Problem 129 - College Algebra Essentials

Question

Exercises \(129-131\) will help you prepare for the material covered in the next section. The function \(C(t)=20+0.40(t-60)\) describes the monthly cost, \(C(t),\) in dollars, for a cellphone plan for \(t\) calling minutes, where \(t>60 .\) Find and interpret \(C(100)\)

Step-by-Step Solution

Answer

The cost of the cellphone plan for 100 calling minutes is $36 per month.

Step 1: Understand the Problem

The problem provides a linear function \(C(t)=20+0.40(t-60)\) that calculates the cost of a cellphone plan based on usage in minutes \(t\), where \(t>60\). We are asked to find the cost for \(t=100\) minutes.

Step 2: Substitute the given value

We replace \(t\) with \(100\) in the cost function: \(C(100) = 20+0.40*(100-60)\)

Step 3: Evaluate the Expression

Calculate the expression inside the parenthesis first: \(100-60=40\), then multiply by 0.40: \(0.40*40=16\). Lastly, add 20: \(20+16=36\). Thus, \(C(100) = 36\)

Step 4: Interpret the Result

The result represents the cost in dollars for a plan with 100 calling minutes. Thus, a plan with 100 calling minutes would cost $36 per month.

Key Concepts

Cost Interpretation

When you're dealing with a function like \(C(t)=20+0.40(t-60)\), you might wonder what each part of this equation really means, especially when it comes to costing you money. Here, \(C(t)\) symbolizes the total monthly cost of your cellphone plan, which is calculated in dollars. The equation is split into two main parts: a fixed base cost and a variable cost.

• The number 20 in the equation is a constant, meaning no matter how many minutes you talk, this fee remains at $20. This could be seen as the basic monthly service fee.
• The term \(0.40(t-60)\) is the variable cost, measured in dollars per additional minute, applicable only when you talk for more than 60 minutes.

Breaking it down, if you talk exactly 60 minutes, you pay only the base fee. As you talk more, each extra minute above 60 costs you 40 cents. This kind of setup helps balance fixed costs while allowing for variability based on usage, making it an example of a linear cost function.

Substitution in Functions

Substitution in functions might sound technical, but it essentially means replacing a variable with a given value. For our cellphone cost problem, we're asked to find \(C(100)\). This means we're substituting \(t\) with 100. Here's how this works:First, write down the original function:

• \(C(t)=20+0.40(t-60)\).

Then, plug in 100 for \(t\):

• \(C(100)=20+0.40(100-60)\).

When you substitute, ensure you perform all calculations inside the parentheses first and then move outward. So in this case, we deduct 60 from 100 before proceeding with further calculations.Substitution is a versatile and powerful tool in algebra, enabling you to compute precise outcomes for given inputs in a function. It helps you see how changes in inputs affect outcomes, which is very helpful in practical scenarios like budgeting for cellphone usage.

Evaluating Expressions

Evaluating expressions involves simplifying them to find a result. In our example, once we've substituted 100 into the function for \(t\), we have the expression \(20+0.40(100-60)\). Let's look at the steps:1. **Calculate inside the parentheses**: Start with \(100-60\), which equals 40.2. **Multiply by the coefficient**: Next, take 40 and multiply it by 0.40, which gives 16.3. **Add the constant**: Finally, add this result to the base number of 20, yielding 36.Each step simplifies the expression further until arriving at the simplest form. The final value, here 36, represents the total cost in dollars when the calling time is 100 minutes.By focusing on these sequential steps, you ensure clarity and accuracy in solving expressions. This technique not only applies to math problems but is also useful for logical problem-solving in other fields. Regular practice in evaluating expressions strengthens your ability to maneuver complex equations with ease.