Problem 8
Question
If y varies directly as \(x\), find the constant of variation and the direct variation equation for each situation. \(y=0.4\) when \(x=2.5\)
Step-by-Step Solution
Verified Answer
The constant of variation is 0.16, and the equation is \( y = 0.16x \).
1Step 1: Understand Direct Variation
Direct variation means that as one quantity increases, the other increases at a constant rate. This is mathematically expressed as \( y = kx \), where \( k \) is the constant of variation.
2Step 2: Substitute Known Values
Given \( y = 0.4 \) when \( x = 2.5 \), substitute these values into the direct variation formula: \( 0.4 = k \times 2.5 \).
3Step 3: Solve for Constant of Variation
To find \( k \), divide both sides of \( 0.4 = k \times 2.5 \) by 2.5: \( k = \frac{0.4}{2.5} \).
4Step 4: Calculate the Constant
Perform the division: \( k = 0.16 \). This is the constant of variation.
5Step 5: Write the Direct Variation Equation
Substitute the constant \( k = 0.16 \) back into the direct variation formula to get the equation: \( y = 0.16x \).
Key Concepts
Constant of VariationProportional RelationshipMathematical Equation
Constant of Variation
In mathematics, the **constant of variation** is a crucial concept when discussing direct variation relationships. It serves as the coefficient of proportionality in the equation of direct variation, expressed as \( y = kx \). Here, \( k \) stands for the constant of variation.
The constant \( k \) indicates how much \( y \) changes for every unit change in \( x \). When given a direct variation problem, like \( y = 0.4 \) when \( x = 2.5 \), finding \( k \) involves substituting these values into the formula \( y = kx \) and solving for \( k \).
The constant \( k \) indicates how much \( y \) changes for every unit change in \( x \). When given a direct variation problem, like \( y = 0.4 \) when \( x = 2.5 \), finding \( k \) involves substituting these values into the formula \( y = kx \) and solving for \( k \).
- If \( y = 0.4 \) and \( x = 2.5 \), the equation becomes \( 0.4 = k \times 2.5 \).
- To determine \( k \), divide both sides by 2.5, resulting in \( k = \frac{0.4}{2.5} = 0.16 \).
Proportional Relationship
A **proportional relationship** exists between two quantities when they increase or decrease at the same rate, maintaining a constant ratio. In mathematical terms, this relationship between \( y \) and \( x \) can be illustrated as \( y = kx \) where \( k \) is the constant of variation.
In our example, because \( y = 0.4 \) changes directly with \( x = 2.5 \), their relationship can be considered proportional. Each increase in \( x \) results in a predictable and fixed increase in \( y \) since the constant of variation \( k \) was calculated as 0.16.
This means that for every 1 unit increase in \( x \), \( y \) increases by 0.16:
In our example, because \( y = 0.4 \) changes directly with \( x = 2.5 \), their relationship can be considered proportional. Each increase in \( x \) results in a predictable and fixed increase in \( y \) since the constant of variation \( k \) was calculated as 0.16.
This means that for every 1 unit increase in \( x \), \( y \) increases by 0.16:
- If you double \( x \), \( y \) doubles.
- If \( x \) is cut in half, \( y \) is halved too.
Mathematical Equation
In the realm of direct variation, a **mathematical equation** succinctly captures the relationship between two variables. This equation is formatted as \( y = kx \). Here \( y \) is the dependent variable, \( x \) is the independent variable, and \( k \) is the constant of variation.
Constructing the mathematical equation involves two primary steps:
Constructing the mathematical equation involves two primary steps:
- First, calculate the constant of variation \( k \) using given data points. For instance, \( y = 0.4 \) and \( x = 2.5 \) gives \( k = 0.16 \) through the calculation \( k = \frac{0.4}{2.5} \).
- Next, substitute this constant back into the formula \( y = kx \) to establish the specific equation connecting \( y \) and \( x \).
Other exercises in this chapter
Problem 8
Graph the solutions of each system of linear inequalities $$ \left\\{\begin{array}{l} y \geq x-5 \\ y \leq-3 x+3 \end{array}\right. $$
View solution Problem 8
Use matrices to solve each system of linear equations. See Example 2. $$ \left\\{\begin{array}{rr} 9 x-3 y= & 6 \\ -18 x+6 y= & -12 \end{array}\right. $$
View solution Problem 9
Solve each system. $$ \left\\{\begin{aligned} 4 x-y+2 z &=5 \\ 2 y+z &=4 \\ 4 x+y+3 z &=10 \end{aligned}\right. $$
View solution Problem 9
Graph the solutions of each system of linear inequalities $$ \left\\{\begin{aligned} x & \geq 3 y \\ x+3 y & \leq 6 \end{aligned}\right. $$
View solution