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Author: Tinku Tara

x-2-1-3-x-3-392-0-

Question Number 66290 by hmamarques1994@gmail.com last updated on 12/Aug/19 $$\:\sqrt[{\sqrt{\mathrm{3}}}]{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\:+\:\boldsymbol{\mathrm{x}}^{\sqrt{\mathrm{3}}} \:−\:\mathrm{392}\:=\:\mathrm{0} \\ $$ Answered by Tanmay chaudhury last updated on 12/Aug/19 $$\left({x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} +{x}^{\sqrt{\mathrm{3}}}…

Question-131827

Question Number 131827 by Salman_Abir last updated on 09/Feb/21 Answered by liberty last updated on 09/Feb/21 $$\left.\:=\:\mathrm{ln}\:\mid\mathrm{tan}\:\mathrm{x}\mid\:\right]_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:=\:\mathrm{ln}\:\mid\mathrm{tan}\:\frac{\pi}{\mathrm{3}}\mid−\mathrm{ln}\:\mid\mathrm{tan}\:\frac{\pi}{\mathrm{4}}\mid \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{3}\right) \\ $$ Terms of…

Question-131820

Question Number 131820 by mr W last updated on 09/Feb/21 Commented by liberty last updated on 09/Feb/21 $$\frac{\mathrm{x}}{\mathrm{2}}=\:\frac{\mathrm{3}}{\mathrm{1}}\Rightarrow\mathrm{x}=\mathrm{6} \\ $$$$\mathrm{x}.\mathrm{1}\:=\:\mathrm{2}.\mathrm{3}\:\Rightarrow\mathrm{x}=\mathrm{6} \\ $$$$\frac{\mathrm{3}}{\mathrm{x}}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{x}=\mathrm{6} \\ $$ Commented…

Question-131823

Question Number 131823 by mr W last updated on 09/Feb/21 Commented by mr W last updated on 10/Feb/21 $${a}\:{small}\:{sphere}\:{of}\:{mass}\:{m}\:{is}\:{released} \\ $$$${from}\:{rest}\:{at}\:{position}\:{as}\:{shown}. \\ $$$${both}\:{the}\:{sphere}\:{and}\:{the}\:{semi}\:{cylinder} \\ $$$${move}\:{without}\:{friction}.…

x-2-2-2-2-2-2-y-2-2-2-2-2-2-wich-statment-is-true-a-xy-4-xy-0-x-y-4-x-y-2-b-x-Z-c-xy-Z-d-x-y-Z-

Question Number 749 by 123456 last updated on 06/Mar/15 $${x}=\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\centerdot\centerdot\centerdot}}}}}} \\ $$$${y}=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\centerdot\centerdot\centerdot}}}}}} \\ $$$${wich}\:{statment}\:{is}\:{true}? \\ $$$${a}.\:{xy}=\mathrm{4}\vee{xy}=\mathrm{0}\vee{x}+{y}=\mathrm{4}\vee{x}+{y}=\mathrm{2} \\ $$$${b}.\:{x}\notin\mathbb{Z} \\ $$$${c}.{xy}\notin\mathbb{Z} \\ $$$${d}.{x}+{y}\notin\mathbb{Z} \\ $$ Answered…