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Ch. 3 - Polynomial and Rational Functions
Lial - College Algebra 13th Edition
Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook
All textbooksLial 13th EditionCh. 3 - Polynomial and Rational FunctionsProblem 29
Chapter 4, Problem 29

Resistance of a Wire The resistance in ohms of a platinum wire temperature sensor varies directly as the temperature in kelvins (K). If the resistance is 646 ohms at a temperature of 190 K, find the resistance at a temperature of 250 K.

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Identify the direct variation relationship: resistance \(R\) varies directly as temperature \(T\), which can be written as \(R = k \times T\), where \(k\) is the constant of proportionality.
Use the given values to find \(k\): substitute \(R = 646\) ohms and \(T = 190\) K into the equation \(646 = k \times 190\).
Solve for \(k\) by dividing both sides of the equation by 190: \(k = \frac{646}{190}\).
Write the general formula for resistance using the found \(k\): \(R = \left( \frac{646}{190} \right) \times T\).
Find the resistance at \(T = 250\) K by substituting into the formula: \(R = \left( \frac{646}{190} \right) \times 250\).

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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 quantity changes proportionally with another. If resistance varies directly with temperature, then resistance = k × temperature, where k is a constant. Understanding this helps set up the equation to find unknown values.
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Finding the Constant of Variation

To solve direct variation problems, first find the constant k by using given values. Substitute the known resistance and temperature into the formula resistance = k × temperature, then solve for k. This constant allows calculation of resistance at any temperature.
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Solving Linear Equations

Once the constant k is found, substitute the new temperature into the equation and solve for resistance. This involves basic algebraic manipulation to isolate the variable and compute the desired value, reinforcing skills in solving linear equations.
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