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2∙ (∛(5x+5x)∙(2^2 ))/(yb-(sin30+yb) )    | sin 30° ??

Hallo Gast!

$2\cdot (∛(5x+5x)\cdot (2^2 ))/(yb-(sin\ 30^{\circ}+yb) )$

$=\dfrac{2\sqrt[3]{5x+5x}\cdot 2^2}{yb-sin\ 30^{\circ}-yb}=\dfrac{8\sqrt[3]{10x}}{-0.5}=\color{blue}-16\sqrt[3]{10x}= -34,471\cdot \sqrt[3]{x}$

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