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3 years ago

|2x| ≤ x + 3

Mathematics

2 answers:

Korolek [52]3 years ago

5 0

Let's solve your inequality step-by-step.

|2x|≤x+3

Solve Absolute Value.

|2x|≤x+3

Let's find the critical points of the inequality.

|2x|=x+3

We know either2x=x+3or2x=−(x+3)

2x=x+3(Possibility 1)

2x−x=x+3−x(Subtract x from both sides)

x=3

2x=−(x+3)(Possibility 2)

2x=−x−3(Simplify both sides of the equation)

2x+x=−x−3+x(Add x to both sides)

3x=−3

−3

(Divide both sides by 3)

x=−1

Check possible critical points.

x=3(Works in original equation)

x=−1(Works in original equation)

Critical points:

x=3 or x=−1

Check intervals in between critical points. (Test values in the intervals to see if they work.)

x≤−1(Doesn't work in original inequality)

−1≤x≤3(Works in original inequality)

x≥3(Doesn't work in original inequality)

Answer:

−1≤x≤3

r-ruslan [8.4K]3 years ago

3 0

Answer:

[-1,3]

Step-by-step explanation:

|2x|<x+3

calculate the absolute value

2*|x|<x+3

move < to left

2*|x|-x<3

split into possible cases

2x-x<3, x<0

2*(-x)-x<3, x<0

solve inequalities

[0,3]

[-1,0]

find union

[-1,3]

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7 0

3 years ago

johnnys baseball team raised $1456 for next month's tournament. They need a total of $2,000 to play. the team has a total of 11

lapo4ka [179]

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4 years ago

Does anyone know how to do this?? Help please!!!!

Doss [256]

Answer:

When we have a rational function like:

$r(x) = \frac{x + 1}{x^2 + 3}$

The domain will be the set of all real numbers, such that the denominator is different than zero.

So the first step is to find the values of x such that the denominator (x^2 + 3) is equal to zero.

Then we need to solve:

x^2 + 3 = 0

x^2 = -3

x = √(-3)

This is the square root of a negative number, then this is a complex number.

This means that there is no real number such that x^2 + 3 is equal to zero, then if x can only be a real number, we will never have the denominator equal to zero, so the domain will be the set of all real numbers.

D: x ∈ R.

b) we want to find two different numbers x such that:

r(x) = 1/4

Then we need to solve:

$\frac{1}{4} = \frac{x + 1}{x^2 + 3}$