The number of students taking German or Spanish is 14 + 6 = 20. Of that group of 20, 2 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 20 - 2 = 18 who are taking at least one language. 27 - 18 = 9 students who are not taking either language.

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:

\( \frac{9 x 7}{3 x 7} \) - \( \frac{3 x 3}{7 x 3} \)

\( \frac{63}{21} \) - \( \frac{9}{21} \)

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

\( \frac{63 - 9}{21} \) = \( \frac{54}{21} \) = 2\(\frac{4}{7}\)

A factorial has the form n! and is the product of the integer (n) and all the positive integers below it. For example, 5! = 5 x 4 x 3 x 2 x 1 = 120.

The amount of flour you need is (3 - 1\(\frac{1}{2}\)) cups. Rewrite the quantities so they share a common denominator and subtract:

(\( \frac{24}{8} \) - \( \frac{12}{8} \)) cups
\( \frac{12}{8} \) cups
1\(\frac{1}{2}\) cups

To divide terms with exponents, the base of both exponents must be the same. In this case they are so divide the coefficients and subtract the exponents:

\( \frac{7x^8}{3x^4} \)
\( \frac{7}{3} \) x^{(8 - 4)}
2\(\frac{1}{3}\)x⁴