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

6\( \sqrt{6} \) x 2\( \sqrt{4} \)
(6 x 2)\( \sqrt{6 \times 4} \)
12\( \sqrt{24} \)

Now we need to simplify the radical:

12\( \sqrt{24} \)
12\( \sqrt{6 \times 4} \)
12\( \sqrt{6 \times 2^2} \)
(12)(2)\( \sqrt{6} \)
24\( \sqrt{6} \)

If a single cook can bake 2 large cakes per hour and the kitchen is available for 2 hours, a single cook can bake 2 x 2 = 4 large cakes during that time. 37 large cakes are needed for the party so \( \frac{37}{4} \) = 9\(\frac{1}{4}\) cooks are needed to bake the required number of large cakes.

If a single cook can bake 15 small cakes per hour and the kitchen is available for 2 hours, a single cook can bake 15 x 2 = 30 small cakes during that time. 390 small cakes are needed for the party so \( \frac{390}{30} \) = 13 cooks are needed to bake the required number of small cakes.

Because you can't employ a fractional cook, round the number of cooks needed for each type of cake up to the next whole number resulting in 10 + 13 = 23 cooks.

Average speed in miles per hour is the number of miles traveled divided by the number of hours:

Solving for distance:

distance = \( \text{speed} \times \text{time} \)
distance = \( 50mph \times 6h \)
300 miles

To subtract these fractions, first find the lowest common multiple of their denominators. The first few multiples of 6 are [6, 12, 18, 24, 30, 36, 42, 48, 54, 60] and the first few multiples of 10 are [10, 20, 30, 40, 50, 60, 70, 80, 90]. The first few multiples they share are [30, 60, 90] making 30 the smallest multiple 6 and 10 share.

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

\( \frac{2 x 5}{6 x 5} \) - \( \frac{6 x 3}{10 x 3} \)

\( \frac{10}{30} \) - \( \frac{18}{30} \)

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

\( \frac{10 - 18}{30} \) = \( \frac{-8}{30} \) = -\(\frac{4}{15}\)

The absolute value is the positive magnitude of a particular number or variable and is indicated by two vertical lines: \(\left|-5\right| = 5\). In the case of a variable absolute value (\(\left|a\right| = 5\)) the value of a can be either positive or negative (a = -5 or a = 5).