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 4 workers at the company now and that's enough to staff 2 crews so there are \( \frac{4}{2} \) = 2 workers on a crew. 4 crews are needed for the busy season which, at 2 workers per crew, means that the roofing company will need 4 x 2 = 8 total workers to staff the crews during the busy season. The company already employs 4 workers so they need to add 8 - 4 = 4 new staff for the busy season.

The first few multiples of 2 are [2, 4, 6, 8, 10, 12, 14, 16, 18, 20] and the first few multiples of 10 are [10, 20, 30, 40, 50, 60, 70, 80, 90]. The first few multiples they share are [10, 20, 30, 40, 50] making 10 the smallest multiple 2 and 10 have in common.

To divide terms with radicals, divide the coefficients and radicands separately:

\( \frac{49\sqrt{25}}{7\sqrt{5}} \)
\( \frac{49}{7} \) \( \sqrt{\frac{25}{5}} \)
7 \( \sqrt{5} \)

To subtract these radicals together their radicands must be the same:

5\( \sqrt{20} \) - 6\( \sqrt{5} \)
5\( \sqrt{4 \times 5} \) - 6\( \sqrt{5} \)
5\( \sqrt{2^2 \times 5} \) - 6\( \sqrt{5} \)
(5)(2)\( \sqrt{5} \) - 6\( \sqrt{5} \)
10\( \sqrt{5} \) - 6\( \sqrt{5} \)

Now that the radicands are identical, you can subtract them:

The least common multiple (LCM) is the smallest positive integer that is a multiple of two or more integers.