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Write an equation that expresses each relationship. Use \(k\) as the constant of variation. \(v\) varies directly as \(r\)

Short Answer

Expert verified
The equation that expresses this relationship is \( v = kr \).

Step by step solution

01

Understand direct variation

In direct variation, as one quantity increases, a directly proportional quantity also increases. This is expressed in the form \( y = kx \), where \( y \) and \( x \) are the quantities, and \( k \) is the constant of variation.
02

Write the equation

The problem states that \( v \) varies directly as \( r \). The constant of variation is mentioned as \( k \). Based on the form given above, substitute \( v \) for \( y \), \( r \) for \( x \), and \( k \) remains as the constant of variation. With these substitutions, the equation that expresses the relationship is \( v = kr \).

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Key Concepts

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

Constant of Variation
Understanding the constant of variation is critical when dealing with direct variation scenarios. In essence, the constant of variation, denoted as \( k \), is the consistent ratio or multiplier that relates two variables which vary directly with one another. For example, if the exercise states that 'Variable \( v \) varies directly as variable \( r \)', it means that for every increase or decrease in \( r \), the value of \( v \) will change at a rate determined by the constant \( k \).

This constant remains unchanged regardless of the values of the related variables and thus provides a straightforward way to predict one variable's behavior given the other. It's like having a speed setting on a treadmill; no matter how long you run, the speed (or constant of variation) determines how fast you go. To find the constant of variation in real problems, you can divide one variable by its corresponding value of the other variable, provided you have a pair of nonzero values.
Algebraic Equations
Algebraic equations are mathematical statements that show the equality between two expressions. When it comes to direct variation and the constant of variation, these algebraic equations take a particular form. For instance, we often see equations like \( v = kr \), which shows that the value of \( v \) is equal to the constant of variation \( k \) multiplied by the value of \( r \).

The power of algebra comes from its ability to represent relationships in a general form that can be manipulated to find unknown values. For example, if you know the values of \( v \) and \( r \), you can solve the equation for \( k \) to find the constant of variation. Here, understanding how to manipulate algebraic equations simplifies complex problems into achievable solutions.
Proportional Relationships
Proportional relationships are the heart of direct variation. These relationships entail that as one quantity changes, another quantity changes at a constant rate, which we refer to as direct proportionality. This constant rate is, as you might have guessed, the constant of variation \( k \).

In the context of our exercise, where \( v \) varies directly with \( r \), this implies a proportional relationship between the two. If you graphed this relationship on a coordinate plane, you would get a straight line passing through the origin, with the slope of the line representing our constant \( k \). It's an incredibly useful concept, especially in real-world applications like calculating the cost of fuel based on the price per gallon (or liter), or determining speed when you know the time and distance traveled.

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Most popular questions from this chapter

The temperature, in degrees Fahrenheit, of a dessert placed in a freezer for \(t\) hours is modeled by $$ \frac{t+30}{t^{2}+4 t+1}-\frac{t-50}{t^{2}+4 t+1} $$ a. Express the temperature as a single rational expression. b. Use your rational expression from part (a) to find the temperature of the dessert, to the nearest hundredth of a degree, after 1 hour and after 2 hours.

add or subtract as indicated. Simplify the result, if possible. $$\frac{6 y^{2}+y}{2 y^{2}-9 y+9}-\frac{2 y+9}{2 y^{2}-9 y+9}-\frac{4 y-3}{2 y^{2}-9 y+9}$$

perform the indicated operations. Simplify the result, if possible. $$\left(\frac{3 x-1}{x^{2}+5 x-6}-\frac{2 x-7}{x^{2}+5 x-6}\right) \div \frac{x+2}{x^{2}-1}$$

Will help you prepare for the material covered in the next section. a. If \(y=k x\) find the value of \(k\) using \(x=2\) and \(y=64\) b. Substitute the value for \(k\) into \(y=k x\) and write the resulting equation. c. Use the equation from part (b) to find \(y\) when \(x=5\)

Anthropologists and forensic scientists classify skulls using $$ \frac{L+60 W}{L}-\frac{L-40 W}{L} $$ where \(L\) is the skull's length and \(W\) is its width. a. Express the classification as a single rational expression. b. If the value of the rational expression in part (a) is less than \(75,\) a skull is classified as long. A medium skull has a value between 75 and \(80,\) and a round skull has a value over \(80 .\) Use your rational expression from part (a) to classify a skull that is 5 inches wide and 6 inches long.

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