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6 months ago

Twice a number, increased by three, is between negative one and nine. Find all the numbers.

Mathematics

1 answer:

pashok25 [27]6 months ago

8 0

The solution set for the obtained inequality is {.......-5, -4, -3, -2, -1, 0, 1, 2}.

Given that, twice a number, increased by three, is between negative one and nine.

<h3>What is inequality?</h3>

The relation between two expressions that are not equal, employing a sign such as ≠ ‘not equal to’, > ‘greater than’, or < ‘less than’.

Let the unknown number be x.

Now, twice a number, increased by three = 2x+3

The given expression is between negative one and nine

-1 < 2x+3 < 9

-1 < 2x+3

Subtract 3 to both the side of the inequality

-1-3<2x+3-3

⇒ -4 < 2x

Divide 2 to both the side of the inequality

-2 < x

Now, 2x+3 < 9

Subtract 3 to both the side of the inequality

2x+3-3<9-3

⇒ 2x<6

Divide 2 to both the side of the inequality

x < 3

So, -2<x<3

Solution set = {.......-5, -4, -3, -2, -1, 0, 1, 2}

Therefore, the solution set for the obtained inequality is {.......-5, -4, -3, -2, -1, 0, 1, 2}.

To learn more about a inequality visit:

brainly.com/question/20383699.

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<u>Step-by-step explanation:</u>

It is given that θ is between 270° and 360°, which means that θ is located in Quadrant IV ⇒ (x > 0, y < 0). Furthermore, the half-angle will be between 135° and 180°, which means the half-angle is in Quadrant II ⇒ $cos\ \dfrac{\theta}{2}$

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Reminder: We previously determined that the half-angle will be negative.

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Twice a​ number, increased by ​three, is between negative one and nine. Find all the numbers.

Twice a number, increased by three, is between negative one and nine. Find all the numbers.