The perimeter of a rectangular piece of plastic is 22 millimeters. The area is 30 square millimeters. What are the dimensions of the piece of plastic?

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2l + 2w = 22

l + w = 11 ---> l = 11 - w

area = 30
lw = 30
(11-w)(w) = 30
11w - w^2 = 30
w^2 - 11w + 30 = 0
(w-5)(w-6) = 0
w = 5 or w = 6

if w = 5 , l = 11-5 = 6
if w = 6, l = 11 - 6 = 5

the dimension is 5 by 6 mm

To find the dimensions of the rectangular piece of plastic, we can use the formulas for perimeter and area of a rectangle.

Let's assume the length of the rectangle is "L" and the width is "W".

Perimeter of a rectangle = 2(L + W)

From the given information, we know that the perimeter of the plastic piece is 22 millimeters. So we can write the equation:

2(L + W) = 22

Area of a rectangle = L * W

From the given information, we also know that the area of the plastic piece is 30 square millimeters. So we can write the equation:

L * W = 30

Now, we have a system of two equations:

1) 2(L + W) = 22
2) L * W = 30

We can use these equations to find the values of L and W.

One way to solve this system of equations is by substitution. Solve the first equation for L or W, and then substitute it into the second equation. Let's solve the first equation for L:

2(L + W) = 22

Divide both sides by 2:

L + W = 11

Now, solve for L:

L = 11 - W

Substitute this value of L into the second equation:

(11 - W) * W = 30

Expand and rearrange the equation:

11W - W^2 = 30

Rearrange the equation to write it in standard form:

W^2 - 11W + 30 = 0

Now we have a quadratic equation, which we can solve by factoring or using the quadratic formula. Factoring the equation, we get:

(W - 5)(W - 6) = 0

Setting each factor equal to zero, we find two possible values for W:

W - 5 = 0 --> W = 5
W - 6 = 0 --> W = 6

Now that we have the possible values for W, we can substitute them back into one of the original equations to find the corresponding values of L.

If we substitute W = 5 into the first equation, we get:

L + 5 = 11

Solving for L, we find:

L = 6

If we substitute W = 6 into the first equation, we get:

L + 6 = 11

Solving for L, we find:

L = 5

So, the possible dimensions of the plastic piece are:
- Length (L) = 6 mm, Width (W) = 5 mm
- Length (L) = 5 mm, Width (W) = 6 mm