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Applications of Reciprocals

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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.

Video Review

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.

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