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Chapter 7: Problem 48

For exercises 39-82, simplify. $$ \frac{9 h k}{40 n^{2}} \div \frac{3 h^{2}}{8 n} $$

Short Answer

Expert verified
\(\frac{3k}{5h}\)

Step by step solution

01

- Rewrite the division as multiplication

To simplify \(\frac{9hk}{40n^2} \div \frac{3h^2}{8n}\), rewrite the division as multiplication by the reciprocal. This means \(\frac{9hk}{40n^2} \div \frac{3h^2}{8n}\) becomes \(\frac{9hk}{40n^2} \times \frac{8n}{3h^2}\).
02

- Multiply the fractions

Multiply the numerators together and the denominators together: \(\frac{9hk \times 8n}{40n^2 \times 3h^2} = \frac{72hkn}{120n^2 h^2}\).
03

- Simplify the fraction

To simplify \(\frac{72hkn}{120n^2 h^2}\), start by canceling out common factors in the numerator and the denominator. Here, both 72 and 120 share a common factor of 24. Additionally, cancel out common variables: \(\frac{72hkn}{120n^2 h^2} = \frac{6kn}{10nh} = \frac{3k}{5h}\).

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Key Concepts

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

algebraic fractions
An algebraic fraction is simply a fraction where the numerator and/or the denominator contain algebraic expressions instead of just numbers. Think of it just like a regular fraction but with variables and constants mixed in. For example, in the exercise given: \( \frac{9hk}{40n^2} \), the numerator \(9hk\) and the denominator \(40n^2\) are both algebraic expressions.
To work with algebraic fractions, you often need to follow the same basic rules that you use with numerical fractions:
  • Adding or subtracting requires a common denominator
  • Multiplication is done across numerators and denominators
  • Division involves multiplying by the reciprocal
Always remember these rules as they help simplify your work with algebraic fractions!
multiplication of fractions
Multiplying fractions, whether they're numerical or algebraic, follows a straightforward rule: multiply the numerators together and the denominators together. In our given problem, after rewriting the division as multiplication, we got: \( \frac{9hk}{40n^2} \times \frac{8n}{3h^2} \).

This means we multiply the numerators (\(9hk \times 8n\)) and the denominators (\(40n^2 \times 3h^2\)). To do the math:
  • Numerators: \(9hk \times 8n = 72hkn\)
  • Denominators: \(40n^2 \times 3h^2 = 120n^2 h^2\)

This gives us \( \frac{72hkn}{120n^2 h^2} \). It looks more complex now, but don't worry, we can simplify it next!
simplifying fractions
Simplifying algebraic fractions follows the same principle as simplifying numerical fractions: find common factors in the numerator and the denominator and cancel them out. For our fraction \(\frac{72hkn}{120n^2 h^2}\):
  • Notice both 72 and 120 share a common factor of 24. Think of it like reducing a fraction like 24/48 to 1/2.
  • The variables can also be simplified by canceling out common factors. For example, \(h\) in the numerator and \(h^2\) in the denominator can be simplified to remove one \(h\).

Breaking it down step-by-step, we get:
1. Cancel numerical common factors: \(\frac{72}{120} = \frac{6}{10}\).
2. Simplify variables: \(\frac{6hk n}{10n^2 h^2} = \frac{6k}{10nh} = \frac{3k}{5h}\).
And that's our simplified fraction: \( \frac{3k}{5h} \)!
reciprocal
A reciprocal is what you get when you flip a fraction upside down. For example, the reciprocal of \( \frac{a}{b} \) is \( \frac{b}{a} \).
When dealing with dividing fractions, you multiply by the reciprocal of the divisor. In our problem, we started with:
  • \( \frac{9hk}{40n^2} \times \frac{8n}{3h^2} \)
  • The reciprocal of \( \frac{3h^2}{8n} \) is \( \frac{8n}{3h^2} \)

Thus, we effectively turned the division problem into a multiplication one. Multiplying by the reciprocal simplifies the fraction and makes solving the problem much easier. Make sure to always double-check the reciprocal to avoid errors!

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Most popular questions from this chapter

For exercises \(67-82\), use the five steps and a proportion. Find the number of adults used to create the ratio "four out of five." Four out of five adults now use the Internet. 184 million adults are online from their homes, offices, schools or other locations. (Source: www.harrisinteractive.com, Nov. 17, 2008)

When a car travels a fixed distance, the relationship between the speed of the car, \(x\), and the time it travels, \(y\), is an inverse variation. When the speed is \(\frac{48 \mathrm{mi}}{1 \mathrm{hr}}\), the time is \(0.75 \mathrm{hr}\). a. Find the constant of proportionality. Include the units of measurement. b. Write an equation that represents this relationship. c. Find the time in hours to travel this distance at a speed of \(\frac{80 \mathrm{mi}}{1 \mathrm{hr}}\). d. Change the time in part \(\mathrm{c}\) to minutes.

Explain why the relationship of the number of bags of leaves per hour that are raked, \(x\), and the hours it takes to rake a yard, \(y\), is an inverse variation.

For exercises 1-10, (a) solve. (b) check. $$ \frac{4}{15} k+\frac{3}{4}=-2 $$

The relationship of \(x\) and \(y\) is an inverse variation. When \(x=2, y=6\). a. Find the constant of proportionality, \(k\). b. Write an equation that represents this inverse variation. c. Find \(y\) when \(x=4\).
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