How do you multiply $\frac{3 z + 12}{8 z^{3}} \cdot \frac{16 z^{3}}{9 z + 36}$ $(3z+12)/(8z^3)*(16z^3)/(9z+36)$ ?

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Answer 1 · 2018-07-25T21:46:26

$\frac{2}{3}$ $2/3$

Given that

$\frac{3 z + 12}{8 z^{3}} \cdot \frac{16 z^{3}}{9 z + 36}$ $\frac{3z+12}{8z^3}\cdot \frac{16z^3}{9z+36}$

$= \frac{3 (z + 4)}{8 z^{3}} \cdot \frac{16 z^{3}}{9 (z + 4)}$ $=\frac{3(z+4)}{8z^3}\cdot \frac{16z^3}{9(z+4)}$

$= (\frac{3 \cdot 16}{8 \cdot 9}) (\frac{z + 4}{z + 4}) (\frac{z^{3}}{z^{3}})$ $=(\frac{3\cdot 16}{8\cdot 9})(\frac{z+4}{z+4}) (\frac{z^3}{z^3})$

$= (\frac{2}{3}) (1) (1)$ $=(2/3)(1)(1)$

$= \frac{2}{3}$ $=2/3$

Answer 2 · 2018-07-26T10:55:22

$\frac{2}{3}$ $2/3$

$\frac{3 z + 12}{8 z^{3}} \cdot \frac{16 z^{3}}{9 z + 36}$ $(3z+12)/(8z^3)*(16z^3)/(9z+36)$

= $\frac{3 z + 12}{9 z + 36} \cdot \frac{16 z^{3}}{8 z^{3}}$ $(3z+12)/(9z+36)*(16z^3)/(8z^3)$

= $\frac{3 \cdot (z + 4)}{9 \cdot (z + 4)} \cdot 2$ $(3*(z+4))/(9*(z+4))*2$

= $\frac{1}{3} \cdot 2$ $1/3*2$

= $\frac{2}{3}$ $2/3$

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