Question Number 198295 by cortano12 last updated on 17/Oct/23 $$\:\:\:\mathrm{x}^{\mathrm{3}} −\sqrt[{\mathrm{3}}]{\mathrm{81x}−\mathrm{8}}\:=\:\mathrm{2x}^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{3}}\mathrm{x}+\mathrm{2}\: \\ $$ Answered by Frix last updated on 17/Oct/23 $${t}=\mathrm{81}{x}−\mathrm{8} \\ $$$${t}^{\mathrm{3}} −\mathrm{138}{t}^{\mathrm{2}}…
Question Number 198321 by necx122 last updated on 17/Oct/23 $${A}\:{certain}\:{preliminary}\:{science}\:{class} \\ $$$${contains}\:\mathrm{50}\:{students}\:{all}\:{of}\:{whom}\:{take} \\ $$$${Mathematics}.\:\mathrm{18}\:{study}\:{Chemistry},\:\mathrm{17} \\ $$$${study}\:{Biology},\:\mathrm{24}\:{study}\:{Physics}.\:{Of} \\ $$$${those}\:{taking}\:{three}\:{subjects},\:\mathrm{5}\:{study} \\ $$$${Physics}\:{and}\:{Chemistry},\:\mathrm{7}\:{study}\:{Physics} \\ $$$${and}\:{Biology}\:{and}\:\mathrm{6}\:{study}\:{Chemistry} \\ $$$${and}\:{Biology}\:{while}\:\mathrm{2}\:{take}\:{all}\:{four}\:{subjects}. \\…
Question Number 198319 by necx122 last updated on 17/Oct/23 $${An}\:{analysis}\:{of}\:\mathrm{100}\:{personal}\:{injury} \\ $$$${claims}\:{made}\:{upon}\:{a}\:{motor}\:{insurance} \\ $$$${company}\:{revealed}\:{that}\:{loss}\:{or}\:{injury} \\ $$$${in}\:{respect}\:{of}\:{an}\:{eye},\:{an}\:{arm}\:{or}\:{leg} \\ $$$${occurred}\:{in}\:\mathrm{30},\:\mathrm{50}\:{and}\:\mathrm{70}\:{cases},\:{respectively}. \\ $$$${Claims}\:{involving}\:{the}\:{loss}\:{or}\:{injury}\:{to} \\ $$$${two}\:{of}\:{these}\:{number}\:\mathrm{44}.\:{How}\:{many} \\ $$$${claims}\:{involved}\:{loss}\:{or}\:{injury}\:{to}\:{all} \\…
Question Number 198313 by sonukgindia last updated on 17/Oct/23 Commented by AST last updated on 17/Oct/23 $${n}^{\mathrm{2}} \equiv\mathrm{34}\left({mod}\:\mathrm{15}\right)?\Rightarrow{n}^{\mathrm{2}} \equiv\mathrm{4}\left({mod}\:\mathrm{15}\right) \\ $$$$\Rightarrow{n}=\mathrm{2},\mathrm{7},\mathrm{8},\mathrm{13}\left({mod}\:\mathrm{15}\right) \\ $$ Terms of…
Question Number 198314 by lapache last updated on 17/Oct/23 $${Solve}\:{the}\:{EDP} \\ $$$${x}^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {u}}{\partial{x}^{\mathrm{2}} }−{y}^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {u}}{\partial{y}^{\mathrm{2}} }=\mathrm{0} \\ $$ Terms of Service Privacy Policy…
Question Number 198309 by BHOOPENDRA last updated on 17/Oct/23 Commented by BHOOPENDRA last updated on 17/Oct/23 $${Sketch}\:{the}\:{partial}\:{auxiliary}\:{view}\:{for}\: \\ $$$${the}\:{auxiliary}\:{surface}\:{object}\:,{given}\: \\ $$$${the}\:{front}\:{and}\:{right}\:{view}. \\ $$ Commented by…
Question Number 198311 by York12 last updated on 17/Oct/23 $${Let}\:\left\{{x}_{{r}} \right\}_{{r}=\mathrm{1}} ^{{n}} {be}\:{n}\:{positive}\:{real}\:{numbers}\:{Show}\:{That}: \\ $$$$\frac{{x}_{\mathrm{1}} }{\mathrm{1}+{x}_{\mathrm{1}} ^{\mathrm{2}} }+\frac{{x}_{\mathrm{2}} }{\mathrm{1}+{x}_{\mathrm{1}} ^{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}} }+…+\frac{{x}_{{n}} }{\mathrm{1}+{x}_{\mathrm{1}} ^{\mathrm{2}}…
Question Number 198304 by universe last updated on 17/Oct/23 $$\:\:\:\mathrm{for}\:\left\{\mathrm{a}_{\mathrm{n}} \right\}\:\mathrm{be}\:\mathrm{a}\:\mathrm{sequence}\:\mathrm{of}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers} \\ $$$$\:\:\:\mathrm{such}\:\mathrm{that}\:\:\mathrm{a}_{\mathrm{1}} =\mathrm{1}\:,\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} ^{\mathrm{2}} −\mathrm{2a}_{\mathrm{n}} \mathrm{a}_{\mathrm{n}+\mathrm{1}} −\mathrm{a}_{\mathrm{n}} \:=\:\mathrm{0}\:,\:\forall\:\mathrm{n}\geqslant\:\mathrm{1} \\ $$$$\:\:\:\mathrm{than}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{series}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{a}_{\mathrm{n}} }{\mathrm{3}^{\mathrm{n}\:} }\:\:\mathrm{lies}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interval}…
Question Number 198269 by SANOGO last updated on 16/Oct/23 $${calcul} \\ $$$$\underset{{k}={o}} {\overset{{n}} {\sum}}{sin}\left({k}\right) \\ $$ Answered by mr W last updated on 16/Oct/23 $${B}=\underset{{k}=\mathrm{0}}…
Question Number 198293 by Mingma last updated on 16/Oct/23 Answered by MM42 last updated on 17/Oct/23 $${B}^{{n}} =\begin{bmatrix}{\left(−\mathrm{1}\right)^{{n}} \:\:\:\:\mathrm{0}}\\{\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{2}^{{n}} }\end{bmatrix}\:\&\:\:\left({ABA}^{−\mathrm{1}} \right)^{{n}} ={AB}^{{n}} {A}^{−\mathrm{1}} \\ $$$$\Rightarrow\left({ABA}^{−\mathrm{1}}…