Question Number 63490 by kumaranjul306@gmail.com last updated on 04/Jul/19 $$\mathrm{A}\:\mathrm{father}\:\mathrm{with}\:\mathrm{8}\:\mathrm{children}\:\mathrm{takes}\:\mathrm{3}\:\mathrm{at}\:\mathrm{a}\: \\ $$$$\mathrm{time}\:\mathrm{to}\:\mathrm{the}\:\mathrm{garden}\:\mathrm{as}\:\mathrm{often}\:\mathrm{as}\:\mathrm{he} \\ $$$$\mathrm{without}\:\mathrm{taking}\:\mathrm{the}\:\mathrm{same}\:\mathrm{3}\:\mathrm{children} \\ $$$$\mathrm{together}\:\mathrm{more}\:\mathrm{than}\:\mathrm{once}.\:\mathrm{The}\:\mathrm{number} \\ $$$$\mathrm{of}\:\mathrm{times}\:\mathrm{he}\:\mathrm{will}\:\mathrm{go}\:\mathrm{to}\:\mathrm{the}\:\mathrm{garden}\:\mathrm{is} \\ $$ Commented by Prithwish sen last…
Question Number 63121 by hovea cw last updated on 29/Jun/19 $${x}^{\mathrm{1}/\mathrm{2}} \:\centerdot\:{x}^{\mathrm{1}/\mathrm{4}} \:\centerdot\:{x}^{\mathrm{1}/\mathrm{8}} \:\centerdot\:{x}^{\mathrm{1}/\mathrm{16}} \:…\:\mathrm{to}\:\infty\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$ Commented by Prithwish sen last updated on 29/Jun/19…
Question Number 63120 by hovea cw last updated on 29/Jun/19 $${x}^{\mathrm{1}/\mathrm{2}} \:\centerdot\:{x}^{\mathrm{1}/\mathrm{4}} \:\centerdot\:{x}^{\mathrm{1}/\mathrm{8}} \:\centerdot\:{x}^{\mathrm{1}/\mathrm{16}} \:…\:\mathrm{to}\:\infty\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 63065 by Enebeli Chinedu Vitalis last updated on 28/Jun/19 $$\mathrm{If}\:{I}=\underset{\:\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{{dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }}\:,\:\mathrm{then} \\ $$ Commented by Enebeli Chinedu Vitalis last updated on…
Question Number 63016 by TB last updated on 27/Jun/19 $$\mathrm{The}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{hexagon}\:\mathrm{are}\:\mathrm{enlarged}\:\mathrm{by}\: \\ $$$$\mathrm{three}\:\mathrm{times}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{the}\:\mathrm{areas} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{new}\:\mathrm{and}\:\mathrm{old}\:\mathrm{hexagon} \\ $$ Answered by MJS last updated on 28/Jun/19 $$\mathrm{no}\:\mathrm{matter}\:\mathrm{which}\:\mathrm{shape}\:\mathrm{you}\:\mathrm{enlarge} \\…
Question Number 62983 by hovea cw last updated on 27/Jun/19 $$\mathrm{If}\:\:\:\mathrm{tan}\:\mathrm{2}\theta\:\mathrm{tan}\:\theta\:=\:\mathrm{1},\:\mathrm{then}\:\theta\:= \\ $$ Answered by Hope last updated on 27/Jun/19 $${tan}\mathrm{2}\theta={cot}\theta={tan}\left(\frac{\pi}{\mathrm{2}}−\theta\right) \\ $$$${formula} \\ $$$$\left[{tan}\alpha={tan}\beta\:\:\:\alpha={n}\pi+\beta\right]…
Question Number 62612 by hovea cw last updated on 23/Jun/19 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{real}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\left(\frac{\mathrm{9}}{\mathrm{10}}\right)^{{x}} =\:−\mathrm{3}+{x}−{x}^{\mathrm{2}} \:\:\:\:\:\mathrm{is} \\ $$ Answered by tanmay last updated on 23/Jun/19 $$−\mathrm{3}+{x}−{x}^{\mathrm{2}}…
Question Number 62610 by hovea cw last updated on 23/Jun/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}\:\mathrm{in} \\ $$$$\frac{\mathrm{1}}{{x}−\mathrm{1}}\:+\:\frac{\mathrm{1}}{{x}−\mathrm{2}}\:=\:\frac{\mathrm{3}}{{x}−\mathrm{3}}\:\:. \\ $$ Commented by mathmax by abdo last updated on 24/Jun/19 $${the}\:{set}\:{of}\:{definition}\:{for}\:{this}\:\left({e}\right)\:{is}\:{D}\:={R}−\left\{\mathrm{1},\mathrm{2},\mathrm{3}\right\}…
Question Number 62611 by hovea cw last updated on 23/Jun/19 $$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of} \\ $$$${x}^{\mathrm{2}} −\left({a}+\mathrm{1}\right){x}+\frac{\mathrm{1}}{\mathrm{2}}\left({a}^{\mathrm{2}} +{a}+\mathrm{1}\right)=\mathrm{0}\:\mathrm{then} \\ $$$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\_\_\_\_\_. \\ $$ Commented by Prithwish sen…
Question Number 62609 by hovea cw last updated on 23/Jun/19 $$\mathrm{If}\:\:\mathrm{5}\mid{x}\mid\:+\:\mathrm{4}\mid{y}\mid\:=\:\mathrm{4}\:\mathrm{and}\:\mathrm{2}\mid{x}\mid\:−\:\mathrm{4}\mid{y}\mid\:=\:\mathrm{10}, \\ $$$$\mathrm{then}\:\mathrm{find}\:{x}\:\mathrm{and}\:{y}. \\ $$ Answered by $@ty@m last updated on 23/Jun/19 $${Let}\:\mid{x}\mid={a}\:\&\mid{y}\mid={b} \\ $$$$\mathrm{5}{a}+\mathrm{4}{b}=\mathrm{4}\:\:…