Study questions platform-wide or filter by specific tests with correct answers revealed.
If $\alpha,\beta\in\mathbb{C}$ are distinct roots of $x^2-x+1=0$, find $\alpha^{101}+\beta^{107}$.
The roots of $x^2-x+1=0$ are $x=\frac{1\pm\sqrt{-3}}{2}=-\omega,-\omega^2$ where $\omega=e^{2\pi i/3}$.
Let $\alpha=-\omega, \beta=-\omega^2$. Then:
$\alpha^{101}=(-\omega)^{101}=-\omega^{101}=-\omega^{101\mod3}=-\omega^2$ (since $101=33\times3+2$)
$\beta^{107}=(-\omega^2)^{107}=-\omega^{214}=-\omega^{214\mod3}=-\omega^1$ (since $214=71\times3+1$)
$\alpha^{101}+\beta^{107}=-\omega^2-\omega=-(w+\omega^2)=-(-1)=\mathbf{1}$
4 g of Mg heated in excess Oโ. Theoretical yield of MgO (g) is: \(2\mathrm{Mg} + \mathrm{O}_2 \rightarrow 2\mathrm{MgO}\)
Molar mass Mg = 24 g/mol: moles Mg = 4/24 = 0.1667 mol. 2 mol Mg โ 2 mol MgO, so mol MgO = 0.1667 mol. Molar mass MgO = 40 g/mol: mass = 0.1667ร40 = 6.67 g โ 6.6 g.
Induced emf in 1m copper rod moving at 20 m/s perpendicular to 0.6 T field:
Sign in to join the conversation and share your thoughts.
Log In to Comment