrictions of the function f(x) = the quantity of x squared plus 6x plus 9, all over 3x plus 15.

Mathematics

1 answer:

dangina [55]3 years ago

6 0

Answer:

x = - 5

Step-by-step explanation:

Given

f(x) = $\frac{x^2+6x+9}{3x+15}$

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be.

solve : 3x + 15 = 0 ⇒ 3x = - 15 ⇒ x = - 5 ← excluded value

Domain x ∈ R , x ≠ - 5

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Enter the correct answer in the box. solve the equation x2 − 16x 54 = 0 by completing the square. fill in the values of a and b

poizon [28]

The roots of the given polynomials exist $x=8+\sqrt{10}$ , and $x=8-\sqrt{10}$ .

<h3>What is the formula of the quadratic equation?</h3>

For a quadratic equation of the form $a x^{2}+b x+c=0$ the solutions are

$x_{1,2}=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Therefore by using the formula we have

$x^{2}-16 x+54=0$$

Let, a = 1, b = -16 and c = 54

Substitute the values in the above equation, and we get

$x_{1,2}=\frac{-(-16) \pm \sqrt{(-16)^{2}-4 \cdot 1 \cdot 54}}{2 \cdot 1}$$

simplifying the equation, we get

$$&x_{1,2}=\frac{-(-16) \pm 2 \sqrt{10}}{2 \cdot 1} \\$

$$&x_{1}=\frac{-(-16)+2 \sqrt{10}}{2 \cdot 1}, x_{2}=\frac{-(-16)-2 \sqrt{10}}{2 \cdot 1} \\$

$$&x=8+\sqrt{10}, x=8-\sqrt{10}$

Therefore, the roots of the given polynomials are $x=8+\sqrt{10}$ , and

$x=8-\sqrt{10}$ .

To learn more about quadratic equations refer to:

brainly.com/question/1214333

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steposvetlana [31]

Answer:

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3 years ago

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Yuri [45]

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Step-by-step explanation:

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Answer:

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d + 0.06d

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Answer:

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Step-by-step explanation:

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