Question Number 198414 by ajfour last updated on 19/Oct/23 $$\frac{\mathrm{3}}{\mathrm{8}}\left(\mathrm{3}{h}−{p}\right)^{\mathrm{2}} +\mathrm{3}{ph}=\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$${and} \\ $$$$\frac{\left(\mathrm{3}{h}−{p}\right)^{\mathrm{3}} }{\mathrm{16}}+{p}\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)={h}^{\mathrm{3}} −{h}−{c} \\ $$$${Find}\:\:{p}\:{and}\:{h}\:\:{in}\:{terms}\:{of}\:\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\centerdot \\ $$ Terms of…
Question Number 198367 by universe last updated on 18/Oct/23 $$\:\:\mathrm{if}\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{{x}^{\mathrm{2}} −\left({b}+\mathrm{1}\right){x}+{b}}{{x}^{\mathrm{2}} −\left({a}+\mathrm{1}\right){x}+{a}}\:\:\left({a}\neq{b}\:\&\:{a},{b}\:\in\:\mathbb{R}\:\sim\:\left\{\mathrm{1}\right\}\right) \\ $$$$\:\:\mathrm{can}\:\mathrm{take}\:\mathrm{all}\:\mathrm{values}\:\mathrm{except}\:\mathrm{two}\:\mathrm{values}\:\alpha\:\&\:\beta \\ $$$$\:\:\mathrm{such}\:\mathrm{that}\:\alpha+\beta\:=\:\mathrm{0}\:\:\mathrm{then}\:\mid\frac{\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} −\mathrm{8}}{\mathrm{ab}}\mid\:\:=\:\:?? \\ $$ Commented by Frix last updated…
Question Number 198372 by cortano12 last updated on 18/Oct/23 $$\:\:\:\frac{\mathrm{1}}{\mathrm{1}×\mathrm{5}}\:+\:\frac{\mathrm{1}}{\mathrm{4}×\mathrm{8}}\:+\:\frac{\mathrm{1}}{\mathrm{7}×\mathrm{11}}\:+\frac{\mathrm{1}}{\mathrm{10}×\mathrm{14}}\:+\:\ldots=? \\ $$ Commented by Frix last updated on 19/Oct/23 $$\frac{\mathrm{1}}{\mathrm{8}}+\frac{\sqrt{\mathrm{3}}\pi}{\mathrm{36}} \\ $$ Answered by mr…
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 $${t}=\mathrm{81}{x}−\mathrm{8} \\ $$$${t}^{\mathrm{3}} −\mathrm{138}{t}^{\mathrm{2}}…
Question Number 198311 by York12 last updated on 17/Oct/23 $${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} \\ $$$$\:\:\:\mathrm{such}\:\mathrm{that}\:\:\mathrm{a}_{\mathrm{1}} =\mathrm{1}\:,\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} ^{\mathrm{2}} −\mathrm{2a}_{\mathrm{n}} \mathrm{a}_{\mathrm{n}+\mathrm{1}} −\mathrm{a}_{\mathrm{n}} \:=\:\mathrm{0}\:,\:\forall\:\mathrm{n}\geqslant\:\mathrm{1} \\ $$$$\:\:\:\mathrm{than}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{series}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{a}_{\mathrm{n}} }{\mathrm{3}^{\mathrm{n}\:} }\:\:\mathrm{lies}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interval}…
Question Number 198293 by Mingma last updated on 16/Oct/23 Answered by MM42 last updated on 17/Oct/23 $${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}} \\ $$$$\Rightarrow\left({ABA}^{−\mathrm{1}}…
Question Number 198266 by Mingma last updated on 15/Oct/23 Answered by mr W last updated on 16/Oct/23 $${a}\left(−\mathrm{4}{x}+\mathrm{3}{y}+\mathrm{4}{z}−\mathrm{3}\right)+{b}\left(−\mathrm{2}{x}+\mathrm{4}{y}+\mathrm{5}{z}−\mathrm{5}\right)=\mathrm{10}{x}−\mathrm{11}{y}+{hz}−{k} \\ $$$$\left(−\mathrm{4}{a}−\mathrm{2}{b}−\mathrm{10}\right){x}+\left(\mathrm{3}{a}+\mathrm{4}{b}+\mathrm{11}\right){y}+\left(\mathrm{4}{a}+\mathrm{5}{b}−{h}\right){z}−\mathrm{3}{a}−\mathrm{5}{b}+{k}=\mathrm{0} \\ $$$$−\mathrm{4}{a}−\mathrm{2}{b}−\mathrm{10}=\mathrm{0} \\ $$$$\mathrm{3}{a}+\mathrm{4}{b}+\mathrm{11}=\mathrm{0} \\…
Question Number 198267 by Fridunatjan08 last updated on 16/Oct/23 $${Find}\:{the}\:{real}\:{values}\:{of}\:{n}:\:{n}^{\mathrm{6}} −{n}^{\mathrm{3}} =\mathrm{2} \\ $$ Answered by Rasheed.Sindhi last updated on 16/Oct/23 $${n}^{\mathrm{6}} −{n}^{\mathrm{3}} =\mathrm{2} \\…
Question Number 198252 by hardmath last updated on 15/Oct/23 Commented by JDamian last updated on 15/Oct/23 what does / / mean? Commented by hardmath last updated on 15/Oct/23 $$\bigtriangleup\mathrm{ABC}…