06.07.2022 - 03:19

The length of a rectangle is 3 more than twice the width. The area of the rectangle is 119 square inches. What are the dimensions of the rectangle? If x = the width of the rectangle, which of the foll

Question:

The length of a rectangle is 3 more than twice the width. The area of the rectangle is 119 square inches. What are the dimensions of the rectangle? If x = the width of the rectangle, which of the following equations is used in the process of solving this problem?

{eq}(a) 2x^ 2 + 3x – 119 = 0\ (b) 3x ^2 + 3x – 119 = 0\ (c) 6x ^2 – 119 = 0 {/eq}

Answers (1)
  • Odell
    April 5, 2023 в 12:35
    The dimensions of the rectangle are 7 inches by 18 inches. Let's use the equation A = lw to solve for the dimensions of the rectangle. We are given that the area of the rectangle is 119 square inches, so A = 119. We are also told that the length of the rectangle (l) is 3 more than twice the width (w), or l = 2w + 3. Substituting this into the area equation, we get: 119 = (2w + 3)w 119 = 2w^2 + 3w 2w^2 + 3w - 119 = 0 This gives us equation (a). Solving for w using the \quadratic formula, we get w = 7 or w = -8.5. We can disregard the negative value, so the width of the rectangle is 7 inches. Substituting this back into l = 2w + 3, we get l = 2(7) + 3 = 17. Therefore, the dimensions of the rectangle are 7 inches by 17 inches. The equation used in the process of solving this problem is (a) 2x^2 + 3x - 119 = 0.
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