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 4 crews so there are \( \frac{12}{4} \) = 3 workers on a crew. 9 crews are needed for the busy season which, at 3 workers per crew, means that the roofing company will need 9 x 3 = 27 total workers to staff the crews during the busy season. The company already employs 12 workers so they need to add 27 - 12 = 15 new staff for the busy season.

guard shots made = shots taken x \( \frac{\text{% made}}{100} \) = 20 x \( \frac{50}{100} \) = \( \frac{50 x 20}{100} \) = \( \frac{1000}{100} \) = 10 shots

The center makes 40% of his shots so he'll have to take:

shots made = shots taken x \( \frac{\text{% made}}{100} \)
shots taken = \( \frac{\text{shots taken}}{\frac{\text{% made}}{100}} \)

to make as many shots as the guard. Plugging in values for the center gives us:

center shots taken = \( \frac{10}{\frac{40}{100}} \) = 10 x \( \frac{100}{40} \) = \( \frac{10 x 100}{40} \) = \( \frac{1000}{40} \) = 25 shots

to make the same number of shots as the guard and thus score the same number of points.

The greatest common factor (GCF) is the greatest factor that divides two integers.

To add or subtract terms with exponents, both the base and the exponent must be the same. In this case they are so add the coefficients and retain the base and exponent:

-3z⁴ + 4z⁴
(-3 + 4)z⁴
z⁴

The distributive property for division helps in solving expressions like \({b + c \over a}\). It specifies that the result of dividing a fraction with multiple terms in the numerator and one term in the denominator can be obtained by dividing each term individually and then totaling the results: \({b + c \over a} = {b \over a} + {c \over a}\). For example, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).