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2 years ago

Which four inequalities can be used to find the solution to this absolute value inequality?

Mathematics

1 answer:

natita [175]2 years ago

6 0

The four inequalities that can be used to find the solution of 3 ≤ |x + 2| ≤ 6 is x + 2 ≤ 6, x + 2 ≥ -6, x + 2 ≥ 3 and x + 2 ≤ -3

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

An equation is an expression that shows the relationship between two or more variables and numbers.

Given the inequality:

3 ≤ |x + 2| ≤ 6

Hence:

x + 2 ≤ 6, -(x + 2) ≤ 6, 3 ≤ x + 2 and 3 ≤ -(x + 2)

This gives:

x + 2 ≤ 6, x + 2 ≥ -6, x + 2 ≥ 3 and x + 2 ≤ -3

The four inequalities that can be used to find the solution of 3 ≤ |x + 2| ≤ 6 is x + 2 ≤ 6, x + 2 ≥ -6, x + 2 ≥ 3 and x + 2 ≤ -3

Find out more on equation at: brainly.com/question/2972832

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Use the Order of Operations to evaluate this expression

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Unfortunately, I don’t knooow :(

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the equation a=1/2(b1+b2)h can be used to determine the area, a, of a trapezoid with height, h, and base lengths, b1 and b2. whi

galina1969 [7]

Answer:

b₁ = (2a – b₂h)/h; b₁ = (2a)/h – b₂; h = (2a)/(b₁ + b₂)

Step-by-step explanation:

A. Solve for b₁

a = ½(b₁ + b₂)h Multiply each side by 2

2a = (b₁ + b₂)h Remove parentheses

2a = b₁h + b₂h Subtract b₂h from each side

2a - b₂h = b₁h Divide each side by h

b₁ = (2a – b₂h)/h Remove parentheses

b₁ = (2a)/h – b₂

B. Solve for h

2a = (b₁ + b₂)h Divide each side by (b₁ + b₂)

h = (2a)/(b₁ + b₂)

5 0

3 years ago

Read 2 more answers

Solve the following system of equations. 2x+y=3 I= 2y - 1 x= y=

marin [14]

Note: I am assuming your second equation is:

x = 2y - 1

Answer:

The solution to the system of equations is:

x = 1

y = 1

Step-by-step explanation:

Given the system of equations

2x+y=3

x = 2y - 1

solving the system of equations using the elimination method

$\begin{bmatrix}2x+y=3\\ x=2y-1\end{bmatrix}$

Arrange equation variables for elimination

$\begin{bmatrix}2x+y=3\\ x-2y=-1\end{bmatrix}$

Multiply x-2y=-1 by2: 2x-4y=-2

$\begin{bmatrix}2x+y=3\\ 2x-4y=-2\end{bmatrix}$

subtracting the equations

$2x-4y=-2$

$-$

$\underline{2x+y=3}$

$-5y=-5$

now solve -5y = -5 for y

$-5y=-5$

Divide both sides by -5

$\frac{-5y}{-5}=\frac{-5}{-5}$

Simplify

$y=1$

For 2x+y=3 plug in y=1

$2x+1=3$

subtract 1 from both sides

$2x+1-1=3-1$

Simplify

$2x=2$

Divide both sides by 2

$\frac{2x}{2}=\frac{2}{2}$

Simplify

$x=1$

Therefore, the solution to the system of equations is:

x = 1

y = 1

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The answer is the second one.
The one that shows 5 square root of 2.

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attashe74 [19]

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