What is (\frac{c^2+4c)/(c^2-4)}{(3c+12)/(c-2)} ?

The solution to (\frac{c^2+4c)/(c^2-4)}{(3c+12)/(c-2)} is c/(3(c+2))

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simplify (\frac{c^2+4c)/(c^2-4)}{(3c+12)/(c-2)}

simplify

\frac{\frac{c ^{2} + 4 c}{c ^{2} - 4}}{\frac{3 c + 12}{c - 2}}

\frac{c}{3 ( c + 2 )}

Solution steps

\frac{\frac{c ^{2} + 4 c}{c ^{2} - 4}}{\frac{3 c + 12}{c - 2}}

Apply the fraction rule:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= \frac{( c ^{2} + 4 c ) ( c - 2 )}{( c ^{2} - 4 ) ( 3 c + 12 )}

Cancel

\frac{( c ^{2} + 4 c ) ( c - 2 )}{( c ^{2} - 4 ) ( 3 c + 12 )} : \frac{c}{( c + 2 ) \cdot 3}

= \frac{c}{( c + 2 ) \cdot 3}

= \frac{c}{3 ( c + 2 )}

cos (θ) (1 + tan^{2} (θ))

\frac{x}{2} \cdot 2

(\frac{1}{3}) (\frac{3}{10}) (\frac{5}{7})

\frac{x + 2}{4 x ^{2} + 5 x + 1} \cdot \frac{4 x + 1}{x ^{2} - 4}

What is (\frac{c^2+4c)/(c^2-4)}{(3c+12)/(c-2)} ?
The solution to (\frac{c^2+4c)/(c^2-4)}{(3c+12)/(c-2)} is c/(3(c+2))