All categories

2 years ago

Use complete sentences to explain why absx + 2 > \absx.

Mathematics

1 answer:

12345 [234]2 years ago

5 0

The value of |x| + 2 is always positive and greater than |x|.

The correct question is as below:-

Use complete sentences to explain why |x| + 2 > |x|.

<h3>What is an expression?</h3>

Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition, subtraction, multiplication and division.

The given expression is as follows:-

|x| + 2 > |x|.

Here x is in the mode so any number either negative or positive it will always give you a positive number. For an example put x = -5 in the above expression.

|-5| + 2 > | -5 |

5+ 2 > 5

7 > 5

Put -1 at the place of x

|-1| + 2 > |-1|

1 + 2 > 1

3 .> 1

As we can see in the above two examples for every value of x there will be a positive number and the value of |x|+2 is always greater than |x|.

Therefore the value of |x| + 2 is always positive and greater than |x|.

To know more about expression follow

brainly.com/question/723406

#SPJ1

You might be interested in

The angles of depression of two ships from the top of the light house are 45° and 30° towards east. If the ships are 200 m apart

azamat

Answer:

273 meters

Step-by-step explanation:

See image attached for the diagram I used to represent this scenario.

The distance between the ships, at angles 30 and 45, is 200 meters. The distance between the left ship and the lighthouse is x meters.

We can use trigonometric ratios to solve this problem. We can use the tangent ratio $\big{(} \frac{\text{opposite}}{\text{adjacent}} \big{)}$ to create an equation with the two angles.

$\displaystyle \text{tan(45)} = \frac{h}{x}$
$\displaystyle \text{tan(30)} = \frac{h}{x+200} }$

Let's take these two equations and solve for x in both of them.

<h2> $\textbf{Equation I}$ </h2>

$\displaystyle \text{tan(45)} = \frac{h}{x}$

tan(45) = 1, so we can rewrite this equation.

$\displaystyle 1=\frac{h}{x}$

Multiply x to both sides of the equation.

$\displaystyle x = h$

<h2> $\textbf{Equation II}$ </h2>

$\displaystyle \text{tan(30)} = \frac{h}{x+200} }$

Multiply x + 200 to both sides and divide h by tan(30).

$\displaystyle \text{x + 200} = \frac{h}{\text{tan (30)}}$

Subtract 200 from both sides of the equation.

$\displaystyle \text{x} = \frac{h}{\text{tan (30)}} - 200$

Simplify h/tan(30).

$x=\sqrt{3}h - 200$

<h2> $\textbf{Equation I = Equation II}$ </h2>

Take Equation I and Equation II and set them equal to each other.

$h=\sqrt{3}h-200$

Subtract √3 h from both sides of the equation.

$h-\sqrt{3}h=-200$

Factor h from the left side of the equation.

$h(1-\sqrt{3}) =-200$

Divide both sides of the equation by 1 - √3.

$\displaystyle h=\frac{-200}{1-\sqrt{3} }$

Rationalize the denominator by multiplying the numerator and denominator by the conjugate.

$\displaystyle h=\frac{-200}{1-\sqrt{3} } \big{(} \frac{1+\sqrt{3} }{1+\sqrt{3}} \big{)}$
$\displaystyle h=\frac{-200+200\sqrt{3} }{1-3}$
$\displaystyle h =\frac{-200+200\sqrt{3} }{-2}$