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 5 are [5, 10, 15, 20, 25, 30, 35, 40, 45, 50]. The first few multiples they share are [15, 30, 45, 60, 75] making 15 the smallest multiple 3 and 5 share.

Next, convert the fractions so each denominator equals the lowest common multiple:

\( \frac{9 x 5}{3 x 5} \) - \( \frac{4 x 3}{5 x 3} \)

\( \frac{45}{15} \) - \( \frac{12}{15} \)

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

\( \frac{45 - 12}{15} \) = \( \frac{33}{15} \) = 2\(\frac{1}{5}\)

The number of students taking German or Spanish is 13 + 13 = 26. Of that group of 26, 7 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 26 - 7 = 19 who are taking at least one language. 25 - 19 = 6 students who are not taking either language.

If the tiger has consumed 48 pounds of food in 8 days that's \( \frac{48}{8} \) = 6 pounds of food per day. The tiger needs to consume 84 - 48 = 36 more pounds of food to reach 84 pounds total. At 6 pounds of food per day that's \( \frac{36}{6} \) = 6 more days.

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 equation for this sequence is:

a_n = a_n-1 + 6

where n is the term's order in the sequence, a_n is the value of the term, and a_n-1 is the value of the term before a_n. This makes the next number:

a₆ = a₅ + 6
a₆ = 25 + 6
a₆ = 31