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

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

9\( \sqrt{12} \) + 3\( \sqrt{3} \)
9\( \sqrt{4 \times 3} \) + 3\( \sqrt{3} \)
9\( \sqrt{2^2 \times 3} \) + 3\( \sqrt{3} \)
(9)(2)\( \sqrt{3} \) + 3\( \sqrt{3} \)
18\( \sqrt{3} \) + 3\( \sqrt{3} \)

Now that the radicands are identical, you can add them together:

Use PEMDAS (Parentheses, Exponents, Multipy/Divide, Add/Subtract):

4 + (3 + 3) ÷ 3 x 2 - 2²
P: 4 + (6) ÷ 3 x 2 - 2²
E: 4 + 6 ÷ 3 x 2 - 4
MD: 4 + \( \frac{6}{3} \) x 2 - 4
MD: 4 + \( \frac{12}{3} \) - 4
AS: \( \frac{12}{3} \) + \( \frac{12}{3} \) - 4
AS: \( \frac{24}{3} \) - 4
AS: \( \frac{24 - 12}{3} \)
\( \frac{12}{3} \)
4

To simplify this fraction, first find the greatest common factor between them. The factors of 24 are [1, 2, 3, 4, 6, 8, 12, 24] and the factors of 68 are [1, 2, 4, 17, 34, 68]. They share 3 factors [1, 2, 4] making 4 their greatest common factor (GCF).

Next, divide both numerator and denominator by the GCF:

\( \frac{24}{68} \) = \( \frac{\frac{24}{4}}{\frac{68}{4}} \) = \( \frac{6}{17} \)

A factorial is the product of an integer and all the positive integers below it. To solve a fraction featuring factorials, expand the factorials and cancel out like numbers:

\( \frac{3!}{2!} \)
\( \frac{3 \times 2 \times 1}{2 \times 1} \)
\( \frac{3}{1} \)
3