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\text{from (2),(3) and (4) }m_2(v+eu)=m_2u+m_3(u+e(v+eu))\Rightarrow \frac{v}{u}= \dfrac{m_2+m_3+e^2m_3-em_2}{m_2-em_3}}
\text{so } }=\dfrac{m_1-em_2}{m_1+m_2}=\dfrac{m_2+m_3+e^2m_3-em_2}{m_2-em_3}}}}
[br]\Rightarrow (m_1-em_2)(m_2-em_3)=(m_2+m_3+e^2m_3-em_2)(m_1+m_2) }}
\Rightarrow m_1m_2-em_2^2-em_1m_3+e^2m_2m_3 }}
\text{If f}(x)=Ax^2\text{exp}\left(-\dfrac{x^2}{2}\right) \text{ is a p.d.f. Then we must have } \displaystyle \int_{-\infty}^\inftyAx^2\text{exp}\left(-\dfrac{x^2}{2}\right)dx=1[br]
\text{larger than 87.3 is given by }\left[-\dfrac{1}{\sqrt{2\pi}}\text{exp}\left(-\dfrac{x^2}{2}\right]_{87.3}^\infty
\mathbf{e}\cdot\mathbf{n} = \mathbf{q}\cdot{\mathbf{n}-(\mathbf{q} \cdot \mathbf{n})\mathbf{n}\cdot\mathbf{n} = 0
|\bf{s}-\bf{x}|^2 = |\bf{s}-\bf{r}-(\bf{x}-\bf{r})|^2 = |\lambda \bf{n} - (\bf{x}-\bf{r})|^2 \\[br]= \lambda^2 |\bf{n}|^2 -2 (\bf{x}-\bf{r})\cdot\bf{n}+|\bf{x}-\bf{r}|^2 = \lambda^2 + 1
\displaystyle m = \int_0^r \frac{2\pi \rho x^2}{r} dx = \left[\frac{2\pi\rhox^3}{3r}\right]_0^r = \frac{2\pi}{3}\rho r^2
\displaystyle I_C = \int_0^r \frac{2\pi \rho x^4}{r}dx = \left[\frac{2\pi\rhox^5}{5r}\right]_0^r = \frac{2\pi}{5} \rho r^4 = \frac{3mr^2}{5}
\text {if the polar coordinates of the point }(x,y) \text { are }(r, \theta) \text{ then f}(x,y)= \phi(r, \theta) \text{ say, where } \Phi(r, \theta)= \dfrac{1}{2\pi} \text{e}^_{-r^2}
\delta \alpha= \tan^{-1}(k+ \delta k)- \tan^{-1}k= \sdelta k \dfrac{ \text{d}}{ \text{d}k} \tan^{-1} k=\dfrac{ \delta k}{1+k^2}
\text {now if } \alpha= \beta \text{ we have }\alpha=\alpha(1-q_0-q_1)+q_0 \Leftrightarrow \alpha(q_0+q_1)=q_0 \Leftrigfhtarrow \alpha= \dfrac{q_0}{q_0+q_1}
\implies q=1- \text{exp} \left( \dfrac{1}{10^6} \ln o=0.5 \right)=1- \text{exp} \lefdt( -\dfrac {\ln 2}{10^6} \right)= \dfrac{\ln2}{10^6}
\text { when }L=L_0 \text { say }q= \dfrac{ \ln2}{10^6} \text { or } \dfrac{\ln2}{10^6}=kL_0 \implies q= \dfrac{\ln2}{10^6L_0^2}L^2 \text { so when }L= \drac{1}{2}L_0, q= \dfrac{\ln2}{4 \times 10^6}
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