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Ipatiy [6.2K]
3 years ago
13

How do you solve 4a/b divided by 2ac/b

Mathematics
2 answers:
joja [24]3 years ago
8 0
\frac{\frac{4a}{b}}{\frac{2ac}{b}}=?\\\\
To\ divide\ fraction\ multiply\ first\ fraction\ by\ inverse\ of\ the\ second:\\\\
\frac{\frac{4a}{b}}{\frac{2ac}{b}}=\frac{4a}{b}*\frac{b}{2ac}=?\\\\
Shortening\ fractions:\\\\
\frac{\frac{4a}{b}}{\frac{2ac}{b}}=\frac{4a}{b}*\frac{b}{2ac}=\frac{2}{c}\\\\Solution\ is\ \frac{2}{c}.
goblinko [34]3 years ago
4 0
\frac{4a}{b} :\frac{ 2ac}{b}= \frac{4a}{b} \cdot \frac{b}{ 2ac}= \frac{2}{c} \\ \\a \ and \ c \neq 0


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pashok25 [27]

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<h3>Further explanation</h3>

Let's recall following formula about Exponents and Surds:

\boxed { \sqrt { x } = x ^ { \frac{1}{2} } }

\boxed { (a ^ b) ^ c = a ^ { b . c } }

\boxed {a ^ b \div a ^ c = a ^ { b - c } }

\boxed {\log a + \log b = \log (a \times b) }

\boxed {\log a - \log b = \log (a \div b) }

<em>Let us tackle the problem!</em>

\texttt{ }

\log \frac{\sqrt{3}(2-x)}{(3x)} = \log \sqrt{3} + \log (2-x) - \log (3x)

\log \frac{\sqrt{3}(2-x)}{(3x)} = \log 3^{1/2} + \log (2-x) - (\log 3 + \log x)

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\log \frac{\sqrt{3}(2-x)}{(3x)} = \log (2-x) - \frac{1}{2}\log 3 - \log x

\texttt{ }

<h3>Learn more</h3>
  • Coefficient of A Square Root : brainly.com/question/11337634
  • The Order of Operations : brainly.com/question/10821615
  • Write 100,000 Using Exponents : brainly.com/question/2032116

<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Exponents and Surds

Keywords: Power , Multiplication , Division , Exponent , Surd , Negative , Postive , Value , Equivalent , Perfect , Square , Factor.

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