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

7\( \sqrt{6} \) x 3\( \sqrt{2} \)
(7 x 3)\( \sqrt{6 \times 2} \)
21\( \sqrt{12} \)

Now we need to simplify the radical:

21\( \sqrt{12} \)
21\( \sqrt{3 \times 4} \)
21\( \sqrt{3 \times 2^2} \)
(21)(2)\( \sqrt{3} \)
42\( \sqrt{3} \)

guard shots made = shots taken x \( \frac{\text{% made}}{100} \) = 30 x \( \frac{60}{100} \) = \( \frac{60 x 30}{100} \) = \( \frac{1800}{100} \) = 18 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 = \( \frac{18}{\frac{40}{100}} \) = 18 x \( \frac{100}{40} \) = \( \frac{18 x 100}{40} \) = \( \frac{1800}{40} \) = 45 shots

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

To subtract these fractions, first find the lowest common multiple of their denominators. The first few multiples of 3 are [3, 6, 9, 12, 15, 18, 21, 24, 27, 30] and the first few multiples of 7 are [7, 14, 21, 28, 35, 42, 49, 56, 63, 70]. The first few multiples they share are [21, 42, 63, 84] making 21 the smallest multiple 3 and 7 share.

Next, convert the fractions so each denominator equals the lowest common multiple:

\( \frac{4 x 7}{3 x 7} \) - \( \frac{5 x 3}{7 x 3} \)

\( \frac{28}{21} \) - \( \frac{15}{21} \)

Now, because the fractions share a common denominator, you can subtract them:

\( \frac{28 - 15}{21} \) = \( \frac{13}{21} \) = \(\frac{13}{21}\)

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

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 42 meters so the equation becomes: 2w + 2h = 42.

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

2w + 2h = 42
w + h = \( \frac{42}{2} \)
w + h = 21
w = 21 - h

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

w = 21 - 2w
3w = 21
w = \( \frac{21}{3} \)
w = 7

Since h = 2w that makes h = (2 x 7) = 14 and the area = h x w = 7 x 14 = 98 m²