The length of a rectangle is 15 and its width is w. The perimeter of the rectangle is, at most, 50. Which inequality can be used to find the longest possible width?
1) 30 + 2w < 50
2) 30 + 2w ≤ 50
3) 30 + 2w > 50
4) 30 + 2w ≥ 50

First, let's find the perimeter of the rectangle in terms of w. We have 2*(15+w) = 30+2w because the perimeter of a rectangle can be written as 2*(length + width). The problem says that the perimeter is at most 50. This means that it cannot be greater than 50, so it would have to be less than or equal to 50. It can be equal to 50 because that is what at most means. Putting this information into an inequality, we get 30 + 2w ≤ 50.

The length of a rectangle is 15 and its width is w. The perimeter of the rectangle is, at most, 50. Which inequality can be used to find the longest possible width? 1) 30 + 2w < 50 2) 30 + 2w ≤ 50 3) 30 + 2w > 50 4) 30 + 2w ≥ 50

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The length of a rectangle is 15 and its width is w. The perimeter of the rectangle is, at most, 50. Which inequality can be used to find the longest possible width?
1) 30 + 2w < 50
2) 30 + 2w ≤ 50
3) 30 + 2w > 50
4) 30 + 2w ≥ 50