Question Number 113261 by Aina Samuel Temidayo last updated on 12/Sep/20 $$\frac{\sqrt{\sqrt{\mathrm{5}}+\mathrm{2}}+\sqrt{\sqrt{\mathrm{5}}−\mathrm{2}}}{\:\sqrt{\sqrt{\mathrm{5}}+\mathrm{1}}}−\sqrt{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}} \\ $$ Answered by som(math1967) last updated on 12/Sep/20 $$\mathrm{let}\:\mathrm{p}=\sqrt{\sqrt{\mathrm{5}}+\mathrm{2}}+\sqrt{\sqrt{\mathrm{5}}−\mathrm{2}} \\ $$$$\mathrm{p}^{\mathrm{2}} =\sqrt{\mathrm{5}}+\mathrm{2}+\sqrt{\mathrm{5}}−\mathrm{2}+\mathrm{2}\sqrt{\left(\sqrt{\mathrm{5}}\right)^{\mathrm{2}}…
Question Number 113097 by nimnim last updated on 11/Sep/20 $$\mathrm{If}\:\mathrm{4}\:\mathrm{women}\:\mathrm{earn}\:\mathrm{as}\:\mathrm{much}\:\mathrm{as}\:\mathrm{9}\:\mathrm{boys};\:\mathrm{4}\:\mathrm{men}\:\mathrm{as}\:\mathrm{much}\:\mathrm{as} \\ $$$$\mathrm{15}\:\mathrm{boys};\:\mathrm{27}\:\mathrm{girls}\:\mathrm{as}\:\mathrm{much}\:\mathrm{as}\:\mathrm{20}\:\mathrm{women},\:\mathrm{how}\:\mathrm{many}\:\mathrm{girls} \\ $$$$\mathrm{will}\:\mathrm{earn}\:\mathrm{the}\:\mathrm{same}\:\mathrm{amount}\:\mathrm{as}\:\mathrm{24}\:\mathrm{men}? \\ $$ Answered by 1549442205PVT last updated on 11/Sep/20 $$\mathrm{4}\:\mathrm{men}\:\mathrm{as}\:\mathrm{much}\:\mathrm{as}\:\mathrm{15}\:\mathrm{boys} \\…
Question Number 47531 by peter frank last updated on 11/Nov/18 Commented by math1967 last updated on 11/Nov/18 $${question}\:{should}\:{be}\:{A}\pm\sqrt{\left({A}+{G}\right)\left({A}−{G}\right)} \\ $$ Commented by peter frank last…
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Question Number 47454 by peter frank last updated on 10/Nov/18 $$\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{locus} \\ $$$$\mathrm{of}\:\mathrm{middle}\:\mathrm{point}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{normal}\:\mathrm{chord}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{parabola}\:\mathrm{y}^{\mathrm{2}} =\mathrm{4ax}\:\mathrm{is} \\ $$$$\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{2a}}+\frac{\mathrm{4a}^{\mathrm{3}} }{\mathrm{y}^{\mathrm{2}} }=\mathrm{x}−\mathrm{2a} \\ $$…
Question Number 47447 by peter frank last updated on 10/Nov/18 Answered by …. last updated on 10/Nov/18 $$\mathrm{turtle}+\mathrm{170}=\mathrm{table}+\mathrm{cat}……\left(\mathrm{1}\right) \\ $$$$\mathrm{cat}+\mathrm{130}=\mathrm{table}+\mathrm{turtle}…..\left(\mathrm{2}\right) \\ $$$$\mathrm{from}..\left(\mathrm{1}\right)\Rightarrow\mathrm{turtle}=\mathrm{table}+\mathrm{cat}−\mathrm{170}….\left(\mathrm{3}\right) \\ $$$$\mathrm{using}\:\left(\mathrm{3}\right)\:\mathrm{in}\:\left(\mathrm{2}\right)… \\…
Question Number 47433 by Tawa1 last updated on 09/Nov/18 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{square}\:\mathrm{root}\:\mathrm{of}\:\:−\:\mathrm{5}\:−\:\mathrm{12i},\:\:\mathrm{hence}\:\mathrm{solve}:\:\:\mathrm{z}^{\mathrm{2}} \:−\:\left(\mathrm{4}\:+\:\mathrm{i}\right)\mathrm{z}\:+\:\left(\mathrm{5}\:+\:\mathrm{6i}\right)\:=\:\mathrm{0} \\ $$ Commented by peter frank last updated on 10/Nov/18 $$\mathrm{let}\:\mathrm{a}+\mathrm{ib}=\sqrt{-\mathrm{5}−\mathrm{12i}} \\ $$$$\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}}…
Question Number 47413 by Tawa1 last updated on 09/Nov/18 $$\mathrm{Calculate}:\:\:\mathrm{1999}^{\mathrm{2}} \:−\:\mathrm{1998}^{\mathrm{2}} \:+\:\mathrm{1997}^{\mathrm{2}} \:−\:\mathrm{1996}^{\mathrm{2}} \:…\:−\:\mathrm{2}^{\mathrm{2}} \:+\:\mathrm{1}^{\mathrm{2}} \\ $$ Answered by mr W last updated on 09/Nov/18…
Question Number 47316 by peter frank last updated on 08/Nov/18 Commented by math1967 last updated on 08/Nov/18 $${man}\:\:{contains}\:\mathrm{2}{shoes}\:{and}\:\mathrm{2}{paper} \\ $$ Commented by MJS last updated…
Question Number 112810 by Aina Samuel Temidayo last updated on 09/Sep/20 Answered by Rasheed.Sindhi last updated on 10/Sep/20 $${t}_{\mathrm{1}} ,{t}_{\mathrm{2}} ,{t}_{\mathrm{3}} ,…,{t}_{{n}} \:;\:{n}\in\mathbb{E}…………\left(\mathrm{1}\right) \\ $$$${Sum}\:{of}\:{odd}\:{numbered}\:{terms}:…