Question

use the area model to multiply \\(\frac{1}{3}(-6 - 9c)\\).

first, find the partial products. write numbers as integers, decimals, or sim
improper fractions.

\\(-6\\) \\(-9c\\)
\\(\frac{1}{3}\\) image of two rectangles, one with -6 and a blue box, one with -9c and a white box

Explanation:

Step1: Multiply $\frac{1}{3}$ by -6

To find the partial product of $\frac{1}{3}$ and -6, we use the multiplication rule for fractions and integers. The formula is $\frac{1}{3} \times (-6)$. We can calculate this as $\frac{-6}{3} = -2$.

Step2: Multiply $\frac{1}{3}$ by -9c

Next, we find the partial product of $\frac{1}{3}$ and -9c. Using the multiplication rule for fractions and terms with variables, the formula is $\frac{1}{3} \times (-9c)$. Calculating this, we have $\frac{-9c}{3} = -3c$.

Step3: Combine the partial products

Now, we combine the two partial products we found. The original expression $\frac{1}{3}(-6 - 9c)$ is equal to the sum of the two partial products, so we add -2 and -3c together.

Answer:

The partial products are -2 (from $\frac{1}{3} \times -6$) and -3c (from $\frac{1}{3} \times -9c$), and the final result of $\frac{1}{3}(-6 - 9c)$ is $-2 - 3c$.