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Question Number 4387 by Rasheed Soomro last updated on 17/Jan/16 Commented by Rasheed Soomro last updated on 17/Jan/16 $$\mathrm{In}\:\mathrm{the}\:\mathrm{trapezium}\:\mathrm{m}\angle\mathrm{A}=\mathrm{m}\angle\mathrm{B}=\frac{\pi}{\mathrm{2}}\:\mathrm{rad}. \\ $$$$\mathrm{m}\overline {\mathrm{AB}}=\mathrm{m}\overline {\mathrm{AD}}=\mathrm{x}\:\mathrm{units}\:\mathrm{and}\:\mathrm{m}\overline {\mathrm{BC}}=\mathrm{2x}\:\mathrm{units}. \\…
Question Number 135459 by Abdoulaye last updated on 13/Mar/21 $${help}\:{me} \\ $$$$ \\ $$$${S}=\left(\mathrm{1}−\mathrm{2}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} +\mathrm{4}\right)+\left(\mathrm{5}^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} \right)+\left(\mathrm{9}^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} −\mathrm{11}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} \right)+ \\…
Question Number 135458 by Dwaipayan Shikari last updated on 13/Mar/21 Commented by Dwaipayan Shikari last updated on 13/Mar/21 Commented by Dwaipayan Shikari last updated on…
Question Number 4384 by Rasheed Soomro last updated on 16/Jan/16 $$\mathrm{A}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{r}_{\mathrm{1}} \:\mathrm{has}\:\mathrm{been}\:\mathrm{divided} \\ $$$$\mathrm{into}\:\mathrm{two}\:\mathrm{parts}\:\mathrm{of}\:\mathrm{equal}\:\mathrm{area}, \\ $$$$\mathrm{by}\:\mathrm{an}\:\mathrm{arc}\:\mathrm{having}\:\mathrm{center}\:\mathrm{on}\:\mathrm{the}\:\mathrm{circle}. \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{radius}\left(\mathrm{r}_{\mathrm{2}} \right)\:\mathrm{of}\:\mathrm{the}\:\mathrm{arc}. \\ $$ Commented by Rasheed Soomro…
Question Number 135452 by mohammad17 last updated on 13/Mar/21 Commented by mohammad17 last updated on 13/Mar/21 $${how}\:{can}\:{two}\:{prove}\:{this}\:? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 135455 by mohammad17 last updated on 13/Mar/21 Commented by mohammad17 last updated on 13/Mar/21 $${solve}\:{the}\: \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 4378 by madscientist last updated on 14/Jan/16 $${what}\:{are}\:{the}\:{formulas}\:{for}\:{functions} \\ $$$$\Phi\left({x}\right)\:{and}\:\Psi\left({x}\right)? \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 69912 by naka3546 last updated on 29/Sep/19 $$\sqrt{{xy}}\:\:+\:\:\mathrm{3}\:\:=\:\:{y} \\ $$$${Find}\:\:\frac{{dy}}{{dx}}\:\:. \\ $$ Answered by MJS last updated on 29/Sep/19 $$\frac{{y}}{\mathrm{2}\sqrt{{xy}}}{dx}+\frac{{x}}{\mathrm{2}\sqrt{{xy}}}{dy}={dy} \\ $$$$\frac{{y}}{\mathrm{2}\sqrt{{xy}}}{dx}=\left(\mathrm{1}−\frac{{x}}{\mathrm{2}\sqrt{{xy}}}\right){dy} \\…
Question Number 4374 by Rasheed Soomro last updated on 14/Jan/16 Commented by Rasheed Soomro last updated on 14/Jan/16 $$\mathcal{I}{n}\:{trapizium}\:\mathrm{ABCD} \\ $$$$\mathrm{A}\:{and}\:\mathrm{B}\:{are}\:{right}\:{angles} \\ $$$$\mathrm{AB}=\mathrm{AD}=\mathrm{x}\:\mathrm{units}\:{and}\:\mathrm{BC}=\mathrm{2x}\:\mathrm{units}. \\ $$$${The}\:{trapizium}\:\mathrm{ABCD}\:{has}\:{been}\:…