Question Number 209856 by SonGoku last updated on 23/Jul/24 Commented by SonGoku last updated on 23/Jul/24 $$\mathrm{How}\:\mathrm{do}\:\mathrm{I}\:\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{this}\:\mathrm{figure}? \\ $$ Commented by mr W last updated…
Question Number 209840 by peter frank last updated on 23/Jul/24 Answered by mahdipoor last updated on 23/Jul/24 $${M}_{{moon}} ={M}_{{earth}} ={M}={cte} \\ $$$${W}_{{moon}} ={Mg}_{{moon}} \:\:\:\:\:\:\:\:\:\:\:\:{W}_{{earth}} ={Mg}_{{earth}}…
Question Number 209822 by SonGoku last updated on 22/Jul/24 $$\mathrm{A}\:\mathrm{ramp}\:\mathrm{is}\:\mathrm{supported}\:\mathrm{by}\:\mathrm{six}\:\mathrm{pillars}\:\mathrm{and}\:\mathrm{the}\:\mathrm{talleste} \\ $$$$\mathrm{on}\:\mathrm{measures}\:\mathrm{6}\:\mathrm{meters}.\:\mathrm{The}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{eachi} \\ $$$$\mathrm{pllar}\:\mathrm{is}\:\mathrm{5}\:\mathrm{meters}.\:\mathrm{What}\:\mathrm{will}\:\mathrm{be}\:\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{third}\:\mathrm{pillar}?\: \\ $$ Commented by mr W last updated on…
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Question Number 209800 by depressiveshrek last updated on 22/Jul/24 Commented by depressiveshrek last updated on 22/Jul/24 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{this}\:\mathrm{series} \\ $$ Answered by mr W last updated…
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Question Number 209813 by mr W last updated on 22/Jul/24 Answered by mahdipoor last updated on 22/Jul/24 $${R}=\:{radius}\:{red}\:{circle} \\ $$$${r}\:=\:{radius}\:{white}\:{circle} \\ $$$$\begin{cases}{\mathrm{2}{R}=\mathrm{8}+\mathrm{2}+\mathrm{2}{r}}\\{{C}_{\mathrm{1}} {C}_{\mathrm{2}} =\sqrt{{r}^{\mathrm{2}} +\left({R}−\left({r}+\mathrm{2}\right)\right)^{\mathrm{2}}…
Question Number 209810 by mnjuly1970 last updated on 22/Jul/24 $$ \\ $$$$\:\:\:\mathrm{lim}_{\:\mathrm{x}\rightarrow\mathrm{0}} \:\frac{\:\left(\mathrm{1}+\:\mathrm{x}\:\right)^{\frac{\mathrm{1}}{\mathrm{x}}} −\mathrm{e}}{\mathrm{x}}\:=\:? \\ $$$$ \\ $$ Answered by mr W last updated on…
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Question Number 209781 by alusto22 last updated on 21/Jul/24 $$\:\:\:\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{40}°−\mathrm{cos}\:\mathrm{40}°}{\mathrm{1}+\mathrm{sin}\:\mathrm{40}°+\mathrm{cos}\:\mathrm{40}°}\:=? \\ $$ Answered by efronzo1 last updated on 21/Jul/24 $$\:\:\:\:\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{40}°−\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{20}°\right)}{\mathrm{1}+\mathrm{sin}\:\mathrm{40}°+\mathrm{2cos}\:^{\mathrm{2}} \mathrm{20}−\mathrm{1}}\: \\ $$$$\:\:=\:\frac{\mathrm{2sin}\:\mathrm{20}°\left(\mathrm{cos}\:\mathrm{20}°+\mathrm{sin}\:\mathrm{20}°\right)}{\mathrm{2cos}\:\mathrm{20}°\left(\mathrm{cos}\:\mathrm{20}°+\mathrm{sin}\:\mathrm{20}°\right)} \\…