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NARA [144]
3 years ago
11

Madison and Tyler each wrote an expression that is equivalent to 14.1 (19.8 + 7.6). The expressions they created are shown in th

e table.
Expressions Madison and Tyler Created
Student
Expression
Madison
14.1 (7.6 + 19.8)
Tyler
14.1 (19.8) + 14.1 (7.6)
Mathematics
2 answers:
diamong [38]3 years ago
6 0

Answer: Tyler used the distributive property.

Step-by-step explanation:

For the expression:  

Madison wrote:  which is by commutative property of addition.

The commutative property of addition says that , for a,b be any real number

Tyler wrote: which is by distributive propery.

The distributive property says that , for a,b,c be any real numbers.

Hence, the last option is correct.

AlladinOne [14]3 years ago
4 0

Answer:

D

Step-by-step explanation:

Took the test ;)

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<h3>log [ (√3)(2 - x)/(3x) ] = log (2 - x) - ¹/₂ log 3 - log x</h3>

<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)

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

\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
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<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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