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Chapter 7: Problem 11

For exercises \(9-24\), evaluate or simplify. $$ \frac{\frac{5 x}{4}}{\frac{2 x^{2}}{15}} $$

Short Answer

Expert verified
\(\frac{25}{8x}\)

Step by step solution

01

- Simplify the complex fraction

To simplify \(\frac{\frac{5 x}{4}}{\frac{2 x^{2}}{15}}\), rewrite it as \(\frac{5 x}{4} \times \frac{15}{2 x^{2}}\).
02

- Multiply the fractions

Multiply the numerators: \(5x \times 15 = 75x\). Multiply the denominators: \(4 \times 2x^2 = 8x^2\).
03

- Simplify the result

Now you have \(\frac{75x}{8x^2}\). Simplify by dividing both the numerator and the denominator by \(x \) and by 3: \(\frac{75 \div 3}{8 \div 3 \times x} = \frac{25}{8x}\).

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Key Concepts

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

Algebraic Fractions
Algebraic fractions are fractions where the numerator or the denominator (or both) contain algebraic expressions. They function similarly to numerical fractions but include variables like \(x\) and constants. An example of an algebraic fraction is \(\frac{5x}{4}\). This means we are dividing the expression \(5x\) by 4.

In the exercise, we dealt with a complex algebraic fraction \(\frac{\frac{5 x}{4}}{\frac{2 x^{2}}{15}}\). To simplify it, we need to apply the same rules we use for basic fractions, with an extra step for handling the variables.
Numerator and Denominator
In a fraction, the numerator is the top part, and the denominator is the bottom. For example, in \(\frac{5x}{4}\), \(5x\) is the numerator and 4 is the denominator.

For complex fractions like \(\frac{\frac{5 x}{4}}{\frac{2 x^{2}}{15}}\), each part itself is a fraction. The numerator is \(\frac{5x}{4}\) and the denominator is \(\frac{2 x^2}{15}\).

To simplify these, you first need to recognize and separate the inner numerators and denominators before multiplying or dividing the fractions.
Multiplying Fractions
To multiply fractions, multiply the numerators together and the denominators together. If we have two fractions \(\frac{a}{b}\) and \(\frac{c}{d}\), we multiply them like this: \(\frac{a \times c}{b \times d}\).

Applying this to our exercise: we had \(\frac{\frac{5 x}{4}}{\frac{2 x^{2}}{15}}\). First, rewrite it as \(\frac{5 x}{4} \times \frac{15}{2 x^{2}}\). Next, multiply the numerators and the denominators:
  • Numerators: \(5x \times 15 = 75x\)
  • Denominators: \(4 \times 2x^2 = 8x^2\)

    • This gives us \(\frac{75x}{8x^2}\).
Simplifying Expressions
Simplifying expressions means reducing them to their simplest form. For \(\frac{75x}{8x^2}\), we need to cancel out common factors in the numerator and the denominator.

Divide both the numerator and the denominator by 3 and by \(x\):
  • Numerator: \(75 \div 3 = 25\)
  • Denominator: \(8 \div 3 = 8\), and \( x^2 \div x = x\)

    • So, \(\frac{75x}{8x^2}\) simplifies to \(\frac{25}{8x}\).

    This is our final answer. Simplifying expressions is crucial as it makes them easier to work with in mathematical problems.

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Most popular questions from this chapter

For exercises 53-56, the formula \(F=\frac{100 S_{u} C_{p}}{S_{p} C_{u}}\) describes the fractional excretion of sodium, \(F\). Is the relationship of the given variables a direct variation or an inverse variation? $$ C_{p}, S_{p} \text {, and } C_{u} \text { are constant; the relationship of } F \text { and } S_{u} $$

For exercises 1-10, (a) solve. (b) check. $$ \frac{2}{3} x+\frac{3}{2}=\frac{1}{3} x+\frac{1}{6} $$

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 \(x=4\).

For exercises 79-82, (a) clear the fractions and solve. (b) check. $$ \frac{3}{2} u+\frac{3}{4}=\frac{9}{2} $$

Find the number of 80,500 children and adolescents who are obese. Round to the nearest hundred. Approximately one in six children and adolescents are obese. (Source: www.cdc.gov, Nov. 2011)
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