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Question Number 64384 by aliesam last updated on 17/Jul/19 Answered by MJS last updated on 17/Jul/19 $$(y=x)\cap (y=\frac{x}{8})=\left(\begin{array}{c}0\\ 0\end{array}\right)$$$$(y=x)\cap (y=\frac{1}{x^2})=\left(\begin{array}{c}1\\ 1\end{array}\right)$$$$\:\:\:\:\:\left[{x}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\Rightarrow\:{x}^{\mathrm{3}} =\mathrm{1}\:\Rightarrow\:{x}=\mathrm{1}\right] \…

Question Number 129892 by mohammad17 last updated on 20/Jan/21 Answered by Ar Brandon last updated on 20/Jan/21 $$\mathrm{GP}\mathrm{with}\mathrm{U}\left(1\right)=\frac{11\mathrm{x}}{12}\mathrm{and}\mathrm{r}=\frac{1}{12}$$$${\mathrm{S}}_{\mathrm{\infty }}=\frac{\frac{11}{12}\mathrm{x}}{1-\frac{1}{12}}=\mathrm{x}$$ Terms of…

Question Number 129890 by 0731619177 last updated on 20/Jan/21 Answered by Olaf last updated on 20/Jan/21 $$\underset{x\to 3}{\lim }\frac{\mathrm{\Gamma }(x+1)-6}{xx-33}$$$$=\underset{x\to 3}{\lim }\frac{\mathrm{\Gamma }(x+1)-6}{11(x-3)}$$$$=\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\Gamma'\left({x}+\mathrm{1}\right)}{\mathrm{11}} \…

Question Number 129887 by zarawan last updated on 20/Jan/21 $$Isthevector\left[\begin{array}{c}1\\ -2\\ 1\end{array}\right]aneigenvectorof\left[\begin{array}{ccc}3& 6& 7\\ 3& 3& 7\\ 5& 6& 5\end{array}\right]?ifspfindthecorrespondingeigenvalue?$$ Answered by Olaf last updated on 20/Jan/21 $$\mathrm{AX}=\left[\begin{array}{ccc}3& 6& 7\\ 3& 3& 7\\ 5& 6& 5\end{array}\right]\left(\begin{array}{c}1\\ -2\\ 1\end{array}\right)=\left(\begin{array}{c}-2\\ 4\\ -2\end{array}\right)$$$$=-2\left(\begin{array}{c}1\\ -2\\ 1\end{array}\right)=-2\mathrm{X}$$$$\exists\lambda\backslash\:\mathrm{AX}\:=\:\lambda\mathrm{X} \…

Question Number 129874 by mohammad17 last updated on 20/Jan/21 Commented by mohammad17 last updated on 20/Jan/21 $$pleassirhelpme$$ Answered by liberty last updated on…