Recent content by Gzyousikai

Isn't it related to density? By setting the equation: Ma=m(t)g solve for m(t), we have m(t)=\frac{1}{5}\left[e^{\sqrt{\frac{g}{L}}t}+e^{-\sqrt{\frac{g}{L}}t}\right]=\frac{1}{5}\left[e^{\sqrt{5}t}+e^{-\sqrt{5}t}\right] if we solve [itex]m(t)=M[\itex] for t, the result would include density

underpinned by Newton's 3rd law.

It is known that the vector in polar coordinate system can be expressed as \mathbf{r}=r\hat{r}. In this formula, we don't see \hat{\theta} appear. But after the derivation yielding speed, \mathbf{v}=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta}. Where does theta come from? And how to define its...

it is a rectangle...

x(t)=v_0\cos\alpha t, y(t)=H+v_0\sin\alpha t-\frac{1}{2}gt^2 combine those two parametric equations, we get y=x\tan\alpha-\frac{g\sec^2\alpha}{2v_0^2}x^2+H let y=0 and solve for x

s(t)=v_0t-1/2gt^2, s(t)_{\text{max}}=h=f(v_0,g), solve s(t)=\frac{h}{2} for t, and then plugin t=10s, get the larger answer

(x^2-x+p)(11y^2-4y+2)=\left[\left(x-\frac{1}{2}\right)^2+p-\frac{1}{4}\right]\left[11\left(y-\frac{2}{11}\right)^2+\frac{18}{11}\right]\ge\left(p-\frac{1}{4}\right)\left(\frac{18}{11}\right)=\frac{9}{4}. solve for p and get p=3

You have too assume that x, y are reals first... divide y both side, and let \frac{x^n}{y^n}=t deduct to proof t^n=1 has only one real solution which is one. which is equivalent to (t-1)\left(\sum_{i=0}^{n}t^i\right)=0 which is relative to the root of unity.

Assume that x,y>0.Let P(x,y)be the polynomial, then P(x,y)=P(-x,-y)\ge P(x,-y)=x^4+y^4-x^3y-xy^3+x^2y^2\ge x^2y^2\ge 0 And the equality cannot hold.