Question Number 143367 by Sawadogo last updated on 13/Jun/21 Answered by TheHoneyCat last updated on 13/Jun/21 $$ \\ $$$$\mathrm{Ceci}\:\mathrm{n}'\mathrm{est}\:\mathrm{pas}\:\mathrm{la}\:\mathrm{reponse}\:{complete}\:\mathrm{mais}\:\mathrm{la}\:\mathrm{traduction}\:\mathrm{en}\:\mathrm{termes}\:\mathrm{plus}\:\mathrm{lisibles}: \\ $$$${svp}\:{completez} \\ $$$$ \\ $$$$\mathrm{Premiere}\:\mathrm{Partie}…
Question Number 12292 by sin (x) last updated on 18/Apr/17 Answered by mrW1 last updated on 18/Apr/17 $${S}=\frac{\mathrm{1}}{\mathrm{6}}×\pi{r}^{\mathrm{2}} +\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{3}}−\mathrm{sin}\:\frac{\pi}{\mathrm{3}}\right) \\ $$$$=\frac{\pi{r}^{\mathrm{2}} }{\mathrm{3}}−\frac{{r}^{\mathrm{2}} }{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\…
Question Number 12291 by 1kanika# last updated on 18/Apr/17 Answered by 433 last updated on 07/May/17 $$\begin{bmatrix}{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{0}}\end{bmatrix}×\begin{bmatrix}{\mathrm{0}}&{−\mathrm{1}}\\{\mathrm{1}}&{\mathrm{0}}\end{bmatrix}=\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{−\mathrm{1}}\end{bmatrix}={A} \\ $$$$\begin{bmatrix}{\mathrm{0}}&{−\mathrm{1}}\\{\mathrm{1}}&{\mathrm{0}}\end{bmatrix}×\begin{bmatrix}{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{0}}\end{bmatrix}=\begin{bmatrix}{−\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}=−{A} \\ $$$$\begin{bmatrix}{\mathrm{0}}&{−\mathrm{1}}\\{\mathrm{1}}&{\mathrm{0}}\end{bmatrix}×\begin{bmatrix}{\mathrm{0}}&{−\mathrm{1}}\\{\mathrm{1}}&{\mathrm{0}}\end{bmatrix}=\begin{bmatrix}{−\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{−\mathrm{1}}\end{bmatrix}=−{I}_{\mathrm{2}} \\ $$$$\begin{bmatrix}{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{0}}\end{bmatrix}^{\mathrm{2}} ={I}_{\mathrm{2}} \\…
Question Number 12284 by tawa last updated on 17/Apr/17 Answered by mrW1 last updated on 18/Apr/17 $${let}\:{t}=\frac{\mathrm{1}}{\mathrm{2}{x}} \\ $$$$\mathrm{log}_{\mathrm{5}} \:\left(\mathrm{5}^{\frac{\mathrm{1}}{{x}}} +\mathrm{125}\right)=\mathrm{log}_{\mathrm{5}} \:\mathrm{6}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}}\:\Rightarrow \\ $$$$\mathrm{log}_{\mathrm{5}} \:\left(\mathrm{5}^{\mathrm{2}{t}}…
Question Number 77819 by aliesam last updated on 10/Jan/20 Answered by lémùst last updated on 10/Jan/20 $${posons}\:{t}=\:^{\mathrm{3}} \sqrt{{r}}\:,\:{ainsi}\:{t}\:−\:\frac{\mathrm{1}}{{t}}\:=\:\mathrm{1} \\ $$$$\Rightarrow{t}^{\mathrm{2}} −{t}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{t}=\pm\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${donc}\:{r}=\pm\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{3}}…
Question Number 77816 by aliesam last updated on 10/Jan/20 $$\int{sin}\left({x}^{\mathrm{2}} \right)\:{sin}\left({x}\right)\:{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 77817 by Dah Solu Tion last updated on 10/Jan/20 $$\int\frac{{dx}}{{x}^{\mathrm{3}} \left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}\right)^{\mathrm{4}} } \\ $$$$ \\ $$ Commented by mathmax by abdo last…
Question Number 12279 by chux last updated on 17/Apr/17 $$\mathrm{i}\:\mathrm{guess}\:\mathrm{i}\:\mathrm{posted}\:\mathrm{this}\:\mathrm{question}\:\mathrm{but} \\ $$$$\mathrm{didnt}\:\mathrm{really}\:\mathrm{get}\:\mathrm{much}\:\mathrm{response}. \\ $$$$\mathrm{pls}\:\mathrm{help}\:\mathrm{out} \\ $$$$ \\ $$$$\mathrm{if}\:\mathrm{8log}_{\mathrm{2}} \:\mathrm{x}\:+\:\left(\mathrm{x}−\mathrm{1}\right)\mathrm{log}_{\mathrm{x}} \:\mathrm{2}\:=\mathrm{3}^{\mathrm{x}} \: \\ $$$$\mathrm{find}\:\mathrm{x}. \\ $$…
Question Number 143348 by gsk2684 last updated on 13/Jun/21 $$\mathrm{if}\:\:\frac{\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{y}}+\frac{\mathrm{sin}\:^{\mathrm{4}} \mathrm{x}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{y}}=\mathrm{1}\:\mathrm{then}\: \\ $$$$\mathrm{find}\:\frac{\mathrm{cos}\:^{\mathrm{4}} \mathrm{y}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}+\frac{\mathrm{sin}\:^{\mathrm{4}} \mathrm{y}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}=? \\ $$ Answered by som(math1967)…
Question Number 143350 by gsk2684 last updated on 13/Jun/21 $$\mathrm{if}\:\mathrm{tan}^{\mathrm{2}} \alpha\mathrm{tan}^{\mathrm{2}} \beta+\mathrm{tan}^{\mathrm{2}} \beta\mathrm{tan}^{\mathrm{2}} \gamma+ \\ $$$$\mathrm{tan}^{\mathrm{2}} \gamma\mathrm{tan}^{\mathrm{2}} \alpha+\mathrm{2tan}^{\mathrm{2}} \alpha\mathrm{tan}^{\mathrm{2}} \beta\mathrm{tan}^{\mathrm{2}} \gamma=\mathrm{1} \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{sin}^{\mathrm{2}} \alpha+\mathrm{sin}^{\mathrm{2}} \beta+\mathrm{sin}^{\mathrm{2}}…