Answer :
Final answer:
The dimensions of a rectangle with a length twice its width and a perimeter 120 feet more than its width, we solve the equation 6w = w + 120 to get a width of 24 feet and a length of 48 feet.
Explanation:
To solve a problem involving the dimensions of a rectangle where the length is twice the width, and the perimeter is 120 feet more than the width, we'll set up a simple algebraic equation. First, we let w represent the width of the rectangle. The perimeter of a rectangle is calculated as 2(width + length). Knowing that the length is twice the width, we can express the length as 2w. Therefore, the perimeter equation in terms of width is 2(w + 2w), which simplifies to 6w.
Given the perimeter is 120 feet more than the width, we set up the equation 6w = w + 120. Solving this equation, we subtract w from both sides to get 5w = 120. Dividing both sides by 5 gives us the width, w = 24 feet.
To find the length, we just multiply width by 2, which yields a length of 48 feet. Therefore, the dimensions of the rectangle are 24 feet by 48 feet.
Answer:
2L + 2w = 120
L = 2w
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2L + 2w = 120
2(2w) + 2w = 120
4w + 2w = 120
6w = 120
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w = 20 ft
L = 40 ft
Step-by-step explanation:
