What is (16a^3b^6c^{-2})/(10ab^{-2)c^3} ?

The solution to (16a^3b^6c^{-2})/(10ab^{-2)c^3} is (8a^2b^8)/(5c^5)

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simplify (16a^3b^6c^{-2})/(10ab^{-2)c^3}

simplify

\frac{16 a ^{3} b ^{6} c ^{- 2}}{10 a b ^{- 2} c ^{3}}

\frac{8 a ^{2} b ^{8}}{5 c ^{5}}

Solution steps

\frac{16 a ^{3} b ^{6} c ^{- 2}}{10 a b ^{- 2} c ^{3}}

Factor the number:

16 = 2 \cdot 8

= \frac{2 \cdot 8 a ^{3} b ^{6} c ^{- 2}}{10 a b ^{- 2} c ^{3}}

Factor the number:

10 = 2 \cdot 5

= \frac{2 \cdot 8 a ^{3} b ^{6} c ^{- 2}}{2 \cdot 5 a b ^{- 2} c ^{3}}

Cancel the common factor:

2

= \frac{8 a ^{3} b ^{6} c ^{- 2}}{5 a b ^{- 2} c ^{3}}

Simplify

\frac{a ^{3}}{a} : a^{2}

= \frac{8 a ^{2} b ^{6} c ^{- 2}}{5 b ^{- 2} c ^{3}}

Simplify

\frac{b ^{6}}{b ^{- 2}} : b^{8}

= \frac{8 a ^{2} b ^{8} c ^{- 2}}{5 c ^{3}}

Simplify

\frac{c ^{- 2}}{c ^{3}} : \frac{1}{c ^{5}}

= \frac{8 a ^{2} b ^{8}}{5 c ^{5}}

lo g_{4} (\frac{1}{64}) = x

- 10 - {9 - [- 8 + 2 (- 7) + 6] + 2} - 4

x^{2} - 8 x + 17 > 0

5 u + 16 = 3 u + 6 u - 12

(\frac{3 i}{n})^{2}

What is (16a^3b^6c^{-2})/(10ab^{-2)c^3} ?
The solution to (16a^3b^6c^{-2})/(10ab^{-2)c^3} is (8a^2b^8)/(5c^5)