To add or subtract terms with exponents, both the base and the exponent must be the same. In this case they are so add the coefficients and retain the base and exponent:

5y⁴ + 6y⁴
(5 + 6)y⁴
11y⁴

The equation for this sequence is:

a_n = a_n-1 + 5

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 + 5
a₆ = 21 + 5
a₆ = 26

First, solve for \( \left|b + 2\right| \):

\( \left|b + 2\right| \) + 6 = 2
\( \left|b + 2\right| \) = 2 - 6
\( \left|b + 2\right| \) = -4

The value inside the absolute value brackets can be either positive or negative so (b + 2) must equal - 4 or --4 for \( \left|b + 2\right| \) to equal -4:

So, b = 2 or b = -6.

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

8\( \sqrt{32} \) + 3\( \sqrt{2} \)
8\( \sqrt{16 \times 2} \) + 3\( \sqrt{2} \)
8\( \sqrt{4^2 \times 2} \) + 3\( \sqrt{2} \)
(8)(4)\( \sqrt{2} \) + 3\( \sqrt{2} \)
32\( \sqrt{2} \) + 3\( \sqrt{2} \)

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

To simplify a radical, factor out the perfect squares:

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