Question Number 201995 by Mingma last updated on 18/Dec/23 Answered by AST last updated on 18/Dec/23 $$\mathrm{2}{b}^{\mathrm{3}} −\mathrm{2}{b}^{\mathrm{2}} =\mathrm{2}{b}^{\mathrm{2}} \left({b}−\mathrm{1}\right)\Rightarrow\left({E}\right) \\ $$ Terms of Service…
Question Number 201991 by MATHEMATICSAM last updated on 18/Dec/23 $$\boldsymbol{\mathrm{Solve}} \\ $$$$\left(\frac{\mathrm{1}}{{x}}\:−\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:+\:\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\mathrm{1} \\ $$ Answered by mr W last updated…
Question Number 202019 by MATHEMATICSAM last updated on 18/Dec/23 $$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\: \\ $$$${ax}^{\mathrm{2}} \:+\:\mathrm{2}{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{and}\:\alpha\:+\:\delta\:\mathrm{and}\:\beta\:+\:\delta\:\mathrm{are} \\ $$$$\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{Ax}^{\mathrm{2}} \:+\:\mathrm{2}{Bx}\:+\:{C}\:=\:\mathrm{0}\:\mathrm{for}\:\mathrm{some}\: \\ $$$$\mathrm{constant}\:\delta\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{{b}^{\mathrm{2}} \:−\:{ac}}{{a}^{\mathrm{2}} }\:=\:\frac{{B}^{\mathrm{2}} \:−\:{AC}}{{A}^{\mathrm{2}} }\:. \\…
Question Number 201977 by Calculusboy last updated on 17/Dec/23 Commented by Calculusboy last updated on 17/Dec/23 $$\boldsymbol{{make}}\:\boldsymbol{{R}}\:\boldsymbol{{subject}}\:\boldsymbol{{of}}\:\boldsymbol{{theformula}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 201965 by ajfour last updated on 17/Dec/23 $${m}−{h}=\mathrm{2}{p} \\ $$$${p}\left({m}−{h}\right)={k}−{q} \\ $$$${mk}−{qh}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${k}−\mathrm{2}{q}=\mathrm{1}+{ph} \\ $$$${Assume}\:{one}\:{find}\:{the}\:{rest}! \\ $$$$\checkmark \\ $$ Answered by Frix…
Question Number 201966 by mnjuly1970 last updated on 17/Dec/23 $$ \\ $$$$\:\:\:\:\:\:\:{solve}\: \\ $$$$\:\: \\ $$$$\:\:\:\:\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{2}}\:\left(\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\:−\sqrt[{\mathrm{3}}]{{x}+\mathrm{2}}\:\right)=\:\mathrm{1} \\ $$$$ \\ $$ Answered by ajfour last…
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Question Number 201912 by York12 last updated on 15/Dec/23 $$ \\ $$ Commented by York12 last updated on 15/Dec/23 Commented by York12 last updated on…
Question Number 201941 by ajfour last updated on 15/Dec/23 $${x}^{\mathrm{4}} −\frac{\mathrm{17}}{\mathrm{18}}{x}^{\mathrm{2}} +\frac{\mathrm{40}}{\mathrm{3}}{x}+\frac{\mathrm{1625}}{\mathrm{144}}=\mathrm{0} \\ $$$${Find}\:{roots}.\:\left({Two}\:{are}\:{real}\:{and}\:{two}\:\right. \\ $$$$\left.{are}\:{complex}\right)\bigstar \\ $$ Commented by Frix last updated on 17/Dec/23…
Question Number 201934 by aurpeyz last updated on 15/Dec/23 Commented by aurpeyz last updated on 15/Dec/23 $${By}\:{complex}\:{analysis}\:{method}. \\ $$ Terms of Service Privacy Policy Contact:…