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Write an absolute value inequality that represents the situation. Then solve the inequality. The difference between the perimeters of the figures is less than or equal to 3

Write an absolute value inequality that represents the situation. Then solve the inequality. The difference between the perimeters of the figures is less than o class=

Answer :

frika

Answer:

[tex]|2x-8|\le 3\\ \\2.5\le x\le 5.5[/tex]

Step-by-step explanation:

You are given the rectangle with sides of (x + 1) units and 3 units and the square with the side of x units.

The perimeter of the rectangle is

[tex]P_r=(x+1)+3+(x+1)+3\\ \\=x+1+3+x+1+3\\ \\=2x+8\ units[/tex]

The perimeter of the square is

[tex]P_S=x+x+x+x\\ \\=4x\ units[/tex]

We do not know whose perimeter is 3 units greater (or equal to), so the absolute value of the difference of these perimeters should be less than or equal to 3.

Hence,

[tex]|4x-(2x+8)|\le 3\\ \\|4x-2x-8|\le 3\\ \\|2x-8|\le 3[/tex]

Solve this inequality:

[tex]-3\le 2x-8\le 3\\ \\-3+8\le 2x-8+8\le 3+8\\ \\5\le 2x\le 11\\ \\2.5\le x\le 5.5[/tex]

abidemiokin
  • The absolute value inequality that represents the situation is [tex]|4x-(2x+8)|\leq 3[/tex]
  • The required solution will be [tex]2.5\leq x\leq5.5[/tex]

The formula for calculating the perimeter of a square is expressed as:

P = 4L

L is the side length of the square

For the square with side length "x"

  • Perimeter of the square = 4x

The area of the rectangle is expressed as 2(L+W)

L is the length

W is the width

For the given rectangle with length 3 and width (x+1)

  • Perimeterof the rectangle = 2(3 + x+1) = 2(x+4) = 2x+8

The difference between the perimeters of the figure will be expressed as:

[tex]|4x-(2x+8)|\leq 3[/tex]

If the modulus value is positive:

[tex]4x-(2x+8)\leq 3\\4x-2x-8\leq3\\2x-8\leq3\\2x\leq11\\x\leq5.5[/tex]

If the modulus value is negative:

[tex]-[4x-(2x+8)]\leq 3\\-[4x-2x-8\leq3\\-4x+2x+8\leq 3\\-2x+8\leq3\\-2x\leq-5\\x\geq2.5[/tex]

Combining both inequalities, the required solution will be [tex]2.5\leq x\leq5.5[/tex]

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