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APPLICATION Let \(x\) represent total fat in grams, and let \(y\) represent saturated fat in grams. Use the model \(y=10+0.5(x-28)\) to predict a. The number of saturated fat grams for a hamburger with a total of 32 grams of fat. b. The total number of fat grams for a hamburger with 15 grams of saturated fat.

Short Answer

Expert verified
For part a, the hamburger has 12 grams of saturated fat. For part b, the hamburger has 38 grams of total fat.

Step by step solution

01

Understand the Given Equation

The provided equation is \(y = 10 + 0.5(x - 28)\), where \(x\) is the total fat in grams and \(y\) is the saturated fat in grams. This means we can use the equation to find one value given the other.
02

Substitute and Solve for Saturated Fat (Part a)

For part a, we are given \(x = 32\) grams (total fat) and need to find \(y\) (saturated fat). Substitute \(x = 32\) into the equation:\(y = 10 + 0.5(32 - 28)\)Calculate the expression inside the parentheses: \(32 - 28 = 4\).Now multiply by 0.5: \(0.5 \times 4 = 2\).Add to 10: \(10 + 2 = 12\).Therefore, \(y = 12\) grams.
03

Rearrange the Equation for Total Fat (Part b)

For part b, we need to find \(x\) when \(y = 15\) grams. Start by rearranging the equation to solve for \(x\):\(y = 10 + 0.5(x - 28)\)Subtract 10 from both sides:\(5 = 0.5(x - 28)\)Divide each side by 0.5:\(10 = x - 28\)Add 28 to both sides to solve for \(x\):\(x = 38\)Thus, there are 38 grams of total fat.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mathematical Modeling
Mathematical modeling involves creating equations to represent real-world situations. This helps us solve problems by translating them into mathematical language. For the exercise at hand, the equation \(y = 10 + 0.5(x - 28)\) models the relationship between total fat \(x\) and saturated fat \(y\) in a hamburger. Using this equation, you're able to predict one variable if the other is known.

The equation is built on the premise that there are certain fixed and variable components affecting the saturated fat content. Here, 10 represents a baseline saturated fat content, while \(0.5(x - 28)\) accounts for additional fat based on the total fat content \(x\).
  • This approach simplifies the complex nature of food composition into a manageable formula.
  • By substituting different values, you can analyze how changes affect the saturated fat content.
Mathematical modeling is essential in various fields, as it provides a structured way to solve practical problems.
Problem Solving
Problem solving in mathematics often involves breaking down a problem into smaller, manageable steps. In the current exercise, each part provides an opportunity to practice this valuable skill. Let’s examine how this is done:

**Part a:** We aimed to find the saturated fat \(y\) for a hamburger with total fat \(x = 32\). Substituting \(x\) in the equation, we calculated \(y\) step by step:
  • Identify the given values and what you need to find.
  • Substitute the known values into the equation.
  • Perform arithmetic operations sequentially to solve for the desired variable.
This logical approach ensures that each step builds upon the last, leading to the solution.

**Part b:** To find total fat \(x\) given \(y = 15\), we reversed the operation, which involved rearranging the equation:
  • Isolate the variable of interest (\(x\)).
  • Perform inverse operations to reverse the equation flow.
  • Solve the simplified equation for \(x\).
Breaking the problem into these steps makes complex questions more approachable and less intimidating for learners.
Substitution Method
The substitution method is a powerful technique in algebra for solving equations. It involves replacing one variable with a given value to find another variable. In this exercise, substitution helped us easily compute saturated fat \(y\) and total fat \(x\).

For part a, where we found \(y\), the process involved substituting the given \(x = 32\) into the equation \(y = 10 + 0.5(x - 28)\). This straightforward substitution transforms the equation into a simple arithmetic problem.
  • Initially plug the known value into the equation.
  • Simplify the equation step by step to find \(y\).
  • Conclude with the solution that meets the real-world condition of the problem.
In part b, we did the reverse with \(y = 15\) to find \(x\). Again, the substitution method shows its versatility:
  • Set \(y\) to 15 in the equation.
  • Rearrange to solve for \(x\).
  • Follow inverse operations to deduce the value of \(x\).
This technique is essential for students as it offers a clear path from an equation to a solution, making algebraic challenges easier to navigate.

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