Question Number 193198 by Mingma last updated on 07/Jun/23 Answered by HeferH last updated on 07/Jun/23 Commented by HeferH last updated on 07/Jun/23 $${x}=\mathrm{60}°−\mathrm{30}°=\mathrm{30}° \\…
Question Number 193149 by Mingma last updated on 04/Jun/23 Answered by ajfour last updated on 05/Jun/23 $$\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{p}^{\mathrm{2}} ={A} \\ $$$${p}={k}\sqrt{{A}}\:\:\:\:{where}\:\:\:{k}^{\mathrm{2}} =\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}} \\ $$$${q}={k}\sqrt{{B}} \\ $$$${B}=\mathrm{9}{A}…
Question Number 193105 by Mingma last updated on 04/Jun/23 Commented by Mingma last updated on 04/Jun/23 Find x Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 193161 by Mingma last updated on 05/Jun/23 Answered by a.lgnaoui last updated on 05/Jun/23 $$\mathrm{Triangle}\:\mathrm{ABC}\:\:\mathrm{equilaterale}\: \\ $$$$\boldsymbol{\mathrm{MN}}=\boldsymbol{\mathrm{NP}}=\boldsymbol{\mathrm{MP}} \\ $$$$\measuredangle\mathrm{BAH}=\mathrm{60}=\measuredangle\mathrm{DAH} \\ $$$$\Rightarrow\:\measuredangle\mathrm{ADH}=\mathrm{30}°\:\:\:\measuredangle\mathrm{AMN}=\mathrm{90}−\mathrm{30}=\mathrm{60} \\ $$$$\mathrm{MNP}\:\:\:\mathrm{Triangle}\:\mathrm{equilaterale}…
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Question Number 64778 by aliesam last updated on 21/Jul/19 Commented by som(math1967) last updated on 21/Jul/19 $${join}\:{O},{Q}\:\:{O},{N}\:{O},{M}\:\:{M},{Q}\:{N},{Q} \\ $$$${now}\:{ON}={NQ}={OQ}={OM}={MQ} \\ $$$$\therefore\angle{NOM}=\mathrm{60}+\mathrm{60}=\mathrm{120}° \\ $$$$\angle{MLN}=\frac{\mathrm{1}}{\mathrm{2}}\angle{NOM}=\mathrm{60}° \\ $$$$\angle{NKM}=\angle{NOM}=\mathrm{120}°\:\:\left[{subtend}\:{on}\:{same}\:{segment}\right]…
Question Number 193093 by Mingma last updated on 04/Jun/23 Answered by Subhi last updated on 04/Jun/23 $$\frac{\mathrm{1}}{\mathrm{2}}.\pi.{r}^{\mathrm{2}} \:=\:\mathrm{3}\pi \\ $$$${r}\:=\:\sqrt{\mathrm{6}} \\ $$$${AB}\:=\:{AD}\:=\:{BE}\:=\:\:\mathrm{2}\sqrt{\mathrm{6}} \\ $$$$\left({CD}\right)^{\mathrm{2}} \:=\:{CE}.{BC}={CE}.\left({CE}+\mathrm{2}\sqrt{\mathrm{6}}\right)…
Question Number 193069 by Mingma last updated on 03/Jun/23 Answered by ajfour last updated on 03/Jun/23 Commented by ajfour last updated on 03/Jun/23 $${a}=\mathrm{17cos}\:\theta \\…
Question Number 193027 by Mingma last updated on 02/Jun/23 Answered by BaliramKumar last updated on 02/Jun/23 $$\alpha\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\sqrt{\mathrm{2}}\right)\:? \\ $$ Answered by ajfour last updated…
Question Number 192966 by Mingma last updated on 01/Jun/23 Answered by aleks041103 last updated on 01/Jun/23 $${at}\:{the}\:{intersection}\:\left(\mathrm{0},\mathrm{0}\right)\in\mathbb{R}^{\mathrm{2}} . \\ $$$${the}\:{whole}\:{circle}\:{is} \\ $$$$\left({x}−\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} +\left({y}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{125}}{\mathrm{16}}=\left(\frac{\mathrm{5}\sqrt{\mathrm{5}}}{\mathrm{4}}\right)^{\mathrm{2}} \\…