Sai is ordering new shelving units for his store. Each unit is 7 feet in length and extends from floor to ceiling. The total length of the walls in Sai's store is 119 feet, which includes a length of 21 feet of windows along the walls . IF the shelving units cannot be placed in front of the windows, which of the following inequalities includes all possible values of r, the number of shelving units that Sai could use?

Options

A. `r le(119-21)/7`
B. `r ge(119+21)/7`
C. `rle119-21+7r`
D. `rge119+21-7r`

Answer

To solve the problem, we need to determine the maximum number of shelving units (r) that Sai can use in his store, given the constraints of the total wall length and the windows.

### Step-by-Step Solution:

1. **Identify the total length of the walls**:
The total length of the walls in Sai's store is given as 119 feet.

2. **Subtract the length of the windows**:
Since there is a length of 21 feet of windows where shelving units cannot be placed, we need to subtract this from the total wall length.
\[ \text{Available length for shelving} = 119 \text{ feet} - 21 \text{ feet} = 98 \text{ feet} \]

3. **Determine the length occupied by each shelving unit**:
Each shelving unit occupies 7 feet of space.

4. **Set up the inequality**:
If r is the number of shelving units, then the total length occupied by r shelving units is \(7r\). The total length occupied must be less than or equal to the available length for shelving.
\[ 7r \leq 98 \]

5. **Solve for r**:
To isolate r, divide both sides of the inequality by 7. Since 7 is a positive number, the direction of the inequality remains the same.
\[ r \leq \frac{98}{7} \]
\[ r \leq 14 \]

6. **Write the final inequality**:
The inequality that includes all possible values of r is:
\[ r \leq 14 \]

### Conclusion:
The inequality that includes all possible values of r, the number of shelving units that Sai could use, is \( r \leq 14 \).