Problem 87
Question
Exercises \(86-88\) will help you prepare for the material covered in the first section of the next chapter. $$ \text { Simplify: } \frac{17 \pi}{6}-2 \pi $$
Step-by-Step Solution
Verified Answer
The simplified form of \(\frac{17 \pi}{6}-2 \pi\) is \(\frac{5 \pi}{6}\)
1Step 1: Identify Values
Identify the values you were given. In this case, we have \(\frac{17 \pi}{6}\) and \(2 \pi\).
2Step 2: Convert Whole Number to Fraction
To simplify this, we must first convert the whole number \(2 \pi\) into a fraction that has the same denominator, 6, as \(\frac{17 \pi}{6}\). To do this, we multiply the numerator and denominator by 6, resulting in the fraction \(\frac{12 \pi}{6}\).
3Step 3: Subtract Fractions
Now subtract the two fractions by subtracting their numerators while keeping the same denominator. \(\frac{17 \pi}{6} - \frac{12 \pi}{6} = \frac{5 \pi}{6}\)
Key Concepts
Fractions SubtractionPi (\(\pi\))Algebraic OperationsConverting Whole Numbers to Fractions
Fractions Subtraction
Understanding how to subtract fractions is essential for working with mathematical expressions, especially when they involve different denominators. When working with fractions that have the same denominator, as in the exercise with \(\frac{17 \pi}{6}-2 \pi\), the process is straightforward. You keep the denominator the same and subtract the numerators. This is possible because the denominators represent equal parts of the whole, so we're just taking away some parts from the others.
Here's the practical rule: When you subtract fractions with the same denominator, subtract the second numerator from the first and keep the denominator fixed. If you have different denominators, you'll first need to find a common denominator, but that's not the case in the given problem.
Here's the practical rule: When you subtract fractions with the same denominator, subtract the second numerator from the first and keep the denominator fixed. If you have different denominators, you'll first need to find a common denominator, but that's not the case in the given problem.
Pi (\(\pi\))
The Greek letter \(\pi\) is a mathematical constant representing the ratio of a circle's circumference to its diameter. It is an irrational number, meaning it has an infinite number of non-repeating decimals. In many mathematical problems, including the one we are discussing, \(\pi\) is treated as a constant term in algebraic expressions.
When performing operations involving \(\pi\), you treat \(\pi\) just like any algebraic variable or constant. For instance, in the given exercise \(\frac{17 \pi}{6}-2 \pi\), \(\pi\) can be factored out, simplifying the numerical part of the expressions first, and then adding \(\pi\) back in at the end. This leads to an easier subtraction of fractions, as seen in the steps to solve the original problem.
When performing operations involving \(\pi\), you treat \(\pi\) just like any algebraic variable or constant. For instance, in the given exercise \(\frac{17 \pi}{6}-2 \pi\), \(\pi\) can be factored out, simplifying the numerical part of the expressions first, and then adding \(\pi\) back in at the end. This leads to an easier subtraction of fractions, as seen in the steps to solve the original problem.
Algebraic Operations
Algebraic operations encompass a variety of mathematical procedures including addition, subtraction, multiplication, division, and exponentiation of algebraic terms. In dealing with expressions such as \(\frac{17 \pi}{6}-2 \pi\), the goal is to simplify or rearrange the terms in a way that makes the expression easier to understand or solve.
To perform these operations with algebraic expressions that include constants like \(\pi\), apply the standard arithmetic rules. For example, if the terms involve addition or subtraction and share a common factor (as \(\pi\) does in this case), you can perform the operation on the numerical coefficients while treating the common factor as a unit that doesn't change until the end of the calculation.
To perform these operations with algebraic expressions that include constants like \(\pi\), apply the standard arithmetic rules. For example, if the terms involve addition or subtraction and share a common factor (as \(\pi\) does in this case), you can perform the operation on the numerical coefficients while treating the common factor as a unit that doesn't change until the end of the calculation.
Converting Whole Numbers to Fractions
Sometimes, algebraic operations require us to work with a mix of fractions and whole numbers. To operate easily, we convert whole numbers to fractions. This involves choosing a denominator that aligns with the fractions in play. In our exercise, to subtract \(2 \pi\) from \(\frac{17 \pi}{6}\), we convert \(2 \pi\) to a fraction with a denominator of 6. This is achieved by multiplying both the whole number and the desired denominator (6 in this case).
The conversion helps to unify the format of the numbers and allows us to apply fraction arithmetic rules. After the conversion, \(\frac{2 \pi \times 6}{6} = \frac{12 \pi}{6}\), we proceed to subtract as we would with any like fractions, culminating in \(\frac{5 \pi}{6}\) as the final simplified form of the given expression.
The conversion helps to unify the format of the numbers and allows us to apply fraction arithmetic rules. After the conversion, \(\frac{2 \pi \times 6}{6} = \frac{12 \pi}{6}\), we proceed to subtract as we would with any like fractions, culminating in \(\frac{5 \pi}{6}\) as the final simplified form of the given expression.
Other exercises in this chapter
Problem 86
In Exercises 81–100, evaluate or simplify each expression without using a calculator. $$ 10^{\log 53} $$
View solution Problem 86
determine whether each statement makes sense or does not make sense, and explain your reasoning. I use the natural base \(e\) when determining how much money I'
View solution Problem 87
Let \(\log _{b} 2=A\) and \(\log _{b} 3=C .\)Write each expression in terms of \(A\) and \(C\). $$ \log _{b} \sqrt{\frac{2}{27}} $$
View solution Problem 87
Solve each logarithmic equation. Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions. Give the exact answer.
View solution