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# Applications of Reciprocals

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# Applications of Reciprocals

Suppose that a car did one lap around a circular race track with a circumference of $1 \frac{4}{7}$ miles. If you use $\frac{22}{7}$ as an approximation for $\pi$ , could you find the diameter of the race track? After completing this Concept, you'll be able to solve real-world problems such as this by using reciprocals.

### Guidance

Using Reciprocals to Solve Real-World Problems

The need to divide rational numbers is necessary for solving problems in physics, chemistry, and manufacturing. The following example illustrates the need to divide fractions in physics.

#### Example A

Newton’s Second Law relates acceleration to the force of an object and its mass: $a = \frac{F}{m}$ . Suppose $F = 7\frac{1}{3}$ and $m= \frac{1}{5}$ . Find $a$ , the acceleration.

Solution: Before beginning the division, the mixed number of force must be rewritten as an improper fraction.

Replace the fraction bar with a division symbol and simplify: $a = \frac{22}{3} \div \frac{1}{5}.$

$\frac{22}{3} \times \frac{5}{1} = \frac{110}{3} = 36 \frac{2}{3}$ . Therefore, the acceleration is $36 \frac{2}{3} \ m/s^2.$

#### Example B

Anne runs a mile and a half in one-quarter hour. What is her speed in miles per hour?

Solution: Use the formula $speed = \frac{distance}{time}$ .

$s = 1.5 \div \frac{1}{4}$

Rewrite the expression and simplify: $s = \frac{3}{2} \cdot \frac{4}{1} = \frac{4 \cdot 3} {2 \cdot 1} = \frac{12}{2} = 6 \ mi/hr.$

#### Example C

For a certain recipe of cookies, you need 3 cups of flour for every 2 cups of sugar. If Logan has 1/2 cup flour, how many cups of sugar will he need to use to make a smaller batch?

Solution: First we need to figure out how many times bigger 3 is than 1/2, by dividing 3 by 1/2:

$3 \div \frac{1}{2}=3 \times \frac{2}{1}=3\times 2=6.$

Since 1/2 goes into 3 six times, then we need to divide the 2 cups of sugar by 6:

$2 \div 6=2\times \frac{1}{6}=\frac{2}{6}=\frac{1}{3}.$

Logan needs 1/3 cup of sugar to make a smaller batch with 1/2 cup flour.

### Guided Practice

1. Newton’s Second Law relates acceleration to the force of an object and its mass: $a = \frac{F}{m}$ . Suppose $F = 5\frac{1}{2}$ and $m= \frac{2}{3}$ . Find $a$ , the acceleration.

2. Mayra runs 3 and a quarter miles in one-half hour. What is her speed in miles per hour?

Solutions:

1. Before we substitute the values into the formula, we must turn the mixed fraction into an improper fraction:

$5\frac{1}{2}=\frac{5\times 2+1}{2}=\frac{11}{2}$

$& a = \frac{F}{m}=\frac{\frac{11}{2}}{\frac{2}{3}}=\\&\frac{11}{2}\div \frac{2}{3}=\frac{11}{2}\times \frac{3}{2}=\\& \frac{11\times 3}{2\times 2}=\frac{33}{4}=8\frac{1}{4}$

Therefore, the acceleration is $8\frac{1}{4}m/s^2$ .

2. Use the formula $speed = \frac{distance}{time}$ :

$& speed = \frac{distance}{time}=3\frac{1}{4}\div \frac{1}{2}=\\& \frac{13}{4}\div \frac{1}{2}= \frac{13}{4}\times 2=\\&\frac{13\times 2}{4}.$

Before we continue, we will simplify the fraction:

$&\frac{13\times 2}{4}=\frac{13\times 2}{2\times 2}=\frac{13}{2}=6\frac{1}{2}.$

Mayra can run 6-and-a-half miles per hour.

### Practice

In 1 – 3, evaluate the expression.

1. $\frac{x}{y}$ for $x = \frac{3}{8}$ and $y= \frac{4}{3}$
2. $4z \div u$ for $u = 0.5$ and $z = 10$
3. $\frac{-6}{m}$ for $m= \frac{2}{5}$
4. The label on a can of paint states that it will cover 50 square feet per pint. If I buy a $\frac{1}{8}$ -pint sample, it will cover a square two feet long by three feet high. Is the coverage I get more, less, or the same as that stated on the label?
5. The world’s largest trench digger, “Bagger 288,” moves at $\frac{3}{8}$ mph. How long will it take to dig a trench $\frac{2}{3}$ -mile long?
6. A $\frac{2}{7}$ Newton force applied to a body of unknown mass produces an acceleration of $\frac{3}{10} \ m/s^2$ . Calculate the mass of the body. Note: $\text{Newton} = kg \ m/s^2$
7. Explain why the reciprocal of a nonzero rational number is not the same as the opposite of that number.
8. Explain why zero does not have a reciprocal.

Mixed Review

Simplify.

1. $199 - (-11)$
2. $-2.3 - (-3.1)$
3. $|16-84|$
4. $|\frac{-11}{4}|$
5. $(4 \div 2 \times 6 + 10-5)^2$
6. Evaluate $f(x)= \frac{1}{9} (x-3); f(21)$ .
7. Define range.