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Question Number 123708 by mnjuly1970 last updated on 27/Nov/20 Commented by mnjuly1970 last updated on 27/Nov/20 $${please}\:{prove}\:\Uparrow\Uparrow \\ $$ Answered by ajfour last updated on…
Question Number 123697 by ajfour last updated on 27/Nov/20 Commented by ajfour last updated on 27/Nov/20 $${Find}\:{radius}\:{of}\:{quarter}\:{circle}\:{in} \\ $$$${terms}\:{of}\:{sides}\:\boldsymbol{{a}},\:\boldsymbol{{b}},\:\boldsymbol{{c}}\:\:{of}\:\bigtriangleup{ABC}. \\ $$ Answered by mr W…
Question Number 189169 by mr W last updated on 12/Mar/23 Commented by mr W last updated on 12/Mar/23 $${find}\:{the}\:{shaded}\:{area}=? \\ $$ Commented by Ari last…
Question Number 189114 by normans last updated on 12/Mar/23 Answered by mr W last updated on 12/Mar/23 Commented by mr W last updated on 12/Mar/23…
Question Number 189110 by normans last updated on 12/Mar/23 Commented by normans last updated on 12/Mar/23 $$\:{proof}\:\mathrm{3}{D}\:{Crossed}\:{Ladders}\:{Theorem}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 189054 by mathlove last updated on 11/Mar/23 Answered by som(math1967) last updated on 11/Mar/23 $$\angle{EBD}=\angle{HBC}\:,\angle{ABC}=\angle{EBF} \\ $$$$\angle{HBF}=\angle{ABD} \\ $$$$\angle{EBD}+\angle{EBF}+\angle{HBF}+\angle{HBC} \\ $$$$+\angle{ABC}+\angle{ABD}=\mathrm{360} \\ $$$$\mathrm{2}\left(\angle{EBD}+\angle{HBF}+\angle{ABC}\right)=\mathrm{360}…
Question Number 57949 by maxmathsup by imad last updated on 14/Apr/19 $$\left(\mathrm{0},{i},{j}\right)\:{is}\:{orthonormal}\:\:\:{A}\:{and}\:\:{B}\:{are}\:{two}\:{points}\:{wich}\:{verify}\:{AB}\:=\mathrm{3} \\ $$$${find}\:\:{the}\:{locus}\:{of}\:{point}\:{M}\:{wich}\:{verify}\:\:{MA}\:+{MB}\:=\mathrm{6}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 188968 by normans last updated on 09/Mar/23 Answered by HeferH last updated on 09/Mar/23 $$\:\alpha\:+\mathrm{132}°\:=\:\mathrm{2}{x}\:+\:\alpha \\ $$$${x}\:=\:\:\frac{\mathrm{132}°}{\mathrm{2}}=\:\mathrm{66}° \\ $$ Commented by HeferH last…
Question Number 188965 by normans last updated on 09/Mar/23 Answered by mr W last updated on 09/Mar/23 $${say}\:{edge}\:{length}\:{of}\:{cube}\:{is}\:\mathrm{1}. \\ $$$${base}\:{of}\:{isosceles}\:{triangle}\:{is}\:\sqrt{\mathrm{2}}. \\ $$$${length}\:{of}\:{its}\:{legs}\:{is}\:\sqrt{\mathrm{1}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}. \\…