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

\( \frac{10\sqrt{15}}{2\sqrt{3}} \)
\( \frac{10}{2} \) \( \sqrt{\frac{15}{3}} \)
5 \( \sqrt{5} \)

A factorial is the product of an integer and all the positive integers below it. To solve a fraction featuring factorials, expand the factorials and cancel out like numbers:

\( \frac{3!}{4!} \)
\( \frac{3 \times 2 \times 1}{4 \times 3 \times 2 \times 1} \)
\( \frac{1}{4} \)
\( \frac{1}{4} \)

The equation for this sequence is:

a_n = a_n-1 + 4(n - 1)

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 + 4(6 - 1)
a₆ = 41 + 4(5)
a₆ = 61

To add these radicals together their radicands must be the same:

4\( \sqrt{27} \) + 9\( \sqrt{3} \)
4\( \sqrt{9 \times 3} \) + 9\( \sqrt{3} \)
4\( \sqrt{3^2 \times 3} \) + 9\( \sqrt{3} \)
(4)(3)\( \sqrt{3} \) + 9\( \sqrt{3} \)
12\( \sqrt{3} \) + 9\( \sqrt{3} \)

Now that the radicands are identical, you can add them together:

To subtract these fractions, first find the lowest common multiple of their denominators. The first few multiples of 8 are [8, 16, 24, 32, 40, 48, 56, 64, 72, 80] and the first few multiples of 10 are [10, 20, 30, 40, 50, 60, 70, 80, 90]. The first few multiples they share are [40, 80] making 40 the smallest multiple 8 and 10 share.

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

\( \frac{5 x 5}{8 x 5} \) - \( \frac{6 x 4}{10 x 4} \)

\( \frac{25}{40} \) - \( \frac{24}{40} \)

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

\( \frac{25 - 24}{40} \) = \( \frac{1}{40} \) = \(\frac{1}{40}\)