/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 Find the constant of variation \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 21

Find the constant of variation \(k\). \(y\) varies directly as \(x\). When \(x\) is \(8, y\) is 20 .

Short Answer

Expert verified
The constant of variation is \( k = 2.5 \).

Step by step solution

01

- Understand direct variation

When two variables vary directly, it means that one variable is equal to a constant multiplied by the other variable. The general form is given by the equation: \( y = kx \) where \( k \) is the constant of variation.
02

- Substitute the given values

Substitute the values provided in the problem into the direct variation equation. Here, when \( x = 8 \), \( y = 20 \). So, we substitute these values into the equation: \( 20 = k \times 8 \)
03

- Solve for the constant of variation

To find the value of \( k \), solve the equation obtained in Step 2: \( 20 = k \times 8 \) Divide both sides by 8: \( k = \frac{20}{8} = 2.5 \)
04

Conclusion

The value of the constant of variation is \( k = 2.5 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Direct Variation
Direct variation describes a relationship between two variables, where one variable is a constant multiple of the other. This is presented by the equation: \( y = kx \).
The constant \(k\) is known as the constant of variation.
Some points to keep in mind:
  • When one variable increases, the other also increases.
  • If one variable decreases, the other also decreases.
  • The ratio of \(y\) to \( x \) is always the same (\( k \)).
For instance, if \( y \) varies directly as \( x \) with \( k = 2 \), then if \( x \) is doubled, \( y \) will also double.
Algebraic Equations
Algebraic equations involve variables and constants. To solve these equations, we often perform operations to isolate the variable.
For example, in the direct variation problem, we start with the equation: \( y = kx \).
To find \( k \), we need to substitute the given values into the equation and then manipulate it.
Substitution means putting the given numbers into the equation.
In our example, we substitute: \( x = 8 \) and \( y = 20 \).
This gives us \( 20 = k \times 8 \).
These steps lead to a simpler equation to solve for \( k \), showing how algebraic operations help us find our desired values.
Solving for Constants
Solving for a constant means finding the value of the unknown, usually represented by a letter like \( k\). This involves basic algebraic manipulation.
Let's revisit our example:
Starting with \( 20 = k \times 8 \), we need to isolate \( k \).
To do this, we divide both sides of the equation by 8: \( k = \frac{20}{8} \).
After simplifying, \( k = 2.5 \).
  • Always remember to perform the same operation on both sides of the equation.
  • Simplify fractions when possible.
  • Check your solution by substituting back into the original equation to verify consistency.
By ensuring all steps are correctly followed, we confirm that \( k = 2.5\) maintains the consistent direct relationship between \( y \) and \( x \).

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The average round trip speed \(S\) (in mph) of a vehicle traveling a distance of \(d\) miles in each direction is given by $$S=\frac{2 d}{\frac{d}{r_{1}}+\frac{d}{r_{2}}}$$ where \(r_{1}\) and \(r_{2}\) are the rates of speed for the initial trip and the return trip, respectively. a. Suppose that a motorist travels \(200 \mathrm{mi}\) from her home to an athletic event and averages \(50 \mathrm{mph}\) for the trip to the event. Determine the speeds necessary if the motorist wants the average speed for the round trip to be at least \(60 \mathrm{mph}\) ? b. Would the motorist be traveling within the speed limit of \(70 \mathrm{mph} ?\)

Explain why \(x=-2\) is not a vertical asymptote of the graph of \(f(x)=\frac{x^{2}+7 x+10}{x+2}\).

Write the domain of the function in interval notation. $$ g(t)=\sqrt{1-t^{2}} $$

The population \(P(t)\) of a bacteria culture is given by \(P(t)=-1500 t^{2}+60,000 t+10,000,\) where \(t\) is the time in hours after the culture is started. Determine the time(s) at which the population will be greater than 460,000 organisms.

A sports trainer has monthly costs of \(\$ 69.95\) for phone service and \(\$ 39.99\) for his website and advertising. In addition he pays a \(\$ 20\) fee to the gym for each session in which he trains a client. a. Write a cost function to represent the \(\operatorname{cost} C(x)\) for \(x\) training sessions. b. Write a function representing the average \(\operatorname{cost} \bar{C}(x)\) for \(x\) sessions. c. Evaluate \(\bar{C}(5), \bar{C}(30),\) and \(\bar{C}(120)\). d. The trainer can realistically have 120 sessions per month. However, if the number of sessions were unlimited, what value would the average cost approach? What does this mean in the context of the problem?
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.