## Problem 12:
Simplify the expression using the distributive property of multiplication over addition:
[tex]\[ \frac{3}{8} \times \left(\frac{-4}{9}\right) + \frac{1}{12} + \frac{7}{16} \times \frac{4}{9} \][/tex]
### Step-by-Step Solution:
1. First, we'll simplify each term individually.
2. Simplify [tex]\(\frac{3}{8} \times \left(\frac{-4}{9}\right)\)[/tex]:
[tex]\[ \frac{3}{8} \times \left(\frac{-4}{9}\right) = \frac{3 \times (-4)}{8 \times 9} = \frac{-12}{72} = -\frac{1}{6} \][/tex]
This term simplifies to:
[tex]\[ -\frac{1}{6} \approx -0.16666666666666666 \][/tex]
3. The next term is already simple:
[tex]\[ \frac{1}{12} \approx 0.08333333333333333 \][/tex]
4. Simplify [tex]\(\frac{7}{16} \times \frac{4}{9}\)[/tex]:
[tex]\[ \frac{7}{16} \times \frac{4}{9} = \frac{7 \times 4}{16 \times 9} = \frac{28}{144} = \frac{7}{36} \][/tex]
This term simplifies to:
[tex]\[ \frac{7}{36} \approx 0.19444444444444442 \][/tex]
5. Finally, sum these simplified terms:
[tex]\[ -\frac{1}{6} + \frac{1}{12} + \frac{7}{36} \approx -0.16666666666666666 + 0.08333333333333333 + 0.19444444444444442 \][/tex]
Adding these values gives us:
[tex]\[ 0.11111111111111109 \][/tex]
So, the simplified result is:
[tex]\[ 0.11111111111111109 \][/tex]
## Problem 13:
A drum full of rice weighs [tex]\(\frac{55}{4} \)[/tex] kg. If the empty drum weighs [tex]\(\frac{21}{6} \)[/tex] kg, find the weight of rice in the drum.
### Step-by-Step Solution:
1. First, convert the weights into decimal form (if necessary):
[tex]\[ \frac{55}{4} = 13.75 \, \text{kg} \][/tex]
[tex]\[ \frac{21}{6} = 3.5 \, \text{kg} \][/tex]
2. Find the weight of the rice by subtracting the weight of the empty drum from the weight of the drum full of rice:
[tex]\[ 13.75 \, \text{kg} - 3.5 \, \text{kg} = 10.25 \, \text{kg} \][/tex]
So, the weight of the rice in the drum is:
[tex]\[ 10.25 \, \text{kg} \][/tex]
