What is (\frac{n-4)/(7n^2-28n)}{(3n)/(n-2)} ?

The solution to (\frac{n-4)/(7n^2-28n)}{(3n)/(n-2)} is (n-2)/(21n^2)

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simplify (\frac{n-4)/(7n^2-28n)}{(3n)/(n-2)}

simplify

\frac{\frac{n - 4}{7 n ^{2} - 28 n}}{\frac{3 n}{n - 2}}

\frac{n - 2}{21 n ^{2}}

Solution steps

\frac{\frac{n - 4}{7 n ^{2} - 28 n}}{\frac{3 n}{n - 2}}

Apply the fraction rule:

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

= \frac{( n - 4 ) ( n - 2 )}{( 7 n ^{2} - 28 n ) \cdot 3 n}

Cancel

\frac{( n - 4 ) ( n - 2 )}{( 7 n ^{2} - 28 n ) \cdot 3 n} : \frac{n - 2}{21 n ^{2}}

= \frac{n - 2}{21 n ^{2}}

\frac{13}{12} - \frac{1}{6}

(s + 3)^{2}

- 5 \cdot (- 5)

\frac{\frac{1}{4} x ^{5} y z ^{2}}{( 0.2 x ^{2} y ^{3} z ^{4} )}

lo g_{10} (0.9)

What is (\frac{n-4)/(7n^2-28n)}{(3n)/(n-2)} ?
The solution to (\frac{n-4)/(7n^2-28n)}{(3n)/(n-2)} is (n-2)/(21n^2)