Номер 12, 13, 14. Заранее спасибо)

Решите задачу:

$\displaystyle \tt 12).\\\\a). \ \ y=16x^{4}; \ \ \ \ x\leq0 \ \ \ \ \ \ \ \ \ x^{4}=\frac{y}{2^{4}} \ \Rightarrow \ \ x=-\frac{1}{2} \sqrt[4]{y};\\\\b). \ \ y=-x^{6}; \ \ \ \ \ x\geq0 \ \ \ \ \ \ \ \ \ x^{6}=-y \ \Rightarrow \ \ x=-\sqrt[6]{(-y)};\\\\$

$\displaystyle \tt c). \ \ y=x^{7} \ \ \ \Rightarrow \ \ x=\sqrt[7]{y};\\\\d). \ \ y=\frac{4}{9}\cdot x^{2}; \ \ \ \ \ x\geq0 \ \ \ \ \ \ \ \ x^{2}=\frac{9}{4}\cdot y \ \ \ \Rightarrow \ \ x=1,5\sqrt{y};\\\\e). \ \ y=-8x^{3} \ \ \ \Rightarrow \ \ x=\sqrt[3]{ \frac{y}{-8}}=-\frac{1}{2}\cdot\sqrt[3]{y};\\\\f). \ \ y=\frac{1}{32}\cdot x^{5} \ \ \ \Rightarrow \ \ x=2\sqrt[5]{y};$

$\displaystyle \tt 13). \ \ C=\frac{5}{9}(F-32) \ \ \ \Rightarrow \ \ F=\frac{9}{5}\cdot C+32\\\\{} \ \ \ \ \ \ \ C=15; \ \ \ \ \ F=\frac{9\cdot 15}{5}+32=27+32=59;\\\\{} \ \ \ \ \ \ \ C=20; \ \ \ \ \ F=\frac{9\cdot 20}{5}+32=36+32=68;\\\\{} \ \ \ \ \ \ \ C=0; \ \ \ \ \ F=\frac{9\cdot 0}{5}+32=0+32=32; \\\\\\$

$\displaystyle \tt 14). \ \ m=(9,37\cdot 10^{-6})\cdot L^{3} \ \ \ \Rightarrow \ \ L=\sqrt[3]{\frac{m}{9,37\cdot 10^{-6}}}=100\cdot \sqrt[3]{\frac{m}{9,37}};\\\\\\{} \ \ \ \ \ \ \ m=0,875; \ \ \ \ \ L=100\cdot \sqrt[3]{\frac{0,875}{9,37}}\approx100\cdot\sqrt[3]{0,09338}\approx45(cm);$