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

2\( \sqrt{112} \) - 6\( \sqrt{7} \)
2\( \sqrt{16 \times 7} \) - 6\( \sqrt{7} \)
2\( \sqrt{4^2 \times 7} \) - 6\( \sqrt{7} \)
(2)(4)\( \sqrt{7} \) - 6\( \sqrt{7} \)
8\( \sqrt{7} \) - 6\( \sqrt{7} \)

Now that the radicands are identical, you can subtract them:

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, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).

By buying two shirts, Ezra will save $46 x \( \frac{10}{100} \) = \( \frac{$46 x 10}{100} \) = \( \frac{$460}{100} \) = $4.60 on the second shirt.

So, his total cost will be
$46.00 + ($46.00 - $4.60)
$46.00 + $41.40
$87.40

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:

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

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:

9a² x 8a³
(9 x 8)a^{(2 + 3)}
72a⁵