Question Number 49680 by peter frank last updated on 09/Dec/18 Answered by math1967 last updated on 09/Dec/18 $${Total}\:{surface}\:{area}\:=\mathrm{2}×\mathrm{6}×\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\mathrm{20}^{\mathrm{2}} +\mathrm{6}×\mathrm{20}×\mathrm{15} \\ $$$$=\mathrm{3878}.\mathrm{46}{cm}^{\mathrm{2}} \left({aprox}\right) \\ $$$${Volume}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\mathrm{6}×\mathrm{20}^{\mathrm{2}} ×\mathrm{15}=\mathrm{233826}.\mathrm{85}{cm}^{\mathrm{3}}…
Question Number 115194 by 1549442205PVT last updated on 24/Sep/20 $$\mathrm{Find}\:\mathrm{a}\:\mathrm{three}−\mathrm{digits}\:\mathrm{number}\:\mathrm{whose} \\ $$$$\mathrm{digits}\:\mathrm{form}\:\mathrm{a}\:\mathrm{geometry}\:\mathrm{progression} \\ $$$$;\mathrm{if}\:\mathrm{you}\:\mathrm{known}\:\mathrm{that}\:\mathrm{after}\:\mathrm{substract}\:\mathrm{that} \\ $$$$\mathrm{number}\:\mathrm{by}\:\mathrm{495}\:,\mathrm{get}\:\mathrm{a}\:\mathrm{number} \\ $$$$\mathrm{written}\:\mathrm{by}\:\mathrm{the}\:\mathrm{same}\:\mathrm{digits}\:\mathrm{as}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{you}\:\mathrm{are}\:\mathrm{looking}\:\mathrm{for}\:\mathrm{but}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{reverse}\:\mathrm{order};\mathrm{if}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{obtained}\:\mathrm{after} \\…
Question Number 180695 by depressiveshrek last updated on 15/Nov/22 $${Find}\:{all}\:{k},\:{m},\:{n}\:\in\:\mathbb{N}\:{such}\:{that} \\ $$$${k}!+\mathrm{3}^{{m}} =\mathrm{3}^{{n}} \\ $$ Answered by nikif99 last updated on 16/Nov/22 $${Trying}\:{an}\:{answer}. \\ $$$${must}\:{be}\:{m}<{n}\:{and}\:\mathrm{3}^{{n}}…
Question Number 115062 by Sudip last updated on 23/Sep/20 $$\mathrm{If}\:\:\mathrm{2cos}\theta−\mathrm{sin}\theta=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\:\left(\mathrm{0}°<\theta<\mathrm{90}°\right) \\ $$$$\mathrm{then}\:\:\mathrm{2sin}\theta+\mathrm{cos}\theta=\:¿ \\ $$ Answered by PRITHWISH SEN 2 last updated on 23/Sep/20 $$\mathrm{let}\:\mathrm{2sin}\:\theta+\mathrm{cos}\:\theta\:=\:\mathrm{k}\:….\left(\mathrm{ii}\right) \\…
Question Number 180554 by cherokeesay last updated on 13/Nov/22 Answered by Rasheed.Sindhi last updated on 13/Nov/22 $${f}\left({x}\right)=\mathrm{2}{x}−\mathrm{5}\:\:,\:\:\:{g}\left({x}\right)={x}+\mathrm{1} \\ $$$$\left({f}\left({g}^{−\mathrm{1}} \right)\right)^{−\mathrm{1}} \left({x}\right)=? \\ $$$$\bullet{g}^{−\mathrm{1}} \left({x}\right)=? \\…
Question Number 49427 by Tawa1 last updated on 06/Dec/18 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{nth}\:\mathrm{term}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sequence}:\:\:\:\mathrm{5},\:\:\mathrm{5},\:\:\mathrm{35},\:\:\mathrm{65},\:\:\mathrm{275},\:… \\ $$ Commented by Tawa1 last updated on 07/Dec/18 $$\mathrm{Answer}:\:\:\:\:\:\mathrm{3}^{\boldsymbol{\mathrm{n}}} \:−\:\left(−\:\mathrm{2}\right)^{\boldsymbol{\mathrm{n}}} \:.\:\:\:\:\mathrm{How}\:??? \\ $$ Terms…
Question Number 180498 by Spillover last updated on 12/Nov/22 Commented by Frix last updated on 13/Nov/22 $$\mathrm{more}\:\mathrm{challenging}\:\mathrm{extra}\:\mathrm{work}: \\ $$$$\omega=\frac{\left(\frac{\mathrm{log}_{{a}^{\mathrm{3}} +{b}^{\mathrm{4}} } ^{\mathrm{2}} \:{z}}{\mathrm{25}}−{R}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \pm\left(\left(\left(\frac{\mathrm{log}_{{a}^{\mathrm{3}}…
Question Number 49357 by peter frank last updated on 06/Dec/18 Commented by maxmathsup by imad last updated on 01/Jan/19 $$\left.\mathrm{2}\right)\:{let}\:{prove}\:{that}\:{y}^{\left({n}\right)} \left({x}\right)=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left(\mathrm{1}+{x}\right)^{{n}} }\:\:{with}\:{y}\left({x}\right)={ln}\left(\mathrm{1}+{x}\right) \\ $$$${by}\:{recurrence}\:{on}\:{n}\:\:{for}\:{n}=\mathrm{1}\:\:\:{y}^{\left(\mathrm{1}\right)}…
Question Number 49355 by peter frank last updated on 06/Dec/18 Answered by tanmay.chaudhury50@gmail.com last updated on 06/Dec/18 Commented by tanmay.chaudhury50@gmail.com last updated on 06/Dec/18 Terms…
Question Number 49340 by Tawa1 last updated on 05/Dec/18 Terms of Service Privacy Policy Contact: info@tinkutara.com