How do you simplify $\frac{{(2 m^{4} n^{2})}^{4}}{- 18 m^{5} n^{4}}$ $(2m^4n^2)^4/(-18m^5n^4)$ ?

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Answer 1 · 2015-12-15T00:14:45

$- \frac{8 m^{11} n^{4}}{9}$ $-(8m^11n^4)/9$

Use the properties of exponents. First, distribute the fourth power in the numerator:

$\frac{{(2 m^{4} n^{2})}^{4}}{- 18 m^{5} n^{4}} = \frac{16 m^{16} n^{8}}{- 18 m^{5} n^{4}}$ $(2m^4n^2)^4/(-18m^5n^4)=(16m^16n^8)/(-18m^5n^4)$

Then, cancel common factors by subtracting exponents in the denominator from those in the numerator:

$\frac{16 m^{16} n^{8}}{- 18 m^{5} n^{4}} = \frac{16 m^{11} n^{4}}{- 18 m^{0} n^{0}}$ $(16m^16n^8)/(-18m^5n^4)=(16m^11n^4)/(-18m^0n^0)$

Finally, simplify the zero exponents and the whole numbers:

$\frac{16 m^{11} n^{4}}{- 18 m^{0} n^{0}} = - \frac{8 m^{11} n^{4}}{9}$ $(16m^11n^4)/(-18m^0n^0) = -(8m^11n^4)/9$

How do you simplify (2m4n2)4−18m5n4(2m^4n^2)^4/(-18m^5n^4)?