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Question 8 (Multiple Choice, Worth 1 point)

The length of a rectangular frame is represented by the expression [tex]2x + 10[/tex], and the width of the rectangular frame is represented by the expression [tex]2x + 6[/tex]. Write an equation to solve for the width of a rectangular frame that has a total area of 140 square inches.

A. [tex]4x^2 + 32x - 80 = 0[/tex]
B. [tex]4x^2 + 32x + 60 = 0[/tex]
C. [tex]2x^2 + 32x - 80 = 0[/tex]
D. [tex]x^2 + 16x + 60 = 0[/tex]

Question 9 (Multiple Choice, Worth 1 point)

Sagot :

To find the correct equation representing the relationship between the dimensions of the rectangular frame and the area provided, we need to set up and solve an equation based on the given information.

1. Identify the expressions for length and width:
- Length, \( L \) = \(2x + 10\)
- Width, \( W \) = \(2x + 6\)

2. Formula for area of a rectangle:
[tex]\[ \text {Area} = \text {Length} \times \text {Width} \][/tex]

3. Given area:
[tex]\[ \text {Area} = 140 \, \text {square inches} \][/tex]

4. Set up the area equation:
[tex]\[ (2x + 10) \times (2x + 6) = 140 \][/tex]

5. Expand the left side of the equation:
[tex]\[ (2x + 10)(2x + 6) = 4x^2 + 20x + 12x + 60 \][/tex]
Simplify the terms:
[tex]\[ 4x^2 + 32x + 60 = 140 \][/tex]

6. Move all terms to one side of the equation to set it to zero:
[tex]\[ 4x^2 + 32x + 60 - 140 = 0 \][/tex]
[tex]\[ 4x^2 + 32x - 80 = 0 \][/tex]

The correct equation is:
[tex]\[ 4x^2 + 32x - 80 = 0 \][/tex]

Therefore, the correct multiple-choice answer is [tex]\( 4x^2 + 32x - 80 = 0 \)[/tex].