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, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).

Calculate the number of vans needed by dividing the number of people that need transported by the capacity of one van:

vans = \( \frac{93}{12} \) = 7\(\frac{3}{4}\)

So, it will take 7 full vans and one partially full van to transport the entire team making a total of 8 vans.

The factors of 52 are [1, 2, 4, 13, 26, 52] and the factors of 80 are [1, 2, 4, 5, 8, 10, 16, 20, 40, 80]. They share 3 factors [1, 2, 4] making 4 the greatest factor 52 and 80 have in common.

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

\( \frac{9\sqrt{15}}{3\sqrt{5}} \)
\( \frac{9}{3} \) \( \sqrt{\frac{15}{5}} \)
3 \( \sqrt{3} \)

guard shots made = shots taken x \( \frac{\text{% made}}{100} \) = 15 x \( \frac{40}{100} \) = \( \frac{40 x 15}{100} \) = \( \frac{600}{100} \) = 6 shots

The center makes 30% 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 = \( \frac{6}{\frac{30}{100}} \) = 6 x \( \frac{100}{30} \) = \( \frac{6 x 100}{30} \) = \( \frac{600}{30} \) = 20 shots

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