Question Number 218088 by ArshadS last updated on 29/Mar/25 $${Solve} \\ $$$$\sqrt{{x}+\mathrm{5}}\:+\sqrt{{x}−\mathrm{3}}\:=\mathrm{4} \\ $$ Answered by A5T last updated on 29/Mar/25 $$\sqrt{\mathrm{x}+\mathrm{5}}=\mathrm{u};\:\sqrt{\mathrm{x}−\mathrm{3}}=\mathrm{v} \\ $$$$\mathrm{u}^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}}…
Question Number 218115 by Hanuda354 last updated on 29/Mar/25 Answered by mr W last updated on 30/Mar/25 Commented by mr W last updated on 30/Mar/25…
Question Number 218076 by malwan last updated on 28/Mar/25 $${How}\:{many}\:{ways}\:{to}\:{arrnge} \\ $$$${the}\:{letters}\:{ABCCCDEFG} \\ $$$$\left(\mathrm{1}\right)\:{in}\:{general}\:. \\ $$$$\left(\mathrm{2}\right)\:{all}\:\mathrm{3}\:{Cs}\:{must}\:{be}\:{together} \\ $$$$\left(\mathrm{3}\right)\:{only}\:\mathrm{2}\:{Cs}\:{must}\:{be}\:{together} \\ $$$$\left(\mathrm{4}\right)\:{no}\:\mathrm{2}\:{or}\:\mathrm{3}\:{Cs}\:{be}\:{together} \\ $$$$\left(\mathrm{5}\right)\:{no}\:{letter}\:{still}\:\:{in}\:{its} \\ $$$${original}\:{place}\:. \\…
Question Number 218066 by ajfour last updated on 28/Mar/25 $${If}\:{Alphaprime}\:{borrows}\:\mathrm{25\$}\:{from} \\ $$$${LYCONTRIX}\:{at}\:{the}\:{rzte}\:{of}\:\mathrm{12\%}\:{p}.{a}. \\ $$$${How}\:{much}\:{does}\:{the}\:{former}\:{owe}\:{to}\:{the} \\ $$$${other}\:{in}\:{a}\:{span}\:{of}\:\mathrm{8}{years}? \\ $$ Commented by Frix last updated on 28/Mar/25…
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Question Number 218063 by alephnull last updated on 27/Mar/25 $$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\mathrm{cos}\:{x}^{\mathrm{2}} \\ $$ Answered by Frix last updated on 28/Mar/25 $$\mathrm{The}\:\mathrm{Fresnel}\:\mathrm{C}\:\mathrm{Integral}: \\ $$$$\mathrm{C}\:\left({x}\right)\::=\:\underset{\mathrm{0}} {\overset{{x}}…
Question Number 218047 by dscm last updated on 27/Mar/25 $${Determine}\:{x}: \\ $$$$\sqrt{{x}+\mathrm{3}}\:+\:\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\:=\mathrm{2} \\ $$ Answered by Rasheed.Sindhi last updated on 27/Mar/25 $$\sqrt{{x}+\mathrm{3}}\:+\:\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\:=\mathrm{2} \\ $$$$\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\:={y}\Rightarrow{x}={y}^{\mathrm{3}} −\mathrm{1}…
Question Number 218057 by ArshadS last updated on 27/Mar/25 $${Solve} \\ $$$$\left({x}−\mathrm{3}\right)^{\mathrm{5}/\mathrm{3}} −\mathrm{6}\left({x}−\mathrm{3}\right)^{\mathrm{2}/\mathrm{3}} −\mathrm{8}=\mathrm{0} \\ $$ Answered by Frix last updated on 27/Mar/25 $$\left({x}−\mathrm{3}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \left({x}−\mathrm{9}\right)=\mathrm{8}…
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