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Ch. 3 - Polynomial and Rational Functions
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
All textbooksBlitzer 8th EditionCh. 3 - Polynomial and Rational FunctionsProblem 14
Chapter 4, Problem 14

Write an equation that expresses each relationship. Then solve the equation for y. x varies directly as the cube root of z and inversely as y.

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1
Identify the given variation relationships: x varies directly as the cube root of z, and inversely as y. This means we can write the equation as \(x = k \frac{\sqrt[3]{z}}{y}\), where \(k\) is the constant of proportionality.
Write the equation explicitly: \(x = k \frac{z^{\frac{1}{3}}}{y}\).
To solve for \(y\), multiply both sides of the equation by \(y\) to get rid of the denominator: \(xy = k z^{\frac{1}{3}}\).
Next, isolate \(y\) by dividing both sides by \(x\): \(y = \frac{k z^{\frac{1}{3}}}{x}\).
The equation is now solved for \(y\): \(y = \frac{k z^{\frac{1}{3}}}{x}\). To find the specific value of \(k\), you would need additional information such as a set of values for \(x\), \(y\), and \(z\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Direct Variation

Direct variation describes a relationship where one variable is proportional to another. If x varies directly as the cube root of z, it means x = k * ∛z for some constant k. This shows that as ∛z increases, x increases proportionally.
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Inverse Variation

Inverse variation means one variable changes in the opposite way to another, expressed as x = k / y. Here, x varies inversely as y, so as y increases, x decreases proportionally, and vice versa, with k being a constant.
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Solving for a Variable in an Equation

Solving for y involves manipulating the equation algebraically to isolate y on one side. This requires understanding operations like multiplication, division, and roots to rewrite the equation in terms of y, making it the subject of the formula.
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