**Problem 5: Solve the Equation** Solve the equation \(\frac{6r + re}{d} = m\) for \(r\). **Question 6:** Express \(\sqrt{99}\) in simplest form. **Solution:** To simplify \(\sqrt{99}\), we need to find the prime factorization of 99. 1. \(99\) can be divided by \(3\): \(99 \div 3 = 33\). 2. \(33\) can also be divided by \(3\): \(33 \div 3 = 11\). 3. \(11\) is a prime number. Thus, the prime factorization of \(99\) is \(3^2 \times 11\). Now, express the square root: \[ \sqrt{99} = \sqrt{3^2 \times 11} = \sqrt{3^2} \times \sqrt{11} = 3\sqrt{11} \] Therefore, the simplest form of \(\sqrt{99}\) is \(3\sqrt{11}\).
**Problem 5: Solve the Equation** Solve the equation \(\frac{6r + re}{d} = m\) for \(r\). **Question 6:** Express \(\sqrt{99}\) in simplest form. **Solution:** To simplify \(\sqrt{99}\), we need to find the prime factorization of 99. 1. \(99\) can be divided by \(3\): \(99 \div 3 = 33\). 2. \(33\) can also be divided by \(3\): \(33 \div 3 = 11\). 3. \(11\) is a prime number. Thus, the prime factorization of \(99\) is \(3^2 \times 11\). Now, express the square root: \[ \sqrt{99} = \sqrt{3^2 \times 11} = \sqrt{3^2} \times \sqrt{11} = 3\sqrt{11} \] Therefore, the simplest form of \(\sqrt{99}\) is \(3\sqrt{11}\).
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
Related questions
Question
Transcribed Image Text:**Problem 5: Solve the Equation**
Solve the equation \(\frac{6r + re}{d} = m\) for \(r\).
![**Question 6:**
Express \(\sqrt{99}\) in simplest form.
**Solution:**
To simplify \(\sqrt{99}\), we need to find the prime factorization of 99.
1. \(99\) can be divided by \(3\): \(99 \div 3 = 33\).
2. \(33\) can also be divided by \(3\): \(33 \div 3 = 11\).
3. \(11\) is a prime number.
Thus, the prime factorization of \(99\) is \(3^2 \times 11\).
Now, express the square root:
\[
\sqrt{99} = \sqrt{3^2 \times 11} = \sqrt{3^2} \times \sqrt{11} = 3\sqrt{11}
\]
Therefore, the simplest form of \(\sqrt{99}\) is \(3\sqrt{11}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0aec33da-7e9a-4f18-8fb9-9cc4b6ca964e%2F959b1929-7202-46dd-8cf2-f30611f4b5f1%2Fm7lop6d.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 6:**
Express \(\sqrt{99}\) in simplest form.
**Solution:**
To simplify \(\sqrt{99}\), we need to find the prime factorization of 99.
1. \(99\) can be divided by \(3\): \(99 \div 3 = 33\).
2. \(33\) can also be divided by \(3\): \(33 \div 3 = 11\).
3. \(11\) is a prime number.
Thus, the prime factorization of \(99\) is \(3^2 \times 11\).
Now, express the square root:
\[
\sqrt{99} = \sqrt{3^2 \times 11} = \sqrt{3^2} \times \sqrt{11} = 3\sqrt{11}
\]
Therefore, the simplest form of \(\sqrt{99}\) is \(3\sqrt{11}\).
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