Question Number 59588 by aliesam last updated on 12/May/19 $${find}\:{x},{y}\:{in}\:{R} \\ $$$$\left({x}+{yi}\right)^{\mathrm{3}} =\frac{\mathrm{1}+\sqrt{\mathrm{15}}\:{i}}{\:\sqrt{\mathrm{5}}\:−\:\sqrt{\mathrm{3}}\:{i}} \\ $$ Commented by maxmathsup by imad last updated on 12/May/19 $$\Rightarrow{x}+{iy}\:=^{\mathrm{3}}…
Question Number 190634 by Mastermind last updated on 07/Apr/23 $$\mathrm{The}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{a}\:\mathrm{right}\:\mathrm{circular}\:\mathrm{cone} \\ $$$$\mathrm{is}\:\mathrm{5litres}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{part}\:\mathrm{into}\:\mathrm{which}\:\mathrm{the}\:\mathrm{cone}\:\mathrm{is} \\ $$$$\mathrm{divided}\:\mathrm{by}\:\mathrm{a}\:\mathrm{plane}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{base}\:\mathrm{one}−\mathrm{third}\:\mathrm{of}\:\mathrm{the}\:\mathrm{way}\:\mathrm{down} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{vertex}\:\mathrm{to}\:\mathrm{the}\:\mathrm{base}\:\mathrm{giving} \\ $$$$\mathrm{your}\:\mathrm{answer}\:\mathrm{to}\:\mathrm{the}\:\mathrm{nearest}\: \\ $$$$\mathrm{millimetres}. \\…
Question Number 125086 by sdfg last updated on 08/Dec/20 Commented by sdfg last updated on 08/Dec/20 $${pls}\:{help} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 59505 by pete last updated on 11/May/19 $$\mathrm{Two}\:\mathrm{particles}\:\mathrm{P}\:\mathrm{and}\:\mathrm{Q}\:\mathrm{move}\:\mathrm{towards}\:\mathrm{each} \\ $$$$\mathrm{other}\:\mathrm{along}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:{MN},\:\mathrm{51}\:\mathrm{meters} \\ $$$$\mathrm{long}.\:\mathrm{P}\:\mathrm{starts}\:\mathrm{from}{M}\:\mathrm{with}\:\mathrm{velocity}\:\mathrm{5}\:\mathrm{ms}^{−\mathrm{1}} \\ $$$$\mathrm{and}\:\mathrm{constant}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{1}\:\mathrm{ms}^{−\mathrm{2}} .\:\mathrm{Q}\:\mathrm{starts} \\ $$$$\mathrm{from}\:\mathrm{N}\:\mathrm{at}\:\mathrm{the}\:\mathrm{same}\:\mathrm{time}\:\mathrm{with}\:\mathrm{velocity}\:\mathrm{6}\:\mathrm{ms}^{−\mathrm{1}} \\ $$$$\mathrm{and}\:\mathrm{at}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{acceleration}\:\mathrm{of}\:\mathrm{3}\:\mathrm{ms}^{−\mathrm{2}} . \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{time}\:\mathrm{when}\:\mathrm{the}: \\…
Question Number 125014 by lontumHans last updated on 07/Dec/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 59478 by subhankar10 last updated on 10/May/19 $$\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} +\mathrm{16}=\mathrm{0} \\ $$ Answered by MJS last updated on 10/May/19 $${x}=\pm\sqrt{{t}} \\ $$$${t}^{\mathrm{2}} +{t}+\mathrm{16}=\mathrm{0}…
Question Number 59442 by Tawa1 last updated on 10/May/19 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{system} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{xy}\:+\:\mathrm{3x}\:+\:\mathrm{2y}\:\:=\:−\:\mathrm{6}\:\:\:\:\:\:\:…..\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{yx}\:+\:\mathrm{y}\:+\:\mathrm{3z}\:\:\:=\:−\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:…..\:\left(\mathrm{ii}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{zx}\:+\:\mathrm{2z}\:+\:\mathrm{x}\:\:\:=\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:…..\:\left(\mathrm{iii}\right) \\ $$ Commented by mr W last updated on…
Question Number 59397 by hhghg last updated on 09/May/19 $$\mathrm{6}^{\mathrm{4}} ×\mathrm{6}^{\mathrm{3}} \\ $$ Commented by Mikael_Marshall last updated on 09/May/19 $$\mathrm{6}^{\mathrm{4}+\mathrm{3}} =\mathrm{6}^{\mathrm{7}} \\ $$$${use}\:{calculatr} \\…
Question Number 124892 by Dwaipayan Shikari last updated on 06/Dec/20 $$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}.\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{5}.\mathrm{5}!}−\frac{\mathrm{1}}{\mathrm{7}.\mathrm{7}!}+… \\ $$ Commented by Dwaipayan Shikari last updated on 06/Dec/20 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sinx}}{{x}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 124891 by Dwaipayan Shikari last updated on 06/Dec/20 $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{24}}} {log}\left({tan}\theta\right){d}\theta \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com