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Use the four-step procedure for solving variation problems given on page 424 to solve Exercises 1-10. \(y\) varies directly as \(x . y=65\) when \(x=5 .\) Find \(y\) when \(x=12 .\)

Short Answer

Expert verified
The value of \(y\) when \(x = 12\) is 156.

Step by step solution

01

Determine the Constant of Variation

Given that \(y = 65\) when \(x = 5\), we can substitute these values into the direct variation formula \(y = kx\) to solve for \(k\). This gives\[65 = k \cdot 5\]From which we find that \(k = 13\).
02

Solve for 'y' when 'x' is 12

Now that we know \(k\) is 13, we can substitute this value, along with \(x = 12\), into the direct variation formula to find the new value of \(y\):\[y = 13 \cdot 12\]This simplifies to \(y = 156\).

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Key Concepts

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

Direct Variation
Direct variation is a fundamental concept in algebra that describes a linear relationship between two variables. When one variable increases, the other variable increases at a constant rate, and vice versa. The general form of direct variation is expressed as \( y = kx \), where \( y \) is the dependent variable, \( x \) is the independent variable, and \( k \) is the constant of variation.

To determine if two variables vary directly, you check for a constant ratio \( \frac{y}{x} = k \) for all corresponding values of \( x \) and \( y \). If this ratio remains consistent, it confirms direct variation. When solving direct variation problems, it's crucial to understand this relationship, as it forms the foundation for setting up and solving algebraic expressions involving directly varying quantities.

An example from everyday life would be the relationship between the distance traveled and time taken at a constant speed. The farther you travel, the longer it takes – assuming the speed is consistent (constant of variation), showcasing direct variation.
Constant of Variation
The constant of variation, often represented as \( k \), is a key component in equations involving variation. In the context of direct variation, it quantifies the consistent ratio between two variables as one changes in respect to the other. To find the constant of variation, you need two corresponding values of variables that exhibit direct variation.

Using the equation \( y = kx \), if you are provided with the values for \( y \) and \( x \) at any particular point, you can solve for \( k \) by rewriting the equation as \( k = \frac{y}{x} \). Once you've determined the constant of variation, you can use it to predict other values or solve for unknowns within the direct variation relationship, as it will remain unchanged.

A practical application could be calculating the cost of items sold in bulk. If the price per item is fixed (constant of variation), then the total cost varies directly with the number of items purchased, allowing a simple calculation for different quantities.
Algebraic Expressions
Algebraic expressions are combinations of variables, numbers, and at least one arithmetic operation, without an equality sign. They can represent a wide range of situations, from simple relationships like direct variation to more complex functions. The ability to manipulate these expressions is essential when solving variation problems.

An example of an algebraic expression in a variation problem would be \( y = kx \), which represents direct variation. When solving for a variable, you often work with these expressions to isolate the variable of interest. This may involve distributive properties, combining like terms, or factoring.

In the context of the exercise provided, we see a simple linear expression. Understanding how to work with algebraic expressions can expand to handling more intricate scenarios in algebra, where expressions become more complex and may include exponents, square roots, and other algebraic concepts. Mastering these expressions is vital for success in algebra and related fields of mathematics.

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Most popular questions from this chapter

The illumination provided by a car's headlight varies inversely as the square of the distance from the headlight. A car's headlight produces an illumination of 3.75 footcandles at a distance of 40 feet. What is the illumination when the distance is 50 feet?

Use a graphing utility to graph \(y=\frac{1}{x^{2}}, y=\frac{1}{x^{4}},\) and \(y=\frac{1}{x^{6}}\) in the same viewing rectangle. For even values of \(n\), how does changing \(n\) affect the graph of \(y=\frac{1}{x^{2}} ?\)

Determine whether each statement makes sense or does not make sense, and explain your reasoning. Using the language of variation, I can now state the formula for the area of a trapezoid, \(A=\frac{1}{2} h\left(b_{1}+b_{2}\right),\) as, "A trapezoid's area varies jointly with its height and the sum of its bases."

Galileo's telescope brought about revolutionary changes in astronomy. A comparable leap in our ability to observe the universe took place as a result of the Hubble Space Telescope. The space telescope was able to see stars and galaxies whose brightness is \(\frac{1}{50}\) of the faintest objects observable using ground-based telescopes. Use the fact that the brightness of a point source, such as a star, varies inversely as the square of its distance from an observer to show that the space telescope was able to see about seven times farther than a ground-based telescope.

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

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