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Chapter 3: Problem 28

Find the constant of variation \(k\). The value of \(y\) equals 24 when \(x\) is \(\frac{1}{2}\). Find \(y\) when \(x=3\) if a. \(y\) varies directly as \(x\). b. \(y\) varies inversely as \(x\).

Short Answer

Expert verified
For direct variation, \(y = 144\). For inverse variation, \(y = 4\).

Step by step solution

01

- Understand Direct Variation

In direct variation, the relationship between two variables is given by the equation: \[ y = kx \] where \(k\) is the constant of variation.
02

- Find the Constant for Direct Variation

Given that \( y = 24 \) when \( x = \frac{1}{2} \), substitute these values into the equation for direct variation: \[ 24 = k \cdot \frac{1}{2} \] Solve for \(k\): \[ k = 24 \div \frac{1}{2} \] This simplifies to: \[ k = 48 \]
03

- Find y for Direct Variation When x = 3

Use the value of \(k\) found in Step 2: \[ y = 48x \] Substitute \( x = 3 \): \[ y = 48 \cdot 3 \] \[ y = 144 \]
04

- Understand Inverse Variation

In inverse variation, the relationship between two variables is given by the equation: \[ y = \frac{k}{x} \] where \(k\) is the constant of variation.
05

- Find the Constant for Inverse Variation

Given that \( y = 24 \) when \( x = \frac{1}{2} \), substitute these values into the equation for inverse variation: \[ 24 = \frac{k}{\frac{1}{2}} \] Solve for \(k\): \[ k = 24 \cdot \frac{1}{2} \] This simplifies to: \[ k = 12 \]
06

- Find y for Inverse Variation When x = 3

Use the value of \(k\) found in Step 5: \[ y = \frac{12}{x} \] Substitute \( x = 3 \): \[ y = \frac{12}{3} \] \[ y = 4 \]

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Key Concepts

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

Direct Variation
Direct variation happens when the relationship between two variables can be represented by a straight line. The equation for direct variation is \( y = kx \), where \( k \) is the constant of variation. This means that as one variable increases, the other does too, and the ratio between them remains constant. Ul, for example:
  • If \( y \) is 24 when \( x \) is \( \frac{1}{2} \), you find the constant of variation by solving \( 24 = k \cdot \frac{1}{2} \).
  • This gives \( k = 48 \).
To find \( y \) when \( x = 3 \), use the equation \( y = 48 \cdot 3 \), which means \( y = 144 \).
Inverse Variation
Inverse variation describes a situation where one variable increases while the other decreases. This relationship is represented by the equation \( y = \frac{k}{x} \), where \( k \) is the constant of variation. Here, the product of the variables is always the same.
  • For example, if \( y = 24 \) when \( x = \frac{1}{2} \), you find \( k \) by solving \( 24 = \frac{k}{\frac{1}{2}} \).
  • This gives \( k = 12 \).
To find \( y \) when \( x = 3 \), use the equation \( y = \frac{12}{3} \), which simplifies to \( y = 4 \).
Constant of Variation
The constant of variation \( k \) is a key factor in both direct and inverse variation. It determines how the variables relate to each other.
  • In direct variation \( y = kx \), the constant of variation \( k \) can be found by dividing \( y \) by \( x \).
  • In inverse variation \( y = \frac{k}{x} \), it can be found by multiplying \( y \) by \( x \).

The constant of variation helps simplify and solve many algebra problems by making it clear how variables change in relation to each other.

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Most popular questions from this chapter

Given the inequality, \(0.552 x^{3}+4.13 x^{2}-1.84 x-3.5<6.7\) a. Write the inequality in the form \(f(x)<0\). b. Graph \(y=f(x)\) on a suitable viewing window. c. Use the Zero feature to approximate the real zeros of \(f(x)\). Round to 1 decimal place. d. Use the graph to approximate the solution set for the inequality \(f(x)<0\)

Determine if the statement is true or false. If 5 is an upper bound for the real zeros of \(f(x)\), then 4 is also an upper bound.

Sometimes it is necessary to use a "friendly" viewing window on a graphing calculator to see the key features of a graph. For example, for a calculator screen that is 96 pixels wide and 64 pixels high, the "decimal viewing window" defined by [-4.7,4.7,1] by [-3.1,3.1,1] creates a scaling where each pixel represents 0.1 unit. The window [-9.4,9.4,1] by [-6.2,6.2,1] defines each pixel as 0.2 unit, and so on. Exercises \(112-113\) compare the use of the standard viewing window to a "friendly" viewing window. a. Identify any vertical asymptotes of the function defined by \(f(x)=\frac{x^{2}-5 x+4}{x-4}\) b. Compare the graph of \(f(x)=\frac{x^{2}-5 x+4}{x-4}\) on the standard viewing window [-10,10,1] by [-10,10,1] and on the window [-9.4,9.4,1] by [-6.2,6.2,1] . Which graph shows the behavior at \(x=4\) more completely?

A monthly phone plan costs \(\$ 59.95\) for unlimited texts and 400 min. If more than 400 min are used the plan charges an addition \(\$ 0.45\) per minute. For \(x\) minutes used, the average cost per minute \(\overline{C_{1}}(x)\) (in \$) is given by $$\overline{C_{1}}(x)=\left\\{\begin{array}{ll} \frac{59.95}{x} & \text { for } 0 \leq x \leq 400 \\ \frac{59.95+0.45(x-400)}{x} & \text { for } x>400 \end{array}\right.$$ a. Find the average cost if a customer talks for \(252 \mathrm{~min}\) \(366 \mathrm{~min},\) and \(400 \mathrm{~min} .\) Round to 2 decimal places. b. Find the average cost if a customer talks for \(436 \mathrm{~min}\) \(582 \mathrm{~min},\) and \(700 \mathrm{~min}\). Round to 2 decimal places. c. Suppose a second phone plan costs \(\$ 79.95\) per month for unlimited minutes and unlimited texts. Write an average cost function to represent the average cost \(\overline{C_{2}}(x)\) (in \$) for \(x\) minutes used. d. Find the average cost \(\overline{C_{2}}(x)\) for the second plan for \(252 \mathrm{~min}, 400 \mathrm{~min},\) and \(700 \mathrm{~min}\)

The procedure to solve a polynomial or rational inequality may be applied to all inequalities of the form \(f(x)>0, f(x)<0,\) \(f(x) \geq 0,\) and \(f(x) \leq 0 .\) That is, find the real solutions to the related equation and determine restricted values of \(x .\) Then determine the sign of \(f(x)\) on each interval defined by the boundary points. Use this process to solve the inequalities. $$ 8 \leq x^{2}+4 x+3<15 $$
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