In order to find how many additional workers are needed to staff the extra crews you first need to calculate how many workers are on a crew. There are 12 workers at the company now and that's enough to staff 3 crews so there are \( \frac{12}{3} \) = 4 workers on a crew. 5 crews are needed for the busy season which, at 4 workers per crew, means that the roofing company will need 5 x 4 = 20 total workers to staff the crews during the busy season. The company already employs 12 workers so they need to add 20 - 12 = 8 new staff for the busy season.

The area of a rectangle is width (w) x height (h). In this problem we know that the rectangle is twice as long as it is wide so h = 2w. The perimeter of a rectangle is 2w + 2h and we know that the perimeter of this rectangle is 30 meters so the equation becomes: 2w + 2h = 30.

Putting these two equations together and solving for width (w):

2w + 2h = 30
w + h = \( \frac{30}{2} \)
w + h = 15
w = 15 - h

From the question we know that h = 2w so substituting 2w for h gives us:

w = 15 - 2w
3w = 15
w = \( \frac{15}{3} \)
w = 5

Since h = 2w that makes h = (2 x 5) = 10 and the area = h x w = 5 x 10 = 50 m²

The commutative property states that, when adding or multiplying numbers, the order in which they're added or multiplied does not matter. For example, 3 + 4 and 4 + 3 give the same result, as do 3 x 4 and 4 x 3.

To fill a 15 gallon tank exactly halfway you'll need 7\(\frac{1}{2}\) gallons of fuel. Each fuel can holds 2\(\frac{1}{2}\) gallons so:

cans = \( \frac{7\frac{1}{2} \text{ gallons}}{2\frac{1}{2} \text{ gallons}} \) = 3

The first few multiples of 6 are [6, 12, 18, 24, 30, 36, 42, 48, 54, 60] and the first few multiples of 14 are [14, 28, 42, 56, 70, 84, 98]. The first few multiples they share are [42, 84] making 42 the smallest multiple 6 and 14 have in common.