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

6\( \sqrt{8} \) + 4\( \sqrt{2} \)
6\( \sqrt{4 \times 2} \) + 4\( \sqrt{2} \)
6\( \sqrt{2^2 \times 2} \) + 4\( \sqrt{2} \)
(6)(2)\( \sqrt{2} \) + 4\( \sqrt{2} \)
12\( \sqrt{2} \) + 4\( \sqrt{2} \)

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

The distributive property for division helps in solving expressions like \({b + c \over a}\). It specifies that the result of dividing a fraction with multiple terms in the numerator and one term in the denominator can be obtained by dividing each term individually and then totaling the results: \({b + c \over a} = {b \over a} + {c \over a}\). For example, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).

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

-3y⁴ x 9y⁵
(-3 x 9)y^{(4 + 5)}
-27y⁹

The equation for this sequence is:

a_n = a_n-1 + 2(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 + 2(6 - 1)
a₆ = 21 + 2(5)
a₆ = 31

The equation for this sequence is:

a_n = a_n-1 + 4

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
a₆ = 17 + 4
a₆ = 21