The equation for this sequence is:

a_n = a_n-1 + 2

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 + 2
a₆ = 9 + 2
a₆ = 11

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:

\( \frac{2 x 7}{3 x 7} \) - \( \frac{4 x 3}{7 x 3} \)

\( \frac{14}{21} \) - \( \frac{12}{21} \)

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

\( \frac{14 - 12}{21} \) = \( \frac{2}{21} \) = \(\frac{2}{21}\)

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.

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:

\( \frac{5c^8}{7c^3} \)
\( \frac{5}{7} \) c^{(8 - 3)}
\(\frac{5}{7}\)c⁵

To simplify a radical, factor out the perfect squares:

\( \sqrt{63} \)
\( \sqrt{9 \times 7} \)
\( \sqrt{3^2 \times 7} \)
3\( \sqrt{7} \)