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Question-163611

Question Number 163611 by Ramjiane last updated on 08/Jan/22 Answered by ajfour last updated on 08/Jan/22 $$\left(\mathrm{1}\right)\:\:{d}=\frac{{h}}{\mathrm{cot}\:\mathrm{25}°−\mathrm{cot}\:\mathrm{50}°} \\ $$$$\left(\mathrm{2}\right)\:{You}\:{jus}\:{cant}\:{do}\:{without}. \\ $$ Terms of Service Privacy…

Question-32534

Question Number 32534 by naka3546 last updated on 27/Mar/18 Answered by MJS last updated on 27/Mar/18 $${m}+\sqrt{{n}}={k} \\ $$$${f}\left({x}\right)={k} \\ $$$${x}^{\mathrm{2}} +{x}−{k}=\mathrm{0} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{4}{k}+\mathrm{1}} \\…

Question-32510

Question Number 32510 by mondodotto@gmail.com last updated on 26/Mar/18 Commented by abdo imad last updated on 26/Mar/18 $${we}\:{have}\:{y}\:=\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{{x}}\:\:}}\:\:\Rightarrow{y}^{\mathrm{2}} \:−\mathrm{1}\:=\left(\mathrm{1}+\sqrt{{x}}\:\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow\mathrm{2}{y}\:{y}^{,} \:=\frac{\mathrm{1}}{\mathrm{2}}\left(\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}\:}\sqrt{\sqrt{{x}}}}\:\:\Rightarrow \\ $$$${y}^{'}…

Question-163522

Question Number 163522 by nurtani last updated on 07/Jan/22 Answered by ajfour last updated on 08/Jan/22 $${since}\:{A}\:{is}\:{not}\:{fixed},\:{let}\:\angle{C}=\mathrm{90}°. \\ $$$${AC}=\mathrm{3}{h},\:{BC}=\mathrm{6} \\ $$$${A}_{\bigtriangleup{ABC}} =\mathrm{9}{h} \\ $$$${eq}.\:{of}\:{AD}:\:\:\frac{{x}}{\mathrm{5}}+\frac{{y}}{\mathrm{3}{h}}=\mathrm{1} \\…