Question Number 186803 by ajfour last updated on 10/Feb/23 Answered by ajfour last updated on 10/Feb/23 $${Curve}\:{is}\:{y}={x}^{\mathrm{3}} −{x} \\ $$$${p}^{\mathrm{3}} −{p}={c} \\ $$$${m}=\mathrm{3}{s}^{\mathrm{2}} −\mathrm{1}=\frac{{p}−{h}}{{k}} \\…
Question Number 121244 by benjo_mathlover last updated on 06/Nov/20 Commented by liberty last updated on 06/Nov/20 Commented by liberty last updated on 06/Nov/20 $$\left(\mathrm{1}\right)\:\mathrm{sin}\:\mathrm{30}°=\frac{\mathrm{2}}{\mathrm{OA}}\:\Rightarrow\:\mathrm{OA}=\mathrm{4} \\…
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Question Number 55138 by Learner last updated on 18/Feb/19 Commented by math1967 last updated on 18/Feb/19 $${PT}^{\mathrm{2}} ={PA}×{PB} \\ $$$$\therefore{PA}=\frac{\mathrm{6}×\mathrm{6}}{\mathrm{3}}=\mathrm{12}{cm} \\ $$$$\therefore{AB}=\mathrm{12}−\mathrm{3}=\mathrm{9}{cm} \\ $$ Terms…
Question Number 55072 by peter frank last updated on 17/Feb/19 Commented by mr W last updated on 17/Feb/19 $$\left({ax}+{by}\right)\left({bx}−{ay}\right)=\mathrm{0} \\ $$$${abx}^{\mathrm{2}} −\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){xy}−{aby}^{\mathrm{2}} =\mathrm{0}…
Question Number 55071 by peter frank last updated on 17/Feb/19 Answered by mr W last updated on 17/Feb/19 $${the}\:{line}\:{through}\:{origin}\:{and}\:{perpendicular}\:{to} \\ $$$${line}\:{lx}+{my}+{n}=\mathrm{0}\:{is}: \\ $$$${mx}−{ly}=\mathrm{0}\:{or}\:{y}=\frac{{m}}{{l}}{x} \\ $$$${let}\:\mathrm{tan}\:\theta_{\mathrm{0}}…
Question Number 186009 by Rupesh123 last updated on 30/Jan/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 120275 by bramlexs22 last updated on 30/Oct/20 $$\:\sqrt[{\mathrm{4}\:}]{\frac{\mathrm{3}+\mathrm{2}\:\sqrt[{\mathrm{4}}]{\mathrm{5}}}{\mathrm{3}−\mathrm{2}\:\sqrt[{\mathrm{4}}]{\mathrm{5}}}}\:=?\: \\ $$ Answered by TANMAY PANACEA last updated on 30/Oct/20 $$\mathrm{3}−\mathrm{2}\left(\mathrm{5}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$=\mathrm{3}−\left(\mathrm{2}^{\mathrm{4}} ×\mathrm{5}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\…
Question Number 120274 by peter frank last updated on 30/Oct/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
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