Question Number 188455 by Rupesh123 last updated on 01/Mar/23 Answered by HeferH last updated on 02/Mar/23 Commented by HeferH last updated on 02/Mar/23 $${Prove}\:{that}:\:{x}\:=\:\mathrm{2}\left({a}\:+\:{b}\right)\:−\:{a} \\…
Question Number 188451 by normans last updated on 01/Mar/23 Answered by mr W last updated on 01/Mar/23 Commented by mr W last updated on 01/Mar/23…
Question Number 188444 by Rupesh123 last updated on 01/Mar/23 Answered by mr W last updated on 01/Mar/23 Commented by mr W last updated on 01/Mar/23…
Question Number 57310 by ANTARES VY last updated on 02/Apr/19 Commented by ANTARES VY last updated on 02/Apr/19 $$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{S}}_{\boldsymbol{\mathrm{A}}'\boldsymbol{\mathrm{B}}'\boldsymbol{\mathrm{C}}'} \\ $$ Terms of Service Privacy…
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Question Number 57296 by mr W last updated on 01/Apr/19 Commented by mr W last updated on 01/Apr/19 $${Find}\:{x}=? \\ $$ Answered by MJS last…
Question Number 188348 by normans last updated on 28/Feb/23 Commented by normans last updated on 28/Feb/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 57283 by ajfour last updated on 01/Apr/19 Commented by ajfour last updated on 01/Apr/19 $$\mathrm{Find}\:\mathrm{maximum}\:\mathrm{area}\:\mathrm{that}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{cut}\:\mathrm{out}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{and} \\ $$$$\mathrm{rectangle}\:\mathrm{from}\:\mathrm{an}\:\mathrm{equilateral}\: \\ $$$$\mathrm{triangular}\:\mathrm{sheet}\:\mathrm{of}\:\mathrm{paper},\:\mathrm{edges}\:\mathrm{a}. \\ $$…
Question Number 188339 by normans last updated on 28/Feb/23 Answered by universe last updated on 28/Feb/23 $${ar}\left(\bigtriangleup{ABC}\right)\:=\:{ar}\left({ABD}\right)\:\:+\:\:{ar}\left({ADC}\right) \\ $$$${AB}.{AC}\mathrm{sin}\:\left(\alpha+\beta\right)\:=\:{AB}.{AD}\mathrm{sin}\left(\alpha\right)\:+\:{AD}.{AC}\mathrm{sin}\left(\beta\right) \\ $$$$\frac{\mathrm{sin}\left(\alpha+\beta\right)\:}{{AD}}\:=\:\frac{\mathrm{sin}\alpha\:}{{AC}}\:+\:\frac{\mathrm{sin}\beta\:}{{AB}}\: \\ $$$$\:\: \\ $$…
Question Number 188336 by normans last updated on 28/Feb/23 Answered by mr W last updated on 16/Mar/23 Commented by mr W last updated on 16/Mar/23…