Question Number 55323 by Kunal12588 last updated on 21/Feb/19 Commented by Kunal12588 last updated on 21/Feb/19 Commented by Kunal12588 last updated on 21/Feb/19 $${is}\:{this}\:{correct}\:{sir} \\…
Question Number 55316 by Kunal12588 last updated on 21/Feb/19 Commented by Kunal12588 last updated on 21/Feb/19 $${pls}\:{help}\:{me}\:{i}\:{found}\:\mathrm{2519}\:{but}\:{don}'{t}\:{know}\:{is}\:{this}\: \\ $$$${smallest}.\:{also}\:{it}\:{takes}\:\mathrm{2}\:{hrs}\:{for}\:{me} \\ $$$$\:{every}\:{answers}\:{are}\:{welcome} \\ $$ Answered by…
Question Number 55306 by Rio Mike last updated on 21/Feb/19 $${A}\:{mass}\:{of}\:\mathrm{6}{kg}\:{lies}\:{on}\:{an}\:{inclined}\:{plane} \\ $$$${which}\:{is}\:{smooth}\:{at}\:{angle}\:\theta\:{to}\:{the}\:{horizontal} \\ $$$${where}\:\:\mathrm{sin}\:\theta=\:\frac{\mathrm{1}}{\mathrm{3}}.{if}\:{it}\:{is}\:{connected}\:{to} \\ $$$${another}\:{mass}\:\mathrm{8}{kg}\:{by}\:{the}\:{same}\:{inelastic}\:{string} \\ $$$${passing}\:{over}\:{a}\:{smooth}\:{fixed}\:{pulley} \\ $$$${at}\:{the}\:{top}\:{of}\:{the}\:{plane}.{the}\:{partices}\:{are} \\ $$$${released}\:{from}\:{rest}\:,{Find}\: \\ $$$$\left.{a}\right)\:{the}\:{acceleration}\:{of}\:{each}\:{mass}…
Question Number 55296 by Otchere Abdullai last updated on 20/Feb/19 $${simplify}\:{the}\:{following}\:{expression}\:{if} \\ $$$${x}<\mathrm{0} \\ $$$$\mid\mathrm{4}{x}−\sqrt{\left(\mathrm{3}{x}−\mathrm{1}\right)^{\mathrm{2}} }\mid \\ $$ Answered by mr W last updated on…
Question Number 186370 by MathsFan last updated on 04/Feb/23 $$ \\ $$A gear with outer radius of r'=5cm move in a clockwise direction causing an…
Question Number 120827 by TITA last updated on 03/Nov/20 Commented by liberty last updated on 03/Nov/20 $$\mathrm{125}×\mathrm{666}×\mathrm{798}×\mathrm{1373}×\mathrm{77777}×\mathrm{111111}\: \\ $$$$\mathrm{7}.\mathrm{882535124571993E20} \\ $$$$ \\ $$ Commented by…
Question Number 120797 by Algoritm last updated on 02/Nov/20 Commented by liberty last updated on 02/Nov/20 $$\begin{cases}{\mathrm{2cos}\:\mathrm{xcos}\:\mathrm{y}=\mathrm{3sin}\:\mathrm{y}}\\{\mathrm{2cos}\:\mathrm{ycos}\:\mathrm{z}=\mathrm{3sin}\:\mathrm{z}}\\{\mathrm{2cos}\:\mathrm{zcos}\:\mathrm{x}=\mathrm{3sin}\:\mathrm{x}}\end{cases}\rightarrow\mathrm{8cos}\:^{\mathrm{2}} \mathrm{xcos}\:^{\mathrm{2}} \mathrm{ycos}\:^{\mathrm{2}} \mathrm{z}=\mathrm{27sin}\:\mathrm{xsin}\:\mathrm{ysin}\:\mathrm{z} \\ $$$$\mathrm{by}\:\mathrm{symetric}\:\begin{cases}{\mathrm{2cos}\:^{\mathrm{2}} \mathrm{x}=\mathrm{3sin}\:\mathrm{x}}\\{\mathrm{2cos}\:^{\mathrm{2}} \mathrm{y}=\mathrm{3sin}\:\mathrm{y}}\\{\mathrm{2cos}\:^{\mathrm{2}} \mathrm{z}=\mathrm{3sin}\:\mathrm{z}}\end{cases}…
Question Number 55240 by Otchere Abdullai last updated on 19/Feb/19 $${make}\:{r}\:{the}\:{subject}\:{of}\:{the}\:{relation} \\ $$$${m}=\frac{\mathrm{4}\sqrt{{u}+{r}}}{{v}−{r}} \\ $$ Answered by MJS last updated on 20/Feb/19 $${m}\left({v}−{r}\right)=\mathrm{4}\sqrt{{u}+{r}} \\ $$$$\mathrm{squaring}\:\mathrm{both}\:\mathrm{sides}…
Question Number 120773 by shaker last updated on 02/Nov/20 Answered by 675480065 last updated on 02/Nov/20 $$\mathrm{let}\:\mathrm{u}=\sqrt{\mathrm{x}} \\ $$$$\Rightarrow\:\mathrm{du}=\frac{\mathrm{1}}{\mathrm{2u}}\mathrm{dx} \\ $$$$\Rightarrow\:\mathrm{dx}=\mathrm{2udu} \\ $$$$\Rightarrow\:\mathrm{I}\:=\:\int\frac{\mathrm{e}^{\mathrm{u}} .\mathrm{e}^{\mathrm{u}^{\mathrm{2}} }…
Question Number 120768 by premsambath last updated on 02/Nov/20 Terms of Service Privacy Policy Contact: info@tinkutara.com