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Question Number 199769 by linlu last updated on 09/Nov/23 $$\mathrm{62}{n}\mathrm{7}{z}\mathrm{7} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 199766 by cortano12 last updated on 09/Nov/23 Answered by mr W last updated on 09/Nov/23 Commented by mr W last updated on 09/Nov/23…
Question Number 199817 by ajfour last updated on 09/Nov/23 Answered by ajfour last updated on 09/Nov/23 $$\left\{{a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \theta\right)−\frac{\mathrm{sin}\:\theta}{\mathrm{2}\left(\mathrm{1}+\mathrm{sin}\:\theta\right)}\right\}^{\mathrm{2}} \\ $$$$\:\:\:={a}\left\{{a}+\frac{\mathrm{sin}\:\theta}{\mathrm{2cos}\:\theta\left(\mathrm{1}+\mathrm{sin}\:\theta\right)}\right\} \\ $$$${b}=\frac{\mathrm{sin}\:\theta}{\mathrm{2cos}\:\theta\left(\mathrm{1}+\mathrm{sin}\:\theta\right)} \\ $$…
Question Number 199688 by cortano12 last updated on 07/Nov/23 Commented by cortano12 last updated on 07/Nov/23 $$\mathrm{prove}\:\mathrm{that}\:\mathrm{formula} \\ $$ Answered by AST last updated on…
Question Number 199625 by cortano12 last updated on 06/Nov/23 $$\mathrm{Given}\:\mathrm{Fibonacci}\:\mathrm{series}\: \\ $$$$\:\mathrm{F}_{\mathrm{1}} =\mathrm{F}_{\mathrm{2}} =\:\mathrm{1}\:\mathrm{and}\:\mathrm{F}_{\mathrm{n}+\mathrm{2}} =\:\mathrm{F}_{\mathrm{n}+\mathrm{1}} +\mathrm{F}_{\mathrm{n}} \\ $$$$\:\mathrm{for}\:\mathrm{n}>\mathrm{0}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\: \\ $$$$\:\mathrm{F}_{\mathrm{2022}} \:\mathrm{divides}\:\mathrm{by}\:\mathrm{5}\: \\ $$ Answered by…
Question Number 199351 by Tawa11 last updated on 01/Nov/23 Answered by MM42 last updated on 02/Nov/23 $${a}+{b}=\mathrm{2} \\ $$$$\left.\int_{\mathrm{0}} ^{\mathrm{4}} \left({a}\sqrt{{x}}+{bx}\right){dx}=\frac{\mathrm{2}{a}}{\mathrm{3}}{x}\sqrt{{x}}+\frac{{b}}{\mathrm{2}}{x}^{\mathrm{2}} \right]_{\mathrm{0}} ^{\mathrm{4}} \:=\mathrm{8} \\…
Question Number 199101 by cortano12 last updated on 28/Oct/23 Answered by ajfour last updated on 28/Oct/23 Commented by ajfour last updated on 28/Oct/23 $${Triangle}\:{side}\:{be}\:\mathrm{2}\sqrt{\mathrm{3}}{s}. \\…
Question Number 198538 by Hridiana last updated on 21/Oct/23 $${hellot} \\ $$$$\sqrt{\mathrm{8888}} \\ $$$$\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{{rainbow}}}}}}}}}} \\ $$$$\mathrm{777} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 198434 by cortano12 last updated on 20/Oct/23 Commented by mr W last updated on 20/Oct/23 $${due}\:{to}\:{symmetry} \\ $$$${max}=\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$ Answered by mr…