The absolute value is the positive magnitude of a particular number or variable and is indicated by two vertical lines: \(\left|-5\right| = 5\). In the case of a variable absolute value (\(\left|a\right| = 5\)) the value of a can be either positive or negative (a = -5 or a = 5).

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

\( \frac{45\sqrt{28}}{9\sqrt{4}} \)
\( \frac{45}{9} \) \( \sqrt{\frac{28}{4}} \)
5 \( \sqrt{7} \)

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

5\( \sqrt{112} \) - 2\( \sqrt{7} \)
5\( \sqrt{16 \times 7} \) - 2\( \sqrt{7} \)
5\( \sqrt{4^2 \times 7} \) - 2\( \sqrt{7} \)
(5)(4)\( \sqrt{7} \) - 2\( \sqrt{7} \)
20\( \sqrt{7} \) - 2\( \sqrt{7} \)

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

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 12 meters so the equation becomes: 2w + 2h = 12.

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

2w + 2h = 12
w + h = \( \frac{12}{2} \)
w + h = 6
w = 6 - h

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

w = 6 - 2w
3w = 6
w = \( \frac{6}{3} \)
w = 2

Since h = 2w that makes h = (2 x 2) = 4 and the area = h x w = 2 x 4 = 8 m²