Chapter 17: Problem 1
If \(y\) varies inversely as \(x,\) and \(y\) is 385 when \(x\) is \(832,\) find \(y\) when \(x\) is 226.
Short Answer
Expert verified
Using the constant of variation, the value of y when x is 226 is approximately 1417.70.
Step by step solution
01
Understand the Concept of Inverse Variation
Inverse variation means that when one variable increases, the other decreases at a rate such that their product is constant. When it's said that 'y varies inversely as x', it is expressed mathematically as y = k/x, where k is the constant of variation.
02
Find the Constant of Variation
Since y is 385 when x is 832, we can find the constant by multiplying them together. Calculate k using the equation k = x * y, which gives us k = 832 * 385.
03
Calculate the Constant of Variation
Multiply the given values to get the constant of variation. So, k = 832 * 385 = 320320.
04
Use the Constant to Find the New Value of y
Now that we have the constant k, we can find y when x is 226 by rearranging the formula to y = k/x. Substitute k = 320320 and x = 226 into the equation to solve for y.
05
Solve for y
Using the formula y = k/x, substitute the known values to get y = 320320 / 226, then compute the value of y.
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 Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Direct and Inverse Proportion
Understanding the relationship between two variables is crucial in mathematics. In the case of direct and inverse proportion, these relationships describe how one variable changes in relation to another.
In direct proportion, as one variable increases, the other also increases at a consistent rate. For example, if the number of hours worked increases, so does the total pay at a fixed hourly rate. This can be expressed algebraically by the formula: \( y = kx \), where \( y \) is the dependent variable, \( x \) is the independent variable, and \( k \) is the constant of variation, signifying the rate of change.
Inverse proportion, on the other hand, involves a reciprocal relationship. As one variable increases, the other variable decreases in such a way that the product of the two remains constant. This situation is well-described in the given exercise, where \( y \) varies inversely as \( x \). The algebraic expression for this is \( y = \frac{k}{x} \), where \( k \) still stands for the constant of variation. If you know the values of \( y \) and \( x \) at one point, you can find \( k \) and then use it to determine either variable when the other is changed.
To get a practical grip on these concepts, it's helpful to visualize scenarios or even use graphs to see how variables change in direct or inverse proportion to each other.
In direct proportion, as one variable increases, the other also increases at a consistent rate. For example, if the number of hours worked increases, so does the total pay at a fixed hourly rate. This can be expressed algebraically by the formula: \( y = kx \), where \( y \) is the dependent variable, \( x \) is the independent variable, and \( k \) is the constant of variation, signifying the rate of change.
Inverse proportion, on the other hand, involves a reciprocal relationship. As one variable increases, the other variable decreases in such a way that the product of the two remains constant. This situation is well-described in the given exercise, where \( y \) varies inversely as \( x \). The algebraic expression for this is \( y = \frac{k}{x} \), where \( k \) still stands for the constant of variation. If you know the values of \( y \) and \( x \) at one point, you can find \( k \) and then use it to determine either variable when the other is changed.
To get a practical grip on these concepts, it's helpful to visualize scenarios or even use graphs to see how variables change in direct or inverse proportion to each other.
Constant of Variation
The constant of variation is a pivotal element in understanding both direct and inverse relationships. It serves as the anchor point, the fixed value that defines the strength and nature of the relationship between the variables.
When dealing with direct variation, the constant of variation (\( k \)) represents the constant multiple that relates the two variables. In inverse variation, as seen in the original exercise, \( k \) represents the constant product of the two variables.
Finding the constant in inverse variation involves multiplying the given values of the two inversely related variables. In our exercise, \( k \) was determined by the initial values of \( x \) and \( y \): \( k = xy = 832 \times 385 = 320320 \). Once this constant is known, it can be used to find unknown values of one variable when the other is given, as in step 4 of the solution.
The constant of variation is not just a number; it is the key to unlocking the relationship between variables. It’s essential for students to grasp that without knowing this constant, solving problems involving direct or inverse proportions would be much more challenging.
When dealing with direct variation, the constant of variation (\( k \)) represents the constant multiple that relates the two variables. In inverse variation, as seen in the original exercise, \( k \) represents the constant product of the two variables.
Finding the constant in inverse variation involves multiplying the given values of the two inversely related variables. In our exercise, \( k \) was determined by the initial values of \( x \) and \( y \): \( k = xy = 832 \times 385 = 320320 \). Once this constant is known, it can be used to find unknown values of one variable when the other is given, as in step 4 of the solution.
The constant of variation is not just a number; it is the key to unlocking the relationship between variables. It’s essential for students to grasp that without knowing this constant, solving problems involving direct or inverse proportions would be much more challenging.
Algebraic Expressions
Central to solving mathematical problems is the ability to work with algebraic expressions. These are combinations of numbers, variables, and operation symbols that represent a specific value or set of values.
Algebraic expressions are foundational in creating formulas and equations that model real-world scenarios. In the context of our problem, the expression \( y = \frac{k}{x} \) succinctly captures the inverse relationship between \( y \) and \( x \).
Working with algebraic expressions involves understanding the role of variables and constants. The variable, in this case \( y \), represents a value that can change, while the constant \( k \) anchors the relationship between \( y \) and \( x \). Solving algebraic expressions often requires substituting known values for variables and performing operations according to the established mathematical hierarchy.
After determining \( k \) in our exercise, we used the expression to calculate a new value of \( y \) when \( x \) changed. This highlights the dynamic utility of algebraic expressions: with just a formula and a single constant, a wide range of problems can be addressed.
Algebraic expressions are foundational in creating formulas and equations that model real-world scenarios. In the context of our problem, the expression \( y = \frac{k}{x} \) succinctly captures the inverse relationship between \( y \) and \( x \).
Working with algebraic expressions involves understanding the role of variables and constants. The variable, in this case \( y \), represents a value that can change, while the constant \( k \) anchors the relationship between \( y \) and \( x \). Solving algebraic expressions often requires substituting known values for variables and performing operations according to the established mathematical hierarchy.
After determining \( k \) in our exercise, we used the expression to calculate a new value of \( y \) when \( x \) changed. This highlights the dynamic utility of algebraic expressions: with just a formula and a single constant, a wide range of problems can be addressed.
