Question Number 203480 by mr W last updated on 20/Jan/24 Commented by mr W last updated on 20/Jan/24 $${is}\:{it}\:{possible}\:{that}\:{the}\:{red}\:{lines}\:{divide} \\ $$$${the}\:{square}\:{into}\:\mathrm{4}\:{parts}\:{with}\:{given}\: \\ $$$${areas}?\:{if}\:{yes},\:{find}\:{the}\:{length}\:{of}\:{the} \\ $$$${red}\:{lines}.…
Question Number 203401 by cherokeesay last updated on 18/Jan/24 Answered by MM42 last updated on 18/Jan/24 $$<{B}=\mathrm{30}^{{o}} \\ $$$${AB}=\mathrm{2}\sqrt{\mathrm{3}}×{cos}\mathrm{30}=\mathrm{3} \\ $$$${R}=\sqrt{\mathrm{21}} \\ $$$${EC}=\mathrm{2}\sqrt{\mathrm{3}}×{cos}\mathrm{60}=\sqrt{\mathrm{3}} \\ $$$${r}=\sqrt{\mathrm{3}}×{tan}\mathrm{30}\Rightarrow{r}=\mathrm{1}…
Question Number 203370 by mr W last updated on 17/Jan/24 Commented by mr W last updated on 17/Jan/24 $${Q}\mathrm{203319} \\ $$ Answered by mr W…
Question Number 203327 by mr W last updated on 16/Jan/24 Answered by ajfour last updated on 16/Jan/24 Commented by ajfour last updated on 16/Jan/24 $${What}\:{would}\:{r}\:{be}\:{in}\:{this}\:{case}?…
Question Number 203319 by ajfour last updated on 16/Jan/24 Commented by mr W last updated on 16/Jan/24 $$\mathrm{4}\:{equal}\:{spheres}\:{inside}\:{a}\:{regular} \\ $$$${tetrahedron}\:{with}\:{edge}\:{length}\:{a}? \\ $$ Commented by ajfour…
Question Number 203282 by mr W last updated on 14/Jan/24 Commented by cortano12 last updated on 14/Jan/24 $$\mathrm{AD}\parallel\mathrm{BC}\:? \\ $$ Commented by mr W last…
Question Number 203269 by ahmetgg last updated on 13/Jan/24 Answered by esmaeil last updated on 13/Jan/24 $${tan}\mathrm{10}=\frac{{AH}}{{BH}} \\ $$$${tan}\mathrm{20}=\frac{{HC}}{{BH}}\rightarrow\frac{{HC}}{{AH}}\approx\mathrm{2}.\mathrm{0642} \\ $$$${tan}\mathrm{50}=\frac{{OH}}{{AH}} \\ $$$${tanx}=\frac{{OH}}{{CH}}\rightarrow\frac{{tan}\mathrm{50}}{{tanx}}\approx\mathrm{2}.\mathrm{0642}\rightarrow \\ $$$${x}={tan}^{−\mathrm{1}}…
Question Number 203232 by Mathstar last updated on 13/Jan/24 Answered by mr W last updated on 13/Jan/24 Commented by mr W last updated on 13/Jan/24…
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Question Number 203187 by Amidip last updated on 12/Jan/24 Answered by mr W last updated on 12/Jan/24 $${supposed}:\:{AB}\bot{AC},\:{AD}\bot{BC} \\ $$$$\frac{\mathrm{45}}{{x}}=\frac{{x}+\mathrm{48}}{\mathrm{45}}\: \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\mathrm{48}{x}−\mathrm{45}^{\mathrm{2}} =\mathrm{0} \\…