Ask Question
9 February, 13:04

The length of a rectangle is 2 more than three times the width. The area of the rectangle is 161 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?

2x^2 + 3x - 161 = 0

3x^2 + 2x - 161 = 0

6x^2 - 161 = 0

+1
Answers (1)
  1. 9 February, 15:53
    0
    Here is the explanation:

    A = L*W

    L = 2+3W

    161 = (2+3W) * (W)

    =2W+2W^2 3W^2+2W-161
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “The length of a rectangle is 2 more than three times the width. The area of the rectangle is 161 square inches. What are the dimensions of ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers