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3 years ago

Alabama was not a slave state. True False

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2 answers:

SCORPION-xisa [38]3 years ago

7 0

The answer is False.

PilotLPTM [1.2K]3 years ago

6 0

Hey there! Your answer is False. Alabama WAS a slave state, and was one of the most recognized out of them all. It was also the one to keep segregation the longest, as of today. Hope this helped!

Thanks, and have a civilistic day!

~Steve

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(x+1/x-1 - x-1/x+1 - 4x/x^2 +1) + 4x/x^4-1

Lunna [17]

Answer:

$\frac{12x}{x^4-1}$

Explanation:

Given

$(\frac{x+1}{x-1} - \frac{x-1}{x+1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}$

Required

Solve:

Take L.C.M of the first two fractions

$(\frac{(x+1)(x+1)-(x-1)(x-1)}{(x-1)(x+1)} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}$

$(\frac{(x+1)(x+1)-(x-1)(x-1)}{x^2-1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}$

$(\frac{x^2+2x+1-(x^2-2x+1)}{x^2-1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}$

$(\frac{x^2+2x+1-x^2+2x-1}{x^2-1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}$

Collect Like Terms

$(\frac{x^2-x^2+2x+2x+1-1}{x^2-1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}$

$(\frac{4x}{x^2-1} - \frac{4x}{x^2 + 1}) + \frac{4x}{x^4 - 1}$

Solve the expression in bracket

$(\frac{4x(x^2+1) - 4x(x^2-1)}{(x^2-1)(x^2+1)} ) + \frac{4x}{x^4 - 1}$

$(\frac{4x(x^2+1) - 4x(x^2-1)}{x^4-1} ) + \frac{4x}{x^4 - 1}$

$(\frac{4x^3+4x - 4x^3+4x)}{x^4-1} ) + \frac{4x}{x^4 - 1}$

$(\frac{4x^3- 4x^3+4x +4x)}{x^4-1} ) + \frac{4x}{x^4 - 1}$

$\frac{8x}{x^4-1} + \frac{4x}{x^4 - 1}$

Take LCM

$\frac{8x+4x}{x^4-1}$

$\frac{12x}{x^4-1}$

<em>The expression can not be further simplified</em>

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