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# 2.11: Applications of Reciprocals

Difficulty Level: Basic Created by: CK-12
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Practice Applications of Reciprocals

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Suppose that a car did one lap around a circular race track with a circumference of 147\begin{align*} 1 \frac{4}{7}\end{align*} miles. If you use 227\begin{align*}\frac{22}{7}\end{align*} as an approximation for π\begin{align*} \pi\end{align*}, 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=Fm\begin{align*}a = \frac{F}{m}\end{align*}. Suppose F=713\begin{align*}F = 7\frac{1}{3}\end{align*} and m=15\begin{align*}m= \frac{1}{5}\end{align*}. Find a\begin{align*}a\end{align*}, 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=223÷15.\begin{align*}a = \frac{22}{3} \div \frac{1}{5}.\end{align*}

223×51=1103=3623\begin{align*}\frac{22}{3} \times \frac{5}{1} = \frac{110}{3} = 36 \frac{2}{3}\end{align*}. Therefore, the acceleration is 3623 m/s2.\begin{align*}36 \frac{2}{3} \ m/s^2.\end{align*}

#### 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=distancetime\begin{align*}speed = \frac{distance}{time}\end{align*}.

s=1.5÷14\begin{align*}s = 1.5 \div \frac{1}{4}\end{align*}

Rewrite the expression and simplify: s=3241=4321=122=6 mi/hr.\begin{align*}s = \frac{3}{2} \cdot \frac{4}{1} = \frac{4 \cdot 3} {2 \cdot 1} = \frac{12}{2} = 6 \ mi/hr.\end{align*}

#### 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 biger 3 is than 1/2, by dividing 3 by 1/2:

3÷12=3×21=3×2=6.\begin{align*}3 \div \frac{1}{2}=3 \times \frac{2}{1}=3\times 2=6.\end{align*}

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

2÷6=2×16=26=13.\begin{align*}2 \div 6=2\times \frac{1}{6}=\frac{2}{6}=\frac{1}{3}.\end{align*}

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=Fm\begin{align*}a = \frac{F}{m}\end{align*}. Suppose F=512\begin{align*}F = 5\frac{1}{2}\end{align*} and m=23\begin{align*}m= \frac{2}{3}\end{align*}. Find a\begin{align*}a\end{align*}, 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:

512=5×2+12=112\begin{align*} 5\frac{1}{2}=\frac{5\times 2+1}{2}=\frac{11}{2}\end{align*}

a=Fm=11223=112÷23=112×32=11×32×2=334=814\begin{align*}& 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}\end{align*}

Therefore, the acceleration is 814m/s2\begin{align*}8\frac{1}{4}m/s^2\end{align*}.

2. Use the formula speed=distancetime\begin{align*}speed = \frac{distance}{time}\end{align*}:

speed=distancetime=314÷12=134÷12=134×2=13×24.\begin{align*}& 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}.\end{align*}

Before we continue, we will simplify the fraction:

13×24=13×22×2=132=612.\begin{align*} &\frac{13\times 2}{4}=\frac{13\times 2}{2\times 2}=\frac{13}{2}=6\frac{1}{2}.\end{align*}

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

### Practice

In 1 – 3, evaluate the expression.

1. xy\begin{align*}\frac{x}{y}\end{align*} for x=38\begin{align*}x = \frac{3}{8}\end{align*} and y=43\begin{align*}y= \frac{4}{3}\end{align*}
2. 4z÷u\begin{align*}4z \div u\end{align*} for u=0.5\begin{align*}u = 0.5\end{align*} and z=10\begin{align*}z = 10\end{align*}
3. 6m\begin{align*}\frac{-6}{m}\end{align*} for m=25\begin{align*}m= \frac{2}{5}\end{align*}
4. The label on a can of paint states that it will cover 50 square feet per pint. If I buy a 18\begin{align*}\frac{1}{8}\end{align*}-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 38\begin{align*}\frac{3}{8}\end{align*} mph. How long will it take to dig a trench 23\begin{align*}\frac{2}{3}\end{align*}-mile long?
6. A 27\begin{align*}\frac{2}{7}\end{align*} Newton force applied to a body of unknown mass produces an acceleration of 310 m/s2\begin{align*}\frac{3}{10} \ m/s^2\end{align*}. Calculate the mass of the body. Note: Newton=kg m/s2\begin{align*}\text{Newton} = kg \ m/s^2\end{align*}
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)\begin{align*}199 - (-11)\end{align*}
2. 2.3(3.1)\begin{align*}-2.3 - (-3.1)\end{align*}
3. |1684|\begin{align*}|16-84|\end{align*}
4. |114|\begin{align*}|\frac{-11}{4}|\end{align*}
5. (4÷2×6+105)2\begin{align*}(4 \div 2 \times 6 + 10-5)^2\end{align*}
6. Evaluate f(x)=19(x3);f(21)\begin{align*}f(x)= \frac{1}{9} (x-3); f(21)\end{align*}.
7. Define range.

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### Vocabulary Language: English Spanish

reciprocal

The reciprocal of a nonzero rational number $\frac{a}{b}$ is $\frac{b}{a}$.

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