/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 473 In the following exercises, solv... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In the following exercises, solve. Terri needs to make some pies for a fundraiser. The number of apples, \(a\), varies directly with number of pies, \(p\). It takes nine apples to make two pies. (a) Write the equation that relates \(a\) and \(p\). (b) How many apples would Terri need for six pies?

Short Answer

Expert verified
\( a = 4.5p \), 27 apples.

Step by step solution

01

- Understanding Direct Variation

The problem states that the number of apples (a) varies directly with the number of pies (p). This means the relationship between a and p can be described by the equation: \[ a = kp \] where k is the constant of proportionality.
02

- Finding the Constant of Proportionality

It is given that it takes 9 apples to make 2 pies. Using the equation from Step 1: \[ 9 = k \times 2 \] Solve for k: \[ k = \frac{9}{2} = 4.5 \]
03

- Writing the Equation Relating a and p

Now that the constant k is known, substitute k back into the equation: \[ a = 4.5p \]
04

- Calculating the Number of Apples for Six Pies

Using the equation \( a = 4.5p \), substitute p with 6 to find the required number of apples: \[ a = 4.5 \times 6 = 27 \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

proportionality
In mathematics, proportionality refers to a specific kind of relationship between two quantities. When we say that two quantities are directly proportional, it means that as one quantity increases, the other quantity increases at the same rate. In the context of our exercise, the number of apples (\(a\)) needed to make pies varies directly with the number of pies (\(p\)). This kind of relationship can be expressed using a linear equation.

In our example, we use the equation \(a = kp\), where \(k\) is known as the constant of proportionality. This equation tells us that for every unit increase in the number of pies, the number of apples used will increase by a constant factor of \(k\). By understanding and applying the concept of proportionality, we can easily predict the number of apples needed for any number of pies.
constants
In algebraic equations, constants are fixed values that do not change. These values are crucial in defining the relationship between variables. In our exercise, the constant \(k\) represents the number of apples required per pie.

To determine this constant, we used the information given: it takes 9 apples to make 2 pies. By substituting these values into our equation (\(a = kp\)), we get \(9 = k \times 2\). Solving this equation for \(k\) gives us \(k = 4.5\), meaning that each pie requires 4.5 apples. This constant value allows us to calculate the number of apples needed for any number of pies, making it a vital part of our equation.
algebraic equations
Algebraic equations are mathematical statements that show the relationship between different variables. They are often used to solve problems where one quantity depends on another. In the given problem, we derived the equation \(a = 4.5p\), which relates the number of apples (\(a\)) to the number of pies (\(p\)).

This equation is linear because it forms a straight line when graphed. The term \(4.5p\) represents a direct variation, showing that as the number of pies increases, the number of apples increases proportionally by a factor of 4.5. Understanding algebraic equations like these helps us solve real-world problems by providing a clear mathematical model of the relationships involved.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

One 12 ounce can of soda has 150 calories. If Josiah drinks the big 32 ounce size from the local mini-mart, how many calories does he get?

In the following exercises, solve. $$ \frac{56}{72}=\frac{y}{9} $$

Kayaking When Trina kayaks upriver, it takes her \(\frac{5}{3-c}\) hours to go 5 miles, where \(c\) is the speed of the river current. It takes her \(\frac{5}{3+c}\) hours to kayak 5 miles down the river. (a) Find an expression for the number of hours it would take Trina to kayak 5 miles up the river and then return by adding \(\frac{5}{3-c}+\frac{5}{3+c}\). (b) Evaluate your answer to part (a) when \(c=1\) to find the number of hours it would take Trina if the speed of the river current is 1 mile per hour.

Write an inverse variation equation to solve the following problems. Paul, a dentist, determined that the number of cavities that develops in his patient's mouth each year varies inversely to the \(\begin{array}{lll}\text { number of } & \text { minutes } & \text { spent }\end{array}\) brushing each night. His patient, Lori, had 4 cavities when brushing her teeth 30 seconds \((0.5\) minutes) each night. (a) Write the equation that relates the number of cavities to the time spent brushing. (b) How many cavities would Paul expect Lori to have if she had brushed her teeth for 2 minutes each night?

Cost of a Road Trip Jesse's car gets 30 miles per gallon of gas. (a) If Las Vegas is 285 miles away, how many gallons of gas are needed to get there and then home? (b) If gas is \(\$ 3.09\) per gallon, what is the total cost of the gas for the trip?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.