Question Number 187903 by Mingma last updated on 23/Feb/23 Answered by HeferH last updated on 23/Feb/23 Commented by HeferH last updated on 23/Feb/23 $${i}.\:\:\frac{{a}+{b}}{{c}}=\:\frac{{d}}{{e}}\:\Rightarrow\:\frac{{e}}{{d}}\:=\:\frac{{c}}{{a}+{b}} \\…
Question Number 187891 by Rupesh123 last updated on 23/Feb/23 Answered by mr W last updated on 24/Feb/23 Commented by mr W last updated on 25/Feb/23…
Question Number 187826 by Rupesh123 last updated on 22/Feb/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 187827 by Rupesh123 last updated on 22/Feb/23 Answered by cortano12 last updated on 23/Feb/23 $$\left(\mathrm{1}\right)\mathrm{tan}\:\mathrm{30}°=\frac{\mathrm{4}}{\mathrm{a}}\Rightarrow\mathrm{a}=\mathrm{4}\sqrt{\mathrm{3}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{tan}\:\mathrm{22},\mathrm{5}^{\mathrm{o}} \:=\frac{\mathrm{4}}{\mathrm{b}}\:;\:\mathrm{tan}\:\mathrm{45}°=\frac{\mathrm{2tan}\:\mathrm{22},\mathrm{5}^{\mathrm{o}} }{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{22},\mathrm{5}^{\mathrm{o}} } \\ $$$$\Rightarrow\mathrm{tan}\:^{\mathrm{2}}…
Question Number 187810 by Rupesh123 last updated on 22/Feb/23 Commented by a.lgnaoui last updated on 23/Feb/23 $$\left.\mathrm{E}\right)\mathrm{30} \\ $$ Commented by mr W last updated…
Question Number 187721 by Mingma last updated on 20/Feb/23 Answered by mr W last updated on 21/Feb/23 Commented by mr W last updated on 21/Feb/23…
Question Number 187701 by Mingma last updated on 20/Feb/23 Answered by HeferH last updated on 20/Feb/23 $${let}\:“{h}''\:{be}\:{the}\:{height}\:{of}\:{the}\:{top}\:{triangle}\:{and}\:“{a}'' \\ $$$${the}\:{base} \\ $$$$\:\frac{\mathrm{7}{ah}}{\mathrm{2}}\:=\:\mathrm{28} \\ $$$$\:{ah}\:=\:\frac{\mathrm{28}\centerdot\mathrm{2}}{\mathrm{7}}\:=\:\mathrm{8}\: \\ $$$$\:{Total}\:{area}\:=\:\mathrm{4}{a}\centerdot\mathrm{4}{h}\centerdot\frac{\mathrm{1}}{\mathrm{2}}\:=\:\mathrm{8}{ah}\:=\:\mathrm{64}{u}^{\mathrm{2}}…
Question Number 187689 by Mingma last updated on 20/Feb/23 Answered by mr W last updated on 21/Feb/23 Commented by mr W last updated on 26/Feb/23…
Question Number 187589 by Mingma last updated on 19/Feb/23 Commented by Mingma last updated on 19/Feb/23 Blue area=? Answered by cortano12 last updated on 19/Feb/23 $$\mathrm{cos}\:\alpha=\frac{\left(\mathrm{5}\sqrt{\mathrm{2}}\right)^{\mathrm{2}}…
Question Number 187587 by Mingma last updated on 19/Feb/23 Commented by Mingma last updated on 19/Feb/23 Blue area=? Answered by mr W last updated on 19/Feb/23…