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91Ó°ÊÓ

Write an equation that expresses each relationship. Then solve the equation for \(y .\) \(x\) varies jointly as \(y\) and the square of \(z\).

Short Answer

Expert verified
\(y = x / (k*z^2)\)

Step by step solution

01

Write the Joint Variation Formula

A joint variation can be expressed in equation form. From the problem statement, it can be derived that \(x\) is equal to \(\(k*y*z^2\)\), where \(k\) is the constant of variation.
02

Isolate y

In order to isolate \(y\) in the equation, divide both sides of the equation by \(k*z^2\). The equation becomes \(y = x / (k*z^2)\).
03

Conclude

From step 1 and 2, we know that when x varies jointly as y and the square of z, the resulting equation solved for y is \(y = x / (k*z^2)\).

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Key Concepts

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

Algebraic Expressions
Algebraic expressions are mathematical phrases that can contain numbers, variables, and operation symbols. They are the backbone of algebra and allow us to represent relationships between quantities. For instance, the given joint variation problem is an example of an algebraic expression. It outlines how one variable, in this case, x, relates to two other variables y and z.

In this scenario, the initial expression derived from the problem statement is set up as x = ky³ú². The expression states that x varies jointly with y and the square of z (³ú²), with a proportionality constant k, also known as the constant of variation. Understanding how to form and manipulate algebraic expressions is key to solving many types of algebra problems.
Isolating Variables
When solving algebraic expressions, the process of isolating variables helps us determine the value of one specific variable. It's like finding out the individual contribution of one variable in the equation.

In the case of our joint variation problem, we are tasked with isolating y. To do this, we initially have the expression x = k*y*³ú². By dividing both sides of the equation by k³ú², we can solve for y. This step is crucial because it transforms the complex relationship into a simpler form allowing us to see how y is influenced by x and z, individually. The equation to isolate y is hence translated into y = x / (k*³ú²). Isolating variables is a fundamental technique applied frequently in algebra to pave the way to solutions.
Constant of Variation
The constant of variation, represented by k in our joint variation equation, is a fixed number that relates the changes in variables to one another. In joint variation, it indicates that one variable changes directly as a product of two or more other variables.

In the given problem, k ties the relationship between x, y, and the square of z (³ú²). To identify this constant in a practical situation, you would need known values of x, y, and z. Keeping k constant ensures that the ratio between x and y * ³ú² remains the same. Understanding the role of this constant is essential for solving joint variation problems, as it provides a stable reference point while different variables adjust accordingly.

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Most popular questions from this chapter

Write an equation that expresses each relationship. Then solve the equation for \(y .\) \(x\) varies directly as \(z\) and inversely as the sum of \(y\) and \(w\).

The average number of daily phone calls, \(C\), between two cities varies jointly as the product of their populations, \(P_{1}\) and \(P_{2}\) and inversely as the square of the distance, \(d\), between them. a. Write an equation that expresses this relationship. b. The distance between San Francisco (population: \(777,000\) ) and Los Angeles (population: \(3,695,000\) ) is 420 miles. If the average number of daily phone calls between the cities is \(326,000,\) find the value of \(k\) to two decimal places and write the equation of variation. c. Memphis (population: \(650,000\) ) is 400 miles from New Orleans (population: \(490,000\) ). Find the average number of daily phone calls, to the nearest whole number, between these cities.

Whe lise a graphing utility to graph $$ f(x)=\frac{x^{2}-4 x+3}{x-2} \text { and } g(x)=\frac{x^{2}-5 x+6}{x-2} $$ What differences do you observe between the graph of \(f\) and the graph of \(g\) ? How do you account for these differences?

Determine whether each statement makes sense or does not make sense, and explain your reasoning. When all is said and done, it seems to me that direct variation equations are special kinds of linear functions and inverse variation equations are special kinds of rational functions.

Determine whether each statement makes sense or does not make sense, and explain your reasoning. My graph of \(y=\frac{x-1}{(x-1)(x-2)}\) has vertical asymptotes at \(x=1\) and \(x=2\)

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