To fill a 20 gallon tank exactly halfway you'll need 10 gallons of fuel. Each fuel can holds 2\(\frac{1}{2}\) gallons so:

cans = \( \frac{10 \text{ gallons}}{2\frac{1}{2} \text{ gallons}} \) = 4

Use PEMDAS (Parentheses, Exponents, Multipy/Divide, Add/Subtract):

2 + (3 + 2) ÷ 2 x 3 - 4²
P: 2 + (5) ÷ 2 x 3 - 4²
E: 2 + 5 ÷ 2 x 3 - 16
MD: 2 + \( \frac{5}{2} \) x 3 - 16
MD: 2 + \( \frac{15}{2} \) - 16
AS: \( \frac{4}{2} \) + \( \frac{15}{2} \) - 16
AS: \( \frac{19}{2} \) - 16
AS: \( \frac{19 - 32}{2} \)
\( \frac{-13}{2} \)
-6\(\frac{1}{2}\)

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 add these radicals together their radicands must be the same:

2\( \sqrt{28} \) + 2\( \sqrt{7} \)
2\( \sqrt{4 \times 7} \) + 2\( \sqrt{7} \)
2\( \sqrt{2^2 \times 7} \) + 2\( \sqrt{7} \)
(2)(2)\( \sqrt{7} \) + 2\( \sqrt{7} \)
4\( \sqrt{7} \) + 2\( \sqrt{7} \)

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

1