Question Number 213530 by Ari last updated on 07/Nov/24 Commented by Ari last updated on 07/Nov/24 $${C}\left({x},{y}\right)=\mathrm{4}{x}+\mathrm{6}{y} \\ $$$${where}\:{x}−{one}\:{type}\:{of}\:{tables} \\ $$$${y}−{other}\:{type} \\ $$$$ \\ $$…
Question Number 213522 by efronzo1 last updated on 07/Nov/24 $$\:\:\mathrm{The}\:\mathrm{two}\:\mathrm{corner}\:\mathrm{points}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}\: \\ $$$$\:\:\mathrm{lie}\:\mathrm{on}\:\mathrm{curve}\:\mathrm{f}\left(\mathrm{x}\right)=\:\mathrm{x}^{\mathrm{2}} −\mathrm{2x}−\mathrm{3}\:\mathrm{and}\: \\ $$$$\:\mathrm{the}\:\mathrm{other}\:\mathrm{two}\:\mathrm{corner}\:\mathrm{points}\:\mathrm{lie}\:\mathrm{on}\: \\ $$$$\:\mathrm{curve}\:\mathrm{g}\left(\mathrm{x}\right)=\:−\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{3}\:.\:\mathrm{It}\:\mathrm{is}\:\mathrm{known} \\ $$$$\:\mathrm{that}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}\:\mathrm{can}\:\mathrm{be}\: \\ $$$$\:\:\mathrm{expressed}\:\mathrm{by}\:\mathrm{p}+\mathrm{q}\sqrt{\mathrm{r}}\:,\:\mathrm{for}\:\mathrm{a}\:\mathrm{natural}\: \\ $$$$\:\mathrm{number}\:\mathrm{p},\mathrm{q}\:,\mathrm{r}\:\mathrm{where}\:\mathrm{r}\:\mathrm{is}\:\mathrm{not}\:\mathrm{divisible}\: \\…
Question Number 213518 by issac last updated on 07/Nov/24 $$\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \:\frac{{z}\centerdot\mathrm{sin}\left({z}\right)}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \left({z}\right)}\:\mathrm{d}{z} \\ $$$$\int_{\:\mid{z}\mid=\mathrm{2}} \:\frac{\mathrm{1}}{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{z} \\ $$$$\int_{\:\mid{z}\mid=\mathrm{2}} \:\frac{\mathrm{sin}\left({z}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{z} \\ $$ Answered by…
Question Number 213519 by RojaTaniya last updated on 07/Nov/24 Answered by mr W last updated on 07/Nov/24 $$\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{{x}} \\ $$$$\mathrm{tan}\:\left(\alpha+\beta\right)=\frac{\mathrm{3}}{{x}} \\ $$$$\mathrm{tan}\:\left(\alpha+\alpha+\beta\right)=\frac{\mathrm{6}}{{x}} \\ $$$$\frac{\mathrm{6}}{{x}}=\frac{\frac{\mathrm{1}}{{x}}+\frac{\mathrm{3}}{{x}}}{\mathrm{1}−\frac{\mathrm{1}}{{x}}×\frac{\mathrm{3}}{{x}}}=\frac{\mathrm{4}{x}}{{x}^{\mathrm{2}} −\mathrm{3}}…
Question Number 213511 by universe last updated on 07/Nov/24 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\:\:\left[\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{r}} }\right] \\ $$$$\:\:\:\mathrm{where}\:\left[\bullet\right]\:\mathrm{greatest}\:\mathrm{integer}\:\mathrm{finction} \\ $$ Answered by issac last updated on 07/Nov/24…
Question Number 213468 by universe last updated on 06/Nov/24 $$\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sequence}\:\left\{\mathrm{a}_{\mathrm{n}} \right\}\:\mathrm{defined}\: \\ $$$$\mathrm{recurssively}\:\mathrm{by}\:\mathrm{a}_{\mathrm{1}} =\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{a}_{\mathrm{n}\:} =\:\sqrt{\mathrm{3a}_{\mathrm{n}−\mathrm{1}\:} \:−\mathrm{2}\:}\:\:\:\:\mathrm{for}\:\mathrm{n}\geqslant\mathrm{2}\:\:\mathrm{converges}\: \\ $$$$\mathrm{and}\:\mathrm{find}\:\mathrm{its}\:\mathrm{limit}. \\ $$ Answered by issac…
Question Number 213503 by Spillover last updated on 06/Nov/24 Answered by mr W last updated on 06/Nov/24 $$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{\mathrm{2}{n}} =\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\:{for}\:\mid{x}\mid<\mathrm{1} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty}…
Question Number 213499 by issac last updated on 06/Nov/24 $$\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \frac{{z}\centerdot\mathrm{sin}\left({z}\right)}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \left({z}\right)}\mathrm{d}{z}\:=?\:\left(\mathrm{contour}\:\mathrm{integral}\right) \\ $$$$\mathrm{pls}\:\mathrm{help}….. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 213467 by efronzo1 last updated on 06/Nov/24 $$\:\:\:\underbrace{\boldsymbol{{B}}} \\ $$ Commented by mr W last updated on 06/Nov/24 �� Commented by mr W…
Question Number 213492 by a.lgnaoui last updated on 06/Nov/24 Commented by a.lgnaoui last updated on 07/Nov/24 $$\mathrm{AH}=\mathrm{4}+\mathrm{2r}\:\:\:,\:\mathrm{6}^{\mathrm{2}} =\left(\mathrm{4}+\mathrm{r}\right)^{\mathrm{2}} −\mathrm{r}^{\mathrm{2}} \\ $$ Commented by A5T last…