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CaHeK987 [17]
3 years ago
5

For what values of $x$ is $$\frac{x^2 + x + 3}{2x^2 + x - 6} \ge 0?$$ note: be thorough and explain why all points in your answe

r are solutions and why all points outside your answer are not solutions.

Mathematics
2 answers:
Rus_ich [418]3 years ago
5 0

To solve the inequality \frac{x^2+x+3}{2x^2+x-6}\ge \:0

\mathrm{Factor\:the\:left\:hand\:side\:}\frac{x^2+x+3}{2x^2+x-6}:

The numerator x^2+x+3 is not factorizable.

so factor the denominator 2x^2+x-6:

2x^2+x-6=\left(2x^2-3x\right)+\left(4x-6\right)=x\left(2x-3\right)+2\left(2x-3\right)\\ \mathrm{Factor\:out\:}x\mathrm{\:from\:}2x^2-3x\mathrm{:\quad }x\left(2x-3\right)\\ \mathrm{Factor\:out\:}2\mathrm{\:from\:}4x-6\mathrm{:\quad }2\left(2x-3\right)\\

Now take \mathrm{Factor\:out\:common\:term\:}\left(2x-3\right)

Then we get factor of the denominator as \left(2x-3\right)\left(x+2\right)

Thus \frac{x^2+x+3}{2x^2+x-6}=\frac{x^2+x+3}{\left(x+2\right)\left(2x-3\right)}

Now \mathrm{Compute\:the\:signs\:of\:the\:factors\:of\:}\frac{x^2+x+3}{\left(x+2\right)\left(2x-3\right)}\\

Signs of x^2+x+3>0

\mathrm{Choosing\:ranges\:that\:satisfy\:the\:required\:condition:}\:\ge \:\:0

x\frac{3}{2} is the required solution of the given inequality.

Anettt [7]3 years ago
3 0

Answer: (-\infty, -2)\cup(3/2,\infty)

Step-by-step explanation:

Here, the given expression,

\frac{x^2 + x + 3}{2x^2 + x - 6} \geq 0

Since, for the value of x,

Denominator ≠ 0,

2x^2+x-6\neq 0

\implies 2x^2+4x - 3x - 6\neq =0

\implies 2x(x+2)-3(x+2)\neq 0

\implies (2x-3)(x+2)\neq 0

\text{ if }2x-3\neq 0\implies x\neq \frac{3}{2}

\text{If } x+2\neq 0\implies x\neq -2

Thus, there is three intervals possible,

1) (-\infty,-2)

2) (-2,\frac{3}{2})

3) (\frac{3}{2},\infty)

In first intervals, \frac{x^2 + x + 3}{2x^2 + x - 6} \geq 0 is  true.

⇒ (-\infty,-2) will contain the value of x.

In second interval, \frac{x^2 + x + 3}{2x^2 + x - 6} \geq 0

is not true,

⇒ (-2, -\frac{3}{2}) will not contain the value of x,

In third interval,

In third interval, \frac{x^2 + x + 3}{2x^2 + x - 6} \geq 0

is true,

⇒ (-2, -\frac{3}{2}) will contain the value of x,

In third interval,

Thus, the value of x is,

(-\infty, -2)\cup(3/2,\infty)

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360

Step-by-step explanation:

To calculate the number of different combinations of 2 different flavours, 1 topping, and 1 cone, we are going to use the rule of multiplication:

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Because first, we have 6 possible options for the flavour, then we only have 5 possible options for the 2nd flavour. Then, we have 4 options for the topping and finally, we have 3 options for the cone.

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Does the following equation determine y to be a function of x? y square=x+3
Phantasy [73]

Answer:

∴ y² = x + 3 is not a function

Step-by-step explanation:

* Lets explain how to solve the problem

- The definition of the function is every input (x) has only one

  output (y)

- Ex:

# y = x + 1 where x ∈ R , is a function because every x has only

  one value of y

# y² = x where x ∈ R , is not a function because y = ±√x, then one

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* Lets solve the problem

∵ y² = x + 3

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∴ y = ± √(x + 3)

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∵ (x + 3) must be greater than or equal zero because there is no

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∴ x + 3 ≥ 0 ⇒ subtract 3 from both sides

∴ x ≥ -3

∴ x must be any number greater than or equal -3

- Let x = 0

∴ y = √(0 + 3) = √3 and y = - √(0 + 3) = -√3

∴ x = 0 has two values of y ⇒ y = √3 and y = -√3

- Any value of x greater than or equal 3 will have two values of y

∴ y² = x + 3 is not a function

5 0
3 years ago
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