The number of students taking German or Spanish is 12 + 5 = 17. Of that group of 17, 4 are taking both languages so they've been counted twice (once in the German group and once in the Spanish group). Subtracting them out leaves 17 - 4 = 13 who are taking at least one language. 22 - 13 = 9 students who are not taking either language.

A rational number (or fraction) is represented as a ratio between two integers, a and b, and has the form \({a \over b}\) where a is the numerator and b is the denominator. An improper fraction (\({5 \over 3} \)) has a numerator with a greater absolute value than the denominator and can be converted into a mixed number (\(1 {2 \over 3} \)) which has a whole number part and a fractional part.

First, solve for \( \left|a + 4\right| \):

\( \left|a + 4\right| \) + 7 = -8
\( \left|a + 4\right| \) = -8 - 7
\( \left|a + 4\right| \) = -15

The value inside the absolute value brackets can be either positive or negative so (a + 4) must equal - 15 or --15 for \( \left|a + 4\right| \) to equal -15:

So, a = 11 or a = -19.

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

2 + (2 + 4) ÷ 4 x 2 - 4²
P: 2 + (6) ÷ 4 x 2 - 4²
E: 2 + 6 ÷ 4 x 2 - 16
MD: 2 + \( \frac{6}{4} \) x 2 - 16
MD: 2 + \( \frac{12}{4} \) - 16
AS: \( \frac{8}{4} \) + \( \frac{12}{4} \) - 16
AS: \( \frac{20}{4} \) - 16
AS: \( \frac{20 - 64}{4} \)
\( \frac{-44}{4} \)
-11

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 6 meters so the equation becomes: 2w + 2h = 6.

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

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

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

w = 3 - 2w
3w = 3
w = \( \frac{3}{3} \)
w = 1

Since h = 2w that makes h = (2 x 1) = 2 and the area = h x w = 1 x 2 = 2 m²