Question

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? A) \(r \leq \frac{119-21}{7}\) B) \(r \leq \frac{119+21}{7}\) C) \(r \leq 119-21+7 r\) D) \(r \geq 119+21-7 r\)

Short answer

The correct inequality, which includes all possible values of \(r\), is \(r \leq \frac{119-21}{7}\).

Step-by-step solution

Calculate the available wall length for shelving units

(Subtract the length of the windows from the total wall length) The available wall length for the shelving units will be the total wall length minus the length of windows, which is: \(119 - 21 = 98\) feet.

Determine the maximum number of shelving units that can fit on the available wall length

(Divide the available wall length by the length of each shelving unit) We need to determine the maximum number of shelving units that can fit in the available wall length. Each shelving unit has a length of 7 feet. To find the maximum number of units, divide the available wall length by the length of each shelving unit: \(\frac{98}{7} = 14\)

Find the inequality that includes all possible values of \(r\)

(Compare each given inequality to Step 2 result) Now we need to find which inequality includes all possible values of \(r\). We found the maximum number of shelving units as 14 in Step 2, so the inequality should express that \(r\) cannot exceed this number. Let's analyze each option: A) \(r \leq \frac{119-21}{7}\) \\ After simplification, we get: \(r \leq \frac{98}{7}\) \\ \(r \leq 14\) B) \(r \leq \frac{119+21}{7}\) \\ After simplification, we get: \(r \leq \frac{140}{7}\) \\ \(r \leq 20\) C) \(r \leq 119-21+7 r\) \\ After simplification, we get: \(r \leq 98+7 r\) \\ This inequality is not able to represent the constraint since it will always give values of \(r\) greater than the maximum of 14. D) \(r \geq 119+21-7 r\) \\ After simplification, we get: \(r \geq 140-7 r\) \\ This inequality also does not represent the constraint on the shelving units' number. Out of the given options, Option A represents the inequality that contains all possible values of \(r\) and indicates that the number of shelving units should not exceed the maximum amount of 14. So, the correct answer is: \(r \leq \frac{119-21}{7}\)