consider the collection of all rectangles that have lengths 2in. less than twice their widths. find the possible widths(in inches) of these rectangles if their perimeters are less than 200 in

Question

consider the collection of all rectangles that have lengths 2in. less than twice their widths. find the possible widths(in inches) of these rectangles if their perimeters are less than 200 in

Answer 1

let width be x

so length y = 2x-2

p = 2(x+y) = 2x+2(2x-2) = 6x-4

we want p < 200

6x-4 < 200
6x < 204
x < 34

so, as long as the width is less than 34, the length will be less than 66, and the perimeter will be less than 200.

Answer 2

Well, well, well, looks like we're trying to find some quirky rectangles with perimeters less than 200 inches! Let's see what we can come up with, shall we?

First, let's set up the equation for the perimeter of a rectangle: P = 2(Length + Width). In this case, we have a special condition where the length is 2 inches less than twice the width. So we can rewrite the equation as: P = 2((2W - 2) + W).

To keep things under a perimeter of 200, we'll insert that into our equation: 200 > 2((2W - 2) + W).

Simplifying the equation, we have: 200 > 2(3W - 2).

Dividing both sides by 2, we get: 100 > 3W - 2.

Adding 2 to both sides, we have: 102 > 3W.

Finally, dividing both sides by 3, we get: W < 34.

So, the possible widths (in inches) of these rectangles are less than 34 inches if their perimeters are less than 200 inches. Just be careful not to trip over any extra long rectangles!

Answer 3

To find the possible widths of the rectangles, we can set up an inequality based on their given conditions:

Let W represent the width of the rectangle (in inches).

According to the given conditions, the length of the rectangle is 2 inches less than twice its width. So, the length (L) can be expressed as L = 2W - 2 (in inches).

The perimeter (P) of a rectangle is given by the formula P = 2(L + W).

Substituting the values of L and W into the perimeter formula, we get:

P = 2((2W - 2) + W)
P = 2(3W - 2)
P = 6W - 4

Since the perimeter of the rectangle should be less than 200 inches, we can set up the inequality:

6W - 4 < 200

Now, let's solve this inequality:

6W < 200 + 4
6W < 204
W < 34

Therefore, the possible widths (W) of these rectangles, where their perimeters are less than 200 inches, are any values less than 34 inches.

Answer 4

To find the possible widths of the rectangles with lengths 2 inches less than twice their widths, you need to set up an inequality based on the given conditions and then solve it.

Let's say the width of the rectangle is represented by "w" (in inches). According to the given conditions, the length of the rectangle is 2 inches less than twice the width, which can be expressed as (2w - 2) inches.

The perimeter of a rectangle can be calculated by adding the lengths of all its sides. For a rectangle with width "w" and length (2w - 2), the perimeter P can be expressed as:

P = 2w + 2(2w - 2) = 6w - 4

Now, you want to find the widths for which the perimeter is less than 200 inches. So, you set up an inequality:

6w - 4 < 200

To solve this inequality, follow these steps:

1. Add 4 to both sides of the inequality to isolate the variable:
6w - 4 + 4 < 200 + 4
6w < 204

2. Divide both sides of the inequality by 6 to solve for w:
(6w) / 6 < 204 / 6
w < 34

The solution to the inequality is w < 34. Therefore, the possible widths of the rectangles, whose perimeters are less than 200 inches, are any values less than 34 inches.