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$8.04 MC) e length of a rectangular frame is represented by the expression 2x+10, and the width of the rectangular frame is represented by the expression 2x+6. Write an equation to solve for the width of a re me that has a total area of 140 square inches. (1 point) {:[4x^(2)+32 x-80=0],[4x^(2)+32 x+60=0],[2x^(2)+32 x-80=0],[x^(2)+16 x+60=0]:}$

$8$ . $04$ MC) $\newline$ e length of a rectangular frame is represented by the expression $2 x+10$ , and the width of the rectangular frame is represented by the expression $2 x+6$ . Write an equation to solve for the width of a re me that has a total area of $140$ square inches. ( $1$ point) $\newline$ $\begin{array}{l} 4 x^{2}+32 x-80=0 \\ 4 x^{2}+32 x+60=0 \\ 2 x^{2}+32 x-80=0 \\ x^{2}+16 x+60=0 \end{array}$

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Math Problems

Grade 6

Solve one-step multiplication and division equations: word problems

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8

04

MC)

\newline

e length of a rectangular frame is represented by the expression

2 x+10

, and the width of the rectangular frame is represented by the expression

2 x+6

. Write an equation to solve for the width of a re me that has a total area of

140

square inches. (

1

point)

\newline

\begin{array}{l} 4 x^{2}+32 x-80=0 \\ 4 x^{2}+32 x+60=0 \\ 2 x^{2}+32 x-80=0 \\ x^{2}+16 x+60=0 \end{array}

Write Equation: $\text{Length} = 2x + 10$ , $\text{Width} = 2x + 6$ , $\text{Area} = 140$ square inches. $\newline$ $\text{Area of rectangle} = \text{Length} \times \text{Width}.$ $\newline$ $140 = (2x + 10)(2x + 6).$
Expand Equation: Expand the equation: $140 = 4x^2 + 12x + 20x + 60$ .
Combine Like Terms: Combine like terms: $140 = 4x^2 + 32x + 60$ .
Subtract and Simplify: Subtract $140$ from both sides to set the equation to zero: $0 = 4x^2 + 32x + 60 - 140$ .
Subtract and Simplify: Subtract $140$ from both sides to set the equation to zero: $0 = 4x^2 + 32x + 60 - 140$ . Simplify the equation: $0 = 4x^2 + 32x - 80$ .