Problem 77 - Algebra A Combined Function

Question

The dosage in milligrams \(D\) of Ivermectin, a heartworm preventive, for a dog who weighs \(x\) pounds is given by $$ D(x)=\frac{136}{25} x $$ Use this function to answer Exercises 77 and \(78 .\) Find the proper dosage for a dog that weighs 30 pounds.

Step-by-Step Solution

Answer

The proper dosage for a 30-pound dog is 163.2 milligrams.

Step 1: Identify the Function

The problem provides a function for dosage, which is given by \( D(x) = \frac{136}{25} x \). This function allows us to calculate the dosage based on the weight of the dog in pounds.

Step 2: Substitute the Dog's Weight into the Function

We need to find the dosage for a dog that weighs 30 pounds. Substitute \( x = 30 \) into the function: \( D(30) = \frac{136}{25} \times 30 \).

Step 3: Compute the Multiplication

First, calculate the multiplication inside the function: \( \frac{136}{25} \times 30 = \frac{136 \times 30}{25} \).

Step 4: Perform the Multiplication

Carry out the multiplication in the numerator: \(136 \times 30 = 4080\). Now our expression is \( \frac{4080}{25} \).

Step 5: Divide the Result

Finally, divide to find the dosage: \( \frac{4080}{25} = 163.2 \).

Step 6: Determine the Final Dosage

The dosage for a 30-pound dog is 163.2 milligrams of Ivermectin.

Key Concepts

Dosage Calculation

Dosage calculation is a critical aspect when administering medication, especially in veterinary medicine. It ensures that each animal receives the proper amount of medication based on its specific requirements. In this case, for heartworm prevention in dogs, the amount of Ivermectin is determined by weight, using the given function: \[ D(x) = \frac{136}{25} x \] Here are a few key points to keep in mind when performing dosage calculations:

• Always use the correct formula provided for the specific medication or treatment.
• Ensure the weight is measured accurately. In this example, the dog weighs 30 pounds.
• Substitute the weight into the function as instructed, to find the exact dosage.

The final step involves arithmetic operations such as multiplication and division. In the example, substituting 30 into the function gives \( D(30) = \frac{136}{25} \times 30 = 163.2 \), indicating that a 30-pound dog should receive 163.2 mg of Ivermectin.

Linear Functions

Linear functions are a fundamental concept in algebra, representing relationships between two variables. They are characterized by a constant rate of change and are graphically represented as straight lines. The general form of a linear function is: \[ y = mx + c \] Where \( m \) is the slope and \( c \) is the y-intercept. In the given exercise, the function \( D(x) = \frac{136}{25} x \) can be viewed as a linear function. Here, the slope, \( \frac{136}{25} \), determines the rate at which the dosage increases with weight. Some features of linear functions include:

• The slope \( m \) indicates how steeply the line rises or falls. In this problem, a positive slope means that as the dog's weight increases, so does the dosage.
• The function passes through the origin \((0,0)\) if there is no constant term added, which means when \( x = 0 \), the function value is 0.

Understanding these properties helps in accurately interpreting and using functions to solve real-world problems, like calculating medication dosages.

Applied Algebra

Applied algebra bridges the gap between theoretical math and real-world applications. It allows us to solve practical problems by modeling situations with mathematical equations. In this exercise, the function \( D(x) = \frac{136}{25} x \) is an excellent example of applied algebra. Let’s delve deeper into why this is applied algebra:

• It translates a real-world scenario (dog medication) into a mathematical model (the linear function).
• It uses algebraic manipulation to perform calculations. Here, substituting the dog's weight into the function is an algebraic step.
• The final computation is another example of applied algebra, transforming the solution from a number in equations to an actionable dosage in mg.

This connection between algebra and real life helps students understand the relevance and importance of algebra in everyday tasks. By learning applied algebra, students can not only interpret mathematical expressions but also use them in practical, day-to-day scenarios, enhancing both learning and problem-solving skills.