Select The Correct Answer From Each Drop-down Menu.Let D ( T D(t D ( T ] Be The Total Number Of Miles Joanna Has Cycled, And Let T T T Represent The Number Of Hours Before Stopping For A Break During Her Ride. D ( T ) = 12 T + 20 D(t) = 12t + 20 D ( T ) = 12 T + 20 So,

Introduction

In mathematics, linear equations are used to model various real-world scenarios. One such scenario is Joanna's cycling ride, where the total number of miles she cycles is represented by the function $d(t) = 12t + 20$. In this article, we will delve into the world of linear equations and explore how they can be used to solve problems in mathematics and real-life situations.

What are Linear Equations?

A linear equation is an equation in which the highest power of the variable(s) is 1. In other words, it is an equation that can be written in the form $ax + b = c$, where $a$, $b$, and $c$ are constants, and $x$ is the variable. Linear equations can be used to model a wide range of real-world scenarios, including the distance traveled by an object, the cost of an item, and the temperature of a substance.

Joanna's Cycling Ride

In the given problem, Joanna's cycling ride is represented by the function $d(t) = 12t + 20$. Here, $d(t)$ represents the total number of miles Joanna has cycled, and $t$ represents the number of hours before stopping for a break during her ride. The function $d(t) = 12t + 20$ indicates that for every hour Joanna cycles, she travels 12 miles, and she starts with a distance of 20 miles.

Solving Linear Equations

To solve linear equations, we need to isolate the variable. In the case of the function $d(t) = 12t + 20$, we can isolate the variable $t$ by subtracting 20 from both sides of the equation and then dividing both sides by 12. This gives us the equation $t = \frac{d(t) - 20}{12}$.

Interpreting the Results

Now that we have isolated the variable $t$, we can interpret the results. The equation $t = \frac{d(t) - 20}{12}$ tells us that for every 12 miles Joanna cycles, she has been cycling for 1 hour. This means that if Joanna has cycled a total of 60 miles, she has been cycling for 5 hours.

Real-World Applications

Linear equations have numerous real-world applications. In the case of Joanna's cycling ride, the function $d(t) = 12t + 20$ can be used to model the distance she travels over time. This can be useful in a variety of situations, such as:

• Planning a cycling trip: By using the function $d(t) = 12t + 20$, Joanna can plan her cycling trip and determine how long it will take her to reach her destination.
• Calculating fuel costs: If Joanna is using a bike with a fuel tank, she can use the function $d(t) = 12t + 20$ to calculate the fuel costs of her trip.
• Determining the time of arrival: By using the function $d(t) = 12t + 20$, Joanna can determine the time of arrival at her destination.

Conclusion

In conclusion, linear equations are a powerful tool for modeling real-world scenarios. By using the function $d(t) = 12t + 20$, we can solve problems in mathematics and real-life situations. Whether it's planning a cycling trip, calculating fuel costs, or determining the time of arrival, linear equations can be used to model a wide range of scenarios.

Example Problems

Problem 1

Joanna has cycled a total of 60 miles. How many hours has she been cycling?

Solution

To solve this problem, we can use the function $d(t) = 12t + 20$. We know that Joanna has cycled a total of 60 miles, so we can set up the equation $60 = 12t + 20$. Subtracting 20 from both sides gives us $40 = 12t$. Dividing both sides by 12 gives us $t = \frac{40}{12} = \frac{10}{3}$. Therefore, Joanna has been cycling for $\frac{10}{3}$ hours.

Problem 2

Joanna starts with a distance of 20 miles and cycles at a rate of 12 miles per hour. How many hours will it take her to reach a distance of 60 miles?

Solution

To solve this problem, we can use the function $d(t) = 12t + 20$. We know that Joanna starts with a distance of 20 miles and cycles at a rate of 12 miles per hour. We can set up the equation $60 = 12t + 20$. Subtracting 20 from both sides gives us $40 = 12t$. Dividing both sides by 12 gives us $t = \frac{40}{12} = \frac{10}{3}$. Therefore, it will take Joanna $\frac{10}{3}$ hours to reach a distance of 60 miles.

Final Thoughts

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Introduction

In our previous article, we explored the concept of linear equations and how they can be used to model real-world scenarios. In this article, we will answer some frequently asked questions about linear equations and provide additional examples to help illustrate the concept.

Q: What is a linear equation?

A: A linear equation is an equation in which the highest power of the variable(s) is 1. In other words, it is an equation that can be written in the form $ax + b = c$, where $a$, $b$, and $c$ are constants, and $x$ is the variable.

Q: How do I solve a linear equation?

A: To solve a linear equation, you need to isolate the variable. This can be done by adding, subtracting, multiplying, or dividing both sides of the equation by the same value.

Q: What is the difference between a linear equation and a quadratic equation?

A: A linear equation is an equation in which the highest power of the variable(s) is 1, while a quadratic equation is an equation in which the highest power of the variable(s) is 2.

Q: Can linear equations be used to model real-world scenarios?

A: Yes, linear equations can be used to model a wide range of real-world scenarios, including the distance traveled by an object, the cost of an item, and the temperature of a substance.

Q: How do I determine the slope of a linear equation?

A: The slope of a linear equation is the ratio of the change in the dependent variable to the change in the independent variable. It can be calculated using the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Q: What is the y-intercept of a linear equation?

A: The y-intercept of a linear equation is the point at which the line intersects the y-axis. It can be calculated using the formula $b = c - ax$.

Q: Can linear equations be used to solve problems in mathematics and real-life situations?

A: Yes, linear equations can be used to solve problems in mathematics and real-life situations. They can be used to model a wide range of scenarios, including the distance traveled by an object, the cost of an item, and the temperature of a substance.

Q: How do I graph a linear equation?

A: To graph a linear equation, you need to plot two points on the graph and then draw a line through them. The two points can be calculated using the x and y intercepts of the equation.

Q: What is the equation of a line in slope-intercept form?

A: The equation of a line in slope-intercept form is $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

Q: Can linear equations be used to solve problems in physics and engineering?

A: Yes, linear equations can be used to solve problems in physics and engineering. They can be used to model a wide range of scenarios, including the motion of objects, the flow of fluids, and the behavior of electrical circuits.

Q: How do I use linear equations to solve problems in finance?

A: Linear equations can be used to solve problems in finance by modeling the cost of an item, the return on investment, and the interest rate.

Q: Can linear equations be used to solve problems in computer science?

A: Yes, linear equations can be used to solve problems in computer science by modeling the behavior of algorithms, the performance of computer systems, and the flow of data.

Conclusion

In conclusion, linear equations are a powerful tool for modeling real-world scenarios. They can be used to solve problems in mathematics, physics, engineering, finance, and computer science. By understanding the concept of linear equations and how to solve them, you can apply this knowledge to a wide range of scenarios and make informed decisions.

Example Problems

Problem 1

Solve the linear equation $2x + 5 = 11$.

Solution

To solve this equation, we need to isolate the variable $x$. We can do this by subtracting 5 from both sides of the equation and then dividing both sides by 2. This gives us the equation $x = \frac{11 - 5}{2} = \frac{6}{2} = 3$.

Problem 2

Graph the linear equation $y = 2x + 3$.

Solution

To graph this equation, we need to plot two points on the graph and then draw a line through them. We can calculate the x and y intercepts of the equation using the formula $y = mx + b$. The x-intercept is the point at which the line intersects the x-axis, and the y-intercept is the point at which the line intersects the y-axis. In this case, the x-intercept is $x = -\frac{3}{2}$ and the y-intercept is $y = 3$.

Problem 3

Solve the linear equation $x + 2 = 7$.

Solution

To solve this equation, we need to isolate the variable $x$. We can do this by subtracting 2 from both sides of the equation and then dividing both sides by 1. This gives us the equation $x = \frac{7 - 2}{1} = \frac{5}{1} = 5$.

Final Thoughts

In conclusion, linear equations are a powerful tool for modeling real-world scenarios. By understanding the concept of linear equations and how to solve them, you can apply this knowledge to a wide range of scenarios and make informed decisions.