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Chapter 7: Problem 38

For exercises 35-38, evaluate. $$ 3 \div \frac{1}{3} $$

Short Answer

Expert verified
9

Step by step solution

01

Understand the Division of Fractions

When dividing by a fraction, recall that dividing by a fraction is equivalent to multiplying by its reciprocal. The reciprocal of \(\frac{1}{3}\) is 3.
02

Multiply the Numbers

Replace the division operation with a multiplication by the reciprocal. So, \( 3 \div \frac{1}{3} = 3 \times 3 \).
03

Simplify the Expression

Perform the multiplication: \( 3 \times 3 = 9 \).
04

Write the Final Answer

The evaluated result of the expression \( 3 \div \frac{1}{3} \) is 9.

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Key Concepts

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

Reciprocal
Understanding the concept of a reciprocal is crucial when dealing with division of fractions. A reciprocal is simply a number that, when multiplied by a given number, results in 1. For instance, the reciprocal of \( \frac{1}{3} \) is 3 because \( \frac{1}{3} \times 3 = 1 \). To find the reciprocal of a fraction, just flip the numerator and the denominator. So, for \( \frac{a}{b} \), the reciprocal is \( \frac{b}{a} \). Remember, every non-zero number has a reciprocal. Even a whole number like 3 has a reciprocal, which is \( \frac{1}{3} \). In our exercise, we found the reciprocal of \( \frac{1}{3} \) to be 3. This step is essential before moving forward to multiplication.
Multiplication
Once you've identified the reciprocal, the next step is to change the division of fractions into a multiplication problem. This transformation is safe since multiplying by the reciprocal is just another way to divide. For example, in our exercise, we transformed \( 3 \div \frac{1}{3} \) into \( 3 \times 3 \). Multiplication is usually more straightforward than division. Thus, converting the problem helps simplify the computation process. Just remember, after finding the reciprocal of the fraction you're dividing by, replace the division symbol with a multiplication symbol and proceed with the operation.
Simplification
Simplification is the final step in evaluating the expression after performing the multiplication. Simplification means reducing the expression to its simplest form. For our problem, this involved calculating \( 3 \times 3 \), which equals 9. Always perform basic arithmetic carefully to ensure accuracy. In multiplication, particularly involving whole numbers and fractions, it's essential to double-check your steps. Simplifying correctly will yield the final answer, making sure all steps align properly. In this case, the problem \( 3 \div \frac{1}{3} \) simplifies correctly to 9.

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Most popular questions from this chapter

For exercises 31-40, (a) solve. (b) check. $$ \frac{n^{2}}{n-9}-\frac{9 n}{n-9}=-9 $$

For exercises 43-58, (a) solve. (b) check. $$ \frac{4}{a+6}=\frac{9}{a-4} $$

For exercises 61-64, the completed problem has one mistake. (a) Describe the mistake in words or copy down the whole problem and highlight or circle the mistake. (b) Do the problem correctly. Problem: In the formula \(A=\frac{10}{B}\), is the relationship between \(A\) and \(B\) a direct variation or an inverse variation? Incorrect Answer: Since as \(B\) increases, \(A\) also increases, this is a direct variation.

For a fixed number of hotel rooms, the number of rooms cleaned per hour, \(x\), and the number of hours it takes to clean the rooms, \(y\), is an inverse variation. If a person can clean 8 rooms per hour, it takes 15 hr to clean the rooms. a. Find the constant of variation, \(k\). Include the units of measurement. b. Write an equation that represents this relationship. c. If a person can clean 6 rooms per hour, find the time needed to clean the rooms.

For exercises \(35-36, T=\frac{336 \mathrm{gm}}{R}\) represents the relationship of tire diameter, \(T\); gear ratio, \(g\); speed, \(m\); and revolutions of the tire per minute, \(R\). Is the relationship of the given variables a direct variation or an inverse variation? $$ g \text { and } R \text { are constant; the relationship of } T \text { and } m \text {. } $$
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