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Anettt [7]
2 years ago
14

What is the denominator of the simplified expression? 2 n

SAT
1 answer:
Shtirlitz [24]2 years ago
7 0

The expression \frac{2n}{6n + 4} * \frac{3n + 2}{3n - 2} is a product expression

The denominator of the simplified fraction is 3n -2

<h3>How to determine the denominator?</h3>

The expression is given as;

\frac{2n}{6n + 4} * \frac{3n + 2}{3n - 2}

Divide the first fraction by 2/2

\frac{n}{3n + 2} * \frac{3n + 2}{3n - 2}\\

Evaluate the above product

\frac{n}{3n - 2}

The above fraction is a simplified fraction of \frac{2n}{6n + 4} * \frac{3n + 2}{3n - 2}, and the denominator is 3n - 2

Hence, the denominator of the simplified fraction is 3n -2

Read more about fractions at:

brainly.com/question/15215592

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Complete question:

<em>What values of c and d would make the following expression represent a real number? i(2 + 3i)(c + di) </em>

<em>choices: A)c = 2, d = 3 B)c = –2, d = 3 C)c = 3, d = –2 D)c = –3, d = –2</em>

<em />

<em>Option D</em><em> is correct. </em> The value of c and d for the expression to be real are -3 and -2 respectively

Given the expression

i(2 + 3i)(c + di)

Expand the expression

=i(2 + 3i)(c + di)\\=(2i+3i^2)(c+di)\\=(2i+3(-1))(c+di)\\=(2i-3)(c+di)\\=2ic+2di^2-3c-3di\\=2ic-2d-3c-3di

Collect the like terms

=2ic-3di-2d-3c

For the resulting function to be real, then the imaginary part must be zero i.e.

2ic-3di=0\\2ci = 3di\\2c = 3d\\c=\frac{3d}{2}\\

If d = -2, then;

c=\frac{3(-2)}{2}\\c=\frac{-6}{2}\\c= -3

Hence the value of c and d for the expression to be real are -3 and -2 respectively.

Learn more here: brainly.com/question/4244460

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