If the recommended adult dosage for a drug is D (in ml), then to determine the appropriate dosage p for a puppy of age a (in weeks), veteranians use the equation
p=0.0417D (a + 1)
Suppose the dosage for an adult dog is 100 ml.
a. Sketch the graph for this function.
b. Find the slope of the graph of p. (Hint: the function for a linear line is y=mx+b, where y=response, x=variable, m=slope, and b=intercept).
c. What is the dosage for a 10-week old puppy?
d. At what age is a dosage of 54.21 ml is used?

Let's break down the problem step by step:

### Given Equation:
The equation provided for the dosage \( p \) for a puppy of age \( a \) (in weeks) is:
\[
p = 0.0417D \times (a + 1)
\]
where \( D \) is the recommended adult dosage (in ml).

The adult dosage \( D \) is given as \( 100 \) ml.

### Substituting \( D = 100 \) ml:
Substitute \( D = 100 \) ml into the equation:
\[
p = 0.0417 \times 100 \times (a + 1)
\]
Simplifying this, we get:
\[
p = 4.17 \times (a + 1)
\]

Now let's solve the sub-questions:

### a. Sketch the Graph for this Function

The function \( p = 4.17 \times (a + 1) \) is a linear function in the form \( y = mx + b \) where:
- \( y \) is the response variable \( p \) (dosage in ml),
- \( x \) is the independent variable \( a \) (age in weeks),
- \( m = 4.17 \) is the slope,
- \( b = 4.17 \) is the y-intercept (since when \( a = 0 \), \( p = 4.17 \times 1 = 4.17 \) ml).

### b. Find the Slope of the Graph of \( p \)

From the function \( p = 4.17(a + 1) \), the slope \( m \) is the coefficient of \( a \), which is \( 4.17 \).

So, the slope \( m \) of the graph is **4.17**.

### c. What is the Dosage for a 10-Week-Old Puppy?

To find the dosage \( p \) for a 10-week-old puppy, substitute \( a = 10 \) into the equation:
\[
p = 4.17 \times (10 + 1) = 4.17 \times 11
\]
\[
p = 45.87 \text{ ml}
\]

So, the dosage for a 10-week-old puppy is **45.87 ml**.

### d. At What Age Is a Dosage of 54.21 ml Used?

To find the age \( a \) when the dosage is \( 54.21 \) ml, we set \( p = 54.21 \) and solve for \( a \):
\[
54.21 = 4.17 \times (a + 1)
\]
Divide both sides by 4.17:
\[
\frac{54.21}{4.17} = a + 1
\]
\[
a + 1 = 13
\]
\[
a = 12 \text{ weeks}
\]

So, a dosage of 54.21 ml is used at **12 weeks** of age.

---

### Summary of Answers:
1. **Graph**: The graph is a straight line with a slope of 4.17 and a y-intercept at 4.17 ml.
2. **Slope**: The slope of the graph is 4.17.
3. **Dosage for a 10-week-old puppy**: 45.87 ml.
4. **Age for 54.21 ml dosage**: 12 weeks.

Would you like a detailed explanation of any part of this solution, or do you have any further questions?

Here are 5 related questions you might find interesting:
1. How would the graph change if the adult dosage \( D \) was different?
2. What would the dosage be for a 5-week-old puppy?
3. What is the significance of the slope in the context of this problem?
4. How would you find the age of a puppy if the dosage required is known?
5. What does the y-intercept represent in the context of this function?

**Tip**: In linear functions, the slope determines the rate of change of the dependent variable (like dosage) with respect to the independent variable (like age).

Explore dosage calculation for puppies using linear functions and slope. Learn how to graph the function p = 4.17 * (a + 1), find slope, determine dosages for specific ages, and interpret the graph. Suitable for high school students.

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