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).

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 54 meters so the equation becomes: 2w + 2h = 54.

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

2w + 2h = 54
w + h = \( \frac{54}{2} \)
w + h = 27
w = 27 - h

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

w = 27 - 2w
3w = 27
w = \( \frac{27}{3} \)
w = 9

Since h = 2w that makes h = (2 x 9) = 18 and the area = h x w = 9 x 18 = 162 m²

First, solve for \( \left|b + 3\right| \):

\( \left|b + 3\right| \) - 2 = 1
\( \left|b + 3\right| \) = 1 + 2
\( \left|b + 3\right| \) = 3

The value inside the absolute value brackets can be either positive or negative so (b + 3) must equal + 3 or -3 for \( \left|b + 3\right| \) to equal 3:

So, b = -6 or b = 0.

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

\( \frac{27\sqrt{16}}{9\sqrt{4}} \)
\( \frac{27}{9} \) \( \sqrt{\frac{16}{4}} \)
3 \( \sqrt{4} \)

To simplify a radical, factor out the perfect squares:

\( \sqrt{45} \)
\( \sqrt{9 \times 5} \)
\( \sqrt{3^2 \times 5} \)
3\( \sqrt{5} \)