Question

The monthly cost \(C\), in dollars, for calls from the United States to Germany on a certain wireless plan is modeled by the function \(C(x)=0.26 x+5,\) where \(x\) is the number of minutes used. (a) What is the cost if you talk on the phone for 50 minutes? (b) Suppose that your monthly bill is \(\$ 21.64 .\) How many minutes did you use the phone? (c) Suppose that you budget \(\$ 50\) per month for calls to Germany.What is the maximum number of minutes that you can talk? (d) What is the domain of \(C\) if there are 30 days in the month?

Short answer

a) $18, b) 64 minutes, c) 173 minutes, d) 0 ≤ x ≤ 43200.

Step-by-step solution

Calculate the cost for 50 minutes

To find the cost for talking on the phone for 50 minutes, substitute 50 in place of x in the function: \(C(x) = 0.26x + 5\). Thus, \(C(50) = 0.26 \cdot 50 + 5\). Then, calculate the value: \(0.26 \cdot 50 = 13\), so \(C(50) = 13 + 5 = 18\). The cost for talking on the phone for 50 minutes is \(\$18\).

Determine the minutes used for a \(21.64 bill

To find the number of minutes used when the cost is \(\$21.64\), set \(C(x) = 21.64\) and solve for x: \(21.64 = 0.26x + 5\). First, subtract 5 from both sides: \(21.64 - 5 = 0.26x\). This simplifies to \(16.64 = 0.26x\). Next, divide both sides by 0.26: \(x = \frac{16.64}{0.26} \approx 64\). Thus, 64 minutes were used.

Calculate the maximum minutes for a \)50 budget

To find the maximum number of minutes for a \(\$50\) budget, set \(C(x) = 50\) and solve for x: \(50 = 0.26x + 5\). Subtract 5 from both sides: \(50 - 5 = 0.26x\). This simplifies to \(45 = 0.26x\). Next, divide both sides by 0.26: \(x = \frac{45}{0.26} \approx 173\). Thus, the maximum number of minutes that can be talked is 173 minutes.

Determine the domain of C(x)

The domain of \(C(x)\) is determined by the possible values of x. Since x represents the number of minutes, and given that there are 30 days in a month, the maximum number of minutes possible would be \( 30 \cdot 24 \cdot 60 = 43200 \) minutes. Thus, the domain is \(0 \leq x \leq 43200\).

Key concepts

Linear equations

A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable. These types of equations create straight lines when graphed. In our exercise, the function for the cost of calls, \(C(x) = 0.26x + 5\), is a linear equation. Here, \(C(x)\) represents the cost, and \(x\) represents the number of minutes used. The coefficient, 0.26, is the cost per minute, and the constant, 5, is the fixed monthly cost. Linear equations are fundamental in algebra because they are straightforward to solve and interpret.

Function domain

The domain of a function refers to all possible input values (usually represented by \(x\)) that the function can accept. For our cost function \(C(x)\), the domain represents the number of minutes someone might talk on the phone during the month. Because there are only so many minutes in a month, the domain is restricted from 0 to 43200 minutes (30 days multiplied by 24 hours multiplied by 60 minutes). Always remember, when setting the domain for a real-world situation, consider only realistic and practical values.

Problem solving in algebra

Problem-solving in algebra involves setting up equations based on a given situation and then solving them step-by-step. For example, in our problem dealing with phone call costs, we replace the variable in the cost function with given values to find specific answers. Whether determining the cost for a certain number of minutes or finding out how long you can talk within a particular budget, each problem can be broken down into manageable steps. This methodical approach ensures clarity and accuracy.

Linear models

Linear models are used to represent relationships with a constant rate of change. These models are built using linear equations and are particularly useful in real-world situations where something changes at a consistent rate. In our exercise, the cost of phone calls is modeled linearly because the cost per minute is constant (0.26 dollars per minute), and there is a fixed monthly fee of 5 dollars. Linear models help in making predictions and decisions, like budgeting or estimating usage.