Question Number 105526 by bramlex last updated on 29/Jul/20 $$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}}+…+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+…+\mathrm{29}} \\ $$ Answered by Dwaipayan Shikari last updated on 29/Jul/20 $$ \\ $$$$ \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}}+….+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+..+\mathrm{29}}\right)−\mathrm{1}…
Question Number 170980 by anas_zaidan last updated on 05/Jun/22 Answered by aleks041103 last updated on 05/Jun/22 $$\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{x}^{\mathrm{2}} {ydxdy}= \\ $$$$=\left(\underset{\mathrm{0}} {\overset{\mathrm{1}}…
Question Number 170914 by mathlove last updated on 03/Jun/22 Commented by infinityaction last updated on 03/Jun/22 $${x}\:=\:\:\frac{\mathrm{1}}{\:^{\mathrm{3}} \sqrt{\mathrm{2}}\:−\mathrm{1}}\:\:\:\:\:\:\:{g}.{p}.\:{sum} \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{{x}}\:=\:^{\mathrm{3}} \sqrt{\mathrm{2}\:} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} \:=\:\mathrm{2} \\…
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Question Number 105018 by bemath last updated on 25/Jul/20 $$\frac{\sqrt{\sqrt{\mathrm{5}}+\mathrm{2}}\:+\:\sqrt{\sqrt{\mathrm{5}}−\mathrm{2}}}{\:\sqrt{\sqrt{\mathrm{5}}+\mathrm{1}}}\:? \\ $$ Commented by bemath last updated on 25/Jul/20 $${thank}\:{you}\:{all} \\ $$ Answered by som(math1967)…
Question Number 170520 by mr W last updated on 25/May/22 $${solve}\:{for}\:{x} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}+\mathrm{1}} =\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{10}}\right)^{\mathrm{10}} \\ $$ Commented by mr W last updated on 26/May/22 $${yes}.\:{show}\:{how}\:{you}\:{got}?…
Question Number 104821 by bobhans last updated on 24/Jul/20 $$\mathrm{3}+\mathrm{8}+\mathrm{15}+\mathrm{24}+\mathrm{35}+…\: \\ $$$${find}\:{sum}\:{of}\:\mathrm{50}^{{th}} −{term}\: \\ $$ Answered by bemath last updated on 24/Jul/20 $${S}_{\mathrm{50}} \:=\:\mathrm{3}{C}_{\mathrm{1}} ^{\:\mathrm{50}}…
Question Number 170270 by cortano1 last updated on 19/May/22 Commented by cortano1 last updated on 19/May/22 $$\:{Find}\:\alpha\: \\ $$ Commented by mr W last updated…
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