+15054 Rechne bitte !
2∙ (∛(5x+5x)∙(2^2 ))/(yb-(sin30+yb) ) | sin 30° ??
Hallo Gast!
\( 2\cdot (∛(5x+5x)\cdot (2^2 ))/(yb-(sin\ 30^{\circ}+yb) ) \)
\(=\dfrac{2\sqrt[3]{5x+5x}\cdot 2^2}{yb-sin\ 30^{\circ}-yb}=\dfrac{8\sqrt[3]{10x}}{-0.5}=\color{blue}-16\sqrt[3]{10x}= -34,471\cdot \sqrt[3]{x}\)
!
