/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 \(\left(\frac{1}{3}\right)^{-1}+... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 46

\(\left(\frac{1}{3}\right)^{-1}+\left(\frac{4}{3}\right)^{-1}\)

Short Answer

Expert verified
\(\frac{15}{4}\)

Step by step solution

01

- Simplify the first term

To simplify \(\frac{1}{3}\right)^{-1}\), apply the property of exponents which states that \((a^{-n} = \frac{1}{a^n})\). Therefore, \(\frac{1}{3})^{-1} = 3\).
02

- Simplify the second term

Similarly, handle \(\frac{4}{3}\right)^{-1}\) by applying the same exponent rule, so \(\frac{4}{3})^{-1} = \frac{3}{4}\).
03

- Add the simplified terms

Now add the simplified terms together: \(3 + \frac{3}{4}\). To add these, express 3 as a fraction: \(\frac{12}{4}\), then use common denominators: \(\frac{12}{4} + \frac{3}{4} = \frac{15}{4}\).

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Key Concepts

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

Exponent Rules
Exponent rules are essential in algebra and help us simplify and solve expressions involving powers. In our exercise, we encounter the negative exponent rule. This rule states that any non-zero number raised to a negative exponent equals the reciprocal of the base raised to the positive exponent: \[a^{-n} = \frac{1}{a^n}\] For example, \[\left(\frac{1}{3}\right)^{-1}\] means we can rewrite the term as the reciprocal of \(\left(\frac{1}{3}\right)\) raised to the power of 1, which simplifies to 3. Similarly, \[\left(\frac{4}{3}\right)^{-1}\] simplifies to \(\frac{3}{4}\). Remembering this rule can help you quickly simplify expressions with negative exponents.
Fraction Addition
Adding fractions is a common task in algebra. To add fractions correctly, they must have a common denominator. If they don鈥檛, you need to find one. In our example, we have simplified terms of 3 (which can be written as \(\frac{12}{4}\) to match the denominator) and \(\frac{3}{4}\).Here鈥檚 a quick guide on adding fractions:
  • Find a common denominator: For fractions \(\frac{a}{c} + \frac{b}{c}\), the common denominator is \(c\).
  • Rewrite each fraction to have the common denominator, if necessary.
  • Add the numerators together while keeping the denominator the same.
  • Simplify the resulting fraction if possible.
So, in our exercise, adding \(\frac{12}{4}\) and \(\frac{3}{4}\) gives us: \(\frac{15}{4}\). This process ensures accurate addition of fractions.
Simplifying Fractions
It's important to simplify fractions where possible to make calculations easier and the results clearer. A fraction is simplified when the numerator and denominator have no common factors other than 1. Here are some steps to simplify fractions:
  • Find the Greatest Common Divisor (GCD) of the numerator and the denominator.
  • Divide both the numerator and the denominator by the GCD.
Our final result, \(\frac{15}{4}\), cannot be simplified further since 15 and 4 have no common factors other than 1. Hence, the fraction is already in its simplest form. Simplifying fractions helps in reducing the complexity of expressions and making them easier to understand.

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Most popular questions from this chapter

The special product \((x+y)(x-y)=x^{2}-y^{2}\) can be used to perform some multiplications.Example: $$\begin{array}{l|l}51 \times 49 & 102 \times 98 \\\=(50+1)(50-1) & =(100+2)(100-2) \\\=50^{2}-1^{2} & =100^{2}-2^{2} \\\=2500-1 & =10,000-4 \\\=2499 & =9996\end{array}$$ Use this method to calculate each product mentally. $$ 103 \times 97 $$

Add or subtract as indicated. \((4 x+2 x y-3)-(-2 x+3 x y+4)\)

Find each product. Recall that \(a^{2}=a \cdot a\) and \(a^{3}=a^{2} \cdot a\). $$ -4 x^{3}\left(3 x^{4}+2 x^{2}-x\right)(-2 x+1) $$

Use scientific notation to calculate the result in each expression. Write answers in scientific notation. 76\. \(\frac{0.000000081(0.000036)}{0.00000048}\)

The special product \((x+y)(x-y)=x^{2}-y^{2}\) can be used to perform some multiplications. Example: $$\begin{array}{l|l}51 \times 49 & 102 \times 98 \\\=(50+1)(50-1) & =(100+2)(100-2) \\\=50^{2}-1^{2} & =100^{2}-2^{2} \\\=2500-1 & =10,000-4 \\\=2499 & =9996\end{array}$$ Use this method to calculate each product mentally. $$ 101 \times 99 $$
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