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Question Number 202436 by MATHEMATICSAM last updated on 26/Dec/23 $$\mathrm{If}\:\alpha,\:\beta\:\mathrm{and}\:\gamma\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${ax}^{\mathrm{3}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{then}\:\mathrm{frame}\:\mathrm{an}\:\mathrm{equation} \\ $$$$\mathrm{whose}\:\mathrm{roots}\:\mathrm{are}\:\alpha^{\mathrm{2}} ,\:\beta^{\mathrm{2}} ,\:\gamma^{\mathrm{2}} \:.\: \\ $$ Answered by aleks041103 last updated…
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Question Number 202400 by MATHEMATICSAM last updated on 26/Dec/23 $$\mathrm{Solve}\:\mathrm{for}\:{a},\:{b}\:\mathrm{and}\:{c} \\ $$$$\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}\:+\:{c}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}\:+\:{a}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\: \\ $$$$\frac{\mathrm{1}}{{c}}\:+\:\frac{\mathrm{1}}{{a}\:+\:{b}}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$ Commented by mr W last updated on…
Question Number 202356 by hardmath last updated on 25/Dec/23 $$\mathrm{Find}: \\ $$$$\mathrm{1}−\left(\mathrm{sin30}°\right)^{\mathrm{2}} \:+\:\left(\mathrm{sin30}°\right)^{\mathrm{4}} \:−\:\left(\mathrm{sin30}°\right)^{\mathrm{6}} \:+\:… \\ $$ Answered by Rasheed.Sindhi last updated on 25/Dec/23 $$\mathrm{1}−\left(\mathrm{sin30}°\right)^{\mathrm{2}}…
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Question Number 202359 by hardmath last updated on 25/Dec/23 $$\mathrm{2}\:,\:\mathrm{8}\:,\:\mathrm{32}\:,\:…\:\mathrm{geometfic}\:\mathrm{serie} \\ $$$$\mathrm{for}\:\:\:\mathrm{b}_{\boldsymbol{\mathrm{m}}} \:>\:\mathrm{1024}\:\:\:\mathrm{find}\:\:\:\mathrm{min}\left(\mathrm{m}\right)\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 25/Dec/23 $$\mathrm{b}_{\mathrm{n}} =\mathrm{ar}^{\mathrm{n}−\mathrm{1}} \:;\:\mathrm{a}=\mathrm{2}\:,\mathrm{r}=\mathrm{8}/\mathrm{2}=\mathrm{4}…
Question Number 202353 by MATHEMATICSAM last updated on 25/Dec/23 $$\mathrm{If}\:\mathrm{2}{x}\:=\:{a}\:+\:\sqrt{\frac{\mathrm{4}{b}^{\mathrm{3}} \:−\:{a}^{\mathrm{3}} }{\mathrm{3}{a}}}\:\mathrm{and} \\ $$$$\mathrm{2}{y}\:=\:{a}\:−\:\sqrt{\frac{\mathrm{4}{b}^{\mathrm{3}} \:−\:{a}^{\mathrm{3}} }{\mathrm{3}{a}}}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$${x}^{\mathrm{3}} \:+\:{y}^{\mathrm{3}} \:=\:{b}^{\mathrm{3}} \:. \\ $$ Answered by…
Question Number 202352 by cortano12 last updated on 25/Dec/23 Commented by cortano12 last updated on 26/Dec/23 Commented by cortano12 last updated on 26/Dec/23 $$\mathrm{it}\:\mathrm{is}\:\mathrm{correct}\:? \\…
Question Number 202383 by MATHEMATICSAM last updated on 25/Dec/23 $$\mathrm{Show}\:\mathrm{that}\:\frac{{a}\sqrt{{b}}\:−\:{b}\sqrt{{a}}}{{a}\sqrt{{b}}\:+\:{b}\sqrt{{a}}}\:=\:\frac{\mathrm{1}}{{a}\:−\:{b}}\left({a}\:+\:{b}\:−\:\mathrm{2}\sqrt{{ab}}\right). \\ $$ Answered by AST last updated on 25/Dec/23 $$=\frac{\left({a}\sqrt{{b}}−{b}\sqrt{{a}}\right)^{\mathrm{2}} }{{ab}\left({a}−{b}\right)}\:=\frac{{a}^{\mathrm{2}} {b}+{b}^{\mathrm{2}} {a}−\mathrm{2}{ab}\sqrt{{ab}}}{{ab}\left({a}−{b}\right)}=\frac{{a}+{b}−\mathrm{2}\sqrt{{ab}}}{{a}−{b}} \\ $$…